Circuit analysis with matlab computing and simulink-modeling

Circuit Analysis I
with MATLAB® Computing and
Simulink®/SimPowerSystems® Modeling
Orchard Publications
www.orchardpublications.com
Steven T. Karris

Circuit Analysis I
Simulink® / SimPowerSystems®
Modeling
Steven T. Karris
Orchard Publications, Fremont, California

Circuit Analysis I with MATLAB® Computing and Simulink® / SimPowerSystems® Modeling
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Library of Congress Cataloging-in-Publication Data
Library of Congress Control Number: 2009923770
ISBN10: 1934404187
ISBN13: 9781934404188
TX 5737590
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The author has made every effort to make this text as complete and accurate as possible, but no warranty is implied.
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Preface
This text is an introduction to the basic principles of electrical engineering. It is the outgrowth of
lecture notes prepared by this author while employed by the electrical engineering and computer
engineering departments as adjunct instructor at various colleges and universities. Many of the
examples and problems are based on the author’s industrial experience. The text is an expansion
of our previous publication, Circuit Analysis I with MATLAB® Applications, ISBN 9780
970951120, and this text, in addition to MATLAB scripts for problem solution, includes
several Simulink® and SimPowerSystems® models. The pages where these models appear are
indicated n the Table of Contents.
The book is intended for students of college grade, both community colleges and universities. It
presumes knowledge of first year differential and integral calculus and physics. While some
knowledge of differential equations would be helpful, it is not absolutely necessary. Chapters 9 and
10 include stepbystep procedures for the solutions of simple differential equations used in the
derivation of the natural and forces responses. Appendices D and E provide a thorough review of
complex numbers and matrices respectively.
In addition to several problems provided at the end of each chapter, this text includes multiple-choice
questions to test and enhance the reader’s knowledge of this subject. Moreover, the
answers to these questions and detailed solutions of all problems are provided at the end of each
chapter. The rationale is to encourage the reader to solve all problems and check his effort for
correct solutions and appropriate steps in obtaining the correct solution. And since this text was
written to serve as a selfstudy, primary, or supplementary textbook, it provides the reader with a
resource to test the reader’s knowledge.
A previous knowledge of MATLAB® would be very helpful. However he material of this text can
be learned without MATLAB, Simulink and SimPowerSystems. This author highly recommends
that the reader studies this material in conjunction with the inexpensive Student Versions of The
MathWorks™ Inc., the developers of these outstanding products, available from:
The MathWorks, Inc.
3 Apple Hill Drive
Natick, MA, 01760
Phone: 508-647-7000,
www.mathworks.com
info@mathworks.com.
Appendix A of this text provides a practical introduction to MATLAB, Appendix B is an
introduction to Simulink, and Appendix C is an introduction to SimPowerSystems. These
packages will be invaluable in later studies such as the design of analog and digital filters.

Preface
Like any other new book, this text may contain some grammar and typographical errors;
accordingly, all feedback for errors, advice and comments will be most welcomed and greatly
appreciated.
39510 Paseo Padre Parkway
Suite 315
Fremont, California 94538
info@orchardpublications.com

Table of Contents
1 Basic Concepts and Definitions 11
1.1 The Coulomb ........................................................................................................11
1.2 Electric Current and Ampere ...............................................................................11
1.3 Two Terminal Devices .........................................................................................14
1.4 Voltage (Potential Difference) .............................................................................15
1.5 Power and Energy .................................................................................................18
1.6 Active and Passive Devices ................................................................................111
1.7 Circuits and Networks ........................................................................................113
1.8 Active and Passive Networks .............................................................................113
1.9 Necessary Conditions for Current Flow .............................................................113
1.10 International System of Units ............................................................................114
1.11 Sources of Energy ...............................................................................................117
1.12 Summary .............................................................................................................119
1.13 Exercises .............................................................................................................121
1.14 Answers / Solutions to EndofChapter Exercises ............................................124
MATLAB Computing: Pages 16 through 18
2 Analysis of Simple Circuits 21
2.1 Conventions .........................................................................................................21
2.2 Ohm’s Law ............................................................................................................21
2.3 Power Absorbed by a Resistor ..............................................................................23
2.4 Energy Dissipated in a Resistor ............................................................................24
2.5 Nodes, Branches, Loops and Meshes ...................................................................25
2.6 Kirchhoff’s Current Law (KCL) ...........................................................................26
2.7 Kirchhoff’s Voltage Law (KVL) ............................................................................27
2.8 Single Mesh Circuit Analysis .............................................................................210
2.9 Single NodePair Circuit Analysis ....................................................................214
2.10 Voltage and Current Source Combinations .......................................................217
2.11 Resistance and Conductance Combinations .....................................................218
2.12 Voltage Division Expressions .............................................................................222
2.13 Current Division Expressions .............................................................................224
2.14 Standards for Electrical and Electronic Devices ................................................226
2.15 Resistor Color Code ...........................................................................................226
2.16 Power Rating of Resistors ...................................................................................228
2.17 Temperature Coefficient of Resistance ..............................................................228
2.18 Ampere Capacity of Wires .................................................................................229
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling i
Copyright © Orchard Publications

2.19 Current Ratings for Electronic Equipment......................................................... 229
2.20 Copper Conductor Sizes for Interior Wiring ......................................................231
2.21 Summary .............................................................................................................236
2.22 Exercises ..............................................................................................................239
2.23 Answers / Solutions to EndofChapter Exercises ............................................247
Simulink / SimPowerSystems models: Pages 224, 226
3 Nodal and Mesh Equations - Circuit Theorems 31
3.1 Nodal, Mesh, and Loop Equations ....................................................................... 31
3.2 Analysis with Nodal Equations ............................................................................ 31
3.3 Analysis with Mesh or Loop Equations ................................................................ 38
3.4 Transformation between Voltage and Current Sources .................................... 320
3.5 Thevenin’s Theorem .......................................................................................... 323
3.6 Norton’s Theorem .............................................................................................. 333
3.7 Maximum Power Transfer Theorem .................................................................. 335
3.8 Linearity.............................................................................................................. 337
3.9 Superposition Principle....................................................................................... 338
3.10 Circuits with Non-Linear Devices...................................................................... 342
3.11 Efficiency ............................................................................................................ 344
3.12 Regulation........................................................................................................... 345
3.13 Summary............................................................................................................. 347
3.14 Exercises ............................................................................................................. 349
3.15 Answers / Solutions to EndofChapter Exercises ............................................ 360
MATLAB Computing: Pages 3-4, 37, 311, 314, 316, 318, 332, 366,
370, 372, 374, 376, 380, 390
ii Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling
Simulink / SimPowerSystems models: Pages 38, 319
4 Introduction to Operational Amplifiers 41
4.1 Signals................................................................................................................... 41
4.2 Amplifiers ............................................................................................................. 41
4.3 Decibels ................................................................................................................ 42
4.4 Bandwidth and Frequency Response.................................................................... 44
4.5 The Operational Amplifier................................................................................... 45
4.6 An Overview of the Op Amp............................................................................... 45
4.7 Active Filters ...................................................................................................... 413
4.8 Analysis of Op Amp Circuits.............................................................................. 416
4.9 Input and Output Resistance ............................................................................. 428
4.10 Summary............................................................................................................. 432
4.11 Exercises ............................................................................................................. 434

4.12 Answers / Solutions to EndofChapter Exercises............................................ 443
MATLAB Computing: Page 4-47
Simulink / SimPowerSystems models: Pages 452
5 Inductance and Capacitance 51
5.1 Energy Storage Devices....................................................................................... 51
5.2 Inductance........................................................................................................... 51
5.3 Power and Energy in an Inductor...................................................................... 511
5.4 Combinations of Inductors in Series and in Parallel......................................... 514
5.5 Capacitance....................................................................................................... 516
5.6 Power and Energy in a Capacitor ...................................................................... 521
5.7 Combinations of Capacitors in Series and in Parallel ....................................... 524
5.8 Nodal and Mesh Equations in General Terms.................................................. 526
5.9 Summary............................................................................................................ 529
5.10 Exercises ............................................................................................................ 531
5.11 Answers / Solutions to EndofChapter Exercises ........................................... 536
MATLAB Computing: Pages 513, 523, 540
6 Sinusoidal Circuit Analysis 61
6.1 Excitation Functions........................................................................................... 61
6.2 Circuit Response to Sinusoidal Inputs ................................................................ 61
6.3 The Complex Excitation Function ..................................................................... 63
6.4 Phasors in R, L, and C Circuits........................................................................... 68
6.5 Impedance......................................................................................................... 614
6.6 Admittance ....................................................................................................... 617
6.7 Summary ........................................................................................................... 623
6.8 Exercises ............................................................................................................ 626
6.9 Solutions to EndofChapter Exercises............................................................ 632
MATLAB Computing: Pages 621, 632, 635
Simulink / SimPowerSystems models: Pages 622, 637, 638
7 Phasor Circuit Analysis 71
7.1 Nodal Analysis .................................................................................................... 71
7.2 Mesh Analysis...................................................................................................... 75
7.3 Application of Superposition Principle ............................................................... 76
7.4 Thevenin’s and Norton’s Theorems ................................................................. 710
7.5 Phasor Analysis in Amplifier Circuits ............................................................... 714
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling iii

7.6 Phasor Diagrams ................................................................................................ 717
7.7 Electric Filters .................................................................................................... 722
7.8 Basic Analog Filters ........................................................................................... 723
7.9 Active Filter Analysis ........................................................................................ 728
7.10 Summary............................................................................................................ 731
7.11 Exercises............................................................................................................. 732
7.12 Answers to EndofChapter Exercises.............................................................. 739
MATLAB Computing: Pages 7-4, 76, 78, 712, 713, 715, 717, 721, 730,
744, 745, 746, 748, 750, 751, 755, 756, 758
iv Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling
Simulink models: Pages 79, 710
8 Average and RMS Values, Complex Power, and Instruments 81
8.1 Periodic Time Functions ...................................................................................... 81
8.2 Average Values .................................................................................................... 82
8.3 Effective Values.................................................................................................... 84
8.4 Effective (RMS) Value of Sinusoids..................................................................... 85
8.5 RMS Values of Sinusoids with Different Frequencies ......................................... 87
8.6 Average Power and Power Factor ........................................................................ 89
8.7 Average Power in a Resistive Load .................................................................... 811
8.8 Average Power in Inductive and Capacitive Loads ........................................... 811
8.9 Average Power in NonSinusoidal Waveforms................................................. 814
8.10 Lagging and Leading Power Factors................................................................... 815
8.11 Complex Power  Power Triangle...................................................................... 816
8.12 Power Factor Correction.................................................................................... 818
8.13 Instruments......................................................................................................... 820
8.14 Summary............................................................................................................. 831
8.15 Exercises ............................................................................................................. 834
8.16 Answers to EndofChapter Exercises .............................................................. 840
MATLAB Computing: Page 8-3 continued on Page 84
9 Natural Response 91
9.1 Natural Response of a Series RL circuit ............................................................... 91
9.2 Natural Response of a Series RC Circuit.............................................................. 99
9.3 Summary ............................................................................................................. 917
9.4 Exercises.............................................................................................................. 918
9.5 Answers to EndofChapter Exercises............................................................... 925
MATLAB Computing: Page 927
Simulink / SimPowerSystems models: Page 924

10 Forced and Total Response in RL and RC Circuits 101
10.1 Unit Step Function .............................................................................................101
10.2 Unit Ramp Function ..........................................................................................10 6
10.3 Delta Function ...................................................................................................10 7
10.4 Forced and Total Response in an RL Circuit ...................................................1014
10.5 Forced and Total Response in an RC Circuit...................................................1021
10.6 Summary............................................................................................................1031
10.7 Exercises ............................................................................................................1033
10.8 Answers to EndofChapter Exercises .............................................................1040
MATLAB Computing: Pages 1018, 1030
Simulink / SimPowerSystems models: Page 1051
A Introduction to MATLAB A1
A.1 Command Window ..............................................................................................A1
A.2 Roots of Polynomials ............................................................................................A3
A.3 Polynomial Construction from Known Roots ......................................................A4
A.4 Evaluation of a Polynomial at Specified Values ..................................................A5
A.5 Rational Polynomials ...........................................................................................A8
A.6 Using MATLAB to Make Plots ..........................................................................A9
A.7 Subplots .............................................................................................................A18
A.8 Multiplication, Division and Exponentiation ...................................................A19
A.9 Script and Function Files ..................................................................................A26
A.10 Display Formats .................................................................................................A31
MATLAB Computations: Entire Appendix A
B Introduction to Simulink B1
B.1 Simulink and its Relation to MATLAB ............................................................... B1
B.2 Simulink Demos ................................................................................................. B20
Simulink Modeling: Entire Appendix B
C Introduction to SimPowerSystems C1
C.1 Simulation of Electric Circuits with SimPowerSystems ...................................... C1
SimPowerSystems Modeling: Entire Appendix C
D A Review of Complex Numbers D1
D.1 Definition of a Complex Number ........................................................................D1
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling v

D.2 Addition and Subtraction of Complex Numbers.................................................D2
D.3 Multiplication of Complex Numbers ...................................................................D3
D.4 Division of Complex Numbers.............................................................................D4
D.5 Exponential and Polar Forms of Complex Numbers ...........................................D4
MATLAB Computing: Pages D6, D7, D8
Simulink Modeling: Page D7
E Matrices and Determinants E1
E.1 Matrix Definition ................................................................................................E1
E.2 Matrix Operations...............................................................................................E2
E.3 Special Forms of Matrices ...................................................................................E6
E.4 Determinants.....................................................................................................E10
E.5 Minors and Cofactors ........................................................................................E12
E.6 Cramer’s Rule ....................................................................................................E17
E.7 Gaussian Elimination Method...........................................................................E19
E.8 The Adjoint of a Matrix....................................................................................E21
E.9 Singular and NonSingular Matrices................................................................E21
E.10 The Inverse of a Matrix.....................................................................................E22
E.11 Solution of Simultaneous Equations with Matrices ..........................................E24
E.12 Exercises ............................................................................................................E31
MATLAB Computing: Pages E3, E4, E5, E7, E8, E9, E10, E12,
E15, E16, E18, E22, E25, E6, E29
Simulink Modeling: Page E3
Excel Spreadsheet: Page E27
References R1
Index IN1
vi Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling

Chapter 1
Basic Concepts and Definitions
his chapter begins with the basic definitions in electric circuit analysis. It introduces the
concepts and conventions used in introductory circuit analysis, the unit and quantities used
in circuit analysis, and includes several practical examples to illustrate these concepts.
T
Throughout this text, a left justified horizontal bar will denote the beginning of an example, and
a right justified horizontal bar will denote the end of the example. These bars will not be shown
whenever an example begins at the top of a page or at the bottom of a page. Also, when one
example follows immediately after a previous example, the right justified bar will be omitted.
1.1 The Coulomb
Two identically charged (both positive or both negative) particles possess a charge of one coulomb
when being separated by one meter in a vacuum, repel each other with a force of newton
where . The definition of coulomb is illustrated in Figure 1.1.
c = velocity of light  3  108 m  s
Vacuum
q 1 m q
– = c2 N
q=1 coulomb
F 107
Figure 1.1. Definition of the coulomb
10–7c2
The coulomb, abbreviated as C
, is the fundamental unit of charge. In terms of this unit, the
charge of an electron is 1.6  10–19 C and one negative coulomb is equal to 6.24  1018
electrons.
Charge, positive or negative, is denoted by the letter or .
q Q
1.2 Electric Current and Ampere
Electric current at a specified point and flowing in a specified direction is defined as the instan-taneous
i
rate at which net positive charge is moving past this point in that specified direction, that
is,
(1.1)
------ q
i dq
= = lim
dt
------
t
t 0
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling 11

Chapter 1 Basic Concepts and Definitions
The unit of current is the ampere abbreviated as and corresponds to charge moving at the
rate of one coulomb per second. In other words,
(1.2)
A q
1 ampere = -----------------------------
1 coulomb
1 second
Note: Although it is known that current flow results from electron motion, it is customary to
think of current as the motion of positive charge; this is known as conventional current flow.
To find an expression of the charge in terms of the current , let us consider the charge trans-ferred
q i q
t0 t
from some reference time to some future time . Then, since
i = dq
------
dt
q
t
t idt
q t0
t0
= 
t
= 
qt – qt0 idt
t0
t
=  + qt0
qt idt
t0
12 Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling
the charge is
or
or
(1.3)
Example 1.1
For the waveform of current i shown in Figure 1.2, compute the total charge q
transferred
between
a. t = 0 and
t = 3 s
b. t = 0 and
t = 9 s

Electric Current and Ampere
i mA
1 2 3 4 5 6 7 8
Figure 1.2. Waveform for Example 1.1
Solution:
We know that
30
20
10
0
10
20
30
9
t s
t
 Area 0
t idt
q t = 0
t = =
0
Then, by calculating the areas, we find that:
a. For 0 < t < 2 s, area = ½  (2  30 mA) = 30 mC
For 2 < t < 3 s, area = 1  30 = 30 mC
Therefore, for 0 < t < 3 s, total charge = total area = 30 mC + 30 mC = 60 mC.
b. For 0 < t < 2 s, area = ½  (2  30 mA) = 30 mC
For 2 < t < 6 s, area = 4  30 = 120 mC
For 6 < t < 8 s, area = ½  (2  30 mA) = 30 mC
For 8 < t < 9 s, we observe that the slope of the straight line for t > 6 s is 30 mA / 2 s, or 15
mA / s. Then, for 8 < t < 9 s, area = ½  {1(15)} = 7.5 mC.
Therefore, for 0 < t < 9 s, total charge = total area = 30 + 120 + 30 7.5 = 172.5 mC.
Convention: We denote the current i
by placing an arrow with the numerical value of the cur-rent
next to the device in which the current flows. For example, the designation shown in Figure
1.3 indicates either a current of 2 A is flowing from left to right, or that a current of –2 A
is
moving from right to left.
2 A 2 A
Device
Figure 1.3. Direction of conventional current flow

Caution: The arrow may or may not indicate the actual conventional current flow. We will see
later in Chapters 2 and 3 that in some circuits (to be defined shortly), the actual direction of the
current cannot be determined by inspection. In such a case, we assume a direction with an arrow
for said current ; then, if the current with the assumed direction turns out to be negative, we
conclude that the actual direction of the current flow is opposite to the direction of the arrow.
Obviously, reversing the direction reverses the algebraic sign of the current as shown in Figure
1.3.
In the case of timevarying currents which change direction from timetotime, it is convenient
to think or consider the instantaneous current, that is, the direction of the current which flows at
some particular instant. As before, we assume a direction by placing an arrow next to the device
in which the current flows, and if a negative value for the current i is obtained, we conclude that
the actual direction is opposite of that of the arrow.
1.3 Two Terminal Devices
In this text we will only consider twoterminal devices. In a twoterminal device the current
entering one terminal is the same as the current leaving the other terminal* as shown in Figure
1.4.
7 A 7 A
Figure 1.4. Current entering and leaving a twoterminal device
Let us assume that a constant value current (commonly known as Direct Current and abbreviated
as DC) enters terminal and leaves the device through terminal in Figure 1.4. The passage of
current (or charge) through the device requires some expenditure of energy, and thus we say that
a potential difference or voltage exists “across” the device. This voltage across the terminals of the
device is a measure of the work required to move the current (or charge) through the device.
Example 1.2
In a twoterminal device, a current enters the left (first) terminal.
a. What is the amount of current which enters that terminal in the time interval ?
b. What is the current at ?
c. What is the charge at given that ?
* We will see in Chapter 5 that a two terminal device known as capacitor is capable of storing energy.
i
Two terminal device
Terminal A Terminal B
A B
it = 20cos100t mA
–10  t  20 ms
t = 40 ms
q t = 5 ms q0 = 0

Voltage (Potential Difference)
Solution:
a.
b.
c.
20 10 3 –  20cos100 20 10–3    20cos100 –10 10–3 = = –   
t 20cos100t
i t0
–10 10 –3 
= 20cos2 – 20cos– = 40 mA
i t = 0.4 ms 20cos100t t = 0.4 ms = = 20cos40 = 20 mA
5 103
5 103
– – =   =   20cos100tdt
+ 0
0.2
qt idt + q0
0
0
------- 2
5 3
 1– 0sin 0.2
------- 100t 0
= = = ------- C


sin---–0 0.2

1.4 Voltage (Potential Difference)
The voltage (potential difference) across a twoterminal device is defined as the work required
to move a positive charge of one coulomb from one terminal of the device to the other terminal.
The unit of voltage is the volt (abbreviated as or ) and it is defined as
(1.4)
V v
1 volt 1 joule
= -----------------------------
1 coulomb
Convention: We denote the voltage v
by a plus (+) minus () pair. For example, in Figure 1.5,
we say that terminal is positive with respect to terminal or there is a potential differ-ence
A 10 V B
of between points and . We can also say that there is a voltage drop of in
10 V A B 10 V
going from point A to point B . Alternately, we can say that there is a voltage rise of 10 V
in
going from to .
Two terminal
device
A
B
+
10 v

Figure 1.5. Illustration of voltage polarity for a twoterminal device
B A
Caution: The (+) and () pair may or may not indicate the actual voltage drop or voltage rise.
As in the case with the current, in some circuits the actual polarity cannot be determined by

inspection. In such a case, again we assume a voltage reference polarity for the voltage; if this ref-erence
polarity turns out to be negative, this means that the potential at the (+) sign terminal is
at a lower potential than the potential at the () sign terminal.
In the case of timevarying voltages which change (+) and () polarity from timetotime, it is
convenient to think the instantaneous voltage, that is, the voltage reference polarity at some partic-ular
instance. As before, we assume a voltage reference polarity by placing (+) and () polarity
signs at the terminals of the device, and if a negative value of the voltage is obtained, we conclude
that the actual polarity is opposite to that of the assumed reference polarity. We must remember
that reversing the reference polarity reverses the algebraic sign of the voltage as shown in Figure
1.6.
B
Figure 1.6. Alternate ways of denoting voltage polarity in a twoterminal device
Example 1.3
The (currentvoltage) relation of a nonlinear electrical device is given by
(10.5)
a. Use MATLAB®* to sketch this function for the interval
b. Use the MATLAB quad function to find the charge at given that
Solution:
a. We use the following script to sketch .
t=0: 0.1: 10; it=0.1.*(exp(0.2.*sin(3.*t))1);
plot(t,it), grid, xlabel('time in sec.'), ylabel('current in amp.')
The plot for is shown in Figure 1.7.
* MATLAB and Simulink are registered marks of The MathWorks, Inc., 3 Apple Hill Drive, Natick, MA, 01760,
www.mathworks.com. An introduction to MATLAB is given in Appendix A, and an introduction to Simulink is given in
Appendix B. Simulink operates in the MATLAB environment. The SimPowerSystems is another product of The
MathWorks and operates in the Simulink environment.
Two terminal device A
+
B

12 v
A Same device
+

12 v

i – v
it = 0.1e0.2 sin3t – 1
0  t  10 s
t = 5 s q0 = 0
it
it

Voltage (Potential Difference)
0 1 2 3 4 5 6 7 8 9 10
time in sec.
it
Figure 1.7. Plot of for Example 1.3
0.025
0.02
0.015
0.01
0.005
0
-0.005
-0.01
-0.015
-0.02
current in amp.
b.The charge is the integral of the current , that is,
(1.6)
qt it
t1 = = 
t1  0.1 e0.2 sin3t – 1dt
qt itdt
t0
0
We will use the MATLAB int(f,a,b) integration function where f is a symbolic expression, and
a and b are the lower and upper limits of integration respectively.
Note:
When MATLAB cannot find a solution, it returns a warning. For this example, MATLAB
returns the following message when integration is attempted with the symbolic expression of
(1.6).
t=sym('t'); % Refer to Appendix A, Page A10, for a discussion on symbolic expressions
s=int(0.1*(exp(0.2*sin(3*t))1),0,10)
When this script is executed, MATLAB displays the following message:
Warning: Explicit integral could not be found.
In C:MATLAB 12toolboxsymbolic@symint.m at line 58
s = int(1/10*exp(1/5*sin(3*t))-1/10,t = 0. . 10)

We will use numerical integration with Simpson’s rule. MATLAB has two quadrature functions
for performing numerical integration, the quad* and quad8. The description of these can be seen
by typing help quad or help quad8. at the MATLAB command prompt. Both of these functions
use adaptive quadrature methods; this means that these methods can handle irregularities such as
singularities. When such irregularities occur, MATLAB displays a warning message but still pro-vides
10–3
p W
Power p dW
= = --------
dt
watt
Power p volts  amperes vi joul
----------- coul
= = = = = ---------- = watts
coul
 ----------- joul
sec
sec
an answer.
For this example, we will use the quad function. It has the syntax q=quad(‘f’,a,b,tol), and per-forms
an integration to a relative error tol which we must specify. If tol is omitted, it is understood
to be the standard tolerance of . The string ‘f’ is the name of a user defined function, and a
and b are the lower and upper limits of integration respectively.
First, we need to create and save a function mfile.† We define it as shown below, and we save it
as CA_1_Ex_1_3.m. This is a mnemonic for Circuit Analysis I, Example 1.3.
function t = fcn_example_1_3(t); t = 0.1*(exp(0.2*sin(3*t))1);
With this file saved as CA_1_Ex_1_3.m, we write and execute the following script.
charge=quad('CA_1_Ex_1_3',0,5)
and MATLAB returns
charge =
0.0170
1.5 Power and Energy
Power is the rate at which energy (or work) is expended. That is,
(1.7)
Absorbed power is proportional both to the current and the voltage needed to transfer one cou-lomb
through the device. The unit of power is the . Then,
(1.8)
and
* For a detailed discussion on numerical analysis and the MATLAB functions quad and quad8, the reader may refer to
Numerical Analysis Using MATLAB® and Excel, ISBN 9781934404034.
† For more information on function mfiles, please refer to Appendix A, Page A26.

Power and Energy
(1.9)
1 watt = 1 volt  1 ampere
Passive Sign Convention: Consider the twoterminal device shown in Figure 1.8.
i
A B
Two terminal device
+  v
Figure 1.8. Illustration of the passive sign convention
In Figure 1.8, terminal A is v volts positive with respect to terminal B
and current i enters the
device through the positive terminal . In this case, we satisfy the passive sign convention and
A
is said to be absorbed by the device.
power = p = vi
The passive sign convention states that if the arrow representing the current i and the (+) ()
pair are placed at the device terminals in such a way that the current enters the device terminal
marked with the (+) sign, and if both the arrow and the sign pair are labeled with the appropri-ate
algebraic quantities, the power absorbed or delivered to the device can be expressed as
. If the numerical value of this product is positive, we say that the device is absorbing
p = vi
power which is equivalent to saying that power is delivered to the device. If, on the other hand,
the numerical value of the product p = vi
is negative, we say that the device delivers power to
some other device. The passive sign convention is illustrated with the examples in Figures 1.9
and 1.10.
2 A 2 A
Two terminal device
B

12 v
A Same device

12 v
=
Power = p = (12)(2) = 24 w Power = p = (12)(2) = 24 w
Figure 1.9. Examples where power is absorbed by a twoterminal device
i=6cos3t
Two terminal device 1
B

A

i=5sin5t
Two terminal device 2
v=18sin3t v=cos5t
p = (cos5t)(5sin5t) = 2.5sin10t w
p = (18sin3t)(6cos3t) = 54sin6t w
Figure 1.10. Examples where power is delivered to a twoterminal device
A
+
B
+
A
+
+ B
In Figure 1.9, power is absorbed by the device, whereas in Figure 1.10, power is delivered to the
device.

Example 1.4
It is assumed a 12volt automotive battery is completely discharged and at some reference time
, is connected to a battery charger to trickle charge it for the next 8 hours. It is also assumed
t = 0
i 8e –t § 3600 =
15000
 8e–t  3600 dt
15000 idt
28800
 8
---------------------e–t  3600
28800
28800
 12 8e–t  3600 
28800
 dt 96
---------------------e–t  3600
28800
that the charging rate is
For this 8hour interval compute:
a. the total charge delivered to the battery
b. the maximum power (in watts) absorbed by the battery
c. the total energy (in joules) supplied
d. the average power (in watts) absorbed by the battery
Solution:
The current entering the positive terminal of the battery is the decaying exponential shown in
Figure 1.11 where the time has been converted to seconds.
Figure 1.11. Decaying exponential for Example 1.4
Then,
a.
b.
Therefore,
c.
it 8e –t  3600 A 0  t  8 hr
0 otherwise   
=
(A)
t (s)
i(t)
8
28800
q t = 0
0
0
–1  3600
0
= = =
–8  3600 e–8 =  – 1  28800 C or 28.8 kC
imax = 8 A (occurs at t=0)
pmax = vimax = 12  8 = 96 w
W  pdt vidt
0
0
–1  3600
0
= = = =
3.456  105 1 e–8 =  –   345.6 KJ.

Active and Passive Devices
d.
Pave
1T
T
 1
--- pdt
= = = --------------------------- = 12 w.
0
28800
 dt 345.6  103
-------------- 12 8e–t  3600 
28800
0
28.8  103
Example 1.5
The power absorbed by a nonlinear device is . If , how much
charge goes through this device in two seconds?
Solution:
The power is
p vi, i pv
= = --
= = --------------------------------------------------- = 3e0.4t – 1 A
9 e0.16t2
------------------------------ 9e0.4t + 1 e 0.4t – 1
 – 1
3e0.4t + 1
then, the charge for 2 seconds is
p 9e0.16t2
=  – 1 v = 3e0.4t + 1
3e0.4t + 1
2
 3
t
 3 e0.4t – 1dt
2 = = = – = 7.5e0.8 – 1 – 6 = 3.19 C
t idt
q t0
t0
0
-------e0.4t
0.4
2
0
3t 0
The twoterminal devices which we will be concerned with in this text are shown in Figure 1.12.
Linear devices are those in which there is a linear relationship between the voltage across that
device and the current that flows through that device. Diodes and Transistors are nonlinear
devices, that is, their voltagecurrent relationship is nonlinear. These will not be discussed in
this text. A simple circuit with a diode is presented in Chapter 3.
1.6 Active and Passive Devices
Independent and dependent voltage and current sources are active devices; they normally (but
not always) deliver power to some external device. Resistors, inductors and capacitors are passive
devices; they normally receive (absorb) power from an active device.

Independent and Dependent Sources
+ Ideal Independent Voltage Source  Maintains same voltage
v or v(t) regardless of the amount of current that flows through it. Its value is either constant (DC) or sinusoidal (AC).
Ideal Independent Current Source  Maintains same current
regardless of the voltage that appears across its terminals.
i or i(t) Its value is either constant (DC) or sinusoidal (AC).
+ Dependent Voltage Source  Its value depends on another
voltage or current elsewhere in the circuit. Here, is a
k1v k2i
or constant and is a resistance as defined in linear devices
below. When denoted as it is referred to as voltage
controlled voltage source and when denoted as k2 i it is
Dependent Current Source  Its value depends on another
current or voltage elsewhere in the circuit. Here, is a
constant and is a conductance as defined in linear devices
Conductance G iG
+ 
dvC
dt
iR iG
vC
k3i k4v
vR
+ 
Figure 1.12. Voltage and current sources and linear devices
Linear Devices
R
C iC
+ 
vR
iR
R = slope
G
+ vG 
vG
G = slope
Resistance R
iC = C
vL
L = slope
diL
dt
Inductance L
iL L
vL = L
diL
dt
vL
iC
C = slope
dvC
dt
Capacitance C

k1
k2
k3
k4
vR = RiR
iG = GvG
below.
k1v
referred to as current controlled voltage source.
When denoted as it is referred to as current
or
k3i
controlled current source and when denoted as k4v i t is
referred to as voltage controlled current source.

Circuits and Networks
1.7 Circuits and Networks
A network is the interconnection of two or more simple devices as shown in Figure 1.13.
+
R L C
vS
Figure 1.13. A network but not a circuit
A circuit is a network which contains at least one closed path. Thus every circuit is a network but
not all networks are circuits. An example is shown in Figure 1.14.
+
L C
vS
R1
R2
Figure 1.14. A network and a circuit
1.8 Active and Passive Networks
Active Network is a network which contains at least one active device (voltage or current
source).
Passive Network is a network which does not contain any active device.
1.9 Necessary Conditions for Current Flow
There are two conditions which are necessary to set up and maintain a flow of current in a net-work
or circuit. These are:
1. There must be a voltage source (potential difference) present to provide the electrical work
which will force current to flow.
2. The circuit must be closed.
These conditions are illustrated in Figures 1.15 through 1.17.
Figure 1.15 shows a network which contains a voltage source but it is not closed and therefore,
current will not flow.

Figure 1.15. A network in which there is no current flow
Figure 1.16 shows a closed circuit but there is no voltage present to provide the electrical work for
current to flow.
Figure 1.16. A closed circuit in which there is no current flow
Figure 1.17 shows a voltage source present and the circuit is closed. Therefore, both conditions
are satisfied and current will flow.
Figure 1.17. A circuit in which current flows
1.10 International System of Units
The International System of Units (abbreviated SI in all languages) was adopted by the General
Conference on Weights and Measures in 1960. It is used extensively by the international scien-tific
community. It was formerly known as the Metric System. The basic units of the SI system are
listed in Table 1.1.
+
R L C
vS
R1
R2
R3
R4
+
R L
C
vS

International System of Units
TABLE 1.1 SI Base Units
Unit of Name Abbreviation
Length Metre m
Mass Kilogram kg
Time Second s
Electric Current Ampere A
Temperature Degrees Kelvin °K
Amount of Substance Mole mol
Luminous Intensity Candela cd
Plane Angle Radian rad
Solid Angle Steradian sr
The SI uses larger and smaller units by various powers of 10 known as standard prefixes. The com-mon
prefixes are listed in Table 1.2 and the less frequently in Table 1.3. Table 1.4 shows some
conversion factors between the SI and the English system. Table 1.5 shows typical temperature
values in degrees Fahrenheit and the equivalent temperature values in degrees Celsius and
degrees Kelvin. Other units used in physical sciences and electronics are derived from the SI
base units and the most common are listed in Table 1.6.
TABLE 1.2 Most Commonly Used SI Prefixes
Value Prefix Symbol Example
Giga G 12 GHz (Gigahertz) = 12 × 10 9 Hz
Mega M 25 MW (Megaohms) = 25 × 10 6 W (ohms)
Kilo K 13.2 KV (Kilovolts) = 13.2 × 10 3 volts
centi c 2.8 cm (centimeters) = 2.8 x 10 –2 meter
milli m 4 mH (millihenries) = 4 × 10 –3 henry
micro μ 6 μw (microwatts) = 6 × 10 –6 watt
nano n 2 ns (nanoseconds) = 2 × 10 –9 second
pico p 3 pF (picofarads) = 3 × 10 –12 Farad
109
106
103
10–2
10–3
10–6
10–9
10–12

TABLE 1.3 Less Frequently Used SI Prefixes
Value Prefix Symbol Example
Exa E 1 Em (Exameter) = 1018 meters
Peta P 5 Pyrs (Petayears) = 5 × 1015 years
Tera T 3 T$ (Teradollars) = 3 × 1012 dollars
femto f 7 fA (femtoamperes) = 7 × 10 –15 ampere
atto a 9 aC (attocoulombs) = 9 × 10 –18 coulomb
TABLE 1.4 Conversion Factors
1 in. (inch) 2.54 cm (centimeters)
1 mi. (mile) 1.609 Km (Kilometers)
1 lb. (pound) 0.4536 Kg (Kilograms)
1 qt. (quart) 946 cm3 (cubic centimeters)
1 cm (centimeter) 0.3937 in. (inch)
1 Km (Kilometer) 0.6214 mi. (mile)
1 Kg (Kilogram) 2.2046 lbs (pounds)
1 lt. (liter) = 1000 cm3 1.057 quarts
1 Å (Angstrom) 10 –10 meter
1 mm (micron) 10 –6 meter
TABLE 1.5 Temperature Scale Equivalents
°F °C °K
–523.4 –273 0
32 0 273
0 –17.8 255.2
77 25 298
98.6 37 310
212 100 373
1018
1015
1012
10–15
10–18

Sources of Energy
TABLE 1.6 SI Derived Units
Unit of Name Formula
Force Newton
Pressure or Stress Pascal
Work or Energy Joule
Power Watt
Voltage Volt
Resistance Ohm
Conductance Siemens or
Capacitance Farad
Inductance Henry
Frequency Hertz
Quantity of Electricity Coulomb
Magnetic Flux Weber
Magnetic Flux Density Tesla
Luminous Flux Lumen
Illuminance Lux
Radioactivity Becquerel
Radiation Dose Gray
Volume Litre
N N = kg  m  s2
Pa Pa = N  m2
J J = N  m
W W = J  s
V V = W  A
  = V  A
S –1   S = A  V
F F = A  s  V
H H = V  s  A
Hz Hz = 1  s
m2
C C = A  s
Wb Wb = V  s
 T  T = Wb 
lm lm = cd  sr
lx lx = lm  m2
Bq Bq s–1 =
Gy S = J  kg
L L m3 10–3 = 
1.11 Sources of Energy
The principal sources of energy are from chemical processes (coal, fuel oil, natural gas, wood
etc.) and from mechanical forms (water falls, wind, etc.). Other sources include nuclear and
solar energy.
Example 1.6
A certain type of wood used in the generation of electric energy and we can get 12,000 BTUs
from a pound (lb) of that wood when burned. Suppose that a computer system that includes a

monitor, a printer, and other peripherals absorbs an average power of 500 w gets its energy from
that burned wood and it is turned on for 8 hours. It is known that 1 BTU is equivalent to 778.3 ft
lb of energy, and 1 joule is equivalent to 0.7376 ftlb.
Compute:
a. the energy consumption during this 8hour interval
b. the cost for this energy consumption if the rate is $0.15 per kwhr
c. the amount of wood in lbs burned during this time interval.
Solution:
a. Energy consumption for 8 hours is
Energy W Pavet 500 w  8 hrs 3600 s
= =  --------------- = 14.4 Mjoules
------------------ 1 kw – hr
=  ----------------------------------------  14.4  106 = $0.60
---------------- 1 BTU
 ------------------------------- 1 lb
  ---------------------------- = 1.137 lb
b. Since ,
c. Wood burned in 8 hours,
1 hr
1 kilowatt – hour = 3.6  106 joules
Cost $0.15
kw – hr
3.6  106 joules
14.4  106 joules 0.7376 f t – lb
joule
778.3 f t – lb
12000 BTU

Summary
1.12 Summary
 Two identically charged (both positive or both negative) particles possess a charge of one cou-lomb
when being separated by one meter in a vacuum, repel each other with a force of
10–7c2
c = velocity of light  3  108 m  s
newton where . Thus, the force with which two
electrically charged bodies attract or repel one another depends on the product of the charges
(in coulombs) in both objects, and also on the distance between the objects. If the polarities
are the same (negative/negative or positive/positive), the socalled coulumb force is repulsive;
if the polarities are opposite (negative/positive or positive/negative), the force is attractive. For
any two charged bodies, the coulomb force decreases in proportion to the square of the
distance between their charge centers.
 Electric current is defined as the instantaneous rate at which net positive charge is moving past
this point in that specified direction, that is,
------ q
i dq
= = lim
dt
------
t
t 0
 The unit of current is the ampere, abbreviated as A, and corresponds to charge q moving at the
rate of one coulomb per second.
 In a twoterminal device the current entering one terminal is the same as the current leaving
the other terminal.
 The voltage (potential difference) across a twoterminal device is defined as the work required
to move a positive charge of one coulomb from one terminal of the device to the other termi-nal.
 The unit of voltage is the volt (abbreviated as V or v) and it is defined as
1 volt 1 joule
= -----------------------------
1 coulomb
 Power p is the rate at which energy (or work) W is expended. That is,
Power p dW
= = --------
dt
 Absorbed power is proportional both to the current and the voltage needed to transfer one
coulomb through the device. The unit of power is the watt and
1 watt = 1 volt  1 ampere
 The passive sign convention states that if the arrow representing the current i and the plus
(+) minus () pair are placed at the device terminals in such a way that the current enters the
device terminal marked with the plus (+) sign, and if both the arrow and the sign pair are
labeled with the appropriate algebraic quantities, the power absorbed or delivered to the

device can be expressed as p = vi
. If the numerical value of this product is positive, we say that
the device is absorbing power which is equivalent to saying that power is delivered to the
device. If, on the other hand, the numerical value of the product p = vi
is negative, we say
that the device delivers power to some other device.
 An ideal independent voltage source maintains the same voltage regardless of the amount of
current that flows through it.
 An ideal independent current source maintains the same current regardless of the amount of
voltage that appears across its terminals.
 The value of an dependent voltage source depends on another voltage or current elsewhere in
the circuit.
 The value of an dependent current source depends on another current or voltage elsewhere in
the circuit.
 Ideal voltage and current sources are just mathematical models. We will discuss practical volt-age
and current sources in Chapter 3.
 Independent and Dependent voltage and current sources are active devices; they normally (but
not always) deliver power to some external device.
 Resistors, inductors, and capacitors are passive devices; they normally receive (absorb) power
from an active device.
 A network is the interconnection of two or more simple devices.
 A circuit is a network which contains at least one closed path. Thus every circuit is a network
but not all networks are circuits.
 An active network is a network which contains at least one active device (voltage or current
source).
 A passive network is a network which does not contain any active device.
 To set up and maintain a flow of current in a network or circuit there must be a voltage source
(potential difference) present to provide the electrical work which will force current to flow
and the circuit must be closed.
 Linear devices are those in which there is a linear relationship between the voltage across that
device and the current that flows through that device.
 The International System of Units is used extensively by the international scientific commu-nity.
It was formerly known as the Metric System.
 The principal sources of energy are from chemical processes (coal, fuel oil, natural gas, wood
etc.) and from mechanical forms (water falls, wind, etc.). Other sources include nuclear and
solar energy.

Exercises
1.13 Exercises
Multiple choice
1. The unit of charge is the
A. ampere
B. volt
C. watt
D. coulomb
E. none of the above
2. The unit of current is the
A. ampere
B. coulomb
C. watt
D. joule
3. The unit of electric power is the
A. ampere
B. coulomb
C. watt
D. joule
4. The unit of energy is the
A. ampere
B. volt
C. watt
D. joule
5. Power is
A. the integral of energy
B. the derivative of energy
C. current times some constant
D. voltage times some constant
k
k
6. Active voltage and current sources
A. always deliver power to other external devices
B. normally deliver power to other external devices
C. neither deliver or absorb power to or from other devices
D. are just mathematical models

7. An ideal independent voltage source
A. maintains the same voltage regardless of the amount of current that flows through it
B. maintains the same current regardless of the voltage rating of that voltage source
C. always delivers the same amount of power to other devices
D. is a source where both voltage and current can be variable
8. An ideal independent current source
A. maintains the same voltage regardless of the amount of current that flows through it
B. maintains the same current regardless of the voltage that appears across its terminals
C. always delivers the same amount of power to other devices
D. is a source where both voltage and current can be variable
9. The value of a dependent voltage source can be denoted as
A. kV
where k is a conductance value
B. kI
where k is a resistance value
C. kV
where k is an inductance value
D. kI
where k is a capacitance value
10. The value of a dependent current source can be denoted as
kV
kI
kV
kI
W (mJ)
10
A. where k is a conductance value
B. where k is a resistance value
C. where k is an inductance value
D. where k is a capacitance value
Problems
1. A two terminal device consumes energy as shown by the waveform below, and the current
through this device is it = 2cos4000t A
. Find the voltage across this device at t = 0.5, 1.5,
4.75 and 6.5 ms. Answers:
2.5 V 0 V 2.5 V –2.5 V
1 t (ms)
0
2 3 4 5 6 7
5

Exercises
2. A household light bulb is rated 75 watts at 120 volts. Compute the number of electrons per
second that flow through this bulb when it is connected to a 120 volt source.
Answer:
3.9  1018 electrons  s
3. An airplane, whose total mass is 50,000 metric tons, reaches a height of 32,808 feet in 20 min-utes
after takeoff.
a. Compute the potential energy that the airplane has gained at this height.
Answer:
1 736 MJ
b. If this energy could be converted to electric energy with a conversion loss of 10%, how
much would this energy be worth at $0.15 per kilowatthour? Answer:
c. If this energy were converted into electric energy during the period of 20 minutes, what
average number of kilowatts would be generated? Answer:
$65.10
1 450 Kw
4. The power input to a television station transmitter is 125 kw and the output is 100 kw which
is transmitted as radio frequency power. The remaining 25 kw of power is converted into
heat.
a. How many BTUs per hour does this transmitter release as heat?
Answer:
b. How many electronvolts per second is this heat equivalent to?
1 electron – volt 1.6 10–19 =  J
Answer:
1 BTU = 1054.8 J
85 234 BTU  hr
1.56  10 23 ------------------------------------------
electron – volts
sec.

1.14 Answers / Solutions to EndofChapter Exercises
Dear Reader:
The remaining pages on this chapter contain answers to the multiplechoice questions and solu-tions
to the exercises.
You must, for your benefit, make an honest effort to answer the multiplechoice questions and
solve the problems without first looking at the solutions that follow. It is recommended that first
you go through and answer those you feel that you know. For the multiplechoice questions and
problems that you are uncertain, review this chapter and try again. If your answers to the prob-lems
do not agree with those provided, look over your procedures for inconsistencies and compu-tational
errors. Refer to the solutions as a last resort and rework those problems at a later date.
You should follow this practice with the multiplechoice and problems on all chapters of this
book.

Answers / Solutions to EndofChapter Exercises
Multiple choice
1. D
2. A
3. C
4. D
5. B
6. B
7. A
8. B
9. B
10. A
Problems
1.
a.
b.
c.
-- dW  dt
v pi
----------------- slope
= = = -------------
i
i
1 ms 5 mJ
slope 0
= ------------ = 5 J  s
1 ms
v t = 0.5 ms
5 J  s
------------------------- 5 J  s
---------------------------------------------------------------- 5 J  s
= = = ------------- = 2.5 V
2 4000 0.5 10–3 cos    A
2cos2 A
2 A
2 ms = 0
slope 1
v t = 1.5 ms
0i
= -- = 0 V
5 ms –5 mJ
slope 4
= --------------- = –5 J  s
1 ms
v t = 4.75 ms
–5 J  s
------------------------------------------------------------------- –5 J  s
= = = = ---------------- = 2.5 V
2 4000 4.75 10–3 cos    A
---------------------------- –5 J  s
2cos19 A
--------------------- –5 J  s
2cos A
–2 A

7 ms –5 mJ
= --------------- = –5 J  s
---------------------------- –5 J  s
---------------------------------------------------------------- –5 J  s
= = = ---------------- = –2.5 V
--------------- 58
= = = -- A
t
= 
1 s
 58
1 s 58
= = = -- C  s
-- C  s 6.24 10 18 electrons
 ----------------------------------------------------- = 3.9  1018 electrons  s
  ----------------------- = 10 000 m = 10 Km
 ----------------- = 1 200 sec.
-------------------------- 25
= = ----- m  s
---------------------------   5 107
 2
   25
= = = 173.61  107 J  1 736 MJ
----- 
d.
2.
3.
where and
Then,
a.
slope 6
1 ms
v t = 6.5 ms
–5 J  s
2 4000 6.5 10–3 cos    A
2cos26 A
2 A
i pv
-- 75 w
120 V
q idt
t0
q t = 1 s
58
--dt
0
--t
0
58
1 C
Wp Wk
12
--mv2 = =
m = mass in kg v = velocity in meters  sec.
33 808 ft 0.3048 m
ft
20 minutes 60 sec.
min
v 10 000 m
1 200 sec.
3
50 000 metric tons 1 000 Kg
metric ton
=  Kg
Wp Wk
12
-- 5 107
3

b.
1 joule = 1 watt-sec
1 736 106J 1 watt-sec
   -------------------------- 1 Kw
1 joule
 --------------------- 1 hr
1 000 w
 ------------------------- = 482.22 Kw-hr
3 600 sec.
and with 10% conversion loss, the useful energy is
c.
4.
a.
b.
482.22  0.9 = 482.22  0.9 = 434 Kw-hr
Cost of Energy $0.15
= ----------------  434 Kw-hr = $65.10
Kw-hr
Pave
----- 1 736 MJ
Wt
= = ---------------------------------------- = 1.45 Mw = 1450 Kw
20 min  ---------------
60 sec
min
1 BTU = 1054.8 J
25 000 watts 1 joule  sec.
-------------------- 3600 sec.
  ------------------------------- 1 BTU
watt
  ---------------------- = 85 234 BTU  hr
1054.8 J
1 hr
1 electron – volt 1.6 10–19 =  J
1 ------------------------------------------- electron – volt
1.6  10 – 19 J
sec.
------------------------------- 1.6 10–19 = =  watt
sec.
25 000 watts 1 electron – volt  sec.
  --------------------------------------------------------- 1.56  10 23 electron – volts
1.6 10–19  watt
= ------------------------------------------
sec.

Chapter 2
Analysis of Simple Circuits
his chapter defines constant and instantaneous values, Ohm’s law, and Kirchhoff’s Current
and Voltage laws. Series and parallel circuits are also defined and nodal, mesh, and loop
analyses are introduced. Combinations of voltage and current sources and resistance com-binations
are discussed, and the voltage and current division formulas are derived.
T
2.1 Conventions
We will use lower case letters such as , , and to denote instantaneous values of voltage, cur-rent,
v i p
and power respectively, and we will use subscripts to denote specific voltages, currents,
resistances, etc. For example, vS and iS
will be used to denote voltage and current sources
respectively. Notations like vR1 and iR2 will be used to denote the voltage across resistance
R1
and the current through resistance R2 respectively. Other notations like vA or v1
will represent
the voltage (potential difference) between point or point with respect to some arbitrarily cho-sen
A 1
reference point taken as “zero” volts or “ground”.
The designations vAB or v12 will be used to denote the voltage between point A or point 1
with
respect to point B or 2 respectively. We will denote voltages as vt and it
whenever we wish
to emphasize that these quantities are time dependent. Thus, sinusoidal (AC) voltages and cur-rents
will be denoted as and respectively. Phasor quantities, to be introduced in Chapter
vt it
6, will be represented with bold capital letters, for phasor voltage and for phasor current.
V I
2.2 Ohm’s Law
We recall from Chapter 1 that resistance is a constant that relates the voltage and the current
as:
(2.1)
R
vR = RiR
This relation is known as Ohm’s law.
The unit of resistance is the Ohm and its symbol is the Greek capital letter . One ohm is the
resistance of a conductor such that a constant current of one ampere through it produces a volt-age
of one volt between its ends. Thus,

Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems  Modeling 21

Chapter 2 Analysis of Simple Circuits
(2.2)
1  1 V
= ---------
1 A
Physically, a resistor is a device that opposes current flow. Resistors are used as a current limiting
devices and as voltage dividers.
In the previous chapter we defined conductance as the constant that relates the current and the
voltage as
(2.3)
G
iG = GvG
This is another form of Ohm’s law since by letting and , we obtain
(2.4)
iG = iR vG = vR
G 1R
= ---
The unit of conductance is the siemens or mho (ohm spelled backwards) and its symbol is or
S
–1
1 –1 1 A
= --------
1 V
A +
Open
Circuit

i = 0
B
i = 0
A+
R = 
G = 0
B
A +
Short
Circuit
vAB = 0 vAB = 0

i
B
i
A+
R = 0
G = 
B
Thus,
(2.5)
Resistances (or conductances) are commonly used to define an “open circuit” or a “short circuit”.
An open circuit is an adjective describing the “open space” between a pair of terminals, and can be
thought of as an “infinite resistance” or “zero conductance”. In contrast, a short circuit is an adjec-tive
describing the connection of a pair of terminals by a piece of wire of “infinite conductance” or
a piece of wire of “zero” resistance.
The current through an “open circuit” is always zero but the voltage across the open circuit termi-nals
may or may not be zero. Likewise, the voltage across a short circuit terminals is always zero
but the current through it may or may not be zero. The open and short circuit concepts and their
equivalent resistances or conductances are shown in Figure 2.1.
Figure 2.1. The concepts of open and short circuits
The fact that current does not flow through an open circuit and that zero voltage exists across the
terminals of a short circuit, can also be observed from the expressions and .
vR = RiR iG = GvG

Power Absorbed by a Resistor
G 1R
That is, since , infinite means zero and zero means infinite . Then, for a finite
voltage, say , and an open circuit,
(2.6)
= --- R G R G
vG
iG G  0
lim GvG G 0
= lim = 0
Likewise, for a finite current, say iR, and a short circuit,
(2.7)
vR R 0
lim RiR R  0
Reminder:
We must remember that the expressions
and
= lim = 0
vR = RiR
iG = GvG
are true only when the passive sign convention is observed. This is consistent with our classifica-tion
of and being passive devices and thus implies the current direction and
R G vR = RiR
voltage polarity are as shown in Figure 2.2.
IR R IR
R
+ 
 +
vR vR
Figure 2.2. Voltage polarity and current direction in accordance with the passive sign convention
But if the voltage polarities and current directions are as shown in Figure 2.3, then,
(2.8)
vR = –RiR
R IR IR
R
 +
+ 
vR vR
Figure 2.3. Voltage polarity and current direction not in accordance to passive sign convention
Note: “Negative resistance,” as shown in (2.8), can be thought of as being a math model that
supplies energy.
2.3 Power Absorbed by a Resistor
A resistor, being a passive device, absorbs power. This absorbed power can be found from Ohm’s
law, that is,

  vR 2
= = = = = -----
I 2 R
t2 = 
and the power relation
Then,
(2.9)
The voltage, current, resistance and power relations are arranged in the pie chart shown in Figure
2.4.
Figure 2.4. Pie chart for showing relations among voltage, current, resistance, and power
Note:
A resistor, besides its resistance rating (ohms) has a power rating in watts commonly referred to as
the wattage of the resistor. Common resistor wattage values are ¼ watt, ½ watt, 1 watt, 2 watts, 5
watts and so on. This topic will be discussed in Section 2.16.
2.4 Energy Dissipated in a Resistor
A resistor, by its own nature, dissipates energy in the form of heat; it never stores energy. The
energy dissipated in a resistor during a time interval, say from to , is given by the integral of
the instantaneous power .Thus,
(2.10)
vR = RiR
pR = vR iR
pR vR iR RiR   iR RiR 2
vR
vR
R
 ----- 
R
P I
PR V R
P
I IR
VI
V 2
R
V
R P
V
P  R
P
I 2
V 2
P
V
I
POWER (Watts)
VOLTAGE (Volts)
CURRENT (Amperes)
RESISTANCE (Ohms)
t1 t2
pR
WR diss pR dt
t1

Nodes, Branches, Loops and Meshes
If the power is constant, say , then (2.10) reduces to
(2.11)
P
WR diss = Pt
Alternately, if the energy is known, we can find the power by taking the derivative of the energy,
that is,
(2.12)
= d WR diss
pR t d
Reminder:
When using all formulas, we must express the quantities involved in their primary units. For
instance in (2.11) above, the energy is in joules when the power is in watts and the time is in sec-onds.
2.5 Nodes, Branches, Loops and Meshes
Definition 2.1
A node is the common point at which two or more devices (passive or active) are connected. An
example of a node is shown in Figure 2.5.
+

Node
Figure 2.5. Definition of node
Definition 2.2
A branch is a simple path composed of one single device as shown in Figure 2.6.
R C vS
Branch
Figure 2.6. Definition of branch
+

Node
Definition 2.3
A loop is a closed path formed by the interconnection of simple devices. For example, the net-work
shown in Figure 2.7 is a loop.

R
A B C
Figure 2.7. Definition of a loop
Definition 2.4
A mesh is a loop which does not enclose any other loops. For example, in the circuit shown in Fig-ure
2.8, is both a loop and a mesh, but is a loop but not a mesh.
Figure 2.8. Example showing the difference between mesh and loop
2.6 Kirchhoff’s Current Law (KCL)
KCL states that the algebraic sum of all currents leaving (or entering) a node is equal to zero. For
example, in Figure 2.9, if we assign a plus (+) sign to the currents leaving the node, we must assign
a minus ( sign to the currents entering the node. Then by KCL,
(2.13)
Figure 2.9. Node to illustrate KCL
But if we assign a plus (+) sign to the currents entering the node and minus () sign to the cur-rents
leaving the node, then by KCL,
(2.14)
or
(2.15)
+ L
C
vS
ABEF ABCDEF
+

F E D
vS
R1
R2
L C
iS
– i1 – i2 + i3 + i4 = 0
i4
i1 i2
i3
i1 + i2 – i3 – i4 = 0
– i1 – i2 + i3 + i4 = 0

Kirchhoff’s Voltage Law (KVL)
We observe that (2.13) and (2.15) are the same; therefore, it does not matter which we choose
as plus (+).
Convention:
In our subsequent discussion we will assign plus (+) signs to the currents leaving the node.
2.7 Kirchhoff’s Voltage Law (KVL)
KVL states that the algebraic sum of the voltage drops (voltages from + to ) or voltage rises
(voltages from  to +) around any closed path (mesh or loop) in a circuit is equal to zero. For
example, in the circuit shown in Figure 2.10, voltages v1 , v2 , v3 , and v4
represent the voltages
across devices 1, 2, 3, and 4 respectively, and have the polarities shown.
+
Device 1
Device 2
 +
v1
v2
v3
v4
 +
Device 4
+
Device 3


A
Figure 2.10. Circuit to illustrate KVL
Now, if we assign a (+) sign to the voltage drops, we must assign a () sign to the voltage rises.
Then, by KVL starting at node and going clockwise we obtain:
(2.16)
A
– v1 – v2 + v3 + v4 = 0
or going counterclockwise, we obtain:
(2.17)
– v4 – v3 + v2 + v1 = 0
Alternately, if we assign a (+) sign to the voltage rises, we must assign a () sign to the voltage
drops. Then, by KVL starting again at node A and going clockwise we obtain:
(2.18)
v1 + v2 – v3 – v4 = 0
or going counterclockwise, we obtain:
(2.19)
v4 + v3 – v2 – v1 = 0
We observe that expressions (2.16) through (2.19) are the same.
Convention:
In our subsequent discussion we will assign plus (+) signs to voltage drops.

Definition 2.5
Two or more devices are said to be connected in series if and only if the same current flows through
them. For example, in the circuit of Figure 2.11, the same current i
flows through the voltage
source, the resistance, the inductance and the capacitance. Accordingly, this is classified as a
series circuit.
R
+
 L
C
vS
vAB
A
A A A A
G L C L G C
i iS S
B
B B B B
R
vR
+ 
Figure 2.11. A simple series circuit
Definition 2.6
Two or more devices are said to be connected in parallel if and only if the same voltage exists across
each of the devices. For example, in the circuit of Figure 2.12, the same voltage exists across
the current source, the conductance, the inductance, and the capacitance and therefore it is clas-sified
as a parallel circuit
Figure 2.12. A simple parallel circuit
Convention:
In our subsequent discussion we will adopt the conventional current flow, i.e., the current that
flows from a higher (+) to a lower () potential. For example, if in Figure 2.13 we are given the
indicated polarity,
Figure 2.13. Device with established voltage polarity
then, the current arrow will be pointing to the right direction as shown in Figure 2.14.

Kirchhoff’s Voltage Law (KVL)
R
vR
+ 
iR
Figure 2.14. Direction of conventional current flow in device with established voltage polarity
Alternately, if current flows in an assumed specific direction through a device thus producing a
voltage, we will assign a (+) sign at the terminal of the device at which the current enters. For
example, if we are given this designation a device in which the current direction has been estab-lished
as shown in Figure 2.15,
iR R
Figure 2.15. Device with established conventional current direction
then we assign (+) and () as shown in Figure 2.16.
iR R
+ 
vR
Figure 2.16. Voltage polarity in a device with established conventional current flow
Note:
Active devices, such as voltage and current sources, have their voltage polarity and current
direction respectively, established as part of their notation. The current through and the voltage
across these devices can easily be determined if these devices deliver power to the rest of the circuit.
Thus with the voltage polarity as given in the circuit of Figure 2.17 (a), we assign a clockwise
direction to the current as shown in Figure 2.17 (b). This is consistent with the passive sign con-vention
since we have assumed that the voltage source delivers power to the rest of the circuit.
+

Rest of the
Circuit
i
(b)
vS vS
+

Rest of the
Circuit
(a)
Figure 2.17. Direction of conventional current flow produced by voltage sources
Likewise, in the circuit of Figure 2.18 (a) below, the direction of the current source is clockwise,
and assuming that this source delivers power to the rest of the circuit, we assign the voltage
polarity shown in Figure 2.18 (b) to be consistent with the passive sign convention.

iS iS
Rest of the
Circuit
v
Figure 2.18. Voltage polarity across current sources
The following facts were discussed in the previous chapter but they are repeated here for empha-sis.
There are two conditions required to setup and maintain the flow of an electric current:
1. There must be some voltage (potential difference) to provide the energy (work) which will force
electric current to flow in a specific direction in accordance with the conventional current flow
(from a higher to a lower potential).
2. There must be a continuous (closed) external path for current to flow around this path (mesh
or loop).
The external path is usually made of two parts: (a) the metallic wires and (b) the load to which the
electric power is to be delivered in order to accomplish some useful purpose or effect. The load
may be a resistive, an inductive, or a capacitive circuit, or a combination of these.
2.8 Single Mesh Circuit Analysis
We will use the following example to develop a stepbystep procedure for analyzing (finding cur-rent,
voltage drops and power) in a circuit with a single mesh.
Example 2.1
For the series circuit shown in Figure 2.19, we want to find:
a. The current i which flows through each device
b. The voltage drop across each resistor
c. The power absorbed or delivered by each device
Rest of the
Circuit
+

(a) (b)

Single Mesh Circuit Analysis
vS1 vS2 vS3
+
4  64 V
6 
+
+

R1
R2
200 V 80 V
R4 R3
10  8 
Figure 2.19. Circuit for Example 2.1
Solution:
a. Step 1: We do not know which voltage source(s) deliver power to the other sources, so let us
assume that the current flows in the clockwise direction* as shown in Figure 2.20.
i
+  + 
vS1 vS2 vS3
+
4  64 V
6 
+
+

R1
R2
i
200 V 80 V
R4 R3
 +  +
10  8 
A
Figure 2.20. Circuit for Example 2.1 with assumed current direction
Step 2: We assign (+) and () polarities at each resistor’s terminal in accordance with the
established passive sign convention.
Step 3: By application of KVL and the adopted conventions, starting at node and going
clockwise, we obtain:
(2.20)
– vS1 + vR1 + vS2 + vR2 + vS3 + vR3 + vR4 = 0
and by Ohm’s law,
vR1 = R1i vR2 = R2i vR3 = R3i vR4 = R4i
Then, by substitution of given values into (2.20), we obtain
or
or
(2.21)
– 200 + 4i + 64 + 6i + 80 + 8i + 10i = 0
28i = 56
i = 2 A
* Henceforth, the current direction will be assumed to be that of the conventional current flow.
A

b. Knowing the current from part (a), we can now compute the voltage drop across each resis-tor
i
v = Ri
vR1 = 4  2 = 8 V vR2 = 6  2 = 12 V
vR3 = 8  2 = 16 V vR4 = 10  2 = 20 V
pR1 = 8  2 = 16 w pR2 = 12  2 = 24 w
pR3 = 16  2 = 32 w pR4 = 20  2 = 40 w
= = = 80  2 = 160 w
–200  2 = –400 w pVS2
64  2 = 128 w pVS3
TABLE 2.1 Power delivered or absorbed by each device on the circuit of Figure 2.19
Device Power Delivered (watts) Power Absorbed (watts)
p = vi
pVS1
200 V Source 400
64 V Source 128
80 V Source 160
4 W Resistor 16
6 W Resistor 24
8 W Resistor 32
10 W Resistor 40
Total 400 400
using Ohm’s law .
(2.22)
c. The power absorbed (or delivered) by each device can be found from the power relation
. Then, the power absorbed by each resistor is
(2.23)
and the power delivered (or absorbed) by each voltage source is
(2.24)
From (2.24), we observe that the 200 volt source absorbs 400 watts of power. This means
that this source delivers (supplies) 400 watts to the rest of the circuit. However, the other two
voltage sources receive (absorb) power from the 200 volt source. Table 2.1 shows that the con-servation
of energy principle is satisfied since the total absorbed power is equal to the power
delivered.
Example 2.2
Repeat Example 2.1 with the assumption that the current flows counterclockwise.
i

Single Mesh Circuit Analysis
Solution:
We denote the current as i ' (i prime) for this example. Then, starting at Node A
and going
counterclockwise, the voltage drops across each resistor are as indicated in Figure 2.21.
+  + 
vS1 vS2 vS3
+
4  64 V
6 
+
+

R1
R2
i
200 V 80 V
R4 R3
 +  +
10  8 
A
Repeating Steps 2 and 3 of Example 2.1, we obtain:
(2.25)
Next, by Ohm’s law,
vR4 + vR3 – vS3 + vR2 – vS2 + vR1 + vS1 = 0
vR1 = R1i' vR2 = R2i' vR3 = R3i' vR4 = R4i'
By substitution of given values, we obtain
or
or
(2.26)
200 + 4i' – 64 + 6i' – 80 + 8i' + 10i' = 0
28i' = –56
i' = –2A
i' = –i
Comparing (2.21) with (2.26) we observe that as expected.
Definition 2.7
A single nodepair circuit is one in which any number of simple elements are connected between
the same pair of nodes. For example, the circuit of Figure 2.22 (a), which is more conveniently
shown as Figure 2.22 (b), is a single nodepair circuit.

A
A A A A
G L C L G C
i iS S
B
B B B B
(a) (b)
Figure 2.22. Circuit with a single nodepair
2.9 Single NodePair Circuit Analysis
We will use the following example to develop a stepbystep procedure for analyzing (finding cur-rents,
voltage drop and power) in a circuit with a single nodepair.
iS1 iS2
iS3
G1 G2 G3
4 –1 6 –1 8 –1
12 A 18 A 24 A
A B
Example 2.3
For the parallel circuit shown in Figure 2.23, find:
a. The voltage drop across each device
b. The current i which flows through each conductance
c. The power absorbed or delivered by each device
Solution:
a. Step 1: We denote the single nodepair with the letters and as shown in Figure 2.24. It is
important to observe that the same voltage (or potential difference) exists across each
device. Node B
is chosen as our reference node and it is convenient to assume that
this reference node is at zero potential (ground) as indicated by the symbol

Single NodePair Circuit Analysis
A A A A A A
iG1 iG2
iS1 iS2 iS3
iG3
G1 G2 G3
vAB
4 –1 6 –1 8 –1
12 A 18 A 24 A
B B B B B B
Figure 2.24. Circuit for Example 2.3 with assumed current directions
Step 2: We assign currents through each of the conductances , , and in accordance
G1 G2 G3
with the conventional current flow. These currents are shown as , , and .
iG1 iG2 iG3
Step 3: By application of KCL and in accordance with our established convention, we choose
node which is the plus (+) reference point and we form the algebraic sum of the
currents leaving (or entering) this node. Then, with plus (+) assigned to the currents
leaving this node and with minus () entering this node we obtain
(2.27)
A
and since
(2.28)
+ + + – + iG3 = 0
– iG1 i
i
S1
iG1 = G1vAB iG2 = G2vAB iG3 = G3vAB
by substitution into (2.27),
(2.29)
– iS1 + G1vAB + iS2 + G2vAB – iS3 + G3vAB = 0
vAB
Solving for , we obtain
(2.30)
S2
iG2 i
S3
vAB
iS1 – iS2 + iS3
G1 + G2 + G3
= --------------------------------
and by substitution of the given values, we obtain
(2.31)
or
(2.32)
b. From (2.28),
(2.33)
vAB
12 – 18 + 24
4 + 6 + 8
= ------------------------------
vAB = 1 V
iG1 = 4 iG2 = 6 iG3 = 8
and we observe that with these values, (2.27) is satisfied.

c. The power absorbed (or delivered) by each device can be found from the power rela-tion
. Then, the power absorbed by each conductance is
(2.34)
p = vi
pG1 = 1  4 = 4 w
pG2 = 1  6 = 6 w
pG3 = 1  8 = 8 w
and the power delivered (or absorbed) by each current source is
(2.35)
pI1 = 1  –12 = –12 w
pI2 = 1  18 = 18 w
pI3 = 1  –24 = –24 w
From (2.35) we observe that the 12 A and 24 A current sources absorb –12 w and
–24 w
respectively. This means that these sources deliver (supply) a total of to the rest of the
circuit. The source absorbs power.
Table 2.2 shows that the conservation of energy principle is satisfied since the absorbed power
is equal to the power delivered.
TABLE 2.2 Power delivered or absorbed by each device of Figure 2.23
Device Power Delivered (watts) Power Absorbed (watts)
12 A Source 12
18 A Source 18
24 A Source 24
4 Conductance 4
6 Conductance 6
8 Conductance 8
Total 36 36
36 w
18 A
–1
–1
–1

Voltage and Current Source Combinations
2.10 Voltage and Current Source Combinations
Definition 2.8
Two or more voltage sources connected in series are said to be series aiding when the plus (+)
terminal of any one voltage source is connected to the minus () terminal of another, or when
the minus () terminal of any one voltage source is connected to the plus () terminal of
another.
Two or more series aiding voltage sources may be replaced by an equivalent voltage source
whose value is the algebraic sum of the individual voltage sources as shown in Figure 2.25.
200 V 64 V 80 V
+
+
+

v1 v2 v3
vAB = v1 + v2 + v3
+ 200 + 64 + 80 = 344 V

A
B
Figure 2.25. Addition of voltage sources in series when all have same polarity
A good example of combining voltage sources as series aiding is when we connect several AA size
batteries each rated at to power up a hand calculator, or a small flashlight.
Definition 2.9
Two or more voltage sources connected in series are said to be series opposing when the plus (+)
terminal of one voltage source is connected to the plus () terminal of the other voltage source
or when the minus () of one voltage source is connected to the minus () terminal of the other
voltage source. Two series opposing voltage sources may be replaced by an equivalent voltage
source whose value is the algebraic difference of the individual voltage sources as shown in Fig-ure
2.26.
1.5 v
200 V 64 V
+

+ 
B
v1 v2
vAB = v1 – v2
+ 
200  64 = 136 V
A
Figure 2.26. Addition of voltage sources in series when they have different polarity
Definition 2.10
Two or more current sources connected in parallel are said to be parallel aiding when the arrows
indicating the direction of the current flow have the same direction. They can be combined into
a single current source as shown in Figure 2.27.

i iT = i1 + i2 + i3 i1 3 i2 iT
Figure 2.27. Addition of current sources in parallel when all have same direction
Definition 2.11
Two or more current sources connected in parallel are said to be parallel opposing when the arrows
indicating the direction of the current flow have opposite direction. They can be replaced by an
equivalent current source whose value is the algebraic difference of the individual current sources
as shown in Figure 2.28.
i1 i2
iT = –i1 + i2
iT
Figure 2.28. Addition of current sources in parallel when they have opposite direction
2.11 Resistance and Conductance Combinations
Often, resistors are connected in series or in parallel. With either of these connections, series or
parallel, it is possible to replace these resistors by a single resistor to simplify the computations of
the voltages and currents. Figure 2.29 shows resistors connected in series.
R1 R2 R3 RN
i
Figure 2.29. Addition of resistances in series
The combined or equivalent resistance is
or
12 A 18 A 24 A 54 A
18 A 24 A 6 A
n
A B
Rest of the circuit
Req
Req
vAB
--------
i
vR1
i
-------
vR2
i
-------
vR3
i
------- 
vRn
i
= = + + + + --------

Resistance and Conductance Combinations
(2.36)
n
Req R1 + R2 + R3 +  + Rn RK
k = 1
= =
For Resistors in Series
Example 2.4
For the circuit of Figure 2.30, find the value of the current i
after combining the voltage sources
to a single voltage source and the resistances to a single resistor.
+  + 
vS1 vS2 vS3
+
4  64 V
6 
+
+

R1
R2
i
200 V 80 V
R4 R3
 +  +
10  8 
A
Solution:
We add the values of the voltage sources as indicated in Definitions 8 and 9, we add the resis-tances
in accordance with (2.36), and we apply Ohm’s law. Then,
(2.37)
= = = ----- = 2 A v
R
i  ------- 200 64 80 +   –
------------------------------------- 56
28
28
Next, we consider the case where resistors are connected in parallel as shown in Figure 2.31.
n
R1
i1
i2
iT iT
A B
R2
Rn
in
vAB
Figure 2.31. Addition of resistances in parallel
By KCL,
(2.38)
iT = i1 + i2 +  + in

The same voltage exists across each resistor; therefore, dividing each term of (2.38) by , we
obtain
(2.39)
and since , then and thus (2.39) can be written as
---------- 1
------ 1
------  1
= + + + ------
----- 1
-----  1
= + + + ------
------ 1
= + ------
n
or
(2.40)
For the special case of two parallel resistors, (2.40) reduces to
or
(2.41)
where the designation indicates that and are in parallel.
Also, since , from (2.38),
(2.42)
that is, parallel conductances combine as series resistors do.
Example 2.5
In the circuit of Figure 2.32,
a. Replace all resistors with a single equivalent resistance
b. Compute the voltage across the current source.
VAB
--------
iT
vAB
--------
i1
vAB
i2
vAB
-------- 
in
vAB
= + + + --------
v  i = R i  v = 1  R
1
RAB
R1
R2
Rn
1
Req
-------- 1
R1
R2
Rn
For Resistors in Parallel
1
Req
-------- 1
R1
R2
Req R1||R2
R1  R2
R1 + R2
= = ------------------
R1||R2 R1 R2
G = 1  R
Geq G1 + G2 +  + Gn Gk
k = 1
= =
Req
vAB

Resistance and Conductance Combinations
R1 R2 R3 R4 R5
3  12  4  20 
+
vAB
11
6
----- A
5 
iS

Solution:
a. We could use (2.40) to find the equivalent resistance Req
. However, it is easier to form
groups of two parallel resistors as shown in Figure 2.33 and use (2.41) instead.
+
vAB
11
6
----- A
R1 R2 R3 R4 R5
3  12  4  20 
5 
iS

Figure 2.33. Groups of parallel combinations for the circuit of Example 2.5.
Then,
Also,
R2||R3
12  4
12 + 4
= --------------- = 3 
R4||R5
20  5
20 + 5
= --------------- = 4 
and the circuit reduces to that shown in Figure 2.34.
R1
+
vAB 3  3  4 
11
6
----- A
iS

Figure 2.34. Partial reduction for the circuit of Example 2.5
Next,
Finally,
3||3 3  3
= ------------ = 1.5 
3 + 3
---------------- 12
Req 1.5||4 1.5  4
= = = ----- 
1.5 + 4
and the circuit reduces to that shown in Figure 2.35
11

Figure 2.35. Reduction of the circuit of Example 2.5 to its simplest form
----- 12
= =  ----- = 2 V

vR2
b. The voltage across the current source is
(2.43)
2.12 Voltage Division Expressions
In the circuit of Figure 2.36, , , and are known.
Figure 2.36. Circuit for the derivation of the voltage division expressions
For the circuit of Figure 2.36, we will derive the voltage division expressions which state that:
These expressions enable us to obtain the voltage drops across the resistors in a series circuit sim-ply
by observation.
Derivation:
By Ohm’s law in the circuit of Figure 2.36 where is the current flowing through i, we obtain
(2.44)
Also,
or
(2.45)
11
6
----- A
vAB
Req
12
11
----- 
iS
+

vAB
vAB IReq
11
6
11
vS R1 R2
+
v  S
R2
R1
+
+

vR1
vR1
R1
R1 + R2
------------------vS and vR2
R2
R1 + R2
= = ------------------vS
i
vR1 = R1i and vR2 = R2i
R1 + R2i = vS
i
vS
R1 + R2
= ------------------

Voltage Division Expressions
and by substitution of (2.45) into (2.44) we obtain the voltage division expressions below.
(2.46)
vR1
R1
R1 + R2
= ------------------vS and vR2
= ------------------vS
R2
R1 + R2
VOLTAGE DIVISION EXPRESSIONS
Example 2.6
In the network of Figure 2.37, the arrows indicate that resistors and are variable and that
the power supply is set for .
a. Compute vR1 and vR2 if R1 and R2 are adjusted for 7  and 5 
respectively.
b. To what values should and be adjusted so that , , and
?
R1 R2
12 V
R1 R2 vR1 = 3 V vR2 = 9 V
R1 + R2 = 12 
c. Using Simulink and SimPowerSystems*, create a model to simulate the voltage
+
Power
Supply
(Voltage
Source)

vR1
12 V
+

+
vR2

R1
R2
Figure 2.37. Network for Example 2.6
Solution:
a. Using the voltage division expressions of (2.46), we obtain
and
vR2
vR1
R1
R1 + R2
= = ------------  12 = 7 V
------------------vS
7
7 + 5
vR2
R2
R1 + R2
= = ------------  12 = 5 V
------------------vS
5
7 + 5
* For an introduction to Simulink and SimPowerSystems, please refer to Appendices B and C respectively.

b. Since , , and the voltage drops are proportional to
the resistances, it follows that if we let and , the voltage drops and
R1
+
-
VM
v
will be and respectively.
c.
Figure 2.38. Simulink / SimPower Systems model for Example 2.6
2.13 Current Division Expressions
In the circuit shown in Figure 2.39, , , and are known.
Figure 2.39. Circuit for the derivation of the current division expressions
For the circuit of Figure 2.39, we will derive the current division expressions which state that
and these expressions enable us to obtain the currents through the conductances (or resistances)
in a parallel circuit simply by observation.
Derivation:
By Ohm’s law for conductances, we obtain
(2.47)
Also,
vR1 + vR2 = 3 + 9 = 12 V R1 + R2 = 12 
R1 = 3  R2 = 9  vR1
vR2 3 V 9 V
R1=7 Ohm, R2=5 Ohm
R2
VM = Voltage Measuremt
Powergui is an environmental block. It is
necessary for simulation of any Simulink
model containg SimPowerSystems blocks
Continuous
powergui
5.00
Display
12V DC
iS G1 G2
v
iS
iG1
iG2
G1 G2
R1 R2
iG1
G1
G1 + G2
-------------------iS and iG2
G2
G1 + G2
= = -------------------iS
iG1 = G1v and iG2 = G2v

Current Division Expressions
or
(2.48)
G1 + G2v = iS
v
and by substitution of (2.48) into (2.47)
(2.49)
Also, since
iG1
G1
G1 + G2
= -------------------iS and iG2
= -------------------iS
R1
by substitution into (2.49) we obtain
(2.50)
iS
G1 + G2
= -------------------
G2
G1 + G2
1
G1
= ------ R2
1
G2
= ------
iR1
R2
R1 + R2
= ------------------iS and iR2
= ------------------iS
R1
R1 + R2
CURRENT DIVISION EXPRESSIONS
Example 2.7
For the circuit inFigure 2.40, compute the voltage drop v
. Verify your answer with a Simulink /
SimPowerSystems model.
+
v
R2
R1
4 
20 
R4
5 
12 
3 A

iS
R3
Solution:
The current source iS divides into currents i1 and i2
as shown in Figure 2.40. We observe that
the voltage v is the voltage across the resistor R1
. Therefore, we are only interested in current
. This is found by the current division expression as
i1
i1
R2 + R3 + R4
R1 + R2 + R3 + R4
--------------------------------------------  iS 4 + 5 + 20
------------------------------------  3 87
= = = ----- A
12 + 4 + 5 + 20
41

+
v
iS
R2
R1
R3
4 
i2 i1
20 
R4
5 
12 
3 A

Figure 2.41. Application of current division expressions for the circuit of Example 2.7
v
–-----  12 1044
v –i1R1 87
= = = –----------- V
41
41
v = –25.46 V
Resistances in Ohms
CCS = Current Controlled Source
-0.88
Display 2
R1= 29 R2 = 12
-3
Constant
CCS
+ -i CM 2
CM 1
+
s
-
CM = Current Measurement
Display 3
v +-
VM
-25.46
Display 4
-2.12
VM = Voltage Measurement
Continuous
powergui
3.00
+ i -
Display 1
+ i -
CM 3
and observing the passive sign convention, the voltage is
or
2.14 Standards for Electrical and Electronic Devices
Standardization of electronic components such as resistors, capacitors and diodes is carried out by
various technical committees. In the United States, the Electronics Industries Association (EIA)
and the American National Standards Institute (ANSI) have established and published several
standards for electrical and electronic devices to provide interchangeability among similar prod-ucts
made by different manufacturers. Also, the U.S. Department of Defense or its agencies issue

Resistor Color Code
standards known as Military Standards, or simply MILstds. All of the aforementioned standards
are updated periodically. The interested reader may find the latest revisions in the Internet or
the local library.
2.15 Resistor Color Code
The Resistor Color Code is used for marking and identifying pertinent data for standard resistors.
Figures 2.43 and 2.44 show the color coding scheme per EIA Standard RS279 and MILSTD
1285A respectively.
1st
2nd
3rd
Significant
Figures
Tolerance
Wider Space to
Identify Direction
of Reading
Left to Right
Multiplier
Figure 2.43. Resistor Color Code per EIA Standard RS279
Significant
Figures
Failure Rate Level
on Established
Reliability Levels
Only
Tolerance
1st
2nd
Multiplier
Figure 2.44. Resistor Color Code per MILSTD1285A
In a color coded scheme, each color represents a single digit number, or conversely, a single digit
number can be represented by a particular color band as shown in Table 2.3 that is based on
MILSTD1285A color code.
As shown in Figure 2.44, the first and second bands designate the first and second significant
digits respectively, the third represents the multiplier, that is, the number by which the first two
digits are multiplied, and the fourth and fifth bands, if they exist, indicate the tolerance and fail-ure
rate respectively. The tolerance is the maximum deviation from the specified nominal value
and it is given as a percentage. The failure rate is the percent probability of failure in a 1000hour
time interval.

TABLE 2.3 Resistor values per MIL-STD-1285A
Let A and B represent the first and second significant digits and C represent the multiplier. Then
the resistance value is found from the expression
(2.51)
Example 2.8
The value of a resistor is coded with the following colored band code, left to right: Brown, Green,
Blue, Gold, Red. What is the value, tolerance, and probability of failure for that resistor?
Solution:
Table 2.3 yields the following data: Brown (1st significant digit) = 1, Green (2nd significant digit)
= 5, and Blue (multiplier) = 1,000,000. Therefore, the nominal value of this resistor is
15,000,000 Ohms or 15 M. Theth band is Gold indicating a ±5% tolerance meaning that the
maximum deviation from the nominal value is 15,000,000 ±5% = 15,000,000 × ±0.05 =
±750,000 Ohms or ±0.75 M. That is, this resistor can have a value anywhere between 14.25
M and 15.75 M. Since the 5th band is Red, there is a 0.1% probability that this resistor will
fail after 1000 hours of operation.
Color Code
1st & 2nd
Digits
Multiplier
(3rd Digit)
Tolerance
(Percent)
Fail Rate
(Percent)
Black 0 1
Brown 1 10 1 1
Red 2 100 2 0.1
Orange 3 1000 0.01
Yellow 4 10000 0.001
Green 5 100000 0.5
Blue 6 1000000 0.25
Violet 7 0.1
Gray 8
White 9
Gold 0.1 5
Silver 0.01 10
No Color 20
R = 10  A + B10C

Power Rating of Resistors
2.16 Power Rating of Resistors
As it was mentioned in Section 2.2, a resistor, besides its resistance rating (ohms) has a power
rating (watts) commonly referred to as the wattage of the resistor, and common resistor wattage
values are ¼ watt, ½ watt, 1 watt, 2 watts, 5 watts and so on. To appreciate the importance of
the wattage of a resistor, let us refer to the voltage divider circuit of Example 2.6, Figure 2.37
where the current is 12 V  12  = 1 A . Using the power relation pR = i2R
, we find that the
wattage of the 7  and 5 
resistors would be 7 watts and 5 watts respectively. We could also
divide the 12 volt source into two voltages of 7 V and 5 V using a 7 k and a 5 k
resistor.
Then, with this arrangement the current would be . The wattage of the
12 V  12 k = 1 mA
7 k 5 k 10–3  2 7 103
  7 103
– =  W = 7 mW
and resistors would then be and
respectively.
10–3  2 5 103
  = 5 mW
2.17 Temperature Coefficient of Resistance
The resistance of any pure metal, such as copper, changes with temperature. For each degree
that the temperature of a copper wire rises above Celsius, up to about , the resis-tance
20 C 200 C
increases 0.393 of 1 percent of what it was at 20 degrees Celsius. Similarly, for each degree
that the temperature drops below 20 C , down to about –50C
, the resistance decreases 0.393 of
1 percent of what it was at . This percentage of change in resistance is called the Tempera-ture
20 C
Coefficient of Resistance. In general, the resistance of any pure metal at temperature T in
degrees Celsius is given by
(2.52)
R = R201 + 20T – 20
where is the resistance at and is the temperature coefficient of resistance at
.
R20 20 C 20
20 C
Example 2.9
The resistance of a long piece of copper wire is 48  at 20 C
.
a. What would the resistance be at 50C
?
b. Construct a curve showing the relation between resistance and temperature.
Solution:
a. The temperature rise is 50 – 20 = 30
degrees Celsius and the resistance increases 0.393% for
every degree rise. Therefore the resistance increases by . This represents
30  0.393 = 11.79%

an increase of in resistance or 5.66 . Therefore, the resistance at 50 degrees
Celsius is .
b. The relation of (2.52) is an equation of a straight line with . This straight line
is easily constructed with the Microsoft Excel spreadsheet shown in Figure 2.45.
From Figure 2.45, we observe that the resistance reaches zero value at approximately .
100
80
60
40
20
Figure 2.45. Spreadsheet for construction of equation (2.52)
2.18 Ampere Capacity of Wires
For public safety, electric power supply (mains) wiring is controlled by local, state and federal
boards, primarily on the National Electric Code (NEC) and the National Electric Safety Code. More-over,
many products such as wire and cable, fuses, circuit breakers, outlet boxes and appliances
are governed by Underwriters Laboratories (UL) Standards which approves consumer products
such as motors, radios, television sets etc.
Table 2.4 shows the NEC allowable currentcarrying capacities for copper conductors based on
the type of insulation.
The ratings in Table 2.4 are for copper wires. The ratings for aluminum wires are typically 84% of
these values. Also, these rating are for not more than three conductors in a cable with tempera-ture
or . The NEC contains tables with correction factors at higher temperatures.
2.19 Current Ratings for Electronic Equipment
There are also standards for the internal wiring of electronic equipment and chassis. Table 2.5
provides recommended current ratings for copper wire based on ( for wires smaller
0.1179  48 
48 + 5.66 = 53.66 
slope = R2020
–235C
Temp Resistance
(deg C) (Ohms)
-250 -2.9328
-240 -1.0464
-230 0.84
-220 2.7264
-210 4.6128
-200 6.4992
-190 8.3856
-180 10.272
-170 12.1584
-160 14.0448
-150 15.9312
-140 17.8176
Resistance of Copper Wire versus Temperature
0
-250 -200 -150 -100 -50 0 50 100 150 200 250
Degrees Celsius
Ohms
30C 86F
45C 40C

Current Ratings for Electronic Equipment
than 22 AWG. Listed also, are the circular mils and these denote the area of the cross section of
each wire size. A circular mil is the area of a circle whose diameter is 1 mil (onethousandth of an
inch). Since the area of a circle is proportional to the square of its diameter, and the area of a cir-cle
one mil in diameter is one circular mil, the area of any circle in circular mils is the square of
its diameter in mils.
TABLE 2.4 Current Ratings for Electronic Equipment and Chassis Copper Wires
Wire Size Maximum Current (Amperes)
AWG Circular Mils
Nominal
Resistance
(Ohms/1000 ft)
at 100 C
Wire in Free
32 63.2 188 0.53 0.32
30 100.5 116 0.86 0.52
28 159.8 72 1.4 0.83
26 254.1 45.2 2.2 1.3
24 404 28.4 3.5 2.1
22 642.4 22 7 5
20 10.22 13.7 11 7.5
18 1624 6.5 16 10
16 2583 5.15 22 13
14 4107 3.2 32 17
12 6530 2.02 41 23
10 10380 1.31 55 33
8 16510 0.734 73 46
6 26250 0.459 101 60
4 41740 0.29 135 80
2 66370 0.185 181 100
1 83690 0.151 211 125
0 105500 0.117 245 150
00 133100 0.092 283 175
000 167800 0.074 328 200
0000 211600 0.059 380 225
† Dry Locations Only ‡ Nickel or nickel-coated copper only
Air
Wire Confined
in Insulation

A milfoot wire is a wire whose length is one foot and has a crosssectional area of one circular
mil.
The resistance of a wire of length can be computed with the relation
(2.53)
l
R l
d2 = -----
where  = resistance per milfoot, l = length of wire in feet, d
= diameter of wire in mils, and
is the resistance at .
R 20 C
Example 2.10
Compute the resistance per mile of a copper conductor inch in diameter given that the resis-tance
1  8
10.4  20 C
1  8 in = 0.125 in = 125 mils
d2 ----- 10.4  5280
R l
1252 = = ---------------------------- = 3.51 
100C
RT = R1001 + 0.004T – 100
RT R100 100C
T
1000 ft AWG 12 30C
1000 ft AWG 12 100C
per milfoot of copper is at .
Solution:
and from (2.53)
Column 3 of Table 2.5 shows the copper wire resistance at . Correction factors must be
applied to determine the resistance at other temperatures or for other materials. For copper, the
conversion equation is
(2.54)
where is the resistance at the desired temperature, is the resistance at for copper,
and is the desired temperature.
Example 2.11
Compute the resistance of of size copper wire at .
Solution:
From Table 2.5 we find that the resistance of of size copper wire at is
. Then, by (2.54), the resistance of the same wire at is
2.02  30C
R30C = 2.021 + 0.00430 – 100 = 1.45 

Copper Conductor Sizes for Interior Wiring
2.20 Copper Conductor Sizes for Interior Wiring
In the design of an interior electrical installation, the electrical contractor must consider two
important factors:
a. The wiring size in each section must be selected such that the current shall not exceed the
current carrying capacities as defined by the NEC tables. Therefore, the electrical contractor
must accurately determine the current which each wire must carry and make a tentative
selection of the size listed in Table 2.4.
b. The voltage drop throughout the electrical system must then be computed to ensure that it
does not exceed certain specifications. For instance, in the lighting part of the system referred
to as the lighting load, a variation of more than in the voltage across each lamp causes an
unpleasant variation in the illumination. Also, the voltage variation in the heating and air
conditioning load must not exceed .
Important!
The requirements stated here are for instructional purposes only. They change from time to
time. It is, therefore, imperative that the designer consults the latest publications of the applica-ble
codes for compliance.
5%
10%
Example 2.12
Figure 2.46 shows a lighting load distribution diagram for an interior electric installation.
kwhr
Meter Panel
Board
L1
L2
L3
Circuit
Breaker
Utility
Company
Switch
Branch
Lines
Main
Lines
Lighting
Load
Figure 2.46. Load distribution for an interior electric installation
The panel board is 200 feet from the meter. Each of the three branches has 12 outlets for 75 w,
120 volt lamps. The load center is that point on the branch line at which all lighting loads may be
considered to be concentrated. For this example, assume that the distance from the panel to the
load center is 60 ft. Compute the size of the main lines. Use T (thermoplastic insulation) type
copper conductor and base your calculations on temperature environment.
25C

Solution:
It is best to use a spreadsheet for the calculations so that we can compute sizes for more and dif-ferent
branches if need be.
The computations for Parts I and II are shown on the spreadsheet in Figures 2.47 and 2.48 where
from the last line of Part II we see that the percent line drop is and this is more than twice
the allowable drop. With the voltage variation the brightness of the lamps would
vary through wide ranges, depending on how many lamps were in use at one time.
A much higher voltage than the rated would cause these lamps to glow far above their
rated candle power and would either burn them immediately, or shorten their life considerably. It
is therefore necessary to install larger than main line. The computations in Parts III
through V of the spreadsheet of Figures 2.47 and 2.48 indicate that we should not use a conduc-tor
less than size .
12.29
5% 12.29%
120 V
12 AWG
6 AWG

Copper Conductor Sizes for Interior Wiring
Figure 2.47. Spreadsheet for Example 2.12, Parts I and II
Part I Sizing in Accordance with NEC (Table 2.2)
Step Description Calculation Value Units
1 Meter to Panel Board Distance 200 ft
2 No. of Branches from Panel Board 3
3 Outlets per branch 12
4 Outlet Voltage Rating 120 V
5 Power drawn from each outlet 75 w
6 Power required by each branch Step 3 x Step 5 900 w
7 Current drawn by each branch Step 6 / Step 4 7.50 A
8 Required wire size for branches * Table 2.4 14 AWG Carries up to 15 A
9 Current required by the main line Step 2 x Step 7 22.50 A
10 Required wire size for main line Table 2.4 12 AWG Carries up to 20 A
Part II Check for Voltage Drops
11 Distance from panel to load center 60 ft
12 Number of wires in each branch 2
13 Total length of wire in each branch Step 11 x Step 12 120 ft
14 Specified Temperature in 0C 25 0C
15 Resistance of 1000 ft 14 AWG wire at 100 0C Table 2.5 3.20 
16 Resistance of 1000 ft 14 AWG wire at Spec. Temp Equation (2.54) 2.24 
17 Resistance of actual length of wire (Step 13 x Step 16) / 1000 0.269 
18 Voltage drop in each branch Step 7 x Step 17 2.02 V
19 Number of wires in each main 2
20 Total length of wire in main line Step 1 x Step 19 400 ft
23 Resistance of actual length of wire (Step 20 x Step 22) /1000 0.57 
24 Voltage drop in main line Step 9 x Step 23 12.73 V
25 Voltage drop from meter to load center Step 18 + Step 24 14.74 V
26 Percent line drop (Step 25 / Step 4) x 100 12.29 %
* No size smaller can be installed; smaller sizes
have insufficient mechanical strength

Figure 2.48. Spreadsheet for Example 2.12, Parts III, IV, and V
Part III Recalculation Using 10 AWG Conductor Size
27 Next larger conductor size for main line Table 2.4 10 AWG Carries up to 30 A
30 Resistance of actual length of wire (Step 20 x Step 29)/1000 0.367 
Part IV Recalculation Using 8 AWG Conductor Size
Part V Recalculation Using 6 AWG Conductor Size

Summary
2.21 Summary
 Ohm’s Law states that the voltage across a device is proportional to the current through that
device and the resistance is the constant of proportionality.
 Open circuit refers to an open branch (defined below) in a network. It can be thought of as a
resistor with infinite resistance (or zero conductance). The voltage across the terminals of an
open may have a finite value or may be zero whereas the current is always zero.
 Short circuit refers to a branch (defined below) in a network that contains no device between
its terminals, that is, a piece of wire with zero resistance. The voltage across the terminals of a
short is always zero whereas the current may have a finite value or may be zero.
 A resistor absorbs power.
 A resistor does not store energy. The energy is dissipated in the form of heat.
 A node is a common point where one end of two or more devices are connected.
 A branch is part of a network that contains a device and its nodes.
 A mesh is a closed path that does not contain other closed paths
 A loop contains two or more closed paths.
 Kirchoff’s Current Law (KCL) states that the algebraic sum of the currents entering (or leav-ing)
a node is zero.
 Kirchoff’s Voltage Law (KVL) states that the algebraic sum of the voltage drops (or voltage
rises) around a closed mesh or loop is zero.
 Two or more devices are said to be connected in series if and only if the same current flows
through them.
 Two or more devices are said to be connected in parallel if and only if the same voltage exists
across their terminals.
 A series circuit with a single mesh can be easily analyzed by KVL.
 A parallel circuit with a single node pair can be easily analyzed by KCL.
 If two or more voltage sources are in series, they can be replaced by a single voltage source
with the proper polarity.
 If two or more current sources are in parallel, they can be replaced by a single current source
with the proper current direction.
 If two or more resistors are connected in series, they can be replaced by an equivalent resis-tance
whose value is

n
Req R1 + R2 + R3 + + Rn RK
 If two or more resistors are connected in parallel, they can be replaced by an equivalent resis-tance
-------- 1
------ 1
------  1
= + + + ------

vR2
whose value is
 For the special case of two parallel resistors, the equivalent resistance is found from the relation
 Conductances connected in series combine as resistors in parallel do.
 Conductances connected in parallel combine as resistors in series do.
 For the simple series circuit below
the voltage division expressions state that:
 For the simple parallel circuit below
the current division expressions state that:
k = 1
= =
1
Req
R1
R2
Rn
Req R1||R2
R1  R2
R1 + R2
= = ------------------
+
v  S
R2
R1
+
+

vR1
vR1
R1
R1 + R2
------------------vS and vR2
R2
R1 + R2
= = ------------------vS
v
iS
iG1
iG2
G1 G2
R1 R2
iR1
R2
R1 + R2
------------------iS and iR2
R1
R1 + R2
= = ------------------iS

Summary
 In the United States, the Electronics Industries Association (EIA) and the American National
Standards Institute (ANSI) have established and published several standards for electrical
and electronic devices to provide interchangeability among similar products made by different
manufacturers.
 The resistor color code is used for marking and identifying pertinent data for standard resis-tors.
Two standards are the EIA Standard RS279 and MILSTD1285A.
 Besides their resistance value, resistors have a power rating.
 The resistance of a wire increases with increased temperature and decreases with decreased
temperature.
 The current ratings for wires and electronic equipment are established by national standards
and codes.

2.22 Exercises
Multiple Choice
1.Ohm’s Law states that
A. the conductance is the reciprocal of resistance
B. the resistance is the slope of the straight line in a voltage versus current plot
C. the resistance is the sum of the voltages across all the devices in a closed path divided by the
sum of the currents through all the devices in the closed path
D. the sum of the resistances around a closed loop is zero
2. Kirchoff’s Current Law (KCL) states that
A. the sum of the currents in a closed path is zero
B. the current that flows through a device is inversely proportional to the voltage across that
R1 R2 R3
RAB
device
C. the sum of the currents through all the devices in a closed path is equal to the sum of the
voltages across all the devices
D. the sum of the currents entering a node is equal to the sum of the currents leaving that node
3. Kirchoff’s Voltage Law (KCL) states that
A. the voltage across a device is directly proportional to the current through that device
B. the voltage across a device is inversely proportional to the current through that device
C. the sum of the voltages across all the devices in a closed path is equal to the sum of the cur-rents
through all the devices
D. the sum of the voltages in a node is equal to the sum of the currents at that node
4. For the three resistors connected as shown below, the equivalent resistance is computed
with the formula
A.
RAB
A B
RAB = R1 + R2 + R3

Exercises
B.
C.
D.
= + 2
2 R2
RAB R1
2 + R3
RAB
R1R2R3
R1 + R2 + R3
= -------------------------------
RAB
R1R2R3
R1 + R2 + R3
= -------------------------------
5. For the three conductances connected as shown below, the equivalent conductance is
computed with the formula
A.
B.
C.
D.
GAB
G1 G2 G3
GAB
A B
GAB = G1 + G2 + G3
GAB = G1
2 + G2
2 +
G3
2 GAB
G1G2G3
G1 + G2 + G3
= --------------------------------
---------- 1
1
GAB
------ 1
= + + ------
G1
------ 1
G2
G3
6. For the three resistances connected as shown below, the equivalent conductance is
A.
B.
C.
GAB
R1 R2
R3
A
GAB 4  12  3 
B
21 –1
1.5 –1
2  3 –1

144  19 –1
D.
7. In the network shown below, when , the voltage . When ,
vR
4 
2 V vX
. When , is
A.
B.
C.
D.
8. The node voltages shown in the partial network below are relative to some reference node not
shown. The value of the voltage is
A.
B.
C.
D.
R = 4  vR = 6 V R = 0 
iR = 2 A R =  vR


Rest of the
Circuit
iR
R
6 V
24 V
8 V
16 V
vX
+ 8 V
2 A
10 V
2 A
3 
20 V
8  6 V
–6 V
16 V
0 V
10 V

Exercises
9. For the network below the value of the voltage is
A.
B.
C.
D.
v
+

8 V
4 
8 V
2 V
–2 V
–8 V
10. For the circuit below the value of the current is
A.
B.
C.
D.
v
+

i
+

8 V
12  i
4 
2 A
0 A
 A
1 A

Problems
1. In the circuit below, the voltage source and both resistors are variable.
Power
Supply
(Voltage
Source)

a. With , , and , compute the power absorbed by .
Answer:
b. With and , to what value should be adjusted so that the power
absorbed by it will be 200 w? Answer:
c. With and , to what value should be adjusted to so that the
power absorbed by will be 100 w? Answer:
2. In the circuit below, represents the load of that circuit.
Compute:
a. Answer:
b. Answer:
c. Answer:
+

+
+
vS 
R1
R2
vS = 120 V R1 = 70  R2 = 50  R2
50 w
vS = 120 V R1 = 0  R2
72 
R1 = 0  R2 = 100  vS
R2 100 V
RLOAD
+

+
75 V
5 A 24 V
3 A
+

iLOAD
vLOAD pLOAD
RLOAD
5 
10 
2 
iLOAD 8 A
vLOAD 20 V
pLOAD 160 w

Exercises
3.For the circuit below, compute the power supplied or absorbed by each device.
24 A 6 A
A
B
12 V C
60 V 36 V
D
+
E
+

+


Answers: , , , ,
pA = 288 w pB = 1152 w pC = –1800 w pD = 144 w pE = 216 w
4. In the circuit below, compute the power delivered or absorbed by the dependent voltage
source.
+
+

50 V
R1 2 
10 
10 A
5iR2
R4
V
R3
R2
6 
10 
iR2
Answer:
62.5 w
5. In the network below, each resistor is 10  Compute the equivalent resistance .
Req
Req
Answer:
360  21 
6. In the network below, and . Compute the current i supplied by the 15
V source.
R1 = 10  R2 = 20 

+
15 V
R1 R1 R1 R1 R1 R1 R1
R2 R2 R2 R2 R2 R2 R1
i
Hint: Begin at the right end and by series and parallel combinations of the resistors, reduce the
circuit to a simple series circuit. This method is known as analysis by network reduction.
Answer:
0.75 A
7. For the circuit below, use the voltage division expression to compute and .
+
5  20 
+

24 V

Answers: ,
8. For the circuit below, use the current division expression to compute and .
5 
20 
Answers: ,
9. A transformer consists of two separate coils (inductors) wound around an iron core as shown in
below. There are many turns in both the primary and secondary coils but, for simplicity, only
few are shown. It is known that the primary coil has a resistance of 5.48  at 20 degrees Cel-sius.
After two hours of operation, it is found that the primary coil resistance has risen to 6.32
. Compute the temperature rise of this coil.
vX vY
+
16 V
10  40 

+
vX
vY
vX = 8  3 V vY = 16  3 V
iX iY
16 A 24 A
10 
40 
iX iY
iX = –16  3 V iY = –8  3 V

Exercises
Answer:
Primary
Coil
Iron Core
Secondary
Coil
36C
10. A new facility is to be constructed at a site which is 1.5 miles away from the nearest electric
utility company substation. The electrical contractor and the utility company have made
load calculations, and decided that the main lines from the substation to the facility will
require several copper conductors in parallel. Each of these conductors must have insulation
type THHN and must carry a maximum current of 220 A in a temperature environ-ment.
20 C
a. Compute the voltage drop on each of these conductors from the substation to the facility
when they carry the maximum required current of 220 A in a temperature environ-ment.
Answer:
20 C
70 V
b. The power absorbed by each conductor under the conditions stated above.
Answer:
15.4 Kw
c. The power absorbed per square cm of the surface area of each conductor under the condi-tions
stated above.
Answer:
0.02 w  cm2
11. For the network below, each of the 12 resistors along the edges of the cube are 1 
each.
Compute the equivalent resistance . Hint: Use any tricks that may occur to you.
Answer:
RAB
A
RAB
B
5  6 

12. A heating unit is rated at 5 Kw
, and to maintain this rating, it is necessary that a voltage of
is applied to establish an initial temperature of . After the heating unit has
220 V 15 C
reached a steady state, it is required that the voltage must be raised to to maintain the
rating. Find the final temperature of the heating element in if the temperature
coefficient is per .
Answer: .
240 V
5 Kw C
 0.0006  1 C
332 C

Multiple Choice
1. B
2. D
3. E
4. E
5. D
6. C
7. B When , the voltage . Therefore, . Also, when
R = 4  vR = 6 V iR = 6  4 = 1.5 A
, , and thus (short circuit). When , but
R = 0  iR = 2 A vR = 0 R =  iR = 0 vR
has a finite value and it is denoted as vR = 
in the figure below. Now, we observe that the
triangles abc and dbe are similar. Then or and thus
be
bc
----- de
= ----- 2.0 – 1.5
ac
-------------------- 6
2.0
= --------------
vR = 
vR =  = 24 V
iR A
vR V
b
d
e
0.5 1.0 1.5 2.0
vR = 
6.0
a
c
8. D We denote the voltage at the common node as shown on the figure below.
vA
10 V
2 A
3 
2 V vX vA
+ 8 V
2 A
4 
20 V
8  6 V

Then, from the branch that contains the resistor, we observe that or
2
R2
= = -------- = 50 w
vS R2

and thus
9. A This is an open circuit and therefore no current flows through the resistor. Accordingly,
there is no voltage drop across the resistor and thus .
10. A The resistor is shorted out by the short on the right side of the circuit and thus the
only resistance in the circuit is the resistor.
Problems
1.
a. With , , and , the circuit is as shown below.
Using the voltage division expression, we obtain
Then,
b. With and , the circuit is as shown below.
We observe that
3 
vA – 10
3
------------------ = 2
vA = 16 vX = – 6 + 16 = 10 V
v = 8 V
12 
4 
vS = 120 V R1 = 70  R2 = 50 
+
vS
R2
R1
50 
70 
120 V
vR2
50
70 + 50
= ------------------  120 = 50 V
pR2
vR2
-------- 502
50
vS = 120 V R1 = 0 
+

R1
72 
0 
120 V
v + R2
vR2
= vs = 120V

and
or
or
2
R2
vR2
-------- = 200 w
1202
R2
----------- = 200 w
R2
1202
200
= ----------- = 72 
c. With and ,the circuit is as shown below.
R1 = 0  R2 = 100 
Then,
or
or
or
0 
R1
vS R2
+
v + R2

100 

100 V
vS 2
R2
------ = 100 w
vS 2
100
-------- = 100 w
vS 2
= 100  100 = 10 000
vS = 10 000 = 100 V
2.
a. Application of KCL at node A of the circuit below yields
+

+
75 V
A
iLOAD
5 A 24 V
3 A
+

iLOAD
2 
vLOAD pLOAD
RLOAD
5 
10 
Mesh 1 Mesh 2
B
iLOAD = 3 + 5 = 8 A

iB iD
24 A 6 A
12 V 60 V 36 V
b. Application of KVL around Mesh 1 yields
or
Application of KVL around Mesh 2 yields
or
or
c.
3. Where not shown, we assign plus (+) and minus () polarities and current directions in accor-dance
with the passive sign convention as shown below.
We observe that and . Also, by KCL at Node X
Then,
By KVL
or
and thus
Also by KVL
or
– 75 + 35 + vAB = 0
vAB = 60 V
– vAB + 24 + 2iLOAD + vLOAD = 0
– 60 + 24 + 2  8 + vLOAD = 0
vLOAD = 20 V
pLOAD = vLOAD  iLOAD = 20  8 = 160 w absorbed power
A
B
C
D
E
+

+

+

 + + 
iA iC iE
vA
vB
vC
vD
vE
X
iA = iB iE = iD
iC = iB + iD = 24 + 6 = 30 A
pA = vA iA = 12  24 = 288 w absorbed
pE = vE iE = 36  6 = 216 w absorbed
pC = vC– iC = 60  –30 = –1800 w supplied
vA + vB = vC
vB = vC – vA = 60 – 12 = 48 V
pB = vBiB = 48  24 = 1152 w absorbed
vD + vE = vC

and thus
pD = vDiD = 24  6 = 144 w absorbed
Check: We must show that
vD = vC – vE = 60 – 36 = 24 V
Power supplied = Power absorbed
pC = pA + pB + pC + pD = 288 + 216 + 1152 + 144 = 1800 w
4. We assign voltages and currents , , , , and as shown in the circuit below.
By KVL,
and by Ohm’s law,
vR2
vR4
iR3
iR4
X
+
+

50 V
R1 2 
10 
10 A
vR4
5iR2
R3
+

vR2
R2
6 
iR2
iR3
iD
vR2
= 50 – 2  10 = 30 V
iR2
vR2
R2
-------- 30
= = ----- = 5 A
6
Therefore, the value of the dependent voltage source is
and
Then,
By KCL at Node X
where
and thus
iD
+

V
iR4
R4
10 
5iR2
= 5  5 = 25 V
vR4
5iR2
= = 25 V
iR4
vR4
R4
------- 25
= = ----- = 2.5 A
10
iD = iR3
–
iR4
iR3
= – = 10 – 5 = 5 A
10 iR2
iD iR3
iR4
= – = 5 – 2.5 = 2.5 A
with the indicated direction through the dependent source. Therefore,
= iD = 25  2.5 = 62.5 w absorbed
pD 5iR2

5. The simplification procedure begins with the resistors in parallel as indicated below.
10
5 15
5 15
Req Req
6. We begin with the right side of the circuit where the last two resistors are in series as shown
below.
5
Req
10 10
10 10
10
10
10
10
10
10
10
10 10
10 10
10
10
10 10
5
5 5
Req
Req
10
10
10
10
10
6
6
10
5
5
Req
16
16
5
5
80  21
80  21
160  21
+
15 V
R1 R1 R1 R1 R1 R1 R1
R2 R2 R2 R2 R2 R2 R1
i

Then,
Next,
R1 + R1 = 10 + 10 = 20 
20  20 = 10 
10 + 10 = 20 
and so on. Finally, addition of the left most resistor with its series equivalent yields
10 + 10 = 20 
i = 15  20 = 0.75 A
and thus
7. We first simplify the given circuit by replacing the parallel resistors by their equivalents. Thus,
and
5  20 5  20
= --------------- = 4 
5 + 20
10  40 10  40
= ------------------ = 8 
10 + 40
The voltage sources are in series opposing connection and they can be replaced by a single
voltage source with value . The simplified circuit is shown below.
24 – 16 = 8 V
i
4 
+
vX

Now, by the voltage division expression,
and
8 
+
8 V

+
vY
vX
4
4 + 8
------------ 8  83
= = -- V
vY
------------  8 16
8
4 + 8
= = ----- V
3
8. We first simplify the given circuit by replacing the series resistors by their equivalents. Thus,
and
5 + 20 = 25 
10 + 40 = 50 
The current sources are in parallel opposing connection and they can be replaced by a single
current source with value . The simplified circuit is shown below.
24 – 16 = 8 A

8 A
i 25  50  X iY
iX
50
25 + 50
------------------  –8 16
= = –----- A
3
iY
25
25 + 50
------------------ 8 –    83
= = –-- A
R20 = 5.48 
RX = 6.32 
T20 = 20C
TX = T
a
234.5 0
R 
b
R20 R0
RX
T C
e d
T20 TX
c
--------
RX
R20
234.5 + + -----------------------------------------
T20 TX
234.5 + T20
234.5 + 20 + TX
234.5 + 20
= = ---------------------------------------
= --------------------------
254.5 + TX
254.5
T= TX
RX
R20
--------  254.5 – 254.5 6.32
= = ----------  254.5 – 254.5 = 36C
5.48
235 A
0.059   1000 ft 100C
0.059 
---------------- 5280
1000ft
 ----------------  1.5 miles = 0.4673  at 100C
1 mile
By the current division expression,
and
9. We construct the resistance versus temperature plot shown below.
From the similar triangles acd and abe, we obtain
or
10.
a. From Table 2.4 we find that the cable size must be 0000 AWG and this can carry up to
. Also, from Table 2.5 we find that the resistance of this conductor is
at . Then, the resistance of this conductor that is 1.5 miles long is
To find the resistance of this cable at , we use the relation of (2.54). Thus,
20C
R20 = R1001 + 0.00420 – 100 = 0.46731 – 0.32 = 0.3178 

and the voltage drop on each of these conductors is
v = iR = 220  0.3178 = 70 V
b. The power absorbed by each conductor is
p = vi = 70  220 = 15 400 w = 15.4 Kw
c. Table 2.5 gives wire sizes in circular mils. We recall that a circular mil is the area of a circle
whose diameter is . To find the diameter in cm, we perform the following conver-sion:
0.001 in
1 circular mil 4
--d2 4
--0.0012 7.854 10–7 = = =  in2
7.854 10–7  in2 2.54 cm2
in2  --------------------------- 5.067 10–6  cm2 = =
From Table 2.5 we find that the cross section of a cable is 211,600 circular
mils. Then, the crosssection of this cable in is
211 600 circular mils 5.067 10 6 –  cm2
  ------------------------------------------ 1.072 cm2 =
Therefore, the cable diameter in cm is
cm2
circular mil
d= 1.072 = 1.035 cm
The crosssection circumference of the cable is
d =   1.035 = 3.253 cm
and the surface area of the cable is
------------------------ 105 cm
Surface area dl 3.253 cm 1.5 miles 1.609 Km
= =    ------------------ = 7.851  105 cm2
cm2
Then, the power absorbed per is
0000 AWG
1 mile
1 Km
p
cm2
Total -------------------------------- power
15 400 w
cm2 = = ---------------------------------------- = 0.02 w  cm2
7.851  105 cm2
11. Let us connect a voltage source of 1 volt
across the corners A and B of the cube as shown
below, and let the current produced by this voltage source be .
I

Since all resistors are equal ( each), the current entering node A will be split into 3
equal currents each. The voltage drop will be regardless of the path from
node A to node B. Arbitrarily, we choose the path through resistors , , and , and the
currents through these resistors are , , and respectively. Then, by KVL,
+ + -- 56
-- I6
--------------- 48400
= = = -------------- = 9.68 
--------------- 57600
= = = -------------- = 11.52 
and since ,
from which
12. The power absorbed by the heating unit when the applied voltage is is and
the resistance is found from the relation or
The power absorbed by the heating unit when the applied voltage is is still
and the resistance is
From relation (2.52),
or
RAB
B
A
+
V
1 volt
I
I
R3
R1
R2
R4
I  3
I  3 R5
I  6
1  I
I  3 VAB 1 volt
R1 R1 R1
I  3 I  6 I  3
I3
--R1
I6
--R4
I3
+ + --R5 = IRAB = V = 1 volt
R1 = R4 = R5 = 1 
I3
-- I3
= --I = IRAB
RAB = 5  6 
P1 220 V 5 Kw
R1 P1 V1 2
=  R1
R1
V1 2
P1
------ 2202
5 Kw
5000
P2 240 V 5 Kw
R2
R2
V1 2
P1
------ 2402
5 Kw
5000
R2 = R1 + R1T2 – T1
R2 – R1 = R1T2 – T1

or
T2 – T1
R2 – R1
R1
------------------ 11.52 – 9.68
= = --------------------------------- = 316.8
9.68  0.0006
Therefore, the final temperature of the heating element is
T2
T2 = 316.8 + T1 = 316.8 + 15  332 C

Chapter 3
Nodal and Mesh Equations  Circuit Theorems
his chapter begins with nodal, loop and mesh equations and how they are applied to the
solution of circuits containing two or more nodepairs and two or more loops or meshes.
Other topics included in this chapter are the voltagetocurrent source transformations
T
and vice versa, Thevenin’s and Norton’s theorems, the maximum power transfer theorem, linear-ity,
superposition, efficiency, and regulation.
3.1 Nodal, Mesh, and Loop Equations
Network Topology is a branch of network theory concerned with the equations required to com-pletely
describe an electric circuit. In this text, we will only be concerned with the following two
theorems.
Theorem 3.1
Let N = number of nodes in a circuit ; then N – 1
independent nodal equations are required to
completely describe that circuit. These equations are obtained by setting the algebraic sum of the
currents leaving each of the N – 1
nodes equal to zero.
Theorem 3.2
Let L = M = number of loops or meshes , B = number of branches ,
N = number of nodes
in a circuit; then independent loop or mesh equations are required to com-pletely
L = M = B – N + 1
describe that circuit. These equations are obtained by setting the algebraic sum of the
voltage drops around each of the L = M = B – N + 1
loops or meshes equal to zero.
3.2 Analysis with Nodal Equations
In writing nodal equations, we perform the following steps:
1.For a circuit containing N nodes, we choose one of these as a reference node assumed to be zero
volts or ground.
2. At each nonreference node we assign node voltages where each of these volt-ages
v1 v2  vn – 1   
is measured with respect to the chosen reference node, i.e., ground.
3. If the circuit does not contain any voltage sources between nodes, we apply KCL and write a
nodal equation for each of the node voltages .
v1 v2  vn – 1   
Circuit Analysis I with MATLAB Computing and Simulink / SimPowerSystems Modeling 31

Chapter 3 Nodal and Mesh Equations  Circuit Theorems
4. If the circuit contains a voltage source between two nodes, say nodes j and k denoted as node
variables and , we replace the voltage source with a short circuit thus forming a com-bined
vj vk
node and we write a nodal equation for this common node in terms of both and ;
then we relate the voltage source to the node variables and .
vj vk
vj vk
Example 3.1
Write nodal equations for the circuit shown in Figure 3.1, and solve for the unknowns of these
equations using matrix theory, Cramer’s rule, or the substitution method. Verify your answers
with Excel® or MATLAB®. Please refer to Appendix A for discussion and examples.
4 
8  10 
6 
12 A 18 A 24 A
G
v1 8  v2 10 
v3
i8 i10
  
i4
i6 4  6 
12 A 18 A 24 A
G
32 Circuit Analysis I with MATLAB Computing and Simulink / SimPowerSystems Modeling
Solution:
We observe that there are 4 nodes and we denote these as , , , and (for ground) as shown
in Figure 3.2.
For convenience, we have denoted the currents with a subscript that corresponds to the resistor
value through which it flows through; thus, the current that flows through the 4 
resistor is
denoted as i4 , the current through the 8  resistor is denoted as i8
, and so on. We will follow
this practice in the subsequent examples.
For the circuit of Figure 3.2, we need N – 1 = 4 – 1 = 3
nodal equations. Let us choose node G
(ground) as our reference node, and we assign voltages v1 v2 , and v3
at nodes , , and 
respectively; these are to be measured with respect to the ground node G. Now, application of
KCL at node  yields

Analysis with Nodal Equations
or
(3.1)
i4 + i8 – 12 = 0
i4 + i8 = 12
where is the current flowing from left to right. Expressing (3.1) in terms of the node voltages,
we obtain
or
or
(3.2)
i8
v1
-----
4
+ ---------------- = 12
-- 18
14
 v1
 + -- 
3v1 – v2 = 96
Next, application of KCL at node  yields
or
(3.3)
v1 – v2
8
18
– --v2 = 12
i8 + i10 + 18 = 0
i8 + i10 = –18
where is the current flowing from right to left * and is the current that flows from left to
right.
Expressing (3.3) in terms of node voltages, we obtain
or
or
i8 i10
v2 – ----------------
v1
8
v2 – v3
10
+ ---------------- = –18
18
-- 1
--v1 – 18
 v2
+ – -----v3 = –18
 + -----
10
1
10
* The direction of the current through the 8 W resistor from left to right in writing the nodal equation at Node 1, and from
right to left in writing the nodal equation at Node 2, should not be confusing. Remember that we wrote independent node
equations at independent nodes and, therefore, any assumptions made in writing the first equation need not be held in
writing the second since the latter is independent of the first. Of course, we could have assumed that the current through
the 8 W resistor flows in the same direction in both nodal equations. It is advantageous, however, to assign a (+) sign to
all currents leaving the node in which we apply KCL. The advantage is that we can check, or even write the node equa-tions
by inspection. With reference to the above circuit and equation (3.1) for example, since G = 1/R, we denote the
coefficients of v1 (1/4 and 1/8 siemens) as self conductances and the coefficient of v2 (1/8) as mutual conductance.
Likewise, in equation (3.3) the coefficients of v2 (1/8 and 1/10 siemens) are the self conductances and the coefficients of
v1 (1/8) and v3 (1/10) are the mutual conductances. Therefore, we can write a nodal equation at a particular node
by inspection, that is, we assign plus (+) values to self conductances and minus () to mutual conductances.

(3.4)
5v1 – 9v2 + 4v3 = 720
i10 + i6 – 24 = 0
i10 + i6 = 24
i10
v3 – ----------------
v2
10
v3
6
+ ----- = 24
1
10
– -----v2 1
----- 16
+   v3 = 24
 + -- 
10
– 3v2 + 8v3 = 720
3 –1 0
5 –9 4
0 –3 8

G
v1
v2
v3
V
96
720
720
I
=


v1 = 20.57 V v2 = –34.29 V
v3 = 77.14 V
Similarly, application of KCL at node  yields
or
where is the current flowing from right to left. Then, in terms of node voltages,
(3.5)
or
or
(3.6)
Equations (3.2), (3.4), and (3.6) constitute a set of three simultaneous equations with three
unknowns. We write them in matrix form as follows:
(3.7)
We can use Cramer’s rule or Gauss’s elimination method as discussed in Appendix A, to solve
(3.7) for the unknowns. Simultaneous solution yields , , and
. With these values we can determine the current in each resistor, and the power
absorbed or delivered by each device.
Check with MATLAB®:
G=[3 1 0; 5 9 4; 0 3 8]; I=[96 720 720]'; V=GI;...
fprintf(' n'); fprintf('v1 = %5.2f volts t', V(1)); ...
fprintf('v2 = %5.2f volts t', V(2)); fprintf('v3 = %5.2f volts', V(3)); fprintf(' n')
v1 = 20.57 volts v2 = -34.29 volts v3 = 77.14 volts

Check with Excel®:
The spreadsheet of Figure 3.3 shows the solution of the equations of (3.7). The procedure is dis-cussed
in Appendix A.
Figure 3.3. Spreadsheet for the solution of (3.7)
Example 3.2
For the circuit of Figure 3.4, write nodal equations in matrix form and solve for the unknowns
using matrix theory, Cramer’s rule, or Gauss’s elimination method. Verify your answers with
Excel or MATLAB. Please refer to Appendix A for procedures and examples. Then construct a
table showing the voltages across, the currents through and the power absorbed or delivered by
each device. Verify your answer with a Simulink / SimPowerSystems model.
8 
+
10 V
4  6 
+

12 V 18 A 24 A
Solution:
We observe that there are 4 nodes and we denote these as as , , , and (for ground) as
shown in Figure 3.5. We assign voltages , and at nodes , , and  respectively; these
are to be measured with respect to the ground node . We observe that is a known voltage,
that is, and thus our first equation is
(3.8)
G
v1 v2 v3
G v1
v1 = 12 V
v1 = 12

v1 v2 v3
  
4  6 
12 V 18 A 24 A
Figure 3.5. Circuit for Example 3.2 with assigned nodes and voltages
Next, we move to node  where we observe that there are three currents flowing out of this
node, one to the left, one to the right, and one down. Therefore, our next nodal equation will
contain three terms. We have no difficulty writing the term for the current flowing from node 
to node , and for the 18 A source; however, we encounter a problem with the third term
because we cannot express it as term representing the current flowing from node  to node . To
work around this problem, we temporarily remove the 10 V voltage source and we replace it with
a “short” thereby creating a combined node (or generalized node or supernode as some textbooks call
it), and the circuit now appears as shown in Figure 3.6.
v1 8 
v2 v3
i8
  
4  6 
Figure 3.6. Circuit for Example 3.2 with a combined node
+
Now, application of KCL at this combined node yields the equation
or
or
* (3.9)
* The combined node technique allows us to combine two nodal equations into one but requires that we use the proper node
designations. In this example, to retain the designation of node 2, we express the current as . Likewise, at node 3,
we express the current as .
8 
+
+

10 V
G
12 V 18 A 24 A
G
Combined Node
 +
10 V
Removed and
replaced by a
short
i6
i8 + 18 + i6 – 24 = 0
i8 + i6 = 6
v2 – ----------------
v1
8
v3
6
+ ----- = 6
i8
v2 – v1
8
----------------
i6
v3
6
-----

or
or
(3.10)
– --v1 18
18
--v2
16
+ + --v3 = 6
–3v1 + 3v2 + 4v3 = 144
To obtain the third equation, we reinsert the 10 V source between nodes  and . Then,
(3.11)
v3 – v2 = 10
In matrix form, equations (3.8), (3.10), and (3.11) are
(3.12)
1 0 0
–3 3 4
0 –1 1

G
v1
v2
v3
V
12
144
10
I
=


Simultaneous solution yields v1 = 12 V , v2 = 20 V , and v3 = 30 V
. From these we can find
the current through each device and the power absorbed or delivered by each device.
Check with MATLAB:
G=[1 0 0; 3 3 4; 0 1 1]; I=[12 144 10]'; V=GI;...
fprintf(' n'); fprintf('v1 = %5.2f volts t', V(1)); ...
fprintf('v2 = %5.2f volts t', V(2)); fprintf('v3 = %5.2f volts', V(3)); fprintf(' n')
v1 = 12.00 volts v2 = 20.00 volts v3 = 30.00 volts
Check with Excel:
A B C D E F G H
Spreadsheet for Matrix Inversion and Matrix Multiplication
1 0 0 12
G= -3 3 4 I= 144
0 -1 1 10
1.000 0.000 0.000 12
G-1= 0.429 0.143 -0.571 V= 20
0.429 0.143 0.429 30
1234567
89
Table 3.1 shows that the power delivered is equal to the power absorbed.

TABLE 3.1 Table for Example 3.2
Power (watts)
Device Voltage (volts) Current (amps) Delivered Absorbed
12 V Source 12 2 24
10 V Source 10 19 190
18 A Source 20 18 360
24 A Source 30 24 720
4 W Resistor 12 3 36
6 W Resistor 30 5 150
8 W Resistor 8 1 8
Total 744 744
v VM 2 +-
VM 1
12.00 V2
R2=8
v +-
VM 3
v +-
20.00
CVS 2
10V
s
-
R3=6
R1=4
-10
Constant 4
24
+
s
-
Constant 3
-18
+
s
-
Constant 2
+
CCS 2
24 A
CCS 1
-18 A
V3
Continuous
powergui
30.00
V1
CVS 1
12 V
12
+
s
-
Constant 1
3.3 Analysis with Mesh or Loop Equations
In writing mesh or loop equations, we follow these steps:
1. For a circuit containing M = L = B – N + 1
meshes (or loops), we assign a mesh or loop cur-rent
i1 i2  in – 1   
for each mesh or loop.
2. If the circuit does not contain any current sources, we apply KVL around each mesh or loop.
3. If the circuit contains a current source between two meshes or loops, say meshes or loops j and
k denoted as mesh variables ij and ik
, we replace the current source with an open circuit thus
forming a common mesh or loop, and we write a mesh or loop equation for this common mesh

Analysis with Mesh or Loop Equations
or loop in terms of both and . Then, we relate the current source to the mesh or loop vari-ables
and .
ij ik
ij ik
Example 3.3
For the circuit of Figure 3.9, write mesh equations in matrix form and solve for the unknowns
using matrix theory, Cramer’s rule, or Gauss’s elimination method. Verify your answers with
Excel or MATLAB. Please refer to Appendix A for procedures and examples. Then construct a
table showing the voltages across, the currents through, and the power absorbed or delivered by
each device.
36 V 8 
2  10 
+ 
4  6  12 
+ 
+
12 V
24 V
Solution:
For this circuit we need mesh or loop equations and we
arbitrarily assign currents , , and all in a clockwise direction as shown in Figure 3.10.
M = L = B – N + 1 = 9 – 7 + 1 = 3
i1 i2 i3
36 V 8 
2  10 
+ 
4  6  12 
i3
+ 
i1 i2
+
12 V
24 V
Applying KVL around each mesh we obtain:
Mesh #1: Starting with the left side of the 2 
resistor, going clockwise, and observing the pas-sive
sign convention, we obtain the equation for this mesh as
or
2i1 + 4i1 – i2 – 12 = 0

(3.13)
6i1 – 4i2 = 12
Mesh #2: Starting with the lower end of the 4 
resistor, going clockwise, and observing the
4i2 – i1 + 36 + 8i2 + 6i2 – i3 = 0
– 4i1 + 18i2 – 6i3 = –36
6 
6i3 – i2 + 10i3 + 12i3 + 24 = 0
– 6i2 + 28i3 = –24
i1
2  4 
2 + 4 = 6 i1
i2 4 
i2 4 
i2 –4i2
4  8  6 
i2 4  6 
passive sign convention, we obtain the equation
or
(3.14)
Mesh #3: Starting with the lower end of the resistor, going clockwise, and observing the
passive sign convention, we obtain:
or
(3.15)
Note: For this example, we assigned all three currents with the same direction, i.e., clockwise.
This, of course, was not mandatory; we could have assigned any direction in any mesh. It is
advantageous, however, to assign the same direction to all currents. The advantage here is that
we can check, or even write the mesh equations by inspection. This is best explained with the fol-lowing
observations:
1. With reference to the circuit of Figure 3.10 and equation (3.13), we see that current flows
through the and resistors. We call these the self resistances of the first mesh. Their
sum, i.e., is the coefficient of current in that equation. We observe that current
also flows through the resistor. We call this resistance the mutual resistance between the
first and the second mesh. Since enters the lower end of the resistor, and in writing
equation (3.13) we have assumed that the upper end of this resistor has the plus (+) polarity,
then in accordance with the passive sign convention, the voltage drop due to current is
and this is the second term on the left side of (3.13).
2. In Mesh 2, the self resistances are the , , and resistors whose sum, 18, is the coef-ficient
of in equation (3.14). The and resistors are also the mutual resistances
between the first and second, and the second and the third meshes respectively. Accordingly,
the voltage drops due to the mutual resistances in the second equation have a minus () sign,
i.e, –4i1 and –6i3
.
3. The signs of the coefficients of i2 and i3
in (3.15) are similarly related to the self and mutual
resistances in the third mesh.

Simplifying and rearranging (3.13), (3.14) and (3.15) we obtain:
(3.16)
(3.17)
(3.18)
and in matrix form
(3.19)
3i1 – 2i2 = 6
2i1–9i2 + 3i3 = 18
3i2–14i3 = 12
3 –2 0
2 –9 3
0 3 –14

R
i1
i2
i3
I
6
18
12
V
=


Simultaneous solution yields i1 = 0.4271 , i2 = –2.3593 , and i3 = –1.3627
where the negative
values for and indicate that the actual direction for these currents is counterclockwise.
Check with MATLAB:
R=[3 2 0; 2 9 3; 0 3 14]; V=[6 18 12]'; I=RV;...
fprintf(' n'); fprintf('i1 = %5.2f amps t', I(1)); ...
fprintf('i2 = %5.2f amps t', I(2)); fprintf('i3 = %5.2f amps', I(3)); fprintf(' n')
i1 = 0.43 amps i2 = -2.36 amps i3 = -1.36 amps
Excel produces the same answers as shown in Figure 3.11.
A B C D E F G H
3 -2 0 6
R= 2 -9 3 V= 18
0 3 -14 12
0.397 -0.095 -0.020 0.4271
R-1= 0.095 -0.142 -0.031 I= -2.3593
0.020 -0.031 -0.078 -1.3627
i2 i3
1234567
89
Table 3.2 shows that the power delivered by the voltage sources is equal to the power absorbed
by the resistors.

Power (watts)
12 V Source 12.000 0.427 5.124
36 V Source 36.000 2.359 84.924
24 V Source 24.000 1.363 32.712
2 W Resistor 0.854 0.427 0.365
4 W Resistor 11.144 2.786 30.964
8 W Resistor 18.874 2.359 44.530
6 W Resistor 5.976 0.996 5.952
10 W Resistor 13.627 1.363 18.570
12 W Resistor 16.352 1.363 22.288
Total 122.760 122.669
Example 3.4
For the circuit of Figure 3.12, write loop equations in matrix form, and solve for the unknowns
using matrix theory, Cramer’s rule, or Gauss’s elimination method. Verify your answers with Excel
or MATLAB. Please refer to Appendix A for procedures and examples. Then, construct a table
showing the voltages across, the currents through and the power absorbed or delivered by each
device.
2  10 
4  6  12 
Solution:
This is the same circuit as that of the previous example where we found that we need 3 mesh or
loop equations. We choose our loops as shown in Figure 3.13, and we assign currents , , and
, all in a clockwise direction.
12 V
24 V
36 V 8 
+
+ 
+ 
i1 i2
i3

2  10 
a b c d
i 4  6  12  1 i2
i3
+ 
h g f e
24 V
36 V 8 
+ 
Figure 3.13. Circuit for Example 3.4 with assigned loops
+
12 V
Applying of KVL around each loop, we obtain:
Loop 1 (abgh): Starting with the left side of the resistor and complying with the passive sign
convention, we obtain:
or
or
(3.20)
2 
2i1 + i2 + i3 + 4i1 – 12 = 0
6i1 + 2i2 + 2i3 = 12
3i1 + i2 + i3 = 6
Loop 2 (abcfgh):As before, starting with the left side of the resistor and complying with the
passive sign convention, we obtain:
or
or
(3.21)
2 
2i1 + i2 + i3 + 36 + 8i2 + i3 + 6i2 – 12 = 0
2i1 + 16i2 + 10i3 = –24
i1 + 8i2 + 5i3 = –12
Loop 3 (abcdefgh): Likewise, starting with the left side of the resistor and complying with
the passive sign convention, we obtain:
2i1 + i2 + i3 + 36 + 8i2 + i3 + 10i3 + 12i3 + 24 – 12 = 0
or
or
(3.22)
2i1 + 10i2 + 32i3 = –48
i1 + 5i2 + 16i3 = –24
and in matrix form
2 

(3.23)
Solving with MATLAB we obtain:
R=[3 1 1; 1 8 5; 1 5 16]; V=[6 12 24]'; I=RV;...
i1 = 2.79 amps i2 = -1.00 amps i3 = -1.36 amps
Excel produces the same answers.
Table 3.3 shows that the power delivered by the voltage sources is equal to the power absorbed by
the resistors and the values are approximately the same as those of the previous example.
Power (watts)
12 V Source 12.000 0.427 5.124
36 V Source 36.000 2.359 84.924
24 V Source 24.000 1.363 32.712
2 W Resistor 0.854 0.427 0.365
4 W Resistor 11.146 2.786 31.053
8 W Resistor 18.872 2.359 44.519
6 W Resistor 5.982 0.997 5.964
10 W Resistor 13.627 1.363 18.574
12 W Resistor 16.352 1.363 22.283
Total 122.760 122.758
Example 3.5
For the circuit of figure 3.14, write mesh equations in matrix form and solve for the unknowns
using matrix theory, Cramer’s rule, or the substitution method. Verify your answers with Excel or
MATLAB. Please refer to Appendix A for procedures and examples.
3 1 1
1 8 5
1 5 16
R
i1
i2
i3
I
6
–12
–24
V
=




2  36 V 8 
10 
+ 
4  6  12 
+
12 V
5 A
Solution:
This is the same circuit as those of the two previous examples except that the 24 V voltage
source has been replaced by a 5 A current source. As in Examples 3.3 and 3.4, we need
mesh or loop equations, and we assign currents , , and
M = L = B – N + 1 = 9 – 7 + 1 = 3 i1 i2
i3
all in a clockwise direction as shown in Figure 3.15.
36 V 8 
+
2  10 
+ 
6  12 
5 A
i1
4 
i2 i3
Figure 3.15. Circuit for Example 3.5 with assigned currents
12 V
For Meshes 1 and 2, the equations are the same as in Example 3.3 where we found them to be
or
(3.24)
and
or
(3.25)
6i1 – 4i2 = 12
3i1 – 2i2 = 6
– 4i1 + 18i2 – 6i3 = –36
2i1–9i2 + 3i3 = 18
For Mesh 3, we observe that the current is just the current of the 5 A current source and thus
our third equation is simply
(3.26)
and in matrix form,
i3
i3 = 5

(3.27)
3 –2 0
2 –9 3
0 0 1

R
i1
i2
i3
I
6
18
5
V
=


Solving with MATLAB we obtain:
R=[3 2 0; 2 9 3; 0 0 1]; V=[6 18 5]'; I=RV;...
i1 = 2.09 amps i2 = 0.13 amps i3 = 5.00 amps
Example 3.6
Write mesh equations for the circuit of Figure 3.16 and solve for the unknowns using MATLAB
or Excel. Then, compute the voltage drop across the source. Verify your answer with a Simu-link
5 A
12 V
5 A
36 V 8 
+
+ 
2 
6 
20 
4 
10  16 
12 
 +
12 V
24 V
+ 
18 
/ SimPowerSystems model.
Solution:
Here, we would be tempted to assign mesh currents as shown in Figure 3.17. However, we will
encounter a problem as explained below.
The currents i3 and i4
for Meshes 3 and 4 respectively present no problem; but for Meshes 1 and
2 we cannot write mesh equations for the currents and as shown because we cannot write a
i1 i2

term which represents the voltage across the 5 A
current source. To work around this problem
we temporarily remove (open) the current source and we form a “combined mesh” (or gener-alized
5 A
mesh or supermesh as some textbooks call it) and the current that flows around this com-bined
mesh is as shown in Figure 3.18.
+
12 V
i1 i2
5 A
36 V 8 
+ 
2 
6 
20 
4 
10  16 
12 
 +
12 V
24 V
+ 
i3 i4
18 
Figure 3.17. Circuit for Example 3.6 with erroneous current assignments
+
12 V
36 V 8 
+ 
2 
6 
20 
4 
Combined
Mesh
10  16 
12 
12 V
+
24 V
+ 
i3 i4
18 
Figure 3.18. Circuit for Example 3.6 with correct current assignments
Now, we apply KVL around this combined mesh. We begin at the left end of the resistor,
and we express the voltage drop across this resistor as since in Mesh 1 the current is essen-tially
.
2 
2i1
i1

Continuing, we observe that there is no voltage drop across the 4 
resistor since no current
flows through it. The current now enters Mesh 2 where we encounter the 36 V drop due to the
voltage source there, and the voltage drops across the and resistors are and
respectively since in Mesh 2 the current now is really . The voltage drops across the and
8  6  8i2 6i2
i2 16 
resistors are expressed as in the previous examples and thus our first mesh equation is
10 
2i1 + 36 + 8i2 + 6i2 + 16i2 – i4 + 10i1 – i3 – 12 = 0
12i1 + 30i2–10i3–16i4 = –24
6i1 + 15i2–5i3–8i4 = –12
i1 – i2 = 5
10i3 – i1 + 12i3 – i4 + 18i3 – 12 = 0
5i1–20i3 + 6i4 = –6
16i4 – i2 + 20i4 – 24 + 12i4 – i3 = 0
4i2 + 3i3–12i4 = –6
6 15 –5 –8
1 –1 0 0
5 0 –20 6
0 4 3 –12

R
i1
i2
i3
i4
I
–12
5
–6
–6
V
=


or
or
(3.28)
Now, we reinsert the 5 A current source between Meshes 1 and 2 and we obtain our second equa-tion
as
(3.29)
For meshes 3 and 4, the equations are
or
(3.30)
and
or
(3.31)
and in matrix form
(3.32)
We find the solution of (3.32) with the following MATLAB script:
R=[6 15 5 8; 1 1 0 0; 5 0 20 6; 0 4 3 12]; V=[12 5 6 6]'; I=RV;...
fprintf(' n');...
fprintf('i1 = %5.4f amps t',I(1)); fprintf('i2 = %5.4f amps t',I(2));...
fprintf('i3 = %5.4f amps t',I(3)); fprintf('i4 = %5.4f amps',I(4)); fprintf(' n')
i1=3.3975 amps i2=-1.6025 amps i3=1.2315 amps i4=0.2737 amps

Now, we can find the voltage drop across the current source by application of KVL around
Mesh 1 using the following relation:
This yields
5 A
2  3.3975 + 4  3.3975 + 1.6025 + v5A + 10  3.3975 – 1.2315 – 12 = 0
v5A = –36.455
We can verify this value by application of KVL around Mesh 2 where beginning with the lower
end of the resistor and going counterclockwise we obtain
6 w
6 + 8  1.6025 – 36 + 4  3.3975 + 1.6025 – 36.455 + 16  1.6025 + 0.2737 = 0
With these values, we can also compute the power delivered or absorbed by each of the voltage
sources and the current source.
36
K 3
CVS 2
36 V
s
-
+
3.40
I 1
2 8
-5
K 2
4 6
-1.60
I 2
+
s
-
CCS 1
5 A
10 16
1.23
18
Continuous
powergui
R1=3
24
K 5
0.27
I 4
I 3
CVS 4
24 V
s
-
+
+
s
-
+
+ i -
CM 1
s
-
i + -
CM 2
20
i + -
CM 4
i + -
CM 3
CVS 1
12 V
12
K 1
12
K 4
CVS 3
12 V1

3.4 Transformation between Voltage and Current Sources
In the previous chapter we stated that a voltage source maintains a constant voltage between its
terminals regardless of the current that flows through it. This statement applies to an ideal voltage
source which, of course, does not exist; for instance, no voltage source can supply infinite current
to a short circuit. We also stated that a current source maintains a constant current regardless of
the terminal voltage. This statement applies to an ideal current source which also does not exist;
for instance, no current source can supply infinite voltage when its terminals are opencircuited.
A practical voltage source has an internal resistance which, to be accounted for, it is represented
with an external resistance RS in series with the voltage source vS
as shown in Figure 3.20 (a).
Likewise a practical current source has an internal conductance which is represented as a resistance
(or conductance ) in parallel with the current source as shown in Figure 3.20 (b).
Rp Gp iS
RS
(a) (b)
Figure 3.20. Practical voltage and current sources
+
In Figure 3.20 (a), the voltage of the source will always be but the terminal voltage will be
if a load is connected at points and . Likewise, in Figure 3.20 (b) the current of
the source will always be but the terminal current will be if a load is con-nected
a a
+ +
RL RL vS iS
 
at points and .
Now, we will show that the networks of Figures 3.20 (a) and 3.20 (b) can be made equivalent to
each other.
In the networks of Figures 3.21 (a) and 3.21 (b), the load resistor is the same in both.
Figure 3.21. Equivalent sources
From the circuit of Figure 3.21 (a),
a
b
a
b
vS iS
Rp
vS vab
vab vS vRs
= – a b
iS iab iab = iS –
iRP
a b
RL
+ (a) (b)
b
RS
vab
iab RP vab
iab
b

Transformation between Voltage and Current Sources
(3.33)
and
(3.34)
From the circuit of Figure 3.21 (b),
(3.35)
and
(3.36)
vab
RL
RS + RL
= -------------------vS
iab
vS
RS + RL
= -------------------
vab
RPRL
Rp + RL
= -------------------iS
iab
RP
Rp + RL
= -------------------iS
Since we want to be the same in both circuits 3.21 (a) and 3.21 (b), from (3.33) and (3.35)
we must have:
(3.37)
vab
vab
RL
RS + RL
= -------------------vS
RPRL
Rp + RL
= -------------------iS
Likewise, we want to be the same in both circuits 3.21 (a) and 3.21 (b). Then, from (3.34)
and (3.36) we obtain:
(3.38)
iab
iab
vS
RS + RL
= -------------------
= -------------------iS
and for any , from (3.37) and (3.38)
(3.39)
and
(3.40)
Rp
Rp + RL
RL
vS = RpiS
Rp = RS
Therefore, a voltage source in series with a resistance can be transformed to a current
source whose value is equal to , in parallel with a resistance whose value is the same
as RS
.
Likewise, a current source in parallel with a resistance can be transformed to a voltage
source whose value is equal to , in series with a resistance whose value is the same as
.
vS RS
iS vS  RS Rp
iS Rp
vS iS  RS
Rp
The voltagetocurrent source or currenttovoltage source transformation is not limited to a
single resistance load; it applies to any load no matter how complex.

Example 3.7
Find the current through the resistor in the circuit of Figure 3.22.
i10
Solution:
This problem can be solved either by nodal or by mesh analysis; however, we will transform the
voltage sources to current sources and we will replace the resistances with conductances except
the resistor. We will treat the resistor as the load of this circuit so that we can com-pute
the current through it. Then, the circuit becomes as shown in Figure 3.23.
i10
Figure 3.23. Circuit for Example 3.7 with voltage sources transformed to current sources
Combination of the two current sources and their conductances yields the circuit shown in Figure
3.24.
i10
Figure 3.24. Circuit for Example 3.7 after combinations of current sources and conductances
Converting the conductance to a resistance and performing currenttovoltage source
transformation, we obtain the circuit of Figure 3.25.
+ 
i10
Figure 3.25. Circuit for Example 3.7 in its simplest form
i10 10 
12 V
32 V
+
2 
4 
+ 
10 
10  10 
i10
6 A 8 A
10 
0.5 –1 0.25 –1
2 A
10 
0.75 –1
0.75 –1
8/3 V
4/3 
10 

Thevenin’s Theorem
Thus, the current through the resistor is
10 
i10
–8  3
10 + 4  3
= ---------------------- = –4  17 A
3.5 Thevenin’s Theorem
This theorem is perhaps the greatest time saver in circuit analysis, especially in electronic* cir-cuits.
It states that we can replace a two terminal network by a voltage source in series with
a resistance as shown in Figure 3.26.
RTH
Load Load
+
Network
to be replaced
by a Thevenin
equivalent
circuit
x
x

vxy vxy

y


(Rest
of the
circuit)
y
VTH
(a) (b)
(Rest
of the
circuit)
Figure 3.26. Replacement of a network by its Thevenin’s equivalent
vTH
RTH
The network of Figure 3.26 (b) will be equivalent to the network of Figure 3.26 (a) if the load is
removed in which case both networks will have the same open circuit voltages and conse-quently,
Therefore,
(3.41)
vxy
vTH = vxy
vTH = vxy open
The Thevenin resistance represents the equivalent resistance of the network being replaced
by the Thevenin equivalent, and it is found from the relation
(3.42)
RTH
RTH
vxy open
ixy short
= ---------------------
= ---------
where stands for shortcircuit current.
vTH
iSC
iSC
* For an introduction to electronic circuits, please refer to Electronic Devices and Amplifier Circuits with MAT-LAB
Applications, ISBN 978-1-934404-13-3.

If the network to be replaced by a Thevenin equivalent contains independent sources only, we
can find the Thevenin resistance by first shorting all (independent) voltage sources, opening
all (independent) current sources, and calculating the resistance looking into the direction that is
opposite to the load when it has been disconnected from the rest of the circuit at terminals and
.
Example 3.8
Use Thevenin’s theorem to find and for the circuit of Figure 3.27.
6  10 
RLOAD
Solution:
We will apply Thevenin’s theorem twice; first at terminals x and y and then at and as shown
in Figure 3.28.
x
6  10 
x' RTH

iLOAD vLOAD
RLOAD
vTH
Figure 3.28. First step in finding the Thevenin equivalent of the circuit of Example 3.8
+
Breaking the circuit at , we are left with the circuit shown in Figure 3.29.
+
Figure 3.29. Second step in finding the Thevenin equivalent of the circuit of Example 3.8
x
Applying Thevenin’s theorem at and and using the voltage division expression, we obtain
RTH
x
y
iLOAD vLOAD
12 V
+
3  3 
5 
7 
8 
+

iLOAD
vLOAD
x' y'
12 V
3 
3 
5 
7 
8 
+


y
+


y 
y'

x – y
x
12 V
3 
6 

y 
RTH
x y

(3.43)
= = ------------  12 = 8 V
RTH Vs 0 =
vTH vxy
6
3 + 6
3  6
3 + 6
= ------------ = 2 
and thus the equivalent circuit to the left of points and is as shown in Figure 3.30.
x x
x
8 V
+
RTH
2 

y 
vTH
Figure 3.30. First Thevenin equivalent for the circuit of Example 3.8
Next, we attach the remaining part of the given circuit to the Thevenin equivalent of Figure
3.30, and the new circuit now is as shown in Figure 3.31.
+
RTH
2 
vLOAD
vTH 
8 V
x

+

3 
5 
7 
10 
iLOAD
RLOAD
8 
y
Figure 3.31. Circuit for Example 3.8 with first Thevenin equivalent
Now, we apply Thevenin’s theorem at points and and we obtain the circuit of Figure 3.32.
x' y'
+
RTH
2 
8 V
x


3 
5 
7 
10 
y
vTH
Figure 3.32. Applying Thevenin’s theorem at points and for the circuit for Example 3.8
Using the voltage division expression, we obtain
x' y'
= = ---------------------------------  8 = 4 V
v'TH vx'y'
10
2 + 3 + 10 + 5
R'TH vTH 0 =
= 2 + 3 + 5 10 + 7 = 12 

This Thevenin equivalent with the load resistor attached to it, is shown in Figure 3.33.
x
+

y
4 V
R'TH
12 
iLOAD
RLOAD
8 
v'TH
vLOAD
Figure 3.33. Entire circuit of Example 3.8 simplified by Thevenin’s theorem
The voltage is found by application of the voltage division expression, and the current
vLOAD
iLOAD
vLOAD
8
12 + 8
= ---------------  4 = 1.6 V
iLOAD
4
12 + 8
= --------------- = 0.2 A
x
y
x

Begin with
this series
combination
250  250 

y
+
50 
240 V
100 
Load
x


y
50 
100 
Then, compute
the equivalent
resistance
looking to the
left of points
x and y
(a) (b)
RTH
= = ------------------------ = 75 
by Ohm’s law as shown below.
It is imperative to remember that when we compute the Thevenin equivalent resistance, we must
always look towards the network portion which remains after disconnecting the load at the and
terminals. This is illustrated with the two examples that follow.
Let us consider the network of Figure 3.34 (a).
Figure 3.34. Computation of the Thevenin equivalent resistance when the load is to the right
This network contains no dependent sources; therefore, we can find the Thevenin equivalent by
shorting the 240 V
voltage source, and computing the equivalent resistance looking to the left of
points and as indicated in Figure 3.34 (b). Thus,
x y
RTH 250 + 50 100 300  100
300 + 100

Now, let us consider the network of Figure 3.35 (a).
x


y
+
50 
250 
240 V
100 
Load
x

250 
R 100  TH

y
50 
Begin with this
series combination
Then, compute the
equivalent resistance
looking to the right of
points x and y
(a) (b)
Figure 3.35. Computation of the Thevenin equivalent resistance when the load is to the left
This network contains no dependent sources; therefore, we can find the Thevenin equivalent by
shorting the 240 V
voltage source, and computing the equivalent resistance looking to the right
of points and as indicated in Figure 3.35 (b). Thus,
x y
RTH 50 + 100 250 150  250
= = ------------------------ = 93.75 
150 + 250
We observe that, although the resistors in the networks of Figures 3.34 (b) and 3.35 (b) have the
same values, the Thevenin resistance is different since it depends on the direction in which we
look into (left or right).
Example 3.9
Use Thevenin’s theorem to find and for the circuit of Figure 3.36.
iLOAD vLOAD
24 V
+ 
3  3 
6  10 
5 
7 
vLOAD
+
12 V
iLOAD
RLOAD
8 
+

Solution:
This is the same circuit as the previous example except that a voltage source of 24 V
has been
placed in series with the 7  resistor. By application of Thevenin’s theorem at points x and y
as
before, and connecting the rest of the circuit, we obtain the circuit of Figure 3.37.

RLOAD

Figure 3.37. Circuit for Example 3.9 with first Thevenin equivalent
Next, disconnecting the load resistor and applying Thevenin’s theorem at points and we
obtain the circuit of Figure 3.38.

+ 
Figure 3.38. Application of Thevenin’s theorem at points and for the circuit for Example 3.9
There is no current flow in the resistor; thus, the Thevenin voltage across the and
points is the algebraic sum of the voltage drop across the resistor and the source, i.e.,
= = ---------------------------------  8 – 24 = –20 V
and the Thevenin resistance is the same as in the previous example, that is,
Finally, connecting the load as shown in Figure 3.39, we compute and as
follows:
Figure 3.39. Final form of Thevenin equivalent with load connected for circuit of Example 3.9
+ 
+
2  3 
5 
10 
7 
8 
+

RTH 24 V
8 V
 x
y
 x

y
iLOAD
vLOAD
VTH
x' y'
+ 
2  3 
5 
10 
7 
RTH 24 V
8 V
 x
y
 x

y
vTH
x' y'
7  x' y'
10  24 V
v'TH vx'y'
10
2 + 3 + 10 + 5
R'TH VTH 0 =
= 2 + 3 + 5 10 + 7 = 12 
RLOAD vLOAD iLOAD
x
y
20 V
+

12 
8 
RTH
RLOAD
v'TH
vLOAD
iLOAD

+

vLOAD
8
12 + 8
= ---------------  –20 = –8 V
iLOAD
–20
12 + 8
= --------------- = –1 A
Example 3.10
For the circuit of Figure 3.40, use Thevenin’s theorem to find and .
iLOAD vLOAD
+

3  3 
20iX
6  10 
5 
4  vLOAD
+
12 V
7 
iLOAD
RLOAD
8 
+

iX
Solution:
This circuit contains a dependent voltage source whose value is twenty times the current
through the resistor. We will apply Thevenin’s theorem at points a and b as shown in Figure
3.39.
+

3  3 
20iX
6  10 
5 
7 
a

+

4  vLOAD
iX

b
Figure 3.41. Application of Thevenin’s theorem for Example 3.10
6 
+
12 V
iLOAD
RLOAD
8 
In the circuit of Figure 3.41, we cannot short the dependent source; therefore, we will find the
Thevenin resistance from the relation
(3.44)
RTH
vOC
iSC
= ---------
= ---------------------------------
vLOAD
RL  
iLOAD RL 0 
To find the open circuit voltage vOC = vab
, we disconnect the load resistor and our circuit now
is as shown in Figure 3.42.

6  10 
Figure 3.42. Circuit for finding of Example 3.10
+
We will use mesh analysis to find which is the voltage across the resistor. We chose
mesh analysis since we only need three mesh equations whereas we would need five equations had
we chosen nodal analysis. Please refer to Exercise 16 at the end of this chapter for a solution
requiring nodal analysis.
Observing that , we write the three mesh equations for this network as
(3.45)
and after simplification and combination of like terms, we write them in matrix form as
(3.46)
Using the spreadsheet of Figure 3.43, we find that
Figure 3.43. Spreadsheet for Example 3.10
Thus, the Thevenin voltage at points a and b is
+

12 V
3  3 
5 
7 
4 
 a
 b
20iX
iX
i1
i2 i3
vOC = vab
vOC 4 
iX = i1 – i2
9i1 – 6i2 = 12
– 6i1 + 24i2 – 10i3 = 0
20i1 – i2 + 4i3 + 10i3 – i2 = 0
3 –2 0
3 –12 5
10 –15 7
R
I1
I2
I3
I
4
0
0
V
=



i3 = –3.53 A
123456789
A B C D E F G H
3 -2 0 4
R= 3 -12 5 V= 0
10 -15 7 0
0.106 -0.165 0.118 0.42
R-1= -0.341 -0.247 0.176 I= -1.36
-0.882 -0.294 0.353 -3.53

vTH = –3.53  4 = –14.12 V
Next, to find the Thevenin resistance , we must first compute the short circuit current .
Accordingly, we place a short across points a and b and the circuit now is as shown in Figure 3.44
and we can find the short circuit current from the circuit of Figure 3.45 where
RTH ISC
iSC iSC = i4
+ 
3  3 
20iX
iX 6  10 
iSC
5 
7 
4 
iSC = iab
Figure 3.44. Circuit for finding in Example 3.10
+
a

b
+ 
+
3  3 
20iX
iX 6  10 
iSC
i1 i2 i3 i4
5 
7 
4 
a

b
iSC = iab
Figure 3.45. Mesh equations for finding in Example 3.10
12 V
12 V
The mesh equations for the circuit of Figure 3.45 are
(3.47)
9i1 – 6i2 = 12
– 6i1 + 24i2 – 10i3 = 0
20i1 – i2 + 4i3 – i4 + 10i3 – i2 = 0
– 4i3 + 11i4 = 0
and after simplification and combination of like terms, we write them in matrix form as
(3.48)
3 –2 0 0
3 –12 5 0
10 –15 7 –2
0 0 –4 11

R
i1
i2
i3
i4
I
4
0
0
0
V
=



We will solve these using MATLAB as follows:
R=[3 2 0 0; 3 12 5 0; 10 15 7 2; 0 0 4 11]; V=[4 0 0 0]'; I=RV;
fprintf(' n');...
fprintf('i1 = %3.4f A t',I(1,1)); fprintf('i2 = %3.4f A t',I(2,1));...
fprintf('i3 = %3.4f A t',I(3,1)); fprintf('i4 = %3.4f A t',I(4,1));...
fprintf(' n');...fprintf(' n')
i1 = 0.0173 A i2 = -1.9741 A i3 = -4.7482 A i4 = -1.7266 A
Therefore,
iSC = i4 = –1.727
RTH
vOC
iSC
--------- –14.12
= = ---------------- = 8.2 
–1.727
vTH
+
RTH
8.2 
14.18 V
 a
b
RLOAD
+
vTH
RTH
8.2 
a

iLOAD
RLOAD
+

vLOAD
14.18 V 8 
b
vLOAD iLOAD
and
The Thevenin equivalent is as shown in Figure 3.46.
Figure 3.46. Final form of Thevenin’s equivalent for the circuit of Example 3.10
Finally, with the load attached to points a and b, the circuit is as shown in Figure 3.47.
Figure 3.47. Circuit for finding and in Example 3.10
Therefore, using the voltage division expression and Ohm’s law we obtain
vLOAD
8
8.2 + 8
= ----------------  –14.18 = –7.00 V
iLOAD
–14.18
8.2 + 8
= ---------------- = –0.875 A

Norton’s Theorem
3.6 Norton’s Theorem
This theorem is analogous to Thevenin’s theorem and states that we can replace everything,
except the load, in a circuit by an equivalent circuit containing only an independent current
source which we will denote as iN in parallel with a resistance which we will denote as RN
, as
shown in Figure 3.48.
Network
to be replaced
by a Norton
equivalent
circuit
x
x
 
vxy RN vxy
y 
Load Load

(Rest
of the
circuit)
y
IN
(a) (b)
(Rest
of the
circuit)
Figure 3.48. Replacement of a network by its Norton equivalent
The current source IN
has the value of the short circuit current which would flow if a short were
connected between the terminals x and y, where the Norton equivalent is inserted, and the resis-tance
is found from the relation
(3.49)
RN
RN
vOC
iSC
= ---------
where vOC
is the open circuit voltage which appears across the open terminals x and y.
Like Thevenin’s, Norton’s theorem is most useful when a series of computations involves chang-ing
the load of a network while the rest of the circuit remains unchanged.
Comparing the Thevenin’s and Norton’s equivalent circuits, we see that one can be derived from
the other by replacing the Thevenin voltage and its series resistance with the Norton current
source and its parallel resistance. Therefore, there is no need to perform separate computations
for each of these equivalents; once we know Thevenin’s equivalent we can easily draw the Nor-ton
equivalent and vice versa.
Example 3.11
Replace the network shown in Figure 3.49 by its Thevenin and Norton equivalents.
+
iX
3  3 
6 
20iX V
x
y

Solution:
We observe that no current flows through the 3  resistor; Therefore, iX = 0
and the dependent
current source is zero, i.e., a short circuit. Thus,
vTH = vOC = vxy = 0
iSC = 0
RTH RN
vOC
iSC
= = ---------
0  0
+
iX
3  3 
6 
x
20iX V i
y
+  1 V
1 V vOC
iSC
RTH RN
vOC
iSC
--------- 1 V
--------- 1 V
= = = = ---------
i
iX
+
iX
v1

3  3 
6 
x
20iX V i
y
+ 
1 V
iX
and also
This means that the given network is some mathematical model representing a resistance, but we
cannot find this resistance from the expression
since this results in the indeterminate form . In this type of situations, we connect an external
source (voltage or current) across the terminals x and y. For this example, we arbitrarily choose to
connect a 1 volt source as shown in Figure 3.50.
Figure 3.50. Network for Example 3.11 with an external voltage source connected to it.
In the circuit of Figure 3.50, the source represents the open circuit voltage and the cur-rent
i represents the short circuit current . Therefore, the Thevenin (or Norton) resistance
will be found from the expression
(3.50)
Now, we can find i from the circuit of Figure 3.51 by application of KCL at Node .

Maximum Power Transfer Theorem
(3.51)
where
(3.52)
v1 – 20iX
----------------------
3
v1
6
+ ----- + iX = 0
iX
v1 – 1
3
= --------------
Simultaneous solution of (3.51) and (3.52) yields v1 = 34  25 and iX = 3  25
. Then, from
(3.50),
= = = -----
RTH RN
------------ 25
1
3  25
3
and the Thevenin and Norton equivalents are shown in Figure 3.52.
RTH
25/3 
RN
25/3 
vTH = 0 IN = 0
Figure 3.52. Thevenin’s and Norton’s equivalents for Example 3.11
3.7 Maximum Power Transfer Theorem
Consider the circuit shown in Figure 3.53. We want to find the value of RLOAD
that will absorb
maximum power from the voltage source whose internal resistance is .
vS RS
+
+
RS
iLOAD
RLOAD vLOAD
vS 
Figure 3.53. Circuit for computation of maximum power delivered to the load
The power delivered to the load is found from
or
(3.53)
RLOAD
pLOAD
= =  
pLOAD vLOAD  iLOAD
  vS
RLOAD
RS + RLOAD
 ------------------------------vS
 ------------------------------ 
RS + RLOAD
pLOAD
RLOAD
2 =
RS + RLOAD2 -------------------------------------vS

To find the value of RLOAD which will make pLOAD
maximum, we differentiate (3.53) with
respect to . Recalling that
RLOAD
d uv
dx
 
 -- 
v
d u – u
d v
= -----------------------------------------
dx
dx
v2 dpLOAD RS RLOAD +  2vS 2
vS
2RLOAD2 R–  S + RLOAD
dRLOAD
RS + RLOAD4 = ---------------------------------------------------------------------------------------------------------------------
RS + RLOAD2vS
2 vS
2RLOAD2 R–  S + RLOAD = 0
RS + RLOAD = 2RLOAD
RLOAD = RS
RS
RN RLOAD RLOAD
and differentiating (3.53), we obtain
(3.54)
and (3.54) will be zero if the numerator is set equal to zero, that is, if
or
or
(3.55)
Therefore, we conclude that a voltage source with internal series resistance or a current
source with internal parallel resistance delivers maximum power to a load when
or . For example, in the circuits of Figure 3.54, the voltage source
and current source deliver maximum power to the adjustable* load when
Figure 3.54. Circuits where is set to receive maximum power
We can use Excel or MATLAB to see that the load receives maximum power when it is set to the
same value as that of the resistance of the source. Figure 3.55 shows a spreadsheet with various
values of an adjustable resistive load. We observe that the power is maximum when
.
* An adjustable resistor is usually denoted with an arrow as shown in Figure 3.54.
RS
RP RLOAD
RLOAD = RS RLOAD = RP vS
iN
RLOAD = RS = RP = 5 
+
5 
5 
5 
5 
vS iS
RLOAD
RLOAD = 5 

Linearity
Maximum Power Transfer - Power vs. Resistance
RLOAD PLOAD
0 0.00
1 2.78
2 4.08
3 4.69
4 4.94
5 5.00
6 4.96
7 4.86
8 4.73
9 4.59
10 4.44
11 4.30
12 4.15
13 4.01
14 3.88
15 3.75
16 3.63
6
5
4
3
2
1
0
Maximum Power Transfer
0 5 10 15 20
Power (watts) vs Resistance (Ohms) - Linear Scale
6
4
2
0
Maximum Power Transfer
1 10 100
Power (watts) vs Resistance (Ohms) - Log Scale
Figure 3.55. Spreadsheet to illustrate maximum power transfer to a resistive load
The condition of maximum power transfer is also referred to as resistance matching or impedance
matching. We will define the term “impedance” in Chapter 6.
The maximum power transfer theorem is of great importance in electronics and communications
applications where it is desirable to receive maximum power from a given circuit and efficiency is
not an important consideration. On the other hand, in power systems, this application is of no
use since the intent is to supply a large amount of power to a given load by making the internal
resistance as small as possible.
3.8 Linearity
A linear passive element is one in which there is a linear voltagecurrent relationship such as
(3.56)
RS
= d iL iC Cdt
vR = RiR vL L t d
= d vC
Definition 3.1
A linear dependent source is a dependent voltage or current source whose output voltage or cur-rent
is proportional only to the first power of some voltage or current variable in the circuit or a

linear combination (the sum or difference of such variables). For example, is a lin-ear
vxy = 2v1 – 3i2

= = = v2  R i ISev nVT
p vi Ri2
relationship but and =
are nonlinear.
Definition 3.2
A linear circuit is a circuit which is composed entirely of independent sources, linear dependent
sources and linear passive elements or a combination of these.
3.9 Superposition Principle
The principle of superposition states that the response (a desired voltage or current) in any branch
of a linear circuit having more than one independent source can be obtained as the sum of the
responses caused by each independent source acting alone with all other independent voltage
sources replaced by short circuits and all other independent current sources replaced by open cir-cuits.
Note: Dependent sources (voltage or current) must not be superimposed since their values depend on
the voltage across or the current through some other branch of the circuit. Therefore, all
dependent sources must always be left intact in the circuit while superposition is applied.
Example 3.12
In the circuit of Figure 3.56, compute by application of the superposition principle.
i6
2  10 
36 V 8 
+ 
4  6  12 
i6
12 V
+
5 A
Solution:
Let represent the current due to the source acting alone, the current due to the
i'6 12 V i''6
source acting alone, and the current due to the source acting alone. Then, by the
36 V i'''6 5 A
i6 = i'6 + i''6 + i'''6
principle of superposition,
First, to find i'6 we short the 36 V voltage source and open the 5 A
current source. The circuit
then reduces to that shown in Figure 3.57.

Superposition Principle
8 
36 V
2  10 
x

4  6  12 
i'6
y
5 A
i'6
+
12 V
Applying Thevenin’s theorem at points x and y of Figure 3.57, we obtain the circuit of Figure
3.58 and from it we obtain
vxy vTH
4  12
2 + 4
= = -------------- = 8 V
12 V
8 
+
2 
36 V
4 
x


y
5 A
i'6
Figure 3.58. Circuit for computing the Thevenin voltage to find in Example 3.12
Next, we will use the circuit of Figure 3.59 to find the Thevenin resistance.
x


y
2  8 
4 
RTH
i'6
Figure 3.59. Circuit for computing the Thevenin resistance to find in Example 3.12
+ ------------ 28
RTH 8 4  2
= = ----- 
4 + 2
3

+
---------------------- 12
= = ----- A
2  10 
+
–---------------------------- 54
= = –----- A
The current is found from the circuit of Figure 3.60 below.
Figure 3.60. Circuit for computing in Example 3.12
(3.57)
Next, the current due to the source acting alone is found from the circuit of Figure 3.61.
and after combination of the and parallel resistors to a single resistor, the circuit simpli-fies
to that shown in Figure 3.62.
Figure 3.62. Simplification of the circuit of Figure 3.61 to compute for Example 3.12
From the circuit of Figure 3.62, we obtain
(3.58)
i'6
8 V
28/3 
6 
VTH
RTH
i'6
i'6
i'6
8
28  3 + 6
23
i''6 36 V
36 V 8 
+ 
4  6  12 
i''6
i''6
2  4 
36 V 8 
+ 
6 
i''6 
43
-- 
i''6
i''6
36
4  3 + 8 + 6
23

Superposition Principle
Finally, to find i'''6 , we short the voltage sources, and with the 5 A
current source acting alone
the circuit reduces to that shown in Figure 3.63.
2  8  10 
i'''6
4  6  12 
5 A
i'''6
Replacing the , , and resistors, and and by single resistors, we obtain
2  4  8  10  12 
------------ + 8 28
2  4
2 + 4
= -----  10 + 12 = 22 
3
and the circuit of Figure 3.63 reduces to that shown in Figure 3.64.
28 6  22 
3
5 A
----- 
i'''6
i'''6
Figure 3.64. Simplification of the circuit of Figure 3.63 to compute for Example 3.12
We will use the current division expression in the circuit of Figure 3.64 to find . Thus,
(3.59)
i'''6
28  3
28  3 + 6
----------------------  –5 70
= = –-----
Therefore, from (3.57), (3.58), and (3.59) we obtain
or
(3.60)
i6 i'6 + i''6 + i'''6 12
– – ----- 112
----- 54
= = = –--------
23
i6 = –4.87 A
and this is the same value as that of Example 3.5.
i'''6
23
----- 70
23
23
23

3.10 Circuits with NonLinear Devices
Most electronic circuits contain nonlinear devices such as diodes and transistors whose i  v
(currentvoltage) relationships are nonlinear. However, for small signals (voltages or currents)
these circuits can be represented by linear equivalent circuit models. A detailed discussion of
these is beyond the scope of this text; however we will see how operational amplifiers can be rep-resented
i – v D
vD iD
+
1.4
1.2
1.0
0.8
0.6
0.4
0.2
by equivalent linear circuits in the next chapter.
If a circuit contains only one nonlinear device, such as a diode, and all the other devices are lin-ear,
we can apply Thevenin’s theorem to reduce the circuit to a Thevenin equivalent in series
with the nonlinear element. Then, we can analyze the circuit using a graphical solution. The
procedure is illustrated with the following example.
Example 3.13
For the circuit of Figure 3.65, the characteristics of the diode are shown in figure 3.66.
We wish to find the voltage across the diode and the current through this diode using a
graphical solution.
Figure 3.66. Diode iv characteristics
1 V
VTH
RTH
Diode; conducts current
only in the indicated direction
vD
iD
1 K
vR
0.0
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
vD (volts)
iD (milliamps)

Circuits with NonLinear Devices
Solution:
or
or
(3.61)
vR + vD = 1 V
RiD = – vD + 1
iD
1R
---vD – 1R
= + ---
We observe that (3.61) is an equation of a straight line and the two points are obtained from it
by first letting vD = 0 , then, iD = 0
. We obtain the straight line shown in Figure 3.67 that is
plotted on the same graph as the given diode characteristics.
i – v
Diode Voltage Diode Current
(Volts) (milliamps)
0.00 0.000
0.02 0.000
0.04 0.000
0.06 0.000
0.08 0.000
0.10 0.000
0.12 0.000
0.14 0.000
0.16 0.000
0.18 0.000
0.20 0.000
0.22 0.000
0.24 0.000
0.26 0.000
0.28 0.000
I-V Relationship for Circuit of
Example 3.13
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
ID=(1/R)VD+1/R
Diode
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
VD (volts)
ID (milliamps)
Figure 3.67. Curves for determining voltage and current in a diode
The intersection of the nonlinear curve and the straight line yields the voltage and the current
of the diode where we find that vD = 0.665V and iD = 0.335 mA
.
Check:
Since this is a series circuit, also. Therefore, the voltage drop across the resis-tor
iR = 0.335 mA vR
is . Then, by KVL
vR = 1 k  0.335 mA = 0.335 V
vR + vD = 0.335 + 0.665 = 1 V

3.11 Efficiency
We have learned that the power absorbed by a resistor can be found from and this
power is transformed into heat. In a long length of a conductive material, such as copper, this lost
power is known as loss and thus the energy received by the load is equal to the energy trans-mitted
minus the loss. Accordingly, we define efficiency as
------------------ Output
= = = ------------------------------------
------------------  100 Output
= = = ------------------------------------  100
60 A
The efficiency is normally expressed as a percentage. Thus,
(3.62)
Obviously, a good efficiency should be close to
Example 3.14
In a twostory industrial building, the total load on the first floor draws an average of 60 amperes
during peak activity, while the total load of the second floor draws 40 amperes at the same time.
The building receives its electric power from a source. Assuming that the total resistance
of the cables (copper conductors) on the first floor is and on the second floor is , com-pute
the efficiency of transmission.
Solution:
First, we draw a circuit that represents the electrical system of this building. This is shown in Fig-ure
3.68.
pR = i 2R
i 2R
i 2R 
Efficiency  Output
Input
Output + Loss

% Efficiency %  Output
Input
Output + Loss
100%
480 V
1  1.6 
480 V
0.8 
+
0.5 
0.5 
1st Floor
Load
2nd Floor
Load
0.8 
vS i2 40 A i1

Regulation
Power supplied by the source:
(3.63)
pS
pS = vS i1 + i2 = 480  60 + 40 = 48 kilowatts
Power loss between source and 1st floor load:
(3.64)
= 0.5  + 0.5  = 60 2  1 = 3.6 kilowatts
ploss1 i1 2
Power loss between source and 2nd floor load:
(3.65)
ploss2 i2 2
Total power loss:
(3.66)
= 0.8  + 0.8  = 40 2  1.6 = 2.56 kilowatts
ploss = ploss1 + ploss2 = 3.60 + 2.56 = 6.16 kilowatts
Total power received by 1st and 2nd floor loads:
(3.67)
(3.68)
pL
pL = pS – ploss = 48.00 – 6.16 = 41.84 kilowatts
% Efficiency %  Output
------------------  100 41.84
= = = -------------  100 = 87.17 %
Input
48.00
3.12 Regulation
The regulation is defined as the ratio of the change in load voltage when the load changes from
no load (NL) to full load (FL) divided by the full load. Thus, denoting the noload voltage as
and the fullload voltage as , the regulation is defined as In other words,
vNL vFL
Regulation
vNL – vFL
vFL
= -----------------------
The regulation is also expressed as a percentage. Thus,
(3.69)
%Regulation
vNL – vFL
vFL
= -----------------------  100
Example 3.15
Compute the regulation for the 1st floor load of the previous example.

Solution:
The current drawn by 1st floor load is given as 60 A and the total resistance from the source to the
load as 1  . Then, the total voltage drop in the conductors is 60  1 = 60 V
. Therefore, the full
load voltage of the load is and the percent regulation is
= = -----------------------  100 = 14.3 %
vFL= 480 – 60 = 420 V
% Regulation
vNL – vFL
vFL
-----------------------  100 480 – 420
420

Summary
3.13 Summary
 When using nodal analysis, for a circuit that contains nodes, we must write indepen-dent
N N – 1
nodal equations in order to completely describe that circuit. When the presence of volt-age
sources in a circuit seem to complicate the nodal analysis because we do not know the cur-rent
through those voltage sources, we create combined nodes as illustrated in Example 3.2.
 When using nodal analysis, for a circuit that contains meshes or loops, branches, and
M L B
N nodes, we must write L = M = B – N + 1
independent loop or mesh equations in order to
completely describe that circuit. When the presence of current sources in a circuit seem to
complicate the mesh or loop analysis because we do not know the voltage across those current
sources, we create combined meshes as illustrated in Example 3.6.
 A practical voltage source has an internal resistance and it is represented by a voltage source
whose value is the value of the ideal voltage source in series with a resistance whose value is
the value of the internal resistance.
 A practical current source has an internal conductance and it is represented by a current
source whose value is the value of the ideal current source in parallel with a conductance
whose value is the value of the internal conductance.
 A practical voltage source vS in series with a resistance RS
can be replaced by a current
source iS whose value is vS  iS in parallel with a resistance RP whose value is the same as
RS
 A practical current source iS in parallel with a resistance RP
can be replaced by a voltage
source vS whose value is equal to iS  RS in series with a resistance RS
whose value is the
same as
RP
 Thevenin’s theorem states that in a two terminal network we can be replace everything
except the load, by a voltage source denoted as in series with a resistance denoted as
vTH
. The value of represents the open circuit voltage where the circuit is isolated from
RTH vTH
the load and is the equivalent resistance of that part of the isolated circuit. If a given cir-cuit
RTH
contains independent voltage and independent current sources only, the value of
RTH
can be found by first shorting all independent voltage sources, opening all independent cur-rent
sources, and calculating the resistance looking into the direction which is opposite to the
disconnected load. If the circuit contains dependent sources, the value of must be com-puted
from the relation
RTH
RTH = vOC  iSC
 Norton’s theorem states that in a two terminal network we can be replace everything except
the load, by a current source denoted as iN in parallel with a resistance denoted as RN
. The
value of represents the short circuit current where the circuit is isolated from the load and
iN

is the equivalent resistance of that part of the isolated circuit. If the circuit contains inde-pendent
voltage and independent current sources only, the value of can be found by first
RN
RN
shorting all independent voltage sources, opening all independent current sources, and calcu-lating
the resistance looking into the direction which is opposite to the disconnected load. If
the circuit contains dependent sources, the value of must be computed from the relation
RN
RN = vOC  iSC
 The maximum power transfer theorem states that a voltage source with a series resistance
RS
or a current source with parallel resistance delivers maximum power to a load when
RS RLOAD
RLOAD = RS RLOAD = RN
or
 Linearity implies that there is a linear voltagecurrent relationship.
 A linear circuit is composed entirely of independent voltage sources, independent current
sources, linear dependent sources, and linear passive devices such as resistors, inductors, and
capacitors.
 The principle of superposition states that the response (a desired voltage or current) in any
branch of a linear circuit having more than one independent source can be obtained as the
sum of the responses caused by each independent source acting alone with all other indepen-dent
voltage sources replaced by short circuits and all other independent current sources
replaced by open circuits.
 Efficiencyis defined as the ratio of output to input and thus it is never greater than unity. It is
normally expressed as a percentage.
 Regulation is defined as the ratio of vNL – vFL to vFL
and ideally should be close to zero. It is
normally expressed as a percentage.

Exercises
3.14 Exercises
Multiple Choice
1. The voltage across the 2 
resistor in the circuit below is
A.
B.
C.
D.
E.
6 V
16 V
–8 V
32 V
none of the above
8 A
2. The current in the circuit below is
A.
B.
C.
D.
E.
+ 
6 V
2 
8 A
i
–2 A
5 A
3 A
4 A
none of the above
+
2 
4 V
+ 
2 
2  2 
10 V i

3. The node voltages shown in the partial network below are relative to some reference node
i
–4 A
8  3 A
–5 A
–6 A
none of the above
6 V
+ 
2 
3 
+ 
12 V
6 
3 
which is not shown. The current is
A.
B.
C.
D.
E.
4. The value of the current for the circuit below is
A.
B.
C.
D.
E.
5. The value of the voltage for the circuit below is
A.
B.
C.
2 
8 V
4 V
i
+  8 V
8 V
6 V 13 V
i
–3 A
–8 A
–9 A
6 A
none of the above
+
6 
3 
12 V 8 A
i
v
4 V
6 V
8 V

Exercises
D.
E.
12 V
none of the above
2 A
2 
vX
+ 
2 
+

+

v
2vX
6. For the circuit below, the value of is dimensionless. For that circuit, no solution is possible if
the value of is
A.
B.
C.
D.
E.
k
k
2
1

0
none of the above
2 A
4 
4 
+

+

v
kv
7. For the network below, the Thevenin equivalent resistance to the right of terminals a
and b is
A.
B.
C.
D.
E.
RTH
1
2
5
10
none of the above

3 
RTH
2  2 
8. For the network below, the Thevenin equivalent voltage across terminals a and b is
A.
B.
C.
D.
E.
2 V
9. For the network below, the Norton equivalent current source and equivalent parallel resis-tance
across terminals a and b are
A.
B.
C.
D.
E.
2 
a
b
2 
2 
4 
VTH
–3 V
–2 V
1 V
5 V
none of the above
+ 
2  2 A
2 
a
b
IN
RN
1 A 2 
1.5 A 25 
4 A 2.5 
0 A 5
none of the above

Exercises
2 A
5 
a
b
5 
2 A
10. In applying the superposition principle to the circuit below, the current due to the
source acting alone is
A.
B.
C.
D.
E.
i 4 V
8 A
–1 A
4 A
–2 A
none of the above
8 A
2 
2 
4 V
+
i
2 
Problems
1. Use nodal analysis to compute the voltage across the 18 A current source in the circuit below.
Answer:
1.12 V
4 –1 5 –1
+
8 –1
4 –1 6  –1

10 –1
v18 A
12 A 18 A 24 A

2. Use nodal analysis to compute the voltage in the circuit below. Answer:
v6  21.6 V
12  15 
12 A 18 A 24 A
3. Use nodal analysis to compute the current through the resistor and the power supplied (or
absorbed) by the dependent source shown below. Answers:
4 
18 A
12  15 
6 
+
4. Use mesh analysis to compute the voltage below. Answer:
+  + 
8  12 
4  6 
4  6 
+

+ 
36 V
v6
6 
–3.9 A –499.17 w
12 A 24 A
36 V
+
iX
5iX
i6
v36A 86.34 V
12 A
240 V
36 A
+

120 V
24 A
4  3 
v36A

Exercises
5. Use mesh analysis to compute the current through the resistor, and the power supplied
i6
(or absorbed) by the dependent source shown below. Answers:
4 
18 A
12  15 
iX
6 
+
5iX
i6
12 A 24 A
+
36 V
6. Use mesh analysis to compute the voltage below. Answer:
–3.9 A –499.33 w
v10 0.5 V
+
12 V
10iX
12  15 
4 
6 
+

+ 
8 
24 V
iX v10
10 
7. Compute the power absorbed by the 10 
resistor in the circuit below using any method.
Answer:
1.32 w
+
12 V
6 
2 
3 
+ 
24 V
10 
+
36 V
8. Compute the power absorbed by the 20 
resistor in the circuit below using any method.
Answer:
73.73 w

9. For the circuit below:
a. To what value should the load resistor should be adjusted to so that it will absorb
maximum power? Answer:
b. What would then the power absorbed by be? Answer:
+ 
12  15 
4  6 
10. Replace the network shown below by its Norton equivalent.
15 
Answers:
11. Use the superposition principle to compute the voltage in the circuit below.
Answer:
12 V
2 
+
6 A
3 
20 
8 A
RLOAD
2.4 
RLOAD 135 w
12 A 18 A
36 V
RLOAD
iN = 0 RN = 23.75 
4  5  iX
5iX
a
b
v18A
1.12 V

Exercises
4 –1 5 –1
+
8 –1
4 –1 6  –1

10 –1
v18 A
12 A 18 A 24 A
12. Use the superposition principle to compute voltage in the circuit below.
Answer:
v6 
21.6 V
+ 
36 V
12  15 
4  6 
+

v6
12 A 18 A 24 A
13. In the circuit below, and are adjustable voltage sources in the range V,
vS1 vS2 –50  V  50
and and represent their internal resistances.
RS1 RS2
RS1 1  RS2 1 
iLOAD
vLOAD 
+
+
+ 

Adjustable
Resistive
Load
vS1 vS2
S1 S2
The table below shows the results of several measurements. In Measurement 3 the load resis-tance
is adjusted to the same value as Measurement 1, and in Measurement 4 the load resis-tance
is adjusted to the same value as Measurement 2. For Measurements 5 and 6 the load
resistance is adjusted to 1 
. Make the necessary computations to fillin the blank cells of
this table.

Measurement Switch Switch (V) (V) (A)
S1 S2 vS1 vS2 iLOAD
1 Closed Open 48 0 16
2 Open Closed 0 36 6
3 Closed Open 0 5
4 Open Closed 0 42
5 Closed Closed 15 18
6 Closed Closed 24 0
–15 V –7 A 11 A –24 V
76.6%
vS
+
480 V
0.8 
100 A
i1 i2
1st Floor
Load
0.8 
0.5 
0.5 
80 A
2nd Floor
Load
36.4%
VS i1 i2
+
480 V
0.8 
100 A
1st Floor
Load
0.8 
0.5 
0.5 
80 A
2nd Floor
Load
iLOAD vLOAD
Answers: , , ,
14. Compute the efficiency of the electrical system below. Answer:
15. Compute the regulation for the 2st floor load of the electrical system below.
Answer:
16. Write a set of nodal equations and then use MATLAB to compute and for the
circuit of Example 3.10, Page 329, which is repeated below for convenience.
Answers:
–0.96 A –7.68 V

Exercises
+

+
12 V
3  3 
20iX
6  10 
5 
7 
iLOAD
RLOAD
8 
+

4  vLOAD
iX

Multiple Choice
1. E The current entering Node A is equal to the current leaving that node. Therefore, there is
no current through the resistor and the voltage across it is zero.
2 
8 A
+ 
6 V
2 
8 A
8 A
8 A 8 A
2. C From the figure below, . Also, and . Then,
VAC = 4 V VAB = VBC = 2 V VAD = 10 V
and . Therefore,
VBD = VAD – VAB = 10 – 2 = 8 V VCD = VBD – VBC = 8 – 2 = 6 V
i = 6  2 = 3 A
A B C
+
2 
4 V
+ 
2 
2  2 
10 V i
D
6 V
6 + 12 = 18 V
VBC = 18 – 6 = 12 V
6 V
12 V
A
C B
.
3. A From the figure below we observe that the node voltage at A is relative to the refer-ence
node which is not shown. Therefore, the node voltage at B is relative
to the same reference node. The voltage across the resistor is and
the direction of current through the 3 
resistor is opposite to that shown since Node B is
at a higher potential than Node C. Thus
i = –12  3 = –4 A
+ 
3 
+

2 
2 
8 V
4 V
i
+
 8 V
8 V
6 V 13 V

4. E We assign node voltages at Nodes A and B as shown below.
At Node A
A B
6 
3 
12 V 8 A
and at Node B
These simplify to
and
+
6 
3 
i
VA – 12
-------------------
6
VA
6
+ + -------------------- = 0
-------
VA – VB
3
VB – VA
--------------------
3
VB
3
+ ------- = 8
23
--VA
13
– --VB = 2
13
–--VA
23
+ --VB = 8
Multiplication of the last equation by 2 and addition with the first yields and
thus .
i = –18  3 = –6 A
5. E Application of KCL at Node A of the circuit below yields
or
Also by KVL
and by substitution
VB = 18
2 A
2 
vX
+ 
2 
+

A
+

v
2vX
v2
--
v 2vX
–
2
+ ------------------ = 2
v – vX = 2
v = vX + 2vX

vX + 2vX – vX = 2
vX = 1
v = vX + 2vX = 1 + 2  1 = 3 V
2 A
4 
4 
+

A
+

v
kv
v4
-- v – kv
+ --------------- = 2
4
14
--2v – kv = 2
k = 2 k  2
2  2 
3 
2 
a
RTH
b
2  2 
2 
2 
4 
2 + 2 = 4 4  4 = 2
2 + 2 = 4 4  3 + 2  2 = 4  3 + 1 = 4  4 = 2 
2 
or
and thus
6. A Application of KCL at Node A of the circuit below yields
or
and this relation is meaningless if . Thus, this circuit has solutions only if .
7. B The two resistors on the right are in series and the two resistors on the left shown
in the figure below are in parallel.
Beginning on the right side and proceeding to the left we obtain , ,
, .
8. A Replacing the current source and its parallel resistance with an equivalent voltage
source in series with a resistance we obtain the network shown below.
2 

+ 
2 V
2  2 A
By Ohm’s law,
and thus
2 
a
b
2 
2 V
 +
 +
2 
4 V
a
b
i
i 4 – 2
= ------------ = 0.5 A
2 + 2
vTH = vab = 2  0.5 + –4 = –3 V
9. D The Norton equivalent current source is found by placing a short across the terminals a
IN
and b. This short shorts out the resistor and thus the circuit reduces to the one shown
below.
5 
a
b
By KCL at Node A,
5 
2 A
5 
2 A
a
b
ISC = IN
2 A
A
5 
2 A
IN + 2 = 2
and thus
IN = 0
The Norton equivalent resistance RN
is found by opening the current sources and looking
to the right of terminals a and b. When this is done, the circuit reduces to the one shown
below.
5 
a
b
Therefore, and the Norton equivalent circuit consists of just a resistor.
RN = 5  5 

10. B With the 4 V
source acting alone, the circuit is as shown below.
A
+

B
i
 +
2 
2 
2 
4 V
 +
We observe that and thus the voltage drop across each of the resistors to
the left of the source is with the indicated polarities. Therefore,
vAB = 4 V 2 
4 V 2 V
i = –2  2 = –1 A
Problems
1. We first replace the parallel conductances with their equivalents and the circuit simplifies to
that shown below.
15 –1
v1 v2 v3
1 2 3
+

12 –1
4 –1 6  –1
v18 A
12 A 18 A 24 A
16v1 – 12v2 = 12
–12v1 + 27v2 – 15v3 = –18
–15v2 + 21v3 = 24
4v1 – 3v2 = 3
–4v1 + 9v2 – 5v3 = –6
–5v2 + 7v3 = 8
Applying nodal analysis at Nodes 1, 2, and 3 we obtain:
Node 1:
Node 2:
Node 3:
Simplifying the above equations, we obtain:
Addition of the first two equations above and grouping with the third yields

6v2 – 5v3 = –3
–5v2 + 7v3 = 8
For this problem we are only interested in . Therefore, we will use Cramer’s rule to
solve for . Thus,
and
v2 = v18 A
v2
v2
D2

= ------ D2
= = – 21 + 40 = 19  6 –5
–3 –5
8 7
= = 42 – 25 = 17
–5 7
v2 = v18 A = 19  17 = 1.12 V
2. Since we cannot write an expression for the current through the source, we form a com-bined
node as shown on the circuit below.
+ 
36 V
v2
2
12  15 
v1
1
v3
4  6 
+

v6
3
12 A 18 A 24 A
At Node 1 (combined node):
and at Node 2,
Also,
v1
-----
4
v1 – v2
12
+ + + ----- – 12 – 24 = 0
----------------
v3 – ----------------
v2
15
v2 – ----------------
v1
12
v2 – v3
15
+ ---------------- = –18
v1 – v3 = 36
Simplifying the above equations, we obtain:
36 V
v3
6
13
--v1
3
20
– -----v2 7
+ -----v3 = 36
30
1
12
+ – -----v3 = –18
v1 –v3 = 36
–-----v1
3
20
-----v2
1
15
Addition of the first two equations above and multiplication of the third by yields
–1  4

4 
18 A
12  15 
6 
+
and by adding the last two equations we obtain
or
Check with MATLAB:
format rat
R=[1/3 3/20 7/30; 1/12 3/20 1/15; 1 0 1];
I=[36 18 36]';
V=RI;
fprintf('n'); disp('v1='); disp(V(1)); disp('v2='); disp(V(2)); disp('v3='); disp(V(3))
v1=
288/5
v2=
-392/5
v3=
108/5
3. We assign node voltages , , , and current as shown in the circuit below.
Then,
and
14
--v1
16
+ --v3 = 18
14
–--v1
14
+ --v3 = –9
5
12
-----v3 = 9
v3 v6 
108
5
= = -------- = 21.6V
v1 v2 v3 v4 iY
12 A 24 A
36 V
+
iX
5iX
i6
iY
v1 v2 v3
v4
v1
-----
4
v1 – v2
12
+ ---------------- + 18 – 12 = 0

v2 – ----------------
v1
12
v2 – v3
12
+ ----------------
+ ---------------- = 0
v2 – v4
6
Simplifying the last two equations above, we obtain
and
13
--v1
1
12
– -----v2 = –6
1
12
– -----v1 19
+ -----v2
– – --v4 = 0
60
1
15
-----v3
16
v1 – v2
12
5
12
= -----v1 – v2
Next, we observe that , and . Then and
by substitution into the last equation above, we obtain
or
iX
= ---------------- v3 = 5iX v4 = 36 V v3
1
12
– -----v1 19
+ – – --36 = 0
-----v2
60
1
15
----- 5
 ----- 16
 12
v1 – v2  19
– --v1 31
+ -----v2 = 6
90
Thus, we have two equations with two unknowns, that is,
13
--v1
1
12
– -----v2 = –6
– --v1 31
19
+ -----v2 = 6
90
Multiplication of the first equation above by and addition with the second yields
or
We find from
Thus,
or
v1
Now, we find from
1  3
19
60
-----v2 = 4
v2 = 240  19
13
--v1
1
12
– -----v2 = –6
13
--v1
1
12
----- 240
–  -------- = –6
19
v1 = –282  19
v3

v3
-----v1 – v2 5
5
12
----- –282
= = = –--------
12
----------- 240
  435
 – --------
19
19
38
v1 = –282  19 V
v2 = 240  19 V
v3 = –435  38 V
v4 = 36 V
6 
i6 
v2 – v4
6
---------------- 240  19 – 36
------------------------------ 74
= = = –----- = –3.9 A
6
19
iY v3
iY – 24 – 18
v3 – v2
15
+ ---------------- = 0
iY = 42 – –-----------------------------------------------
435  38 – 240  19
15
42 915  38
+ ------------------ 1657
= = -----------
15
38
= = = –-------------- = –499.17 w
p v3iY
–-------- 1657
435
38
 ----------- 72379
38
145
36 A
Therefore, the node voltages of interest are:
The current through the resistor is
To compute the power supplied (or absorbed) by the dependent source, we must first find the
current . It is found by application of KCL at node voltage . Thus,
or
and
that is, the dependent source supplies power to the circuit.
4. Since we cannot write an expression for the current source, we temporarily remove it
and we form a combined mesh for Meshes 2 and 3 as shown below.

12 A
+  + 
4  i 3  6 i5
8  12 
120 V
4  6 
i1 i2 i3
Mesh 1:
Combined mesh (2 and 3):
or
240 V
24 A
i4
i1 = 12
– 4i1 + 12i2 + 18i3 – 6i4 – 8i5 – 12i6 = 0
– 2i1 + 6i2 + 9i3 – 3i4 – 4i5 – 6i6 = 0
36 A
We now reinsert the current source and we write the third equation as
Mesh 4:
Mesh 5:
or
Mesh 6:
or
i2 – i3 = 36
i4 = –24
–8i2 + 12i5 = 120
–2i2 + 3i5 = 30
–12i3 + 15i6 = –240
–4i3 + 5i6 = –80
Thus, we have the following system of equations:
and in matrix form
i1 = 12
– 2i1 + 6i2 + 9i3 – 3i4 – 4i5 – 6i6 = 0
i2 – i3 = 36
i4 = –24
–2i2 + 3i5 = 30
–4i3 + 5i6 = –80

We find the currents through with the following MATLAB script:
R=[1 0 0 0 0 0; 2 6 9 3 4 6;...
+  + 
8  12 
4  6 
0 1 1 0 0 0; 0 0 0 1 0 0;...
0 2 0 0 3 0; 0 0 4 0 0 5];
V=[12 0 36 24 30 80]';
I=RV;
fprintf('n');...
fprintf('i1=%7.2f A t', I(1));...
fprintf('i2=%7.2f A t', I(2));...
fprintf('i3=%7.2f A t', I(3));...
fprintf('n');...
fprintf('i4=%7.2f A t', I(4));...
fprintf('i5=%7.2f A t', I(5));...
fprintf('i6=%7.2f A t', I(6));...
fprintf('n')
i1= 12.00 A i2= 6.27 A i3= -29.73 A
i4= -24.00 A i5= 14.18 A i6= -39.79 A
Now, we can find the voltage by application of KVL around Mesh 3.
Thus,
1 0 0 0 0 0
–2 6 9 –3 –4 –6
0 1 –1 0 0 0
0 0 0 1 0 0
0 –2 0 0 3 0
0 0 –4 0 0 5
R
i1
i2
i3
i4
i5
i6
I

12
0
36
–24
30
–80
V
=



i1 i6
v36 A
12 A
240 V
36 A
+

120 V
24 A
4  3 
v36A
i3

or
v36 A = v12  + v6  = 12  –29.73 – –39.79 + 6  –29.73 – 24.00
v36 A = 86.34 V
To verify that this value is correct, we apply KVL around Mesh 2. Thus, we must show that
v4  + v8  + v36 A = 0
By substitution of numerical values, we find that
4  6.27 – 12 + 8  6.27 – 14.18 + 86.34 = 0.14
5. This is the same circuit as that of Problem 3. We will show that we obtain the same answers
using mesh analysis.
We assign mesh currents as shown below.
Mesh 1:
Mesh 2:
or
Mesh 3:
4 
18 A
12  15 
iX
6 
+
5iX
i6
i5
i3 i4
12 A 24 A
i1
i2
i1 = 12
– 4i1 + 22i2 – 6i3 – 12i5 = –36
– 2i1 + 11i2 – 3i3 – 6i5 = –18
–6i2 + 21i3 – 15i5 + 5iX = 36
and since , the above reduces to
or
Mesh 4:
+
36 V
iX = i2 – i5
–6i2 + 21i3 – 15i5 + 5i2 – 5i5 = 36
– i2 + 21i3 – 20i5 = 36
i4 = –24

Mesh 5:
Grouping these five independent equations we obtain:
and in matrix form,
We find the currents through with the following MATLAB script:
R=[1 0 0 0 0 ; 2 11 3 0 6; 0 1 21 0 20; ...
0 0 0 1 0; 0 0 0 0 1];
V=[12 18 36 24 18]';
I=RV;
fprintf('n');...
fprintf('i1=%7.2f A t', I(1));...
fprintf('i2=%7.2f A t', I(2));...
fprintf('i3=%7.2f A t', I(3));...
fprintf('n');...
fprintf('i4=%7.2f A t', I(4));...
fprintf('i5=%7.2f A t', I(5));...
fprintf('n')
i1= 12.00 A i2= 15.71 A i3= 19.61 A
i4= -24.00 A i5= 18.00 A
By inspection,
Next,
i5 = 18
i1 = 12
– 2i1 + 11i2–3i3 –6i5 = –18
– i2 + 21i3 – 20i5 = 36
i4 = –24
i5 = 18
1 0 0 0 0
–2 11 –3 0 –6
0 –1 21 0 –20
0 0 0 1 0
0 0 0 0 1
R
i1
i2
i3
i4
i5
I

12
–18
36
–24
18
V
=



i1 i5
i6  = i2 – i3 = 15.71 – 19.61 = –3.9 A

p5iX
= 5iXi3 – i4 = 5i2 – i5i3 – i4
= 515.71 – 18.0019.61 + 24.00 = –499.33 w
These are the same answers as those we found in Problem 3.
6. We assign mesh currents as shown below and we write mesh equations.
Mesh 1:
or
Mesh 2:
Mesh 3:
or
Mesh 4:
or
12 V
4 
i4
6 
10iX
12  15 
+
+ 
8 
i2
24 V
i1
24i1 – 8i2 – 12i4 – 24–12 = 0
6i1 – 2i2 – 3i4 = 9
–8i1 + 29i2 – 6i3 – 15i4 = –24
–6i2 + 16i3 = 0
–3i2 + 8i3 = 0
i4= 10iX = 10i2 – i3
10i2 – 10i3 – i4 = 0
Grouping these four independent equations we obtain:
and in matrix form,
+

iX v10
10 
i3
6i1 – 2i2 – 3i4 = 9
–8i1 + 29i2 – 6i3 – 15i4 = –24
–3i2 + 8i3 = 0
10i2 – 10i3 – i4 = 0

6 –2 0 –3
–8 29 –6 –15
0 –3 8 0
0 10 –10 –1

R
i1
i2
i3
i4
I

9
–24
0
0
V
=


We find the currents i1 through i4
with the following MATLAB script:
R=[6 2 0 3; 8 29 6 15; 0 3 8 0 ; 0 10 10 1];
V=[9 24 0 0]'; I=RV;
fprintf('n');...
fprintf('i1=%7.2f A t', I(1));...
fprintf('i2=%7.2f A t', I(2));...
fprintf('i3=%7.2f A t', I(3));...
fprintf('i4=%7.2f A t', I(4));...
fprintf('n')
i1= 1.94 A i2= 0.13 A i3= 0.05 A i4= 0.79 A
v10
v10 = 10i3 = 10  0.05 = 0.5 V
6 
v6 = 6i2 – i3 = 60.13 – 0.05 = 0.48 V
2  3  10 
6 A 8 A 6 A
6 
Now, we find by Ohm’s law, that is,
The same value is obtained by computing the voltage across the resistor, that is,
7. Voltagetocurrent source transformation yields the circuit below.
By combining all current sources and all parallel resistors except the 10 
resistor, we obtain
the simplified circuit below.
1  10 
4 A

Applying the current division expression, we obtain
and thus
i10 
1
1 + 10
---------------  4 4
= = ----- A
11
 2
2 10 4
= = = = -------- = 1.32 w
p10  i10 
 ----- 
11
 10 16
--------  10 160
121
121
8. Currenttovoltage source transformation yields the circuit below.
12 V
From this series circuit,
and thus
2  12 V
20  3 
+
+ 
+ 
i 24 V
------- 48
i v
= = ----- A
R
25
 2
=  20 2304
p20  i220 48
= = -----------  20 = 73.73 w
 ----- 
25
625
9. We remove from the rest of the rest of the circuit and we assign node voltages , ,
and . We also form the combined node as shown on the circuit below.
v3
Node 1:
or
RLOAD v1 v2
+ 
36 V
1
3
v1 2
v2 v3
12  15 
4  6 
12 A 18 A
x

y

v1
-----
4
v1 – v2
12
+ + + ----- = 0
---------------- – 12
v3 – ----------------
v2
15
v3
6

– -----v2 7
+ -----v3 = 12
+ -----v3 = 12
–----- 7
-----
–----- 3
----- 1
–-----
Node 2:
or
Also,
For this problem, we are interested only in the value of which is the Thevenin voltage ,
and we could find it by Gauss’s elimination method. However, for convenience, we will group
these three independent equations, express these in matrix form, and use MATLAB for their
solution.
and in matrix form,
We find the voltages through with the following MATLAB script:
G=[1/3 3/20 7/30; 1/12 3/20 1/15; 1 0 1];
I=[12 18 36]'; V=GI;
fprintf('n');...
fprintf('v1=%7.2f V t', V(1)); fprintf('v2=%7.2f V t', V(2)); fprintf('v3=%7.2f V t', V(3));
fprintf('n')
v1= 0.00 V v2= -136.00 V v3= -36.00 V
Thus,
13
--v1
3
20
30
v2 – ----------------
v1
12
v2 – v3
15
+ ---------------- = –18
1
12
–-----v1
3
20
-----v2
1
15
+ –-----v3 = –18
v1 – v3 = 36
v3 vTH
13
--v1
3
20
– -----v2 7
30
1
12
–-----v1
3
20
-----v2
1
15
+ –-----v3 = –18
v1 – v3 = 36
13
-- 3
20
30
1
12
20
15
1 0 –1
G
v1
v2
v3
V

12
–18
36
I
=



v1 v3
vTH = v3 = –36 V

To find RTH
we short circuit the voltage source and we open the current sources. The circuit
then reduces to the resistive network below.
12  15 
4  6 
x

y

RTH
We observe that the resistors in series are shorted out and thus the Thevenin resistance is the
parallel combination of the and resistors, that is,
4  6 
4   6  = 2.4 
and the Thevenin equivalent circuit is as shown below.
2.4 
+ 
36 V
Now, we connect the load resistor at the open terminals and we obtain the simple
series circuit shown below.
36 V
a. For maximum power transfer,
RLOAD
RLOAD 2.4 
2.4 
+ 
RLOAD = 2.4 
b. Power under maximum power transfer condition is
pMAX i2RLOAD
 2
36
2.4 + 2.4
= =  2.4 = 7.52  2.4 = 135 w
 --------------------- 

10. We assign a node voltage Node 1 and a mesh current for the mesh on the right as shown
v1 iX
15 
1
4  5  iX
5iX
a
b
v1
4
----- + iX = 5iX
15 + 5iX = v1
20iX
4
----------- + iX = 5iX
6iX = 5iX
iX = 0
iN
vOC
RN
= ---------
= = = ----------- = 0
vab
RN
-------
5  iX
RN
-------------- 5  0
RN
a
b
RN
below.
At Node 1:
Mesh on the right:
and by substitution into the node equation above,
or
but this can only be true if .
Then,
Thus, the Norton current source is open as shown below.
To find the value of we insert a current source as shown below.
RN 1 A

At Node A:
But
vA iX
15 
A
vB
B
iX 4  5 
5iX
iX 1 A
vA
4
------
vA – vB
15
+ ------------------ = 5iX
vB = 5   iX = 5iX
and by substitution into the above relation
or
At Node B:
or
a
b
vA
------
4
vA – vB
15
+ ------------------ = vB
19
-----60
vA
16
– -----15
vB = 0
vB – vA
15
------------------
vB
5
+ ----- = 1
1
15
– -----vA 4
+ -----vB = 1
15
For this problem, we are interested only in the value of which we could find by Gauss’s
elimination method. However, for convenience, we will use MATLAB for their solution.
and in matrix form,
vB
19
60
-----vA
16
15
– -----vB = 0
– -----vA 4
1
15
+ -----vB = 1
15
19
60
----- 16
–-----
15
1
15
–----- 4
-----
15

G
vA
vB
V

0
1
I
=


We find the voltages and with the following MATLAB script:
v1 v2

RN
Vab
ISC
= = ------- = 23.75 
--------
VB
1
v'18A 12 A
12 A
v1 v2 v3
+
12 –1
–1 v'–1
4 18A
6  
15 –1
16v1 – 12v2 = 12
– 12v1 + 27v2 – 15v3 = 0
–15v2 + 21v3 = 0
4v1 – 3v2 = 3
– 4v1 + 9v2 – 5v3 = 0
–5v2 + 7v3 = 0
v2 = v'18A v2
6v2 – 5v3 = 3
–5v2 + 7v3 = 0
G=[19/60 16/15; 1/15 4/15];
I=[0 1]'; V=GI;
fprintf('n');...
fprintf('vA=%7.2f V t', V(1)); fprintf('vB=%7.2f V t', V(2));
fprintf('n')
vA= 80.00 V vB= 23.75 V
Now, we can find the Norton equivalent resistance from the relation
11. This is the same circuit as that of Problem 1. Let be the voltage due to the current
source acting alone. The simplified circuit with assigned node voltages is shown below where
the parallel conductances have been replaced by their equivalents.
The nodal equations at the three nodes are
or
Since , we only need to solve for . Adding the first 2 equations above and group-ing
with the third we obtain
Multiplying the first by and the second by we obtain
7 5

and by addition of these we obtain
42v2 – 35v3 = 21
–25v2 + 35v3 = 0
= = ----- V
v2 v'18A
21
17
Next, we let be the voltage due to the current source acting alone. The simpli-fied
v''18A 18 A
circuit with assigned node voltages is shown below where the parallel conductances
have been replaced by their equivalents.
12 –1
vA vB vC
4 –1 6  –1
18 A
or
+

15 –1
v''18A
16vA – 12vB = 0
– 12vA + 27vB – 15vC = –18
–15vB + 21vC = 0
4vA – 3vB = 0
– 4vA + 9vB – 5vC = –6
–5vB + 7vC = 0
Since vB = v''18A , we only need to solve for vB
. Adding the first 2 equations above and
grouping with the third we obtain
6vB – 5vC = –6
–5vB + 7vC = 0
7 5
42vB – 35vC = –42
–25vB + 35vC = 0

Finally, we let be the voltage due to the current source acting alone. The simpli-fied
circuit with assigned node voltages is shown below where the parallel conductances have
12 –1
vX vY vZ
+ + ----- 19
-------- 40
----- –42
= = = ----- = 1.12 V
been replaced by their equivalents.
or
Since , we only need to solve for . Adding the first 2 equations above and
grouping with the third we obtain
and thus
vB v''18A
–42
17
= = -------- V
v'''18A 24 A
24 A
+

15 –1
–1 v'''–1
4 18A
6  16vX – 12vY = 0
– 12vA + 27vY – 15vZ = 0
–15vB + 21vZ = 24
4vX – 3vY = 0
– 4vX + 9vY – 5vZ = 0
–5vY + 7vZ = 8
vY = v'''18A vY
6vY – 5vZ = 0
–5vY + 7vZ = 0
7 5
42vY – 35vZ = 0
–25vY + 35vZ = 40
vY v'''18A
40
17
= = ----- V
v18A v'18A + v''18A + v'''18A 21
17
17
17
17

This is the same answer as in Problem 1.
12. This is the same circuit as that of Problem 2. Let be the voltage due to the cur-rent
v'6  12 A
source acting alone. The simplified circuit is shown below.
12 A
12  15 
4  6 
+

v'6 
The 12  and 15 
resistors are shorted out and the circuit is further simplified to the one
shown below.
12 A
4  6 
+

v'6 
The voltage v'6 
is computed easily by application of the current division expression and
multiplication by the resistor. Thus,
6 
v'6 
   6 144
4
4 + 6
= = -------- V
------------  12
5
Next, we let be the voltage due to the current source acting alone. The simpli-fied
v''6  18 A
circuit is shown below. The letters A, B, and C are shown to visualize the circuit simpli-fication
process.
12  15 
A B A
4  6 
+

v''6 
18 A
C
+

v''6 
6 
A
15 
12 
4 
B
18 A
+

v''6 
C 6 
A
12  15 
4 
B
18 A
C

The voltage is computed easily by application of the current division expression and
multiplication by the resistor. Thus,
------------  –18  6 –216
= = ----------- V
Now, we let be the voltage due to the current source acting alone. The simplified
circuit is shown below.
12  15 
The and resistors are shorted out and voltage is computed by application
of the current division expression and multiplication by the resistor. Thus,
   6 288
= = -------- V
Finally, we let be the voltage due to the voltage source acting alone. The simpli-fied
12  15 
12 
4 
viv
= = –--------
circuit is shown below.
By application of the voltage division expression we find that
Therefore,
v''6 
6 
v''6 
4
4 + 6
5
v'''6  24 A
24 A
4  6 
+

v'''6 
12  15  v'''6 
6 
v'''6 
4
4 + 6
------------  24
5
viv
6  36 V
4  6 
+

+ 
36 V
viv
6 
+
15  36 V
6 
A
C
B
A
B
C +

6 
viv
6 
6
4 + 6
------------  –36 108
5

+ – -------- 108
-------- 108
– -------- 288
-------- 216
+ + + 6  144
v6  v'6  v''6  v'''6  viv
= = = -------- = 21.6 V
5
This is the same answer as that of Problem 2.
13. The circuit for Measurement 1 is shown below.
+
48 V
Let . Then,
5
5
5
5
RS1
1 
iLOAD1
16 A
RLOAD1
vS1
Req1 = RS1 + RLOAD1
Req1
vS1
iLOAD1
----------------- 48
= = ----- = 3 
16
For Measurement 3 the load resistance is the same as for Measurement 1 and the load cur-rent
is given as . Therefore, for Measurement 3 we find that
–5 A
vS1 = Req1–5 = 3  –5 = –15 V
and we enter this value in the table below.
The circuit for Measurement 2 is shown below.
+
36 V
Let . Then,
RS2
1 
iLOAD2
6 A
RLOAD2
vS2
Req2 = RS1 + RLOAD2
Req2
vS2
iLOAD2
----------------- 36
= = ----- = 6 
6
For Measurement 4 the load resistance is the same as for Measurement 2 and is given as
. Therefore, for Measurement 4 we find that
vS2
–42 V

iLOAD2
vS2
Req2
---------- 42
= = –----- = –7 A
6
RS1 RS2
1  1 
+
+
vLOAD RLOAD

+
vS1 vS2
iLOAD
1 
15 V 18 V
0.5 
iLOAD
RLOAD 1 
33 A
iLOAD
0.5
0.5 + 1
= ----------------  33 = 11 A
A
vA
RS1 RS2
1  1 
RLOAD 1  +
 +
vS1 vS2
iLOAD
24 V
Replacing the voltage sources with their series resistances to their equivalent current sources
with their parallel resistances and simplifying, we obtain the circuit below.
Application of the current division expression yields
We observe that iLOAD will be zero if vA = 0 and this will occur when vS1 = –24
. This can
be shown to be true by writing a nodal equation at Node A. Thus,

or
14.
vA – –24
--------------------------
1
vA – 24
1
+ ------------------ + 0 = 0
vA = 0
Measurement Switch Switch
S1 S2 vS1 vS2 iL
1 Closed Open 48 0 16
2 Open Closed 0 36 6
3 Closed Open -15 0 -5
4 Open Closed 0 -42 -7
5 Closed Closed 15 18 11
6 Closed Closed -24 24 0
vS
+
480 V
0.8 
100 A
i1 i2
1st Floor
Load
0.8 
0.5 
0.5 
The power supplied by the voltage source is
pS = vS i1 + i2 = 480100 + 80 = 86 400 w = 86.4 Kw
The power loss on the 1st floor is
= 20.5 + 0.5 = 100 2  1 = 10 000 w = 10 Kw
pLOSS1 i1
The power loss on the 2nd floor is
= 20.8 + 0.8 = 80 2  1.6 = 10 240 w = 10.24 Kw
pLOSS2 i2
and thus the total loss is
Then,
(V) (V) (A)
80 A
2nd Floor
Load
Total loss = 10 + 10.24 = 20.24 Kw

Output power = Input power – power losses = 86.4 – 20.24 = 66.16Kw
% Efficiency  Output
------------------  100 66.16
= = = -------------  100 = 76.6%
Input
86.4
vS
+
480 V
0.8 
100 A
i1 i2
1st Floor
Load
0.8 
0.5 
0.5 
80 A
2nd Floor
Load
vcond = RT i2 = 1.6  80 = 128 V
vFL = 480 – 128 = 352 V
% Regulation
vNL – vFL
vFL
-----------------------  100 480 – 352
= = -----------------------  352
100 = 36.4% +
v1 v2 v3 v4
+
6  10 
RLOAD
and
This is indeed a low efficiency.
15.
The voltage drop on the second floor conductor is
and thus the fullload voltage is
Then,
This is a very poor regulation.
16. We assign node voltages and we write nodal equations as shown below.

12 V
3  3 
5 
7 
8 
+

4  vLOAD
iLOAD
iX
20iX
combined node
v5
v1 = 12

v3 – ----------------
v2
3
where and thus
v2 – ----------------
v1
3
v2
6
+ + ---------------- = 0
-----
v2 – v3
3
v3 – v5
10
+ ----------------
+ + ---------------- = 0
v4 – ----------------
v5
4
v3 – v4 = 20iX
iX = v2  6
v5
10
3
= -----v2
v5
-----
5
v5 – v3
10
+ ----------------
+ + ---------------- = 0
v5 – ----------------
v4
4
Collecting like terms and rearranging we obtain
and in matrix form
v1 = 12
–1
3
-----v1
56
+ --v2
+ -----v3 = 0
-----–1
3
v2
+ + – -----v5 = 0
10
3
– -----v2 + v3 – v4 = 0
– -----v3 19
56
1 0 0 0 0
----- –1
-- –1
3
----- 0 0
3
0 –1
----- 13
3
----- 19
30
----- 19
60
0 10
–----- 1 –1 0
3
0 0 1
–----- 37
–----- 19
10
60

G
We use MATLAB to solve the above.
v4 – v5
7 + 8
v5 – v4
7 + 8
–1
3
13
30
-----v3
19
60
-----v4
19
60
1
10
– -----v4 37
60
+ -----v5 = 0
60
–-----
60
-----
60
v1
v2
v3
v4
v5
V

12
0
0
0
0
I
=



= = ------------------------------------------ = –0.96 A
G=[1 0 0 0 0;...
1/3 5/6 1/3 0 0;...
0 1/3 13/30 19/60 19/60;...
0 10/3 1 1 0;...
0 0 1/10 19/60 37/60];
I=[12 0 0 0 0]'; V=GI;
fprintf('n');...
fprintf('v1 = %7.2f V n',V(1));...
fprintf('v2 = %7.2f V n',V(2));...
fprintf('v3 = %7.2f V n',V(3));...
fprintf('v4 = %7.2f V n',V(4));...
fprintf('v5 = %7.2f V n',V(5));...
fprintf('n'); fprintf('n')
v1 = 12.00 V
v2 = 13.04 V
v3 = 20.60 V
v4 = -22.87 V
v5 = -8.40 V
Now,
and
iLOAD
v4 – v5
8 + 7
---------------- – 22.87 – –8.40
15
vLOAD = 8iLOAD = 8  –0.96 = –7.68 V

Chapter 4
Introduction to Operational Amplifiers
his chapter is an introduction to amplifiers. It discusses amplifier gain in terms of decibels
(dB) and provides an overview of operational amplifiers, their characteristics and applica-tions.
Numerous formulas for the computation of the gain are derived and several practical
T
examples are provided.
4.1 Signals
A signal is any waveform that serves as a means of communication. It represents a fluctuating
electric quantity, such as voltage, current, electric or magnetic field strength, sound, image, or
any message transmitted or received in telegraphy, telephony, radio, television, or radar. A typical
signal which varies with time is shown in figure 4.1 where f t
can be any physical quantity such
as voltage, current, temperature, pressure, and so on.
f t
Figure 4.1. A signal that changes with time
t
4.2 Amplifiers
An amplifier is an electronic circuit which increases the magnitude of the input signal. The sym-bol
of a typical amplifier is a triangle as shown in Figure 4.2.
vin
vout
Electronic Amplifier
Figure 4.2. Symbol for electronic amplifier
An electronic (or electric) circuit which produces an output that is smaller than the input is
called an attenuator. A resistive voltage divider is a typical attenuator.

Chapter 4 Introduction to Operational Amplifiers
An amplifier can be classified as a voltage amplifier, current amplifier, or power amplifier.
The gain of an amplifier is the ratio of the output to the input. Thus for a voltage amplifier,
----------------------------------------= -
or
(4.1)
The current gain and power gain are defined similarly.
Note 1: Throughout this text, the common (base 10) logarithm of a number x will be denoted
as while its natural (base e) logarithm will be denoted as .
4.3 Decibels
The ratio of any two values of the same quantity (power, voltage or current) can be expressed in
. For instance, we say that an amplifier has power gain or a transmission
line has a power loss of (or gain  If the gain (or loss) is , the output is equal to
the input.
We must remember that a negative voltage or current gain or indicates that there is a
phase difference between the input and the output waveforms. For instance, if an amplifier
has a gain of 100 (dimensionless number), it means that the output is 180 degrees outofphase
with the input. Therefore, to avoid misinterpretation of gain or loss, we use absolute values of
power, voltage and current when these are expressed in dB.
By definition,
(4.2)
Therefore,
It is useful to remember that
Also,
Voltage Gain Output Voltage
Input Voltage
Gv
vout
vin
= ---------
Gi Gp
logx lnx
decibels dB 10 dB
7 dB –7 dB 0 dB
Gv Gi
180
dB 10
pout
pin
= log ---------
10 dB represents a power ratio of 10
10n dB represents a power ratio of 10 n

Decibels
1 dB represents a power ratio of approximately 1.25
3 dB represents a power ratio of approximately 2
7 dB represents a power ratio of approximately 5
From these, we can estimate other values. For instance, which is equiva-lent
4 dB = 3 dB + 1 dB
to a power ratio of approximately 2  1.25 = 2.5 .Likewise, 27 dB = 20 dB + 7 dB
and this
is equivalent to a power ratio of approximately .
Since and , if we let , the dB values for voltage and
current ratios become:
(4.3)
and
(4.4)
100  5 = 500
y = logx2 = 2logx p = v 2  R = i2R R = 1
= = log ---------
dBv 10
vout
vin
log 20
---------
2
vout
vin
dBi 10
iout
iin
log 20
= --------
= log --------
2
iout
iin
Example 4.1
Compute the gain in for the amplifier shown in Figure 4.3.
pin pout
1 w 10 w
Figure 4.3. Amplifier for Example 4.1
Solution:
dBw
= = log----- = 10log10 = 10  1 = 10 dBw
dBw 10
pout
pin
log--------- 10 10
1
Example 4.2
Compute the gain in for the amplifier shown in Figure 4.4, given that .
dBv log2 = 0.3
vin vout
1 v 2 v
Figure 4.4. Amplifier for Example 4.2

Gain Amplifier Gain Amplifier
Feedback Path
Solution:
4.4 Bandwidth and Frequency Response
Like electric filters, amplifiers exhibit a band of frequencies over which the output remains nearly
constant. Consider, for example, the magnitude of the output voltage of an electric or elec-tronic
circuit as a function of radian frequency as shown in Figure 4.5.
Figure 4.5. Typical bandwidth of an amplifier
As shown above, the bandwidth is where and are the lower and upper cutoff
frequencies respectively. At these frequencies, and these two points are
known as the 3dB down or halfpower points. They derive their name from the fact that power
, and for and or , the power is
, that is, the power is “halved”. Alternately, we can define the bandwidth as the frequency
band between halfpower points.
Most amplifiers are used with a feedback path which returns (feeds) some or all its output to the
input as shown in Figure 4.6.
Figure 4.6. Gain amplifiers used with feedback
dBv 20
vout
vin
--------- log 20 21
= = log-- = 20log0.3 = 20  0.3 = 6 dBv
vout

1
0.707

Bandwidth
vout
1 2
BW = 2 – 1 1 2
vout = 2  2 = 0.707
p = v 2  R = i2R R = 1 v = 2  2 = 0.707 i = 2  2 = 0.707
1  2
 In Out
+
Partial Output Feedback
 In Out
+
Entire Output Feedback
Feedback Circuit

The Operational Amplifier
In Figure 4.6, the symbol  (Greek capital letter sigma) inside the circle denotes the summing
point where the output signal, or portion of it, is combined with the input signal. This summing
point may be also indicated with a large plus (+) symbol inside the circle. The positive (+) sign
below the summing point implies positive feedback which means that the output, or portion of it,
is added to the input. On the other hand, the negative () sign implies negative feedback which
means that the output, or portion of it, is subtracted from the input. Practically, all amplifiers use
used with negative feedback since positive feedback causes circuit instability.
4.5 The Operational Amplifier
The operational amplifier or simply op amp is the most versatile electronic amplifier. It derives it
name from the fact that it is capable of performing many mathematical operations such as addi-tion,
multiplication, differentiation, integration, analogtodigital conversion or vice versa. It
can also be used as a comparator and electronic filter. It is also the basic block in analog com-puter
design. Its symbol is shown in Figure 4.7.
1
2
3

+
Figure 4.7. Symbol for operational amplifier
As shown above the op amp has two inputs but only one output. For this reason it is referred to
as differential input, single ended output amplifier. Figure 4.8 shows the internal construction of a
typical op amp. This figure also shows terminals VCC and VEE
. These are the voltage sources
required to power up the op amp. Typically, is +15 volts and is 15 volts. These termi-nals
VCC VEE
are not shown in op amp circuits since they just provide power, and do not reveal any other
useful information for the op amp’s circuit analysis.
4.6 An Overview of the Op Amp
The op amp has the following important characteristics:
1. Very high input impedance (resistance)
2. Very low output impedance (resistance)
3. Capable of producing a very large gain that can be set to any value by connection of external
resistors of appropriate values
4. Frequency response from DC to frequencies in the MHz range
5. Very good stability
6. Operation to be performed, i.e., addition, integration etc. is done externally with proper selec-tion
of passive devices such as resistors, capacitors, diodes, and so on.

1 NON-INVERTING INPUT
2 INVERTING INPUT
3 OUTPUT
Figure 4.8. Internal Devices of a Typical Op Amp
An op amp is said to be connected in the inverting mode when an input signal is connected to the
inverting () input through an external resistor whose value along with the feedback resistor
determine the op amp’s gain. The noninverting (+) input is grounded through an external
resistor R as shown in Figure 4.9.
For the circuit of Figure 4.9, the voltage gain is
(4.5)
VCC
VEE
1 2
3
Rin
Rf
Gv
Gv
vout
---------
vin
Rf
Rin
= = –--------

An Overview of the Op Amp

Rf
Rin
 +
+
R
+

vin
vout
Figure 4.9. Circuit of Inverting op amp
Note 2: The resistor R connected between the noninverting (+) input and ground serves only
as a current limiting device, and thus it does not influence the op amp’s gain. It will be
omitted in our subsequent discussion.
Note 3: The input voltage and the output voltage as indicated in the circuit of Figure
vin vout
4.9, should not be interpreted as open circuits; these designations imply that an input
voltage of any waveform may be applied at the input terminals and the corresponding
output voltage appears at the output terminals.
As shown in the formula of (4.5), the gain for this op amp configuration is the ratio
–Rf  Rin
where Rf
is the feedback resistor which allows portion of the output to be fed back to the input.
The minus () sign in the gain ratio –Rf  Rin
implies that the output signal has opposite polarity
from that of the input signal; hence the name inverting amplifier. Therefore, when the input sig-nal
is positive (+) the output will be negative () and vice versa. For example, if the input is +1
volt DC and the op amp gain is 100, the output will be 100 volts DC. For AC (sinusoidal) sig-nals,
the output will be 180 degrees outofphase with the input. Thus, if the input is 1 volt AC
and the op amp gain is 5, the output will be 5 volts AC or 5 volts AC with 180 degrees outof
phase with the input.
Example 4.3
Compute the voltage gain Gv and then the output voltage vout
for the inverting op amp circuit
shown in Figure 4.10, given that vin = 1 mV . Plot vin and vout as mV
versus time on the same
set of axes.
Solution:
This is an inverting amplifier and thus the voltage gain is
Gv
Gv
Rf
Rin
–-------- 120 K
= = –--------------------
20 K

120 K
20 K
+ 
Rf
vin vout
Rin
 +
+ 
Gv = –6
Gv
vout
vin
= ---------
vout = Gvvin = –6  1
vout = –6 mV
vin vout
vin = 1 mv
v (mv)
0 t
vout = –6 mv
or
and since
the output voltage is
or
The voltages and are plotted as shown in Figure 4.11.
Figure 4.11. Input and output waveforms for the circuit of Example 4.3
Example 4.4
for the inverting op amp circuit
shown in Figure 4.12, given that vin = sint mV . Plot vin and vout as mV
versus time on the
same set of axes.

120 K
20 K
+ 
Rf
vin vout
Rin
 +
+ 
Solution:
This is the same circuit as that of the previous example except that the input is a sine wave with
unity amplitude and the voltage gain is the same as before, that is,
and the output voltage is
Gv
Gv
Rf
Rin
–-------- 120 K
= = –-------------------- = –6
20 K
vout = Gvvin = –6  sint = –6sint mV
vin vout
v (mv)
vin = sint
vout = –6sint
0 2 4 6 8 10 12
6
4
2
0
-2
-4
-6
An op amp is said to be connected in the noninverting mode when an input signal is connected
to the noninverting () input through an external resistor R which serves as a current limiter,
and the inverting () input is grounded through an external resistor Rin
In our subsequent discussion, the resistor R will represent the internal resistance of the applied
voltage .
vin

Rin Rf

+
+ vout vin
+ R 

Figure 4.14. Circuit of noninverting op amp
Gv
Gv
vout
vin
= --------- = 1
+ --------
Rf
Rin
20 K Rf

(4.6)
As indicated by the relation of (4.6), the gain for this op amp configuration is and
therefore, in the noninverting mode the op amp output signal has the same polarity as the input
signal; hence, the name noninverting amplifier. Thus, when the input signal is positive (+) the
output will be also positive and if the input is negative, the output will be also negative. For
example, if the input is and the op amp gain is , the output will be . For
AC signals the output will be inphase with the input. For example, if the input is and
the op amp gain is , the output will be and inphase with
the input.
Example 4.5
Compute the voltage gain and then the output voltage for the noninverting op amp
circuit shown in Figure 4.15, given that . Plot and as versus time on
the same set of axes.
1 + Rf  Rin
+1 mV DC 75 +75 mV DC
0.5 V AC
Gv= 1 + 19 K  1 K = 20 10 V AC
Gv vout
vin = 1 mV vin vout mV
+
+ R


+ vout
vin
Rin
120 K

Solution:
The voltage gain is
and thus
Gv
Gv
vout
vin
= --------- = 1
= + -------------------- = 1 + 6 = 7
Rf
Rin
+ -------- 1 120 K
20 K
vout = Gvvin = 7  1 mV = 7 mV
vin vout
v (mv) vout = 7 mv
vin = 1 mv
0 t
Example 4.6
for the noninverting op amp
circuit shown in Figure 4.17, given that vin = sint mV . Plot vin and vout as mV
versus time on
the same set of axes.
120 K
20 K Rf

+
Rin
+ R
+ vout


vin
Solution:
This is the same circuit as in the previous example except that the input is a sinusoid. Therefore,
the voltage gain is the same as before, that is,
Gv

= = = + -------------------- = 1 + 6 = 7
8
6
4
2
0
-2
-4
-6
v (mv) vout = 7sint
+
and the output voltage is
Quite often an op amp is connected as shown in Figure 4.19.
Figure 4.19. Circuit of unity gain op amp
(4.7)
and thus
(4.8)
Gv
vout
vin
--------- 1
Rf
Rin
+ -------- 1 120 K
20 K
vout = Gvvin = 7  sint = 7sint mV
vin vout
0 2 4 6 8 10 12
-8
vin = sint
+ R 

vin
vout
Gv
Gv
vout
vin
= --------- = 1
vout = vin

Active Filters
For this reason, the op amp circuit of Figure 4.19 it is called unity gain amplifier. For example, if
the input voltage is the output will also be , and if the input voltage is
5 mV DC 5 mV DC
, the output will also be . The unity gain op amp is used to provide a very
2 mV AC 2 mV AC
high resistance between a voltage source and the load connected to it. An example will be given
in Section 4.8.
4.7 Active Filters
An active filter is an electronic circuit consisting of an amplifier and other devices such as resis-tors
and capacitors. In contrast, a passive filter is a circuit which consists of passive devices such
as resistors, capacitors and inductors. Operational amplifiers are used extensively as active filters.
A lowpass filter transmits (passes) all frequencies below a critical (cutoff ) frequency denoted as
, and attenuates (blocks) all frequencies above this cutoff frequency. An op amp lowpass fil-ter
is shown in Figure 4.20 and its frequency response in Figure 4.21.
R3
R1 R2
C2
vin C1
vout
Figure 4.20. A lowpass active filter
Low Pass Filter Frequency Respone
1
0.8
0.6
0.4
0.2
0
c
Radian Frequency (log scale)
|vOUT / vIN|
Ideal
Half-Pow er Point
Realizable

Figure 4.21. Frequency response for amplitude of a lowpass filter
C
In Figure 4.21, the straight vertical and horizontal lines represent the ideal (unrealizable) and
the smooth curve represents the practical (realizable) lowpass filter characteristics. The vertical

scale represents the magnitude of the ratio of outputtoinput voltage , that is, the gain
vout  vin
. The cutoff frequency is the frequency at which the maximum value of which is
Gv c vout  vin
unity, falls to 0.707  Gv , and as mentioned before, this is the half power or the –3 dB
point.
A highpass filter transmits (passes) all frequencies above a critical (cutoff) frequency  and
attenuates (blocks) all frequencies below the cutoff frequency. An op amp highpass filter is
shown in Figure 4.22 and its frequency response in Figure 4.23.
C2 R2
vin vout
Figure 4.22. A highpass active filter
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Figure 4.23. Frequency response for amplitude of a highpass filter
c
In Figure 4.23, the straight vertical and horizontal lines represent the ideal (unrealizable) and the
smooth curve represents the practical (realizable) highpass filter characteristics. The vertical
scale represents the magnitude of the ratio of outputtoinput voltage , that is, the gain
. The cutoff frequency is the frequency at which the maximum value of which is
unity, falls to , i.e., the half power or the point.
A bandpass filter transmits (passes) the band (range) of frequencies between the critical (cutoff)
frequencies denoted as and  where the maximum value of which is unity, falls to
C1 R1
C3
High-pass Filter Frequency Response
0.0
|vOUT / vIN|
c
Ideal
Realizable

Half-Pow er Point
vout  vin
Gv c vout  vin
0.707  Gv –3 dB
1 2 Gv

Active Filters
, while it attenuates (blocks) all frequencies outside this band. An op amp bandpass
0.707  Gv
filter is shown in Figure 4.24 and its frequency response in Figure 4.25.
C1
R1
R3 C2
vin R2
vout
Figure 4.24. An active bandpass filter
Band Pass Filter Frequency Response
c
Half-Pow er Points
Half-Pow er Point


1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1 2
Ideal
Realizable
|vOUT / vIN|
Figure 4.25. Frequency response for amplitude of a bandpass filter
A bandelimination or bandstop or bandrejection filter attenuates (rejects) the band (range) of
frequencies between the critical (cutoff) frequencies denoted as and  where the maxi-mum
1 2
value of which is unity, falls to , while it transmits (passes) all frequencies
Gv 0.707  Gv
outside this band. An op amp bandstop filter is shown in Figure 4.26 and its frequency response
in Figure 4.27.

C1 C2
Figure 4.26. An active bandelimination filter
Half-Pow er Points
Half-Pow er Point
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Figure 4.27. Frequency response for amplitude of a bandelimination filter
4.8 Analysis of Op Amp Circuits
The procedure for analyzing an op amp circuit (finding voltages, currents and power) is the same
as for the other circuits which we have studied thus far. That is, we can apply Ohm’s law, KCL
and KVL, superposition, Thevenin’s and Norton’s theorems. When analyzing an op amp circuit,
we must remember that in any op amp:
a. The currents into both input terminals are zero
b. The voltage difference between the input terminals of an op amp is zero
c. For circuits containing op amps, we will assume that the reference (ground) is the common
terminal of the two power supplies. For simplicity, the power supplies will not be shown.
vin
v R out 1
C3
R2
c

Band-Elimination Filter Frequency Response
0
|vOUT / vIN|
1 2

Ideal
Realizable

Analysis of Op Amp Circuits
We will provide several examples to illustrate the analysis of op amp circuits without being con-cerned
about its internal operation; this is discussed in electronic circuit analysis books.
Example 4.7
The op amp circuit shown in Figure 4.28 is called inverting op amp. Prove that the voltage gain
is as given in (4.9) below, and draw its equivalent circuit showing the output as a dependent
Gv
source.

Rf
Rin
 +
+
R
+

vin
vout
Figure 4.28. Circuit for deriving the gain of an inverting op amp
(4.9)
Gv
vout
vin
= ---------
= –--------
Rf
Rin
Proof:
No current flows through the () input terminal of the op amp; therefore the current which
flows through resistor flows also through resistor . Also, since the (+) input terminal is
grounded and there is no voltage drop between the () and (+) terminals, the () input is said to
be at virtual ground. From the circuit of Figure 4.28,
where
and thus
or
i
Rin Rf
vout = –Rf i
i
vin
Rin
= --------
vout
Rf
Rin
= –--------vin
Gv
vout
vin
= ---------
= –--------
Rf
Rin

The input and output parts of the circuit are shown in Figure 4.29 with the virtual ground being
the same as the circuit ground.
+ +

 
Figure 4.29. Input and output parts of the inverting op amp
These two circuits are normally drawn with the output as a dependent source as shown in Figure
4.30. This is the equivalent circuit of the inverting op amp and, as mentioned in Chapter 1, the
dependent source is a Voltage Controlled Voltage Source (VCVS).
+ 
Rf
 Rin

Figure 4.30. Equivalent circuit of the inverting op amp
Example 4.8
The op amp circuit shown in Figure 4.31 is called noninverting op amp. Prove that the voltage
gain is as given in (4.10) below, and draw its equivalent circuit showing the output as a
dependent source.
Rin Rf

+vout
Figure 4.31. Circuit of noninverting op amp
(4.10)
+
+
 i
i
vin Rf vout Rin
+
+
vin
Rin vout
-------vin
Gv
+
+ 

vin
Gv
vout
vin
--------- 1
Rf
Rin
= = + --------

Proof:
Let the voltages at the () and (+) terminals be denoted as v1 and v2
respectively as shown in
Figure 4.32.
+ 

+
+vout
Rf
v1 v2
i2
+ 

Rin
vin
i1
Figure 4.32. Noninverting op amp circuit for derivation of (4.10)
By application of KCL at
or
(4.11)
v1
i1 + i2 = 0
--------
v1
Rin
v1 – vout
Rf
+ --------------------- = 0
There is no potential difference between the () and (+) terminals; therefore, or
. Relation (4.11) then can be written as
v1 = v2 = vin
or
Rearranging, we obtain
v1 – v2 = 0
--------
vin
Rin
vin – vout
Rf
+ ----------------------- = 0
-------- 1
1
Rin
 vin
 + ----- 
Rf
vout
Rf
= ---------
Gv
vout
vin
= --------- = 1
+ --------
Rf
Rin
and its equivalent circuit is as shown in Figure 4.33. The dependent source of this equivalent cir-cuit
is also a VCVS.

Rf
Rin
+ +

+
   1
+ ------- 
vin vout vin

Figure 4.33. Equivalent circuit of the noninverting op amp
Example 4.9
If, in the noninverting op amp circuit of the previous example, we replace with an open cir-cuit
Rin
( ) and with a short circuit ( ), prove that the voltage gain is
(4.12)
Rin   Rf Rf  0 Gv
Gv
vout
vin
= --------- = 1
vout = vin
Rin Rf

+
+ vout vin
+ 

Rin Rf
and thus
(4.13)
Proof:
With open and shorted, the noninverting amplifier of the previous example reduces to
the circuit of Figure 4.34.
Figure 4.34. Circuit of Figure 4.32 with open and shorted
The voltage difference between the (+) and () terminals is zero; then vout = vin
.
We will obtain the same result if we consider the noninverting op amp gain Gv = 1 + Rf  Rin
.
Then, letting Rf  0 , the gain reduces to Gv = 1
and for this reason this circuit is called unity
gain amplifier or voltage follower. It is also called buffer amplifier because it can be used to “buffer”
(isolate) one circuit from another when one “loads” the other as we will see on the next example.

Example 4.10
For the circuit of Figure4.35
a.With the load RLOAD disconnected, compute the open circuit voltage
vab
b.With the load connected, compute the voltage vLOAD across the load
RLOAD
c.Insert a buffer amplifier between a and b and compute the new voltage across the same
load
+

12 V
 a
7 K
5 K RLOAD
5 K

b
vin
vLOAD
RLOAD
Solution:
a. With the load disconnected the circuit is as shown in Figure 4.36.
RLOAD
+

12 V
 a
7 K
5 K 5K

RLOAD
b
vin
Figure 4.36. Circuit for Example 4.10 with the load disconnected
The voltage across terminals a and b is
vab
5 K
= -----------------------------------  12 = 5 V
7 K + 5 K
b. With the load reconnected the circuit is as shown in Figure 4.37. Then,
RLOAD
vLOAD
5 K || 5 K
= -------------------------------------------------------  12 = 3.16 V
7 K + 5 K || 5 K
Here, we observe that the load “loads down” the load voltage from to
and this voltage may not be sufficient for proper operation of the load.
RLOAD 5 V 3.16 V

 a
7 K
5 K 5 K
Figure 4.37. Circuit for Example 4.10 with the load reconnected
c. With the insertion of the buffer amplifier between points a and b and the load, the circuit now
a

7 K
5 K 5 K
RLOAD vLOAD = vab = 5 V
Figure 4.38. Circuit for Example 4.10 with the insertion of a buffer op amp
From the circuit of Figure 4.38, we observe that the voltage across the load is as desired.
Example 4.11
The op amp circuit shown in Figure 4.39 is called summing circuit or summer because the output is
the summation of the weighted inputs.

Rf
Figure 4.39. Twoinput summing op amp circuit
+

12 V

b
vin
RLOAD
+
12 V

b
+
5 V
+
v  in
5 V
+

+

+
vout
+ Rin1 Rin2
vin2 vin1

Prove that for this circuit,
(4.14)
vout Rf
vin1
Rin1
 
----------
vin2
Rin2
+ ----------
 
 
= –
Proof:
We recall that the voltage across the () and (+) terminals is zero. We also observe that the (+)
input is grounded, and thus the voltage at the () terminal is at “virtual ground”. Then, by appli-cation
of KCL at the () terminal, we obtain
vin1
Rin1
----------
vin2
Rin2
+ ----------
+ --------- = 0
vout
Rf
and solving for vout
we obtain (4.14). Alternately, we can apply the principle of superposition to
derive this relation.
Example 4.12
Compute the output voltage for the amplifier circuit shown in Figure 4.40.
vout
Rf 1 M

+
Rin3
20 K 30 K
vin3
vin1
Rin2
vin2
Rin1
10 K
1 mV 
+ +
 
4 mV 10 mV
+
vout
+
Solution:
Let be the output due to acting alone, be the output due to acting alone,
and be the output due to acting alone. Then by superposition,
vout1 vin1 vout2 vin2
vout3 vin3
vout = vout1 + vout2 + vout3
First, with vin1 acting alone and vin2 and vin3
shorted, the circuit becomes as shown in Figure
4.41.

Rf 1 M

Rin3
Figure 4.41. Circuit for Example 4.12 with acting alone
We recognize this as an inverting amplifier whose voltage gain is

Rin1
10 K
Rin2 20 K
and thus
(4.15)
Next, with acting alone and and shorted, the circuit becomes as shown in Figure
4.42.
The circuit of Figure 4.42 as a noninverting op amp whose voltage gain is
and the voltage at the plus (+) input is computed from the voltage divider circuit shown in Fig-ure
4.43.
+

+
+
1 mV
Rin1
10 K
vin1
Rin2
20 K
30 K
vout1

vin1
Gv
Gv = 1 M  10 K = 100
vout1 = 100–1 mV = –100 mV
vin2 vin1 vin3
+
+
+

4 mV
Rf 1 M
vout2
Rin3
30 K
vin2
vin2
Gv
Gv = 1 + 1 M  10 K = 101

Rin2 20 K
+

vin2
4 mV
+
30 K

Rin3
To v+
v+ vin2
Figure 4.43. Voltage divider circuit for the computation of with acting alone
Then,
and thus
(4.16)
v+
Rin3
Rin2 + Rin3
---------------------------  vin2 30 K
= = -----------------  4 mV = 2.4 mV
50 K
vout2 = 101  2.4 mV = 242.4 mV
Finally, with acting alone and and shorted, the circuit becomes as shown in Fig-ure
4.44.
vin3 vin1 vin2
Rin3
+
Rf 1 M

+
+

Rin2 20 K
vin3
10 mV
Rin1
10 K
30 K
vout3
vin3
The circuit of Figure 4.44 is also a noninverting op amp whose voltage gain is
Gv
Gv = 1 + 1 M  10 K = 101
and the voltage at the plus (+) input is computed from the voltage divider circuit shown in Fig-ure
4.45.

Rin3
+
Rin2 20 K
vin3
10 mV
+
30 K

To v+
Figure 4.45. Voltage divider circuit for the computation of with acting alone
= = -----------------  10 mV = 4 mV
R Rf 1

Then,
and thus
(4.17)
Therefore, from (4.15), (4.16) and (4.17),
Example 4.13
For the circuit shown in Figure 4.46, derive an expression for the voltage gain in terms of the
external resistors , , , and .
Solution:
We apply KCL at nodes and as shown in Figure 4.47.
v+ vin3
v+
Rin2
Rin2 + Rin3
---------------------------  vin2 20 K
50 K
vout3 = 101  4 mV = 404 mV
vout = vout1 + vout2 + vout3 = – 100 + 242.4 + 404 = 546.4 mV
Gv
R1 R2 R3 Rf
+

+
+

R2
R3
vout
vin
v1 v2

+

R Rf 1

+
+

R2
v1
v2
R3
vout
vin
Figure 4.47. Application of KCL for the circuit of Example 4.13
At node :
or
or
or
(4.18)
v1
At node :
or
(4.19)
v1 – vin
R1
------------------
1
R1
------ 1
 v1
 + ----- 
R1 + Rf
R1Rf
 
 ------------------

 
v1
= -------------------------------------
v2
v2
and since , we rewrite (4.19) as
(4.20)
v1 – vout
Rf
+ --------------------- = 0
Rf
vin
R1
= ------
+ ---------
v1
Rf vin + R1vout
= -------------------------------------
R1Rf
Rf vin + R1vout
R1 + Rf
v2 – vin
R2
------------------
v2
R3
+ ------ = 0
R3vin
R2 + R3
= ------------------
v2 = v1
v1
R3vin
R2 + R3
= ------------------
Equating the right sides of (4.18) and (4.20) we obtain
vout
Rf

Rf vin + R1vout
-------------------------------------
R1 + Rf
R3vin
R2 + R3
= ------------------
Rf vin + R1vout
R3vin
R2 + R3
= ------------------R1 + Rf
R1vin
vout
vin
---------
R3R1 + Rf
R1R2 + R3
= ------------------------------
Rf
R1
– ------
Gv
vout
vin
= = -------------------------------
---------
R1R3 – R2Rf
R1R2 + R3
Rin vS
iS
Rin
vS
iS
= -----
iS
vS iS
Rin
or
Dividing both sides of the above relation by and rearranging, we obtain
and after simplification
(4.21)
4.9 Input and Output Resistance
The input and output resistances are very important parameters in amplifier circuits.
The input resistance of a circuit is defined as the ratio of the applied voltage to the current
drawn by the circuit, that is,
(4.22)
Therefore, in an op amp circuit the input resistance provides a measure of the current which
the amplifier draws from the voltage source . Of course, we want to be as small as possible;
accordingly, we must make the input resistance as high as possible.
Example 4.14
Compute the input resistance Rin
of the inverting op amp amplifier shown in Figure 4.48 in
terms of and .
R1 Rf

Input and Output Resistance
R Rf 1
+ 
+
vS
vout
iS
Solution:
By definition,
(4.23)
+


Rin
vS
iS
= -----
and since no current flows into the minus () terminal of the op amp and this terminal is at vir-tual
ground, it follows that
(4.24)
From (4.23) and (4.24) we observe that
(4.25)
iS
vS
R1
= ------
Rin = R1
It is therefore, desirable to make R1 as high as possible. However, if we make R1
very high such
as , for a large gain, say , the value of the feedback resistor should be . Obvi-ously,
10 M 100 Rf 1 G
this is an impractical value. Fortunately, a large gain can be achieved with the circuit of
Problem 8 at the end of this chapter.
Example 4.15
Compute the input resistance of the op amp shown in Figure 4.49.
Rin
Rf

+
+
vout

+
vin
100 K


Solution:
In the circuit of Figure 4.49, vin
is the voltage at the minus () terminal; not the source voltage
. Therefore, there is no current drawn by the op amp. In this case, we apply a test (hypo-thetical)
vS iS
current as shown in Figure 4.49, and we treat as the source voltage.
iX vin

+
+
Figure 4.50. Circuit for Example 4.15 with a test current source
= = ---- = 0
We observe that is zero (virtual ground). Therefore,
By definition, the output resistance is the ratio of the open circuit voltage to the short circuit cur-rent,
that is,
(4.26)
The output resistance is not the same as the load resistance. The output resistance provides
a measure of the change in output voltage when a load which is connected at the output termi-nals
draws current from the circuit. It is desirable to have an op amp with very low output resis-tance
as illustrated by the following example.
Example 4.16
The output voltage of an op amp decreases by when a load is connected at the out-put
terminals. Compute the output resistance .
Solution:
Consider the output portion of the op amp shown in Figure 4.51.

Rf
vout
vin 100 K
iX
vin
Rin
vin
iX
------- 0
iX
Rout
Rout
vOC
iSC
= ---------
Rout
10% 5 K
Rout

Input and Output Resistance

+
+
vout

Rout
Figure 4.51. Partial circuit for Example 4.16
With no load connected at the output terminals,
(4.27)
vout = vOC = Gvvin
With a load connected at the output terminals, the load voltage is
(4.28)
RLOAD vLOAD
and from (4.27) and (4.28)
(4.29)
Therefore,
vLOAD
vLOAD
and solving for we obtain
RLOAD
Rout + RLOAD
= ----------------------------------  vout
RLOAD
Rout + RLOAD
= ----------------------------------  Gvvin
vLOAD
vOC
---------------- 0.9 5 K
= = -------------------------------
Rout + 5 K
Rout
Rout = 555 
We observe from (4.29) that as , relation (4.29) reduces to and by
comparison with (4.27), we see that
Rout  0 vLOAD = Gvvin
vLOAD = vOC

4.10 Summary
 A signal is any waveform representing a fluctuating electric quantity, such as voltage, current,
electric or magnetic field strength, sound, image, or any message transmitted or received in
telegraphy, telephony, radio, television, or radar. al that changes with time.
 An amplifier is an electronic circuit which increases the magnitude of the input signal.
 The gain of an amplifier is the ratio of the output to the input. It is normally expressed in deci-bel
dB = 10log pout  pin
(dB) units where by definition
 Frequency response is the band of frequencies over which the output remains fairly constant.
 The lower and upper cutoff frequencies are those where the output is of its maximum
value. They are also known as halfpower points.
 Most amplifiers are used with feedback where the output, or portion of it, is fed back to the
input.
 The operational amplifier (op amp) is the most versatile amplifier and its main features are:
1. Very high input impedance (resistance)
2. Very low output impedance (resistance)
3. Capable of producing a very large gain that can be set to any value by connection of exter-nal
resistors of appropriate values
4. Frequency response from DC to frequencies in the MHz range
5. Very good stability
6. Operation to be performed, i.e., addition, integration etc. is done externally with proper
selection of passive devices such as resistors, capacitors, diodes, and so on.
 The gain of an inverting op amp is the ratio where is the feedback resistor which
allows portion of the output to be fed back to the minus () input. The minus () sign implies
that the output signal has opposite polarity from that of the input signal.
 The gain of an noninverting op amp is where is the feedback resistor which
allows portion of the output to be fed back to the minus () input which is grounded through
the resistor. The output signal has the same polarity from that of the input signal.
 In a unity gain op amp the output is the same as the input. A unity gain op amp is used to pro-vide
a very high resistance between a voltage source and the load connected to it.
 Op amps are also used as active filters.
0.707
–Rf  Rin Rf
1 + Rf  Rin Rf
Rin

Summary
 A lowpass filter transmits (passes) all frequencies below a critical (cutoff) frequency denoted
as C
and attenuates (blocks) all frequencies above this cutoff frequency.
 A highpass filter transmits (passes) all frequencies above a critical (cutoff) frequency c
 and
attenuates (blocks) all frequencies below the cutoff frequency.
 A bandpass filter transmits (passes) the band (range) of frequencies between the critical (cut-off)
frequencies denoted as and  where the maximum value of which is unity, falls
1 2 Gv
to 0.707  Gv
, while it attenuates (blocks) all frequencies outside this band.
 A bandelimination or bandstop or bandrejection filter attenuates (rejects) the band
(range) of frequencies between the critical (cutoff) frequencies denoted as and  where
the maximum value of which is unity, falls to , while it transmits (passes) all fre-quencies
Gv 0.707  Gv
outside this band.
1 2
 A summing op amp is a circuit with two or more inputs.
 The input resistance is the ratio of the applied voltage to the current drawn by the cir-cuit,
that is,
vS iS
Rin = vS  iS
 The output resistance (not to be confused with the load resistance) is the ratio of the open cir-cuit
voltage when the load is removed from the circuit, to the short circuit current which is
the current that flows through a short circuit connected at the output terminals, that is,
Ro = vOC  iSC

4.11 Exercises
Multiple Choice
1. In the op amp circuit below vin = 2 V , Rin = 1 K , and it is desired to have vout = 8 V
.
This will be obtained if the feedback resistor has a value of
A.
B.
C.
D.
E.
Rin Rf

+ vout vin
2. In the circuit below , , and . Then will be
A.
B.
C.
D.
E.
+ 
 + Rin
Rf
1 K
2 K
3 K
4 K
none of the above
+
+ R 

vin = 6 V Rin = 2 K Rf = 3 K vout
–9 V
9 V
–4 V
4 V
none of the above
+

Rf
vin vout

Exercises
3. In the circuit below iS = 2 mA and Rf = 5 K . Then vout
will be
A.
B.
C.
D.
E.
 V
0 V
10 V
–10 V
none of the above

+
+

Rf
iS
vout
4. In the circuit below iS = 4 mA and R = 3 K . Then vout
will be
A.
 V
B.
0 V
C. indeterminate
D.
E.
–12 V
none of the above

+
+

R
iS
vout

5. In the circuit below , , , and . Then
will be
+ 
i


+ 

A.
B.
C.
D.
E.
6. In the circuit below and all resistors have the same value. Then will be
A.
B.
C.
D.
E.
vin = 4 V Rin = 12 K Rf = 18 K RLOAD = 6 K i
–1 mA
1 mA
–4  3 mA
4  3 mA
none of the above
+


+
Rin Rf
vin
RLOAD vout
vin = 1 V vout
–2 V
2 V
–4 V
4 V
none of the above
+
+
vout
Rin1
Rf11
vin
+
Rin22
Rf22

Exercises
7. In the circuit below vin1
= 2 V , vin2
= 4 V , and Rin = Rf = 1 K . Then vout
will be
A.
B.
C.
D.
E.
–2 V
2 V
–8 V
8 V
none of the above
Rin
+ 
Rf
vin11 + vout
 +
+
vin22
8. In the circuit below . Then will be
A.
B.
C.
D.
E.

vin = 30 mV vout
–5 mV
–10 mV
–15 mV
–90 mV
none of the above
10 K
20 K
10 K
10 K
+ 
 
vin +
+
vout
10 K

+ 

vX 10 
9. For the circuit below the input resistance is
A.
B.
C.
D.
E.
10. For the circuit below the current is
A.
B.
C.
D.
E.
Rin
1 K
2 K
4 K
8 K
none of the above
+
vin
4 K
+

vout
4 K
2 K
1 K
2 K
Rin
i
–40 A
40 A
–400 A
400 A
none of the above
+
2 A
i
40vX
+

5 

Exercises
Problems
1. For the circuit below compute . Answer:
vout2 –0.9 V
27 K
vin1 +

+
+
10 mV
90 K

10 K
vin2
+ +
vout1


3 K
vout2

i5K 4A
+
60 mV

+
4 K
3 K
i5K
6 K 5 K
3. For the circuit below , , and represent the internal resistances of the input volt-ages
Rin1 Rin2 Rin3
, , and respectively. Derive an expression for in terms of the input
vin1 vin2 vin3 vout
voltage sources and their internal resistances, and the feedback resistance .
Answer:
Rf
+

+
+

Rin1
Rin2 Rin3
+

vout

+
vin1
vin2 vin3
Rf
=  
vout Rf
vin3
Rin3
 ----------
– – ---------- 
vin2
Rin2
----------
vin1
Rin1

vout –40 mV
+
40 mV
50 K

+
+

vout
10 K
20 K
40 K
RL 3.75 K
+ 
+
+ 
 
+ +


+

5. The op amp circuit (a) below can be represented by its equivalent circuit (b). For the circuit
(c), compute the value of so that it will receive maximum power. Answer:
6. For the circuit below compute using Thevenin’s theorem. Answer:
+
(a) (b)
R1
R2
R1
R2
R1
-----vin
vin v vin out
vout
2 K
20 K
vin

+
5 K
15 K
(c)
vout +
RLOAD

v5K 20 mV

Exercises
100 K

+
+

+

72 mV
84 K
v5K 20 K
12 K
4 K
5 K
7. For the circuit below compute the gain . Answer:
Gv = vout  vin –2  37
+
vout
vin
R2
R1
R3
40 K
R4
200 K
R5
50 K
50 K
40 K
 
8. For the circuit below, show that the gain is given by
+
vout

+
Gv
vout
vin
--------- 1
= = +  
–------ R4 R2
R1
R4
R3
 ------ + 1
+
+
vin
R1
R2
R3
R4
 vout

9. Create a Simulink / SimPowerSystems model for the equivalent circuit of the inverting op
amp shown below.

+ 
Rf
 Rin

+
+
vin
Rin vout
-------vin
vin = 1 v peak f = 0.2 Hz Phase = 0 deg
Rin = 1 K Rf = 10 K

Multiple Choice
1. C For vin = 2 and vout = 8 , the gain must be Gv = 4 or 1 + Rf  Rin = 4 and
Rf = 3 K
2. A
vout = –Rf  Rin  vin = –9 V
3. D All current flows through Rf and the voltage drop across it is
–2 mA  5 K = –10 V
4. E All current flows through R and the voltage drop across it is 4 mA  3 K = 12 V
. Since
this circuit is a unity gain amplifier, it follows that also.
vout = 12 V
5. C . Therefore, .
vout = –18  12  4 = –6 V iLOAD = vout  RLOAD = –6 V  6 K = –1 mA
Applying KCL at the plus (+) terminal of we obtain
vout
– -- 43
----------------------------------------- + 1 – 13
-------------- – 6 V – 4 V
i –6 V
= = = –-- mA
6 K
18 K + 12 K
6. D The gain of each of the noninverting op amps is 2. Thus, the output of the first op amp is
and the output of the second is .
2 V 4 V
7. E By superposition, due to acting alone is and due to acting alone
vout1
vin1
is . Therefore,
–2 V vout2
vin2
8 V vout = – 2 + 8 = 6 V
8. B We assign node voltage as shown below and we replace the encircled part by its equiv-alent.
vA
10 K
20 K
10 K
10 K
+ vA

 
vin +
+
vout
10 K
We now attach the remaining resistors and the entire equivalent circuit is shown below.
10 K 10 K
A
+ +
+ 

vA 5 K
vout
+
2vA
 
vin 30 mV

vA – 30
10
------------------
vA
5
------
vA – –2vA
+ + ----------------------------- = 0
10
vA = 30  6 = 5 mV
vout = –2vA = –10 mV
4 K  4 K = 1 vout = –vin v
4 K
4 K
+ i v


Rin 2 K
+
vin
+

vout
2 K
1 K
v – vin
4
---------------
v – –vin
+ ----------------------- = 0
4
v = 0
i
vin
4 K
= --------------
Rin
vin
= ------------------------- = 4 K
vin  4 K
Application of KCL at Node A yields
and thus
Therefore,
NOTE: For this circuit, the magnitude of the voltage is less than the magnitude of the
input voltage. Therefore, this circuit is an attenuator, not an amplifier. Op amps
are not configured for attenuation. This circuit is presented just for instructional
purposes. A better and simpler attenuator is a voltage divider circuit.
9. C The voltage gain for this circuit is and thus . The voltage
at the minus () input of the op amp is zero as proved below.
or
Then,
and
10. A For this circuit, and thus . Then,
Problems
1.
vX = –10 V 40vX = –400 V i = –400  10 = –40 A
vout1 = –27  3  10 = –90 mV

and thus
Then,
vin2 = vout1 = –90 mV
vout2 1 90
=    –90 = –0.9 V
 + ----- 
10
2. We assign , , and as shown below.
RLOAD v1 vLOAD
+
60 mV
4 K
3 K
3 K  6 K = 2 K
and by the voltage division expression
+

6 K 5 K
v1
2 K
= -----------------------------------  60 mV = 20 mV
4 K + 2 K
and since this is a unity gain amplifier, we obtain
Then,
i5K
3. By superposition
where

+
i5K
v1
+

vLOAD
vLOAD = v1 = 20 mV
vLOAD
RLOAD
----------------- 20 10 3 – 
----------------- 20 mV
– = = = =  A = 4 A
5 K
---------------------- 4 106

5 103
vout = vout1 + vout2 + vout3
vout1 vin2 0 =
vin3 = 0
Rf
Rin1
= –----------vin1
We observe that the minus () is a virtual ground and thus there is no current flow in
and . Also,
and
Rin1
Rin2
vout2 vin1 0 =
vin3 = 0
Rf
Rin2
= –----------vin2

vout3 vin1 = 0
=  
+
50 K

v
v+
--------------------v
--------------------vout – 4 106
--------------------v+ 2 106
= ----------------------
-------------------- 80 10 3 – 
  1
 ---------------------- 
--------------------vout – 4 106
Then,
4. We assign voltages and as shown below.
At the minus () terminal
or
At the plus (+) terminal
or
or
Since we equate the nodal equations and we obtain
Multiplication by yields
vin2 = 0
Rf
Rin3
= –----------–vin3
vout Rf
vin3
Rin3
----------
vin2
Rin2
----------
vin1
Rin1
 – – ---------- 
v v+
40 mV
+
+

vout
10 K
20 K
40 K
v – 40 mV
10 K
-----------------------------
v – vout
50 K
+ --------------------- = 0
6
50  103
1
50  103
– = 
v+ – 40 mV
20 K
-----------------------------
v+
40 K
+ ----------------- = 0
3
40  10 3
– = 
v+
80 10–3 
3
v+ = v
6
50  103
3
50  103
– = 
50  103

or
2 80 103
  –  50  103
----------------------------------------------------------- vout – 4 106
50  103
– =   50  103
vout = –40 mV
Check using MATLAB:
R1=10000; R2=20000; R3=40000; Rf=50000; Vin=40*10^(-3);
Vout=(R1*R3R2*Rf)*Vin/(R1*(R2+R3))
Vout =
-0.0400
5. We attach the , , and resistors to the equivalent circuit as shown below.
5 K 15 K RLOAD
+
v  in
By Thevenin’s theorem
or
2 K
5 K
15 K
10vin

+
a

 b
= = = --------------------------------------–10vin
vTH vOC vab
15 K
5 K + 15 K
vTH = –7.5vin
Because the circuit contains a dependent source, we must compute the Thevenin resistance
using the relation where is found from the circuit below.
RTH = vTH  iSC iSC
5 K
15 K

10vin
+
a

b

iSC
We observe that the short circuit shorts out the and thus
Then,
15 K
iSC
–10vin
5 K
---------------- 2 10–3= = –  vin

RTH
–7.5vin
2 103
= ------------------------------ = 3.75 K
– –  vin

+
vTH
RTH
3.75 K
RLOAD
RLOAD = RTH = 3.75 K
Rf
Rin
+ +
+
  1
+ ------- 
vin vout vin
 
vin = v5K
 
1
+ --------
 
Rf
Rin
 
v5K 1 100
=   v5K = 6v5K
 + -------- 
20
84 K
+
v  5K
12 K 4 K
a

+
 
+
72 mV
5 K
b

6v5K
and the Thevenin equivalent circuit is shown below.
Therefore, for maximum power transfer we must have
6. This is a noninverting op amp whose equivalent circuit is shown below.
For this circuit and the value of the VCVS is
Attaching the external resistors to the equivalent circuit above we obtain the circuit below.
To find the Thevenin equivalent at points a and b we disconnect the 5 K
resistor. When
this is done there is no current in the and the circuit simplifies to the one shown below.
4 K

By KVL
or
Also,
or
= = = = –  
vTH vab v5K 72 mV – 12 Ki 72 mV 12 K
and thus
+
+
6v5K
72 mV
12 K 84 K
i
a
vab
b
+

12 K + 84 Ki + 6v5K = 72 mV
i
72 mV – 6v5K
12 K + 84 K
= ----------------------------------------------
72 mV – 6v5K
---------------------------------------
96 K
72 mV – 9 mV 34
= + --v5K
v5K
34
– --v5K = 63 mV
vTH = vab = v5K = 252 mV
The Thevenin resistance is found from where is computed with the ter-minals
RTH = vOC  iSC iSC
a and b shorted making and the circuit is as shown on the left below. We
v5K = 0
also perform voltagesource to currentsource transformation and we obtain the circuit on
the right below.
72 mV
Now,
+
12 K
84 K
ab
4 K
iSC
12 K  84 K = 10.5 K
and by the current division expression
Therefore,
6 A
12 K
4 K
84 K a
iSC b
= = = -------- A
iSC iab
10.5 K
------------------------------------------  6 A 126
10.5 K + 4 K
29
RTH
vOC
iSC
--------- 252
= = ------------------ = 58 K
126  29

and the Thevenin equivalent circuit with the resistor is shown below.
a
RTH
----------------- 1
----------------- 1
-------------------- 1
 v1
 + + + ----------------- 
vout
vin
R2
R1
R3
40 K
R4
200 K
v1 v2
R5
50 K
50 K
40 K
Finally,
7. We assign node voltages and as shown below and we write node equations observing
that (virtual ground).
Node 1:
or
Multiplication of each term by and simplification yields
Node 2:
or
Equating the right sides we obtain
5 K
+
 5 K
b
12 K
252 mV
vTH
v5K
5
58 + 5
= ---------------  252 = 20 mV
v1 v2
v2 = 0
v1 – --------------------
vin
200 K
v1 – vout
40 K
--------------------
-----------------
v1 – 0
50 K
v1
50 K
+ + + ----------------- = 0
1
200 K
40 K
50 K
50 K
--------------------
vin
200 K
vout
40 K
= + -----------------
+
+
 
200 K
v1
1
14
= -----vin + 5vout
0 – -----------------
v1
50 K
0 – vout
40 K
+ ------------------ = 0
v1
54
= –--vout

or
1
14
----- vin 5vout +   54
= –--vout
37
28
-----vout
1
14
= –-----vin
Simplifying and dividing both sides by we obtain
vin
Gv
vout
vin
--------- 2
= = –-----
37
8. We assign node voltages and as shown below and we write node equations observing
v1 v2
that (virtual ground).
v1 = 0
Node 1:
or
Node 2:
or
or
+
vin
R1
R2
R3
R4
v1
v2
 vout
v2 – 0
--------------
R2
1
R2
------ 1
 v2
 + + ------ 
v2
= -------------------------------------------------vout
Equating the right sides we obtain
0 – ---------------
vin
R1
0 – v2
R2
+ -------------- = 0
v2
R2
R1
= –------vin
v2
R3
+ ------
+ -------------------- = 0
v2 – vout
R4
------ 1
R3
R4
= ---------
1
R4  R2 + R4  R3 + 1
1
-------------------------------------------------vout
R4  R2 + R4  R3 + 1
= –------vin
Simplifying and dividing both sides by we obtain
+

vout
R4
R2
R1
vin

= = +  
–------ R4 R2
+ 
Rf
 Rin

CVS=Controlled Voltage Source
+
9.
Gv
vout
vin
--------- 1
R1
R4
R3
 ------ + 1
+
+
vin
Rin vout
-------vin
vin = 1 v peak f = 0.2 Hz Phase = 0 deg Rin = 1 K Rf = 10 K
Vin
Vout
Vin = 1 volt peak, frequency 0.2 Hz
Rin=1 K, Rf= 10 K, entered at the MATLAB command prompt
Continuous
powergui
Vin
v +-
VM 2
v +-
VM 1
Scope
Rin Product
-Rf/Rin
Constant
s
-
CVS Bus
Creator

Chapter 5
Inductance and Capacitance
his chapter is an introduction to inductance and capacitance, their voltagecurrent rela-tionships,
power absorbed, and energy stored in inductors and capacitors. Procedures for
analyzing circuits with inductors and capacitors are presented along with several examples.
T
5.1 Energy Storage Devices
In the first four chapters we considered resistive circuits only, that is, circuits with resistors and
constant voltage and current sources. However, resistance is not the only property that an elec-tric
circuit possesses; in every circuit there are two other properties present and these are the
inductance and the capacitance. We will see through some examples that will be presented later
in this chapter, that inductance and capacitance have an effect on an electric circuit as long as
there are changes in the voltages and currents in the circuit.
The effects of the inductance and capacitance properties can best be stated in simple differential
equations since they involve the changes in voltage or current with time. We will study induc-tance
first.
5.2 Inductance
Inductance is associated with the magnetic field which is always present when there is an electric
current. Thus, when current flows in an electric circuit the conductors (wires) connecting the
devices in the circuit are surrounded by a magnetic field. Figure 5.1 shows a simple loop of wire
and its magnetic field represented by the small loops.
Figure 5.1. Magnetic field around a loop of wire
The direction of the magnetic field (not shown) can be determined by the lefthand rule if con-ventional
current flow is assumed, or by the righthand rule if electron current flow is assumed.
The magnetic field loops are circular in form and are referred to as lines of magnetic flux. The unit
of magnetic flux is the weber (Wb).

Chapter 5 Inductance and Capacitance
In a loosely wound coil of wire such as the one shown in Figure 5.2, the current through the
wound coil produces a denser magnetic field and many of the magnetic lines link the coil several
times.
Figure 5.2. Magnetic field around several loops of wire
The magnetic flux is denoted as and, if there are N turns and we assume that the flux passes
through each turn, the total flux, denoted as  is called flux linkage. Then,
(5.1)
 

 = N
Now, we define a linear inductor one in which the flux linkage is proportional to the current
through it, that is,
(5.2)
 = Li
where the constant of proportionality is called inductance in webers per ampere.
We also recall Faraday’s law of electromagnetic induction which states that
(5.3)
L
v d
= -----
dt
v Ldi
= ----
dt
L
RLOAD
vL
iL
 
L
and from (5.2) and (5.3),
(5.4)
Alternately, the inductance is defined as the constant which relates the voltage across and the
current through a device called inductor by the relation of (5.4).
The symbol and the voltagecurrent* designations for the inductor are shown in Figure 5.3.
Figure 5.3. Symbol for inductor
* In the first four chapters we have used the subscript LOAD to denote a voltage across a load, a current through a load,
and the resistance of a such load as to avoid confusion with the subscript L which henceforth will denote induc-tance.
We will continue using the subscript LOAD for any load connected to a circuit.

Inductance
For an inductor, the voltagecurrent relationship is
(5.5)
= -------
vL L
diL
dt
vL iL vL
where and have the indicated polarity and direction. Obviously, has a nonzero value
only when changes with time.
The unit of inductance is the Henry abbreviated as . Since
(5.6)
iL
H
L
vL
------- volts
= = ----------------------
diL
dt
amperes
---------------------
seconds
we can say that one henry is the inductance in a circuit in which a voltage of one volt is induced
by a current changing at the rate of one ampere per second.
By separation of the variables we rewrite (5.5) as
(5.7)
and integrating both sides we obtain:
or
or
(5.8)
diL
1L
= --vLdt
it
 1L
diL
it0
t
= 
-- vLdt
t0
iLt – iLt0 1L
t
= 
--- vLdt
t0
t
 iL t0 = +  
iLt 1L
-- vLdt
t0
where , more often denoted as , is the current flowing through the inductor at some
reference time usually taken as , and it is referred to as the initial condition.
We can also express (5.8) as
(5.9)
iLt0 iL0
t = 0
iL t   1L
t
 1L
= --- vLdt
= + 
–
0
 1L
--- vLdt
–
t
--- vLdt
0
where the first integral on the right side represents the initial condition.

Example 5.1
The current passing through a inductor is shown in Figure 5.4.
a. Compute the flux linkage at
b. Compute and sketch the voltage for the time interval
3 8
25
20
15
10
5
Solution:
a. The flux linkage is directly proportional to the current; then from (5.1) and (5.2)
Therefore, we need to compute the current i at , , , and
For time interval , where is the slope of the straight line segment,
and b is the intercept which, by inspection, is . The slope is
= ---------------------- = –15
3 ms = –15t + 25
and thus
(5.10)
At , (5.10) yields . Then, the flux linkage is
and
(5.11)
iLt 50 mH
 t = 2 5 9 and 11 ms
vLt –  t  14 ms
(mA)
0
5
10
15
20
t (ms)
6 10 12
14
iLt

 = N = Li
t = 2 ms t = 5 ms t = 9 ms
t = 11 ms
0  t  3 ms i = mt + b m
i – axis 25 mA m
m – 20 – 25
3 – 0
i t = 0
t = 2 ms i = –5 mA
 Li 50 10–3   –5 10–3 = = 
 t = 2 ms = –250 Wb

Inductance
For the time interval , where
and thus
3  t  6 ms i = mt + b
m 15 – –20
------------------------- 35
= = -----
3 – 0
3
i 35
= -----t + b
3
To find b we use the fact that at , as seen in Figure 5.4. Then,
t = 3 ms i = –20 mA
–20 35
= -----  3 + b
3
from which .
Thus, the straight line equation for the time interval is
(5.12)
b = –55
3  t  6 ms
6 ms 35
i t = 3 ms
= -----t – 55
3
and therefore at , , and the flux linkage is
or
(5.13)
t = 5 ms i = 10  3 mA
 Li 50 10–3  10
 ----- 10–3 = = 
 t = 5 ms
Using the same procedure we find that
(5.14)
Also,
(5.15)
and with (5.15),
(5.16)
Likewise,
(5.17)
and with (5.17),
(5.18)
b. Since
3
500
3
= -------- Wb
8 ms = –12.5t + 90
i t = 6 ms
10 ms = 7.5t – 70
i t = 8 ms
 t = 9 ms = Li = –125 Wb
12 ms = –2.5t + 30
i t = 10 ms
 t = 11 ms = Li = 125 Wb
= ------
vL L
diL
dt
to compute and sketch the voltage vLt for the time interval –  t  14 ms
, we only need
to differentiate, that is, compute the slope of the straight line segments for this interval.
These were found in part (a) as (5.10), (5.12), (5.14), (5.15), and (5.17). Then,

(5.19)
(5.20)
(5.21)
(5.22)
(5.23)
(5.24)
(5.25)
= = ----------  –15 A  s = –750 mV
We now have all values given by (5.19) through (5.25) to sketch as a function of time. We
can do this easily with a spreadsheet such as Excel as shown in Figure 5.5.
Example 5.2
The voltage across a inductor is as shown on the waveform of Figure 5.6, and it is given
that the initial condition is . Compute and sketch the current which
flows through this inductor in the interval
slope –  t  0 = 0
vL  –  t  0
= L  slope = 0
slope 0  t  3 ms = –15 mA  ms = –15 A  s
vL 0 t 3 ms  
L  slope 50 10–3  v
A  s
slope 3  t  6 ms = 35  3 mA  ms = 35  3 A  s
vL 3 t 6 ms  
L  slope 50 10–3 = =   35  3 = 583.3 mV
slope 6  t  8 ms = –12.5 mA  ms = –12.5 A  s
vL 6 t 8 ms  
L  slope 50 10–3 = =   –12.5 = –625 mV
slope 8  t  10 ms = 7.5 mA  ms = 7.5 A  s
vL 8 t 10 ms  
L  slope 50 10–3 = =   7.5 = 375 mV
slope 10  t  12 ms = –2.5 mA  ms = –2.5 A  s
vL 10 t 12 ms  
L  slope 50 10–3 = =   –2.5 = –125 mV
slope 12  t  14 ms = 0
vL 12 t 14 ms  
= L  slope = 0
vL
50 mH
iLt0 = iL0 = 25 mA
–5  t  5 ms

Inductance
vLt (V)
1.75  3
4 6 8 10 12 14
2
Figure 5.5. Voltage waveform for Example 5.1
vLt 1.75  3
4 6 8 10 12 14
2
0
0.625
0.500
0.375
0.250
0.125
0.125
0.250
0.375
0.500
0.625
0.750
t (ms)
(V)
0
0.625
0.500
0.375
0.250
0.125
0.125
0.250
0.375
0.500
0.625
0.750
t (ms)
Solution:
The current iLt in an inductor is related to the voltage vLt
by (5.8) which is repeated here
for convenience.
iL t   1L
t
= ---  vLdt
+ iLt0
t0
where is the initial condition, that is,
iLt0 = iL0 = 25 mA
iL  –  t  0
= 25 mA

vL
0  t  3 ms
3 ms
= ----------------------  –0.75dt
+ 25 
10–3 3 10 3 – 
  25 10–3 +  20 2.25 10–3 –   20  0 25 10–3 = = + + 
25
20
15
10
5
=  3 ms + initial condition
20 0.750 3 10–3 –    25 10–3 = +  = –20 mA
=  6 ms + initial condition
  20 10–3 = –  = 15 mA
---------- 3 103
–    
From the given waveform,
Then,
that is, the current has dropped linearly from at to at as shown
in Figure 5.7.
Figure 5.7. Inductor current for , Example 5.2
The same result can be obtained by graphical integration. Thus,
and the value of now becomes our initial condition for the time interval
.
Continuing with graphical integration, we obtain
= –0.75 V
iL 0 t 3 ms  
1
50 10–3 
0
20 0.75t 0
– 
45 10–3 –  25 10–3 +  –20 10–3 = =  = –20 mA
25 mA t = 0 –20 mA t = 3 ms
(mA)
0
5
10
15
20
t (ms)
3
iLt
0  t  3 ms
iL t 3 ms =
1L
--- Area t = 0
iL t 3 ms =
= –20 mA
3  t  6 ms
iL t 6 ms =
1L
--- Area t = 3
20 1.75
3

Inductance
and now the current has increased linearly from –20 mA at t = 3 ms to 15 mA at t = 6 ms
as
(mA)
iLt
25
20
15
10
5
0
5
10
15
20
t (ms)
6
3
0  t  6 ms
For the time interval , we obtain
6  t  8 ms
iL t 8 ms =
1L
= 3
0---  Area 8 ms initial condition
t = 6
+ 20 – 0.625 2 1–     15 10–3 = +  = –10 mA
Therefore, the current has decreased linearly from at to at
15 mA t = 6 ms –10 mA t = 8 ms
(mA)
iLt
25
20
15
10
5
0
5
10
15
20
3 8
6
t (ms)
0  t  8 ms
For the time interval we obtain
8 ms  t  10 ms

iL t = 10 ms
1L
= ---  Area 10 ms + initial condition
t = 8
= 20  0.375  2  10–3 –10  10–3 = 5 mA
that is, the current has increased linearly from –10 mA at t = 8 ms to 5 mA at t = 10 ms
as
(mA)
iLt
25
20
15
10
5
0
5
10
15
20
t (ms)
3 8
6 10
0  t  10 ms
10 ms  t  12 ms
iL t 12 ms =
1L
= 3
0---  Area 12 ms initial condition
t = 10
+ 20 – 0.125 2 1–     5 103
– = +  = 0
5 mA t = 10 ms 0 mA t = 12 ms
t  12 ms
Finally, for the time interval we obtain
that is, the current has decreased linearly from at to at and
remains at zero for as shown in Figure 5.11.
Example 5.2 confirms the well known fact that the current through an inductor cannot change
instantaneously. This can be observed from the voltage and current waveforms for this and the
previous example. We observe that the voltage across the inductor can change instantaneously as
shown by the discontinuities at t = 0 3 6 8 10 and 12 ms
. However, the current through the
inductor never changes instantaneously, that is, it displays no discontinuities since its value is
explicitly defined at all instances of time.

Power and Energy in an Inductor
(mA)
iLt
25
20
15
10
5
0
5
10
15
20
t (ms)
3 8
6 10 12
14
0  t  12 ms
5.3 Power and Energy in an Inductor
Power in an inductor with inductance is found from
(5.26)
L
diL = = =
diL
 
  iL LiL t d
pL vLiL L t d
and the energy in an inductor, designated as is the integral of the power, that is,
or
or
WL
it
 L iLdiL
t
 L iLdt
t pLdt
WL t0
= = = 
t0
diL t d
it0
it
it0
t 12
WL t0
--LiL 2
i t   12
= =  – t0
it0
--L iL 2
t   iL 2
WLt – WLt0 12
--L iL 2
t   iL 2
=  – t0
and letting at , we obtain the energy stored in an inductor as
(5.27)
iL = 0 t = 0
WL t   12
--LiL 2
= t
Unlike the resistor which dissipates energy (in the form of heat), the (ideal) inductor is a physical
device capable of storing energy in analogy to the potential energy of a stretched spring.
Electric circuits which contain inductors can be simplified if the applied voltage and current
sources are constant as shown by the following example.

Example 5.3
For the circuit shown in Figure 5.12, compute , , and , after steadystate*conditions have
been reached. Then, compute the power absorbed and the energy consumed by the induc-tor.
24 V 35 mH 15 A
20 mH
Solution:
Since both the voltage and the current sources are constant, the voltages and the currents in all
branches of the circuit will be constant after steadystate conditions have been reached.
Since
= = d cons tant = 0
then, all voltages across the inductors will be zero and therefore we can replace all inductors by
short circuits. The given circuit then reduces to the one shown in Figure 5.13 where the and
parallel resistors have been combined into a single resistor.
Figure 5.13. Circuit for Example 5.3 after steadystate conditions have been reached
* By steady state conditions we mean the condition (state) where the voltages and currents, after some transient disturbances,
have subsided. Transients will be in Chapter 10.
v1 v2 v3
5 mH
+
+ 
+

+ 
5 mH
30 mH
25 mH 40 mH
15 mH
60 mH
10 
4 
5 
6 
3 
8 
12 
v2
v1
v3
9 
vL L
diL
dt
------- L
dt
3 
6  2 
+

+ 
+

+ 
v1
24 V
4 
9 
5 
v2
15 A
v3 8 
vA 2  vB

Power and Energy in an Inductor
Now, in Figure 5.13, by inspection, since the resistor was shorted out by the
v1 = 0 12 
inductor. To find and , let us first find and using nodal analysis.
60 mH v2 v2 vA vB
At Node ,
or
(5.28)
At Node
or
(5.29)
vA vA – 24
------------------
4
vA
------
9
vA – vB
5 + 2
+ + ------------------ = 0
-- 17
-- 19
14
 vA
 + + --
17
– --vB = 6
vB
vB – vA
5 + 2
------------------ – 15
vB
8
+ ----- = 0
17
-- 18
--vA – 17
+   vB = 15
 + --
We will use the MATLAB script below to find the solution of (5.28) and (5.29).
format rat % Express answers in rational form
G=[1/4+1/9+1/7 1/7; 1/7 1/7+1/8]; I=[6 15]'; V=GI;
disp('vA='); disp(V(1)); disp('vB='); disp(V(2))
vA=
360/11
vB=
808/11
Therefore,
and
that is,
Also,
or
vA = 360  11 V
vB = 808  11 V
v2 = vA – v2 = –448  11 V
v3 = v2 = 808  11 V
p5 mH = v5 mH  i5 mH = 0  i5 mH = 0
p5 mH = 0 watts
W5 mH
12
2 12
--Li5 mH
--L
v3
8
 2
-----
 2
0.5 5 10–3   808  11
------------------
8
= = = 
W5 mH = 0.211 J

5.4 Combinations of Inductors in Series and in Parallel
Consider the circuits of figures 5.14 (a) and 5.14 (b) where the source voltage vS
is the same for
both circuits. We wish to find an expression for the equivalent inductance which we denote as
in terms of in Figure 5.14 (a) so that the current i will be the same for both
LSeq L1 L2 LN
+

+
+

L1 L2
LN
LN – 1
i i

+
+


+
vS
(a) (b)
vS
LSeq
L1
di
dt
---- L2
di
dt
+ ---- +  + LN – 1
+ ---- = vS
di
dt
---- LN
di
dt
L1 + L2 + + LN – 1 + LN di
---- = vS
dt
LSeq
di
dt
---- = vS
LSeq = L1 + L2 + + LN – 1 + LN
circuits.
Figure 5.14. Circuits for derivation of equivalent inductance for inductors in series
or
(5.30)
(5.31)
Equating the left sides of (5.30) and (5.31) we obtain:
(5.32)
Thus, inductors in series combine as resistors in series do.
Next, we will consider the circuits of Figures 5.15 (a) and 5.15 (b) where the source current is
the same for both circuits. We wish to find an expression for the equivalent inductance which we
denote as in terms of in Figure 5.15 (a) so that the voltage v will be the same
for both circuits.
Figure 5.15. Circuits for derivation of equivalent inductance for inductors in parallel
iS
LPeq L1 L2  LN
+

(a)
+

(b)
iS
v
L1 L2 LN
LPeq
i1 i2 iN – 1 v
LN – 1
iN

Combinations of Inductors in Series and in Parallel
From the circuit of Figure 5.15 (a)
or
or
(5.33)
1
L1
t
 1
----- vdt
–
t
  1
+ ----- vdt
+ + +  = iS
L2
–
----- 1
1
L1
-----  1
t
 = iS
  vdt
 + + + + ------ 
L2
From the circuit of Figure 5.15 (b)
(5.34)
i1 + i2 + + iN – 1 + iN = iS
t
 1
------------- vdt
LN – 1
------------- 1
LN – 1
LN
1
LPeq
t
 = iS
---------- vdt
–
(5.35)
---------- 1
1
LPeq
----- 1
= + + + ------
L1
-----  1
L2
and for the special case of two parallel inductors
(5.36)
–
–
LN
LPeq
L1L2
L1 + L2
= -------------------
Thus, inductors in parallel combine as resistors in parallel do.
t
------ vdt
LN
–
Example 5.4
For the network of Figure 5.16, replace all inductors by a single equivalent inductor.
120 mH
Leq 125 mH 60 mH
45 mH
35 mH 40 mH
30 mH
15 mH
90 mH
Solution:
Starting at the right end of the network and moving towards the left end, we find that
, , , and also
. The network then reduces to that shown in Figure 5.17.
60 mH || 120 mH = 40 mH 30 mH || 15 mH = 10 mH 40 mH + 35 mH = 75 mH
45 mH || 90 mH = 30 mH

Leq 125 mH 40 mH
30 mH
75 mH
10 mH
Figure 5.17. First step in combination of inductances
Finally, with reference to Figure 5.17, , and
40 mH + 35 mH + 10 mH || 125 mH = 62.5 mH
Leq = 30 mH + 62.5 mH = 92.5 mH
Leq 92.5 mH
C
+
C
S
vS
RS
Figure 5.18. Network showing the equivalent inductance of Figure 5.16
5.5 Capacitance
In Section 5.2 we learned that inductance is associated with a magnetic field which is created
whenever there is current flow. Similarly, capacitance is associated with an electric field. In a sim-ple
circuit we can represent the entire capacitance with a device called capacitor, just as we con-sidered
the entire inductance to be concentrated in a single inductor. A capacitor consists of two
parallel metal plates separated by an air space or by a sheet of some type of insulating material
called the dielectric.
Now, let us consider the simple series circuit of Figure 5.19 where the device denoted as , is the
standard symbol for a capacitor.
Figure 5.19. Simple circuit to illustrate a charged capacitor
When the switch S
closes in the circuit of Figure 5.19, the voltage source will force electrons
from its negative terminal through the conductor to the lower plate of the capacitor and it will
accumulate negative charge. At the same time, electrons which were present in the upper plate of
the capacitor will move towards the positive terminal of the voltage source. This action leaves the

Capacitance
upper plate of the capacitor deficient in electrons and thus it becomes positively charged. There-fore,
an electric field has been established between the plates of the capacitor.
The distribution of the electric field set up in a capacitor is usually represented by lines of force
similar to the lines of force in a magnetic field. However, in an electric field the lines of force
start at the positive plate and terminate at the negative plate, whereas magnetic lines of force are
always complete loops.
Figure 5.20 shows the distribution of the electric field between the two plates of a capacitor.
+

+ + + + +
    
Figure 5.20. Electric field between the plates of a capacitor
We observe that the electric field has an almost uniform density in the area directly between the
plates, but it decreases in density beyond the edges of the plates.
The charge on the plates is directly proportional to the voltage between the plates and the
capacitance is the constant of proportionality. Thus,
(5.37)
q
C
q = Cv
and recalling that the current i is the rate of change of the charge q, we have the relation
or
(5.38)
i dq
= ------
= d Cv
dt
dt
= ---------
iC C
dvC
dt
where and in (5.38) obey the passive sign convention.
The unit of capacitance is the Farad abbreviated as F and since
(5.39)
iC vC
C
iC
-------- amperes
= = ---------------------
dvC
dt
volts
seconds
--------------------

we can say that one farad is the capacitance in a circuit in which a current of one ampere flows
when the voltage is changing at the rate of a one volt per second.
By separation of the variables we rewrite (5.38) as
(5.40)
dvC
1C
= --- iC dt
vCt
 1C
dvC
vCt0
t
= 
--- iCdt
t0
vCt – vCt0 1C
t
= 
--- iCdt
t0
vC t   1C
t
= ---  iCdt
+ vCt0
t0
vCt0
t = 0 vC0
vC t   1C
t
 1C
= --- iCdt
= + 
–
0
 1C--- iCdt
--- iCdt
–
t
0
1 103
–   vC0
and integrating both sides we obtain:
or
or
(5.41)
where is the initial condition, that is, the voltage across a capacitor at some reference time
usually taken as , and denoted as .
We can also write (5.41) as
where the initial condition is represented by the first integral on the right side.
Example 5.5
The waveform shown in Figure 5.21 represents the current flowing through a 1 F
capacitor.
Compute and sketch the voltage across this capacitor for the time interval 0  t  4 ms
given that
the initial condition is vC0 = 0
.
Solution:
The initial condition vC0 = 0 , establishes the first point at the coordinates 0 0 on the
vCt
versus time plot of Figure 5.22.
Next,
vC t 1 ms =
1C
--- iC dt
0
0
= +


Capacitance
(mA)
2 3 4
1
0 t (ms)
iCt
1.00
0.75
0.50
0.25
0.25
0.50
0.75
1.00
(V)
0
t (ms)
vCt
1.00
0.75
0.50
0.25
0.25 1
2 3 4
0.50
0.75
1.00
0  t  1 ms
Figure 5.22. Straight line segment for of the voltage waveform for Example 5.5
or
vC t 1 ms =
1C
3
 1 0
1–  1
– = =     = 1 volt
--- Area t = 0
 
------------------- 1 103
1 
1– 6
0–  1 103
and this value establishes the second point of the straight line segment passing through the origin
This value of 1 volt at t = 1 ms becomes our initial condition for the time interval 1  t  2
.
Continuing, we obtain
vC t 2 ms =
1C
1 
1– 3
0=   + 1
--- Area t = 0
 
1
------------------- –1 10–3  2 – 1 10–3 =     + 1 = 0 volts
1 
1– 6
0Thus, the capacitor voltage then decreases linearly from at to at
1 volt t = 1 ms 0 volts
t = 2 ms

1.00
0.75
0.50
0.25
Figure 5.23. Voltage waveform for of Example 5.5
There is no need to calculate the values of the capacitor voltage at and at
because the waveform of the current starts repeating itself at , and the initial condi-tions
and the areas are the same as before. Accordingly, the capacitor voltage waveform of fig-ure
(b) starts repeating itself also as shown in Figure 5.24.
1.00
0.75
0.50
0.25
Figure 5.24. Voltage waveform for of Example 5.5
Example 5.5 has illustrated the well known fact that the voltage across a capacitor cannot change
instantaneously. Referring to the current and voltage waveforms for this example, we observe
that the current through the capacitor can change instantaneously as shown by the discontinui-ties
at in Figure 5.21. However, the voltage across the capacitor never
changes instantaneously, that is, it displays no discontinuities since its value is explicitly defined
at all instances of time as shown in Figure 5.24.
(V)
0 t (ms)
0.25 1
2 3 4
0.50
0.75
1.00
vCt
0  t  2 ms
vc t = 3 ms t = 4 ms
ic t = 2 ms
vc
(V)
0 t (ms)
0.25 1
2 3 4
0.50
0.75
1.00
vCt
0  t  4 ms
t = 1 2 3 and 4 ms

Power and Energy in a Capacitor
5.6 Power and Energy in a Capacitor
Power in a capacitor with capacitance C is found from
dvC
 
= =  
WC
pC vC iC vC Cdt
and the energy in a capacitor, denoted as is the integral of the power, that is,
or
dvC t d
vt
 C vCdvc
t
 C vCdt
t pCdt
WC t0
= = = 
t0
vt0
vt
vt0
12
= --Cvc
2
=  – 2t0
i t   12
it0
2t vc
--C vc
WCt – WCt0 12
= --C  vc
2t – vc
2t0
and letting at , we obtain the energy stored in a capacitor as
(5.42)
vC = 0 t = 0
WC t   12
--CvC 2
= t
Like an inductor, a capacitor is a physical device capable of storing energy.
It was stated earlier that the current through an inductor and the voltage across a capacitor can-not
change instantaneously. These facts can also be seen from the expressions of the energy in an
inductor and in a capacitor, equations (5.27) and (5.42) where we observe that if the current in
an inductor or the voltage across a capacitor could change instantaneously, then the energies
and would also change instantaneously but this is, of course, a physical impossibility.
WL WC
Example 5.6
In the circuit of figure 5.25, the voltage and current sources are constant.
a. Compute iL1 and
vC2
b. Compute the power and energy in the 2 F
capacitor.
Solution:
a. The voltage and current sources are constant; thus, after steadystate conditions have been
reached, the voltages across the inductors will be zero and the currents through the capacitors
will be zero. Therefore, we can replace the inductors by short circuits and the capacitors by
open circuits and the given circuit reduces to that shown in Figure 5.26.

1 F 30 mH
3 F
4  2 
7  5 
R6
4  2 
7  5 
Figure 5.26. First simplification of the circuit of Example 5.6
We can simplify the circuit of figure 5.26 by first exchanging the current source and
resistor for a voltage source of in series with as shown in Figure 5.27.
We also combine the seriespar a l l e l r e s i s tors through . Thus,
.But now we observe that the branch in which the current
flows has disappeared; however, this presents no problem since we can apply the current divi-sion
expression once i, shown in Figure 5.27, is found. The simplified circuit then is as shown
in Figure 5.27.
We can apply superposition here. Instead, we will write two mesh equations and we will solve
using MATLAB. These in matrix form are
+
 +

40 mH
25 mH
60 mH 20 mH
24 V
15 A
iL1
vC2
10 
6 
5  8 
2 F
C1
R1
C2
C3
L1
L2
L3
L5
R5 L4
R8
R4
R3
R2
R7
+

+
24 V 
15 A
iL1
vC2
10 
6 
8 
R1
R5
R6
R8
R4
R3
R2
R7
15 A
R8 15  8 = 120 V R8
R1 R4
Req = 4 + 2 || 7 + 5 = 4  iL1
20 –6
–6 14
i1
i2
24
–120
=

Power and Energy in a Capacitor
+
+
i
Req
4 
R6
vC2
i2
24 V 
+

120 V
R8
8 
R5
10 
6 
i1
Figure 5.27. Final simplification of the circuit of Example 5.6
Solution using MATLAB:
format rat; R=[20 6; 6 14]; V=[24 120]'; I=RV; disp(‘i1=’); disp(I(1)); disp(‘i2=’);
disp(I(2))
i1=
-96/61
i2=
-564/61
Therefore, with reference to the circuit of Figure 5.28 below, we obtain
+
R3
4  2 
iL1
vC2
R1
R4
R2
7  5 
6 
R6
24 V 
15 A
R5
10 
R7
iL1 vC2
Figure 5.28. Circuit for computation of and for Example 5.6
and
b.
and
+

8 
R8
iL1
4 + 2
   32
---------------------------------------- 96
= = –----- = –0.525 A
4 + 2 + 7 + 5
–----- 
61
61
vC2 6 96
  2808
– ----- 564
= = ----------- = 46.03 V
 + -------- 
61
61
61
p2 F = v2 F  i2 F = vC2  0 = 0
W2 F
12
 2
2 0.5 2 10–6   2808
--Cv2 F
-----------
= =  = 2 mJ
61

5.7 Combinations of Capacitors in Series and in Parallel
Consider the circuits of figures 5.29 (a) and 5.29 (b) in which the source voltage is the same
for both circuits. We want to find an expression for the equivalent capacitance which we denote
as in terms of in Figure 5.29 (a) so that the current i will be the same in
both circuits.
vC1 vC2 vCN – 1
vS i i
+
Figure 5.29. Circuits for derivation of equivalent capacitance for capacitors in series
t
 1
t
  1
+ ------ idt
+ + +  = vS
t
 1
-------------- idt
t
------- idt
------ 1
------  1
-------------- 1
t
 = vS
  idt
 + + + + ------- 
t
 = vS
----------- 1
------ 1
------  1
-------------- 1
= + + + + -------
or
or
(5.43)
From the circuit of Figure 5.29 (b)
(5.44)
(5.45)
and for the special case of two capacitors in series
(5.46)
Thus capacitors in series combine as resistors in parallel do.
vS
CSeq C1 C2  CN – 1 CN
+

+ 
+
(a) (b)
CSeq
vS
C1 C2
CN
+ +
+
 
 
CN – 1
vCN
vC1 + vC2 + + vCN – 1 + vCN = vS
1
C1
------ idt
–
C2
–
CN – 1
–
CN
–
1
C1
C2
CN – 1
CN
–
1
CSeq
----------- idt
–
1
CSeq
C1
C2
CN – 1
CN
CSeq
C1C2
C1 + C2
= ------------------

Combinations of Capacitors in Series and in Parallel
Next, we will consider the circuits of figures 5.30 (a) and 5.30 (b) where the source current is
the same for both circuits. We wish to find an expression for the equivalent capacitance which
we denote as CPeq in terms of C1 C2  CN in Figure 5.30 (a) so that the voltage v
will be the
same in both circuits.
+
v CPeq

i1 i2 iN – 1
iN
CN C1 C2 CN – 1
(a)
+
v

(b)
iS
iS
Figure 5.30. Circuits for derivation of equivalent capacitance for capacitors in parallel
or
or
iS
(5.47)
i1 + i2 + + iN – 1 + iN = iS
C1
dv
dt
------ C2
+ + + + ------ = iS
C1 + C2 + + CN – 1 + CN dv
(5.48)
dv
dt
------  CN – 1
dv
dt
------ CN
CPeq
dv
dt
------ = iS
(5.49)
------ = iS
dt
CPeq = C1 + C2 + + CN – 1 + CN
Thus, capacitors in parallel combine as resistors in series do.
dv
dt
Example 5.7
For the network of Figure 5.31, replace all capacitors by a single equivalent capacitor.
Solution:
Beginning at the right of the network and moving towards the left, we find that
3 F || 1 F = 4 F
2 F || 4 F = 6 F

15 F 30 F
1 F 3 F
560
31
-------- F
Ceq
8 F 2 F
4 F
12 F
15 F in series with 30F = 10F
8 F || 12 F = 20 F
10 F
560
31
-------- F
20 F 6 F
Ceq 4 F
10 4 and 6F capacitors 60  31 F
60  31 F || 560  31 F = 20 F 20 F 20 F
Ceq = 10 F
Ceq 20 F
The network then reduces to that shown in Figure 5.32.
Figure 5.32. First step in combination of capacitances
Next, the series combination of yields and
. Finally, the series combination of and yields
Figure 5.33. Network showing the equivalent inductance of Figure 5.16
5.8 Nodal and Mesh Equations in General Terms
In Examples 5.3 and 5.6 the voltage and current sources were constant and therefore, the steady
state circuit analysis could be performed by nodal, mesh or any other method of analysis as we
learned in Chapter 3. However, if the voltage and current sources are timevarying quantities we
must apply KCL or KVL in general terms as illustrated by the following example.

Nodal and Mesh Equations in General Terms
Example 5.8
Write nodal and mesh equations for the circuit shown in Figure 5.34.
+
+ 
C
L
vS1
vS2
R1
R2
Solution:
a. Nodal Analysis:
We assign nodes as shown in Figure 5.35. Thus, we need nodal equations.
N – 1 = 5 – 1 = 4
v1 v2 v3
+
+ 
C
L
vS1
v4
vS2
R1
R2
v0
Figure 5.35. Nodal analysis for the circuit of Example 5.8
At Node 1:
At Node 2:
At Node 3:
At Node 4:
v1 = vS1
v2 – v1
R1
---------------- C d
---- v2 v4 –   1L
+ +  = 0
dt
t
--- v2 – v3dt
–
1L
t
 v3
--- v3 – v2dt
–
+ ------ = 0
R2
v4 = –vS2
b. Mesh Analysis:
We need M = B – 1 = 6 – 5 + 1 = 2 mesh equations. Thus, we assign currents i1 and i2
as

+
+ 
C
L
vS1
vS2
R1
Figure 5.36. Mesh analysis for the circuit of Example 5.8
t
+  – vS1 – vS2 = 0
t
+  = 0
For Mesh 1:
For Mesh 2:
In both the nodal and mesh equations, the initial conditions are included in the limits of integra-tion.
Alternately, we can add the initial condition terms and in the integrodifferential equations
above, replace the lower limit of integration with zero.
R2
i1 i2
R1i1
1C
--- i1 – i2dt
–
L d
dt
----i2 + R2i2 + vS2 1C
--- i2 – i1dt
–
–

Summary
5.9 Summary
 Inductance is associated with a magnetic field which is created whenever there is current flow.
 The magnetic field loops are circular in form and are called lines of magnetic flux. The unit of
magnetic flux is the weber (Wb).
 The magnetic flux is denoted as  and, if there are N turns and we assume that the flux

passes through each turn, the total flux, denoted as   is called flux linkage. Then,
 = N
 For an inductor, the voltagecurrent relationship is
vL = LdiL  dt
 The unit of inductance is the Henry abbreviated as H.
 Unlike the resistor which dissipates energy (in the form of heat), the (ideal) inductor is a phys-ical
device capable of storing energy in analogy to the potential energy of a stretched spring.
WL t   1 2   LiL 2
C 2
 The energy stored in an inductor is
= t
 The current through an inductor cannot change instantaneously.
 In circuits where the applied voltage source or current source are constants, after steadystate
conditions have been reached, an inductor behaves like a short circuit.
 Inductors in series combine as resistors in series do.
 Inductors in parallel combine as resistors in parallel do.
 Capacitance is associated with an electric field.
 A capacitor consists of two parallel metal plates separated by an air space or by a sheet of some
type of insulating material called the dielectric.
 The charge q
on the plates of a capacitor is directly proportional to the voltage between the
plates and the capacitance C is the constant of proportionality. Thus,
q = Cv
 In a capacitor, the voltagecurrent relationship is
iC = CdvC  dt
 The unit of capacitance is the Farad abbreviated as F.
 Like an inductor, a capacitor is a physical device capable of storing energy.
 The energy stored in a capacitor is
WC  t  =  1  2 Cvt
 The voltage across a capacitor cannot change instantaneously.
 In circuits where the applied voltage source or current source are constants, after steadystate
conditions have been reached, a capacitor behaves like an open circuit.
 Capacitors in series combine as resistors in parallel do.

 Capacitors in parallel combine as resistors in series do.
 In a circuit that contains inductors and/or capacitors, if the applied voltage and current sources
are timevarying quantities, the nodal and mesh equations are, in general, integrodifferential
equations.

Exercises
5.10 Exercises
Multiple Choice
1. The unit of inductance is the
A. Farad
B. Ohm
C. mH
D. Weber
E. None of the above
2. The unit of capacitance is the
A.
F
B. Ohm
C. Farad
D. Coulomb
3. Faraday’s law of electromagnetic induction states that
A.
 = N
B.
 = Li
C.
v = Ldi  dt
D.
v = d  dt
4. In an electric field of a capacitor, the lines of force
A. are complete loops
B. start at the positive plate and end at the negative plate
C. start at the negative plate and end at the positive plate
D. are unpredictable
5. The energy in an inductor is
1  2Li2
1  2Lv2
vLiL
A.
B.
C.
D. dissipated in the form of heat

1  2Ci2
1  2Cv2
vCiC
6. The energy in a capacitor is
A.
B.
C.
D. dissipated in the form of heat
7. In an inductor
A. the voltage cannot change instantaneously
B. the current cannot change instantaneously
C. neither the voltage nor the current can change instantaneously
D. both the voltage and the current can change instantaneously
8. In a capacitor
A. the voltage cannot change instantaneously
B. the current cannot change instantaneously
C. neither the voltage nor the current can change instantaneously
D. both the voltage and the current can change instantaneously
9. In the circuit below, after steadystate conditions have been established, the current
through the inductor will be
A.
B.
C.
D.
10.In the circuit below, after steadystate conditions have been established, the voltage
across the capacitor will be
iL

iS
iL
5 A
5 
5 mH
0 A
 A
2.5 A
5 A
vC

Exercises
A.
B.
C.
D.
+

+ 
C
vS
R
10 V
5 
2 F
0 V
 V
–10 V
10 V
Problems
1. The current flowing through a 10 mH inductor is shown by the waveform below.
iL
60
50
10 20 30
iL
10
0
10
40
(mA)
t (ms)
a. Compute and sketch the voltage across this inductor for
b. Compute the first time after when the power absorbed by this inductor is
Answer:
vL t  0
t = 0 pL
pL = 50 w t = 5 ms
c. Compute the first time after when the power absorbed by this inductor is
Answer:
t = 0 pL
pL = –50 w t = 25 ms
2. The current flowing through a capacitor is given as , and it is
iC 1 F iC t = cos100t mA
vC 0 = 0
known that
a. Compute and sketch the voltage across this capacitor for
b. Compute the first time after when the power absorbed by this capacitor is
. Answer:
vC t  0
t = 0 pC
pC = 5 mw 7.85 ms
c. Compute the first time after when the power absorbed by this capacitor is
. Answer:
t = 0 pC
pC = –5 mw 23.56 ms

3. For the network below, compute the total energy stored in the series combination of the resis-tor,
5  0.4 mH
+ 
100 F
3 mH
capacitor, and inductor at if:
a. and it is known that . Answer:
b. and it is known that . Answer:
4. For the circuit below, compute the energy stored in the inductor at given that
. Answer:
5. For the circuit below, replace all capacitors with an equivalent capacitance and then com-pute
the energy stored in at given that in all capacitors.
Answer:
t = 10 ms
it 0.1e–100t = mA vC 0 = –10 V 3.4 mJ
it = 0.5cos5t mA vC 0 = 0 50 J
R L C
Rest of the Network
it
vCt
5 mH t = 1 s
i 0 = 0 1 mJ
+

5 mH
10 mH 7 mH
vSt
it
10e–t mV
Ceq
Ceq t = 1 ms vC 0 = 0
10 pJ
6 F
3 F
8 F
10 F
iSt 10 A

Exercises
6. Write nodal equations for the circuit below.
+

vSt C2
7.Write mesh equations for the circuit below.
C1
R1
R2
L
+
R1
L2
L1
R2
C1
C2
vS t

Multiple Choice
1. E Henry
2. C
3. D
4. B
5. A
6. B
7. B
8. A
9. E
–5 A
10. D
Problems
1.
a. In an inductor the voltage and current are related by . Thus,
vL = LdiL  dt = L  slope
we need to compute the slope of each segment of the given waveform and multiply it by .
10 ms L  slope L
-------- 10 10–3  10 10 3 –  A
= = =  ----------------------------- = 10 mV
20 ms = L  slope = L  0 = 0 mV
40 ms L  slope 10 10–3  –10–10 10 –3  A
= =  --------------------------------------------------- = –10 mV
50 ms = L  slope = L  0 = 0 mV
10 ms L  slope 10 10–3  0 – –10 10 –3  A
= =  --------------------------------------------------- = 10 mV
Likewise,
The current, voltage, and power waveforms are shown below.
L
vL 0
iL
t
10 10–3  s
vL 10
vL 20
40 – 20 10–3  s
vL 40
vL 0
60 – 50 10–3  s

60
50
(mA)
10 20 30
iL
10
0
10
40
t (ms)
10 mH
iLt

vL
10
(mV)
+ 
vLt 10 20 30 50 60

0 40
10
t (ms)
pL  w
A
0 40 t (ms)
30 50 60
20
10
100
100
pL = vLiL
B
b. From the power waveform above, we observe that occurs for the first
time at point A where
pL = vLiL = 50 w
t = 5 ms
c. From the power waveform above, we observe that occurs for the
first time at point A where
pL = vLiL = –50 w
t = 25 ms
2.
a. For this problem and the current is a sinusoid given as
– = = F iC
C 1 F 106
as shown below. The voltage across this capacitor is found
iCt = cos100t mA vCt
from

vC t   1C
t
 + vC0 106 10–3   cos100d + 0
= --- iC d
= 
0
t
0
t
 103
103 cos100d
= = = 10sin100t
0
-------- sin100
100
t
0
t s
iCt
mA cos100t
0 2 4 6
1
0
-1
vCt
vCt
V

100
--------
10sin100t
0 1 2 3 4 5 6
101
0.5
0
-0.5
–10
-1
T
-----

50
T = 2  = 2  T 10 2
sin------t = 10sin100t 2  T = 100
T
T = 2  100 T =   50
vCt iCt
and the waveform of is shown below.
Now, or . Then, or and
or .
b. Since is a sine function and a cosine function, the first time after zero that their
product will be positive is in the interval where we want
or
0  t    200 pC = vCiC = 5 mw

or
Recalling that
it follows that
or
or
pC 10sin100t 10–3 100t cos   5 103
– = =  w
pC = 10sin100t cos100t = 5 w
sin2x = 2sinxcosx
pC = 5sin200t = 5 w
sin200t = 1
t 1 –1 sin
---------------   2
= = = -------- = 0.00785 s = 7.85 ms
200
--------- 
200
400
c. The time where will occur for the first time is after or
pC = –5 mw 7.85 ms t =   200 s
after . Therefore, will occur for the first time at
t = 1000  200 ms = 5 ms pC = –5 mw
t = 7.85 + 5 = 7.85 + 15.71 = 23.56 ms
3.
a. There is no energy stored in the resistor; it is dissipated in the form of heat. Thus, the total
energy is stored in the capacitor and the inductor, that is,
where
and
or
Then,
WT WL + WC 12
2 12--CvC 2
--LiL
= = +
iL it 0.1e–100t = =
vC t   1C
t
 + vC0 104 0.1e–100 d – 10
= --- iC d
= 
0
104  0.1
t
0
---------------------e–100
10 10e–100t = = = – – 10
–100
t
– 10 10e–100
0
0
t
vCt 10e–100t = –
WT t 10 ms =
-- 0.4 10–3   0.1e–100t  2  12
12
-- 10–4  10e–100t – 2 = + 
2.5 10–4 0.1e–1  2 10e–1 – 2
=   +  = 3.4 mJ

We’ve used MATLAB as a calculator to obtain the answer, that is,
WT=2.5*10^(4)*((0.1*exp(1))^2+((10)*exp(1))^2);
fprintf(' n'); fprintf('WT=%7.4f J',WT); fprintf(' n')
WT= 0.0034 J
iL = it = 0.5cos5t mA
vC t   1C
t
 + vC0 104 10–4  5cos5d + 0
= = = t = sin5t
--- iC d
0
t
 sin5 0
0
WT WL + WC 12
-- 0.4 10–4   0.5 5t cos  2  12
-- 10–4 = = +    sin5t2
0.5 10–4 cos25t + sin25t
=  
= 0.05 mJ = 50 J
1
7 + 3 = 10 10  10 = 5 5 + 5 = 10 mH
iLt
+
 10 mH
vSt
10e–t mV
iLt
iLt 1L
t
 + iL0 1
1 e–t = = = = = –
--- vS d
0
t
 e– – 0
---------------------- 10 10–3   e– d
10 10–3 
0
t
e–
0
t
W5 mH t 1 s =
12
= --  5  1–  1 – e–t 2
= 2.5  10–3   1 – e–1 2  1 mJ
t = 3
01 s 540 Circuit Analysis I with MATLAB Computing and Simulink / SimPowerSystems Modeling
b. For this part,
and
Then,
We observe that the total power is independent of time.
4.
Beginning with the right side and proceeding to the left, the seriesparallel combination of
, , and reduces the given circuit to the one shown
below.
The current is
Then,
5.
Beginning with the right side and proceeding to the left, the seriesparallel combination
reduces the given circuit to the one shown below.

The current is
Then,
5 F
iSt
10 A
+
vCt

vCt
vCt 1C
t
 + vC0 1
= = = t = 2t
--- iS d
0
t
 2 0
------------------- 10 10–6   d
5 
1– 6
0W5 F t 1 ms =
12
-- 5 106
–   2t2
t = 1 ms 2.5 10–6 4 106
– = =    = 10 pJ
6.
We assign node voltages , , and as shown below.
Then,
v1 v2 v3
+

C1
v1
v2 v3
vSt C2
R1
v1 = vS
C1
d
dt
----v2 – v1 C2
d
dt
----v2 – v3
+ + ------ = 0
v3 – v1
R2
+ +  = 0
---------------- C2
---- v3 v2 –   1L
d
dt
t
--- v3 d
–
7.
We assign mesh currents , , and as shown below.
0
R2
L
v2
R1
i1 i2 i3

L1
t
 R1i1 – i2 1
t
+ + ------  i1 – i3 d
= vS
t
 L2
Then,
+
R1
L2
R2
C1
C2
vS t i1
i2
i3
1
C1
------ i1 d
–
C2
–
R1i2 – i1 L1
di2
dt
------- L2
d
dt
+ + ----i2 – i3 = 0
1
C2
------ i3 – i1 d
–
d
dt
+ ----i3 – i2 + R2i3 = 0

Chapter 6
Sinusoidal Circuit Analysis
his chapter is an introduction to circuits in which the applied voltage or current are sinu-soidal.
The time and frequency domains are defined and phasor relationships are developed
for resistive, inductive and capacitive circuits. Reactance, susceptance, impedance and
T
admittance are also defined. It is assumed that the reader is familiar with sinusoids and complex
numbers. If not, it is strongly recommended that Appendix B is reviewed thoroughly before read-ing
this chapter.
6.1 Excitation Functions
The applied voltages and currents in electric circuits are generally referred to as excitations or driv-ing
functions, that is, we say that a circuit is “excited” or “driven” by a constant, or a sinusoidal, or
an exponential function of time. Another term used in circuit analysis is the word response; this
may be the voltage or current in the “load” part of the circuit or any other part of it. Thus the
response may be anything we define it as a response. Generally, the response is the voltage or cur-rent
at the output of a circuit, but we need to specify what the output of a circuit is.
In Chapters 1 through 4 we considered circuits that consisted of excitations (active sources) and
resistors only as the passive devices. We used various methods such as nodal and mesh analyses,
superposition, Thevenin’s and Norton’s theorems to find the desired response such as the voltage
and/or current in any particular branch. The circuit analysis procedure for these circuits is the
same for DC and AC circuits. Thus, if the excitation is a constant voltage or current, the
response will also be some constant value; if the excitation is a sinusoidal voltage or current, the
response will also be sinusoidal with the same frequency but different amplitude and phase.
In Chapter 5 we learned that when the excitation is a constant and steadystate conditions are
reached, an inductor behaves like a short circuit and a capacitor behaves like an open circuit.
However, when the excitation is a timevarying function such as a sinusoid, inductors and capac-itors
behave entirely different as we will see in our subsequent discussion.
6.2 Circuit Response to Sinusoidal Inputs
We can apply the circuit analysis methods which we have learned in previous chapters to circuits
where the voltage or current sources are sinusoidal. To find out how easy (or how difficult) the
procedure becomes, we will consider the simple series circuit of Example 6.1.
Example 6.1
For the circuit shown in Figure 6.1, derive an expression for in terms of , , , and
vC t Vp R C 
Circuit Analysis I with MATLAB Computing and Simulink / SimPowerSystem Modeling 61

Chapter 6 Sinusoidal Circuit Analysis
where the subscript is used to denote the peak or maximum value of a time varying function,
and the sine symbol inside the circle denotes that the excitation is a sinusoidal function.
Solution:
By KVL,
(6.1)
where
and
Then,
and by substitution into (6.1) we obtain
(6.2)
As we know, differentiation (and integration) of a sinusoid of radian frequency results in
another sinusoid of the same frequency . Accordingly, the solution of (6.2) must have the form
(6.3)
where the amplitude and phase angle are constants to be determined from the circuit
parameters of , , , and . Substitution of (6.3) into (6.2) yields
(6.4)
and recalling that
and
we rewrite (6.4) as
p
R
+
C 
vCt
it
vS = VP cost
vS
vR + vC = vS
vR = Ri = RiC
iC C
dvC
dt
= --------
vR RC
dvC
dt
= ---------
RC
dvC
dt
--------- + vC = vS = Vp cost


vCt = Acost  
A 
Vp R C 
–ARCsint +  + Acost +  = Vp cost
sinx + y = sinxcosy + cosxsiny
cosx + y = cosxcosy – sinxsiny
– ARCsint cos – ARCcost sin + Acost cos – Asint sin = Vp cost

The Complex Excitation Function
Collecting sine and cosine terms, equating like terms and, after some more tedious work, solving
for amplitude and phase angle we obtain:
(6.5)
A 
vCt
Vp
--------------------------------- t tan1
– = cos – RC
1 RC2 +
Obviously, analyzing circuits with sinusoidal excitations when they contain capacitors and/or
inductors, using the above procedure is impractical. We will see on the next section that the
complex excitation function greatly simplifies the procedure of analyzing such circuits. Complex
numbers are discussed in Appendix B.
The complex excitation function does not imply complexity of a circuit; it just entails the use of
complex numbers. We should remember also that when we say that the imaginary part of a com-plex
number is some value, there is nothing “imaginary” about this value. In other words, the
imaginary part is just as “real” as the real part of the complex number but it is defined on a differ-ent
axis. Thus we display the real part of a complex function on the axis of the reals (usually the
xaxis), and the imaginary part on the imaginary axis or the yaxis.
6.3 The Complex Excitation Function
We recall that the derivatives and integrals of sinusoids always produce sinusoids of the same fre-quency
but different amplitude and phase since the cosine and sine functions are 90 degrees out
ofphase. Thus, if
then
and if
then
vt = Acost + 
dv = –Asint + 
dt
t = Bsint + 
di = Bcost + 
dt
Let us consider the network of Figure 6.2 which consists of resistors, inductors and capacitors,
and it is driven (excited) by a sinusoidal voltage source .
vS t
iLOADt

LOAD

Linear Network
Consisting of
Resistors,
Inductors and
Capacitors
Excitation
vS t
vLOADt
Figure 6.2. General presentation of a network showing excitation and load
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystem Modeling 63

Let us also define the voltage across the load as * as the response. As we know from
Chapter 5, the nodal and mesh equations for such circuits are integrodifferential equations, and it
is shown in differential equations textbooks† that the forced response or particular solution of these
circuits have the form
We also know from Euler’s identity that
(6.6)
and therefore, the real component is the response due to and the imaginary component is
the response to We will use Example 6.2 to illustrate the ease by which we can obtain the
response of a circuit, which is excited by a sinusoidal source, using the complex function
approach. In this text, we will represent all sinusoidal variations in terms of the cosine
function.
Example 6.2
Repeat Example 6.1, that is, find the capacitor voltage for the circuit of Figure 6.3 using
the complex excitation method.
Solution:
Since
we let the excitation be
* Some textbooks denote the voltage across and the current through the load as and respectively. As we stated previ-ously,
in this text, we use the and notations to avoid confusion with the voltage across and the current
through an inductor.
† This topic is also discussed in Circuit Analysis II with MATLAB Computing and Simulink / SimPowerSystems Mod-eling,
ISBN 9781934404195.
vLOADt
vL iL
vLOAD iLOAD vL
iL
vLDt = Acost + Bsint
Acost + jAsint Ae jt =
cost
sint
Ae jt
vC t
R
+
C 
vCt
it
vS = VP cost
vS
cost = Reejt
vSt Vpe jt =

and thus the response will have the form
As in Example 6.1,
(6.7)
or
or
Ct VC e jt +  =
RC
dvC
dt
--------- + vC Vpe j =
C d
dt
----VCe jt +  VCe jt +  + Vpe j =
jRC + 1VCe jt +  Vpejt =

The last expression above shows that radian frequency is the same for the response as it is for
the excitation; therefore we only need to be concerned with the magnitude and the phase angle
of the response. Accordingly, we can eliminate the radian frequency by dividing both sides of
that expression by and thus the inputoutput (excitationresponse) relation reduces to
from which

e jt
jRC + 1VCe j = Vp
VCe j Vp
–1 = -----------------------
= =
--------------------------------e–j  tan RC  jRC + 1
Vp
-----------------------------------------------------------------
1 + 2R2C2e j RC –1  tan 
Vp
1 + 2R2C2
This expression above shows the response as a function of the maximum value of the excitation,
its radian frequency and the circuit constants R and C
.
If we wish to express the response in complete form, we simply multiply both sides by e jt
and
we obtain
VCe jt +  Vp
--------------------------------e j t RC –1  – tan  =
1 + 2R2C2
Finally, since the excitation is the real part of the complex excitation, we use Euler’s identity on
both sides and equating reals parts, we obtain
----------------------------- t –1RC = = cos – tan 
vC t VC cost + 
Vp
1 + 2R2C2
The first part of the above procedure where the excitationresponse relation is simplified to
amplitude and phase relationship is known as timedomain to frequencydomain transformation;
the second part where the excitationresponse is put back to its sinusoidal form is known as fre-

quencydomain to timedomain transformation. For brevity, we will denote the time domain as the
t – domain j – domain
mt = Asint +  = Acost +  – 90
j – domain
M Aej  90  –  A = =  – 90
M V I
vt = 10sin100t – 60 j – domain
vt = 10sin100t – 60 = 10cos100t – 60 – 90
 
v t   10 100t 150 –   cos = Re 10e j 100t 150 
 – 
 
 
=
j – domain
, and the frequency domain as the .
If a sinusoid is given in terms of the sine function, we must first convert it to a cosine function.
Thus,
(6.8)
and in the it is expressed as
(6.9)
where represents a phasor (rotating vector) voltage or current .
In summary, the , to transformation procedure is as follows:
1. Express the given sinusoid as a cosine function
2. Express the cosine function as the real part of the complex excitation using Euler’s identity
3. Extract the magnitude and phase angle from it.
Example 6.3
Transform the sinusoid to its equivalent expression.
Solution:
For this example, we have
or
Since the contains only the amplitude and phase, we extract these from the brack-eted
term on the right side of the above expression, and we obtain the phasor as
The to transformation procedure is as follows:
V
V 10e–j150
= = 10–150
j – domain t – domain

1.Convert the given phasor from polar to exponential form
2.Add the radian frequency  multiplied by t
to the exponential form
3. Extract the real part from it.
Example 6.4
Transform the phasor I = 120–90
to its equivalent timedomain expression.
Solution:
First, we express the given phasor in exponential form, that is,
I 120–90 120e–j90
= =
 t
Next, adding the radian frequency multiplied by to the exponent of the above expression we
obtain
0 i  t  =
120e 
j  t – 9 and finally we extract the real part from it. Then,
 
j  
i  t  Re 120e 0 t – 9
= = 120cost – 90 = 120sint
 
 
We can add, subtract, multiply and divide sinusoids of the same frequency using phasors as illus-trated
by the following example.
Example 6.5
It is given that and . Compute the sum
i1t = 10cos120t + 45 i2t = 5sin120t–45
.
it = i1t + i2t
Solution:
As a first step, we express as a cosine function, that is,
i2t
i2t = 5sin120t–45 = 5cos120t–45 – 90 = 5cos120t–135
Next, we perform the to transformation and we obtain the phasors

I1 = 1045 and I2 = 5–135
------ j 2
– ------ j 2
  5 2
I I1 + I2 1045 + 5–135 10 2
= = = +  
 + ------
2
2
 – ------
2
2
------ j 2
I 5 2
=   = 545
 + ------
2
2
t – domain
it = 5cos120t + 45
R L C
R L C
V I V
I j – domain
j – domain
V I R
+
vR t

+
VR IR

vS t
iR t

vR t = RiR t = Vp cost + 

VR = RIR
VS
a t – domain network b j – domain phasor network
vRt = RiRt R
and by addition,
or
and finally transforming the phasor I into the , we obtain
Also, for brevity, in our subsequent discussion we will designate resistive, inductive and capaci-tive
circuits as , , and respectively.
6.4 Phasors in R, L, and C Circuits
The circuit analysis of circuits containing , , and devices, and which are excited by sinusoi-dal
sources, is considerably simplified with the use of phasor voltages and phasor currents which
we will represent by the boldface capital letters and respectively. We will now derive and
phasor relationships in the . We must always remember that phasor quantities
exist only in the .
1. and phasor relationship in branches
Consider circuit 6.4 (a) below where the load is purely resistive.
Figure 6.4. Voltage across a resistive load in and
We know from Ohm’s law that where the resistance is a constant. We will
show that this relationship also holds for the phasors VR and IR
shown in circuit 6.4 (b), that
is, we will prove that

Phasors in R, L, and C Circuits
VR = RIR
Proof:
In circuit 6.4 (a) we let be a complex voltage, that is,
(6.10)
vR t
Vpe jt +  = Vp cost +  + jVp sint + 
R 
and since is a constant, it will produce a current of the same frequency and the same
phase * whose form will be

and by Ohm’s law,
(6.11)
Ipe jt +  = Ip cost +  + jIp sint + 
Vpe jt +  RIpe jt +  =
j – domain
Vpe j RIpe j
Transforming (6.11) to the , we obtain the phasor relationship
= or Vp = RIp
I V I V
Since the phasor current is inphase with the voltage (both and have the same phase
), we let

and it follows that
Vp = VR and Ip = IR
VR = RIR
V I
Therefore, the phasor and relationship in resistors, obeys Ohm’s law also, and the current
through a resistor is always inphase with the voltage across that resistor.
Example 6.6
For the network in Figure 6.5, find iR t when vR t = 40sin377t – 75
.
Solution:
We first perform the to i.e., transformation as follows:
t – domain j – domain vRtVR
vRt = 40sin377t – 75 = 40cos377t – 165VR = 40–165
* The phase will be the same since neither differentiation nor integration is performed here.

iR t

R = 5 
Figure 6.5. Voltage across the resistive load of Example 6.6
= = ------------------------- = 8–165 A
= = -------------------------------------------- = 8sin377t – 75 A
 

VL = jLIL
VS
VL IL

= -------
vLt L
diL
dt
Then,
Therefore,
Alternately, since the resistance is a constant, we can compute directly from the
expression for , that is,
Consider circuit 6.6 (a) below where the load is purely inductive.
Figure 6.6. Voltage across an inductive load in and
We will prove that the relationship between the phasors and shown in circuit 6.6 (b) is
(6.12)
Proof:
+

vS t
vR t
vR t = 40sin377t – 75
IR
VR
R
------- 40–165
5
IR = 8–165 A
j – domain
iRt = 8cos377t – 165 = 8sin377t – 75 A
t – domain



R iR t
t – domain vR t
iRt
vRt
R
------------ 40sin377t – 75
5
V I L
+

+

vS t
iL t vL t
VL IL
VL = jLIL
vLt

(6.13)
xt = sint +  then dx  dt = cost + 
and recalling that if , that is, differentiation (or
integration) does not change the radian frequency or the phase angle , the current
through the inductor will have the form
(6.14)
and since
then,
(6.15)
 
= -------
vLt L
diL
dt
Vpe jt +  L d
= ----Ipe jt +  =
jLIpe jt +  dt
j – domain
Next, transforming (6.16) to the , we obtain the phasor relationship
and letting
we obtain
(6.16)
Vpe j jLIpe j
= or Vp = jLIp
Vp = VL and Ip = IL
VL = jLIL
j
90
The presence of the operator in (6.17) indicates that the voltage across an inductor leads the
current through it by .
Example 6.7
For the network in Figure 6.7, find iL t when vLt = 40sin2t – 75
.
Solution:
t – domain j – domain vLtVL
+
iL t vL t


vS t

L = 5 mH
vLt = 40sin2t – 75
Figure 6.7. Voltage across the inductive load in Example 6.7

--------- 40–165 10 –3 
------------------------------------------------ 40–165
= = = ------------------------- = 4–255 = 4105 A

IC = jCVC
VS
VC IC

iCt C
dvC
dt
= ---------
----Vpe jt +  jCVpe jt +  = =
and
Therefore,
Consider circuit 6.8 (a) below where the load is purely capacitive.
Figure 6.8. Voltage across a capacitive load in and
We will prove that the relationship between the phasors and shown in the network in
Figure 6.8 (b) is
(6.17)
Proof:
then the current through the capacitor will have the form
and since
It follows that
(6.18)
vLt = 40sin2t – 75 = 40cos2t – 165VL = 40–165 mV
IL
VL
jL
j10 10–3 
1090
IL = 4105 A
j – domain
iLt = 4cos2t + 105 = 4sin2t – 165 A
t – domain



V I C
+

+

vS t
i vC t C t
VC IC
IC = jCVC
vCt
iCt
dvC
dt
= ---------
Ipe jt +  C d
dt

j – domain
Next, transforming (6.18) to the , we obtain the phasor relationship
and letting
we obtain
(6.19)
Ipe j jCVpe j = or Ip = jCVp
Ip = IC and Vp = VC
IC = jCVC
j
90
The presence of the operator in (6.19) indicates that the current through a capacitor leads
the voltage across it by .
Example 6.8
For the circuit shown below, find when .
iC t vCt = 170cos60t – 45
+
i vC t C t

vS t

C=106 nF
vCt = 170cos60t – 45
Figure 6.9. Voltage across the capacitive load of Example 6.8
Solution:
Then,
IC jCVC j 60 106 10–9     170–45 190 3.4 10–3 = = =    1–45
Therefore,
t – domain j – domain vCtVC
vCt = 170cos60t – 45VC = 170–45
3.4 10–3 =  45 = 3.445 mA
IC = 3.445 mA
j – domain
iCt = 3.4cos60 + 45 mA

t – domain
 

6.5 Impedance
Consider the circuit in Figure 6.10 (a) and its equivalent phasor circuit shown in
6.10 (b).
R L
R L + 
+ 
+ +
 
vR t vL t
= ----
t
---- 1C
t
+ +  = vS t

+ + ----------I = VS
Algebraic Equation

C
+

Figure 6.10. The and relationships in a series RLC circuit
The last equation on the right side of the phasor circuit may be written as
(6.20)
  + + ---------- 
I = VS
I
and dividing both sides of (6.20) by we obtain the impedance which, by definition, is
(6.21)
= = = = + + ----------
-------------------------------------
Expression (6.21) is referred to as Ohm’s law for AC Circuits.
Like resistance, the unit of impedance is the Ohm (
We can express the impedance as the sum of a real and an imaginary component as follows:
t – domain
VR
C +

I
VL
vRt = RiRt
vLt Ldi
dt
vC t   1C
--- idt
–
= 
VR = IR
VL = jLI
VC
1
jC
= ----------I
vRt + vLt + vCt = vS t VR + VL + VC = VS
Rit Ldi
dt
--- idt
–
Integrodifferential Equation
Very difficult to work with
RI jLI 1
jC
Much easier to work with
vS t
i t
vC t
VS
VC
R jL 1
jC
Impedance Z Phasor Voltage
Phasor Current
VS
I
------- R jL 1
jC
Z

Impedance
Since
it follows that
and thus
(6.22)
-- jj
-- 1j
1j
= = ---- = –j
j2 --------- 1
j1
jC
Z R j L 1
= +  
We can also express (6.22) in polar form as
(6.23)
 - j
= – -------
C
 – -------
C
= +  2
tan
–1  L 1
  R Z R2 L 1
 – -------
C
 – ------- 
C
We must remember that the impedance is not a phasor; it is a complex quantity whose real part is
the resistance and the imaginary part is  that is,
(6.24)
R L – 1  C
ReZ = R and ImZ L 1
= – -------
C
Z X
The imaginary part of the impedance is called reactance and it is denoted with the letter .
The two components of reactance are the inductive reactance XL and the capacitive reactance XC
,
i.e.,
(6.25)
(6.26)
(6.27)
X = XL + XC = L – 1  C
XL = L
XC = 1  C
The unit of the inductive and capacitive reactances is also the Ohm ().
In terms of reactances, the impedance can be expressed as
(6.28)
Z R jX + R j XL+  – XC R2 XL – XC2 +  XL – XC  R –1 = = = tan
By a procedure similar to that of Chapter 2, we can show that impedances combine as resistances
do.
Example 6.9
For the circuit in Figure 6.11, find the current given that .
it vS t = 100cos100t – 30


Solution:
If we attempt to solve this problem in the timedomain directly, we will need to solve an inte-grodifferential
equation. But as we now know, a much easier solution is with the transformation
of the given circuit to a phasor circuit. Here, and thus
= = j100  0.1 = j10 
--------- j1
– ------- –jXC j 1
= = = – ----------------------------------------- = –j100
V j10  S
= = ------------------------------------- = 1.11– 30 – –86.82
and
Also,
and the phasor circuit is as shown in Figure 6.12.
Figure 6.12. Phasor circuit for Example 6.9
From the phasor circuit in Figure 6.12,
and
Therefore,
+ R  L
C +
+ 

vS t
i t
5  100 mH
100 F
vS t = 100cos100t – 30
 = 100 rad  s
jL jXL
1
jC
C
10 2  10 2 10–6 
VS = 100–30
+ 
+
+ 

5 
I j100 
Z 5 + j10 – j100 5 – j90 52 902 + –1–90  5 = = = tan = 90.14–86.82
I
VS
Z
------- 100–30
90.14–86.82
I = 1.1156.82it = 1.11cos100t + 56.82

Admittance
6.6 Admittance
Consider the t – domain
circuit in Figure 6.13 (a) and its equivalent phasor circuit shown in
Figure 6.13 (b).
+
v t
iS t iR t iL t iC t

+
R L C R L C
V
IS IR IC IL

iGt = Gvt
iCt = Cdv
------
dt
iL t   1L
t
--- vdt
–
= 
IG = GV
IC = jCV
IL
1
jL
= ---------V
iGt + iLt + iCt = iSt IR + IL + IC = IS
Gvt Cdv
------ 1L
+ +  = iSt
dt
t
--- vdt
–

Integrodifferential Equation
Very difficult to work with
GV jCV 1
+ + ---------V jL
= IS
Algebraic Equation

Much easier to work with
Figure 6.13. The and relationships in a parallel RLC circuit
The last equation of the phasor circuit may be written as
(6.29)
G 1
  + --------- + jCV = jL

IS
V
Dividing both sides of (6.29) by , we obtain the admittance, that is, by definition
(6.30)
Admit tance Y Phasor Current
= = = = = ---
--------------------------------------
Phasor Voltage
IS
V
----- G 1
+ --------- 1Z
+ jC jL
Y Z G
Here we observe that the admittance is the reciprocal of the impedance as conductance
is the reciprocal of the resistance R
.
Like conductance, the unit of admittance is the Siemens or mho  –1 
.
As with the impedance Z , we can express the admittance Y
as the sum of a real component and
an imaginary component as follows:

(6.31)
Y G j C 1
= +  
 – -------
L
–1
 2
+ C 1
Y G2 C 1
= tan
 – -------
L
   G
 – -------
L
Z Y
G C 1
– -------
L
ReY = G and ImY C 1
= – -------
L
Y B
BC
BL
B = BC + BL = C – 1  L
BC = C
BL = 1  L
BC BL –1  
Y
+  – BL G2 BC – BL2 + BC – BL  G –1 = = = tan
Y G jB + G j BC
and in polar form
(6.32)
Like the impedance , the admittance it is not a phasor; it is a complex quantity whose real part
is the conductance and the imaginary part is  that is,
(6.33)
The imaginary part of the admittance is called susceptance and it is denoted with the letter .
The two components of susceptance are the capacitive susceptance and the inductive suscep-tance
, that is,
(6.34)
(6.35)
(6.36)
The unit of the susceptances and is also the Siemens .
In terms of susceptances, the admittance can be expressed as
(6.37)
By a procedure similar to that of Chapter 2, we can show that admittances combine as conduc-tances
do.
Duality is a term meaning that there is a similarity in which some quantities are related to others.
The dual quantities we have encountered thus far are listed in Table 6.1.

Admittance
TABLE 6.1 Dual quantities
Series Parallel
Voltage Current
Resistance Conductance
Thevenin Norton
Inductance Capacitance
Reactance Susceptance
Impedance Admittance
Example 6.10
Consider the series and parallel networks shown in Figure 6.14. How should their real and imag-inary
terms be related so that they will be equivalent?
R L
G L
C
Z Y
Figure 6.14. Networks for Example 6.10
C
Solution:
For these circuits to be equivalent, their impedances or admittances must be equal. There-fore,
Y 1Z
---------------- R – jX
---------------- G + jB 1
--- 1
= = = = = = – -------------------
R + jX
R + jX
and equating reals and imaginaries we obtain
(6.38)
= ------------------- and B –X
G R
R2 X2 +
Relation (6.38) is worth memorizing.
------------------- j X
------------------- R
 ---------------- R – jX
R – jX
Example 6.11
Compute and for the network in Figure 6.15.
Z Y
R2 X2 +
R2 X2 +
R2 X2 +
= -------------------
R2 X2 +
Z Y

Solution:
Since this is a parallel network, it is easier to compute the admittance first. Thus,
= + --------- + jC = 4 – j2 + j5 = 4 + j3 = 536.9
Since the impedance is the reciprocal of admittance, it follows that
= = -------------------- = 0.2–36.9 = 0.16 – j0.12
Example 6.12
Compute and for the circuit shown below. Verify your answers with MATLAB.
Solution:
Let the given network be represented as shown in Figure 6.17 where ,
+ ----------------------------------------------- j5 11.1826.625.61–38.7
= = = + ----------------------------------------------------------------------------
, and
Then,
and
G L C
Z, Y
4  j5 –1 –1 –j2 –1
Y
Y G 1
jL
Z 1Y
---- 1
536.9
Z Y
C1
Z, Y
20 
j16 
j13 
R1 R2
L1
L2
C2
10 
j5 
j8 
Z1 = j13 – j8 = j5
Z2 = 10 + j5 Z3 = 20 – j16
Z Z1
Z2Z3
Z2 + Z3
+ ------------------ j5 10 + j520 – j16
10 + j5 + 20 – j16
31.95–20.1
= j5 + 8.968 = j5 + 8.87 + j1.25 = 8.87 + j6.25 = 10.8535.2

Admittance
Z, Y
Z1
Z2 Z3
Figure 6.17. Simplified network for Example 6.12
Y 1Z
--- 1
= = ------------------------------- = 0.092–35.2 = 0.0754 – j0.531
10.8535.2
Check with MATLAB:
z1=j*5; z2=10+j*5; z3=20j*16; z=z1+(z2*z3/(z2+z3)), y=1/z % Impedance z, Admittance y
z =
8.8737 + 6.2537i
y =
0.0753 -0.0531i
As we found out in Example 6.1, analyzing circuits with sinusoidal excitations when they contain
capacitors and/or inductors, using the procedure in that example is impractical. However, we
can use a Simulink / SimPowerSystems model to display sinusoidal voltages and currents in
branches of a circuit as illustrated in the Example 6.13 below.
Example 6.13
Create a Simulink / SimPowerSystems model to display the potential difference in the
circuit of Figure 6.18.
5 A peak AC
va vb
4 
1 F
2 
1 mH
8 
2 F
10 A peak AC
Figure 6.18. Circuit to be analyzed with a Simulink / SimPowerSystems model
Solution:
The model is shown in Figure 6.19, and the waveforms in Figure 6.20 where
va – vb

va = 5sint vb = –15sint vab = 20sint  = 2f = 0.4
Vs = AC Voltage Source, Is = AC Current Source, f= 0.2 Hz
VA VB
Figure 6.19. Simulink / SimPowerSystems model for the circuit in Example 6.13
Figure 6.20. Waveforms for
4
1
2 1
8
2
Resistances in Ohms
Capacitors in microfarads
Inductor in millihenries
VM = Voltage Mezasurement
Continuous
powergui
Vs
v +-
VM 2
v +-
VM 1
Subtract
Scope1
VA
Scope 3
VA , VB
VA-VB
Scope 2
VB
Is
Bus
Creator
va vb and va – vb

Summary
6.7 Summary
 Excitations or driving functions refer to the applied voltages and currents in electric circuits.
 A response is anything we define it as a response. Typically response is the voltage or current
in the “load” part of the circuit or any other part of it.
 If the excitation is a constant voltage or current, the response will also be some constant
value.
 If the excitation is a sinusoidal voltage or current, in general, the response will also be sinusoi-dal
with the same frequency but with different amplitude and phase.
 If the excitation is a timevarying function such as a sinusoid, inductors and capacitors do not
behave like short circuits and open circuits respectively as they do when the excitation is a
constant and steadystate conditions are reached. They behave entirely different.
 Circuit analysis in circuits where the excitation is a timevarying quantity such as a sinusoid is
difficult and time consuming and thus impractical in the .
 The complex excitation function greatly simplifies the procedure of analyzing such circuits
when excitation is a timevarying quantity such as a sinusoid.
 The procedure where the excitationresponse relation is simplified to amplitude and phase
relationship is known as timedomain to frequencydomain transformation.
 The procedure where the excitationresponse is put back to its sinusoidal form is known as
frequencydomain to timedomain transformation.
 For brevity, we denote the time domain as the , and the frequency domain as the
.
t – domain
t – domain
j – domain
 If a sinusoid is given in terms of the sine function, it is convenient to convert it to a cosine
function using the identity before con-verting
mt = Asint +  = Acost +  – 90
j – domain
it to the .
The to transformation procedure is as follows:
1. Express the given sinusoid as a cosine function
2. Express the cosine function as the real part of the complex excitation using Euler’s identity
3. Extract the magnitude and phase angle from it.
 The j – domain to t – domain
transformation procedure is as follows:
1. Convert the given phasor from polar to exponential form
2. Add the radian frequency multiplied by to the exponential form
 t

3. Extract the real part from it.
 The circuit analysis of circuits containing R , L , and C
devices, and which are excited by
sinusoidal sources, is considerably simplified with the use of phasor voltages and phasor cur-rents
V I
which we represent by the boldface capital letters and respectively.
j – domain
 Phasor quantities exist only in the
 In the j – domain
the current through a resistor is always inphase with the voltage across
that resistor
 In the the current through an inductor lags the voltage across that inductor by
j – domain
j – domain
j – domain Z
Impedance Z Phasor Voltage
= = = = + + ----------
-------------------------------------
Phasor Current
VS
I
------ R jL 1
jC
R
L – 1  C
ReZ = R and ImZ L 1
= – -------
C
= +  2
tan
–1  L 1
  R Z R2 L 1
 – -------
C
 – -------
C
Z X
XL
XC
X XL – XC L 1
= = – -------
C
90
 In the the current through a capacitor leads the voltage across that capacitor by
90
 In the the impedance is defined as
 Like resistance, the unit of impedance is the Ohm (
 Impedance is a complex quantity whose real part is the resistance , and the imaginary part is
 that is,
 In polar form the impedance is expressed as
 The imaginary part of the impedance is called reactance and it is denoted with the letter .
The two components of reactance are the inductive reactance and the capacitive reac-tance
, i.e.,
 The unit of the inductive and capacitive reactances is also the Ohm ().
 In the the admittance is defined as
j – domain Y

Summary
Admit tance Y Phasor Current
= = = = = ---
--------------------------------------
Phasor Voltage
IS
V
--------- jC + + 1Z
---- G 1
jL
Y Z G
R
 The admittance is the reciprocal of the impedance as conductance is the reciprocal of
the resistance .
 The unit of admittance is the siemens or mho .
 The admittance is a complex quantity whose real part is the conductance and the imag-inary
Y G
C 1
part is  that is,
–1  
– -------
L
ReY = G and ImY C 1
= – -------
L
Y
 The imaginary part of the admittance is called susceptance and it is denoted with the letter
B BC
. The two components of susceptance are the capacitive susceptance and the inductive
susceptance , that is,
BL
B BC – BL C 1
= = – ------
 In polar form the admittance is expressed as
L
–1
 2
+ C 1
Y G2 C 1
= tan
 – -------
L
   G
 – -------
L
BC BL –1  
 The unit of the susceptances and is also the siemens .
 Admittances combine as conductances do.
 In phasor circuit analysis, conductance is not necessarily the reciprocal of resistance, and sus-ceptance
is not the negative reciprocal of reactance. Whenever we deal with resistance and
reactance we must think of devices in series, and when we deal with conductance and suscep-tance
we must think of devices in parallel. However, the admittance is always the reciprocal of
the impedance
 The ratio V  I of the phasor voltage to the phasor current exists only in the j – domain
and
it is not the ratio vt  it in the t – domain . Although the ratio vt  it
could yield some
value, this value is not impedance. Similarly, the ratio it  vt
is not admittance.
 Duality is a term meaning that there is a similarity in which some quantities are related to oth-ers.

6.8 Exercises
Multiple Choice
1. Phasor voltages and phasor currents can be used in the if a circuit contains
A. independent and dependent sources with resistors only
B. independent and dependent sources with resistors and inductors only
C. independent and dependent sources with resistors and capacitors only
D. independent and dependent sources with resistors, inductors, and capacitors
2. If the excitation in a circuit is a single sinusoidal source with amplitude , radian frequency
, and phase angle , and the circuit contains resistors, inductors, and capacitors, all voltages
and all currents in that circuit will be of the same
A. amplitude but different radian frequency and different phase angle
B. radian frequency but different amplitude and different phase angle
C. phase angle but different amplitude and different radian frequency
D. amplitude same radian frequency and same phase angle
3. The sinusoid in the is expressed as
A.
B.
C.
D.
4. A series RLC circuit contains two voltage sources with values and
. We can transform this circuit to a phasor equivalent to find the cur-rent
by first replacing these with a single voltage source whose value is
A.
t – domain
A
 
A  
 A 
 A 
A  
vt = 120sint + 90 j – domain
V 120e jt + 90 =
V 120e jt =
V 120e j90 =
V 120e j0 =
v1t = 100cos10t + 45
v2t = 200sin5t–60
vt = v1t + v2t
vt = 300cos15t–15

Exercises
vt = 100cos5t + 105
t = 150cos7.5t–15
vt = 150cos7.5t + 15
B.
C.
D.
5.The equivalent impedance Zeq
of the network below is
A.
B.
C.
D.

j0.5 
j2 
2 
2 
Zeq
Figure 6.21. Network for Questions 5 and 6
1 + j1
1 – j1
–j1
2 + j0
Yeq
6. The equivalent admittance of the network in Figure 6.18 is
A.
B.
C.
4–j1.5
16
73
----- j6
+ -----
73
----- j2
12
37
+ -----
37
2 – j2
D.
7.The resistance of a coil is R = 1.5  and the inductance of that coil is L = 5.3 mH
. If a
current of flows through that coil and operates at the frequency of
it = 4cost A

f = 60Hz V
1053.1 V
60 V
5.3 10–3  90 V
6.845 V
R = 5 
 –jXC = –5  I = 80 A
t – domain
80cost
80sint
56.6cost – 45
56.6cost + 45
G 0.3 –1 =
 jBC j0.3 –1 =
V = 100 t – domain
, the phasor voltage across that coil is
A.
B.
C.
D.
8. A resistor with value is in series with a capacitor whose capacitive reactance at
some particular frequency is . A phasor current with value is
flowing through this series combination. The voltage across this series combina-tion
is
A.
B.
C.
D.
9. A conductance with value is in parallel with a capacitor whose capacitive sus-ceptance
at some particular frequency is . A phasor voltage with value
is applied across this parallel combination. The total current
through this parallel combination is
A.
3cost + j3sint
B.
3cost–j3sint
C.
5sint + 53.2
D.
5cost + 53.2
10. If the phasor I = je j  2 , then in the t – domain it
is
A.
cost +   2

Exercises
sint +   2
–cost
–sint
B.
C.
D.
Problems
1.Express the sinusoidal voltage waveform shown below as , that is, find
A ,  , and  . Answer:
vt = 2cos1000t + 36.1
vt = Acost + 
1.62 V
v (V)
2.2 ms 0.94 ms
0 t (ms)
it
2. The current through a device decays exponentially as shown by the waveform below, and
two values are known as indicated. Compute i1 , that is, the current at t = 1 ms
.
Answers: ,
it 50e–750t = mA 23.62 mA
i = 15.00 mA at 1.605 ms
I = 5.27 mA at 3.000 ms
i (mA)
0 1 2 3 4 5 6
I
t (ms)
f
3. At what frequency is the network shown below operating if it is known that
vS = 120cost V and i = 12cost – 36.9 A ? Answer:
f = 5.533 KHz

R L
8  C
4. In the circuit below, and the symbols V and A inside the circles
denote an AC voltmeter* and ammeter respectively. Assume that the ammeter has negligible
internal resistance. The variable capacitor C is adjusted until the voltmeter reads 25 V and the
ammeter reads 5 A. Find the value of the capacitor. Answer:
2 
5. In the circuit shown below, is it possible to adjust the variable resistor and the variable
capacitor so that and have the same numerical value regardless of the operating
frequency? If so, what are these values? Answer: Yes, if and
* Voltmeters and Ammeters are discussed in Chapter 8. For this exercise, it will suffice to say that these instru-ments
indicate the magnitude (absolute) values of voltage and current.
F
1 mH
vS i
vS = Vcos2000t +  V
C = 89.6 F
R
L
C
Other Part
of the
Network
vS
A
V 0.5 mH
R1
C ZIN YIN
C = 1 F R1 = 1 

Exercises
1 
R1 R2
ZIN
YIN L 1 H
C
Other Part
of the
Network
vS
BC
6. Consider the parallel RLC circuit below. As we know, the are the capacitive susceptance
and the inductive susceptance are functions of frequency, that is, , and
BL BC = 2fC
BL = 1  2fL
R L C
Y
1  1 mH 1 F
Find the frequency* at which the capacitive susceptance cancels the inductive susceptance,
that is, the frequency at which the admittance Y, generally computed from the relation
Y G2 BC BL2 is reduced to Y G= 2 = + – = G . Answer:
f  5 KHz
* This frequency is known as the resonance frequency. It is discussed in detail in Circuit Analysis II with MAT-LAB
Computing and Simulink / SimPowerSystems Modeling, ISBN 9781934404195.

6.9 Solutions to EndofChapter Exercises
Multiple Choice
1. E Phasors exist in the j – domain
only
2. B
3. D
4. E The voltage sources and operate at different frequencies. Therefore, to find the
v1t v1t
3 – j0.5
 = 2f = 2  60 = 377 r  s jXL jL j 377 5.3 10–3 = =    = j2 
Z = 1.5 + j2 = 2.553.13 V = ZI = 2.553.13  40 = 1053.13
t – axis T  2 = 2.2 + 0.94 = 3.14 ms
T = 6.28 ms f = 1  T = 103  6.28 = 103  2
 = 2f = 2  103  2 = 1000 r  s
   2 +  = 2.2 ms
1.62 V
v (V)
current we must apply superposition.
5. E This value is obtained with the MATLAB script z1=2+0.5j; z2=2*(2j)/(22j);
z=z1+z2
z = 3.00000.5000i
6. C
7. A ,
,
8. C
9. D
10. C
Problems
1. The crossings define half of the period T. Thus, , and
one period is . The frequency is . Then,
or
Next, we find the phase angle from the figure above observing that
2.2 ms 0.94 ms
0 t (ms)


Solutions to EndofChapter Exercises
or
– -- 2.2 10–3  s 2 rad
 2.2ms 2
------------------------------- 180
-------------   2
= = 6.28  10–3 s
 rad
– --
2.2  2  180
---------------------------------- 2
= – -- = 126.1 – 90 = 36.1
6.28
A t = 0 v = 1.62 V
Finally, we find the amplitude by observing that at , , that is,
or
Therefore,
v0 = 1.62 = Acos0 + 36.1
A 1.62
= --------------------- = 2 V
cos36.1
vt = 2cos1000t + 36.1
it Ae–t = mA ms
2. The decaying exponential has the form where the time is in and thus
for this problem we need to compute the values of and using the given values. Then,
and
i t = 1.605 ms 15 mA Ae 1.605 10 3 – –   = =
i t = 3.000 ms 5.27 mA Ae 3.000 10 3 – –   = =
Division of the first equation by the second yields
or
or
or
or
and thus
A 
Ae 1.605 10 3 – –  
Ae 3.000 10 3 – –  
--------------------------------------- 15 mA
= ---------------------
5.27 mA
e 1.605 10 3 – –    3.000 10–3 +    15
= ----------
5.27
e1.395 10 3 –   15
= ----------
5.27
15
5.27
ln  1.395 10–3=  
 ---------- 
 ln15  5.27 10 3
= --------------------------------------------- = 750
1.395

To find the value of we make use of the fact that . Then,
– = 
= = --------------------------------- = 1036.9
or
or
Therefore,
and
3. The equivalent phasor circuit is shown below.
In the , , , and
Then,
and
or
or
or
or
it Ae–750t = mA
A i t = 3 ms = 5.27 mA
– 0 5.27 =
Ae – 3
750  3  1A 5.27 10 3 – 
e–2.25 = ---------------------------
A = 0.050 A = 50mA
it 50e–750t = mA
i t = 1 ms 50e 750 10 3 – –  = = 23.62 mA
R L
8  C
F
vS
I
jL
j
C
–--------
j domain – VS 120 0 V  = I 12 36.9 A –  = jL j103
–j  C j106 = –  
Z
VS
I
------ 1200 V
12–36.9 A
Z 10 R2
L – 1  C2 = = +
R2 L – 1  C2 + = 100
82 L – 1  C2 + = 100
L – 1  C2 = 36
L – 1  C = 6

or
or
2 6L
– --- 1
– ------- = 0
LC
2 6 103
–   109 – = 0

Solving for and ignoring the negative value, we obtain
and
09
 + 36  106 + 4 
1 6 103
= ------------------------------------------------------------------------- = 34 765 r  s
f 
------ 34 765 r  s
= = ---------------------------- = 5 533 Hz = 5.533 KHz
2
jL = j34.765 –j  C = –j28.765
Z = R + jL – 1  C = 8 + j34.765 – 28.765 = 8 + j6 = 1036.9
Check: ,
and
2
2
I 1200
= ----------------------- = 12–36.9
1036.9
4. Since the instruments read absolute values, we are only need to be concerned the magnitudes
of the phasor voltage, phasor current, and impedance. Thus,
or
V Z I 25 R2 L – 1  C2 = = = +  5
V 2 252 R2 L – 1  C2 +   25  4 1 5 104
= = = +  25
100 25 250 10 4 – 
– ------------------------- 625 10 –8 
= + + ------------------------- = 625
C
C2 and after simplification we obtain
 2
– 
C
 – ------------------- 
500C2 250 10–4 C 625 10–8 + –  = 0
Using MATLAB, we obtain
p=[500 250*10^(4) 625*10^(8)]; r=roots(p)
and this yields
C = 89.6 F
The second root of this polynomial is negative and thus it is discarded.
5. We group the series devices as shown below.

Z R2 1  IN
YIN
C
1 H
Z1 Z2
R1
L
Z1 = R1 + j Z2 = 1 – j  C
ZIN
Z1  Z2
Z1 + Z2
= ------------------
= ---------------------------------------------------
R1 + j1 – j  C
R1 + j + 1 – j  C

YIN
1
ZIN
= --------
= ---------------------------------------------------
YIN = ZIN ZIN
R1 + j + 1 – j  C
R1 + j1 – j  C
YIN
R1 + j1 – j  C
R1 + j + 1 – j  C
---------------------------------------------------
R1 + j + 1 – j  C
R1 + j1 – j  C
= ---------------------------------------------------
R1 + j1 – j  C 2 R1 + j + 1 – j  C 2 =
R1 + j1 – j  C = R1 + j + 1 – j  C
R1 j
R1
C
-------- – j 1C
+ + --- R1 1 j  1
= + +  
 – -------- 
C
+   + j R 1 + 1 j  1
  j 
 R1
+ --- 
1C
R1
C
 – -------- 
= +  
 – -------- 
C
R1
1C
+ --- = R1 + 1 
R1
C
– --------  1
= – --------
C
C = 1 F
Thus , , and
and at any frequency
Therefore, if the condition is to hold for all frequencies, the right sides of
and must be equal, that is,
Equating reals and imaginaries we obtain
From the first equation above we obtain and by substitution of this value into the
second equation we obtain .
R1 = 1 

6.
R L C
Z, Y
1  1 mH 1 F
Y G2 BC – BL2 = + Y G= 2 = G BC – BL = 0
The admittance is reduced to when ,
or , or , from which , and with the given val-ues,
BC = BL 2fC = 1  2fL f = 1  2 LC
f 1
= -------------------------------------------  5000 Hz
2 10–3 10–6  
Z = Y = 1
and since the resistive branch is unity, at this frequency and the phase is zero
degrees.
The magnitude and phase at other frequencies can be plotted with a spreadsheet or MAT-LAB,
but it is easier with the Simulink / SimPowerSystems model shown in Figure 6.22.
RLC Branch Z
IM
IM = Impedance Measurement
Continuous
powergui
Figure 6.22. SimPowerSystems model for impedance measurement
After the simulation command is executed, we must click the Powergui block, and on the pop
up window we must select the Impedance vs Frequency Measurement option to display the
magnitude and phase of the impedance function shown in Figure 6.23.

Figure 6.23. Magnitude and Phase plots for the SimPowerSystems model in Figure 6.22
We observe that the maximum value of the impedance, i.e., , occurs at approximately
, and at this frequency the phase is zero degrees.
1 
5 KHz

Chapter 7
Phasor Circuit Analysis
his chapter begins with the application of nodal analysis, mesh analysis, superposition, and
Thevenin’s and Norton’s theorems in phasor circuits. Then, phasor diagrams are intro-duced,
and the inputoutput relationships for an RC lowpass filter and an RC highpass
T
filter are developed.
7.1 Nodal Analysis
The procedure of analyzing a phasor* circuit is the same as in Chapter 3, except that in this chap-ter
we will be using phasor quantities. The following example illustrates the procedure.
Example 7.1
Use nodal analysis to compute the phasor voltage for the circuit of Figure 7.1.
VAB = VA – VB
VA VB
4 
–j6 
2 
j3 
8 
–j3 
 
  
100 A
50 A
Solution:
As in Chapter 3, we choose a reference node as shown in Figure 7.2, and we write nodal equa-tions
at the other two nodes and . Also, for convenience, we designate the devices in series
A B
as as shown, and then we write the nodal equations in terms of these impedances.
Z1 Z2 and Z3
* A phasor is a rotating vector
Z1 = 4 – j6 = 7.211–56.3
Z2 = 2 + j3 = 3.60656.3
Z3 = 8 – j3 = 8.544–20.6

Chapter 7 Phasor Circuit Analysis
j3 
Z2
Figure 7.2. Nodal analysis for the circuit for Example 7.1
----- 1
 VA
 + ----- 
 VA
-------------------------------------------------------------------------VA
– -------------------------------VB = 50
----------------------------------VA – 0.277–56.3VB = 50
--------------------
----- 1
+  VB = –100
 + ----- 
72 Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems  Modeling
By application of KCL at ,
(7.1)
or
and by substitution for we obtain
(7.2)
Next, at :
(7.3)
In matrix form (7.1) and (7.3) are written as follows:
100 A
50 A
VB VA
4 
–j6 
2 
8 
–j3 
Z1 Z3
VA
VA
-------
Z1
VA – VB
Z2
+ -------------------- = 50
1
Z1
Z2
1
Z2
– -----VB = 50
Z1 + Z2
Z1Z2
------------------
1
Z2
– ------VB = 50
Z1 and Z2
4 – j6 + 2 + j3
7.211–56.33.60656.3
1
3.60656.3
6 – j3
26.00
--------------------VA – 0.277–56.3VB = 50
6.708–26.6
260
0.258–26.6VA – 0.277–56.3VB = 50
VB VB – VA
Z2
VB
Z3
+ ------- = –100
1
Z2
– -----VA 1
Z2
Z3

Nodal Analysis
(7.4)
----- 1
1
Z1
  1
 + ----- 
Z2
–-----
Z2
1
Z2
–----- 1
----- 1
 
 + ----- 
Z2
Z3
VA
VB
5
–10
=
We will follow a stepbystep procedure to solve these equations using Cramer’s rule, and we
will use MATLAB®* to verify the results.
We rewrite (7.3) as
(7.5)
1
Z2
– -----VA
Z2 + Z3
Z2Z3
+  VB = 10180
------------------
-------------------------------1
VA 2 + j3 + 8 – j3
– 3.60656.3
+ ----------------------------------------------------------------------------VB = 10180
3.60656.38.544–20.6
– 0.277–56.3VA 10
+ ----------------------------------VB = 10180
30.81035.7
– 0.277–56.3VA + 0.325–35.7VB = 10180
and thus with (7.2) and (7.5) the system of equations is
(7.6)
0.258–26.6VA – 0.277–56.3VB = 50
– 0.277–56.3VA + 0.325–35.7VB = 10180
We find and from
(7.7)
and
(7.8)
VA VB

The determinant is
Also,
VA
= ------
VB
= ------
 0.258–26.6 –0.277–56.3
–0.277–56.3 0.325–35.7
=
= 0.258–26.6  0.325–35.7 – 0.277–56.3  –0.277–56.3
= 0.084–62.3 – 0.077–112.6 – 0.039 – j0.074 – – 0.023 – j0.071
= 0.062 – j0.003 = 0.062–2.8
* If unfamiliar with MATLAB, please refer to Appendix A
D1

D2


D1
50 –0.277–56.3
10180 0.325–35.7
=
= 500.325–35.7 – 10180–0.277–56.3
= 1.625–35.7 + 2.770123.7 = 1.320 – j0.948 + – 1.537 + j2.305
= – 0.217 + j1.357 = 1.37499.1
D2
0.258–26.6 50
–0.277–56.3 10180
=
= 0.258–26.610180 – –0.277–56.350
= 2.580153.4 + 1.385–56.3 = – 2.307 + j1.155 + 0.769 – j1.152
= – 1.358 + j0.003 = 1.358179.9
VA
D1

------ 1.37499.1
= = ------------------------------- = 22.161101.9 = – 4.570 + j21.685
0.062–2.8
VB
D2

------ 1.358179.9
= = ---------------------------------- = 24.807–177.3 = – 24.780 – j1.169
0.062–2.8
VAB = VA – VB = – 4.570 + j21.685 – – 24.780 – j1.169
= 20.21 + j22.85 = 30.548.5
and
Therefore, by substitution into (7.7) and (7.8), we obtain
and
Finally,
Check with MATLAB:
z1=4j*6; z2=2+j*3; z3=8j*3; % Define z1, z2 and z3
Z=[1/z1+1/z2 1/z2; 1/z2 1/z2+1/z3]; % Elements of matrix Z
I=[5 10]'; % Column vector I
V=ZI; Va=V(1,1); Vb=V(2,1); Vab=VaVb; % Va = V(1), Vb = V(2) are also acceptable
% With fprintf only the real part of each parameter is processed so we will use disp
fprintf(' n'); disp('Va = '); disp(Va); disp('Vb = '); disp(Vb); disp('Vab = '); disp(Vab);
fprintf(' n');
Va = -4.1379 + 19.6552i
Vb = -22.4138 - 1.0345i
Vab = 18.2759 + 20.6897i
These values differ by about 10% from the values we obtained with Cramer’s rule where we
rounded the values to three decimal places. MATLAB performs calculations with accuracy of 15
decimal places, although it only displays four decimal places in the short (default) number display
format. Accordingly, we should accept the MATLAB values as more accurate.

Mesh Analysis
7.2 Mesh Analysis
Again, the procedure of analyzing a phasor circuit is the same as in Chapter 3 except that in this
chapter we will be using phasor quantities. The following example illustrates the procedure.
Example 7.2
For the circuit of Figure 7.3, use mesh analysis to find the voltage , that is, the voltage
across the current source.
4 
–j6 
2 
j3 
8 
–j3 
V10A
100
+
V10A

100 A
50 A
Solution:
As in the previous example, for convenience, we denote the passive devices in series as
, and we write mesh equations in terms of these impedances. The circuit then is as
Z1 Z2 and Z3
shown in Figure 7.4 with the mesh currents assigned in a clockwise direction.
We observe that the voltage across the current source is the same as the voltage across
the and series combination.
By inspection, for Mesh 1,
(7.9)
100
I1 = 50
+
V10A

100 A
50 A
Z2
4 
–j6 
2 
j3 
8 
Z1 Z3
I2
–j3 
I1
I3
Figure 7.4. Mesh analysis for the circuit of Example 7.2
8  –j3 
By application of KVL around Mesh 2,

(7.10)
–Z1I1 + Z1 + Z2 + Z3I2–Z3I3 = 0
–4 – j6I1 + 14 – j6I2 – 8 – j3I3 = 0
I3 = 100
1 0 0
–4 – j6 14 – j6 ––8 – j3
0 0 1
I1
I2
I3
5
0
10
=
100 A
V10A = Z3I2 – I3 = 8 – j37.586 – j1.035 – 10 = – 22.417 – j1.038
VB
C2
Also, by inspection for Mesh 3,
(7.11)
and in matrix form, (7.9), (7.10), and (7.11) are written as
(7.12)
We use MATLAB for the solution of 7.12.*
Z=[1 0 0; (4j*6) 14j*6 (8j*3); 0 0 1];
V=[5 0 10]';
I=ZV; i1=I(1); i2=I(2); i3=I(3); fprintf(' n');
disp('i1 = '); disp(i1); disp('i2 = '); disp(i2); disp('i3 = '); disp(i3); fprintf(' n');
i1 = 5 i2 = 7.5862 - 1.0345i i3 = 10
Therefore, the voltage across the current source is
We observe that this is the same value as that of the voltage in the previous example.
7.3 Application of Superposition Principle
As we know from Chapter 3, the superposition principle is most useful when a circuit contains
two or more independent voltage or current sources. The following example illustrates the appli-cation
of the superposition principle in phasor circuits.
Example 7.3
Use the superposition principle to find the phasor voltage across capacitor in the circuit of
Figure 7.5.
* As we experienced with Example 7.1, the computation of phasor voltages and currents becomes quite tedious. Accordingly,
in our subsequent discussion we will use MATLAB for the solution of simultaneous equations with complex coefficients.

Application of Superposition Principle
4 
–j6 
2 
j3 
8 
–j3 
C2
100 A
50 A
Solution:
Let the phasor voltage across due to the current source acting alone be denoted as
C2 50 A
, and that due to the current source as . Then,
V'C2 100 A V''C2
VC2 = V'C2 + V''C2
With the current source acting alone, the circuit reduces to that shown in Figure 7.6.
50 A
50 A
4 
–j6 
2 
j3 
8 
–j3 
V 'C2
C2
50 A
Figure 7.6. Circuit for Example 7.3 with the current source acting alone
By application of the current division expression, the current through is
I 'C2 C2
I 'C2
-------------------------------------------------------4 – j6
50 7.211–56.3
= = -------------------------------------50 = 2.367–33.1
4 – j6 + 2 + j3 + 8 – j3
15.232–23.2
The voltage across with the current source acting alone is
(7.13)
C2 50
V'C2 = –j32.367–33.1 = 3–902.367–33.1
= 7.102–123.1 = – 3.878 – j5.949
Next, with the 100 A
current source acting alone, the circuit reduces to that shown in Figure
7.7.

j3 
V''C2
Figure 7.7. Circuit for Example 7.3 with the current source acting alone
and by application of the current division expression, the current through is
= -------------------------------------------------------–100
The voltage across with the current source acting alone is
(7.14)
Addition of (7.13) with (7.14) yields
or
(7.15)
Check with MATLAB:
z1=4-6j; z2=2+3j; z3=8-3j; Is=5; i1=z1*Is/(z1+z2+z3);...
i1, magI1=abs(i1), phaseI1=angle(i1)*180/pi, v1=-3j*i1,...
magV1=abs(v1), phaseV1=angle(v1)*180/pi,...
Is2=-10; i2=(z1+z2)*Is2/(z1+z2+z3); magI2=abs(i2), phaseI2=angle(i2)*180/pi,...
v2=-3j*i2, magV2=abs(v2), phaseV1=angle(v2)*180/pi,...
vC=v1+v2, magvC=abs(vC), phasevC=angle(vC)*180/pi
i1 =
1.9828 - 1.2931i
magI1 =
2.3672
phaseI1 =
-33.1113
v1 =
-3.8793 - 5.9483i
100 A
4 
–j6 
2 
8 
–j3 
C2
100 A
I ''C2 C2
I ''C2
4 – j6 + 2 + j3
4 – j6 + 2 + j3 + 8 – j3
6.708–26.6
15.232–23.2
= -------------------------------------10180 = 4.404176.6
C2 100
V''C2 = –j34.404176.6 = 3–904.404176.6
= 13.21386.6 = 0.784 + j13.189
VC2 = V'C2 + V''C2 = – 3.878 – j5.949 + 0.784 + j13.189
VC2 = – 3.094 + j7.240 = 7.873113.1

Application of Superposition Principle
magV1 =
7.1015
phaseV1 =
-123.1113
magI2 =
4.4042
phaseI2 =
176.6335
v2 =
0.7759 +13.1897i
magV2 =
13.2125
phaseV1 =
86.6335
vC =
-3.1034 + 7.2414i
magvC =
7.8784
phasevC =
113.1986
The Simulink models for the computation of V'C2 and V''C2
are shown in Figures 7.8 and
7.9respectively.
Figure 7.8. Model for the computation of , Example 7.3
V'C2
The final step is to add V'C2 with V''C2
. This addition is performed with the model of Figure
7.10 where the models of Figures 7.8 and 7.9 have been converted to Subsystems 1 and 2 respec-tively.

Figure 7.9. Model for the computation of , Example 7.3
V''C2
Figure 7.10. Model for the addition of with , Example 7.3
V'C2 V''C2
The model in Figure 7.10 can now be used with the circuit of Figure 7.5 for any values of the cur-rent
sources and the impedances.
7.4 Thevenin’s and Norton’s Theorems
These two theorems also offer a very convenient method in analyzing phasor circuits as illustrated
by the following example.
Example 7.4
For the circuit of Figure 7.11, apply Thevenin’s theorem to compute IX
and then draw Norton’s
equivalent circuit.

Thevenin’s and Norton’s Theorems
1700 V
85  –j100 
50 
IX
100 
j200 
Solution:
With the resistor disconnected, the circuit reduces to that shown in Figure 7.12.
100 
1700 V
85  –j100 
V1 V2
j200  50 
100 
Figure 7.12. Circuit for Example 7.4 with the resistor disconnected
By application of the voltage division expression,
(7.16)
and
(7.17)
V1
-----------------------1700 20090
j200
85 + j200
= = -----------------------------1700 = 156.4623 = 144 + j61.13
V2
----------------------50
1700 50
= = ---------------------------------------1700 = 7663.4 = 34 + j68
50 – j100
Then, from (7.16) and (7.17),
(7.18)
217.3167
111.8–63.4
VTH = VOC = V12 = V1 – V2 = 144 + j61.13 – 34 + j68
VTH = 110 – j6.87 = 110.21–3.6
Next, we find the Thevenin equivalent impedance ZTH by shorting the 1700 V
voltage
source. The circuit then reduces to that shown in Figure 7.13.

85  –j100 
X Y
j200  50 
85  –j100 
X Y ZTH
j200  50 
j200 
Figure 7.13. Circuit for Example 7.4 with the voltage source shorted
We observe that the parallel combinations and are in series as shown in Fig-ure
X
ZTH
= + -------------------------------
7.14.
Figure 7.14. Network for the computation of for Example 7.4
From Figure 7.14,
and with MATLAB,
Zth=85*200j/(85+200j) + 50*(100j)/(50100j)
Zth =
1.1200e+002 + 1.0598e+001i
or
The Thevenin equivalent circuit is shown in Figure 7.15.
85 
50 
X
Y
–j100 
j200 || 85 50 || j100
j200  85
50 j100 
Y
ZTH
ZTH
85  j200
85 + j200
----------------------- 50  –j100
50 – j100
ZTH = 112.0 + j10.6 = 112.55.4 

Thevenin’s and Norton’s Theorems
ZTH
112 j10.6 
VTH 110.213.6
X
Y
110j6.87
Figure 7.15. Thevenin equivalent circuit for Example 7.4
With the resistor connected at XY, the circuit becomes as shown in Figure 7.16.
j10.6 
100
112 
VTH 110j6.87
X
IX
Y
IX
Figure 7.16. Simplified circuit for computation of in Example 7.4
100 
We find using MATLAB:
Vth=1106.87j; Zth=112+10.6j; Ix=Vth/(Zth+100);
fprintf(' n'); disp('Ix = '); disp(Ix); fprintf(' n');
Ix = 0.5160 - 0.0582i
that is,
(7.19)
IX
IX
VTH
= --------------------------------- = 0.516 – j0.058 = 0.519–6.4 A
ZTH + 100 
The same answer is found in Example C.18 of Appendix where we applied nodal analysis to
find .
Norton’s equivalent is obtained from Thevenin’s circuit by exchanging and its series
with in parallel with as shown in Figure 7.14. Thus,
and
C
IX
VTH ZTH
IN ZN
IN
VTH
ZTH
---------- 110.21–3.6
= = ---------------------------------- = 0.98–9 A
112.55.4
ZN = ZTH = 112.55.4 

IN
Figure 7.17. Norton equivalent circuit for Example 7.4
7.5 Phasor Analysis in Amplifier Circuits
Other circuits such as those who contain op amps and op amp equivalent circuits can be analyzed
using any of the above methods.
Example 7.5
Compute for the circuit in Figure 7.18 where .
2 
8 
iX t
Solution:
As a first step, we perform the , to transformation. Thus,
------- –j 1
= = -------------------------------------------------- = –j10
----- 10–6  
and
Also,
and the phasor circuit is shown in Figure 5.19.
ZN
iX t vint = 2cos30000t V
+

+

0.2 mH
10 
50 
4 
vC t
vint
10  3 F 5vC t
jXL jL j0.2 103
– = =   30  103 = j6
–jXC –j 1
C
30  103 10
3
VIN = 20

Phasor Analysis in Amplifier Circuits
50 
V2 V3

+

+

V1

2  10 
8 
j10 

20
5VC
VC
j6 
Figure 7.19. Phasor circuit for Example 7.5
At Node :
(7.20)
and since
+

VIN
V1 – 20
--------------------------
2
V1
8 + j6
+ + + ----------------------- = 0
--------------
-------------- 1
1
8 + j6
-------------- 8–j6
= = = – -----
8 + j6
the nodal equation of (7.20) simplifies to
(7.21)
At Node :
or
(7.22)
At Node :
IX
4 
V1 – VC
10
-------------------
V1 – 5VC
50
 ----------- 8–j6
8–j6
----- j3
----------- 4
100
50
50
----- j3
35
50
 V1
 – ----- 
50
15
– --VC = 10
VC – V1
10
-------------------
VC
–j10
+ ---------- = 0
1
10
----- j1
– -----V1 1
+  VC = 0
 + -----
10
10
V3 = 5VC
We use MATLAB to solve (7.21) and (7.22).
G=[35/50 j*3/50; 1/5 1/10+j*1/10]; I=[1 0]'; V=GI;
Ix=5*V(2,1)/4; % Multiply Vc by 5 and divide by 4 to obtain current Ix
magIx=abs(Ix); theta=angle(Ix)*180/pi; % Convert current Ix to polar form
fprintf(' n'); disp(' Ix = ' ); disp(Ix);...
fprintf('magIx = %4.2f A t', magIx); fprintf('theta = %4.2f deg t', theta);...
fprintf(' n'); fprintf(' n');
Ix = 2.1176 - 1.7546i magIx = 2.75 A theta = -39.64 deg
Therefore,
I = 2.75–39.6 it = 2.75cos30000t – 39.6

5 
j5 
5 

--------------------------
-----------------------
-----------------------
----- j3
 V1
 + ----- 
-- j15
–  Vout = 10
+ -----------------------
+ ----------------------- = 0
Example 7.6
Compute the phasor for the op amp circuit of Figure 7.20.
Solution:
We assign phasor voltages and as shown in Figure 7.21, and we apply KCL at these nodes,
while observing that
Figure 7.21. Application of KCL for the circuit of Example 7.6
At Node :
or
(7.23)
At Node ,
and thus,
or
Vout
4 
j5 
j10 
10 
Vout
Vin = 40 V
V1 V +
Vout V + =
4 
j10 
10 
V1
V +
Vin = 40 V 
Vout
V1 – 40
4
V1 – Vout
–j5
V1 – Vout
5
V1
–j10
+ + + ---------- = 0
9
20
10
15
 + -- 
V2 = V + = Vout
Vout
10
-----------
Vout – V1
5
Vout – V1
–j5

Phasor Diagrams
(7.24)
-- j15
15
– V1
 + -- 
----- j15
3
10
+  Vout = 0
 + -- 
Solving (7.23) and (7.24) with MATLAB we obtain:
format rat
G=[9/20+j*3/10 1/5j*1/5; 1/5j*1/5 3/10+j*1/5]; I=[1 0]'; V=GI;
fprintf(' n');disp(‘V1 = ’); disp(V(1,1)); disp(‘Vout = ’); disp(V(2,1));
format short
magV=abs(V(2,1)); thetaV=angle(V(2,1))*180/pi;
fprintf('magIx = %5.3f A t', magIx); fprintf('theta = %4.2f deg t', theta);...
fprintf(' n'); fprintf(' n')
V1 = 68/25 - 24/25i Vout = 56/25 - 8/25i
magIx = 2.750 A theta = -39.64 deg
Therefore,
(7.25)
Vout = 2.263–8.13
7.6 Phasor Diagrams
A phasor diagram is a sketch showing the magnitude and phase relationships among the phasor
voltages and currents in phasor circuits. The procedure is best illustrated with the examples
below.
Example 7.7
Compute and sketch all phasor quantities for the circuit of Figure 7.22.
VS
j3 
+  + 
VR VL
VC +

j5 
2 
I
Solution:
Since this is a series circuit, the phasor current I is common to all circuit devices. Therefore, we
assign to this phasor current the value I = 10
and use it as our reference as shown in the
phasor diagram of Figure 7.23 where:

VR = 2 10 = 20 V
VL = j3 10 = j3 = 390 V
VC = –j5 10 = –j5 = 5–90 V
VS = VR + VL + VC = 2–j2 = 2 2–45
Figure 7.23. Phasor diagram for the circuit of Example 7.7
Example 7.8
Compute and sketch all phasor quantities for the circuit of Figure 7.24.
Solution:
Since this is a parallel circuit, the phasor voltage V is common to all circuit devices. Therefore let
us assign this phasor voltage the value and use it as our reference phasor as shown in
the phasor diagram of Figure 7.25 where:
VR I = 10
VL
VC
VL+VC
VS=VR+(VL+VC)
IS
IC
10  j20  j10 
V
IR IL
+

V = 10
IR = 10  10 = 1000 mA
IL = 10  j20 = 10  2090 = 50–90 m
IC = 10  –j10 = 10  10–90 = 10090 mA
IC + IL = 5090 mA
IS = IR + IC + IL = 100 + j50 = 111.826.6

Phasor Diagrams
IS=IR+(IC+IL)
V = 10
IR
IC
IL
IC+IL
Figure 7.25. Phasor diagram for Example 7.8
We can draw a phasor diagram for other circuits that are neither series nor parallel by assigning
any phasor quantity as a reference.
Example 7.9
Compute and sketch all phasor voltages for the circuit of Figure 7.26. Then, use MATLAB to
plot these quantities in the .
t – domain
VS
j3 
2 
+ VR1  + VL 
VC
+

j5 
5 
+

IR2
VR2
Solution:
We will begin by selecting as our reference as shown on the phasor diagram of
Figure 7.27. Then,
and
IR2 = 10 A
VR2 = 5   IR2 = 5  10 = 50
VL = j3   IR2 = 390  10 = 390
VC = VL + VR2 = 50 + 390 = 5 + j3 = 5.8331
= = = =  
VR1 2   IR1 2IC + IR2 2
VC
–j5
  2 5.8331
------- + IR2
----------------------- + 50 
5–90
= 2.33121 + 100 = – 1.2 + j2 + 10 = 8.8 + j2 = 912.8
VS = VR1 + VC = 8.8 + j2 + 5 + j3 = 13.8 + j5 = 14.720

VC = 5.8331
VL = 390
VS = 14.720
VR2 = 50 IR2 = 10 A
Figure 7.27. Phasor diagram for Example 7.9
Now, we can transform these phasors into timedomain quantities and use MATLAB to plot
them. We will use the voltage source as a reference with the value , and we will apply
nodal analysis with node voltages V1, V2, and V3 assigned as shown in Figure 7.28.
2 
+ VR1 + VL
IR2
Figure 7.28. Circuit for Example 7.9 with the voltage source taken as reference
  1
-- – 12
------- 1
 + + ---- 
–----
–---- 1
---- 15
 
 + -- 
The node equations are shown below in matrix form.
The MATLAB script is as follows:
% Enter the nonzero values of the G matrix
G(1,1)=1;
G(2,1)=1/2;
G(2,2)=1/21/5j+1/3j;
G(2,3)=1/3j;
G(3,2)=1/3j;
G(3,3)=1/3j+1/5;
%
% Enter all values of the I matrix
I=[1 0 0]';
%
% Compute node voltages
V=GI;
%
VR1 = 912.8
VS = 10
VS
VC
+

j3 
j5 
5 
+

VR2
V V3 1 V2
10 V
1 0 0
12
-- 1
–j5
j3
j3
0 1
j3
j3
G
V1
V2
V3
V
1
0
0
I
=




Phasor Diagrams
VR1=V(1)V(2);
VL=V(2)V(3);
% Compute magnitudes and phase angles of voltages
magV1=abs(V(1)); magV2=abs(V(2)); magV3=abs(V(3));
phaseV1=angle(V(1))*180/pi; phaseV2=angle(V(2))*180/pi; phaseV3=angle(V(3))*180/pi;
magVR1=abs(VR1); phaseVR1=angle(VR1)*180/pi;
magVL=abs(VL); phaseVL=angle(VL)*180/pi;
%
% Denote radian frequency as w and plot wt for 0 to 2*pi range
wt=linspace(0,2*pi);
V1=magV1*cos(wtphaseV1);
VR1t=magVR1*cos(wtphaseVR1);
VLt=magVL*cos(wtphaseVL);
%
% Convert wt to degrees
deg=wt*180/pi;
%
% Print phasor voltages, magnitudes, and phase angles
fprintf(' n');
% With fprintf only the real part of each parameter is processed so we will use disp
disp('V1 = '); disp(V(1)); disp('V2 = '); disp(V(2)); disp('V3 = '); disp(V(3));
disp('VR1 = '); disp(VR1); disp('VL = '); disp(VL);
fprintf('magV1 = %4.2f V t', magV1); fprintf('magV2 = %4.2f V t', magV2);
fprintf('magV3 = %4.2f V', magV3); fprintf(' n'); fprintf(' n');
fprintf('phaseV1 = %4.2f deg t', phaseV1);
fprintf('phaseV2 = %4.2f deg t', phaseV2); fprintf('phaseV3 = %4.2f deg', phaseV3);
fprintf('magVR1 = %4.2f V t', magVR1); fprintf('phaseVR1 = %4.2f deg ', phaseVR1);
fprintf('magVL = %4.2f V t', abs(VL)); fprintf('phaseVL = %4.2f deg ', phaseVL);
fprintf(' n');
%
plot(deg,V1,deg,V2,deg,V3,deg,VR1t,deg,VLt)
fprintf(' n');
V1 = 1
V2 = 0.7503 - 0.1296i
V3 = 0.4945 - 0.4263i
VR1 = 0.2497 + 0.1296i
VL = 0.2558 + 0.2967i
magV1 = 1.00 V magV2 = 0.76 V magV3 = 0.65 V
phaseV1 = 0.00 deg phaseV2 = -9.80 deg phaseV3 = -40.76 deg
magVR1 = 0.28 V phaseVR1 = 27.43 deg
magVL = 0.39 V phaseVL = 49.24 deg
and with these values we have
vSt = v1t = cost v2t = 0.76cost – 9.8 v3t = 0.65cost – 40.8

1
0 .8
0 .6
0 .4
0 .2
0
-0 .2
-0 .4
-0 .6
-0 .8
These are plotted with MATLAB as shown in Figure 7.29.
Figure 7.29. The plots for Example 7.9
7.7 Electric Filters
The characteristics of electric filters were introduced in Chapter 4 but are repeated below for con-venience.
Analog filters are defined over a continuous range of frequencies. They are classified as lowpass,
highpass, bandpass and bandelimination (stopband). Another, less frequently mentioned filter,
is the allpass or phase shift filter. It has a constant amplitude response but is phase varies with fre-quency.
This is discussed in Signals and Systems with MATLAB Computing and Simulink Modeling,
ISBN 9781934404119.
The ideal amplitude characteristics of each are shown in Figure 7.30. The ideal characteristics are
not physically realizable; we will see that practical filters can be designed to approximate these
characteristics. In this section we will derive the passive RC low and highpass filter characteris-tics
and those of an active lowpass filter using phasor analysis.
A digital filter, in general, is a computational process, or algorithm that converts one sequence of
numbers representing the input signal into another sequence representing the output signal.
Accordingly, a digital filter can perform functions as differentiation, integration, estimation, and,
of course, like an analog filter, it can filter out unwanted bands of frequency. Digital filters are dis-cussed
in Signals and Systems with MATLAB Computing and Simulink Modeling, ISBN 9781
934404119.
vR1t = 0.28cost + 27.4 vLt = 0.39cost + 49.2
0 50 100 150 200 250 300 350 400
-1
v1t = vSt
v2t
v3t
vR1t
vLt
t – domain

Basic Analog Filters
---------- Vout
----------
Vin
 
PASS
BAND
STOP
BAND
(CUTOFF)
PASS
BAND
STOP
BAND
c
 
c
Ideal lowpass filter Ideal high pass filter
STOP
BAND
PASS
BAND
STOP
BAND
PASS
BAND
PASS
STOP
BAND BAND

Vout
----------
Vin
 
1 2 1 2
Ideal band pass Filter Ideal band  elimination filter
Figure 7.30. Amplitude characteristics of the types of filters

Vout
Vin
Vout
----------
Vin
7.8 Basic Analog Filters
An analog filter can also be classified as passive or active. Passive filters consist of passive devices
such as resistors, capacitors and inductors. Active filters are, generally, operational amplifiers
with resistors and capacitors connected to them externally. We can find out whether a filter,
passive or active, is a lowpass, highpass, etc., from its the frequency response that can be
obtained from its transfer function. The procedure is illustrated with the examples that follow.
Example 7.10
Derive expressions for the magnitude and phase responses of the series RC network of Figure
7.31, and sketch their characteristics.
+
+

Vin C Vout

R
Figure 7.31. Series RC network for Example 7.10
Solution:
By the voltage division expression,

----------------------------------------------------------------------- 1
----------------------- 1
= = = ---------------------------------–atanRC
= ---------- 1
= ---------------------------------
= arg = arg  = –atanRC
and denoting the ratio as , we obtain
(7.26)
The magnitude of (7.26) is
(7.27)
and the phase angle , also known as the argument, is
(7.28)
We can obtain a quick sketch for the magnitude versus by evaluating (7.27) at
, , and . Thus,
as ,
for ,
and as ,
The magnitude, indicated as versus radian frequency for several values of is shown in
Figure 7.32 where, for convenience, we have let . The plot shows that this circuit is an
approximation, although not a good one, to the amplitude characteristics of a lowpass filter.
We can also obtain a quick sketch for the phase angle, i.e., versus by evaluat-ing
of (11.3) at , , , and . Thus,
as ,
for ,
for ,
as ,
and as ,
Vout
1  jC
R + 1  jC
= ---------------------------Vin
Vout  Vin Gj
Gj
Vout
Vin
= ---------- 1
1 + jRC
1 2R2C2  + atanRC
1 2R2C2 +
Gj Vout
Vin
1 2R2C2 +

 Gj
Vout
Vin
----------
Gj 
 = 0  = 1  RC   
  0 Gj  1
 = 1  RC Gj = 1  2 = 0.707
   Gj  0
Gj 
RC = 1
 = argGj 
 = 0  = 1  RC  = –1  RC   –   
  0   –atan0  0
 = 1  RC  = –atan1 = –45
 = –1  RC  = –atan–1 = 45
  –  = –atan– = 90
    = –atan = –90

Figure 7.32. Amplitude characteristics of a series RC lowpass filter
Figure 7.33 shows the phase characteristic of an RC lowpass filter where, again for conve-nience,
we have let .
RC = 1
Figure 7.33. Phase characteristics of a series RC lowpass filter

Example 7.11
The network of Figure 7.31 is also a series RC circuit, where the positions of the resistor and
capacitor have been interchanged. Derive expressions for the magnitude and phase responses,
and sketch their characteristics.
+

Figure 7.34. RC network for Example 7.11
= ---------------------------Vin
------------------------ jRC 2R2C2 +
----------------------------------------- RCj + RC
= = = = --------------------------------------
--------------------------------------------- 1
--------------------------------------------------------------------------------------- 1
= = atan 
 ------------ 
= ---------------------------------------------
= = atan 
 ------------ 
Solution:
or
(7.29)
The magnitude of (7.29) is
(7.30)
and the phase angle or argument, is
(7.31)
We can obtain a quick sketch for the magnitude versus by evaluating (7.30) at ,
, and . Thus,
as ,
for ,
and as ,
Figure 7.35 shows versus radian frequency for several values of where . The
plot shows that this circuit is an approximation, although not a good one, to the amplitude char-acteristics
of a highpass filter.
+

C
Vin Vout
Vout
R
R + 1  jC
Gj
Vout
Vin
---------- jRC
1 + jRC
1 2R2C2 +
1 2R2C2 +
RC 1 2R2C2 + atan1  RC
1 2R2C2 +
1 + 1  2R2C2
RC
Gj 1
1 + 1  2R2C2
 argGj 1
RC
Gj   = 0
 = 1  RC   
  0 Gj  0
 = 1  RC Gj = 1  2 = 0.707
   Gj  1
Gj  RC = 1

Figure 7.35. Amplitude characteristics of a series RC highpass filter
 = argGj 
We can also obtain a quick sketch for the phase angle, i.e., versus , by eval-uating
(7.31) at , , , , and . Thus,
 = 0  = 1  RC  = –1  RC   –   
as   0 ,
  –atan0  0
for  = 1  RC ,
 = –atan1 = –45
for  = –1  RC ,
 = –atan–1 = 45
as   – ,
 = –atan– = 90
and as    ,
 = –atan = –90
Figure 7.36 shows the phase angle versus radian frequency for several values of , where
.
 
RC = 1

Figure 7.36. Phase characteristics of an RC highpass filter
We should remember that the RC lowpass filter in Figure 7.31 and the RC highpass filter in
Figure 7.34 behave as filters only when the excitation (input voltage) is sinusoidal at some fre-quency.
If the excitation is any input, the RC network in Figure 7.28 behaves as an integrator
provided that , while the RC network in Figure 7.31 behaves as a differentiator pro-vided
that . The proofs are left as exercises for the reader at the end of this chapter.
7.9 Active Filter Analysis
We can analyze active filters, such as those we discussed in Chapter 4, using phasor circuit analy-sis.
Example 7.12
Compute the approximate cutoff frequency of the circuit of Figure 7.37 which is known as a
Multiple Feed Back (MFB) active lowpass filter.
Solution:
We assign two nodes as shown in Figure 7.38, and we write the phasor circuit nodal equations as
follows:
vOUT « vIN
vOUT « vIN

Active Filter Analysis
40 k
R3
C2
50 k
R1
R2
200 k
vin vout
25 nF
10 nF
C1
Figure 7.37. Lowpass filter for Example 7.12
40 k
R3
C2
v1 v2
50 k
R1
R2
200 k 
vin vout
25 nF
10 nF

C1
Figure 7.38. Circuit for nodal analysis, Example 7.12
At Node :
(7.32)
At node :
(7.33)
v1 – vin
-------------------
R1
v1
------------------------
+ + + ----------------- = 0
1  jC1
v1 – vout
---------------------
R2
v1 – v2
R3
v2 – v1
R3
-----------------
C2
= ------------------------
1  jC2
and since (virtual ground), relation (7.33) reduces to
(7.34)
v2 = 0
v1 = –jR3C2vout
and by substitution of (7.34) into (7.32), rearranging, and collecting like terms, we obtain:
(7.35)
or
(7.36)
 ------ 1
 R –j3C2 1
1
R1
------ 1
 + + ------ + jC1 
R2
R3
– ------ vout
R2
1
R1
= ------vin
vout
vin
---------- 1
= ------------------------------------------------------------------------------------------------------------------
R1
 ------ 1
 R –j3C2 1
1
R1
------ 1
 + + ------ + jC1 
R2
R3
– ------
R2
By substitution of given values of resistors and capacitors, we obtain

vout
vin
---------- 1
= ----------------------------------------------------------------------------------------------------------------------------------------------------------------------
 1
  j5  10 4 10–8 –   1
–  +   
2 105
--------------------- j2.5 108
20  10 3
– ------------------

4 104
Gj
vout
vin
---------- –1
= = -------------------------------------------------------------------------
2.5  10–62 j5 10–3 –   + 5
700 rad  s 0.2  0.707 = 0.141
Magnitude Vout/Vin vs. Radian Frequency
or
(7.37)
and now we can use MATLAB to find and plot the magnitude of (7.37) with the following script.
w=1:10:10000; Gjw=1./(2.5.*10.^(6).*w.^25.*j.*10.^(3).*w+5);
semilogx(w,abs(Gjw)); grid; hold on
xlabel('Radian Frequency w'); ylabel('|Vout/Vin|');
title('Magnitude Vout/Vin vs. Radian Frequency')
The plot is shown in Figure 7.39 where we see that the cutoff frequency occurs at about
. We observe that the halfpower point for this plot is .
Figure 7.39. Plot for the magnitude of the lowpass filter circuit of Example 7.12

Summary
7.10 Summary
 In Chapter 3 we were concerned with constant voltage and constant current sources, resis-tances
and conductances. In this chapter we were concerned with alternating voltage and
alternating current sources, impedances, and admittances.
 Nodal analysis, mesh analysis, the principle of superposition, Thevenin’s theorem, and Nor-ton’s
theorem can also be applied to phasor circuits.
 The use of complex numbers make the phasor circuit analysis much easier.
 MATLAB can be used very effectively to perform the computations since it does not require
any special procedures for manipulation of complex numbers.
 Whenever a branch in a circuit contains two or more devices in series or two or more devices
in parallel, it is highly recommended that they are grouped and denoted as z1 , z2
, and so on
before writing nodal or mesh equations.
 Phasor diagrams are sketches that show the magnitude and phase relationships among sev-eral
phasor voltages and currents. When constructing a phasor diagram, the first step is to
select one phasor as a reference, usually with zero phase angle, and all other phasors must be
drawn with the correct relative angles.
 The RC lowpass and RC highpass filters are rudimentary types of filters and are not used in
practice. They serve as a good introduction to electric filters.

V
2 + j0 V
1 + j0 V
1–j0 V
1 + j V
IS +
V j 

10 A
1 
1  1 
7.11 Exercises
Multiple Choice
1. In the circuit below the phasor voltage is
A.
B.
C.
D.
2. In the circuit below the phasor current is
A.
B.
C.
D.
3. In the circuit below the voltage across the capacitor is
A.
B.
C.
1 
j0.5 
I
0 + j2 A
0 – j2 A
1 + j0 A
2 + j2 A
VS
20 V j1 
j1 
I
C2
8 104
–  sin2000t + 90 V
50cos2000t – 45 V
50cos2000t + 45 V

Exercises
D.
50cos2000t + 90 V
R
4 
L1
3 mH L2
500 F 2 mH
8sin2000t + 90
vSt
C2
C1
100 F
4. In the circuit below the current through the capacitor is
A.
B.
C.
D.
iCt
4sin2000t
4sin2000t + 180
32cos2000t–45
32cos2000t + 90
1 
iC
iSt = 4cos2000t
500 F 0.5 mH
iSt
5. The Thevenin equivalent voltage at terminals A and B in the circuit below is
A.
B.
C.
D.
VTH
10–90 V
10–53.13 V
1053.13 V
10–45 V
VS
j5 
100 V
4  j2 
A
B

6. The Thevenin equivalent impedance ZTH
at terminals A and B in the circuit above is
A.
2 + j4 
B.
4 + j2 
C.
4–j2 
D.
–j5 
7. In the circuit below the phasor voltage VC
is
A.
5–90 V
B.
5–45 V
C.
4–53.1 V
D.
453.1 V
VS
+ IX 4 
4IX V 
j4 
+
4 
VX 5 
A.
B.
C.
D.
j4 
200 V j3 
+

VR5 
20 + j0 V
0 + j20 V
20 + j20 V
80–j80 V
IS
40 A 
2VX A
+

VR5 

Exercises
A.
B.
C.
D.
VOUT 2
2 + j0 V
4 + j0 V
4–j0 V
1 + j1 V
10 

j5  –j5 
VOUT 1
VIN
10
10. In the circuit below the voltage is
A.
B.
C.
D.
10 
VOUT 2
t – domain vABt
1.89cost + 45 V
0.53cost–45 V
2cost V
0.5cost + 53.1 V
A
VS j  j2 
2 
20 V
2 
B
+


Problems
1. For the circuit below . Compute and .
iS t = 2cos1000t A vABt iC t
20 mH
2.Write nodal equations and use MATLAB to compute for the circuit below given that
2 
.
3.Write mesh equations and use MATLAB to compute for the circuit below given that
.
4. For the circuit below it is given that
and
. Use superposition to find
8 
A
6 
5 
7 
2 
B
iC t
iS t
1000  6 F
iC t
vS t = 12cos1000t + 45 V
4 
10  5 
20  5 mH
100 F
+

vS t
iC t
iL t
vS t = 100cos10000t + 60 V
4 
10  5 
20  2 mH
2 
10 F
+

vS t
iL t
vS1t = 40cos5000t + 60 V
vS2t = 60sin5000t + 60 V vC t

Exercises
L1 R1 R2 L2
10  25  +
2 mH 5 mH
+

vS1 t vC t vS2 t

+

20 F
5. For the circuit below find if , , and
vC t vS1 = 15 V vS2t = 20cos1000t V
. Plot using MATLAB or Excel.
iS t = 4cos2000t A vC t
1 mH 2 mH
+
 5 
10 
+
500 F
+

vC t
vS1
vS2 t
iS t
6. For the circuit below find the value of which will receive maximum power.
ZLD
vS
+

ZS
ZLD
7. For the circuit below, to what value should the load impedance ZLD
be adjusted so that it will
receive maximum power from the voltage source?
4 
ZLD
10  5 
+
 1700
20 
j5 j10 8. For the circuit below draw a phasor diagram that shows the voltage and current in each
branch.

9. For the op amp circuit below . Find .
R1
1 K
C
R2
3 K
10. Prove that the RC network below, for any input it behaves as an integrator if , that
+
= ----vIN
+
is, show that
11. Prove that the RC network below, for any input it behaves as a differentiator if ,
that is, show that
4  10  5 
20 
+

VS j5 j10
vint = 3cos1000t V voutt
vint vout t
vOUT « vIN
vOUT
1
RC
 –--------vIN
R
C
vIN

+

vOUT
vOUT « vIN
vOUT –RC d
dt
R
C
vIN


+

vOUT

Answers to EndofChapter Exercises
7.12 Answers to EndofChapter Exercises
Multiple Choice
1. E
IS +
VL
j0.5 
V j 

10 A
1 
VC
where and is found from the nodal equation
V = VL + VC VL = 10  j1  2 = j1  2 V VC
VC
---------- 1
1 – j
1
1 + j
-- j12
---------- 12
 ---------- 1 – j
= = = – -- V
or or .
-------
VCj
+ ------- = 1 + j0 1 + jVC = 1 VC
–
Therefore,
2. C
1 – j
2
V = j1  2 + 1  2 – j1  2 = 1  2 + j0 V
VS
1 
1  1 
I
20 V j1 
j1 
Denoting the resistor in series with the voltage source as , the resistor in series with the
capacitor as , and the resistor in series with the capacitor as , the equivalent impedance
is
and
3. B
z1
z2 z3
+ ---------------- 1 1 – j11 + j1
-------------------------------------- + 1 22
= = = + -- = 2 + j0
1 – j1 + 1 + j1
------ 2 + j0
= = -------------- = 1 + j0 A
2 + j0
, , , ,
Zeq z1
z2  z3
z2 + z3
I
VS
Z
8sin2000t + 90 = 8cos2000t80 V jL1 = j6 jL2 = j4 –j  C1 = –j1
–j  C1 = –j1
and the phasor equivalent circuit is shown below.

VS
L1
80 V j5 
-------------- 4 – j4
-------------- 8 + j0
 ------------- 32 – j32
, , and
= = = = ------------------- = 1 – j1
= = ------------------------- = j1  40 = 190  40 = 490 A
-------------------------  100 5–90  100
------------------------------------------ 50–90
= = = = ------------------------ = 10–53.1 V
thus
4. D
, , , , and the phasor
equivalent circuit is shown below.
Denoting the parallel combination of the conductance and inductance as and
using the current division expression for admittances we obtain
and thus
5. B
By the voltage division expression
4  j6 
C2
C1
L2
j1  j4 
I
+

R
Z = 4 + j6 – j1 + j4 – j5 = 4 + j4 I
VS
Z
------ 8 + j0
4 + j4
4 + j4
4 – j4
32
VC2
= –j5  1 – j = 5 – j5 = 50–45 50cos2000t – 45 V
4cos2000t40 G 1  R 1 –1 = = jC j1 –1 = –j  L j1 –1 = –
IS
40 A
IC
1 –1 j1  –1 j1 –1 –
Y1 = 1 – j1
IC
jC
jC + Y1
-----------------------  IS j1
j1 + 1 – j1
iSt = 4cos2000t + 90 A
VS
j5 
100 V
4  j2 
A
B
VTH VAB
–j5
4 + j2 – j5
4 – j3
5–36.9

6. C
We short the voltage source and looking to the left of points A and B we observe that the
capacitor is in parallel with the series combination of the resistance and inductance. Thus,
7. D
------------------- 4 + j3
 -------------- 100 – j50
= = = = ---------------------- = 4 – j2
4 + j3
+

4IX V 
,
and
8. E
and
ZTH
–j54 + j2
4 + j2 – j5
------------------- 10 – j20
-------------------------------- 10 – j20
4 – j3
4 – j3
25
VS
j4 
+ IX 4 
200 V j3 
IX
---------------- 200
200
4 + j3
= = -------------------- = 4–36.9 4IX = 16–36.9
536.9
IC
4IX
–j4
-------- 16–36.9
= = --------------------------- = 453.1
4–90
IS
j4 
+
4 
VX 5 
40 A 
2VX A
+

VR5 
VX
4
4 + j4
----------------------- 64  32
--------------  40  j4 6490
= = = ----------------------45 = 2 3245
3245
32
VR5 
------ j 2
2VX  5 20  3245 20 32 2
= = =   = 80 + j80
 + ------
2
2

------------------------------
 VAB = 290
 + + -------------- 
---------------------------- 290
------------------------------------------------ 290
= = = -----------------------
9. B
and
10. A
We write the nodal equation at Node A for as
or and in the
10 

10 
VOUT 1
VIN
10
VOUT 2
j5  –j5 
VOUT 1
10
j5
= –-----  10 = j2  10 = 290  10 = 290
VOUT 2
10
–j5
= –-------  VOUT 1 = –j2  20 = 2–90  290 = 40 = 4 + j0
VS j  j2 
2 
20 V
2 
A
B
+

VAB
VAB – 20
–j
VAB
2
----------
VAB
2 + j2
+ + -------------- = 0
12
-- j 1
2 + j2
VAB
290
1  2 + j + 1  4 – j  4
3  4 + j3  4
1.0645
VAB = 1.8945 t – domain vABt = 1.89cost + 45

Problems
1. We transform the current source and its parallel resistance to a voltage source series resis-tance,
we combine the series resistors, and we draw the phasor circuit below.
A
6 
5 
+

IC
–j6 
z1
j20 
C
15  8 
VS
100 V
B
z2
z3
VS 2 0  5  10 0 V  = = jL j103
20 10–3 =   = j20 
For this phasor circuit, , and
= –    103  6  10–6  = –j6 , z1 = 5  , z2 = 15 + j20  , and
z3 = 8–j6 
j – C  j 103
We observe that and . At Node A,
and
VA = VAB = VAC + VCB = VAC + 100 V VB = 0
VA – VB
z2
--------------------
VA – 100
+ ------------------------------
+ -------------------- = 0
z1
VA – VB
z3
---- 1
1
z1
---- 1
 VA
 + + ---- 
z2
z3
100
z1
= ----------------
-------------------- 1
-- 1
15
 VA
 + + ----------- 
15 + j20
8–j6
10 0 5
= ---------------- = 20
VA
------------------------------------------------------------------------ 20
20
= = -----------------------------------------------------------------------------
----------------------- 1
0.2 1
+ + ---------------------------
2553.1
10–36.9
0.2 + 0.04–53.1 + 0.136.9
20
= -----------------------------------------------------------------------------------------------------------------------------------------------------------------
0.2 + 0.04cos53.1 – j0.04sin53.1 + 0.1cos36.9 + j0.1sin36.9
------------------------------------------------------------------------------------------------------------------------------ 20
20
= = -----------------------------------
0.2 + 0.04  0.6– j0.04  0.8 + 0.1  0.8 + j0.1  0.6
20
= ------------------------------- = 6.55–5.26
0.3055.26
Then, in the .
Also,
0.304 + j0.028
t – domain vABt = 6.55cos1000 + 5.26

= = ------------------------------- = 0.65531.7
2 
z1
V2 V V1 V3 S
-------------------
---- 1
---- 1
---- 1
 V1
 + + + ---- 
– ----V2 1
– ----V3 1
= ----VS
-------------------
1
– +  ---- ---- 1
1
V2  + + ---- 
– ----V3 = 0
-------------------
---- 1
---- 1
– +  V3 = 0
 + + ---- 
and
Check with MATLAB:
z1=5; z2=15+20j; z3=86j; VA=(10+0j)/(z1*(1/z1+1/z2+1/z3)); fprintf(' n');...
fprintf('magVA = %5.2f V t',abs(VA));...
fprintf('phaseVA = %5.2f deg t',angle(VA)*180/pi); fprintf(' n'); fprintf(' n');
magVA = 6.55 V phaseVA = -5.26 deg
2. The equivalent phasor circuit is shown below where and
Node :
or
Node :
or
Node :
or
and in matrix form
IC
VA
z3
------- 6.55–5.26
10–36.9
iCt = 0.655cos1000 + 31.7
jL j103
5 103
– =   = j5
j – C  j 103
10–4 = –     = –j10
4 
10  5 
20 
+

1245
z2
z3
z4
z5
z6
IC
z7
j5  –j10 
V1 V1 – VS
z1
V1 – V2
z3
-------------------
V1
z2
------
V1 – V3
z7
+ + + -------------------
1
z1
z2
z3
z7
1
z3
z7
z1
V2 V2 – V1
z3
V2
z4
------
V2 – V3
z5
+ + ------------------- = 0
1
z3
----V1
1
z3
z4
z5
z5
V3 V3 – V2
z5
V3 – -------------------
V1
z7
V2 – V3
z6
+ + ------------------- = 0
1
z7
----V1
1
z5
– ----V2 1
z5
z6
z7

1
z1
  1
---- 1
---- 1
 + + + ---- 
z2
---- 1
z3
z7
–---- 1
z3
–----
z7
1
z3
–---- 1
---- 1
  1
 + + ---- 
z3
---- 1
z4
z5
–----
z5
1
z7
–---- 1
–---- 1
z5
---- 1
 
 + + ---- 
z5
---- 1
z6
z7
V1
V2
V3

1
z1
----VS
0
0
=
3
04
0Shown below is the MATLAB script to solve this system of equations.
Vs=12*(cos(pi/4)+j*sin(pi/4)); % Express Vs in rectangular form
z1=4; z2=20; z3=10; z4=5j; z5=5; z6=10j; z7=2;...
Y=[1/z1+1/z2+1/z3+1/z7 1/z3 1/z7;...
1/z3 1/z3+1/z4+1/z5 1/z5;...
1/z7 1/z5 1/z5+1/z6+1/z7];...
I=[Vs/z1 0 0]'; V=YI; Ic=V(3)/z6;...
magIc=abs(Ic); phaseIc=angle(Ic)*180/pi;...
disp('V1='); disp(V(1)); disp('V2='); disp(V(2));...
disp('V3='); disp(V(3)); disp('Ic='); disp(Ic);...
format bank % Display magnitude and angle values with two decimal places
disp('magIc='); disp(magIc); disp('phaseIc='); disp(phaseIc);...
fprintf(' n');
V1 = 5.9950 - 4.8789i
V2 = 5.9658 - 0.5960i
V3 = 5.3552 - 4.4203i
Ic = 0.4420 + 0.5355i
magIc = 0.69
phaseIc = 50.46
Therefore,
IC = 0.6950.46iCt = 0.69cos1000t + 50.46 A
3. The equivalent phasor circuit is shown below where jL = j1 2  1– = j20
and
 10 10–6 = –     = –j10
j – C  j 104

4 
2 
z7
z1
I2
+
z3
10  IL
z5
5 
z2
 –j10 
20 
z4
I1
VS = 10060 j20 
z6
I3 I4
I1 z1 + z2I1 – z2I3 = VS
I2 z1 + z2 + z7I2 – z3I3 – z5I4 = 0
I3
– z2I1–z3I2 + z2 + z3 + z4I3 – z4I4 = 0
I4
– z5I2–z4I3 + z4 + z5 + z6I4 = 0
z1 + z2 0 –z2 0
0 z1 + z2 + z7 –z3 –z5
–z2 –z3 z2 + z3 + z4 –z4
0 –z5 –z4 z4 + z5 + z6
I1
I2
I3
I4

VS
0
0
0
=
Mesh :
Mesh :
Mesh :
Mesh :
and in matrix form
Shown below is the MATLAB script to solve this system of equations.
Vs=100*(cos(pi/3)+j*sin(pi/3)); % Express Vs in rectangular form
z1=4; z2=20; z3=10; z4=20j; z5=5; z6=10j; z7=2;...
Z=[z1+z2 0 z2 0;...
0 z3+z5+z7 z3 z5;...
z2 z3 z2+z3+z4 z4;...
0 z5 z4 z4+z5+z6];...
V=[Vs 0 0 0]'; I=ZV; IL=I(3)I(4);...
magIL=abs(IL); phaseIL=angle(IL)*180/pi;...
disp('I1='); disp(I(1)); disp('I2='); disp(I(2));...
disp('I3='); disp(I(3)); disp('I4='); disp(I(4));...
disp('IL='); disp(IL);...
format bank % Display magnitude and angle values with two decimal places
disp('magIL='); disp(magIL); disp('phaseIL='); disp(phaseIL);...
fprintf(' n');
I1 = 5.4345 - 3.4110i
I2 = 4.5527 + 0.7028i

I3 = 4.0214 + 0.2369i
I4 = 7.4364 + 1.9157i
IL= -3.4150 - 1.6787i
magIL = 3.81
phaseIL = -153.82
Therefore,
4. The equivalent phasor circuit is shown below where
IL = 3.81–153.82iLt = 3.81cos104t – –153.82
jL1 j5 103  2 103
– =   = j10
jL2 j5 103  5 103
– =   = j25
  20 10–6 = –     = –j10
j – C  j 5 103
j10  j25 
10  25 
+

+

+

VS1 –j10  VS2
VC
4060 V 60–30 V
'We let VC VC
where is the capacitor voltage due to acting alone, and is
the capacitor voltage due to acting alone. With acting alone the circuit reduces to
that shown below.
By KCL
V''C + = V'C
VS1 V''C
VS2 VS1
j10  j25 
10  25 
+

+

V'C
z1
z2
VS1 –j10 
4060 V
z3
V'C
– VS1
z1
----------------------
V'C
z2
+ ------
+ ------ = 0
V'C
z3
1
z1
---- 1
---- 1
 V'C
 + + ---- 
z2
z3
VS1
z1
= ---------

V'C
VS1
----------------------------------------------
= = ----------------------------------
z1
---- 1
1
z1
---- 1
  
 + + ---- 
z2
z3
VS1
 
1
z1
z2
----
and with MATLAB,
Vs1=40*(cos(pi/3)+j*sin(pi/3)); z1=10+10j; z2=10j; z3=25+25j; V1c=Vs1/(1+z1/z2+z1/z3)
V1c = 36.7595 - 5.2962i
Therefore,
Next, with acting alone the circuit reduces to that shown below.
10  25 
j10  j25 
–j10  VS2
+ + ----------------------- = 0
---- 1
---- 1
 V''C
 + + ---- 
----------------------------------------------
= = ----------------------------------
---- 1
---- 1
  
 + + ---- 
 
By KCL
and with MATLAB
Vs2=60*(cos(pi/6)j*sin(pi/6));...
z1=10+10j; z2=10j; z3=25+25j; V1c=36.75955.2962j;...
V2c=Vs2/(z3/z1+z3/z2+1); Vc=V1c+V2c; fprintf(' n');...
disp('V1c = '); disp(V1c); disp('V2c = '); disp(V2c);...
disp('Vc=V1c+V2c'); fprintf(' n'); disp('Vc = '); disp(Vc);...
fprintf('magVc = %4.2f V t',abs(Vc));...
fprintf('phaseVc = %4.2f deg t',angle(Vc)*180/pi);...
V1c = 36.7595 - 5.2962i
z1
z3
 + + ---- 
V'C
= 36.76 – j5.30 V
VS2
+

+

V''C
60–30 V
z1
z2
z3
V''-------
C
z1
V''-------
C
z2
V'C' – VS2
z3
1
z1
z2
z3
VS2
z3
= ---------
V''C
VS2
z3
1
z1
z2
z3
VS2
----
z3
z3
z1
z2
 + ---- + 1 

V2c = -3.1777 - 22.0557i
Vc = V1c+V2c
Vc = 33.5818 - 27.3519i
magVc = 43.31 V phaseVc = -39.16 deg
Then,
and
VC V'C
= + V''C = 33.58 – j27.35 = 43.3127.35
vCt = 43.31cos5000t – 27.35
5. This circuit is excited by a DC (constant) voltage source, an AC (sinusoidal) voltage source,
and an AC current source of different frequency. Therefore, we will apply the superposition
principle.
Let be the capacitor voltage due to acting alone, the capacitor voltage due to
V'C
vS1 V''C
acting alone, and the capacitor voltage due to acting alone. Then, the
vS2t V'C'' iS t
capacitor voltage due to all three sources acting simultaneously will be
With the DC voltage source acting alone, after steadystate conditions have been reached
the inductors behave like short circuits and the capacitor as an open circuit and thus the cir-cuit
is simplified as shown below.
10 
+
15 V
By the voltage division expression
and
VC V'C
= + V'C' + V''C'
+

5 
+

V'C VR5 
V'C
= = ---------------  15 = 5 V DC
VR5 
5
10 + 5
v'Ct = 5 V DC
Next, with the sinusoidal voltage source acting alone the reactances are
vS2t
j1L1 j103 1 103
– =   = j1 
j1L2 j103 2 103
– =   = j2 

j – 1C  j – 103 5 104
– =      = –j2 
z1 z3
z2
j1  j2 
+

25 
10 
+

V''C
VS2 –j10 
200 V
V'C' – VS2
-----------------------
z1
V''C
z2
+ -------
+ ------- = 0
V''C
z3
1
z1
---- 1
---- 1
 V''C
 + + ---- 
z2
z3
VS2
z1
= ---------
V''C
VS2
----------------------------------------------
= = -----------------------------------
z1
---- 1
1
z1
---- 1
  
 + + ---- 
z2
z3
VS2
 
 1
+ + ---- 
----
z1
z2
z1
z3
V''C = 1.81 – j3.54 = 3.97–62.9
v'C' t = 3.97cos1000t – 62.9
 5 104
– =      = –j1 
and the equivalent phasor circuit is as shown below.
By KCL
and with MATLAB
Vs2=20+0j; z1=10+j; z2=2j; z3=5+2j; V2c=Vs2/(1+z1/z2+z1/z3); fprintf(' n');...
disp('V2c = '); disp(V2c); fprintf('magV2c = %4.2f V t',abs(V2c));...
fprintf('phaseV2c = %4.2f deg t',angle(V2c)*180/pi); fprintf(' n'); fprintf(' n');
V2c = 1.8089 - 3.5362i
magV2c = 3.97 V phaseV2c = -62.91 deg
Then,
and
Finally, with the sinusoidal current source acting alone the reactances are
iS t
j2L1 j2 103  1 103
– =   = j2 
j2L2 j2 103  2 103
– =   = j4 
j – 2C  j – 2 103

and the equivalent phasor circuit is as shown below where the current source and its parallel
resistance have been replaced with a voltage source with a series resistor.
By KCL
z1 z3
z2
j2  j4 
–j1  VS3
10  5 
+

+

V'''C
200 V
V'''--------
C
z1
V'''C
z2
+ --------
+ ------------------------ = 0
V''C' – VS3
z3
---- 1
1
z1
---- 1
 V'''C
 + + ---- 
z2
z3
VS3
z3
= ---------
V'''C
VS3
----------------------------------------------
= = ----------------------------------
z3
---- 1
1
z1
---- 1
  
 + + ---- 
z2
z3
VS3
z3
z1
 
 ----
+ ---- + 1 
z3
z2
and with MATLAB
Vs3=20+0j; z1=10+2j; z2=j; z3=5+4j; V3c=Vs3/(z3/z1+z3/z2+1); fprintf(' n');...
disp('V3c = '); disp(V3c); fprintf('magV3c = %4.2f V t',abs(V3c));...
fprintf('phaseV3c = %4.2f deg t',angle(V3c)*180/pi); fprintf(' n'); fprintf(' n');
V3c = -1.4395 - 3.1170i
magV3c = 3.43 V phaseV3c = -114.79 deg
Then,
V''C' = – 1.44 – j3.12 = 3.43–114.8
or
v''C' t = 3.43cos2000t – 114.8
and
vCt = v'C + v'C' t + v''C' t = 5 + 3.97cos1000t – 62.9 + 3.43cos2000t – 114.8
These waveforms are plotted below using the following MATLAB script:
wt=linspace(0,2*2*pi); deg=wt*180/pi; V1c=5;
V2c=3.97.*cos(wt62.9.*pi./180);
V3c=3.43.*cos(2.*wt114.8.*pi./180); plot(deg,V1c,deg,V2c,deg,V3c, deg,V1c+V2c+V3c)

vCt
v'C = 5 V DC
v'C'' t v'C' t
vS
+

ZS
ZLD
ZS ZLD ZS = ReZS + jImZS
ZLD = ReZLD + jImZLD Re Im
vS 2 ZS + ZLD2 = = -----------------------------  ZLD
pLD i2
LD  ZLD
vS 2
 ZLD
ReZS + jImZS + jReZLD + jImZLD2 = ------------------------------------------------------------------------------------------------------------------------------
ReZLD ImZLD
pLD
ImZLD = –ImZS
6.
Since and are complex quantities, we will express them as
and where and denote the real and imaginary compo-nents
respectively.
We want to maximize the expression
The only quantities that can vary are and and we must consider them
independently from each other.
From the above expression we observe that will be maximum when the denominator is
minimum and this occurs when , that is, when the imaginary parts of
and cancel each other. Under this condition, simplifies to
ZLD ZS pLD

pLD
vS 2
 RLD
RS + RLD2 = ------------------------------
and, as we found in Chapter 3, for maximum power transfer RLD = RS
. Therefore, the load
impedance will receive maximum power when
ZLD
ZLD ZS= 
that is, when ZLD is adjusted to be equal to the complex conjugate of ZS
.
7.
4 
ZLD
10  5 
+
 1700
20 
j5 j10 For this, and other similar problems involving the maximum power transfer theorem, it is best
to replace the circuit with its Thevenin equivalent. Moreover, we only need to compute .
For this problem, to find we remove and we short the voltage source. The remain-ing
ZTH ZLD
circuit then is as shown below.
ZTH
z1
X Y
z2
z3
z4
z5
z6
We observe that is in parallel with and this combination is shown as in the simpli-fied
z1 z2 z12
circuit below.

But this circuit cannot be simplified further unless we perform Wye to Delta transformation
which we have not discussed. This and the Delta to Wye transformation are very useful in
threephase circuits and are discussed in Circuit Analysis II with MATLAB Applications, ISBN
9781934404195. Therefore, we will compute using the relation
where is the open circuit voltage, that is, and is the current that would flow
between the terminals when the load is replaced by a short. Thus, we will begin our computa-tions
+
 j5 j10
1700
1 V1 2 V2 z VY 1
z2
-------------------
with the Thevenin voltage.
We disconnect from the circuit at points X and Y as shown below.
We will replace the remaining circuit with its Thevenin equivalent. Thus, with discon-nected
the circuit simplifies to that shown below.
Now, we will find
At Node 1:
X Y
z12
z3
z4
z5
z6
ZTH ZTH = VOC  ISC
VOC VTH ISC
ZLD
4 
10  5 
20 
X Y
ZLD
4  5 
20 
+

1700
X 10  Y
j5  –j10 
z3
z4
z5
VS
VTH VXY VX – VY V1 V2 VR5 
= = = –  – 
V1 – VS
z1
V1
z2
------
V1 – V2
z3
+ + ------------------- = 0

At Node 2:
---- 1
1
z1
---- 1
 V1
 + + ---- 
z2
z3
1
z3
– ----V2
VS
z1
= ------
V2 – -------------------
V1
z3
V2
z4
+ ------
+ ------ = 0
V2
z5
1
z3
– ----V1 1
---- 1
+  V2
 + + ---- 
z3
---- 1
z4
z5
and with MATLAB,
Vs=170; z1=4; z2=20; z3=10; z4=5j; z5=510j;...
Y=[1/z1+1/z2+1/z3 1/z3; 1/z3 1/z3+1/z4+1/z5]; I=[Vs/z1 0]'; V=YI; V1=V(1); V2=V(2);...
VX=V1; VY=(5/z5)*V2; VTH=VXVY; fprintf(' n');...
disp('V1 = '); disp(V1); disp('V2 = '); disp(V2);...
disp('VTH = '); disp(VTH); fprintf('magVTH = %4.2f V ',abs(VTH));...
fprintf('phaseVTH = %4.2f deg ',angle(VTH)*180/pi); fprintf(' n'); fprintf(' n');
V1 = 1.1731e+002 + 1.1538e+001i
V2 = 44.2308+46.1538i
VTH = 1.2692e+002 - 1.5385e+001i
magVTH = 127.85 V phaseVTH = -6.91 deg
Thus,
VTH = 127.85–6.91
Next, we must find from the circuit shown below.
ISC
4 
ISC
X Y
I4
10  5 
z1
a
+
b
 j5 j10
z5
I1 20 
z6 I2 I3
1700
VS
z2
z3
z4
We will write four mesh equations as shown above but we only are interested in phasor cur-rent
. Observing that a and b are the same point the mesh equations are
I4

I4
I20
I10
and in matrix form
With MATLAB,
Vs=170; VTH=126.9215.39j; z1=4; z2=20; z3=10; z4=5j; z5=5; z6=10j;...
Z=[z1+z2 z2 0 0; z2 z2+z3+z4 z4 z3; 0 z4 z4+z5+z6 z5; 0 z3 z5 z3+z5];...
V=[Vs 0 0 0]'; I=ZV; I1=I(1); I2=I(2); I3=I(3); I4=I(4);...
ZTH=VTH/I4; fprintf(' n'); disp('I1 = '); disp(I1); disp('I2 = '); disp(I2);...
disp('I3 = '); disp(I3); disp('I4 ='); disp(I4); disp('ZTH ='); disp(ZTH); fprintf(' n');
I1 = 15.6745 - 2.6300i
I2 = 10.3094 - 3.1559i
I3 = -1.0520 + 10.7302i
I4 = 6.5223 + 1.4728i
ZTH = 18.0084 - 6.4260i
Thus, and by Problem 6, for maximum power transfer there must be
or
8. We assign phasor currents as shown below.
We choose as a reference, that is, we let
z1 + z2 I1 – z2 I2 = VS
– z2 I1 + z2 + z3 + z4 I2 – z4 I3 – z3 I4 = 0
– z4 I2 + z4 + z5 + z6 I3 – z5 I4 = 0
– z3 I2 – z5 I3 + z3 + z5 I4 = 0
z1 + z2 –z2 0 0
–z2 z2 + z3 + z4 –z4 –z3
0 –z4 z4 + z5 + z6 –z5
0 –z3 –z5 z3 + z5
I1
I2
I3
I4

VS
0
0
0
=
ZTH = 18.09 – j6.43 
ZLD = ZTH
ZLD = 18.09 + j6.43 
4  10  5 
20 
+

VS j5 j10
IL
I5 IC
I5

Then,
and since
Next,
and
Now,
and
IC = I5
VC = IC  –j10 = 10  10–90 = 10–90 V
VL = V5 + VC = 50 + 10–90 = 5 + –j10 = 5 – j10 = 11.18–63.4 V
IL = VL  j5 = 11.18–63.4  590 = 2.24–153.4 = – 2 – j A
I10 = IL + I5 = – 2 – j + 1 = – 1 – j = 2–135 A
V10 = 10  2–135 = 10  – 1 – j = – 10 – j10 V
Continuing we find that
and
Also,
and
Finally,
I5 = 10 A
V5 = 50 V
V20 = V10 + VL = – 10 – j10 + 5 – j10 = – 5 – j20 V
I20 = V20  20 = – 5 – j20  20 = – 0.25 – j A
I4 = I20 + I10 = – 0.25 – j – 1 – j = – 1.25 – j2 A
V4 = 4I4 = 4  – 1.25 – j2 = – 5 – j8 V
VS = V4 + V20 = – 5 – j8– 5 – j20 = – 10 – j28 = 29.73–109.7 V
The magnitudes (not to scale) and the phase angles are shown below.
IL
I5 = IC
VL
VC
V4
I10 I4
VS
V10
V20
I20
The phasor diagram above is acceptable. However, it would be more practical if we rotate it by
to show the voltage source as reference at as shown below.
109.7 VS 0

VS
VC
V20
VL
I5 = IC
I4
I10
V4
V10 IL
I20
9. The equivalent phasor circuit is shown below where , , and
z1 = R1 = 1 K z2 = R2 = 3 K
z3 –j  C –j 10 3 0.25 10–6 = =      = –j4 K
R1
1 K
C
R2
z3
z2
3 K
z1
VIN = 30 V
V
VOUT
V – VIN
z1
-------------------
V – VOUT
+ ------------------------
+ ------------------------ = 0
z2
V – VOUT
z3
V = 0
---- 1
1
z2
 VOUT
 + ---- 
z3
–VIN
z1
= ------------
VOUT
–VIN
-----------------------------------
= = -------------------------
z1
---- 1
1
z2
  
 + ---- 
z3
–VIN
z1
z2
 
 ----
+ ---- 
z1
z3
Application of KCL at the inverting input yields
and since the above relation reduces to
or
and with MATLAB,
Vin=3; z1=1000; z2=3000; z3=4000j; Vout=Vin/(z1/z2+z1/z3);...
fprintf(' n'); disp('Vout = '); disp(Vout); fprintf('magVout = %5.2f V t',abs(Vout));...
fprintf('phaseVout = %5.2f deg t',angle(Vout)*180/pi); fprintf(' n'); fprintf(' n');
Vout = -5.7600 + 4.3200i

magVout = 7.20 V phaseVout = 143.13 deg
Thus,
and
10.
VOUT = – 5.76 + j4.32 = 7.2143.13 V
voutt = 7.2cos1000t + 143.13 V
R
iC
C
+
vIN

iR
+

vOUT
iC = iR
C
dvC
dt
--------- vOUT – vIN
= --------------------------
R
dvOUT
---------------
dt
vOUT – vIN
= --------------------------
RC
and since , by integrating both sides of the expression above, we obtain
11.
vOUT « vIN
+
vIN


and since , we obtain
vOUT
1
RC
 –--------vIN
iR
R
C
iC
+

vOUT
iR = iC
vOUT
R
------------ C d
= ----vOUT – vIN
dt
vOUT « vIN
vOUT –RC d
= ----vIN
dt

Chapter 8
Average and RMS Values, Complex Power, and Instruments
his chapter defines average and effective values of voltages and currents, instantaneous and
average power, power factor, the power triangle, and complex power. It also discusses elec-trical
instruments that are used to measure current, voltage, resistance, power, and energy.
T
8.1 Periodic Time Functions
A periodic time function satisfies the expression
(8.1)
ft = ft + nT
where n is a positive integer and T
is the period of the periodic time function. The sinusoidal and
sawtooth waveforms of Figure 8.1 are examples of periodic functions of time.
cost cost + 
T T

T T
T T T T
Figure 8.1. Examples of periodic functions of time
Other periodic functions of interest are the square and the triangular waveforms.

Chapter 8 Average and RMS Values, Complex Power, and Instruments
8.2 Average Values
The average value of any continuous function such as that shown in Figure 8.2 over an inter-val
b
 1
= =  b
----------- area a
VP
,
Figure 8.2. A continuous time function
is defined as
(8.2)
The average value of a periodic time function is defined as the average of the function over one
period.
Example 8.1
Compute the average value of the sinusoid shown in Figure 8.3, where denotes the peak
(maximum) value of the sinusoidal voltage.
Solution:
By definition,
f t
a  t  b
a b
t
f t
f t
ftave
1
b – a
----------- f tdt
a
b – a
f t
Vp
0   2  3  2 2 5  2
T
0
–VP
VP sint
trad

Average Values
Vave
1T
T
 1
= --- vtdt
= = 
0
Vp
2
T
 Vp
------- Vp sintdt
T
0
2
------ sintdt
2
0
= ------–cost
= = ------1–1 = 0
2 Vp
0
------ cost
2
0 Vp
2
2
as expected since the net area of the positive and negative half cycles is zero.
Example 8.2
Compute the average value of the halfwave rectification waveform shown in Figure 8.4.
Ipsint
 2
T
Radians
Current (i)
Solution:
This waveform is defined as
(8.3)
it
=
Then, its average value is found from
(8.4)
Ip sint 0  t  
0  t 2     
Iave
1
2
2
 1
= ------ Ip sintdt
= 
0
Ip
2
2
d + 

------ Ip sint t 0 dt
2

0
–cos 
------ t 0
Ip
2
cos 0
------ t 
Ip
2
------1 – –1
Ip

= = = = ----
In other words, the average value of the halfwave rectification waveform is equal to its peak
value divided by 
.

8.3 Effective Values
The effective current of a periodic current waveform is defined as the current which pro-duces
Ieff it
heat in a given resistance at the same average rate as a direct (constant) current ,
R Idc
2 = = =
2 RIdc
Average Power Pave RIeff
it
pt R i
= 2t
Pave
= = = 
1T
T
 1T
--- ptdt
0
T
 RT
--- Ri2 dt
0
T
--- i2 dt
0
= 
2 RT
RIeff
T
--- i2 dt
0
2 1T
Ieff
T
--- i 2dt
0
= 
Ieff
T
 IRoot Mean Square IRMS Ave i= = = =  2
1T
--- i 2 dt
0
that is,
(8.5)
Also, in a periodic current waveform , the instantaneous power is
(8.6)
and
(8.7)
Equating (8.5) with (8.7) we obtain
or
or
(8.8)
Caution 1:
In general, avei2  iave2 since the expression avei2
implies that the function i must first be
squared and the average of the squared value is then to be found. On the other hand,
iave2
implies that the average value of the function must first be found and then the average must be
squared. The waveforms in Figure 8.5 illustrate this point.

Effective (RMS) Value of Sinusoids
sqrtavei2  0 avei2  0
i i 0 2
iave = 0
2 = 0 sqrt iave
iave
 2  = 0
avei2  iave2
Figure 8.5. Waveforms to illustrate that
Caution 2:
In general, . For example, if and , then
Pave  Vave  Iave vt = Vp cost it = Ip cost + 
, and also . Thus, . However,
Vave = 0 Iave = 0 Pave = 0
Pave
= = =   0
1T
T
 1T
--- ptdt
0
T
 1T
--- vtitdt
0
T
--- Vp costIp cost + dt
0
8.4 Effective (RMS) Value of Sinusoids
Now, we will derive an expression for the Root Mean Square (RMS) value of a sinusoid in terms of
its peak (maximum) value. We will denote the peak values of voltages and currents as and
respectively. The value from positive to negative peak will be denoted as Vp and – p Ip – p
, and
the RMS values as and . Their notations and relationships are shown in Figure 8.6.
VRMS IRMS
RMS Value = 0.707 × Peak = 120 V
Peak (Max) Value = 170 V
180 270 360 Time (Degrees)
Peak (Max) Value = 170 V
90
PeaktoPeak Value = 340 V
Vp – p Ip – p VRMS IRMS Vp Ip
Figure 8.6. Definitions of , , , and in terms of and
Vp Ip

i = Ip cost – 
2 1T
IRMS
T
 1
--- i2dt
= = 
0
2 cos2t – dt
2
------ Ip
2
0
 2 cos 12
= -- cos2 + 1
2
4
2 Ip
IRMS
2
+ 
2
= 
Ip
------ cos2t – dt dt
0
0
2
4
------ sin2t – 
= ------------------------------
= ---------------------------------------------------- + 2
2
2
0
+  2
t 0
Ip
2
4
------ sin4 –  – sin–
2
sinx – y = sinxcosy – cosxsiny
–sin– = sin
2
4
2 Ip
IRMS
0 1
------ sin4cos – cos4sin + sin
--------------------------------------------------------------------------- + 2
2
Ip
2
4
------2
2
2
Let
then,
and using the identity
we obtain
(8.9)
Using the trigonometric identities
and
by substitution into (8.9), we obtain
and therefore,
(8.10)
We observe that the value of a sinusoid is independent of the frequency and phase angle,
in other words, it is dependent on the amplitude of the sinusoid only.
Example 8.3
Compute the and for the sawtooth waveform shown in Figure 8.7.
Ip
= = = -----
IRMS
Ip
2
------ 0.707Ip
= =
FOR SINUSOIDS ONLY
RMS
Iave IRMS

RMS Values of Sinusoids with Different Frequencies
10 A
it
Solution:
By inspection, the period is as shown in Figure 8.8.
t
T
T
10 A
t
it
Figure 8.8. Defining the period for the waveform of Example 8.3
The average value is
Iave
----------------- 1  2  10  T
Area
Period
= = ------------------------------------ = 5 A
T
To find we cannot use (8.10); this is for sinusoids only. Accordingly, we must use the defi-nition
IRMS
of the value as derived in (8.8). Then,
or
RMS
2 1T
IRMS
T
 1T
--- i2tdt
= = = = = --------
0
--- 100
T2 -------- t 3
T
 1T
 2
--- 10
-----t
T
dt
0
 
 ----
 
3
 
T
0
--- 100
T2 -------- T3
1T
  100
 ------
 
3
 
3
IRMS
100
3
-------- 10
= = ------ = 5.77 A
3
8.5 RMS Values of Sinusoids with Different Frequencies
The value of a waveform which consists of a sum of sinusoids of different frequencies, is
equal to the square root of the sum of the squares of the values of each sinusoid. Thus, if
(8.11)
RMS
RMS
i = I0 + I1 cos1t  1 + I2 cos2t  2 +  + IN cosN t  N
where represents a constant current, and represent the amplitudes of the sinu-soids.
I0 I1 I2 IN
Then, the value of i is found from
RMS

(8.12)
= + + + + 2
2 I1 RMS
2 12
2 12
2
2
2 1T
T
 1
2
 1
------ i 2 dt
2

 –1 2 dt
------ 1 2 dt
2  +  1
 t 
= = ------ + 2 –  = 1
or
(8.13)
Example 8.4
Find the value of the square waveform of Figure 8.9 by application of (8.12)
Solution:
By inspection, the period is as shown in Figure8.10.
Figure 8.10. Determination of the period to the waveform of Example 8.4
Then,
a.
IRMS I0
2 I2 RMS
2  IN RMS
IRMS I0
--I
1 p
--I
2 p
 12
--I
N p
= + + + +
IRMS
1
1
t
T = 2
1
1
 2 t
T
IRMS
--- i 2dt
0
2
0
2
0

= = = + 
1
2
------ t 0
2

Average Power and Power Factor
or
(8.14)
IRMS = 1
b. Fourier series analysis textbooks* show that the square waveform above can be expressed as
(8.15)
i t   4
-- t sin 13
-- 3t sin 15
=  
 + + -- sin5t + 
and as we know, the value of a sinusoid is a real number independent of the frequency
and the phase angle, and it is equal to 0.707 times its peak value, that is, IRMS = 0.707  Ip
.
Then from (8.12) and (8.15),
(8.16)
RMS
IRMS
-- 0 12
4
 2 12
-- 13
-- 1  2 12
 -- 
 2
-- 15
 -- 
= + + + +  = 0.97
The numerical accuracy of (8.16) is good considering that higher harmonics have been
neglected.
8.6 Average Power and Power Factor
Consider the network shown in Figure 8.11.
R
+
Rest of
the Circuit Load

vS t
iLDt
vLDt
Figure 8.11. Network where it is assumed that and are outofphase
iLDt vLDt
We will assume that the load current is degrees outofphase with the voltage ,
i.e., if vLDt = Vp cost , then iLDt = Ip cost + 
. We want to find an expression for the
average power absorbed by the load.
We know that
that is,
iLDt  vLDt
p = vi
ins tan taneous power = ins tan taneous voltage  ins tan taneous current
and the instantaneous power absorbed by the load is
pLDt
* Refer to Signals and Systems with MATLAB Computing and Simulink Modeling, ISBN 9780974423998.

(8.17)
T
 1T
  dt
0
T

-----------  cos2t + dt
T
= = = --------------------------------------------
Using the trigonometric identity
we express (8.17) as
(8.18)
and the average power is
(8.19)
We observe that the first integral on the right side of (8.19) is zero, and the second integral, being
a constant, has an average value of that constant. Then,
(8.20)
and using the relations
and
we can express (8.19) as
(8.21)
and it is imperative that we remember that these relations are valid for circuits with sinusoidal
excitations.
The term in (8.20) and (8.21) is known as the power factor and thus
(8.22)
pLDt = vLDt  iLDt = VpIp cost  cost + 
x y cos cos 12
= -- cosx + y + cosx – y
pLOADt
VpIp
2
= ----------- cos2t +  + cos
Pave LD
1T
--- pLDdt
0
---
VpIp
2
 ----------- cos2t +  + cos
T
= = 
VpIp
2T
0
VpIp
2T
----------- cosdt
0
= + 
Pave LD
VpIp
2
= ----------- cos
VRMS
Vp
2
= ------
IRMS
Ip
2
= ------
Pave LD = VRMS LD IRMS LDcos
cos
Power FactorLD PFLD cosLD
Pave LD
VRMS LD IRMS LD

Average Power in a Resistive Load
8.7 Average Power in a Resistive Load
The voltage and current in a resistive branch of a circuit are always in phase, that is, the phase
angle . Therefore, denoting that resistive branch with the subscript we have:
(8.23)
or
(8.24)
 = 0 R
Pave R = VRMS R IRMS R cos0 = VRMS R IRMS R
Pave R
2
R
VRMS R
2 R 12
= = = = 2 R
------------------ IRMS R
--
2
R
Vm R
------------ 12
--Ip R
8.8 Average Power in Inductive and Capacitive Loads
With inductors and capacitors there is a 90
phase difference between the voltage and current,
that is,  = 90
and therefore, denoting that inductive or capacitive branch with the subscript
we obtain:
X
Pave X = VRMS X IRMS X cos90 = 0
Of course, the instantaneous power is zero only at specific instants.
Obviously, if the load of a circuit contains resistors, inductors and capacitors, the phase angle

between and will be within ,and the power factor will
be within .
VRMS LOAD IRMS LOAD 0    90 cos
0  cos  1
Example 8.5
For the circuit of Figure 8.11, find the average power supplied by the voltage source, the average
power absorbed by the resistor, the inductor, and the capacitor.
10  j20 
I
VS = 1700
–j10 
Solution:
Since this is a series circuit, we need to find the current I
and its phase relation to the source
voltage . Then,
VS

(8.25)
---------------------------------- 1700
-------------------- 1700
= = = = --------------------------- = 12–45
Relation (8.25) indicates that , ,and the power factor is
Therefore, using (8.24) we find that the average power absorbed by the resistor is
(8.26)
2 R 12
= = --12210 = 720 w
The average power absorbed by the inductor and the capacitor is zero since the voltages and cur-rents
= = ------------------------0.707 = 721 w
VA VB
in these devices are outofphase with each other.
Check: The average power delivered by the voltage source is
(8.27)
and we observe that (8.26) and (8.27) are in close agreement.
Example 8.6
For the circuit of Figure 8.13, find the power absorbed by each resistor, and the power supplied
(or absorbed) by the current sources.
Solution:
This is the same circuit as in Example 7.1 where we found that
(8.28)
and
(8.29)
Then,
I
VS
Z
------ 1700
10 + j20 – j10
10 + j10
10 245
Ip = 12 A  = –45
cos = cos–45 = 0.707
Pave R
12
--Ip R
90
Pave SOURCE
VpIp
2
----------- cos 17012
2
8 
2 
4 
j3 
50 A j6  j3  100 A
VA = – 4.138 + j19.655 = 20.086101.9
VB = – 22.414 – j1.035 = 22.440–177.4

Average Power in Inductive and Capacitive Loads
and
(8.30)
Also,
and
(8.31)
Likewise,
and
(8.32)
I2 
VA – VB
2 + j3
----------------------------------------- 32.430145.0
-------------------- 18.276 + j20.690
= = = ------------------------------------- = 8.98388.7
3.6156.3
3.6156.3
Pave 2 
12
2
= = --  8.9832  2 = 80.70 w
--I
p 2 
2    12
I4 
VA
4 – j6
------------- 20.086101.9
= = ------------------------------------- = 2.786158.2
7.21–56.3
Pave 4 
12
2
= = --  2.786 2  4 = 15.52 w
--I
p 4 
4    12
I8 
VB
8 – j3
------------- 22.440–177.4
= = ---------------------------------------- = 2.627–156.7
8.54–20.6
Pave 8 
12
2
= = --  2.6272  8 = 27.61 w
--I
p 8 
8    12
The voltages across the current sources are the same as and but they are and
VA VB 101.9
outofphase respectively with the current sources as shown by (8.28) and (8.29).
–177.4
Therefore, we let and  Then, the power absorbed by the source
is
(8.33)
1 = 101.9 2 = –177.4 5 A
Pave 5 A
VpIp
2
----------- cos1
= = ----------------------- cos101.9
20.086  5
= -------------------------  –0.206 = –10.35 w
2
and the power absorbed by the source is
(8.34)
VA 5 A
2
10 A
Pave 10 A
VpIp
2
= ----------- cos2
= -------------------------- cos–177.4
22.440  10
VB 10 A
2
= ----------------------------  –0.999 = –112.08 w
2
The negative values in (8.33) and (8.34) indicate that both current sources supply power to the
rest of the circuit.
Check: Total average power absorbed by resistors is
80.70 + 15.52 + 27.61 = 123.83 w
and the total average power supplied by current sources is
112.08 + 10.35 = 122.43 w

Thus, the total average power supplied by the current sources is equal to the total average power
absorbed by the resistors. The small difference is due to rounding of fractional numbers.
8.9 Average Power in NonSinusoidal Waveforms
If the excitation in a circuit is nonsinusoidal, we can compute the average power absorbed by a
resistor from the relations
(8.35)
T
 1T
T
 1T
= = = 
------dt
0
T
Example 8.7
Compute the average power absorbed by a resistor when the voltage across it is the half
wave rectification waveform shown in Figure 8.14.
Solution:
We first need to find the numerical value of . It is found as follows:
–  s T 2  2
------ 103
= = = = = 
10–3
 0
------------------- 10 2 10 3sin2 t
T
v 2
1
 – = ---  = +  dt
------dt
0
----------------------------------
2 10 3
10–3
 5 103
3
0 10–3
10–3
 2 1= =  – 
 
=
03
 t 2 15 110 03
– 3 2 03
1 
– ------------------------------
10–3
=  – ---------------------------------------------
 
 
and thus
Then,
or
Pave
1T
--- pdt
0
--- v 2
R
--- i2R t d
0
5 
1
Voltage (v)
t (ms)
10sint
2

T 2 ms 2 103
T
10sint = 10sin103
Pave
1T
R
2 
1– 3
05
0
10–3
Pave
100
10 10–3 
---------------------- 12
--1 2 103
 – cos  tdt
0
dt
0
cos  tdt
0
5 103
sin  t
2 103
 
0
 10–3 sin  
2 103
 

Lagging and Leading Power Factors
and since for , the last term of the expression above reduces to
sin2n = 0 n = integer
Pave = 5 w
8.10 Lagging and Leading Power Factors
By definition an inductive load is said to have a lagging power factor. This refers to the phase angle
of the current through the load with respect to the voltage across this load as shown in Figure
8.15.
VLOAD
ILOAD
1
Figure 8.15. Lagging power factor
In Figure 8.15, the cosine of the angle 1 , that is, cos1
is referred to as lagging power factor and
it is denoted as pf lag.
The term “inductive load” means that the load is more “inductive” (with some resistance) than
it is “capacitive”. But in a “purely inductive load” and thus the power factor is
1 = 90
cos1 = cos90 = 0
By definition a capacitive load is said to have a leading power factor. Again, this refers to the
phase angle of the current through the load with respect to the voltage across this load as shown
in Figure 8.16.
VLOAD
ILOAD
2
Figure 8.16. Leading power factor
In Figure 8.16, the cosine of the angle 2 , that is, cos2
is referred to as leading power factor and
it is denoted as pf lead.
The term “capacitive load” means that the load is more “capacitive” (with some resistance) than
it is “inductive”. But in a “purely capacitive load” and thus the power factor is
2 = 90
cos2 = cos90 = 0

Pave
12
= --VpIp cos = VRMS IRMS cos
 Pa
Q
Pa

(a) Power Triangle for Inductive Load (b) Power Triangle for Capacitive Load
8.11 Complex Power  Power Triangle
We recall that
(8.36)
This relation can be represented by the socalled power triangle. Figure 8.17 (a) shows the power
triangle of an inductive load, and Figure 8.16 (b) shows the power triangle for both a capacitive
load.
Figure 8.17. Power triangles for inductive and capacitive loads
In a power triangle, the product is referred to as the apparent power, and it is
denoted as . The apparent power is expressed in or . The product
is referred to as the reactive power, and it is denoted as . The reactive
power is expressed in or . Thus, for either triangle of Figure 8.17,
(8.37)
(8.38)
(8.39)
The apparent power is the vector sum of the real and reactive power components, that is,
(8.40)
where the (+) sign is used for inductive loads and the () sign for capacitive loads. Because rela-tion
of (8.40) consists of a real part and an imaginary part, it is known as the complex power.
Example 8.8
For the circuit shown in Figure 8.18, find:
Q
P real = Pave
P real = Pave
VRMS  IRMS
Pa volt – amperes VA
VRMS  IRMS  sin Q
volt – amperes reactive VAR
Preal = Pave = VRMS IRMS cos = (in watts)
Q = Reactive Power = VRMS IRMS sin = (in VARs)
Pa = Apparent Power = VRMS IRMS (in VAs)
Pa
Pa = Preal power  jQ = Pave  jQ

Complex Power  Power Triangle
a. the average power delivered to the load
b. the average power absorbed by the line
c. the apparent power supplied by the voltage source
d. the power factor of the load
e. the power factor of the line plus the load
Rline = 1 
VS
Load
10 + j10
Rline = 1 
4800 V RMS
Solution:
For simplicity, we redraw the circuit as shown in Figure 8.19 where the line resistances have been
combined into a single resistor.
2 
Rline = 2 
VS
4800 V RMS
ZLD
Load
IRMS 10 + j10
Figure 8.19. Circuit for Example 8.8 with the line resistances combined
From the circuit of Figure 8.19, we find that
IRMS
VS RMS
Rline + ZLD
----------------------------- 4800
---------------------------- 4800
= = = ------------------------------- = 30.73–39.8
2 + 10 + j10
15.6239.8
and therefore, the current lags the voltage as shown on the phasor diagram of Figure 8.20.
VS –39.8
I
Figure 8.20. Phasor diagram for the circuit of Example 8.8
Then,
a. The average power delivered to the load is

= 2 ReZLD = 30.732  10 = 9443 w = 9.443 Kw
Pave LD IRMS
= 2 Rline = 30.732  2 = 1889 w = 1.889 Kw
Pave line IRMS
Pa source VS RMS IRMS = = 480  30.73 = 14750 w = 14.750 Kw
pfLD cosLD
Pave LD
Pa LD
----------------- 9443
= = = ----------------------------------------
VRMS LD IRMS
------------------------------------------------------------------------------------------------ 9443
9443
= = = -------------- = 0.707
4800 – 230.73–39.8  30.73
--------------------------------------- 9443
434.5630.73
13354
pfline + LD cosline + LD
Pave total
Pa source
= = ---------------------
= = ------------------------------ = 0.77
Pave line + Pave LD
---------------------------------------------- 1889 + 9443
Pa source
14750

i2R
0.85
b. The average power absorbed by the line is
c. The apparent power supplied by the voltage source is
d. The power factor of the load is
e. The power factor of the line plus the load is
8.12 Power Factor Correction
The consumer pays the electric utility company for the average or real power, not the apparent
power and, as we have seen, a low power factor (larger angle ) demands more current. This
additional current must be furnished by the utility company which must provide larger current
carrying capacity if the voltage must remain constant. Moreover, this additional current creates
larger losses in the utility’s transmission and distribution system. For this reason, electric util-ity
companies impose a penalty on industrial facility customers who operate at a low power factor,
typically lower than . Accordingly, facility engineers must install the appropriate equipment
to raise the power factor.
The power factor correction procedure is illustrated with the following example.
Example 8.9
In the circuit shown in Figure 8.21, the resistance of the lines between the voltage source and the
load and the internal resistance of the source are considered small, and thus can be neglected.

Power Factor Correction
1 Kw Load
@ pf =0.8 lag
VS
4800 V RMS
60 Hz
It is desired to “raise” the power factor of the load to 0.95 lagging. Compute the size and the rat-ing
of a capacitor which, when added across the load, will accomplish this.
Solution:
The power triangles for the existing and desired power factors are shown in Figure 8.22.
1 Kw 1 Kw
1
2
Q2
Q1
1 –10.8 = cos = 36.9 2 0.95 –1 = cos = 18.2
This is what we have This is what we want
Figure 8.22. Power triangles for existing and desired power factors
Since the voltage across the given load must not change (otherwise it will affect the operation of
it), it is evident that a load, say , in opposite direction of must be added, and must be con-nected
Q3 Q1
in parallel with the existing load. Obviously, the Q3
load must be capacitive. Accord-ingly,
the circuit of Figure 8.21 must be modified as shown in Figure 8.23.
1 Kw Load
@ pf =0.8 lag
IC
Capacitive
Load with
Leading pf
VS
4800 V RMS 60 Hz
Figure 8.23. Circuit for power factor correction
For the existing load,
Q1 = 1 Kw tan36.9 = 750 VAR
and for the desired , the VAR value of must be reduced to
pf = cos2 = 0.95 Q2
Q2 = 1 Kw tan18.2 = 329 VAR
Therefore, the added capacitive load must be a vector such that
Q3

Q3 = Q1 – Q2 = 750 – 329 = 421 VAR
IC
Q3 = ICVC = ICVS
IC
Q3
VS RMS
------------------ 421
= = -------- = 0.88 A
480
XC
VC
IC
------- 480
= = ---------- = 547 
0.88
C 1
----------- 1
= = = ------------------------------- = 4.85 F
XC
---------------- 1
2fXC
260547
4.85 F
VC max = 2  480 = 679 V
5 F
The current through the capacitive load is found from
Then,
and
Therefore, the capacitive load must consist of a capacitor with the value
However, not any capacitor will do; the capacitor must be capable of withstanding a
maximum voltage of
and for all practical purposes, we can choose a capacitor rated at 700 volts or higher.
8.13 Instruments
Ammeters are electrical instruments used to measure current in electric circuits, voltmeters mea-sure
voltage, ohmmeters measure resistance, wattmeters measure power, and watthour meters
measure electric energy. Voltmeters, Ohmmeters, and Milliammeters (ammeters which measure
current in milliamperes) are normally combined into one instrument called VOM. Figure 8.24
shows a typical analog type VOM, and Figure 8.25 shows a typical digital type VOM. We will see
how a digital VOM can be constructed from an analog VOM equivalent at the end of this sec-tion.
An oscilloscope is an electronic instrument that produces an instantaneous trace on the
screen of a cathoderay tube corresponding to oscillations of voltage and current. A typical oscil-loscope
is shown in Figure 8.26. DC ammeters and voltmeters read average values whereas AC
ammeters and voltmeters read RMS values.
The basic meter movement consists of a permanent horse shoe magnet, an electromagnet which
typically is a metal cylinder with very thin wire wound around it which is referred to as the coil,
and a control spring. The coil is free to move on pivots, and when there is current in the coil, a
torque is produced that tends to rotate the coil. Rotation of the coil is restrained by a helical
spring so that the motion of the coil and the pointer which is attached to it, is proportional to the
current in the coil.

Instruments
Figure 8.24. The Triplett Analog Multimeter Model 60
Figure 8.25. The Voltcraft Model 3850 Digital Multimeter

Figure 8.26. The Agilent Technologies Series 5000 Portable Oscilloscope
An ammeter measures current in amperes. For currents less than one ampere, a milliammeter or
microammeter may be used where the former measures current in milliamperes and the latter in
microamperes.
Ammeters, milliammeters, and microammeters must always be connected in series with the cir-cuits
5 10 0
1
IT IM RM IT
in which they are used.
Often, the electric current to be measured, exceeds the range of the instrument. For example, we
cannot directly measure a current of to milliamperes with a milliammeter whose range is
to milliampere. In such a case, we can use a low range milliammeter with a shunt (parallel)
resistor as shown in Figure 8.27, where the circle with represents an ideal milliammeter (a
milliammeter with zero resistance). In Figure 8.27 is the total current to be measured, is
the current through the meter, is the current through the shunt resistor, is the milliamme-ter
internal resistance, and is the shunt resistance.
mA
IT IM
IS RM
RS
mA
RS
IS

Instruments
Figure 8.27. Milliammeter with shunt resistor
RS
From the circuit of Figure 8.27, we observe that the sum of the current flowing through the mil-liammeter
and the current through the shunt resistor is equal to the total current , that
is,
IM IS IT
(8.41)
IT = IM + IS
Also, the shunt resistor RS
is in parallel with the milliammeter branch; therefore, the voltages
across these parallel branches are equal, that is,
RM IM = RS IS
and since we normally need to calculate the shunt resistor, then
(8.42)
RS
IM
IS
= -----RM
Example 8.10
In the circuit of Figure 8.28, the total current entering the circuit is and the milliammeter
range is 0 to 1 milliampere, that is, the milliammeter has a fullscale current Ifs of 1 mA
, and
its internal resistance is . Compute the value of the shunt resistor .
40  RS
IM=Maximum allowable current
through the milliammeter
Ifs = 1 mA
IT IM RM IT
40 
mA
RS
IS
5 mA
Solution:
The maximum current that the milliammeter can allow to flow through it is and since the
total current is milliamperes, the remaining milliamperes must flow through the shunt resis-tor,
that is,
5 4
IS = IT – IM = 5 – 1 = 4 mA
The required value of the shunt resistor is found from (8.42), i.e.,
1 mA

Check:The calculated value of the shunt resistor is ; this is onefourth the value of the mil-liammeter
internal resistor of . Therefore, the resistor will allow four times as much
IT IM
+
- RS3
current as the milliammeter to flow through it.
A multirange ammeter/milliammeter is an instrument with two or more scales. Figure 8.29
shows the circuit of a typical multirange ammeter/milliammeter.
Figure 8.29. Circuit for a multirange ammeter/milliammeter
A voltmeter, as stated earlier, measures voltage in volts. Typically, a voltmeter is a modified mil-liammeter
where an external resistor is connected in series with the milliammeter as shown
in Figure 8.30 where
RS
IM
IS
-----RM
14
= = --  40 = 10 
10 
40  10 
A
IT
RM
RS1
IS
RS2
RV
I = current through circuit
RM = internal resis tance of milliameter
RV = external resistor in series with RM
VM = voltmeter full scale reading

Instruments
RV = Voltmeter internal resistance
VM = Voltmeter range
IM RM RV
mA
+ 
VM
Figure 8.30. Typical voltmeter circuit
For the circuit of Figure 8.30,
or
(8.43)
IMRM + RV = VM
RV
VM
IM
= -------- – RM
Voltmeters must always be connected in parallel with those devices of the circuit whose voltage
is to be measured.
Example 8.11
Design a voltmeter which will have a 1 volt fullscale using a milliammeter with 1
milliampere
fullscale and internal resistance 100 
.
Solution:
The voltmeter circuit consists of the milliammeter circuit and the external resistance as
IM RM RV
mA
100 
VM + 
RV
Here, we only need to compute the value of the external resistor so that the voltage across
the series combination will be full scale. Then, from (8.43),
(8.44)
RV
1 volt
RV
VM
IM
-------- – RM 1
10 –3 = = ---------- – 100 = 1000 – 100 = 900 

Therefore, to convert a 1 milliampere fullscale milliammeter with an internal resistance of
to a fullscale voltmeter, we only need to attach a resistor in series with
100  1 volt 900 
mA
100 
RM
100 V
VM
+

99.9 k
9.9 k
900 
I
IM = 1 mA fs
10 V
1 V
RX
 0
I RM
mA
Zero
Adjust
+

VS
RX
that milliammeter.
Figure 8.32 shows a typical multirange voltmeter.
Figure 8.32. Circuit for a multirange voltmeter
An Ohmmeter measures resistance in Ohms. In the series type Ohmmeter, the resistor whose
resistance is to be measured, is connected in series with the Ohmmeter circuit shown in Figure
8.33.
Figure 8.33. Circuit for a series type Ohmmeter
We observe from Figure 8.33 that for the series type Ohmmeter, the current I
is maximum when
the resistor RX is zero (short circuit), and the current is zero when RX
is infinite (open circuit).
For this reason, the 0
(zero) point appears on the rightmost point of the Ohmmeter scale, and
the infinity symbol appears on the leftmost point of the scale.
Figure 8.34 shows the circuit of a shunt (parallel) type Ohmmeter where the resistor RX
whose
value is to be measured, is in parallel with the Ohmmeter circuit.

Instruments
VS
+
 Adjust
mA
I
RM
Zero
RX
0 
Figure 8.34. Circuit for a parallel type Ohmmeter
From Figure 8.34 we see that, for the shunt type Ohmmeter, the current through the milliamme-ter
circuit is zero when the resistor is zero (short circuit) since all current flows through that
RX
short. However, when is infinite (open circuit), the current through the milliammeter
branch is maximum. For this reason, the 0
(zero) point appears on the leftmost point of the
Ohmmeter scale, and the infinity symbol appears on the rightmost point of the scale.
An instrument which can measure unknown resistance values very accurately is the Wheatstone
Bridge shown in Figure 8.35.
R3
R1
VS A
VA A
VB  +
R4 R2
Figure 8.35. Wheatstone Bridge Circuit
RX

0
One of the resistors, say R4
, is the unknown resistor whose value is to be measured, and another
resistor, say R3
is adjusted until the bridge is balanced, that is, until there is no current flow
through the meter of this circuit. This balance occurs when
R1
R2
------
R3
R4
= ------
from which the value of the unknown resistor is found from
(8.45)
R4
R2
R1
= ------R3

Example 8.12
In the Wheatstone Bridge circuit of Figure 8.36, resistor is adjusted until the meter reads zero,
and when this occurs, its value is . Compute the value of the unknown resistor .
R3
VS R1 A
A
Solution:
When the bridge is balanced, that is, when the current through the meter is zero, relation (8.45)
holds. Then,
When measuring resistance values, the voltage sources in the circuit to which the unknown resis-tance
is connected must be turned off, and one end of the resistor whose value is to be measured
must be disconnected from the circuit.
Because of their great accuracy, Wheatstone Bridges are also used to accept or reject resistors
whose values exceed a given tolerance.
A wattmeter is an instrument which measures power in watts or kilowatts. It is constructed with
two sets of coils, a current coil and a voltage coil where the interacting magnetic fields of these
coils produce a torque which is proportional to the product.
A watthour meter is an instrument which measures electric energy , where is the product
of the average power in watts and time in hours, that is, in watthours. Electric util-ity
companies use kilowatthour meters to bill their customers for the use of electricity.
Digital meters include an additional circuit called analogtodigital converter (ADC). There are
different types of analogtodigital converters such as the flash converter, the timewindow con-verter,
slope converter and tracking converter. Shown in Figure 8.37 is a flash converter ADC.
R3
120  R4
600 
200 

0
 +
120 
VA VB
R2 R4
R4
R2
R1
------R3
200
600
= = --------  120 = 40 
V  I
W W
P t W = Pt

Instruments
ANALOGTODIGITAL CONVERTER
12 V Supply

+

+
Overflow

+

+

+

+

+

+

+
8to3
Encoder
A8
A7
A6
A5
A4
A3
A2
A1
A0
Analog Input
B2
B1
B0
Comparator

+
VX
VY
Inputs Output
VX = VY Previous Value
VX  VY Ai = 0
VX  VY
Ai = 1
For Example, if Analog Input = 5.2 V,
then A0 = A1 = A2 = A3 = 1
and A4 = A5 = A6 = A7 = A8 = 0
Figure 8.37. Typical analogtodigital converter
12 V
10.5 V
9 V
7.5 V
6 V
4.5 V
3 V
1.5 V
0 V
Ai
Analog Input A8 A7 A6 A5 A4 A3 A2 A1 A0 B2 B1 B0
Less than 0 V 0 0 0 0 0 0 0 0 0 x x x†
0 to less than 1.5 V 0 0 0 0 0 0 0 0 1 0 0 0
1.5 to less than 3.0 V 0 0 0 0 0 0 0 1 1 0 0 1
3.0 to less than 4.5 V 0 0 0 0 0 0 1 1 1 0 1 0
4.5 to less than 6.0 V 0 0 0 0 0 1 1 1 1 0 1 1
6.0 to less than 7.5 V 0 0 0 0 1 1 1 1 1 1 0 0
7.5 to less than 9.0 V 0 0 0 1 1 1 1 1 1 1 0 1
9.0 to less than 10.5 V 0 0 1 1 1 1 1 1 1 1 1 0
10.5 to 12 V 0 1 1 1 1 1 1 1 1 1 1 1
Greater than 12 V 1 1 1 1 1 1 1 1 1 x x x‡
† Underflow
‡ Overflow
As shown in Figure 8.37, the flash type ADC consists of a resistive network, comparators (denoted
as triangles), and an eighttothree line encoder.
A digitaltoanalog converter (DAC) performs the inverse operation, that is, it converts digital
values to equivalent analog values.
Figure 8.38 shows a fourbit R2R ladder network and an opamp connected to form a DAC.

DIGITALTOANALOG CONVERTER
2R R R R
2R
2R 2R 2R 2R
B0 B1 B2 B3
(lsb) (msb)
Negative reference voltage is used so that the
+

Figure 8.38. A typical digitaltoanalog converter
 1 Volt

+
Switch Settings: For Logic “0” (ground) positioned to the right
For Logic “1” (+5 V) positioned to the left
With the switches positioned as shown, B3 B2 B1 B0 = 0100
inverting op amp’s output will be positive.
Vout
lsb = least significant bit
msb = most significant bit

Summary
8.14 Summary
 A periodic time function is one which satisfies the relation where n is a posi-tive
ft = ft + nT
integer and T is the period of the periodic time function.
 The average value of any continuous function over an interval ,is defined as
f t a  t  b
ftave
1
b – a
b
 1
= =  b
----------- f tdt
a
----------- area a
b – a
 The average value of a periodic time function f t
is defined as the average of the function
over one period.
 A halfwave rectification waveform is defined as
ft
Asint 0  t  
0  t 2     
=
 The effective current of a periodic current waveform is defined as
T
 IRoot Mean Square IRMS Ave i= = = =  2
IRMS = Ip  2 = 0.707Ip
Ieff
 For sinusoids only,
Ieff it
1T
--- i 2 dt
0
2 = + + + +
2 I1 RMS
2 I2 RMS
2  IN RMS
 For sinusoids of different frequencies,
IRMS I0
 For circuits with sinusoidal excitations the average power delivered to a load is
Pave LD
VpIp
2
= ----------- cos = VRMS LD IRMS LDcos
where is the phase angle between and and it is within the range ,and
 VLD ILD 0    90
is known as the power factor defined within the range .
cos 0  cos  1
 The average power in a resistive load is
Pave R
2
R
VRMS R
= = 2 R
------------------ IRMS R
 The average power in inductive and capacitive loads is
Pave X = VRMS X IRMS X cos90 = 0

 If the excitation in a circuit is nonsinusoidal, we can compute the average power absorbed by
a resistor from the relations
Pave
1T
T
 1T
= --- pdt
= = 
0
T
 1T
--- v 2
------dt
0
R
T
 An inductive load is said to have a lagging power factor and a capacitive load is said to have a
leading power factor.
 In a power triangle
 The apparent power , also known as complex power, is the vector sum of the real and reac-tive
power components, that is,
where the (+) sign is used for inductive loads and the () sign for capacitive loads.
 A power factor can be corrected by placing a capacitive load in parallel with the load of the
circuit.
 Ammeters are instruments used to measure current in electric circuits. Ammeters, milliamme-ters,
and microammeters must always be connected in series with the circuits in which they
are used.
 Voltmeters are instruments used to measure voltage. Voltmeters must always be connected in
parallel with those devices of the circuit whose voltage is to be measured.
 Ohmmeters are instruments used to measure resistance. When measuring resistance values,
the voltage sources in the circuit to which the unknown resistance is connected must be
turned off, and one end of the resistor whose value is to be measured must be disconnected
from the circuit.
 A Wheatstone Bridge is an instrument which can measure unknown resistance values very
accurately.
 Voltmeters, Ohmmeters, and Milliammeters (ammeters which measure current in milliam-peres)
are normally combined into one instrument called VOM.
 Wattmeters are instruments used to measure power.
--- i2Rdt
0
Preal = Pave = VRMS IRMS cos (in watts)
Q = Reactive Power = VRMS IRMS sin (in VARs)
Pa = Apparent Power = VRMS IRMS (in VAs)
Pa
Pa = Preal power  jQ = Pave  jQ

Summary
 WattHour meters are instruments used to measure energy.
 An oscilloscope is an electronic instrument that produces an instantaneous trace on the
screen of a cathoderay tube corresponding to oscillations of voltage and current.
 DC ammeters and DC voltmeters read average values
 AC ammeters and AC voltmeters read RMS values.
 Digital meters include an additional circuit called analogtodigital converter (ADC).

8.15 Exercises
Multiple Choice
1. The average value of a constant (DC) voltage of 12 V is
A.
6 V
B.
12 V
C.
12  2 V
D.
12  2 V
2. The average value of i = 5 + cos100t A
is
A.
5 + 2  2 A
B.
5  2 A
C.
5  2 A
D.
5 A
3. The RMS value of a constant (DC) voltage of 12 V
is
A.
12  2 V
B.
6  2  2 V
C.
12 V
D.
12  2 V
4. The RMS value of i = 5 + cos100t A
is
A.
B.
C.
D.
5 + 2  2 A
5  2 A
5  2 A
5 A

Exercises
5. The voltage across a load whose impedance is is 115 V RMS. The average
power absorbed by that load is
A.
176.33 w
B.
157.44 w
C.
71.3 w
D.
352.67 w
6. The average value of the waveform below is
24
4 8 12
v V
A.
B.
C.
D.
7. The RMS value of the waveform below is
A.
B.
C.
D.
Z = 75 + j38 
24 V
16 V
12 V
6 V
10
1 3
i A
t s
10  2 V
10  2 V
10  3 V
10  3 V

8. A current with a value of is flowing through a load that consists of the
series combination of , , and . The average power absorbed by
this load is
A.
B.
C.
D.
9. If the average power absorbed by a load is and the reactive power is , the
apparent power is
A.
B.
C.
D.
10. A load with a leading power factor of can be corrected to a lagging power factor of
by adding
A. a capacitor in parallel with the load
B. an inductor in parallel with the load
C. an inductor is series with the load
D. a capacitor in series with the load
Problems
1. The current through a inductor is given as . Compute:
a. The average values of the current, voltage and power for this inductor.
b. The values of the current and voltage.
i = 5cos10000t A
R = 2  L = 1 mH C = 10 F
25 w
10 w
5 w
0 w
500 watts 500 VAR
0 VA
500 VA
250 VA
500  2 VA
0.60 0.85
iLt 0.5 H iLt = 5 + 10sint A
RMS

Exercises
2. Compute the average and RMS
values of the voltage waveform below.
vt V
15
5
0
3. Compute the value of the voltage waveform below.
ts
RMS
vt
A
0
V
ts
4. Compute the RMS value of it = 10 + 2cos100t + 5sin200t
.
5. A radar transmitter sends out periodic pulses. It transmits for 5 s
and then rests. It sends out
one of these pulses every . The average output power of this transmitter is . Com-pute:
1 ms 750 w
a. The energy transmitted in each pulse.
b. The power output during the transmission of a pulse.
6. For the circuit below, vst = 100cos1000t V.
Compute the average power delivered (or
absorbed) by each device.
2 
5 
3 mH
200 F
vS t
7. For the circuit below, the input impedance of the PCB (Printed Circuit Board) is
and the board must not absorb more that of power; otherwise it
ZIN = 100–j100  200 mw

will be damaged. Compute the largest RMS value that the variable voltage source can be
adjusted to.
8. For the multirange ammeter/milliammeter shown below, the meter full scale is . Com-pute
the values of so that the instrument will display the indicated values.
980 
100 mA
10 mA
9. The circuit below is known as fullwave rectifier. The input and output voltage waveforms are
shown in Figure 8.47. During the positive input half cycle, current flows from point to point
, through to point , through the resistor to point , through diode to point ,
and returns to the other terminal point of the input voltage source. During the negative
input half cycle, current flows from point F to point E, through diode to point , through
the resistor to point , through the diode to point , and returns to the other terminal
point of the input voltage source. There is a small voltage drop across each diode* but it
can be neglected if . Compute the value indicated by the DC voltmeter.
* For silicon type diodes, the voltage drop is approximately 0.7 volt.
VS
VS
ZIN PCB
1 mA
R1 R2 R3 and R4
mA
IT
IM
IT
R1
+

R2
R3
RM =20 
1 A
A
B D2 C R D D3 E
F
D4 C
R D D1 B
A vD
vin » vD

Exercises
Diode  Allows current to flow in the indicated direction only
A
B
D1 D2
vOUT
R
 +
D C
E
F
V
DC Voltmeter
I
vIN
D4
D3
Vp
Vint
Vp sint
0 t (r)
Vp
t (r)
Voutt Vp sint

Multiple Choice
1. B
2. D
3. C
4. A
5. E
, and thus
Z = 75 + j38 = 84.0826.87 IRMS = 1150  84.0826.87 = 1.37–26.87
= = ----5 + 10sint = 5cost
T  1T--- 5 10 t sin +   t d
T = = 
T  = 0
T  1T
T =  = 0
T
 1T
T
 1T
T
 1T
T = = = = 
6. B
7. C
8. A
9. D
10. B
Problems
1. ,
a.
and since
it follows that
Likewise,
Also,
Pave = VRMS  IRMS cos = 115  1.37  cos–26.87 = 140.54 w
iL = 5 + 10sint vL L
diL
dt
------- 0.5 d
dt
iL ave
1T
--- iLdt
0
0
1T--- 10sintdt
0
1T
--- 5dt
0
= --- 5T = 5 A
vL ave
1T
--- 5costdt
0
pL ave
1T--- pL t d
0
--- vLiLdt
0
--- 5cost5 + 10sintdt
0
--- 25cost + 50sint costdt
0

and using it follows that and thus
b.
sin2x = 2sinxcosx 50sint cost = 25sin2t
pL ave
1T
= ---  T 25cost + 25sin2tdt
= 0
0
2 1T
IL RMS
T  1T
L --- 
T = idt
= 2
0
--- 5 + 10sint2dt
0
1T
T  25
--- T = 51 + 2sint2dt
= 
0
----- 1 4sint 4 2t  + + sin dt
T
0
= ------------------------ 1T
sin2x 1 – cos2x
T
 = 0 1T
--- 4sintdt
T
 = 0
--- cos2tdt
Using and observing that and we obtain
and
2
0
0
2 25
IL RMS
  25
T 42
T  + 
= = -----T + 2T = 75
----- t 0
T
--t 0
T
IL RMS = 75 = 8.66 A
For sinusoids and since it follows that
VRMS = Vp   2 = 0.707Vp Vp = 5
VRMS = 0.707  5 = 3.54 V
2. From the waveform below we observe that and since
Also,
Period = T = 5
Vave = Area  Period = 15 + 20  5 = 7 V
vt V
15
5
0
ts
 4
T
2 1T
VRMS
T  1
--- v2dt
= = = -----225 + 125 – 25 = 65
0

 52dt
----- 152dt
5
0
5
+  1

5

VRMS = 65 = 8.06 V
T
vt
A
0
V
ts
T
vt = -------
2A
T
y = mx + b 0  t  T  2 vt 2A
= -------t
2 1T
T
 1T
+  4A2
T  2
 1T
 2
= = -------t
= 
4A2
---------t3
3T3 T
T  2
T  2
= = ------A = 0.41A
and thus
3. We choose the period as shown below.
Using the straight line equation we find that for , . Then,
and
4. The effective (RMS) value of a sinusoid is a real number that is independent of frequency and
phase angle and for current it is equal to . The RMS value of sinusoids with dif-ferent
frequencies is given by (8.13). For this problem
5. The waveform representing the transmitter output pulses is shown below.
T
VRMS
--- v2dt
0
--- 2A
T
dt
0
--- 0  dt
T  2
--------- t2dt
T3 0
0
4A2
24
= = --------- = A2  6
VRMS A2  6 6
6
IRMS = Ip  2
IRMS 102 12
--22 12--52 = + + = 100 + 2 + 12.5 = 10.7 A
5 s
ts
A
1 2

For this problem we do no know the amplitude of each pulse but we know the average
power of one period . Since
T = 1 s
----------------- Area
Pave 750 w Area
= = = ------------
it follows that:
a. Energy transmitted during each pulse is
A 5 s
Period
Area of each pulse = 750 w  s
b. The power during the transmission of a pulse is
1 s
P W t  750 w s  5 s  750 w s  5 106
– = = =   = 150  106 w = 150 Mw
jL j103
3 103
– =   = j3 
6. The phasor equivalent circuit is shown below where and
j – C  j – 103 2 104
– =    = –j5 
By application of KCL
Also,
2 
5 
VS
z1
VC
I2  IC
1000
j3 
z2
IL
–j5 
z3
VC – --------------------
VS
z1
VC
z2
+ -------
+ ------- = 0
VC
z2
1
z1
---- 1
---- 1
 VC
 + + ---- 
z2
z3
VS
z1
= ------
VC
VS
= --------------------------------------------------
1 + z1  z2 + z1  z3 
I2 
VS – VC
z1
= -------------------- IC
VC
z1
= ------- IL
VC
z3
= -------
and with MATLAB
Vs=100; z1=2; z2=5j; z3=5+3j;...
Vc=Vs/(1+z1/z2+z1/z3); I2=(VsVc)/z1; Ic=Vc/z2; IL=Vc/z3; fprintf(' n');...
disp('Vc = '); disp(Vc); disp('magVc = '); disp(abs(Vc));...
disp('phaseVc = '); disp(angle(Vc)*180/pi);...
disp('I2 = '); disp(I2); disp('magI2 = '); disp(abs(I2));...
disp('phaseI2 = '); disp(angle(I2)*180/pi);...

disp('Ic = '); disp(Ic); disp('magIc = '); disp(abs(Ic));...
disp('phaseIc = '); disp(angle(Ic)*180/pi);...
disp('IL = '); disp(IL); disp('magIL = '); disp(abs(IL));...
disp('phaseIL = '); disp(angle(IL)*180/pi);
Vc = 75.0341-12.9604i
magVc = 76.1452
phaseVc = -9.7998
I2 = 12.4829 + 6.4802i
magI2 = 14.0647
phaseI2 = 27.4350
Ic = 2.5921 + 15.0068i
magIc = 15.2290
phaseIc = 80.2002
IL = 9.8909 - 8.5266i
magIL = 13.0588
phaseIL = -40.7636
The average power delivered by the voltage source is computed from the relation
= 2R2  = 0.5  14.072  2 = 197.97 w
= 2R5  = 0.5  13.062  5 = 426.41 w
where as shown by the phasor diagram below.
Therefore,
Also,
and
VS
Pave VRMSIRMS  cos 12--VpIp  cos = =
 = 27.43
VS
I2 
 = 27.43
PS ave
12
= --  VS  I2  cos = 0.5  100  14.07  cos27.43 = 624.4 w
P2  ave
12
--Ip
P5  ave
12
--IL

Check:
P2  ave + P5  ave = 197.97 + 426.41 = PS ave = 624.4 w
The average power in the capacitor and the inductor is zero since and .
7. Let us consider the network below.
Let
and
Then,
and using
we obtain
 = 90 cos = 0
t – domain
VS
it
ZIN PCB
vSt
vS = Vp cost
i = Ip cost + 
p = vSi = VpIp cost  cost + 
x y cos  cos 12
= -- cosx + y + cosx – y
p
VpIp
2
= ----------- cos2t +  + cos
We require that the power p does not exceed 200 mw or 0.2 w
, that is, we must satisfy the
condition
p
VpIp
2
= ----------- cos2t +  + cos  0.2 w
and therefore we must find the phase angle . Since appears also in the , we
can find its value from the given input impedance, that is, or
and in the
  j – domain
ZIN = 100 – j100 
ZIN ZIN  100 2 100 2 + –100 –1
= = tan--------------------- = 100 2–45
100
t – domain
p
VpIp
2
= ----------- cos2t – 45 + cos–45
The maximum power occurs when , that is,
p cos2t – 45 = 1
pmax
VpIp
2
----------- 1 2
=   = 0.2 w
 + ------
2

VpIp = 0.4  1.707
Ip Vp ZIN = 100 2
Ip = Vp  ZIN
2 0.4  100 2
Vp
= ------------------------------ = 33.14
1.707
Vp = 33.14 = 5.76
VRMS
Vp
2
------ 5.76
= = ------------- = 4.07 V
1.414
10 mA
9 mA 1 mA
mA
980 
R1
10 mA
+

R2
R3
20 
10 mA
9 103
–  R1 + R2 + R3 980 + 20 10–3 = 
R1 + R2 + R3 = -----------
1000
9
980 
1 mA 1 mA
Then,
and now we can express in terms of using the relation and
and by substitution
or
and
8. With the switch at the position, the circuit is as shown below.
Then,
or
(1)
With the switch at the position, the circuit is as shown below.
100 mA
mA
R1
+

R2
R3
20 
100 mA
100 mA
99 mA

Then,
or
99 10–3  R2 + R3 R1 + 980 + 20 10–3 = 
(2)
– R1 + 99R2 + 99R3 = 1000
With the switch at the position, the circuit is as shown below.
Then,
or
980 
999 10–3 R3 R1 + R2 + 980 + 20 10–3 = 
(3)
1 A
+

Addition of (1) and (3) yields
or
1 mA
1 A 999 mA
– R1 – R2 + 999R3 = 1000
----------- + 1000 10000
= = --------------
(4)
1000R3
Addition of (1) and (2) yields
or
R1
R2
R3
----------- + 1000 10000
= = --------------
(5)
1 A
100R2 + 100R3 1000
Substitution of (4) into (5) yields
R2 + R3 = --------
100
(6)
1000
9
R3
10
9
= ----- 
R2 = 10 
and substitution of (4) and (6) into (1) yields
(7)
1 mA
mA
20 
9
9
9
9
R1 = 100 

9. DC instruments indicate average values. Therefore, the DC voltmeter will read the average
value of the voltage across the resistor. The period of the fullwave rectifier waveform is
taken as .
Then,

 Vp
------–cost

0 Vp
------1 + 1
As expected, this average is twice the average value of the halfwave rectifier waveform in
Example 8.2.
vOUT

Vp
t (r)
Voutt Vp sint
 2
vOUT ave
1
-- Vp sintdt
0

t = 0
= =
Vp

------ cost


2Vp

= = = ---------

Chapter 9
Natural Response
his chapter discusses the natural response of electric circuits.The term natural implies that
there is no excitation in the circuit, that is, the circuit is sourcefree, and we seek the cir-cuit’s
natural response. The natural response is also referred to as the transient response.
T
9.1 Natural Response of a Series RL circuit
Let us find the natural response of the circuit of Figure 9.1 where the desired response is the cur-rent
i, and it is given that at , , that is, the initial condition is .
t = 0 i = I0 i0 = I0
+

R L
+
i 
Figure 9.1. Circuit for determining the natural response of a series RL circuit
Application of KVL yields
or
(9.1)
vL + vR = 0
Ldi
---- dt
+ Ri = 0
Here, we seek a value of i which satisfies the differential equation of (9.1), that is, we need to find
the natural response which in differential equations terminology is the complementary function. As
we know, two common methods are the separation of variables method and the assumed solution
method. We will consider both.
1. Separation of Variables Method
Rearranging (9.1), so that the variables i and t are separated, we obtain
---- RL
di
i
= –---dt
Next, integrating both sides and using the initial condition, we obtain
i
 RL
1i
--di
I0
t
= – 
--- d
0

Chapter 9 Natural Response
i RL
t = –
---- ln RL
where is a dummy variable. Integration yields
or
or
Recalling that implies , we obtain
(9.2)
Substitution of (9.2) into (9.1) yields and that at , . Thus, both the differ-ential
equation and the initial condition are satisfied.
2. Assumed Solution Method
Relation (9.1) indicates that the solution must be a function which, when added to its first deriv-ative
will become zero. An exponential function will accomplish that and therefore, we assume a
solution of the form
(9.3)
where and are constants to be determined. Now, if (9.3) is a solution, it must satisfy the dif-ferential
equation (9.1). Then, by substitution, we obtain:
or
The left side of the last expression above will be zero if , or if , or if . But,
if or , then every response is zero and this represents a trivial solution. Therefore,
is the only logical solution, and by substitution into (9.3) we obtain
We must now evaluate the constant . This is done with the use of the initial condition
. Thus, or and therefore,

lni I0
--- 0
lni – lnI0 RL
= –---t
i
I0
= –---t
x = lny y ex =
it I0 e –R  Lt =
0 = 0 t = 0 i0 = I0
it Aest =
A s
RAest sLAest + = 0
s RL--- +  
 Aest = 0
A = 0 s = – s = –R  L
A = 0 s = –
s = –R  L
it Ae –R  Lt =
A
i0 = I0 I0 Ae0 = A = I0
it I0e –R  Lt =

Natural Response of a Series RL circuit
as before. Next, we rewrite it as
(9.4)
and sketch it as shown in Figure 9.2.
it
I0
-------- e –R  Lt =
e–R  Lt
it  I0 = – R  Lt + 1
13.5%
36.8%
Time constants
5%
it  I0
Figure 9.2. Plot for in a series RL circuit
Percent i(t)/I0
From Figure 9.2 we observe that at t = 0 , i  I0 = 1 , and i  0 as t  
.
The initial rate (slope) of decay is found from the derivative of evaluated at , that is,
i  I0 t = 0
---- d
i
dt
I0
 
 ---- 
t = 0
RL
–---e –R  Lt
= = –---
t = 0
RL
and thus the slope of the initial rate of decay is
Next, we define the time constant as the time required for to drop from unity to zero
assuming that the initial rate of decay remains constant. This constant rate of decay is repre-sented
by the straight line equation
and at , . Then,
or
(9.5)
–R  L
 i  I0
-------- RL
it
I0
= – ---t + 1
t =  i  I0 = 0
0 RL
= – --- + 1
 LR
= ---
Time Cons tant for RL Circuit

Evaluating (9.4) at , we obtain
or
(9.6)
Therefore, in one time constant, the response has dropped to approximately 36.8% of its initial
value.
If we express the rate of decay in time constant intervals as shown in Figure 9.2, we find that
after , that is, it reaches its final value after five time constants.
Example 9.1
For the circuit shown in Figure 9.3, in how many seconds after has the
a. current has reached ½ of its initial value?
b. energy stored in has reached ¼ of its initial value?
c. power dissipated in has reached ¾ of its initial value?
Solution:
From (9.2),
where . Then,
a. The current will have reached ½ of its initial value when
or
or
t =  = L  R
i
I0
--------- e–R  L e–R  LL  R e–1 = = = = 0.368
i = 0.368I0
it  I0  0 t = 5
t = 0
it
L
R
+
+
R
i 

10 
L
10 mH
it I0 e –R  Lt =
I0 = iL0
it
0.5I0 I0e
10 10 10–3 –   t
I0e–1000t = =
e–1000t = 0.5
–1000t = ln0.5 = –0.693

and therefore,
t = 693 s
b.To find the energy stored in L
which reaches ¼ of its initial value, we begin with
and at , . Then,
and
t = 0 I0 = iL0
Therefore,
or
and
WL t   12
= --Li2t
WL 0   12
2 =
--LI0
-- 12
--WL 0   14
14
=  
--LI0
2
 
14
--WL t   12
--Li2 t   12
--L I0 e –R  Lt  
2 14
-- 12
= = =  
--LI0
2
 
e –2R  Lt = 1  4
e–2000t = 1  4
–2000t = ln0.25 = –1.386
t = 693 s
This is the same answer as in part (a) since the energy is proportional to the square of the cur-rent.
c. To find the power dissipated in when it reaches ¾ of its initial value, we start with the fact
that the instantaneous power absorbed by the resistor is , and since for the given
circuit
then,
R
pR iR 2
= R
it = iRt =
I0 e –R  Lt pR =
I0
2Re –2R  Lt and the energy dissipated (in the form of heat) in the resistor is

 I0
2 = = = =
WR pRdt
0
Also, from part (b) above,
and thus

 I0
2R e–2R  Lt dt
0
2R L
 e –2R  Lt
–-------
2R
 12
0
--LI0
WL 0   12
2 =
--LI0

34
--WR
34
--WL 0   12
--Li2 t   12
--L I0e –R  Lt  
2 34
-- 12
= = = =  
--LI0
2
 
e–2R  Lt = 3  4
e–2000t = 3  4
–2000t = ln0.75 = –0.288
t = 144 s
iL0 = iL0 = iL0+
iL0 iL0+
S
t = 0 iLt t  0 vR0 vR0+
+
32 V
+

t = 0
1 mH
S
20 
iLt
10 
vRt
or
and
In some examples and exercises that follow, the initial condition may not be given directly but it
can be found from the fact that the current through an inductor cannot change instantaneously
and therefore,
(9.7)
where will be used to denote the time just before a switch is opened or closed, and
will be used to denote the time just after the change has occurred.
Also, in our subsequent discussion, the expression “long time” will mean that sufficient time has
elapsed so that the circuit has reached its steadystate conditions. As we know from Chapter 5,
when the excitations are constant, at steady state conditions the inductor behaves as a short cir-cuit,
and the capacitor behaves as an open circuit.
Example 9.2
In the circuit of Figure 9.4, the switch has been in the closed position for a long time and opens
at . Find for , , and
Solution:
We are not given an initial condition for this example; however, at t =
0 the inductor acts as a
short thereby shorting also the 20 
resistor. The circuit then is as shown in Figure 9.5.

Circuit at t 0 =
32 V
+
10 
 +
vRt
iLt
Figure 9.5. Circuit for Example 9.2 at
From the circuit of Figure 9.5, we observe that
t 0 =
iL0 = iL0 = iL0+ = 32  10 = 3.2A
and thus the initial condition has now been established as . We also observe that
I0 = 3.2 A
vR0 = 0
t 0+ = 32 V 10 
At ,the source and the resistor are disconnected from the circuit which now is
Circuit at t 0+ =
+
vRt

1 mH
20  iLt
For the circuit of Figure 9.6,
or
and
or
We observe that
t 0+ =
iLt I0e –R  Lt 3.2e
20 10–3 –  t
= =
iLt 3.2e–20000t =
vR0+ = 20–I0 = 20–3.2
vR0+ = –64 V
vR0+  vR0

Example 9.3
In the circuit shown in Figure 9.7, the switch has been closed for a long time and opens at
S
t = 0
iLt t  0
i60t t = 100 s
i48t t = 200 s
72 V
+
4  30 
1 mH
S
t = 0
24 
48 
60 
i60t iLt i48t
t 0 = 24  48 
72 V
+
Circuit at t 0 =
4  30 
60 
iL0
iT 0
t 0 =
iT 0 72 V
---------------------------- 72 V
= = --------------- = 3 A
4 + 60 || 30
4 + 20
iL0 60
------------------ iT 0    69
= = --  3 = 2 A
30 + 60
. Find:
a. for
b. at
c. at
Solution:
a. At the inductor acts as a short thereby shorting also the and resistors. The
circuit then is as shown in Figure 9.8.
Then,
and by the current division expression,
and thus the initial condition has been established as
I0 = 2 A
At t = 0+ ,the 72 V source and the 4 
resistor are disconnected from the circuit which now

Natural Response of a Series RC Circuit
Circuit at t 0+ =
30  24 
1 mH
i60 t iLt
i48 t
48 
60 
From (9.2),
where
and thus
or
Also,
or
and
or
t 0+ =
iLt I0e
–Req  Lt
=
Req = 60 + 30 || (24+48) = 40 
iLt 2e 40 10 –3 –  t =
iLt 2e–40000t =
i60t
t = 100 s
24 + 48
----------------------------------------------------–iLt
30 + 60 + 24 + 48
t = 100 s
=
i60t
t = 100 s
12
27
= ----- – 2e–40000t 
= – --e–4 = –16.3 mA
t = 100 s
89
i48t
t = 200 s
30 + 60
----------------------------------------------------–iLt
30 + 60 + 24 + 48
t = 200 s
=
i48t
t = 200 s
15
27
= ----- – 2e–40000t 
= – -----e–8 = –0.373 mA
t = 200 s
10
9
9.2 Natural Response of a Series RC Circuit
In this section, we will find the natural response of the RC
circuit shown in Figure 9.10 where
the desired response is the capacitor voltage vC , and it is given that at t = 0 , vC = V0
, that is,
the initial condition is .
v0 = V0

Figure 9.10. Circuit for determining the natural response of a series RC circuit
+ ---------- = 0
 Aest = 0
 + --------
By KCL,
(9.8)
and with
and
by substitution into (9.8), we obtain the differential equation
(9.9)
As before, we assume a solution of the form
and by substitution into (9.9)
or
(9.10)
Following the same reasoning as with the circuit, (9.10) will be satisfied when
and therefore,
The constant is evaluated from the initial condition, i.e., or .
Therefore, the natural response of the circuit is
(9.11)
We express (9.11) as
R
+
C
iC
vCt
iR
iC + iR = 0
iC C
dvC
dt
= ---------
iR
vC
R
= -----
dvC
dt
---------
vC
RC
+ -------- = 0
vCt Aest =
Asest Aest
RC
s 1
RC
RL s = –1  RC
vC t Ae –1  RCt =
A vC0 V0 Ae0 = = A = V0
RC
vCt V0e –1  RCt =
vCt
V0
------------ e–1  RCt =

and we sketch it as shown in Figure 9.11.
Percent vCt   V0
e–1  RCt
vCt  V0 = – 1  RCt + 1
13.5%
36.8%
Time constants
5%
Figure 9.11. Circuit for determining the natural response of a series RC circuit
From Figure 9.11 we observe that at t = 0 , vC  V0 = 1 , and i  0 as
t  
The initial rate (slope) of decay is found from the derivative of vCt  V0 evaluated at t = 0
,
that is,
d
dt
----
vC
V0
 
 ------ 
t = 0
1
RC
–--------e –1  RCt
= = –--------
t = 0
1
RC
and thus the slope of the initial rate of decay is
Next, we define the time constant  as the time required for vCt  V0
to drop from unity to zero
assuming that the initial rate of decay remains constant. This constant rate of decay is repre-sented
by the straight line equation
(9.12)
and at , . Then,
or
(9.13)
vCt
V0
------------- 1
= – --------t + 1
t =  vCt  V0 = 0
0 1
= – -------- + 1
 = RC
Time Cons tant for RC Circuit
Evaluating (9.11) at , we obtain
–1  RC
RC
RC
t =  = RC

vC
V0
------------- e–  RC e–RC  RC e–1 = = = = 0.368
vC = 0.368V0
vCt  V0  0 t = 5
vC0 = vC0 = vC0+
S
t = 0 vC t t  0 i0 i0+
t = 0
or
(9.14)
Therefore, in one time constant, the response has dropped to approximately 36.8% of its initial
value.
If we express the rate of decay in time constant intervals as shown in Figure 9.11, we find that
after , that is, it reaches its final value after five time constants.
In the examples that follow, we will make use of the fact that
(9.15)
Example 9.4
In the circuit of Figure 9.12, the switch has been in the closed position for a long time, and
opens at . Find for , , and .
Solution:
At the capacitor acts as an open. The circuit then is as shown in Figure 9.13.
From the circuit of Figure 9.13 we observe that
60 V
10 K
50 K
+
+

S
i t 10 F vC t
t 0 =
60 V
10 K
50 K
+
Circuit at t 0 =
i t
vC t
t 0 =

vC0 vC0+ 50 K  i0 50 60 V
= = =  ----------------------------------------- = 50 V
10 K + 50 K
and thus the initial condition has been established as . We also observe that
V0 = 50 V
i0 60 V
= ----------------------------------------- = 1 mA
10 K + 50 K
t 0+ = 60 V 10 K
At the source and the resistor are disconnected from the circuit which now
From (9.11),
vCt V0e –1  RCt =
Circuit at t 0+ =
+
 10 F
i t vC t
50 K
where
Then,
and
t 0+ =
RC 50  10 3 10 10–6 =   = 0.5
vCt 50e–1  0.5t 50e–2t = =
i0+
V0
R
------ 50 V
= = ----------------- = 1 mA
50 K
i0+ = i0
We observe that . This is true because the voltage across the capacitor cannot
change instantaneously; hence, the voltage across the resistor must be the same at and at
.
t 0 =
t 0+ =
Example 9.5
In the circuit of Figure 9.15, the switch S
has been in the closed position for a long time and
opens at t = 0
. Find:
a. for
b. at
vC t t  0
v60 t t = 100 s

v10 t t = 200 s
72 V
30 K
S 20 K
t = 0 + +
60 K
+
6 K
v60 t vC t v10 t
  
10 K
+
40
9
----- F
t 0 =
72 V
Circuit at t 0 =
30 K
6 K
iT t i10 t
60 K +
20 K
+ +
v60 t vC t 10 K
v10 t
+
  
t 0 =
------------------------------------------------------------- 72 V
iT 0 72 V
= = -------------------------------------- = 2 mA
6 K + 60 K  60 K
6 K + 30 K
i10 0 60 K
----------------------------------------- iT 0    12
= = --  2 = 1 mA
60 K + 60 K
vC 0 = 20 K + 10 K  i10 0 = 30 V
c. at
Solution:
a. At the capacitor acts as an open and the circuit then is as shown in Figure 9.16.
From the circuit of Figure 9.16,
and using the current division expression, we obtain
Then,
and thus the initial condition has been established as V0 = 30 V
.
At t = 0+ , the 72 V source and the 6 K
 resistor are disconnected from the circuit which
now is as shown in Figure 9.17.

Circuit at t 0+ =
30 K
20 K
+
+ +
40 vC t
9
v60 t ----- F v10 t
  
60 K 10 K
From (9.11),
where
Then,
and
b.
or
c.
or
t 0+ =
vCt V0e
–1  ReqCt
=
Req = 60 K + 30 K || 20 K + 10 K = 22.5 K
 40
ReqC 22.5 10 3
----- 10–6 =   = 0.1
9
vCt 30e –1  0.1t 30e–10t = =
v60t
t = 100 ms
60 K
-----------------------------------------  vCt
30 K + 60 K
t = 100 ms
=
v60t
t = 100 ms
23
-- 30e–10t  
20e–1 = = = 7.36 V
t = 100 ms
v10t
t = 200 ms
10 K
-----------------------------------------  vCt
10 K + 20 K
t = 200 ms
=
v10t
t = 200 ms
13
10e–2 = = = 1.35 V
-- 30e–10t  
t = 200 ms
Example 9.6
For the circuit of Figure 9.18, it is known that .
a. To what value should the resistor be adjusted so that the initial rate of change would be

vC 0 = V0 = 25 V
R
–200 V  s
b. What would then the energy in the capacitor be after two time constants?

C
+

10 F
vCt R
vCt V0e–1  RCt =
vCt 25e–100000  Rt =
–200 V  s
dvC
dt
---------
t = 0
100000
R
  25e–100000  Rt
= = –---------------------- = –200
–----------------- 
t = 0
– = =   3.382 = 57.2 J
Solution:
a. The capacitor voltage decays exponentially as
and with the given values,
Now, if the initial rate (slope) is to be  then
and solving for we obtain
b. After two time constants the capacitor voltage will drop to the value of
Therefore, the energy after two time constants will be
2.5  106
R
R R = 12.5 K
vC2 25e–1  RC2 25e–2RC  RC 25e–2 = = = = 3.38 V
WC t = 2
12
--CvC 2
5 10 6

Summary
9.3 Summary
 The natural response of the inductor current in a simple circuit has the form
iLt RL
iLt = I0 e –R  Lt where I0 denotes the value of the current in the inductor at
t = 0
 In a simple RL circuit the time constant  is the time required for iLt  I0
to drop from unity
to zero assuming that the initial rate of decay remains constant, and its value is
 = L  R
 In one time constant the natural response of the inductor current in a simple RL
circuit has
dropped to approximately 36.8%
of its initial value.
 The natural response of the inductor current in a simple RL
circuit reaches its final value,
that is, it decays to zero, after approximately 5
time constants.
 The initial condition I0
can be established from the fact that the current through an inductor
cannot change instantaneously and thus
iL0 = iL0 = iL0+
 The natural response of the capacitor voltage in a simple circuit has the form
vCt RC
vCt = V0 e –1  RCt where V0 denotes the value of the voltage across the capacitor at
t = 0
 In a simple RC circuit the time constant  is the time required for vCt  V0
to drop from
unity to zero assuming that the initial rate of decay remains constant, and its value is
 = RC
 In one time constant the natural response of the capacitor voltage in a simple RC
circuit has
dropped to approximately 36.8%
of its initial value.
 The natural response of capacitor voltage in a simple RC
circuit reaches its final value, that
is, it decays to zero after approximately 5
time constants.
 The initial condition V0
can be established from the fact that the voltage across a capacitor
cannot change instantaneously and thus
vC0 = vC0 = vC0+

RL 
s–1
S1
t = 0
9.4 Exercises
Multiple Choice
1. In a simple circuit the unit of the time constant is
A. dimensionless
B. the millisecond
C. the microsecond
D. the reciprocal of second, i.e.,
2. In a simple RC circuit the unit of the term 1  RC
is
A. the second
B. the reciprocal of second, i.e.,
s–1
C. the millisecond
D. the microsecond
3. In the circuit below switch S1 has been closed for a long time while switch S2
has been open
for a long time. At t = 0 . switch S1 opens and switch S2 closes. The current iLt for all
t  0
is
A.
2 A
B.
2e–100t A
C.
2e–50t A
D.
e–50t A
t = 0
S2 iLt
5 
5 
2 A
100 mH

Exercises
4. In the circuit below switch has been closed for a long time while switch has been open
for a long time. At . switch opens and switch closes. The voltage for all
is
t = 0 S1 S2 vCt
t  0
A.
B.
C.
D.
S1 S2
10 V
10e–10t V
10e–t V
10e–0.1t V
50 K 50 K
S1
t = 0
+
+ 20 F
 
t = 0
S2
vCt
10 V
5. In the circuit below switch has been closed for a long time while switch has been open
S1 S2
for a long time. At t = 0 . switch S1 opens and switch S2
closes. The power absorbed by the
inductor at t = +
will be
A.
0 w
B.
1 w
C.
2 w
D.
0.2 w
S1
t = 0
t = 0
5 
S2 iLt
5 
2 A
100 mH

6. In the circuit below, switch S1 has been closed for a long time while switch S2
has been open
for a long time. At . switch opens and switch closes. The power absorbed by the
capacitor at will be
A.
B.
C.
D.
50 K 50 K
S1
t = 0
+ 20 F
7. In a simple circuit where and the time constant is
A.
B.
C.
D.
8. In a simple circuit where and the time constant is
A.
B.
C.
D.
t = 0 S1 S2
t = +
0 w
10 w
5 w
10 mw
t = 0
S2
vCt
10 V
+
 
RL R = 10 M L = 10 H 
1 s
100 s
1012 s
10–12 s
RC R = 10 M C = 10 F 
100 s
0.01 s
100 s
0.01 s

Exercises
9. In a simple RL
circuit the condition(s) ___ are always true.
A. and
iL0 = iL0 = iL0+ vL0 = vL0 = vL0+
iL0 = iL0 = iL0+ iR0 = iR0 = iR0+
iL0 = iL0 = iL0+ vR0 = vR0 = vR0+
iL0 = iL0 = iL0+
B. and
C. and
D.
E. none of the above.
10. In a simple RC
circuit the condition(s) ___ are always true.
vC0 = vC0 = vC0+ iC0 = iC0 = iC0+
vC0 = vC0 = vC0+ vR0 = vR0 = vR0+
vC0 = vC0 = vC0+
vC0 = vC0 = vC0+ iR0 = iR0 = iR0+
A. and
B. and
C.
D. and
E. none of the above.
Problems
1. In the circuit below, switch S1 has been closed for a long time and switch S2
has been open
for a long time. Then, at t = 0 switch S1 opens while S2 closes. Compute the current
iS2t
through switch for .
S2 t  0
15 V
8 
3 
5 
iS2t
S1
t = 0
+  6 
10 
2.5 mH
t = 0
S2
2. In the circuit below, both switches S1 and S2
have been closed for a long time and both are
opened at . Compute and sketch the current for the time interval
t = 0 iLt 0  t  1 ms

12 V
2 
3. In a series circuit, the voltage across the inductor is and the current
at is . Compute the values of and for that circuit.
4. In the circuit below both switches and have been closed for a long time, while switch
has been open for a long time. At and are opened and is closed. Compute the
current for .

1 K 2 K
iLt
vin2 +
vin1 vout
5. In the circuit below switch has been closed and has been open for a long time. At
switch is opened and is closed. Compute the voltage for .
t = 0
S1 S2
t = 0
vC1t vC2t
24 V
16  8 
+
4 
10 
12 
+
6 
iLt
t = 0
t = 0
S1
S2
10  3 mH
RL vL vL 0.2e–2000t = V
iL t = 0 iL0 = 10 mA R L
S1 S2 S3
t = 0 S1 S2 S3
iLt t  0
+
+
+
10 mV 20 mV

3 mH
10 K
5 K
t = 0
t = 0
S3
S1 S2
t = 0
10 K
S1 S2 t = 0
S1 S2 vC2t t  0
12 V
10 K 50 K
+
+

6 F
10 K
3 F
+


Exercises
6. In the circuit below switch S has been in the A position for a long time and at t = 0
is
thrown in the B position. Compute the voltage vC t across the capacitor for t  0
, and the
energy stored in the capacitor at .
t = 1 ms
24 V
4 K 16 K
+
+

S
5 F
2 K A
B
8 K
6 K
t = 0
vCt
7. In the circuit below switch has been open for a long time and closes at . Compute
for .
S t = 0
iSWt t  0
100 
6 mH 10 F
3 
6 
+

iSWt
t = 0
36 V S

All resistor values in Ohms
v +-
VM 2
72
Display 1
s
-
Manual
Switch in
2
Simulink Continuous
CVS = Controlled Voltage Source
4 6
Ground in
SimPower Systems
v +-
VM 1
Ground in
Simulink
CM=Current Measurement
powergui
+
+
CVS = Controlled Voltage Source
72
Display 1
s
-
Manual
Switch in
Simulink Continuous
+
+
8.
a. In the Simulink / SimPowerSystems model shown below, are the values shown in the Dis-play
blocks justified after the simulation command is issued?
b. When the Manual Switch block is double-clicked the model is as shown below. Are the
values shown in the Display blocks justified after the simulation command is issued?
0.01824
Display 4
18
Display 3
36
Display 2
+
CVS
i
-
CM 1
8
72V DC
1 mH 3
i
-
CM 2
2
4 6
Ground in
SimPower Systems
All resistor values in Ohms
Ground in
Simulink
CM=Current Measurement
powergui
v +-
VM 2
v +-
VM 1
0
Display 4
0
Display 3
0
Display 2
+
CVS
i
-
CM 1
8
72V DC
1 mH 3
i
-
CM 2

Multiple Choice
1. E
2. B
3. D
4. C
5. A
6. A
7. D
8. B
9. D
10. C
Problems
1. The circuit at is as shown below.
 = L  R = volt  ampere  second  volt  ampere = second s
t 0 =
3 
6  +
15 V
8 
x

10 
5 
y iL0

Replacing the circuit above with its Thevenin equivalent to the left of points and we find
that and and attaching the rest of the cir-cuit
vTH
6
3 + 6
= ------------  15 = 10 V RTH
to it we obtain the circuit below.
x y
3  6
3 + 6
= ------------ + 8 = 10 
10 
RTH
10  +
10 V
vTH
5 
iL0
By voltagesource to currentsource transformation we obtain the circuit below.

10  10 
5 
and by inspection, , that is, the initial condition has been established as
The circuit at is as shown below.
We observe that the closed shorts out the and resistors and the circuit simplifies to
that shown below.
Thus for ,
2. The circuit at is as shown below and the mesh equations are
Then,
iL0
5 
1 A
iL0 = 0.5 A
iL0 = iL0 = iL0+ = I0 = 0.5 A
t 0+ =
8 
6 
2.5 mH
5 
iS2t
Closed
Switch
iL0+ = 0.5 A
6  8 
2.5 mH
5 
iS2t
iL0+ = I0 = 0.5 A
t  0 iS2t –iLt I0e–R  Lt – 0.5e 5 2.5 10 3 – –   t – 0.5e–2000t = = = = – A
t 0 =
20i1 – 4i3 = 24
16i2 – 6i3 – 8i4 = –12
– 4i1 – 6i2 + 20i3 – 10i4 = 0
8i2 – 10i3 + 30i4 = 0
iL0 = i3 – i4

24 V
8 
16 
+
+
12 V
i2
6 
2 
iL0
4  10  12 
i i4 3
i1
and with MATLAB
R=[20 0 4 0; 0 16 6 8; 4 6 20 10; 0 8 10 30];...
V=[24 12 0 0]'; I=RV; iL0=I(3)I(4); fprintf(' n');...
fprintf('i1 = %4.2f A t', I(1)); fprintf('i2 = %4.2f A t', I(2));...
fprintf('i3 = %4.2f A t', I(3)); fprintf('i4 = %4.2f A t', I(4));...
fprintf('iL0 = %4.2f A t', I(3)I(4)); fprintf(' n'); fprintf(' n');
i1 = 1.15 A i2 = -1.03 A i3 = -0.26 A i4 = -0.36 A iL0 = 0.10 A
Therefore,
iL0 = iL0 = iL0+ = I0 = 0.1 A
t 0+ =
Shown below is the circuit at and the steps of simplification.
6 
4 
iL0+
10 
Thus for ,
and
8 
10
3
-----mH
iL0+ -----mH
20/3 
12  10  20 
10 
10
3
iL0+
20/3 
10
3
-----mH
t  0
iLt I0e–R  Lt 0.1e–5000t = = A
iL t 0.4 ms =
0.1e–2 = = 0.0137 A = 13.7 mA
iLt 0  t  1 ms
To compute and sketch the current for the time interval we use MATLAB
as shown below.
t=(0: 0.01: 1)*10^(3);...
iLt=0.1.*10.^(3).*exp(5000.*t);...
plot(t,iLt); grid

+ +
 
– =  = 20  R  L = 2000 L = 20  2000 = 0.01 = 10 mH
 
 
-------------- 2 102
–   10 2 10–2 = = –   = –0.2 V
3. From the figure below for
and with , by substitution
or Also, from ,
4. The circuit at is as shown below and using the relation
that was developed in Example 4.11 we have
and
vL RiL 0.2e–2000t = = t  0
iL vL
R L
iL0 = 10 mA R 10 10–3    0.2e–0 = = 0.2
R 0.2 102
t 0 =
vout Rf
vin1
Rin1
----------
vin2
Rin2
+ ----------
 
= –
vout 10 K 10 2 –
1 K
– 
2 K
 + ------------------- 
iL0 I0 i5 K
–0.2 V
5 K
---------------- 40 106
– = = = = –  A = –40 A

1 K 2 K
+
10 K

+
+

vin2 +

vin1 20 mV
vout
iLt
5 K
t 0+ = iL0+ = 40 A
The circuit at is as shown below where with the direction shown.
Then for
3 mH
+
10 K

iL0+
5 K

0.2V
+
 
iL0+
+ 3 mH
+
15 K
t  0
iLt I0eR  Lt 40 10–6  e 15 103  3 103
–    t 40e 5 106
–  t = = = A
with the direction shown.
5. The circuit at t =
0 is as shown below. As we’ve learned in Chapter 5, when a circuit is
excited by a constant (DC) source, after sufficient time has elapsed the capacitor behaves as
an open and thus the voltage across the capacitor is as shown.
C1 12 V
12 V
10 K
C1
6 F
10 K
+
+
 vC10 = 12V
t 0+ = 12 V
The circuit at is as shown below where the represents the voltage across capaci-tor
.
C1

12 V
3 F
vC1
C1 C2
50 K
+
+

6 F
+

vC2t
vC2t vC1e –1  RCeq
= vC1 = 12 V Ceq
= = ------------ = 2 F
=     2  1– 6
 = 10 vC2t =
12e–10t t =
0 0= = = -----------------------------------  24 = 12 V
8 K 6 K vC0+
= = --------------- + 16 = 4 + 16 = 20 K
Now, where and
Then, and thus
Because the capacitor behaves as an open, there is no current in the . Then,
The circuit at is as shown below where .
Series and parallel resistances reduction yields
and the circuit for reduces to the one shown below.
C1  C2
C1 + C2
------------------ 6  3
6 + 3
1 RCeq  1 5 104
24 V
4 K 16 K
+
+
5 F 
2 K
6 K vC0
16 K
vC0 V0 v6K
6 K
6 K + 6 K
t 0+ = vC0+ = 12 V
4 K 16 K
+
5 F 
Req 8 K + 4 K  6 K + 16 K 12  6
12 + 6
t  0

+
20 K vCt
5 F 
ReqC 2 104
=   5  1– 6
= 0.1 1  ReqC = 10 vCt = V0e–10t 0=
12e–10t Now, , and . Also,
WC 1 ms
12
--CvC 2
= t
=
0.5  5  10–6  144e–20t t = 1 ms 1 ms
360  10–6 e–20t
t = 1 ms = = 0.35 mJ
t 0 =
Then,
and
+

36 V
100 
3 
6 
iL 0 +
vC0

iL 0 36 V
= ----------------------- = 4 A
6 + 3 
vC 0 = 3  iL 0 = 3  4 = 12 V
t 0+ = iSWt
The circuit at is as shown below and the current through the switch is the sum
of the currents due to the voltage source, due to , and due to
.
36 V iL0+ = 4 A
vC0+ = 12 V
36 V
100 
6 mH
3 
6 
+
iL0+ = 4 A
iSWt vC0+ = 12 V
10 F
+


We will apply superposition three times. Thus for t  0
:
I. With the 36 V voltage source acting alone where iL = 0 (open) and vC = 0
(shorted), the
circuit is as shown below.
36 V
100 
3 
6 
+
i'SWt
Since the is shorted out, we have
II.With the current source acting alone the circuit is as shown below where we
observe that the and resistors are shorted out and thus where
, , , , and thus
III. With the voltage source acting alone the circuit is as shown below
where we observe that the resistor is shorted out.
vCt = 0
iLt = 0
100 
i'SWt = 36  6 = 6 A
iL0+ = 4 A
6  100  i''SWt = –iLt
iLt I0e–R  Lt = I0 4 A = R 3  = L 6 mH = R L  3 6 103
– =     = 500
i''SWt –iLt 4e–500t = = –
100 
3 
6 
6 mH
iL0+ = 4 A iSWt vCt = 0 V
vC0+ = V0 = 12 V
6 

and thus
i'''SWt vC0+ = 12 V
.
Then,
100 
iLt = 0
3 
6 
vCt V0e–1  RCt 12e 1 100 10 5 – –    t 12e–1000t = = =
i'''SWt vCt  100  0.12e–1000t = =
Therefore, the total current through the closed switch for is
8.
t  0
iSWt i'SWt + i''SWt + i'''SWt 6 4e–500t – 0.12e–1000t = = + A
a. With the Manual Switch block in the upper position, all resistors are in parallel with the
72 V voltage source and thus the voltages across the 8, 2, and 4 Ohm resistors are 72
volts. Thus, the current through the 2 Ohm resistor is 72  2 = 36 amps
, and the current
through the 4 Ohm resistor is 72  4 = 18 amps
. It is observed that immediately after the
simulation command is issued, the current through the inductor resists any change, and
finally stabilizes at . The 6 and 3 Ohm resistors are shorted by the inductor.
18 amps
b. With the Manual Switch block in the lower position, all resistors and the inductor to the
right of the switch are grounded and thus all readings are zero. The 8 Ohm resistor is still
in parallel with the 72 V voltage source and thus the voltages across it is 72 volts.

Chapter 10
Forced and Total Response in RL and RC Circuits
his chapter discusses the forced response of electric circuits.The term “forced” here implies
that the circuit is excited by a voltage or current source, and its response to that excitation
is analyzed. Then, the forced response is added to the natural response to form the total
T
response.
u0t
10.1 Unit Step Function
A function is said to be discontinuous if it exhibits points of discontinuity, that is, if the function
jumps from one value to another without taking on any intermediate values.
A wellknown discontinuous function is the unit step function * which is defined as
(10.1)
u0t
0 t  0
1 t 0    
=
It is also represented by the waveform in Figure 10.1.
1
0
u0t
Figure 10.1. Waveform for
u0t
u0t
u0t
In the waveform of Figure 10.1, the unit step function changes abruptly from 0 to 1 at
. But if it changes at instead, its waveform and definition are as shown in Figure
t = 0 t = t0
10.2.
u0t – t0
t0
t
u0t – t0
Figure 10.2. Waveform and definition of

0 t t0
   
1 t t0
=
1
0
u0t – t0
* In some books, the unit step function is denoted as , that is, without the subscript 0. In this text we will reserve this des-ignation
for any input.
ut

Chapter 10 Forced and Total Response in RL and RC Circuits
Likewise, if the unit step function changes from 0 to 1 at t = –t0
as shown in Figure 10.3, it is
denoted as
u0t + t0
u0t + t0
 –
0 t t0
–    
1 t t0
=
1 u0t + t0
t t 0 0
u0t + t0
0
t
(a) (b) (c)
A A A
–Au0t –Au0t – T –Au0t + T
Au0– t + T Au0– t – T
A A A
t
 
t t
Au0–t
0
0 0
0 
0
t t
0
(d) (e) (f)
t

 
0 0 t t
(g) (h) (i)
A A A
–Au0–t –Au0– t + T –Au0– t – T
u0t – u0t – 1
1
t t t
0 1
0 0
Figure 10.3. Waveform and definition of
Other forms of the unit step function are shown in Figure 10.4.
Figure 10.4. Other forms of the unit step function
Unit step functions can be used to represent other timevarying functions such as the rectangular
pulse shown in Figure 10.5. This pulse is represented as .
Figure 10.5. A rectangular pulse expressed as the sum of two unit step functions
1
u0t
–u0t – 1
u0t – u0t – 1

Unit Step Function
The unit step function offers a convenient method of describing the sudden application of a volt-age
or current source. For example, a constant voltage source of applied at , can be
24 V t = 0
denoted as 24u0t V . Likewise, a sinusoidal voltage source vt = Vmcost V
that is applied to
a circuit at t = t0 , can be described as vt = Vmcostu0t – t0 V
. Also, if the excitation in a
circuit is a rectangular, or triangular, or sawtooth, or any other recurring pulse, it can be repre-sented
as a sum (difference) of unit step functions.
Example 10.1
Express the square waveform of Figure 10.6 as a sum of unit step functions. The vertical dotted
lines indicate the discontinuities at , and so on.
T 2T 3T
A
0
A

T 2T 3T t
vt



Figure 10.6. Square waveform for Example 10.1
Solution:
The line segment  has height , starts at , and terminates at on the time axis.
Then, as in Figure 10.5, this segment can be expressed as
v1t = Au0t – u0t – T
–A t = T t = 2T
(10.2)
A t = 0 t = T
The line segment  has height , starts at , on the time axis, and terminates at .
This segment can be expressed as
(10.3)
v2t = –Au0t – T – u0t – 2T
A t = 2T t = 3T
v3t = Au0t – 2T – u0t – 3T
–A t = 3T t = 4T
v4t = –Au0t – 3T – u0t – 4T
Line segment  has height , starts at , and terminates at . This segment can be
expressed as
(10.4)
Line segment  has height , starts at , and terminates at . This segment can
be expressed as
(10.5)
Thus, the square waveform of Figure 10.6 can be expressed as the summation of (10.2) through
(10.5), that is,

(10.6)
vt = v1t + v2t + v3t + v4t
= Au0t – u0t – T–Au0t – T – u0t – 2T
+Au0t – 2T – u0t – 3T–Au0t – 3T – u0t – 4T
vt = Au0t – 2u0t – T + 2u0t – 2T – 2u0t – 3T + 
A
T/2
t
it
0 T/2
A t = –T  2 t = T  2
i t   Au0 t T2
  Au0 t T2
= = –  
 + --- 
–   A u0 t T2
 – --- 
  u0 t T2
 + --- 
 – --- 
1
T/2
t
vt
0 T/2
Combining like terms, we obtain
(10.7)
Example 10.2
Express the symmetric rectangular pulse of Figure 10.7 as a sum of unit step functions.
Figure 10.7. Symmetric rectangular pulse for Example 10.2
Solution:
This pulse has height , it starts at , and terminates at . Therefore, with refer-ence
to Figures 10.3 and 10.4 (b), we obtain
(10.8)
Example 10.3
Express the symmetric triangular waveform shown in Figure 10.8 as a sum of unit step functions.
Figure 10.8. Symmetric triangular waveform for Example 10.3

Unit Step Function
Solution:
As a first step, we derive the equations of the linear segments  and  shown in Figure 10.9.
vt
--- t + 1 2T
1
T/2
–--- t + 1
t
 
0 T/2
2T
Figure 10.9. Equations for the linear segments of Figure 10.8
For line segment ,
(10.9)
and for line segment ,
(10.10)
v1 t   2T
  u0 t T2
=   – u0t
--- t + 1
v2 t   2T
  u0t u0 t T2
= –  
–--- t + 1
Combining (10.9) and (10.10), we obtain
(10.11)
 + --- 
 – --- 
vt = v1t + v2t
2T
  u0 t T2
= + –  
--- t + 1
2T
   + --- 
– u0  t    u0t u0 t T2
–--- t + 1
 – --- 
Example 10.4
Express the waveform shown in Figure 10.10 as a sum of unit step functions.
t
vt
3
2
1
0 1 2 3

Solution:
As in the previous example, we first find the equations of the linear segments  and  shown in
Figure 10.11.
3
2
vt

2t + 1 – t + 3

Figure 10.11. Equations for the linear segments of Figure 10.10
Following the same procedure as in the previous examples, we obtain
Multiplying the values in parentheses by the values in the brackets, we obtain
t
or
and combining terms inside the brackets, we obtain
(10.12)
Two other functions of interest are the unit ramp function and the unit impulse or delta function. We
will discuss the unit ramp function first.
10.2 Unit Ramp Function
The unit ramp function, denoted as , is defined as
(10.13)
where  is a dummy variable.
t
1
0 1 2 3
vt = 2t + 1u0t – u0t – 1 + 3u0t – 1 – u0t – 2
+ – t + 3u0t – 2 – u0t – 3
vt = 2t + 1u0t – 2t + 1u0t – 1 + 3u0t – 1
– 3u0t – 2 + – t + 3u0t – 2 – – t + 3u0t – 3
vt = 2t + 1u0t + – 2t + 1 + 3u0t – 1
+ – 3 + – t + 3u0t – 2 – – t + 3u0t – 3
vt = 2t + 1u0t–2t – 1u0t – 1–tu0t – 2 + t – 3u0t – 3
u1t
u1t
u1t u0d
–
= 

Delta Function
We can evaluate the integral of (10.13) by considering the area under the unit step function
from to as shown in Figure 10.12.
Area = 1 =  = t
t 
1
– t
Figure 10.12. Area under the unit step function from to
u0t – t
Therefore,
(10.14)
u1t
0 t  0
t t 0    
=
and since is the integral of , then must be the derivative of , i.e.,
(10.15)
u1t u0t u0t u1t
d
dt
----u1t = u0t
Higher order functions of can be generated by repeated integration of the unit step function.
For example, integrating twice and multiplying by , we define as
(10.16)
Similarly,
(10.17)
and in general,
(10.18)
Also,
(10.19)
t
u0t 2 u2t
u2t
0 t  0
t2 t 0    
t
= 
= or u2t 2 u1d
–
u3t
0 t  0
t3 t 0    
t
= 
= or u3t 3 u2d
–
unt
0 t  0
t n t 0    
t
= 
= or unt n un – 1d
–
un 1 – t   1n
-- d
dt
= ----unt
t
10.3 Delta Function
The unit impulse or delta function, denoted as t , is the derivative of the unit step u0t
. It is
generally defined as

(10.20)
t
 d
= u0t
–
t = 0 for all t  0
t u0t

0
 
t
Area =1 Figure (b)
0
 
1
2
Figure (a)
t
  0
  0 1  2
1 t
1
ftt = f0t
ftt – a = fat
f t t

 ftt – dt
= f
–
where
(10.21)
To better understand the delta function , let us represent the unit step as shown in Fig-ure
10.13 (a).
Figure 10.13. Representation of the unit step as a limit.
The function of Figure 10.13 (a) becomes the unit step as . Figure 10.13 (b) is the deriva-tive
of Figure 10.13 (a), where we see that as , becomes unbounded, but the area of
the rectangle remains . Therefore, in the limit, we can think of as approaching a very large
spike or impulse at the origin, with unbounded amplitude, zero width, and area equal to .
Two useful properties of the delta function are the sampling property and the sifting property.
The Sampling Property of the Delta Function states that
(10.22)
or
(10.23)
that is, multiplication of any function by the delta function results in sampling the func-tion
at the time instants where the delta function is not zero. The study of discretetime systems
is based on this property.
The Sifting Property of the Delta Function states that
(10.24)
that is, if we multiply any function f t by t – 
and integrate from to , we will obtain
the value of evaluated at .
f t t – 

Delta Function
The proofs of (10.22) through (10.24) and additional properties of the delta function are beyond
the scope of this book. They are provided in Signals and Systems with MATLAB Computing and
Simulink Modeling, ISBN 9781934404119.
MATLAB has two builtin functions for the unit step and the delta functions. These are desig-nated
by the names of the mathematicians who used them in their work. The unit step is
u0t
called Heavyside(t) and the delta function t
is called Dirac(t). Shown below are examples of
how they are being used.
syms k a t
u=k*sym('Heaviside(ta)') % Create unit step function at t=a
u =
k*Heaviside(t-a)
d=diff(u) % Compute the derivative of the unit step function
d =
k*Dirac(t-a)
int(d) % Integrate the delta function
ans =
Heaviside(t-a)*k
Example 10.5
For the circuit shown in Figure 10.14, the inputs are applied at different times as indicated.
50 K

+
6 K
5 K
vin1 vin2
iin
vin1 = 0.8u0t – 3 V
vin2 = 0.5u0t – 1 V
iin = 0.14u0t + 1 + u0t – 2 mA
+

+
vout

3 K
+
Rf

vout
t = –0.5 s
t = 1.5 s
t = 5 s
1 2 3
t(s)
0
t(s)
t(s)
iin = 0.14u0t + 1 + u0t – 2 mA
1
1
u0t – 1
1 2
0
0
vin1 = 0.8u0t – 3 V
u0t – 3
vin2 = 0.5u0t – 1 V
u0t + 1 + u0t – 2
t = –0.5 s iin
To op amp’s inverting input
5 K
0.7 V
+
3 K
6 K
iin
Compute at:
a.
b.
c.
Solution:
Let us first sketch the step functions for each of the inputs.
a. At only the signal due to is active; therefore, exchanging the current source and
its parallel resistance with an equivalent voltage source with a series resistance, the input cir-cuit
becomes as shown in Figure 10.15.
Figure 10.15. Input to the circuit of Example 10.5 when is acting alone
Replacing the circuit of Figure 10.15 with its Thevenin equivalent, we obtain the network of
Figure 10.16.

Delta Function
2 K
0.7 V
+

3 K 6 K = 2 K
vTH1 V2 K
2 K
= = -----------------------------------–0.7 = –0.2 V
2 K + 5 K
2 K  5 K
= ----------------------------------- = 10  7 K
7 K
5 K
10  7 K
0.2 V
+

RTH1
vTH1
iin
Figure 10.16. Simplified input to the circuit of Example 10.5 when is acting alone
RTH1
Now, we can compute with the circuit of Figure 10.17.
vout1
50 K
Rf

+
+

vout1
0.2 V 
10  7 K
vTH1
Figure 10.17. Circuit for computation of
(10.25)
vout1
50
10  7
= –  vTH1 = –35  –0.2 mV = 7 V
------------
b. At t = 1.5 s
the active inputs are
and
+
vout1
iin = 0.14u0t + 1 + u0t – 2 mA
vin2 = 0.5u0t – 1 V
Since we already know the output due to iin acting alone, we will find the output due to
vin2
acting alone and then apply superposition to find the output when both of these inputs are
present. Thus, with the input acting alone, the input circuit is as shown in Figure 10.18.
vin2

Replacing this circuit of Figure 10.18 with its Thevenin equivalent, we obtain the network of
Figure 10.19.
15  8
15  8 + 6
----------------------0.5 5
= = = ----- V
RTH2 10  7 K
+


–  vTH2 35 5
–    25
-----
= = = –----- V
– ----- 17
= = ----- V
(10.26)
Therefore, from (10.25) and (10.26) the op amp’s output voltage at is
(10.27)
0.5 V
+
3 K
5 K
6 K
vin2
0.5 V
+

vTH2 v15  8 K 15/8 K
42
3 K 5 K = 15  8 K
RTH2 = RTH1 = 10  7 K
6 K
5
42
----- V
vTH2
vin2
vout2
+
+

+

50 K
Rf
vTH2
5
42
----- V
10  7 K
vout2
RTH2
vout2
vout2
50
10  7
------------
42
6
t = 1.5 s
vout1 + vout2 7 25
6
6

Delta Function
c. At the active inputs are
and
t = 5 s
vin1 = 0.8u0t – 3 V
vin2 = 0.5u0t – 1 V
Since we already know the output due to vin2 acting alone, we will find the output due to
vin1
acting alone and then apply superposition to find the output when both of these inputs are
present. Thus, with the input acting alone, the input circuit is as shown in Figure 10.21.
vin1
3 K
0.8 V
+
6 K
5 K
vin1
Replacing this circuit of Figure 10.21 with its Thevenin equivalent, we obtain the network of
Figure 10.22.
3 K
0.8 V
+

6 K 5 K = 30  11 k
30  11
30  11 + 3
-------------------------0.8 10
vTH3 = v  30 30 /11 K  11 K
= = ----- V
21
RTH3 = RTH2 = RTH1 = 10  7 K
10  7 K
+
 10
----- V
21
vin1
(10.28)
vout3
vout3
50
10  7
–  vTH3 35 10
 ------------ 
–    50
 ----- 
= = = –----- V
21
3
Therefore, from (10.26) and (10.28) the op amp’s output voltage at is
(10.29)
t = 5 s
vout2 + vout3 25
– ----- 50
= = –-------- V
6
– ----- 125
3
6

+

it it
10.4 Forced and Total Response in an RL Circuit
For the circuit shown in Figure 10.24(a), is constant. We will derive an expression for the
inductor current for given that the initial condition is . Here, the
inductor current will be referred to as the total response.
The switch in Figure 10.24 (a) can be omitted if we multiply the excitation by the unit step
function as shown in Figure 10.24 (b).
Figure 10.24. Circuits for derivation of the total response
We begin by applying KVL, that is,
(10.30)
The initial condition states that ; thus for ,
For , we must solve the differential equation
(10.31)
+

+

50 K
Rf
vTH3
10
21
----- V
10  7 K
vout3
vout3
VS
iLt = it t  0 iL0 = 0
iLt
VS
u0t
+
R
L
R
L
t = 0
VS VS u0t
(a) (b)
iLt = it
Ldi
dt
---- + Ri = VS u0t
iL0 = 0 t  0 it = 0
t  0
Ldi
dt
---- + Ri = VS

Forced and Total Response in an RL Circuit
It is shown in differential equations textbooks that a differential equation such as the above, can
be solved by the method of separation of the variables. Thus, rearranging (10.31), separating the
variables, and integrating we obtain:
or
or
Ldi
dt
---- = VS – Ri
Ldi
VS – Ri
------------------ = dt
Ldi
VS – Ri
------------------ = dt
and referring to a table of integrals, we obtain
(10.32)
LR
–--- lnVS – Ri = t + k
The constant in (10.32) represents the constant of integration of both sides and it can be eval-uated
k
from the initial condition, and as we stated in the previous chapter
(10.33)
Therefore, at
iL0 = iL0 = iL0+
t 0+ =
LR
–--- lnVS – 0 = 0 + k
k LR
and by substitution into (10.32), we obtain
= –--- lnVS
LR
--- VS Ri –   ln – t LR
= –--- lnVS
LR–--- lnVS – Ri – lnVS = t
--- VS – Ri
LR
– ln------------------ = t
VS
VS – Ri
VS
------------------ ln RL
= –--- t
VS – Ri
VS
------------------- –=
e R  L t

Ri VS VSe–R  L t
Short
Circuit
as t  VS u0t if
The general expression for all t is
(10.34)
We observe that the right side of (10.34) consists of two terms, which is constant called
the forced response, and the exponential term that has the same form as that of the
previous chapter which we call the natural response.
The forced response is a result of the application of the excitation (forcing) function
applied to the circuit. This value represents the steadystate condition reached as
since the inductor at this state behaves as a short circuit.
The amplitude of the natural response is and depends on the values of and .
The summation of the forced response and the natural response constitutes the total response or
complete response, that is,
or
(10.35)
Now, let us return to the circuit of Figure 10.24 to find the complete (total) response by
the summation of the forced and the natural responses as indicated in (10.35).
The forced response is found from the circuit of Figure 10.25 where we let
Figure 10.25. Circuit for derivation of the forced response
Then, from the circuit of Figure 10.25,
= –
it
VS
------
R
VS
R
------e –R  L t
= –
it
VS
R
------
VS
R
------e –R  L t  – 
=   u0t
VS  R
VS
R
–------e–R  Lt
VS  R
VS u0t RL
t   L
–VS  R VS R
ittotal = it forced response + it natural response
itotal = if + in
RL itotal
if t  
+
R
L
if

(10.36)
if
VS
R
= ------
Next, we need to find the natural response. This is found by letting the excitation (forcing func-tion)
go to zero as shown in the circuit of Figure 10.26.
R
VS u0t = 0 L in
Figure 10.26. Circuit for derivation of the natural response
VS u0t
in
We found in Chapter 9 that the natural response has the exponential form
(10.37)
Therefore, the total response is
(10.38)
in
in Ae–R  Lt =
itotal if + in
VS
R
------ Ae–R  Lt
= = +
A iL0 = iL0 = iL0+
where the constant is evaluated from the initial condition
Substitution of the initial condition into (10.38) yields
or
------ Ae0 = = +
i0 0
A
= –------
and with this substitution (10.38) is rewritten as
(10.39)
VS
R
VS
R
itotal
VS
R
=  ------
 u0t
VS
R
–------e–R  Lt
 
and this is the same as (10.34).
We can sketch easily if we sketch VS
itotal
------
and VS
R
–
------e –R  Lt separately and then add these. This
R
is done with MATLAB and the plots are shown in Figure 10.27.

it VS  R VS  Re–Rt  L = –
it = VS  Lt
Figure 10.27. Curves for forced, natural, and total responses in a series RL circuit
The curves in Figure 10.27 were created with the following MATLAB script:
x=0:0.01:5; Vs=1; R=1; L=1; y=(Vs./R).*exp(R.*x./L); z=Vs./R+y; plot(x,y,x,z)
The time constant is defined as before, and its numerical value can be found from the circuit
constants and as follows:
The equation of the straight line with is found from
Assuming constant rate of change as shown in Figure 10.27, at 
–-----e–R  LL  R VS
----- 1 e–1  – 
and thus
or
as before. Also, from (10.39)
or
(10.40)
Time Constants
Percent VS/R
VS  Re–Rt  L –
0.632VS  R


R L
slope = VS  L
d
dt
----itotal
t = 0
RL
---
VS
R
-----e –R  Lt 
t = 0
VS
L
= = -----
t = 
it
VS
R
= -----
VS
R
------
VS
L
= -----
 LR
= ---
i
VS
R
----- VS
R
R
VS
R
= = = -----1 – 0.368
i 0.632
VS
R
= ------

Therefore, the current in a series RL
circuit which has been excited by a constant source, in one
time constant has reached of its final value.
63.2%
Example 10.6
For the circuit of Figure 10.28, compute the energy stored in the inductor at
.
3
6
10 mH
5u0t A
iLt
10 mH
t = 100 ms
+

12 V
Solution:
For t  0 , the circuit is as shown in Figure 10.29 where the 3 
resistor is shorted out by the
inductor.
+
12 V
6
iLt
Figure 10.29. Circuit of Example 10.6 for
From the circuit of Figure 10.29,
iL0 12
= ----- = 2 A
and this value establishes our initial condition as
(10.41)
iL0+ = 2 A
For , the circuit is as shown in Figure 10.30.
6
+
iLt
3
6
12 V
10 mH
Figure 10.30. Circuit of Example 10.6 for
t  0
t  0
5 A
t  0

iLt
iLt = if + in
if
10 mH if
2 A 5 A
if = 2 – 5
if = –3 A
in
6
10 mH 3 2
We will compute from the relation
The forced component is found from the circuit at steady state conditions. It is shown in Figure
10.31 where the voltage source and its series resistance have been exchanged for an equivalent
current source with a parallel resistor. The resistors have been shorted out by the inductor.
Figure 10.31. Circuit of Example 10.6 under steadystate conditions
By inspection, or
(10.42)
To find we short the voltage source and open the current source. The circuit then reduces to
that shown in Figure 10.32.
Figure 10.32. Circuit of Example 10.6 for determining the natural response
The natural response of circuit of Figure 10.32 is
or
(10.43)
The total response is the summation of (10.42) and (10.43), that is,
(10.44)
Using the initial condition of (10.42), we obtain
or
Finally, by substitution into (10.44) we obtain
10 mH
3  || 6  = 2 
i in n
RL
in Ae–R  Lt Ae 2 10 10 3 – –   t = =
in Ae–200t =
itotal if + in – 3 Ae–200t = = +
iL0+ 2 – 3 Ae–0
= = +
A = 5

Forced and Total Response in an RC Circuit
(10.45)
itotal – 3 5e–200t =  + u0t
and the energy stored in the inductor at is
(10.46)
t = 100 ms
WL t 100 ms =
12
--LiL 2
= =
t = 100 ms
--10  10–3 – 3 5e 200 100 10 –3 –  
12
 
 + 
2
 – 3 – 5e 20 –
5 10 3
 + 
2
= = 45 mJ
10.5 Forced and Total Response in an RC Circuit
For the circuit shown in Figure 10.33 (a), VS
is constant. We will derive an expression for the
capacitor voltage for given that the initial condition is . Here, the capac-itor
vCt t  0 vC0 = 0
voltage will be referred to as the total response.
+
R
+ 
vRt
C C +
+
R
+

t = 0
vCt vCt
VS VSu0t
(a) (b)
Figure 10.33. Circuits for derivation of the total response
vCt

vCt
The switch in Figure 10.33 (a) can be omitted if we multiply the excitation by the unit step
function as shown in Figure 10.33 (b).
We begin by applying KVL, that is,
(10.47)
u0t
and since
we can express as
vR + vC = VSu0t
vR
= = --------
vR Ri RC
By substitution into (10.47), we obtain
(10.48)
VS
= = --------
i iC C
dvC
dt
dvC
dt
RC
dvC
dt
--------- + vC = VSu0t

vC0 = 0 t  0 vCt = 0
t  0
RC
dvC
dt
--------- + vC = VS
RCdvC = VS – vCdt
dvC
vC – VS
----------------- 1
= –--------dt
RC
dvC
vC – VS
-----------------dt 1
= –--------dt
RC
lnvC – VS 1
= –--------t + k
RC
vC – VS e–1  RC t + k eke–1  RC t k1e–1  RC t = = =
vC VS k1e–1  RC t = –
k1 vC 0+ = vC 0 = 0
vC0+ 0 VS k1e0 = = –
k1 = VS
vCt VS VS e –1  RCt
=  – u0t
RL
VS
t   C
The initial condition states that ; thus for ,
For , we must solve the differential equation
(10.49)
Rearranging, separating variables and integrating, we obtain:
(10.50)
or
where k represents the constant of integration of both sides of (10.51). Then,
or
(10.51)
The constant can be evaluated from the initial condition where by sub-stitution
into (10.51) we obtain
or
Therefore, the solution of (10.49) is
(10.52)
As with the circuit of the previous section, we observe that the solution consists of a forced
response and a natural response. The constant term is the voltage attained across the capaci-tor
as and represents the steadystate condition since the capacitor at this state behaves
as an open circuit.
The amplitude of the exponential term natural response is 
–VS

The summation of the forced response and the natural response constitutes the total response,
i.e.,
or
(10.53)
vCt complete response = vCt forced response + vCt natural response
vCtotal vCf = + vCn
Now, let us return to the RC
circuit of Figure 10.33(b) to find the complete (total) response by
summing the forced and the natural responses indicated in (10.53).
The forced response is found from the circuit of Figure 10.34 where we let .
vCf t  
+

R Open
Circuit
vCf
V as t  Su0t
Figure 10.34. Circuit for derivation of the forced response
Then, from the circuit of Figure 10.34,
(10.54)
vCf
vCf = VS
Next, we need to find the natural response and this is found by letting the excitation (forcing
function) go to zero as shown in Figure 10.35.
C
R
VSu0t = 0 vCn
Figure 10.35. Circuit for derivation of the natural response
VSu0t
vCn
We found in Chapter 9 that the natural response has the exponential form
and thus the total response is
(10.55)
vCn
vCn Ae–1  RCt =
vCt vCf + vCn VS Ae–1  RCt = = +
A vC0 = vC0 = vC0+ = 0
where the constant is evaluated from the initial condition

vCt VS VSe–1  RC = –
vCt = VS  RCt
Substitution of the initial condition into (10.55) yields
or
With this substitution (10.55) is rewritten as
(10.56)
and this is the same as (10.52).
We can sketch easily if we sketch and separately and then add these.
This is done with MATLAB and the plots are shown in Figure 10.36.
Figure 10.36. Curves for forced, natural, and total responses in a series RC circuit
The time constant is defined as before, and its numerical value can be found from the circuit
constants and as follows:
The equation of the straight line with is found from
Assuming constant rate of change as shown in Figure 10.36, at 
and thus
vC0+ 0 VS Ae0 = = –
A = –VS
vCt VS VSe–1  RCt =  – u0t
vCtotal VS VSe –1  RCt –
Time Constants
Percent VC/VS
VSe–1  RC –
0.632VC  VS


R C
slope = VS  RC
d
dt
----vCt
t = 0
1
RC
-------- VS e –1  RCt 
t = 0
VS
RC
= = --------
t = 
vCt = VS

or
as before. Also, from (10.56)
or
(10.57)
VS
VS
RC
= --------
 = RC
vC VS–VSe–1  RCRC VS 1 e–1 = =  –  = VS1 – 0.368
vC = 0.632VS
Therefore, the voltage across a capacitor in a series RC circuit which has been excited by a con-stant
source, in one time constant has reached of its final value.
Example 10.7
For the circuit shown in Figure 10.37 find:
a. and
b. and
c. and
d. for
63.2%
vC 1 iC 1
vC 1+ iC 1+
vCt = 10 min. iCt = 10 min.
iC t t  1
+
iCt
 C
9u0t – 1 mA
60 K
vCt
10 F
20 K
10 K
Solution:
a. No initial condition is given so we must assume that sufficient time has elapsed for steady
state conditions to exist for all t  1 s.
We assume time is in seconds since we are not told
otherwise. Then, since there is no voltage or current source present to cause current to flow,
we obtain
vC 1 = 0

+
10 K
+
= ---------------------------------- = 3 mA
= = ----------------------------------------------  60 K = 60 V
and
b. Exchanging the current source and the resistor with a voltage source with a
series resistor, the circuit at is as shown in Figure 10. 38.
Now, since , no current flows through the resistor at ; if it did,
the voltage across the capacitor would change instantaneously, and as we know, this is a physi-cal
impossibility. Instead, the current path is through the capacitor which at exactly
acts as a short circuit since . Therefore,
(10.58)
c. The time is the essentially the same as , and at this time the capacitor volt-age
is constant and equal to the voltage across the resistor, i.e.,
Also,
d. For
where from part (c)
and
With the voltage source shorted in the circuit of Figure 10.38, the equivalent resistance is
or
iC 1 = 0
10 K 10 K
t 1+ =

60 K 20 K
iCt
C 10 F
vCt
9u0t – 1 V
t 1+ =
vC 1+ = vC 1 60 k t 1+ =
t 1+ =
vC 1+ = vC 1 = 0
iC 1+ 90 V
20 + 10 K
t = 10 min t = 
vCt = 10 min 60 K
vCt = 10 min= vC v60 K
90 V
20 + 10 + 60 K
iCt t =  C
dvC
dt
= -------- = 0
t  1
iCt
t  1
= iC f + iCn
iCf  = 0
iCn Ae
–1  ReqC t
=
Req = 10 K + 20 K || 60 K = 20 K

Therefore,
(10.59)
ReqC 20 103
 10 10–6 =   = 0.2 s
iCn Ae–1  0.2 t Ae–5t = =
A
We can evaluate the constant using (10.59) where
or
and by substitution into (10.59),
(10.60)
iC 1+ 3 mA Ae–5 = =
A 3 103
– 
e–5 = ------------------- = 0.445
iCt
t  1
iCf i+ Cn iCn 0.445e 5t – u0 t 1 = = =  –  mA
Example 10.8
In the circuit shown in Figure 10.39, the switch is actually an electronic switch and it is open for
and closed for . Initially, the capacitor is discharged, i.e., . Compute and
15 s 15 s vC0 = 0
sketch the voltage across the capacitor for two repetitive cycles.
VSt
+

C
+

1 K
6 V
350 
250 
0.02 F vCt
Solution:
With the switch in the open position the circuit is as shown in Figure 10.40.
+

C
+

VS
6 V
1 K 250 
0.02 F
Switch
vCt
open
Figure 10.40. Circuit for Example 10.8 with the switch in the open position

For the time period the time constant for the circuit of Figure 10.40 is
Thus, at the end of the first period when the switch is open, the voltage across the capacitor is
(10.61)
Next, with the switch closed for the circuit is as shown in Figure 10.41.
+
250 

Figure 10.41. Circuit for Example 10.8 with the switch in the closed position
Replacing the circuit to the left of points x and y by its Thevenin equivalent, we obtain the circuit
Figure 10.42. Thevenin equivalent circuit for the circuit of Figure 10.41
The time constant for the circuit of Figure 10.42 where the switch is closed, is
The capacitor voltage for the circuit of Figure 10.42 is
(10.62)
and the constant is evaluated from initial condition at which by (10.62) is
Then,
0  topen  15 s
open ReqC 1 K + 0.25 K 0.02 10–6 = =   = 25 s
vCt t = 15 s vCf + vCn VS VS e–t  RC – 6 6e 4 104
–  t – 6 6e–0.6 = = = = – = 2.71 V
15  tclosed  30 s
+
C

6 V
1 K
350 
0.02 F
x
Switch y

vCt
closed
VS
+

C

+
1.56 V
259  250 
0.02 F
VTH
350
1350
= -----------  6 = 1.56 V
RTH
350  1000
1350
= --------------------------- = 259 
vCt
VTH
RTH
closed ReqC 259  + 250  0.02 10–6 = =   = 10.2 s
vCt
vCt vCf v+ Cn VTH A1e –1  RCt – 15s
+ 1.56 A1e –1  10.2st – 15s
= = = +
A1 t = 15 s
vCt t = 15 s
= 2.71 V
vCt = 2.71 – = 1.56 +
A1e 1 10.2s15 – 15s
15  t  30 s

or
(10.63)
A1 = 1.15
vCt = 1.56 – +
1.15e 1 10.2st – 15s
15  t  30 s At the end of the first period when the switch is closed, the voltage across the capacitor is
(10.64)
vCt t = 30 s 1.56 1.15e –1  10.2s30 – 15s
= + = 1.82 V
30  topen  45 s
For the next cycle, that is, for when the switch is open, the time constant
is the same as before, i.e., and the capacitor voltage is
(10.65)
open open = 25 s
vCt vv6 A2e –1  = + 25st – 30s
Cf Cn = +
The constant is computed with (10.65) as
A2
vCt t = 30 s 1.82 6 A2e –1  25s30 – 30s
or
(10.66)
= = +
A2 = –4.18
vCt
30  t  45 s
6 4.18e –1  25st – 30s
= –
At the end of the second period when the switch is open, the voltage across the capacitor is
(10.67)
vCt
t = 45 s
6–4.18e –1  25s45 – 30s = = 3.71 V
The second period when the switch is closed is . Then,
(10.68)
vCt
45  t  60 s
vCf + vCn VTH A3e–1  RCt – 45 + 1.56 A3e–1  10.2st – 45s = = = +
and with (10.67) we obtain
Therefore,
(10.69)
and
(10.70)
45  tclosed  60 s
A3 = 2.15
vCt
45  t  60 s
1.56 2.15e–1  10.2st – 45s = +
vCt
t = 60 s
1.56 2.15e–1  10.2s60 – 45s = + = 2.05 V

Repeating the above steps for the third open and closed switch periods, we obtain
(10.71)
vCt
60  t  75 s
6 3.95e–1  25st – 60s = –
vCt
t = 75 s
= 3.83 V
vCt
75  t  90 s
1.56 2.27e–1  10.2st – 75s = +
vCt
t = 90 s
= 2.08 V
and
(10.72)
Likewise,
(10.73)
and
(10.74)
and using the MATLAB script below we obtain the waveform shown in Figure 10.43.
t0=(0:0.01:15)*10^(6);
v0=66.*exp(4.*10.^4.*t0);
t1=(15:0.01:30)*10^(6);
v1=1.56+1.15.*exp((1./(10.2.*10.^(6))*(t115.*10.^(6))));
t2=(30:0.01:45)*10^(6);
v2=64.18.*exp((1./(25.*10.^(6))*(t230.*10.^(6))));
t3=(45:0.01:60)*10^(6);
v3=1.56+2.15.*exp((1./(10.2.*10.^(6))*(t345.*10.^(6))));
t4=(60:0.01:75)*10^(6);
v4=63.95.*exp((1./(25.*10.^(6))*(t460.*10.^(6))));
t5=(75:0.01:90)*10^(6);
v5=1.56+2.27.*exp((1./(10.2.*10.^(6))*(t575.*10.^(6))));
plot(t0,v0,t1,v1,t2,v2,t3,v3,t4,v4,t5,v5)
Figure 10.43. Voltage across the capacitor for the circuit of Example 10.8

Summary
10.6 Summary
 The unit step function is defined as
u0t
u0t
0 t  0
1 t 0    
=
and it is represented by the waveform below.
1
0
u0t
 Unit step functions can be used to represent other timevarying functions.
 The unit step function offers a convenient method of describing the sudden application of a
voltage or current source.
 The unit ramp function , is defined as the integral of the unit step function, that is,
u1t
t
= 
u1t u0d
–
where  is a dummy variable. It is also expressed as
u1t
0 t  0
t t 0    
=
 The unit impulse or delta function, denoted as , is the derivative of the unit step . It
is defined as
or
and
t u0t
 d
= ----u0t
dt
t
 d
= u0t
–
t = 0 for all t  0
 In a simple circuit that is excited by a voltage source the current is
RL VS u0t
= =   u0t
it if + in
VS
R
 ------
– ------e –R  L t 
VS
R

where the forced response if represents the steadystate condition reached as t  
. Since the
inductor L at this state behaves as a short circuit, if = VS  R . The natural response in
is the
second term in the parenthesis of the above expression, that is,
in =
–VS  Re –R  L t  In a simple circuit that is excited by a voltage source the voltage across the capac-itor
RC VS u0t
vCt vCf + vCn VS Ae–1  RCt = =  + u0t
vCf t  
is
where the forced response represents the steadystate condition reached as . Since
the capacitor C at this state behaves as an open circuit, vCf = VS . The natural response
vCn
is the second term in the parenthesis of the above expression, that is, vCn =
Ae–1  RCt . The
constant must be evaluated from the total response.
A

Exercises
10.7 Exercises
Multiple Choice
1. For the circuit below the time constant is
A.
0.5 ms
B.
71.43 s
C.
2 000 s
D.
0.2 ms
+

12u0t V
2. For the circuit below the time constant is
A.
B.
C.
D.
4 
12 
2 
1 mH
50 ms
100 ms
190 ms
78.6 ms
5u0t A
4 K
10 K
5 K
6 K 10 F
3. The forced response component of the inductor current for the circuit below is
A.
B.
C.
D.
iLf
16 A
10 A
6 A
2 A

4. The forced response component of the capacitor voltage for the circuit below is
A.
B.
C.
D.
5. For the circuit below . For the total response of is
A.
B.
C.
D.
16u0t A
4  12 
5 
1 mH
vCf
10 V
2 V
32  3 V
8 V
16u0t V
4 K
12 K
2 K
1 F
iL0 = 2 A t  0 iLt
6 A
6e–5000t A
6 6e–5000t + A
6–4e –5000t A
16u0t A
4  12 
5 
1 mH
iLt
vC0 = 5 V t  0 vCt

Exercises
A.
12 V
B.
10 – 5e–500t V
C.
12 – 7e–200t V
D.
12 + 7e–200t V
+
16u0t V
4 K
12 K
2 K
1 F
+

vCt
iL0 = 2 A t  0 vLt
20e–5000t V
20e–5000t V
32e–8000t – V
32e–8000t V
A.
B.
C.
D.
16u0t A
4  12 
5 
+

1 mH vLt
vC0 = 5 V t  0 iCt
8. For the circuit below . For the total response is
1400e–200t A
1.4e–200t A
3500e–500t A
3.5e–500t A
A.
B.
C.
D.

9. The waveform below can be expressed as
A.
B.
C.
D.
10. The waveform below can be expressed as
A.
B.
C.
D.
+
16u0t V
4 K
12 K
2 K
1 F iCt
3tu0t A
3u0t – 3u0t – 3 A
3tu0t – u0t – 1 + – 1.5t + 4.5u0t – 1 – u0t – 3 A
3tu0t – u0t – 3 + – 1.5t + 4.5u0t – u0t – 3 A
iLt A
t s
1
1 2
2
3
3
2 1 e–t – e–t  – u0t V
2 2e
– t  – u0t – u0t – 2 2e–t +  u0t – 2 – u0t – 3 V
2 2e
– t  – u0t – u0t – 2 2e–t – u0t – 2 – u0t – 3 V
2 2e
– t  – u0t 2e–t – u0t – 3 V

Exercises
vCt V
t s
2
1
2 2e–t –
1 2
2e–t
3
Problems
1. In the circuit below, the voltage source varies with time as shown by the waveform
below it. Compute, sketch, and express as a sum of unit step functions for
vS t
vLDt 0  t  5 s.
+
+
vLDt

6 6
12
10
vS t

vSt
120
60
0
60
1 2 3 4 5 6
(V)
t(s)
2. In the circuit below . Compute for .
vSt = 15u0t – 30u0t – 2 V iLt t  0
+

R
L
3 K
1 mH
iLt
vSt

3. In the circuit below the excitation vS t
is a pulse shown next to it.
a. Compute iLt for
0  t  0.3 ms
b. Compute and sketch for all
iLt t  0
3 
iLt
1 mH
+
 6
vS t V
24
0.3 t (ms)
6
(a) (b)
vS t
4. In the circuit below switch S has been open for a very long time and closes at t = 0
. Compute
and sketch and for .
iLt iSWt t  0
vS
+

6
4 
1 H
iSWt
t = 0 S
20 V 8
iLt
vCt t  0
10u0t A
2 
38  +

50 F
vS
+

24 V
vCt
5. For the circuit below compute for .
6. For the op amp circuit below compute voutt for t  0 in terms of R , C , and vinu0t
given
that
vC0 = 0

Exercises
R C
vinu0t
voutt
7. In the circuit of Figure 10.61, switch S has been open for a very long time and closes at t = 0
.
Compute and sketch and for .
vCt vR3t t  0
+

vS2
R1 R2
+
50 K 25K
1 F
C
R3
100K
+ +
vCt
vR3t


S
50 V
t = 0
vS1
100u0–t V
vC0 = 5 V iCt t  0
8. For the circuit below it is given that . Compute for . Hint: Be careful
in deriving the time constant for this circuit.
18 
+
C R2

R1
vCt 12 
1 F
+

iCt
10iCt
9. A 12V DC , a 1 M , and a 1 F
are connected in series. Create a Simulink / SimPower Sys-tems
modedisplay the waveform of the voltage across the capacitor as a function of time.

Multiple Choice
1. D
2. B
3. C
4. E
5. E
6. C
7. D
8. A
9. C
10. B
Problems
1.We replace the given circuit shown below with its Thevenin equivalent.
6 6
x

10
120
60
For the Thevenin equivalent voltage at different time intervals is as shown below.
12 V
6 4e–8000t – A
+

12
10
+
vLDt
vS t

y
+

10
+
vLDt
vTH t
 
0
60
1 2 3 4 5 6
(V)
t(s)
vSt

and
vTHt
12
18
-----vs t   23
-- 60  40 V t 4  s = =           
= --  60 = 40 V 0  t  1 s
-----vs t   23
12
18
= --  120 = 80 V 1  t  3 s
12
18
-----vs t   23
= --  –60 = –40 V 3  t  4 s
-----vs t   23
12
18
=
vLDt 10
-----vTHt 0.5vTHt
= = =
20
0.5  40 = 20 V 0  t  1 s
0.5  80 = 40 V 1  t  3 s
0.5 40  20 V t 4  s =      
0.5  –40 = –20 V 3  t  4 s
The waveform of the voltage across the load is as shown below.
vLDt V
40
20
0
–20
1 2 3 4 5 t s
The waveform above can now be expressed as a sum of unit step functions as follows:
vLDt = 20u0t – 20u0t – 1 + 40u0t – 1 – 40u0t – 3 + 20u0t – 4
–20u0t – 3 + 20u0t – 4 + 20u0t – 4
= 20u0t + 20u0t – 1 – 60u0t – 3 + 40u0t – 4
t 0 =
2. The circuit at is as shown below and since we are not told otherwise, we will assume
that
iL0 = 0
3 K
1 mH
vS0 = 0 iL0 = 0

For we let be the inductor current when the voltage source acts alone and
t  0 iL1t 15u0t
iL2t 30u0t – 2 iL TOTALt = iL1t + iL2t
0  t  2 s
+
3 K
1 mH iL1t
15 V
iL1t = iL1f + iL1n iL1f
15
3 K
= -------------- = 5 mA iL1n A1e–R  Lt Ae 3 106
–  t = + mA iL0 = iL0+ = 0
–  t = –

–  t – 2 = + mA
when the voltage source acts alone. Then,
For the circuit is as shown below.
Then, where and
Thus, and using the initial condition , we
obtain or . Therefore,
(1)
Next, with the voltage source acts alone the circuit is as shown below.
Then,
and
Thus,
and the initial condition at is found from (1) above as
–  t = =
iL1 t   5 A1e 3 106
iL10 = 5 + A1e0 mA A1 = –5
iL1 t   5 5e 3 106
30u0t – 2
+
3 K
1 mH iLt
30 V
iL2t = iL2f + iL2n
iL2f
–30
3 K
= -------------- = –10 mA
iL2n A2e–R  Lt – 2 Be 3 106
–  t – 2 = =
iL2 t   10 – A2e 3 106
t = 2

Therefore,
or
and
–  2 – 2 = = = + mA
5 10 – A2e 3 106
(2)
iL1 t = 2 s
– 6  106
= 5 – 5e t  5 mA
iL2 t 2 s =
iL1 t 2 s =
A2 = 15
–  t – 2 = + mA
iL2 t   10 – 15e 3 106
Thus, the total current when both voltage sources are present is the summation of (1) and (2),
that is,
–  t – 3  106
= = – – 10 + 15e t – 2 mA
iL TOTAL t   iL1 t   iL2 t   + 5 5e 3 106
– 3  106
= – 5 – 5e t + 15e – 3  1t – 2 06
mA
3.
a. For this circuit and since we are not told otherwise, we will
vS t = 24u0t – u0t – 0.3
iL0 = 0 0  t  0.3 ms
assume that . For the circuit and its Thevenin equivalent are as
shown below.
Then,
and at
or

+ 

vS t
vS t = 24u0t – u0t – 0.3
t = 0
and thus for
3  6
6 1 mH iLt
 +
vTH t
vTH t = 16u0t – u0t – 0.3
iLt iLf + iLn 16  8 A1e–R  Lt + 2 Ae–8000t = = = +
1
AiL  0  = iL  0  = 0 = 2 +
e0 (1)
8 
1 mH iLt
A1 = –2
0  t  0.3 ms
iLt 2–2e–8000t =
b. For t  0.3 ms
the circuit is as shown below. For this circuit

(2)
iLt A2e–R  Lt – 0.3 A2e–8000t – 0.3 = =
and is found from the initial condition at , that is, with (1) above we obtain
A2 t = 0.3 ms
3  6

6 1 mH iLt

0.3 10–3 –    – 2 2e2.4 = = – = 1.82 A
and by substitution into (2) above
or
Therefore for
The waveform for the inductor current for all is shown below.
4. At the circuit is as shown below where and thus the initial
condition has been established.
8  1 mH iLt
vS t = 0
iL t 0.3 ms =
2 2e 8 103
iL t 0.3 ms =
1.82 A2e–80000.3 – 0.3 = =
A2 = 1.82
t  0.3 ms
iLt 1.82e–t – 0.3 ms =
iLt t  0
t ms
iL A
1.82
0.3
t 0 = iL0 = 20  4 + 6 = 2 A

+
6
4 
20 V
vS
iL0
For the circuit and its Thevenin equivalent are as shown below where
and
t  0
Then,
vTH
8
4 + 8
= ------------  20 = 40  3 V
RTH
8  4
8 + 4
= ------------ + 6 = 26  3 
+

6
4 
1 H
Closed
Switch
20 V 8
vS
iLt
iLt iLf + iLn 40  3
------------ Ae–R  Lt + 20  13 Ae–26  3t = = = +
26  3
and is evaluated from the initial condition, i.e.,
A
iL0 iL0+ 2 20  13 Ae0 = = = +
from which and thus for
RTH
26  3 
1 H
(1)
+
vTH
iLt
40  3 V
A = 6  13 t  0
----- 6
iLt 20
-----e–26  3t + 1.54 0.46e–8.67t = = + A
13
13
Next, to find iSWt we observe that this current flows also through the 8 
resistor and this
can be found from shown on the circuit below.
v8 

+
6
4 
1 H
vS
iSWt
20 V 8
iLt
+

v8 
= = + -------
v8  v6  + vLt 6iLt L
diL
dt
6 1.54 0.46e–8.67t  +  1 d
= +  ----  1.54 0.46e–8.67t 
dt
+ = 9.24 + 2.76e–8.67t – 8.67 
0.46e–8.67t = 9.24 –
1.23e–8.67t ----------------------------------------- 1.16 0.15e–8.67t = = = = – A
iSWt i8 
---------- 9.24 1.23e 8.67t – –
v8 
8
8
iL0+ = 2 iL = 1.54 iSW0+ = 1.16 – 0.15 = 1.01 iSW = 1.16
Now,
and
(2)
Therefore, from the initial condition, (1) and (2) above we have
and with these values we sketch and as shown below.
iLt iSWt
iLt 1.54 0.46e–8.67t = + A
iSWt 1.16 0.15e–8.67t = – A

t = 0 vC0 = 24 V
5. At the circuit is as shown below where and thus the initial condition
has been established.
+

38 
50 F
2 
+

24 V
vS
vCt
The circuit for is shown below where the current source has been replaced with a voltage
source.
Now,
t  0
+

38 
+ 
20 V
50 F
2 
vS
+

24 V
vCt
+

40 
50 F
+

4 V
vCt
vCt vCf + vCn 4 Ae–1  RCt + 4 Ae–500t = = = +
vC0 vC0+ 24 V 4 Ae0 = = = + A = 20
and with the initial condition from which we
obtain
vCt 4 20e–500t = +
6. For t  0
the op amp circuit is as shown below.
+ 
R C
+
+
v
vint voutt
 
Application of KCL at the minus () input yields
and since
v – vin
+ --------- = 0
------------------- C
R
dvC
dt
v = 0

C
dvC
dt
---------
vin
R
= -------
dvC
dt
---------
vin
RC
= --------
voutt = –vCt
voutt
vin
RC
= – --------t + k
k
vC0 = vC0+ = 0 = 0 + k
k = 0
voutt
vin
RC
–  u= 0t
 -------- t 
vin  RC
slope = –vin  RC
t 0 = vC0 = 150 V
+

175 K
1 F
150 V
+
vC0

or
Integrating both sides and observing that we obtain
where is the constant of integration of both sides and it is evaluated from the given initial
condition. Then,
or . Therefore,
and is the slope as shown below.
7.At the circuit is as shown below where and thus the initial condition
has been established.
The circuit for t  0 is shown below where the voltage source vS1
is absent for all positive time
and the is shorted out by the closed switch.
50 K

For the circuit above
vCt vCf + vCn 50 Ae–t  RC + 50 Ae–8t = = = +
and with the initial condition
vC0 vC0+ 150 50 Ae0 = = = +
from which and thus for
+
125 K
1 F
50 V
+
vCt

A = 100 t  0
vCt 50 100e–8t = + V
To find we will first find from the circuit below where
Then,
vR3t iCt
– e–8t = =  –  
iCt C
dvC
dt
--------- 10–6 8 104
+
25 K
1 F
+ +
 vCt
vR3t
50 V 100K 
vR3 t   100 K  iC 105 8 104
– e–8t –   80e–8t = = = – V
The sketches below show and as they approach their final values.
vCt vR3t

8. For this circuit we cannot short the dependent source and therefore we cannot find by
combining the resistances and in parallel combination in order to find the time con-stant
. Instead, we will derive the time constant from the differential equation of (9.9)
+
vC
R1
vCt 12 
R2
------ 1
 vC
 + ------ 
  R1 + R2
  C
 – ------ 
+   vC = 0
------------------
of the previous chapter, that is,
From the given circuit shown below,
or
or
or
vCt 50 100e–8t = +
vR3t –80e–8t =
Req
R1 R2
 = RC
dvC
---------
dt
vC
RC
+ -------- = 0
C
18 
1 F 
+

iCt
10iCt
iC
vC – -----------------------
10iC
R1
vC
R2
+ + ------
C
dvC
dt
--------- 1
R1
R2
10
R1
------C
dvC
dt
+ – --------- = 0
1 10
R1
dvC
dt
 ---------
R1  R2

dvC
---------
dt
R1 + R2
R1  R2
 
------------------
+ ------------------------------vC = 0
1 10
   C
 – ------ 
R1
and from this differential equation we see that the coefficient of is
and thus
vC
-- 30  216
------------ 1
1
ReqC
------------------ 15  9
----------------------------- 30  216
= = = = = = ----- = 0.3125
1 10
   1
 – ----- 
18
8  18
-------- 5
------------------ 135
4  108
432
16
vCt Ae–0.3125t =
vC0 = V0 = A = 5 V
vCt 5e–0.3125t =
and with the given initial condition we obtain
Then, using the relation
we find that for
= ---------
iC C
dvC
dt
t  0
iCt 1 0.3125 5e–0.3125t –   1.5625e–0.3125t = = –
and the minus () sign indicates that the iCt
direction is opposite to that shown.
9.
Step time: 0
Initial value:0
Final value: 12
s
-
CVS 10^(6)
+
10^(-6)
CVS=Controlled Voltage Source
Continuous
+
-
VM
v powergui
Scope
VM=Voltage Measurement
Step

Appendix A
Introduction to MATLAB®
his appendix serves as an introduction to the basic MATLAB commands and functions,
procedures for naming and saving the user generated files, comment lines, access to MAT-LAB’s
Editor / Debugger, finding the roots of a polynomial, and making plots. Several exam-ples
T
are provided with detailed explanations.
A.1 MATLAB® and Simulink®
MATLAB and Simulink are products of The MathWorks,™ Inc. These are two outstanding soft-ware
packages for scientific and engineering computations and are used in educational institu-tions
and in industries including automotive, aerospace, electronics, telecommunications, and
environmental applications. MATLAB enables us to solve many advanced numerical problems
rapidly and efficiently.
A.2 Command Window
To distinguish the screen displays from the user commands, important terms, and MATLAB
functions, we will use the following conventions:
Click: Click the left button of the mouse
Courier Font: Screen displays
Helvetica Font: User inputs at MATLAB’s command window prompt >> or EDU>>*
Helvetica Bold: MATLAB functions
Times Bold Italic: Important terms and facts, notes and file names
When we first start MATLAB, we see various help topics and other information. Initially, we are
interested in the command screen which can be selected from the Window drop menu. When the
command screen, we see the prompt >> or EDU>>. This prompt is displayed also after execution
of a command; MATLAB now waits for a new command from the user. It is highly recommended
that we use the Editor/Debugger to write our program, save it, and return to the command screen
to execute the program as explained below.
To use the Editor/Debugger:
1. From the File menu on the toolbar, we choose New and click on MFile. This takes us to the
Editor Window where we can type our script (list of statements) for a new file, or open a previ-ously
saved file. We must save our program with a file name which starts with a letter.
* EDU>> is the MATLAB prompt in the Student Version
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling A1

Appendix A Introduction to MATLAB®
Important! MATLAB is case sensitive, that is, it distinguishes between upper and lowercase let-ters.
Thus, t and T are two different letters in MATLAB language. The files that we create are
saved with the file name we use and the extension .m; for example, myfile01.m. It is a good prac-tice
to save the script in a file name that is descriptive of our script content. For instance, if the
script performs some matrix operations, we ought to name and save that file as matrices01.m or
any other similar name. We should also use a floppy disk or an external drive to backup our files.
2. Once the script is written and saved as an mfile, we may exit the Editor/Debugger window by
clicking on Exit Editor/Debugger of the File menu. MATLAB then returns to the command
window.
3. To execute a program, we type the file name without the .m extension at the >> prompt;
then, we press <enter> and observe the execution and the values obtained from it. If we have
saved our file in drive a or any other drive, we must make sure that it is added it to the desired
directory in MATLAB’s search path. The MATLAB User’s Guide provides more information
on this topic.
Henceforth, it will be understood that each input command is typed after the >> prompt and fol-lowed
A2 Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems  Modeling
by the <enter> key.
The command help matlabiofun will display input/output information. To get help with other
MATLAB topics, we can type help followed by any topic from the displayed menu. For example,
to get information on graphics, we type help matlabgraphics. The MATLAB User’s Guide con-tains
numerous help topics.
To appreciate MATLAB’s capabilities, we type demo and we see the MATLAB Demos menu.
We can do this periodically to become familiar with them. Whenever we want to return to the
command window, we click on the Close button.
When we are done and want to leave MATLAB, we type quit or exit. But if we want to clear all
previous values, variables, and equations without exiting, we should use the command clear. This
command erases everything; it is like exiting MATLAB and starting it again. The command clc
clears the screen but MATLAB still remembers all values, variables and equations that we have
already used. In other words, if we want to clear all previously entered commands, leaving only
the >> prompt on the upper left of the screen, we use the clc command.
All text after the % (percent) symbol is interpreted as a comment line by MATLAB, and thus it is
ignored during the execution of a program. A comment can be typed on the same line as the func-tion
or command or as a separate line. For instance,
conv(p,q) % performs multiplication of polynomials p and q
% The next statement performs partial fraction expansion of p(x) / q(x)
are both correct.

Roots of Polynomials
One of the most powerful features of MATLAB is the ability to do computations involving com-plex
numbers. We can use either , or to denote the imaginary part of a complex number, such as
i j
3-4i or 3-4j. For example, the statement
z=34j
displays
z = 3.00004.0000i
In the above example, a multiplication (*) sign between 4 and was not necessary because the
complex number consists of numerical constants. However, if the imaginary part is a function, or
variable such as , we must use the multiplication sign, that is, we must type cos(x)*j or
j*cos(x) for the imaginary part of the complex number.
A.3 Roots of Polynomials
In MATLAB, a polynomial is expressed as a row vector of the form . These
are the coefficients of the polynomial in descending order. We must include terms whose coeffi-cients
are zero.
j
cosx
an an – 1  a2 a1 a0
We find the roots of any polynomial with the roots(p) function; p is a row vector containing the
polynomial coefficients in descending order.
Example A.1
Find the roots of the polynomial
p1x x4 10x3 – 35x2
= + – 50x + 24
Solution:
The roots are found with the following two statements where we have denoted the polynomial as
p1, and the roots as roots_ p1.
p1=[1 10 35 50 24] % Specify and display the coefficients of p1(x)
p1 =
1 -10 35 -50 24
roots_ p1=roots(p1) % Find the roots of p1(x)
roots_p1 =
4.0000
3.0000
2.0000
1.0000
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems  Modeling A3

We observe that MATLAB displays the polynomial coefficients as a row vector, and the roots as a
column vector.
p2x x5 7x4 – 16x 2
= + + 25x + 52
1 2 3 and 4
A = a + jb A = a – jb
Example A.2
Find the roots of the polynomial
Solution:
There is no cube term; therefore, we must enter zero as its coefficient. The roots are found with
the statements below, where we have defined the polynomial as p2, and the roots of this polyno-mial
as roots_ p2. The result indicates that this polynomial has three real roots, and two complex
roots. Of course, complex roots always occur in complex conjugate* pairs.
p2=[1 7 0 16 25 52]
p2 =
1 -7 0 16 25 52
roots_ p2=roots(p2)
roots_p2 =
6.5014
2.7428
-1.5711
-0.3366 + 1.3202i
-0.3366 - 1.3202i
A.4 Polynomial Construction from Known Roots
We can compute the coefficients of a polynomial, from a given set of roots, with the poly(r) func-tion
where r is a row vector containing the roots.
Example A.3
It is known that the roots of a polynomial are . Compute the coefficients of this
polynomial.
* By definition, the conjugate of a complex number is

Polynomial Construction from Known Roots
Solution:
We first define a row vector, say r3
, with the given roots as elements of this vector; then, we find
the coefficients with the poly(r) function as shown below.
r3=[1 2 3 4] % Specify the roots of the polynomial
r3 =
1 2 3 4
poly_r3=poly(r3) % Find the polynomial coefficients
poly_r3 =
1 -10 35 -50 24
We observe that these are the coefficients of the polynomial of Example A.1.
p1x
Example A.4
It is known that the roots of a polynomial are  Find the coeffi-cients
of this polynomial.
–1 –2 –3 4 + j5 and 4 – j5
Solution:
We form a row vector, say r4
, with the given roots, and we find the polynomial coefficients with
the poly(r) function as shown below.
r4=[ 1 2 3 4+5j 45j ]
r4 =
Columns 1 through 4
-1.0000 -2.0000 -3.0000 -4.0000+ 5.0000i
Column 5
-4.0000- 5.0000i
poly_r4=poly(r4)
poly_r4 =
1 14 100 340 499 246
Therefore, the polynomial is
p4x x5 14x4 100x3 340x2 = + + + + 499x + 246

A.5 Evaluation of a Polynomial at Specified Values
The polyval(p,x) function evaluates a polynomial px
at some specified value of the indepen-dent
x
p5x x6 3x5 – 5x3 4x2 = + – + 3x + 2
x = –3
p1 x5 3x4 = – + 5x2 + 7x + 9
p2 2x6 8x4 = – + 4x2 + 10x + 12
variable .
Example A.5
Evaluate the polynomial
(A.1)
at .
Solution:
p5=[1 3 0 5 4 3 2]; % These are the coefficients of the given polynomial
% The semicolon (;) after the right bracket suppresses the
% display of the row vector that contains the coefficients of p5.
%
val_minus3=polyval(p5, 3) % Evaluate p5 at x=3; no semicolon is used here
% because we want the answer to be displayed
val_minus3 =
1280
Other MATLAB functions used with polynomials are the following:
conv(a,b)  multiplies two polynomials a and b
[q,r]=deconv(c,d) divides polynomial c by polynomial d and displays the quotient q and
remainder r.
polyder(p)  produces the coefficients of the derivative of a polynomial p.
Example A.6
Let
and
Compute the product using the conv(a,b) function.
p1  p2

Evaluation of a Polynomial at Specified Values
Solution:
p1=[1 3 0 5 7 9]; % The coefficients of p1
p2=[2 0 8 0 4 10 12]; % The coefficients of p2
p1p2=conv(p1,p2) % Multiply p1 by p2 to compute coefficients of the product p1p2
p1p2 =
2 -6 -8 34 18 -24 -74 -88 78 166 174 108
Therefore,
Example A.7
Let
and
p1  p2 2x11 6x10 8x9 – – 34x8 18x7 24x6 = + + –
–74x5–88x4 + 78x3 + 166x2 + 174x + 108
p3 x7 3x5 = – + 5x3 + 7x + 9
p4 2x6 8x5 = – + 4x2 + 10x + 12
Compute the quotient using the [q,r]=deconv(c,d) function.
Solution:
% It is permissible to write two or more statements in one line separated by semicolons
p3=[1 0 3 0 5 7 9]; p4=[2 8 0 0 4 10 12]; [q,r]=deconv(p3,p4)
q =
0.5000
r =
p3  p4
0 4 -3 0 3 2 3
Therefore,
Example A.8
Let
q 0.5 = r 4x5
3x4 = – + 3x2 + 2x + 3
p5 2x6 8x4 = – + 4x2 + 10x + 12
d
dx
------p5
Compute the derivative using the polyder(p) function.

Solution:
p5=[2 0 8 0 4 10 12]; % The coefficients of p5
der_p5=polyder(p5) % Compute the coefficients of the derivative of p5
der_p5 =
d
dx
------p5 = 12x5 – 32x3 + 4x2 + 8x + 10
Rx Numx
= = ------------------------------------------------------------------------------------------------------------------------
--------------------
Denx
bnxn bn – 1xn – 1 bn – 2xn – 2 + + +  + b1x + b0
amxm am – 1xm – 1 am – 2xm – 2 + + +  + a1x + a0
Rx
pnum
pden
------------ x5 3x4 – + 5x2 + 7x + 9
= = ---------------------------------------------------------
x6 4x4 – + 2x2 + 5x + 6
12 0 -32 0 8 10
Therefore,
A.6 Rational Polynomials
Rational Polynomials are those which can be expressed in ratio form, that is, as
(A.2)
where some of the terms in the numerator and/or denominator may be zero. We can find the roots
of the numerator and denominator with the roots(p) function as before.
As noted in the comment line of Example A.7, we can write MATLAB statements in one line, if
we separate them by commas or semicolons. Commas will display the results whereas semicolons
will suppress the display.
Example A.9
Let
Express the numerator and denominator in factored form, using the roots(p) function.
Solution:
num=[1 3 0 5 7 9]; den=[1 0 4 0 2 5 6]; % Do not display num and den coefficients
roots_num=roots(num), roots_den=roots(den) % Display num and den roots
roots_num =
2.4186 + 1.0712i 2.4186 - 1.0712i -1.1633
-0.3370 + 0.9961i -0.3370 - 0.9961i

Rational Polynomials
roots_den =
1.6760 + 0.4922i 1.6760 - 0.4922i -1.9304
-0.2108 + 0.9870i -0.2108 - 0.9870i -1.0000
As expected, the complex roots occur in complex conjugate pairs.
For the numerator, we have the factored form
pnum = x–2.4186 – j1.0712x–2.4186 + j1.0712x + 1.1633
x + 0.3370 – j0.9961x + 0.3370 + j0.9961
and for the denominator, we have
pden = x–1.6760 – j0.4922x–1.6760 + j0.4922x + 1.9304
x + 0.2108–j0.9870x + 0.2108 + j0.9870x + 1.0000
We can also express the numerator and denominator of this rational function as a combination of
linear and quadratic factors. We recall that, in a quadratic equation of the form
whose roots are x1 and x2 , the negative sum of the roots is equal to the coefficient b of the
x
term, that is, –x1 + x2 = b , while the product of the roots is equal to the constant term c
, that
is, x1  x2 = c . Accordingly, we form the coefficient b
by addition of the complex conjugate roots
and this is done by inspection; then we multiply the complex conjugate roots to obtain the con-stant
term using MATLAB as follows:
c
(2.4186 + 1.0712i)*(2.4186 1.0712i)
ans = 6.9971
(0.3370+ 0.9961i)*(0.33700.9961i)
ans = 1.1058
(1.6760+ 0.4922i)*(1.67600.4922i)
ans = 3.0512
(0.2108+ 0.9870i)*(0.21080.9870i)
ans = 1.0186
Thus,
x2 + bx + c = 0
Rx
pnum
pden
------------ x2 – 4.8372x + 6.9971 x 2 + 0.6740x + 1.1058x + 1.1633
= = ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
x2 – 3.3520x + 3.0512x2 + 0.4216x + 1.0186x + 1.0000x + 1.9304

We can check this result of Example A.9 above with MATLAB’s Symbolic Math Toolbox which is
a collection of tools (functions) used in solving symbolic expressions. They are discussed in detail
in MATLAB’s Users Manual. For the present, our interest is in using the collect(s) function that
is used to multiply two or more symbolic expressions to obtain the result in polynomial form. We
must remember that the conv(p,q) function is used with numeric expressions only, that is, poly-nomial
A
R1
R2
V L
C
coefficients.
Before using a symbolic expression, we must create one or more symbolic variables such as x, y, t,
and so on. For our example, we use the following script:
syms x % Define a symbolic variable and use collect(s) to express numerator in polynomial form
collect((x^24.8372*x+6.9971)*(x^2+0.6740*x+1.1058)*(x+1.1633))
ans =
x^5-29999/10000*x^4-1323/3125000*x^3+7813277909/
1562500000*x^2+1750276323053/250000000000*x+4500454743147/
500000000000
and if we simplify this, we find that is the same as the numerator of the given rational expression
in polynomial form. We can use the same procedure to verify the denominator.
A.7 Using MATLAB to Make Plots
Quite often, we want to plot a set of ordered pairs. This is a very easy task with the MATLAB
plot(x,y) command that plots y versus x, where x is the horizontal axis (abscissa) and y is the ver-tical
axis (ordinate).
Example A.10
Consider the electric circuit of Figure A.1, where the radian frequency  (radians/second) of the
applied voltage was varied from 300 to 3000 in steps of 100 radians/second, while the amplitude
was held constant.
Figure A.1. Electric circuit for Example A.10

Using MATLAB to Make Plots
The ammeter readings were then recorded for each frequency. The magnitude of the impedance
|Z| was computed as and the data were tabulated on Table A.1.
Z = V  A
TABLE A.1 Table for Example A.10
 (rads/s) |Z| Ohms  (rads/s) |Z| Ohms
300 39.339 1700 90.603
400 52.589 1800 81.088
500 71.184 1900 73.588
600 97.665 2000 67.513
700 140.437 2100 62.481
800 222.182 2200 58.240
900 436.056 2300 54.611
1000 1014.938 2400 51.428
1100 469.83 2500 48.717
1200 266.032 2600 46.286
1300 187.052 2700 44.122
1400 145.751 2800 42.182
1500 120.353 2900 40.432
1600 103.111 3000 38.845
Plot the magnitude of the impedance, that is, |Z| versus radian frequency 
.
Solution:
We cannot type 
(omega) in the MATLAB Command prompt, so we will use the English letter
w instead.
If a statement, or a row vector is too long to fit in one line, it can be continued to the next line by
typing three or more periods, then pressing <enter> to start a new line, and continue to enter
data. This is illustrated below for the data of w and z. Also, as mentioned before, we use the semi-colon
(;) to suppress the display of numbers that we do not care to see on the screen.
The data are entered as follows:
w=[300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900....
2000 2100 2200 2300 2400 2500 2600 2700 2800 2900 3000];
%
z=[39.339 52.789 71.104 97.665 140.437 222.182 436.056....
1014.938 469.830 266.032 187.052 145.751 120.353 103.111....
90.603 81.088 73.588 67.513 62.481 58.240 54.611 51.468....
48.717 46.286 44.122 42.182 40.432 38.845];
Of course, if we want to see the values of w or z or both, we simply type w or z, and we press

<enter>. To plot z (yaxis) versus w
(xaxis), we use the plot(x,y) command. For this example,
we use plot(w,z). When this command is executed, MATLAB displays the plot on MATLAB’s
graph screen and MATLAB denotes this plot as Figure 1. This plot is shown in Figure A.2.
0 500 1000 1500 2000 2500 3000
1200
1000
800
600
400
200
0
z 
Figure A.2. Plot of impedance versus frequency for Example A.10
This plot is referred to as the magnitude frequency response of the circuit.
To return to the command window, we press any key, or from the Window pulldown menu, we
select MATLAB Command Window. To see the graph again, we click on the Window pulldown
menu, and we choose Figure 1.
We can make the above, or any plot, more presentable with the following commands:
grid on: This command adds grid lines to the plot. The grid off command removes the grid. The
command grid toggles them, that is, changes from off to on or vice versa. The default* is off.
box off: This command removes the box (the solid lines which enclose the plot), and box on
restores the box. The command box toggles them. The default is on.
title(‘string’): This command adds a line of the text string (label) at the top of the plot.
xlabel(‘string’) and ylabel(‘string’) are used to label the x and yaxis respectively.
The magnitude frequency response is usually represented with the xaxis in a logarithmic scale.
We can use the semilogx(x,y) command which is similar to the plot(x,y) command, except that
the xaxis is represented as a log scale, and the yaxis as a linear scale. Likewise, the semil-ogy(
x,y) command is similar to the plot(x,y) command, except that the yaxis is represented as a
* A default is a particular value for a variable that is assigned automatically by an operating system and remains
in effect unless canceled or overridden by the operator.

log scale, and the xaxis as a linear scale. The loglog(x,y) command uses logarithmic scales for
both axes.
Throughout this text it will be understood that log is the common (base 10) logarithm, and ln is
the natural (base e) logarithm. We must remember, however, the function log(x) in MATLAB is
the natural logarithm, whereas the common logarithm is expressed as log10(x), and the logarithm
to the base 2 as log2(x).
Let us now redraw the plot with the above options by adding the following statements:
semilogx(w,z); grid; % Replaces the plot(w,z) command
title('Magnitude of Impedance vs. Radian Frequency');
xlabel('w in rads/sec'); ylabel('|Z| in Ohms')
After execution of these commands, the plot is as shown in Figure A.3.
If the yaxis represents power, voltage or current, the xaxis of the frequency response is more
often shown in a logarithmic scale, and the yaxis in dB (decibels).
1200
1000
800
600
400
200
0
Magnitude of Impedance vs. Radian Frequency
102 103 104
w in rads/sec
Figure A.3. Modified frequency response plot of Figure A.2.
|Z| in Ohms
To display the voltage v
in a dB scale on the yaxis, we add the relation dB=20*log10(v), and we
replace the semilogx(w,z) command with semilogx(w,dB).
The command gtext(‘string’)* switches to the current Figure Window, and displays a crosshair
that can be moved around with the mouse. For instance, we can use the command gtext(‘Imped-ance
|Z| versus Frequency’), and this will place a crosshair in the Figure window. Then, using
* With the latest MATLAB Versions 6 and 7 (Student Editions 13 and 14), we can add text, lines and arrows directly into
the graph using the tools provided on the Figure Window. For advanced MATLAB graphics, please refer to The Math-
Works Using MATLAB Graphics documentation.

the mouse, we can move the crosshair to the position where we want our label to begin, and we
press <enter>.
The command text(x,y,’string’) is similar to gtext(‘string’). It places a label on a plot in some
specific location specified by x and y, and string is the label which we want to place at that loca-tion.
We will illustrate its use with the following example which plots a 3phase sinusoidal wave-form.
The first line of the script below has the form
linspace(first_value, last_value, number_of_values)
This function specifies the number of data points but not the increments between data points. An
alternate function is
x=first: increment: last
and this specifies the increments between points but not the number of data points.
The script for the 3phase plot is as follows:
x=linspace(0, 2*pi, 60); % pi is a builtin function in MATLAB;
% we could have used x=0:0.02*pi:2*pi or x = (0: 0.02: 2)*pi instead;
y=sin(x); u=sin(x+2*pi/3); v=sin(x+4*pi/3);
plot(x,y,x,u,x,v); % The xaxis must be specified for each function
grid on, box on, % turn grid and axes box on
text(0.75, 0.65, 'sin(x)'); text(2.85, 0.65, 'sin(x+2*pi/3)'); text(4.95, 0.65, 'sin(x+4*pi/3)')
These three waveforms are shown on the same plot of Figure A.4.
0 1 2 3 4 5 6 7
Figure A.4. Threephase waveforms
1
0.5
0
-0.5
-1
sin(x) sin(x+2*pi/3) sin(x+4*pi/3)

In our previous examples, we did not specify line styles, markers, and colors for our plots. How-ever,
MATLAB allows us to specify various line types, plot symbols, and colors. These, or a com-bination
of these, can be added with the plot(x,y,s) command, where s is a character string con-taining
one or more characters shown on the three columns of Table A.2. MATLAB has no
default color; it starts with blue and cycles through the first seven colors listed in Table A.2 for
each additional line in the plot. Also, there is no default marker; no markers are drawn unless
they are selected. The default line is the solid line. But with the latest MATLAB versions, we can
select the line color, line width, and other options directly from the Figure Window.
TABLE A.2 Styles, colors, and markets used in MATLAB
Symbol Color Symbol Marker Symbol Line Style
b blue  point  solid line
g green o circle  dotted line
r red x xmark  dashdot line
c cyan + plus  dashed line
m magenta * star
y yellow s square
k black d diamond
w white  triangle down
 triangle up
 triangle left
 triangle right
p pentagram
h hexagram
For example, plot(x,y,'m*:') plots a magenta dotted line with a star at each data point, and
plot(x,y,'rs') plots a red square at each data point, but does not draw any line because no line was
selected. If we want to connect the data points with a solid line, we must type plot(x,y,'rs'). For
additional information we can type help plot in MATLAB’s command screen.
The plots we have discussed thus far are twodimensional, that is, they are drawn on two axes.
MATLAB has also a threedimensional (threeaxes) capability and this is discussed next.
The plot3(x,y,z) command plots a line in 3space through the points whose coordinates are the
elements of x, y and z, where x, y and z are three vectors of the same length.
The general format is plot3(x1,y1,z1,s1,x2,y2,z2,s2,x3,y3,z3,s3,...) where xn, yn and zn are vectors
or matrices, and sn are strings specifying color, marker symbol, or line style. These strings are the
same as those of the twodimensional plots.

z 2x3
= – + x + 3y2
– 1
-10
-5
0
5
10
-10
-5
0
5
3000
2000
1000
0
-1000
-2000
10
Example A.11
Plot the function
(A.3)
Solution:
We arbitrarily choose the interval (length) shown on the script below.
x= 10: 0.5: 10; % Length of vector x
y= x; % Length of vector y must be same as x
z= 2.*x.^3+x+3.*y.^21; % Vector z is function of both x and y*
plot3(x,y,z); grid
The threedimensional plot is shown in Figure A.5.
Figure A.5. Three dimensional plot for Example A.11
In a twodimensional plot, we can set the limits of the x and yaxes with the axis([xmin xmax
ymin ymax]) command. Likewise, in a threedimensional plot we can set the limits of all three
axes with the axis([xmin xmax ymin ymax zmin zmax]) command. It must be placed after the
plot(x,y) or plot3(x,y,z) commands, or on the same line without first executing the plot com-mand.
This must be done for each plot. The threedimensional text(x,y,z,’string’) command will
place string beginning at the coordinate (x,y,z) on the plot.
For threedimensional plots, grid on and box off are the default states.
* This statement uses the so called dot multiplication, dot division, and dot exponentiation where the multiplication, division,
and exponential operators are preceded by a dot. These important operations will be explained in Section A.9.

We can also use the mesh(x,y,z) command with two vector arguments. These must be defined as
and where . In this case, the vertices of the mesh
lengthx = n lengthy = m m n = sizeZ
lines are the triples xj yi Zi j
. We observe that x corresponds to the columns of Z, and y
corresponds to the rows.
To produce a mesh plot of a function of two variables, say z = f x y
, we must first generate the
X and Y matrices that consist of repeated rows and columns over the range of the variables x and
y. We can generate the matrices X and Y with the [X,Y]=meshgrid(x,y) function that creates the
matrix X whose rows are copies of the vector x, and the matrix Y whose columns are copies of the
vector y.
Example A.12
The volume of a right circular cone of radius and height is given by
(A.4)
V r h
V 13
= --r2h
Plot the volume of the cone as r and h vary on the intervals 0  r  4 and 0  h  6
meters.
Solution:
The volume of the cone is a function of both the radius r and the height h, that is,
V = f r h
The threedimensional plot is created with the following MATLAB script where, as in the previ-ous
example, in the second line we have used the dot multiplication, dot division, and dot expo-nentiation.
This will be explained in Section A.9.
[R,H]=meshgrid(0: 4, 0: 6); % Creates R and H matrices from vectors r and h;...
V=(pi .* R .^ 2 .* H) ./ 3; mesh(R, H, V);...
xlabel('xaxis, radius r (meters)'); ylabel('yaxis, altitude h (meters)');...
zlabel('zaxis, volume (cubic meters)'); title('Volume of Right Circular Cone'); box on
The threedimensional plot of Figure A.6 shows how the volume of the cone increases as the
radius and height are increased.
The plots of Figure A.5 and A.6 are rudimentary; MATLAB can generate very sophisticated
threedimensional plots. The MATLAB User’s Manual and the Using MATLAB Graphics Man-ual
contain numerous examples.

Volume of Right Circular Cone
0
1
2
0
2
4
x-axis, radius r (meters)
z-axis, volume (cubic meters)
y-axis, altitude h (meters)
Figure A.6. Volume of a right circular cone.
3
4
150
100
50
6 0
A.8 Subplots
MATLAB can display up to four windows of different plots on the Figure window using the com-mand
subplot(m,n,p). This command divides the window into an m  n matrix of plotting areas
and chooses the pth area to be active. No spaces or commas are required between the three inte-gers
m, n and p. The possible combinations are shown in Figure A.7.
We will illustrate the use of the subplot(m,n,p) command following the discussion on multiplica-tion,
111
Full Screen Default
211
212
221 222
223 224
121 122
221 222
212
211
223 224
221
223
122 121
222
224
division and exponentiation that follows.
Figure A.7. Possible subplot arrangements in MATLAB
A.9 Multiplication, Division, and Exponentiation
MATLAB recognizes two types of multiplication, division, and exponentiation. These are the
matrix multiplication, division, and exponentiation, and the elementbyelement multiplication,
division, and exponentiation. They are explained in the following paragraphs.

Multiplication, Division, and Exponentiation
In Section A.2, the arrays , such a those that contained the coefficients of polynomi-als,
a b c 
consisted of one row and multiple columns, and thus are called row vectors. If an array has
one column and multiple rows, it is called a column vector. We recall that the elements of a row
vector are separated by spaces. To distinguish between row and column vectors, the elements of a
column vector must be separated by semicolons. An easier way to construct a column vector, is to
write it first as a row vector, and then transpose it into a column vector. MATLAB uses the single
quotation character () to transpose a vector. Thus, a column vector can be written either as
b=[1; 3; 6; 11]
or as
b=[1 3 6 11]'
As shown below, MATLAB produces the same display with either format.
b=[1; 3; 6; 11]
b =
-1
3
6
11
b=[1 3 6 11]' % Observe the single quotation character (‘)
b =
-1
3
6
11
We will now define Matrix Multiplication and ElementbyElement multiplication.
1. Matrix Multiplication (multiplication of row by column vectors)
Let
and
A = a1 a2 a3  an
B = b1 b2 b3  bn'
A B
be two vectors. We observe that is defined as a row vector whereas is defined as a col-umn
vector, as indicated by the transpose operator (). Here, multiplication of the row vector
by the column vector , is performed with the matrix multiplication operator (*). Then,
(A.5)
A B
A*B = a1b1 + a2b2 + a3b3 +  + anbn = single value

For example, if
and
the matrix multiplication produces the single value 68, that is,
and this is verified with the MATLAB script
A=[1 2 3 4 5]; B=[ 2 6 3 8 7]'; A*B % Observe transpose operator (‘) in B
ans =
68
Now, let us suppose that both and are row vectors, and we attempt to perform a rowby
row multiplication with the following MATLAB statements.
A=[1 2 3 4 5]; B=[2 6 3 8 7]; A*B % No transpose operator (‘) here
When these statements are executed, MATLAB displays the following message:
??? Error using ==> *
Inner matrix dimensions must agree.
Here, because we have used the matrix multiplication operator (*) in A*B, MATLAB expects
vector to be a column vector, not a row vector. It recognizes that is a row vector, and
warns us that we cannot perform this multiplication using the matrix multiplication operator
(*). Accordingly, we must perform this type of multiplication with a different operator. This
operator is defined below.
2. ElementbyElement Multiplication (multiplication of a row vector by another row vector)
Let
and
be two row vectors. Here, multiplication of the row vector by the row vector is per-formed
with the dot multiplication operator (.*). There is no space between the dot and the
multiplication symbol. Thus,
(A.6)
A = 1 2 3 4 5
B = –2 6 –3 8 7'
A*B
AB = 1  –2 + 2  6 + 3  –3 + 4  8 + 5  7 = 68
A B
B B
C = c1 c2 c3  cn
D = d1 d2 d3  dn
C D
C.D = c1d1 c2d2 c3d3  cndn

This product is another row vector with the same number of elements, as the elements of
and .
As an example, let
and
D
C = 1 2 3 4 5
D = –2 6 –3 8 7
Dot multiplication of these two row vectors produce the following result.
C.D = 1  –2 2  6 3  –3 4  8 5  7 = –2 12 –9 32 35
Check with MATLAB:
C=[1 2 3 4 5]; % Vectors C and D must have
D=[2 6 3 8 7]; % same number of elements
C.*D % We observe that this is a dot multiplication
ans =
-2 12 -9 32 35
C
Similarly, the division (/) and exponentiation (^) operators, are used for matrix division and
exponentiation, whereas dot division (./) and dot exponentiation (.^) are used for element
byelement division and exponentiation, as illustrated in Examples A.11 and A.12 above.
We must remember that no space is allowed between the dot (.) and the multiplication, divi-sion,
and exponentiation operators.
Note: A dot (.) is never required with the plus (+) and minus () operators.
Example A.13
Write the MATLAB script that produces a simple plot for the waveform defined as
(A.7)
y ft 3e –4t cos5t 2e –3t – sin2t t2
= = + ----------
t + 1
in the 0  t  5
seconds interval.
Solution:
The MATLAB script for this example is as follows:
t=0: 0.01: 5; % Define taxis in 0.01 increments
y=3 .* exp(4 .* t) .* cos(5 .* t)2 .* exp(3 .* t) .* sin(2 .* t) + t .^2 ./ (t+1);
plot(t,y); grid; xlabel('t'); ylabel('y=f(t)'); title('Plot for Example A.13')
The plot for this example is shown in Figure A.8.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
5
4
3
2
1
0
-1
Figure A.8. Plot for Example A.13
Had we, in this example, defined the time interval starting with a negative value equal to or less
than , say as  MATLAB would have displayed the following message:
Warning: Divide by zero.
This is because the last term (the rational fraction) of the given expression, is divided by zero
when . To avoid division by zero, we use the special MATLAB function eps, which is a
number approximately equal to . It will be used with the next example.
The command axis([xmin xmax ymin ymax]) scales the current plot to the values specified by
the arguments xmin, xmax, ymin and ymax. There are no commas between these four argu-ments.
This command must be placed after the plot command and must be repeated for each plot.
The following example illustrates the use of the dot multiplication, division, and exponentiation,
the eps number, the axis([xmin xmax ymin ymax]) command, and also MATLAB’s capability
of displaying up to four windows of different plots.
Example A.14
Plot the functions
in the interval using 100 data points. Use the subplot command to display these func-tions
on four windows on the same graph.
t
y=f(t)
Plot for Example A.13
–1 –3  t  3
t = –1
2.2 10–16 
y = sin2x z = cos2x w = sin2x  cos2x v = sin2x  cos2x
0  x  2

Solution:
The MATLAB script to produce the four subplots is as follows:
x=linspace(0,2*pi,100); % Interval with 100 data points
y=(sin(x).^ 2); z=(cos(x).^ 2);
w=y.* z;
v=y./ (z+eps);% add eps to avoid division by zero
subplot(221);% upper left of four subplots
plot(x,y); axis([0 2*pi 0 1]);
title('y=(sinx)^2');
subplot(222); % upper right of four subplots
plot(x,z); axis([0 2*pi 0 1]);
title('z=(cosx)^2');
subplot(223); % lower left of four subplots
plot(x,w); axis([0 2*pi 0 0.3]);
title('w=(sinx)^2*(cosx)^2');
subplot(224); % lower right of four subplots
plot(x,v); axis([0 2*pi 0 400]);
title('v=(sinx)^2/(cosx)^2');
These subplots are shown in Figure A.9.
y=(sinx)2
0 2 4 6
z=(cosx)2
0 2 4 6
1
0.5
0
w=(sinx)2*(cosx)2
0 2 4 6
v=(sinx)2/(cosx)2
0 2 4 6
400
200
0
Figure A.9. Subplots for the functions of Example A.14
1
0.5
0
0.2
0.1
0
The next example illustrates MATLAB’s capabilities with imaginary numbers. We will introduce
the real(z) and imag(z) functions that display the real and imaginary parts of the complex quan-tity
z = x + iy, the abs(z), and the angle(z) functions that compute the absolute value (magni-tude)
and phase angle of the complex quantity z = x + iy = rWe will also usethe
polar(theta,r) function that produces a plot in polar coordinates, where r is the magnitude, theta

is the angle in radians, and the round(n) function that rounds a number to its nearest integer.
a
Zab 10 F
b
10 
10 
0.1 H
Zab
Zab Z 10 104 j 106
–   
= = + --------------------------------------------------------
10 + j 0.1 – 105   
Example A.15
Consider the electric circuit of Figure A.10.
Figure A.10. Electric circuit for Example A.15
With the given values of resistance, inductance, and capacitance, the impedance as a func-tion
of the radian frequency  can be computed from the following expression:
(A.8)
a. Plot ReZ
(the real part of the impedance Z) versus frequency .
b. Plot ImZ
(the imaginary part of the impedance Z) versus frequency .
c. Plot the impedance Z versus frequency  in polar coordinates.
Solution:
The MATLAB script below computes the real and imaginary parts of Zab
which, for simplicity,
are denoted as z
, and plots these as two separate graphs (parts a & b). It also produces a polar
plot (part c).
w=0: 1: 2000; % Define interval with one radian interval;...
z=(10+(10 .^ 4 j .* 10 .^ 6 ./ (w+eps)) ./ (10 + j .* (0.1 .* w 10.^5./ (w+eps))));...
%
% The first five statements (next two lines) compute and plot Re{z}
real_part=real(z); plot(w,real_part);...
xlabel('radian frequency w'); ylabel('Real part of Z'); grid

0 200 400 600 800 1000 1200 1400 1600 1800 2000
1200
1000
800
600
400
200
0
radian frequency w
Real part of Z
Figure A.11. Plot for the real part of the impedance in Example A.15
% The next five statements (next two lines) compute and plot Im{z}
imag_part=imag(z); plot(w,imag_part);...
xlabel('radian frequency w'); ylabel('Imaginary part of Z'); grid
0 200 400 600 800 1000 1200 1400 1600 1800 2000
600
400
200
0
-200
-400
-600
radian frequency w
Imaginary part of Z
Figure A.12. Plot for the imaginary part of the impedance in Example A.15
% The last six statements (next five lines) below produce the polar plot of z
mag=abs(z); % Computes |Z|;...
rndz=round(abs(z)); % Rounds |Z| to read polar plot easier;...
theta=angle(z); % Computes the phase angle of impedance Z;...
polar(theta,rndz); % Angle is the first argument
ylabel('Polar Plot of Z'); grid

1500
1000
500
30
150
180 0
210
60
120
240
90
270
300
330
Polar Plot of Z
Figure A.13. Polar plot of the impedance in Example A.15
Example A.15 clearly illustrates how powerful, fast, accurate, and flexible MATLAB is.
A.10 Script and Function Files
MATLAB recognizes two types of files: script files and function files. Both types are referred to as
mfiles since both require the .m extension.
A script file consists of two or more builtin functions such as those we have discussed thus far.
Thus, the script for each of the examples we discussed earlier, make up a script file. Generally, a
script file is one which was generated and saved as an mfile with an editor such as the MAT-LAB’s
Editor/Debugger.
A function file is a userdefined function using MATLAB. We use function files for repetitive
tasks. The first line of a function file must contain the word function, followed by the output argu-ment,
the equal sign ( = ), and the input argument enclosed in parentheses. The function name
and file name must be the same, but the file name must have the extension .m. For example, the
function file consisting of the two lines below
function y = myfunction(x)
y=x.^ 3 + cos(3.* x)
is a function file and must be saved as myfunction.m
For the next example, we will use the following MATLAB functions:
fzero(f,x)  attempts to find a zero of a function of one variable, where f is a string containing the
name of a realvalued function of a single real variable. MATLAB searches for a value near a
point where the function f changes sign, and returns that value, or returns NaN if the search fails.

Script and Function Files
Important: We must remember that we use roots(p) to find the roots of polynomials only, such as
those in Examples A.1 and A.2.
fplot(fcn,lims) plots the function specified by the string fcn between the xaxis limits specified
by lims = [xmin xmax]. Using lims = [xmin xmax ymin ymax] also controls the yaxis limits.
The string fcn must be the name of an mfile function or a string with variable .
NaN (NotaNumber) is not a function; it is MATLAB’s response to an undefined expression
such as 0  0 ,   
or inability to produce a result as described on the next paragraph.We can
avoid division by zero using the eps number, which we mentioned earlier.
Example A.16
Find the zeros, the minimum, and the maximum values of the function
(A.9)
x
fx 1
---------------------------------------- 1
= – ---------------------------------------- – 10
x – 0.12 + 0.01
x – 1.22 + 0.04
in the interval
Solution:
We first plot this function to observe the approximate zeros, maxima, and minima using the fol-lowing
script.
–1.5  x  1.5
x=1.5: 0.01: 1.5;
y=1./ ((x0.1).^ 2 + 0.01) 1./ ((x1.2).^ 2 + 0.04) 10;
plot(x,y); grid
The plot is shown in Figure A.14.
100
80
60
40
20
0
-20
-40
-1.5 -1 -0.5 0 0.5 1 1.5
Figure A.14. Plot for Example A.16 using the plot command

The roots (zeros) of this function appear to be in the neighborhood of x = –0.2 and x = 0.3
. The
maximum occurs at approximately x = 0.1 where, approximately, ymax = 90
, and the minimum
occurs at approximately x = 1.2 where, approximately, ymin = –34
.
Next, we define and save f(x) as the funczero01.m function mfile with the following script:
function y=funczero01(x)
% Finding the zeros of the function shown below
y=1/((x0.1)^2+0.01)1/((x1.2)^2+0.04)10;
To save this file, from the File drop menu on the Command Window, we choose New, and when
the Editor Window appears, we type the script above and we save it as funczero01. MATLAB
appends the extension .m to it.
Now, we can use the fplot(fcn,lims) command to plot as follows:
fplot('funczero01', [1.5 1.5]); grid
This plot is shown in Figure A.15. As expected, this plot is identical to the plot of Figure A.14
which was obtained with the plot(x,y) command as shown in Figure A.14.
100
80
60
40
20
0
-20
-40
Figure A.15. Plot for Example A.16 using the fplot command
We will use the fzero(f,x) function to compute the roots of in Equation (A.9) more precisely.
The MATLAB script below will accomplish this.
x1= fzero('funczero01', 0.2);
x2= fzero('funczero01', 0.3);
fprintf('The roots (zeros) of this function are r1= %3.4f', x1);
fprintf(' and r2= %3.4f n', x2)
fx
-1.5 -1 -0.5 0 0.5 1 1.5
fx

Script and Function Files
MATLAB displays the following:
The roots (zeros) of this function are r1= -0.1919 and r2= 0.3788
The earlier MATLAB versions included the function fmin(f,x1,x2) and with this function we
could compute both a minimum of some function or a maximum of since a maximum of
fx fx
is equal to a minimum of . This can be visualized by flipping the plot of a function
fx –fx fx
upsidedown. This function is no longer used in MATLAB and thus we will compute the maxima
and minima from the derivative of the given function.
From elementary calculus, we recall that the maxima or minima of a function can be
found by setting the first derivative of a function equal to zero and solving for the independent
variable x
. For this example we use the diff(x) function which produces the approximate deriva-tive
of a function. Thus, we use the following MATLAB script:
y = fx
syms x ymin zmin; ymin=1/((x0.1)^2+0.01)1/((x1.2)^2+0.04)10;...
zmin=diff(ymin)
zmin =
-1/((x-1/10)^2+1/100)^2*(2*x-1/5)+1/((x-6/5)^2+1/25)^2*(2*x-12/5)
When the command
solve(zmin)
is executed, MATLAB displays a very long expression which when copied at the command
prompt and executed, produces the following:
ans =
0.6585 + 0.3437i
ans =
0.6585 - 0.3437i
ans =
1.2012
The real value 1.2012 above is the value of x at which the function y
has its minimum value as
we observe also in the plot of Figure A.15.
To find the value of y corresponding to this value of x, we substitute it into fx
, that is,
x=1.2012; ymin=1 / ((x0.1) ^ 2 + 0.01) 1 / ((x1.2) ^ 2 + 0.04) 10
ymin = -34.1812
We can find the maximum value from –f x
whose plot is produced with the script
x=1.5:0.01:1.5; ymax=1./((x0.1).^2+0.01)1./((x1.2).^2+0.04)10; plot(x,ymax); grid
and the plot is shown in Figure A.16.

40
20
0
-20
-40
-60
-80
-100
-1.5 -1 -0.5 0 0.5 1 1.5
–fx
Figure A.16. Plot of for Example A.16
Next we compute the first derivative of and we solve for to find the value where the max-imum
–fx x
of occurs. This is accomplished with the MATLAB script below.
ymax
syms x ymax zmax; ymax=(1/((x0.1)^2+0.01)1/((x1.2)^2+0.04)10); zmax=diff(ymax)
zmax =
1/((x-1/10)^2+1/100)^2*(2*x-1/5)-1/((x-6/5)^2+1/25)^2*(2*x-12/5)
solve(zmax)
When the command
solve(zmax)
is executed, MATLAB displays a very long expression which when copied at the command
prompt and executed, produces the following:
ans =
0.6585 + 0.3437i
ans =
0.6585 - 0.3437i
ans =
1.2012
ans =
0.0999
From the values above we choose x = 0.0999
which is consistent with the plots of Figures A.15
and A.16. Accordingly, we execute the following script to obtain the value of .
ymin

Display Formats
x=0.0999; % Using this value find the corresponding value of ymax
ymax=1 / ((x0.1) ^ 2 + 0.01) 1 / ((x1.2) ^ 2 + 0.04) 10
ymax = 89.2000
A.11 Display Formats
MATLAB displays the results on the screen in integer format without decimals if the result is an
integer number, or in short floating point format with four decimals if it a fractional number. The
format displayed has nothing to do with the accuracy in the computations. MATLAB performs all
computations with accuracy up to 16 decimal places.
The output format can changed with the format command. The available MATLAB formats can
be displayed with the help format command as follows:
help format
FORMAT Set output format.
All computations in MATLAB are done in double precision.
FORMAT may be used to switch between different output display formats
as follows:
FORMAT Default. Same as SHORT.
FORMAT SHORT Scaled fixed point format with 5 digits.
FORMAT LONG Scaled fixed point format with 15 digits.
FORMAT SHORT E Floating point format with 5 digits.
FORMAT LONG E Floating point format with 15 digits.
FORMAT SHORT G Best of fixed or floating point format with 5 digits.
FORMAT LONG G Best of fixed or floating point format with 15 digits.
FORMAT HEX Hexadecimal format.
FORMAT + The symbols +, - and blank are printed for positive, negative,
and zero elements.Imaginary parts are ignored.
FORMAT BANK Fixed format for dollars and cents.
FORMAT RAT Approximation by ratio of small integers.
Spacing:
FORMAT COMPACT Suppress extra line-feeds.
FORMAT LOOSE Puts the extra line-feeds back in.
Some examples with different format displays age given below.
format short 33.3335 Four decimal digits (default)
format long 33.33333333333334 16 digits
format short e 3.3333e+01 Four decimal digits plus exponent
format short g 33.333 Better of format short or format short e
format bank 33.33 two decimal digits
format + only + or - or zero are printed

format rat 100/3 rational approximation
The disp(X) command displays the array X without printing the array name. If X is a string, the
text is displayed.
The fprintf(format,array) command displays and prints both text and arrays. It uses specifiers to
indicate where and in which format the values would be displayed and printed. Thus, if %f is
used, the values will be displayed and printed in fixed decimal format, and if %e is used, the val-ues
will be displayed and printed in scientific notation format. With this command only the real
part of each parameter is processed.
This appendix is just an introduction to MATLAB.* This outstanding software package consists
of many applications known as Toolboxes. The MATLAB Student Version contains just a few of
these Toolboxes. Others can be bought directly from The MathWorks, Inc., as addons.
* For more MATLAB applications, please refer to Numerical Analysis Using MATLAB and Excel, ISBN 978
1934404034.

Appendix B
Introduction to Simulink
his appendix is a brief introduction to Simulink. This author feels that we can best intro-duce
Simulink with a few examples. Some familiarity with MATLAB is essential in under-standing
Simulink, and for this purpose, Appendix A is included as an introduction to
T
MATLAB.
B.1 Simulink and its Relation to MATLAB
The MATLAB and Simulink environments are integrated into one entity, and thus we can
analyze, simulate, and revise our models in either environment at any point. We invoke Simulink
from within MATLAB. We will introduce Simulink with a few illustrated examples.
Example B.1
For the circuit of Figure B.1, the initial conditions are iL0 = 0 , and vc0 = 0.5 V
. We will
compute .
+
vCt

R L
+
1  4 H
1  C
it
vst = u0t
4  3 F
Figure B.1. Circuit for Example B.1
vct
For this example,
(B.1)
= = = --------
i iL iC C
and by Kirchoff’s voltage law (KVL),
(B.2)
+ ------- + vC = u0t
RiL L
Substitution of (B.1) into (B.2) yields
dvC
dt
diL
dt
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling B1

(B.3)
d2vC
dt + --------2--- + vC = u0t
Substituting the values of the circuit constants and rearranging we obtain:
(B.4)
(B.5)
--d2vC
dt2 ----------- 43
To appreciate Simulink’s capabilities, for comparison, three different methods of obtaining the
solution are presented, and the solution using Simulink follows.
First Method  Assumed Solution
Equation (B.5) is a secondorder, nonhomogeneous differential equation with constant coeffi-cients,
and thus the complete solution will consist of the sum of the forced response and the natu-ral
response. It is obvious that the solution of this equation cannot be a constant since the deriva-tives
of a constant are zero and thus the equation is not satisfied. Also, the solution cannot
contain sinusoidal functions (sine and cosine) since the derivatives of these are also sinusoids.
However, decaying exponentials of the form where k and a are constants, are possible candi-dates
since their derivatives have the same form but alternate in sign.
It can be shown* that if and where and are constants and and are the
roots of the characteristic equation of the homogeneous part of the given differential equation,
the natural response is the sum of the terms and . Therefore, the total solution will
be
(B.6)
The values of and are the roots of the characteristic equation
* Please refer to Circuit Analysis II with MATLAB Applications, ISBN 0970951159, Appendix B for a
B2 Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling
thorough discussion.
RC
dvC
dt
-------- LC
13
--dvC
dt
+ -------- + vC = u0t
d2vC
dt2 ----------- 4
dvC
dt
+ -------- + 3vC = 3u0t
d2vC
dt2 ----------- 4
dvC
dt
+ -------- + 3vC = 3 t  0
ke–at
k1e –s1t k2e –s2t k1 k2 s1 s2
k1e –s1t k2e –s2t
vct natural response + forced response vcnt + vcft k1e –s1t k2e –s2t = = = + + vcft
s1 s2

Simulink and its Relation to MATLAB
(B.7)
s2 + 4s + 3 = 0
Solution of (B.7) yields of and and with these values (B.6) is written as
(B.8)
s1 = –1 s2 = –3
vct k1e–t = + k2e–3t + vcft
The forced component is found from (B.5), i.e.,
(B.9)
vcft
d2vC
dt2 ----------- 4
dvC
dt
+ -------- + 3vC = 3 t  0
Since the right side of (B.9) is a constant, the forced response will also be a constant and we
denote it as . By substitution into (B.9) we obtain
or
(B.10)
vCf = k3
0 0 3k3
+ + = 3
vCf = k3 = 1
Substitution of this value into (B.8), yields the total solution as
(B.11)
vCt= vCnt + vCf k1e–t = + k2e–3t + 1
The constants and will be evaluated from the initial conditions. First, using
and evaluating (B.11) at , we obtain
k1 k2 vC0 = 0.5 V
(B.12)
Also,
and
(B.13)
t = 0
vC0 = k1e0 + k2e0 + 1 = 0.5
k1 + k2 = –0.5
= = ---------
iL iC C
dvC
dt
dvC
dt
--------- iL
 = ----
C
dvC
dt
---------
t = 0
iL0
C
------------ 0C
= = --- = 0
Next, we differentiate (B.11), we evaluate it at , and equate it with (B.13). Thus,
(B.14)
t = 0
dvC
dt
---------
t = 0
= –k1–3k2
By equating the right sides of (B.13) and (B.14) we obtain

(B.15)
–k1–3k2 = 0
Simultaneous solution of (B.12) and (B.15), gives and . By substitution into
(B.8), we obtain the total solution as
(B.16)
k1 = –0.75 k2 = 0.25
vCt –0.75e–t =  + 0.25e–3t + 1u0t
Check with MATLAB:
syms t % Define symbolic variable t
y0=0.75*exp(t)+0.25*exp(3*t)+1; % The total solution y(t), for our example, vc(t)
y1=diff(y0) % The first derivative of y(t)
y1 =
3/4*exp(-t)-3/4*exp(-3*t)
y2=diff(y0,2) % The second derivative of y(t)
y2 =
-3/4*exp(-t)+9/4*exp(-3*t)
y=y2+4*y1+3*y0 % Summation of y and its derivatives
y =
3
Thus, the solution has been verified by MATLAB. Using the expression for in (B.16), we
find the expression for the current as
(B.17)
  e–t e–3t = = = = – A
+
Second Method  Using the Laplace Transformation
The transformed circuit is shown in Figure B.2.
Figure B.2. Transformed Circuit for Example B.1
vCt
i= iL iC C
dvC
dt
--------- 43
-- 34
--e t – 34
–--e–3t
 

R L
+

 C
Vss = 1  s VCs
Is
0.25s
3  4s
+
VC0
0.5  s

By the voltage division* expression,
VCs 3  4s
+ ------- 0.5s2 + 2s + 3
--------------------------------- 0.5
=    + 0.5
------- 1.5
---------------------------------------------- 1s
1 + 0.25s + 3  4s
-- 0.5
 – -------
Using partial fraction expansion,† we let
(B.18)
0.5s2 + 2s + 3
ss + 1s + 3
-----------------------------------
and by substitution into (B.18)
s
s
= = -----------------------------------
r1
s
= ----
+ + ----------------
----------------
r2
s + 1
r1
0.5s2 + 2s + 3
s + 1s + 3
---------------------------------
= = 1
r2
0.5s2 + 2s + 3
ss + 3
= = –0.75
---------------------------------
r3
0.5s2 + 2s + 3
ss + 1
---------------------------------
= = 0.25
VCs 0.5s2 + 2s + 3
----------------------------------- 1s
= = + + ----------------
ss + 1s + 3
Taking the Inverse Laplace transform‡ we find that
vC t   1 0.75et
Third Method  Using State Variables
ss2 + 4s + 3
r3
s + 3
-- –0.75
---------------- 0.25
s + 1
– 0.25e–3t = – +
**
s
ss + 1s + 3
s = 0
s = –1
s = –3
s + 3
+ ------- + vC = u0t
RiL L
diL
dt
* For derivation of the voltage division and current division expressions, please refer to Circuit Analysis I with
MATLAB Applications, ISBN 0970951124.
† Partial fraction expansion is discussed in Chapter 3, this text.
‡ For an introduction to Laplace Transform and Inverse Laplace Transform, please refer Chapters 2 and 3, this
text.
** Usually, in StateSpace and State Variables Analysis, ut
denotes any input. For distinction, we will denote
the Unit Step Function as u0t
. For a detailed discussion on StateSpace and State Variables Analysis, please
refer to Chapter 5, this text.

--diL
dt
14
------- = –1iL – vC + 1
diL
dt
------- = – 4iL – 4vC + 4
x1 = iL x2 = vC
x·
1
diL
dt
= -------
x· x x·
By substitution of given values and rearranging, we obtain
or
(B.19)
Next, we define the state variables and . Then,
* (B.20)
and
(B.21)
Also,
and thus,
or
(B.22)
Therefore, from (B.19), (B.20), and (B.22), we obtain the state equations
and in matrix form,
(B.23)
Solution† of (B.23) yields
* The notation (x dot) is often used to denote the first derivative of the function , that is, = dx  dt
.
† The detailed solution of (B.23) is given in Chapter 5, Example 5.10, Page 523, this text.
x·
2
dvC
dt
= --------
iL C
dvC
dt
= --------
x1 iL C
dvC
dt
-------- Cx·2
43
--x·
= = = = 2
x·
2
34
= --x1
x·
1 = – 4x1 – 4x2 + 4
x·
2
34
= -- x1
x·
1
x·
2
–4 –4
3  4 0
x1
x2
4
0
= + u0t

Then,
(B.24)
and
(B.25)
x1
x2
e–t –e –3t
1–0.75e–t 0.25e–3t +
=
x1 iL e–t –e –3t = =
x2 vC 1 – 0.75e –t 0.25e–3t = = +
Modeling the Differential Equation of Example B.1 with Simulink
To run Simulink, we must first invoke MATLAB. Make sure that Simulink is installed in your sys-tem.
In the MATLAB Command prompt, we type:
simulink
Alternately, we can click on the Simulink icon shown in Figure B.3. It appears on the top bar on
MATLAB’s Command prompt.
Figure B.3. The Simulink icon
Upon execution of the Simulink command, the Commonly Used Blocks appear as shown in Fig-ure
B.4.
In Figure B.4, the left side is referred to as the Tree Pane and displays all Simulink libraries
installed. The right side is referred to as the Contents Pane and displays the blocks that reside in
the library currently selected in the Tree Pane.
Let us express the differential equation of Example B.1 as
(B.26)
d2vC
dt2 ----------- 4
dvC
dt
= – -------- – 3vC + 3u0t
A block diagram representing relation (B.26) above is shown in Figure B.5. We will use Simulink
to draw a similar block diagram.*
* Henceforth, all Simulink block diagrams will be referred to as models.

Figure B.4. The Simulink Library Browser
-------- vC
u0t 3  dt dt
Figure B.5. Block diagram for equation (B.26)
To model the differential equation (B.26) using Simulink, we perform the following steps:
1. On the Simulink Library Browser, we click on the leftmost icon shown as a blank page on the
top title bar. A new model window named untitled will appear as shown in Figure B.6.
4
3
d2vC
dt2 ----------- dvC
dt

Figure B.6. The Untitled model window in Simulink.
The window of Figure B.6 is the model window where we enter our blocks to form a block dia-gram.
We save this as model file name Equation_1_26. This is done from the File drop menu of
Figure B.6 where we choose Save as and name the file as Equation_1_26. Simulink will add
the extension .mdl. The new model window will now be shown as Equation_1_26, and all
saved files will have this appearance. See Figure B.7.
Figure B.7. Model window for Equation_1_26.mdl file
2. With the Equation_1_26 model window and the Simulink Library Browser both visible, we
click on the Sources appearing on the left side list, and on the right side we scroll down until
we see the unit step function shown as Step. See Figure B.8. We select it, and we drag it into
the Equation_1_26 model window which now appears as shown in Figure B.8. We save file
Equation_1_26 using the File drop menu on the Equation_1_26 model window (right side of
Figure B.8).
3. With reference to block diagram of Figure B.5, we observe that we need to connect an ampli-fier
with Gain 3 to the unit step function block. The gain block in Simulink is under Com-monly
Used Blocks (first item under Simulink on the Simulink Library Browser). See Figure
B.8. If the Equation_1_26 model window is no longer visible, it can be recalled by clicking on
the white page icon on the top bar of the Simulink Library Browser.
4. We choose the gain block and we drag it to the right of the unit step function. The triangle on
the right side of the unit step function block and the > symbols on the left and right sides of
the gain block are connection points. We point the mouse close to the connection point of the
unit step function until is shows as a cross hair, and draw a straight line to connect the two

blocks.* We doubleclick on the gain block and on the Function Block Parameters, we
change the gain from 1 to 3. See Figure B.9.
Figure B.8. Dragging the unit step function into File Equation_1_26
Figure B.9. File Equation_1_26 with added Step and Gain blocks
* An easy method to interconnect two Simulink blocks by clicking on the source block to select it, then hold down
the Ctrl key and leftclick on the destination block.

5. Next, we need to add a theeinput adder. The adder block appears on the right side of the
Simulink Library Browser under Math Operations. We select it, and we drag it into the
Equation_1_26 model window. We double click it, and on the Function Block Parameters
window which appears, we specify 3 inputs. We then connect the output of the of the gain
block to the first input of the adder block as shown in Figure B.10.
Figure B.10. File Equation_1_26 with added gain block
6. From the Commonly Used Blocks of the Simulink Library Browser, we choose the Integra-tor
block, we drag it into the Equation_1_26 model window, and we connect it to the output
of the Add block. We repeat this step and to add a second Integrator block. We click on the
text “Integrator” under the first integrator block, and we change it to Integrator 1. Then, we
change the text “Integrator 1” under the second Integrator to “Integrator 2” as shown in Fig-ure
B.11.
Figure B.11. File Equation_1_26 with the addition of two integrators
7. To complete the block diagram, we add the Scope block which is found in the Commonly
Used Blocks on the Simulink Library Browser, we click on the Gain block, and we copy and
paste it twice. We flip the pasted Gain blocks by using the Flip Block command from the For-mat
drop menu, and we label these as Gain 2 and Gain 3. Finally, we doubleclick on these
gain blocks and on the Function Block Parameters window, we change the gains from to 4
and 3 as shown in Figure B.12.
Figure B.12. File Equation_1_26 complete block diagram

= = 0 vc0 = 0.5 V
8. The initial conditions , and are entered by double
clicking the Integrator blocks and entering the values for the first integrator, and for the
second integrator. We also need to specify the simulation time. This is done by specifying the
simulation time to be seconds on the Configuration Parameters from the Simulation drop
menu. We can start the simulation on Start from the Simulation drop menu or by clicking on
the icon.
9. To see the output waveform, we double click on the Scope block, and then clicking on the
Autoscale icon, we obtain the waveform shown in Figure B.13.
Figure B.13. The waveform for the function for Example B.1
Another easier method to obtain and display the output for Example B.1, is to use State
Space block from Continuous in the Simulink Library Browser, as shown in Figure B.14.
Figure B.14. Obtaining the function for Example B.1 with the StateSpace block.
iL0 C
dvC
---------
dt
t = 0
0 0.5
10
vCt
vCt
vCt

The simout To Workspace block shown in Figure B.14 writes its input to the workspace. The
data and variables created in the MATLAB Command window, reside in the MATLAB Work-space.
This block writes its output to an array or structure that has the name specified by the
block's Variable name parameter. This gives us the ability to delete or modify selected variables.
We issue the command who to see those variables. From Equation B.23, Page B6,
The output equation is
or
x·
1
x·
2
–4 –4
3  4 0
x1
x2
4
0
= + u0t
y = Cx + du
y 0 1 x1
= + 0u
x2
We doubleclick on the StateSpace block, and in the Functions Block Parameters window we
enter the constants shown in Figure B.15.
Figure B.15. The Function block parameters for the StateSpace block.

The initials conditions x1 x2'
are specified in MATLAB’s Command prompt as
x1=0; x2=0.5;
As before, to start the simulation we click clicking on the icon, and to see the output wave-form,
we double click on the Scope block, and then clicking on the Autoscale icon, we
vCt
d 4y
dt4 --------- a3
d 3y
dt3 --------- a2
d2y
dt2 -------- a1
dy
dt
+ + + ------ + a0 yt = ut
obtain the waveform shown in Figure B.16.
Figure B.16. The waveform for the function for Example B.1 with the StateSpace block.
The statespace block is the best choice when we need to display the output waveform of three or
more variables as illustrated by the following example.
Example B.2
A fourthorder network is described by the differential equation
(B.27)
where yt is the output representing the voltage or current of the network, and ut
is any input,
and the initial conditions are y0 = y'0 = y''0 = y'''0 = 0
.
a. We will express (B.27) as a set of state equations

b. It is known that the solution of the differential equation
(B.28)
d4y
dt4 -------- 2d2y
dt + ------2-- + yt = sint
y0 = y'0 = y''0 = y'''0 = 0
subject to the initial conditions , has the solution
(B.29)
yt 0.125 3 t2 =  –  – 3t cost
In our set of state equations, we will select appropriate values for the coefficients
so that the new set of the state equations will represent the differential equa-tion
a3 a2 a1 and a0
of (B.28), and using Simulink, we will display the waveform of the output .
yt
1. The differential equation of (B.28) is of fourthorder; therefore, we must define four state vari-ables
that will be used with the four firstorder state equations.
We denote the state variables as , and , and we relate them to the terms of the
given differential equation as
(B.30)
x1 = yt x2
We observe that
(B.31)
and in matrix form
(B.32)
x1 x2 x3 x4
dy
dt
= ------ x3
x·
1 = x2
x·
2 = x3
x·
3 = x4
d 4y
dt4 --------- x·= 4 = –a0x1–a1x2 – a2x3–a3x4 + ut
x·
1
x·
2
x·
3
x·
4
0 1 0 0
0 0 1 0
0 0 0 1
–a0 –a1 –a2 –a3
= + ut
In compact form, (B.32) is written as
(B.33)
Also, the output is
(B.34)
where
d 2y
dt2 = --------- x4
d 3y
dt3 = ---------
x1
x2
x3
x4
0
0
0
1
x·
= Ax + bu
y = Cx + du

(B.35)
x·
x·
1
x·
2
= A
x·
3
x·
4
0 1 0 0
0 0 1 0
0 0 0 1
–a0 –a1 –a2 –a3
= x
= A
and since the output is defined as
relation (B.34) is expressed as
(B.36)
2. By inspection, the differential equation of (B.27) will be reduced to the differential equation of
(B.28) if we let
and thus the differential equation of (B.28) can be expressed in statespace form as
(B.37)
where
(B.38)
Since the output is defined as
in matrix form it is expressed as
x1
x2
x3
x4
= b
0
0
0
1
   =  and u = ut
yt = x1
y 1 0 0 0
x1
x2
x3
x4
=  + 0ut
a3 = 0 a2 = 2 a1 = 0 a0 = 1 ut = sint
x·
1
x·
2
x·
3
x·
4
0 1 0 0
0 0 1 0
0 0 0 1
–a0 0 –2 0
x1
x2
x3
x4
0
0
0
1
= + sint
x·
x·
1
x·
2
x·
3
x·
4
0 1 0 0
0 0 1 0
0 0 0 1
–a0 0 –2 0
= x
x1
x2
x3
x4
= b
0
0
0
1
   =  and u = sint
yt = x1

(B.39)
y 1 0 0 0
x1
x2
x3
x4
=  + 0 sint
We invoke MATLAB, we start Simulink by clicking on the Simulink icon, on the Simulink
Library Browser we click on the Create a new model (blank page icon on the left of the top
bar), and we save this model as Example_1_2. On the Simulink Library Browser we select
Sources, we drag the Signal Generator block on the Example_1_2 model window, we click
and drag the StateSpace block from the Continuous on Simulink Library Browser, and we
click and drag the Scope block from the Commonly Used Blocks on the Simulink Library
Browser. We also add the Display block found under Sinks on the Simulink Library
Browser. We connect these four blocks and the complete block diagram is as shown in Figure
B.17.
Figure B.17. Block diagram for Example B.2
We now doubleclick on the Signal Generator block and we enter the following in the Func-tion
Block Parameters:
Wave form: sine
Time (t): Use simulation time
Amplitude: 1
Frequency: 2
Units: Hertz
Next, we doubleclick on the statespace block and we enter the following parameter values
in the Function Block Parameters:
A: [0 1 0 0; 0 0 1 0; 0 0 0 1; a0 a1 a2 a3]
B: [0 0 0 1]’
C: [1 0 0 0]
D: [0]
Initial conditions: x0

Absolute tolerance: auto
Now, we switch to the MATLAB Command prompt and we type the following:
>> a0=1; a1=0; a2=2; a3=0; x0=[0 0 0 0]’;
We change the Simulation Stop time to , and we start the simulation by clicking on the
icon. To see the output waveform, we double click on the Scope block, then clicking on the
Autoscale icon, we obtain the waveform shown in Figure B.18.
Figure B.18. Waveform for Example B.2
The Display block in Figure B.17 shows the value at the end of the simulation stop time.
Examples B.1 and B.2 have clearly illustrated that the StateSpace is indeed a powerful block. We
could have obtained the solution of Example B.2 using four Integrator blocks by this approach
would have been more time consuming.
Example B.3
Using Algebraic Constraint blocks found in the Math Operations library, Display blocks found
in the Sinks library, and Gain blocks found in the Commonly Used Blocks library, we will create
a model that will produce the simultaneous solution of three equations with three unknowns.
The model will display the values for the unknowns , , and in the system of the equations
(B.40)
25
z1 z2 z3
a1z1 + a2z2 + a3z3 + k1 = 0
a4z1 + a5z2 + a6z3 + k2 = 0
a7z1 + a8z2 + a9z3 + k3 = 0

The model is shown in Figure B.19.
Figure B.19. Model for Example B.3
Next, we go to MATLAB’s Command prompt and we enter the following values:
a1=2; a2=3; a3=1; a4=1; a5=5; a6=4; a7=6; a8=1; a9=2;...
k1=8; k2=7; k3=5;
After clicking on the simulation icon, we observe the values of the unknowns as ,
, and .These values are shown in the Display blocks of Figure B.19.
z1 = 2
z2 = –3 z3 = 5
The Algebraic Constraint block constrains the input signal to zero and outputs an algebraic
state z
. The block outputs the value necessary to produce a zero at the input. The output must
affect the input through some feedback path. This enables us to specify algebraic equations for
index 1 differential/algebraic systems (DAEs). By default, the Initial guess parameter is zero. We
can improve the efficiency of the algebraic loop solver by providing an Initial guess for the alge-braic
state z that is close to the solution value.
fz

An outstanding feature in Simulink is the representation of a large model consisting of many
blocks and lines, to be shown as a single Subsystem block.* For instance, we can group all blocks
and lines in the model of Figure B.19 except the display blocks, we choose Create Subsystem
from the Edit menu, and this model will be shown as in Figure B.20† where in MATLAB’s Com-mand
prompt we have entered:
a1=5; a2=1; a3=4; a4=11; a5=6; a6=9; a7=8; a8=4; a9=15;...
k1=14; k2=6; k3=9;
Figure B.20. The model of Figure B.19 represented as a subsystem
The Display blocks in Figure B.20 show the values of , , and for the values specified in
MATLAB’s Command prompt.
B.2 Simulink Demos
At this time, the reader with no prior knowledge of Simulink, should be ready to learn Simulink’s
additional capabilities. It is highly recommended that the reader becomes familiar with the block
libraries found in the Simulink Library Browser. Then, the reader can follow the steps delineated
in The MathWorks Simulink User’s Manual to run the Demo Models beginning with the thermo
model. This model can be seen by typing
thermo
at the MATLAB Command prompt.
* The Subsystem block is described in detail in Chapter 2, Section 2.1, Page 22, Introduction to Simulink with
Engineering Applications, 9781934404096.
† The contents of the Subsystem block are not lost. We can doubleclick on the Subsystem block to see its con-tents.
The Subsystem block replaces the inputs and outputs of the model with Inport and Outport blocks. These
blocks are described in Section 2.1, Chapter 2, Page 22, Introduction to Simulink with Engineering Applica-tions,
ISBN 9781934404096.
z1 z2 z3

Appendix C
Introduction to SimPowerSystems
his appendix is a brief introduction to SimPowerSystems blockset that operates in the
Simulink environment. An introduction to Simulink is presented in Appendix B. For
additional help with Simulink, please refer to the Simulink documentation.
T
C.1 Simulation of Electric Circuits with SimPowerSystems
As stated in Appendix B, the MATLAB and Simulink environments are integrated into one
entity, and thus we can analyze, simulate, and revise our models in either environment at any
point. We can invoke Simulink from within MATLAB or by typing simulink at the MATLAB
command prompt, and we can invoke SimPowerSystems from within Simulink or by typing pow-erlib
at the MATLAB command prompt. We will introduce SimPowerSystems with two illus-trated
examples, a DC electric circuit, and an AC electric circuit
Example C.1
For the simple resistive circuit in Figure C.1, , , and . From the volt-age
vS = 12v R1 = 7 R2 = 5
division expression, and from Ohm’s law,
.
vR2 = R2  vS  R1 + R2 = 5  12  12 = 5v
+
v  S
R2
R1
i
Figure C.1. Circuit for Example C.1
i = vS  R1 + R2 = 1A
To model the circuit in Figure C.1, we enter the following command at the MATLAB prompt.
powerlib
and upon execution of this command, the powerlib window shown in Figure C.2 is displayed.
From the File menu in Figure C.2, we open a new window and we name it Sim_Fig_C3 as shown
in Figure C.3.
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling C1

Figure C.2. Library blocks for SimPowerSystems
Figure C.3. New window for modeling the circuit shown in Figure C.1
The powergui block in Figure C.2 is referred to as the Environmental block for SimPowerSys-tems
models and it must be included in every model containing SimPowerSystems blocks.
Accordingly, we begin our model by adding this block as shown in Figure C.4.
We observe that in Figure C.4, the powergui block is named Continuous. This is the default
method of solving an electric circuit and uses a variable step Simulink solver. Other methods are
the Discrete method used when the discretization of the system at fixed time steps is desired, and
the Phasors method which performs phasor simulation at the frequency specified by the Phasor
frequency parameter. These methods are described in detail in the SimPowerSystems documen-tation.
C2 Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling

Simulation of Electric Circuits with SimPowerSystems
Figure C.4. Window with the addition of the powergui block
Next, we need to the components of the electric circuit shown in Figure C.1. From the Electrical
Sources library in Figure C.2 we select the DC Voltage Source block and drag it into the model,
from the Elements library we select and drag the Series RLC Branch block and the Ground
block, from the Measurements library we select the Current Measurement and the Voltage
Measurement blocks, and from the Simulink Sinks library we select and drag the Display block.
The model now appears as shown in Figure C.5.
Figure C.5. The circuit components for our model
From the Series RLC Branch block we only need the resistor, and to eliminate the inductor and
the capacitor, we double click it and from the Block Parameters window we select the R compo-nent
with value set at 7  as shown in Figure C.6.

Figure C.6. The Block Parameters window for the Series RLC Branch
We need two resistors for our model and thus we copy and paste the resistor into the model, using
the Block Parameters window we change its value to , and from the Format drop window we
click the Rotate block option and we rotate it clockwise. We also need two Display blocks, one
for the current measurement and the second for the voltage measurement and thus we copy and
paste the Display block into the model. We also copy and paste twice the Ground block and the
model is now as shown in Figure C.7 where we also have renamed the blocks to shorter names.
Figure C.7. Model with blocks renamed
From Figure C.7 above, we observe that both terminals of the voltage source and the resistors are
shown with small square ( ) ports, the left ports of the CM (Current Measurement), and VM
(Voltage Measurement) are also shown with ports, but the terminals on the right are shown with
the Simulink output ports as >. The rules for the SimPowerSystems electrical terminal ports
and connection lines are as follows:
5 

1. We can connect Simulink ports (>) only to other Simulink ports.
2. We can connect SimPowerSystems ports ( ) only to other SimPowerSystems ports.*
3. If it is necessary to connect Simulink ports (>) to SimPowerSystems ports ( ), we can use
SimPowerSystems blocks that contain both Simulink and SimPowerSystems ports such as the
Current Measurement (CM) block and the Voltage Measurement (VM) block shown in Fig-ure
C.7.
The model for the electric circuit in Figure C.1 is shown in Figure C.8.
Figure C.8. The final form of the SimPowerSystems model for the electric circuit in Figure C.1
For the model in Figure C.8 we used the DC Voltage Source block. The SimPowerSystems doc-umentation
states that we can also use the AC Voltage Source block as a DC Voltage Source
block provided that we set the frequency at 0 Hz and the phase at 90 degrees
in the Block
Parameters window as shown in Figure C.9.
* As in Simulink, we can autoconnect two SimPowerSystems blocks by selecting the source block, then holding
down the Ctrl key, and left-clicking the destination block.

Figure C.9. Block parameter settings when using an AC Voltage Source block as a DC Voltage Source
Figure C.10. Model with AC Voltage Source used as DC Voltage Source
A third option is to use a Controlled Voltage Source block with a Constant block set to the
numerical value of the DC voltage Source as shown in the model of Figure C.11.

Figure C.11. Model with Controlled Voltage Source block
Example C.2
Consider the AC electric circuit in Figure C.12
VS
I
R L
0.2H
C
1200 V
1 
10–3 F
60 Hz
Figure C.12. Electric circuit for Example C.2
The current I and the voltage Vc across the capacitor are computed with MATLAB as follows:
Vs=120; f=60; R=1; L=0.2; C=10^(3); XL=2*pi*f*L; XC=1/(2*pi*f*C);...
Z=sqrt(R^2+(XLXC)^2); I=Vs/Z, Vc=XC*I
I =
1.6494
Vc =
4.3752
The SimPowerSystems model and the waveforms for the current I and the voltage Vc are shown
in Figures C.13 and C.14 respectively.

Figure C.13. SimPowerSystems model for the electric circuit in Figure C.12
Figure C.14. Waveforms for the current I and voltage Vc across the capacitor in Figure C.12
The same results are obtained if we replace the applied AC voltage source block in the model of
Figure C.13 with a Controlled Voltage Source (CVS) block as shown in Figure C.15.
Figure C.15. The model in Figure C.13 with the AC Voltage Source block replaced with a CVS block

Appendix D
A Review of Complex Numbers
his appendix is a review of the algebra of complex numbers. The basic operations are
defined and illustrated by several examples. Applications using Euler’s identities are pre-sented,
and the exponential and polar forms are discussed and illustrated with examples.
T
D.1 Definition of a Complex Number
In the language of mathematics, the square root of minus one is denoted as , that is, .
In the electrical engineering field, we denote as to avoid confusion with current . Essentially,
is an operator that produces a 90degree counterclockwise rotation to any vector to which it is
applied as a multiplying factor. Thus, if it is given that a vector has the direction along the
right side of the xaxis as shown in Figure D.1, multiplication of this vector by the operator will
result in a new vector whose magnitude remains the same, but it has been rotated counter-clockwise
90
by .
i j i
x
y
jA
jjA = j2A = –A
A
j–jA –j2= A = A
j–A = j 3A = –jA
Figure D.1. The j operator
i i = –1
j
A
j
jA
jA j 90
Also, another multiplication of the new vector by will produce another counterclock-wise
A 180 –A
direction. In this case, the vector has rotated and its new value now is . When
90 270 j–A = –jA
this vector is rotated by another for a total of , its value becomes . A
fourth 90 rotation returns the vector to its original position, and thus its value is again A
.
Therefore, we conclude that , , and .
j 2 = –1 j 3 = –j j 4 = 1
Circuit Analysis I with MATLAB  Computing and Simulink / SimPower Systems Modeling D1

Note: In our subsequent discussion, we will denote the xaxis (abscissa) as the real axis, and the
yaxis (ordinate) as the imaginary axis with the understanding that the “imaginary” axis is just as
“real” as the real axis. In other words, the imaginary axis is just as important as the real axis.*
An imaginary number is the product of a real number, say r , by the operator j . Thus, r
is a real
number and jr
is an imaginary number.
A complex number is the sum (or difference) of a real number and an imaginary number. For
example, the number A = a + jb where a and b
are both real numbers, is a complex number.
Then, a = ReA and b = ImA where ReA denotes real part of A, and
b = ImA
the imaginary part of A
.
By definition, two complex numbers A and B where A = a + jb and B = c + jd
, are equal if
and only if their real parts are equal, and also their imaginary parts are equal. Thus, if and
only if and .
D.2 Addition and Subtraction of Complex Numbers
The sum of two complex numbers has a real component equal to the sum of the real components,
and an imaginary component equal to the sum of the imaginary components. For subtraction, we
change the signs of the components of the subtrahend and we perform addition. Thus, if
D2 Circuit Analysis I with MATLAB  Computing and Simulink / SimPower Systems Modeling
and
then
and
Example D.1
It is given that , and . Find and
Solution:
and
* We may think the real axis as the cosine axis and the imaginary axis as the sine axis.
A = B
a = c b = d
A = a + jb B = c + jd
A + B = a + c + jb + d
A – B = a – c + jb – d
A = 3 + j4 B = 4 – j2 A + B A – B
A + B = 3 + j4 + 4 – j2 = 3 + 4 + j4 – 2 = 7 + j2
A – B = 3 + j4 – 4 – j2 = 3 – 4 + j4 + 2 = – 1 + j6

Multiplication of Complex Numbers
D.3 Multiplication of Complex Numbers
Complex numbers are multiplied using the rules of elementary algebra, and making use of the
fact that . Thus, if
A = a + jb and
B = c + jd
then
j 2 = –1
A  B = a + jb  c + jd = ac + jad + jbc + j2bd
j2 = –1
and since , it follows that
(D.1)
A  B = ac + jad + jbc–bd
= ac – bd + jad + bc
Example D.2
It is given that and . Find
Solution:
A = 3 + j4 B = 4 – j2 A  B
A  B = 3 + j4  4 – j2 = 12 – j6 + j16 – j28 = 20 + j10
A
The conjugate of a complex number, denoted as , is another complex number with the same
real component, and with an imaginary component of opposite sign. Thus, if , then
.
A = a + jb
A = a–jb
Example D.3
It is given that A = 3 + j5 . Find
A
Solution:
The conjugate of the complex number has the same real component, but the imaginary com-ponent
has opposite sign. Then,
A
A = 3–j5
If a complex number is multiplied by its conjugate, the result is a real number. Thus, if
, then
A
A = a + jb
A  A a + jba – jb a2 – jab + jab – j 2b2 a2 b2 = = = +

------------- a + jbc – jd
---- a + jb
------------------------------------- AB
 ------ ac + bd + jbc – ad
= = = = ------------------------------------------------------
ac + bd
---------------------- jbc – ad
= + ----------------------
---- 3 + j4
-------------- 3 + j44 – j3
-------------------------------------- 12 – j9 + j16 + 12
------------------------------------------- 24 + j7
----- j 7
----------------- 24
= = = = = + ----- = 0.96 + j0.28
Example D.4
It is given that . Find
Solution:
D.4 Division of Complex Numbers
When performing division of complex numbers, it is desirable to obtain the quotient separated
into a real part and an imaginary part. This procedure is called rationalization of the quotient, and it
is done by multiplying the denominator by its conjugate. Thus, if and ,
then,
(D.2)
In (D.2), we multiplied both the numerator and denominator by the conjugate of the denomina-tor
to eliminate the j operator from the denominator of the quotient. Using this procedure, we see
that the quotient is easily separated into a real and an imaginary part.
Example D.5
It is given that , and . Find
Solution:
Using the procedure of (D.2), we obtain
D.5 Exponential and Polar Forms of Complex Numbers
The relations
(D.3)
A = 3 + j5 A  A
A  A = 3 + j53 – j5 = 32 + 52 = 9 + 25 = 34
A = a + jb B = c + jd
AB
c + jd
c + jdc – jd
---- B
B
c2 d 2 +
c2 d 2 +
c2 d 2 +
A = 3 + j4 B = 4 + j3 A  B
AB
4 + j3
4 + j34 – j3
42 32 +
25
25
25
e j = cos + j sin

Exponential and Polar Forms of Complex Numbers
and
(D.4)
e– j = cos–j sin
are known as the Euler’s identities.
Multiplying (D.3) by the real positive constant C we obtain:
(D.5)
Ce j = Ccos + jCsin
a + jb
This expression represents a complex number, say , and thus
(D.6)
Ce j = a + jb
where the left side of (D.6) is the exponential form, and the right side is the rectangular form.
Equating real and imaginary parts in (D.5) and (D.6), we obtain
(D.7)
a = Ccos and b = Csin
Squaring and adding the expressions in (D.7), we obtain
Then,
or
(D.8)
Also, from (D.7)
or
(D.9)
a2 b2 + Ccos2 Csin2 + C2 2 cos 2 = =  + sin  = C2
C2 a2 b2 = +
C a2 b2 = +
ba
-- Csin
= --------------- = tan
Ccos
 ba
–1  = tan
 -- 
To convert a complex number from rectangular to exponential form, we use the expression
(D.10)
a + jb a2 b2 + e
j tan 1 – ba
 
 --
=
To convert a complex number from exponential to rectangular form, we use the expressions
(D.11)
Ce j = Ccos + jCsin
Ce– j = Ccos–jCsin

The polar form is essentially the same as the exponential form but the notation is different, that
is,
(D.12)
where the left side of (D.12) is the exponential form, and the right side is the polar form.
We must remember that the phase angle is always measured with respect to the positive real
axis, and rotates in the counterclockwise direction.
Example D.6
Convert the following complex numbers to exponential and polar forms:
a.
b.
c.
d.
Solution:
a. The real and imaginary components of this complex number are shown in Figure D.2.
 
Figure D.2. The components of
Then,
Check with MATLAB:
x=3+j*4; magx=abs(x); thetax=angle(x)*180/pi; disp(magx); disp(thetax)
5
53.1301
Ce j = C

3 + j4
– 1 + j2
– 2 – j
4 – j3
Re
Im
4
3
5
53.1
3 + j4
3 + j4 32 42 + e
j 43
--
–1
 tan 
5e = = j53.1 = 553.1

Check with the Simulink Complex to MagnitudeAngle block* shown in the Simulink
model of Figure D.3.
Figure D.3. Simulink model for Example D.6a
b. The real and imaginary components of this complex number are shown in Figure D.4.
116.6
Re
Im
2
5
63.4
1
Then,
– 1 + j2
5e j116.6 5116.6 = = =
– 1 + j2 12 22 + e
–1
j 2
 
tan -----

–1
Check with MATLAB:
y=1+j*2; magy=abs(y); thetay=angle(y)*180/pi; disp(magy); disp(thetay)
2.2361
116.5651
c. The real and imaginary components of this complex number are shown in Figure D.5.
Re
Im
2
26.6 153.4Measured
5 Clockwise)
1
206.6
– 2 – j
* For a detailed description and examples with this and other related transformation blocks, please refer to Intro-duction
to Simulink with Engineering Applications, ISBN 9781934404096.

j –1
= = 5206.6 5ej 153.4 –  5 153.4 = = = – 
–2–j1 22 12 + e
  tan –1 ----- 

–2
5e j206.6
Re
Im
4
3
323.1×
36.9×
5
4 – j3
= = 5323.1 5e –j36.9 5 36.9 = = = – 
4–j3 42 32 + e
–1
j –3
 
-----
tan 
4
5e j323.1
–230
–1 = j2 j 90 –230
230 180
–230 = 230 + 180 = 2210 = 2–150
Then,
Check with MATLAB:
v=2j*1; magv=abs(v); thetav=angle(v)*180/pi; disp(magv); disp(thetav)
2.2361
-153.4349
d. The real and imaginary components of this complex number are shown in Figure D.6.
Then,
Check with MATLAB:
w=4j*3; magw=abs(w); thetaw=angle(w)*180/pi; disp(magw); disp(thetaw)
5
-36.8699
Example D.7
Express the complex number in exponential and in rectangular forms.
Solution:
We recall that . Since each rotates a vector by counterclockwise, then is
the same as rotated counterclockwise by .Therefore,
The components of this complex number are shown in Figure D.7.

Re
Im
1.73
150Measured
1
210
30
2
Clockwise)
Then,
2–150
2–150 2e –j150
= = 2 cos150 – j sin150 = 2– 0.866 – j0.5 = – 1.73 – j
Note: The rectangular form is most useful when we add or subtract complex numbers; however,
the exponential and polar forms are most convenient when we multiply or divide complex
numbers.
To multiply two complex numbers in exponential (or polar) form, we multiply the magnitudes
and we add the phase angles, that is, if
then,
(D.13)
AB MN +  Me jNe j MNe j +  = = =
Example D.8
Multiply by
Solution:
Multiplication in polar form yields
A = M and B = N
A = 1053.1 B = 5–36.9
AB = 10  553.1 + –36.9 = 5016.2
and multiplication in exponential form yields
AB 10e j53.15e –j36.9 50e j 53.1 – 36.9 50e j16.2 = = =
To divide one complex number by another when both are expressed in exponential or polar
form, we divide the magnitude of the dividend by the magnitude of the divisor, and we subtract
the phase angle of the divisor from the phase angle of the dividend, that is, if
A = M and B = N

----- –  Me j
---- MN
AB
---- 1053.1
= ------------------------ = 253.1 – –36.9 = 290
---- 10e j53.1
53.1e j36.9 2e j90 = = =
then,
(D.14)
Example D.9
Divide by
Solution:
Division in polar form yields
Division in exponential form yields
Ne j ------------- MN
----e j –  = = =
A = 1053.1 B = 5–36.9
AB
5–36.9
AB
5e–j36.9 --------------------- 2ej

Appendix E
Matrices and Determinants
his appendix is an introduction to matrices and matrix operations. Determinants, Cramer’s
rule, and Gauss’s elimination method are reviewed. Some definitions and examples are not
applicable to the material presented in this text, but are included for subject continuity,
T
and academic interest. They are discussed in detail in matrix theory textbooks. These are
denoted with a dagger (†) and may be skipped.
E.1 Matrix Definition
A matrix is a rectangular array of numbers such as those shown below.
2 3 7
1 –1 5
In general form, a matrix A is denoted as
(E.1)
or
1 3 1
–2 1 –5
4 –7 6
A
a11 a12 a13  a1n
a21 a22 a23  a2n
a31 a32 a33  a3n
    
am1 am2 am3  amn
=
The numbers are the elements of the matrix where the index indicates the row, and indi-cates
aij i j
the column in which each element is positioned. For instance, indicates the element
a43
positioned in the fourth row and third column.
A matrix of m rows and n columns is said to be of m  n
order matrix.
If m = n , the matrix is said to be a square matrix of order m (or n
). Thus, if a matrix has five
rows and five columns, it is said to be a square matrix of order 5.
Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems  Modeling E1

Appendix E Matrices and Determinants
In a square matrix, the elements a11 a22 a33  ann
are called the main diagonal elements.
Alternately, we say that the matrix elements a11 a22 a33  ann
, are located on the main
diagonal.
† The sum of the diagonal elements of a square matrix A is called the trace* of A
.
† A matrix in which every element is zero, is called a zero matrix.
E.2 Matrix Operations
Two matrices and are equal, that is, , if and only if
(E.2)
A = aij B = bij A = B
aij = bij i = 1 2 3 m j = 1 2 3  n
Two matrices are said to be conformable for addition (subtraction), if they are of the same order
m  n
A = aij B = bij
C A B C
A B
C = A  B = aij  bij
A + B A – B
A 1 2 3
= B 2 3 0
0 1 4
–1 2 5
=
A + B 1 + 2 2 + 3 3 + 0
= =
0 – 1 1 + 2 4 + 5
3 5 3
–1 3 9
A – B 1 – 2 2 – 3 3 – 0
= =
0 + 1 1 – 2 4 – 5
–1 –1 3
1 –1 –1
E2 Circuit Analysis I with MATLAB  Computing and Simulink / SimPowerSystems Modeling
.
If and are conformable for addition (subtraction), their sum (difference) will
be another matrix with the same order as and , where each element of is the sum (dif-ference)
of the corresponding elements of and , that is,
(E.3)
Example E.1
Compute and given that
and
Solution:
and
* Henceforth, all paragraphs and topics preceded by a dagger ( † ) may be skipped. These are discussed in matrix
theory textbooks.

Matrix Operations
Check with MATLAB:
A=[1 2 3; 0 1 4]; B=[2 3 0; 1 2 5]; % Define matrices A and B
A+B, AB % Add A and B, then Subtract B from A
ans =
3 5 3
-1 3 9
ans =
-1 -1 3
1 -1 -1
Check with Simulink:
Note: The elements of matrices
A and B are specified in
MATLAB's Command prompt
Sum 1
Sum 2
3
-1
-1
1
5
3
Display 1 (A+B)
-1
-1
3
9
3
-1
Display 2 (A-B)
A
Constant 1
B
Constant 2
If is any scalar (a positive or negative number), and not which is a matrix, then mul-tiplication
k k 1  1
of a matrix by the scalar is the multiplication of every element of by .
Example E.2
Multiply the matrix
by
a.
b.
A k A k
A 1 –2
2 3
=
k1 = 5
k2 = – 3 + j2

 5  1 5  –2
k1  A 5 1 –2
 – 3 + j2  1 – 3 + j2  –2
Shows that A and B are conformable for multiplication
A B
Solution:
a.
b.
Check with MATLAB:
k1=5; k2=(3 + 2*j); % Define scalars k1 and k2
A=[1 2; 2 3]; % Define matrix A
k1*A, k2*A % Multiply matrix A by scalars k1 and k2
ans =
5 -10
10 15
ans =
-3.0000+ 2.0000i 6.0000- 4.0000i
-6.0000+ 4.0000i -9.0000+ 6.0000i
Two matrices and are said to be conformable for multiplication in that order, only
when the number of columns of matrix is equal to the number of rows of matrix . That is, the
product (but not ) is conformable for multiplication only if is an matrix and
matrix is an matrix. The product will then be an matrix. A convenient way
to determine if two matrices are conformable for multiplication is to write the dimensions of the
two matrices sidebyside as shown below.
For the product we have:
2 3
5  2 5  3
5 –10
10 15
= = =
k2  A – 3 + j2 1 –2
2 3
– 3 + j2  2 – 3 + j2  3
– 3 + j2 6 – j4
– 6 + j4 – 9 + j6
= = =
A B A  B
A B
A  B B  A A m  p
B p  n A  B m  n
m  p p  n
Indicates the dimension of the product A  B
B  A

Matrix Operations
Here, B and A are not conformable for multiplication
B A
p  n m  p
For matrix multiplication, the operation is row by column. Thus, to obtain the product , we
multiply each element of a row of by the corresponding element of a column of ; then, we
add these products.
Example E.3
Matrices and are defined as
A B
and
A  B
C D
C = 2 3 4 D
1
–1
2
=
Compute the products C  D and
D  C
Solution:
The dimensions of matrices and are respectively ; therefore the product is
feasible, and will result in a , that is,
C D 1  3 3  1 C  D
1  1
C  D 2 3 4
1
–1
2
= = 2  1 + 3  –1 + 4  2 = 7
The dimensions for D and C are respectively 3  1 1  3 and therefore, the product D  C
is
also feasible. Multiplication of these will produce a matrix as follows:
3  3
D  C
1
–1
2
= = =
2 3 4
1  2 1  3 1  4
–1  2 –1  3 –1  4
2  2 2  3 2  4
2 3 4
–2 –3 –4
4 6 8
Check with MATLAB:
C=[2 3 4]; D=[1 1 2]’; % Define matrices C and D. Observe that D is a column vector
C*D, D*C % Multiply C by D, then multiply D by C
ans =
7

A
A
a11 a12 a13  a1n
0 a22 a23  a2n
0 0   
  0  
0 0 0  amn
=
B
B
a11 0 0  0
a21 a22 0  0
   0 0
    0
am1 am2 am3  amn
=
ans =
2 3 4
-2 -3 -4
4 6 8
Division of one matrix by another, is not defined. However, an analogous operation exists, and it
will become apparent later in this chapter when we discuss the inverse of a matrix.
E.3 Special Forms of Matrices
† A square matrix is said to be upper triangular when all the elements below the diagonal are
zero. The matrix of (E.4) is an upper triangular matrix. In an upper triangular matrix, not
all elements above the diagonal need to be nonzero.
(E.4)
† A square matrix is said to be lower triangular, when all the elements above the diagonal are
zero. The matrix of (E.5) is a lower triangular matrix. In a lower triangular matrix, not all
elements below the diagonal need to be nonzero.
(E.5)
† A square matrix is said to be diagonal, if all elements are zero, except those in the diagonal. The
matrix of (E.6) is a diagonal matrix.
C

Special Forms of Matrices
(E.6)
C
a11 0 0  0
0 a22 0  0
0 0  0 0
0 0 0  0
0 0 0  amn
=
† A diagonal matrix is called a scalar matrix, if a11 = a22 = a33 =  = ann = k where k
is a
scalar. The matrix of (E.7) is a scalar matrix with .
(E.7)
D k = 4
D
4 0 0 0
0 4 0 0
0 0 4 0
0 0 0 4
=
A scalar matrix with , is called an identity matrix . Shown below are , , and
k = 1 I 2  2 3  3
identity matrices.
(E.8)
4  4
1 0
0 1
1 0 0
0 1 0
0 0 1
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
The MATLAB eye(n) function displays an identity matrix. For example,
eye(4) % Display a 4 by 4 identity matrix
ans =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
n  n
Likewise, the eye(size(A)) function, produces an identity matrix whose size is the same as matrix
A . For example, let matrix A
be defined as
A=[1 3 1; 2 1 5; 4 7 6] % Define matrix A
A =
1 3 1
-2 1 -5
4 -7 6

A AT
A
A 1 2 3
= then AT
4 5 6
1 4
2 5
3 6
=
A AT = A A
A
A
1 2 3
2 4 –5
3 –5 6
= AT
1 2 3
2 4 –5
3 –5 6
= = A
Then,
eye(size(A))
displays
ans =
1 0 0
0 1 0
0 0 1
† The transpose of a matrix , denoted as , is the matrix that is obtained when the rows and
columns of matrix are interchangeE. For example, if
(E.9)
In MATLAB, we use the apostrophe () symbol to denote and obtain the transpose of a matrix.
Thus, for the above example,
A=[1 2 3; 4 5 6] % Define matrix A
A =
1 2 3
4 5 6
A' % Display the transpose of A
ans =
1 4
2 5
3 6
† A symmetric matrix is a matrix such that , that is, the transpose of a matrix is the
same as . An example of a symmetric matrix is shown below.
(E.10)
† If a matrix A has complex numbers as elements, the matrix obtained from A
by replacing each
element by its conjugate, is called the conjugate of , and it is denoted as , for example,
A A

Special Forms of Matrices
A 1 + j2 j
= A 1 – j2 –j
3 2 – j3
3 2 + j3
=
MATLAB has two builtin functions which compute the complex conjugate of a number. The
first, conj(x), computes the complex conjugate of any complex number, and the second,
conj(A), computes the conjugate of a matrix . Using MATLAB with the matrix defined
as above, we obtain
A = [1+2j j; 3 23j] % Define and display matrix A
A =
A A
1.0000 + 2.0000i 0 + 1.0000i
3.0000 2.0000 - 3.0000i
conj_A=conj(A) % Compute and display the conjugate of A
conj_A =
1.0000 - 2.0000i 0 - 1.0000i
3.0000 2.0000 + 3.0000i
A AT = –A
† A square matrix such that is called skew-symmetric. For example,
A
0 2 –3
–2 0 –4
3 4 0
= AT
0 –2 3
2 0 4
–3 –4 0
= = –A
Therefore, matrix above is skew symmetric.
† A square matrix such that is called Hermitian. For example,
A
A AT = A
A
1 1 – j 2
1 + j 3 j
2 –j 0
AT 1 1 + j 2
= = = = A
1 – j 3 –j
2 j 0
AT* 1 1 + j 2
1 – j 3 –j
2 j 0
Therefore, matrix A
above is Hermitian.
† A square matrix A such that AT = –A
is called skewHermitian. For example,
A
j 1 – j 2
= = = = –A
– 1 – j 3j j
–2 j 0
AT j – 1 – j –2
1 – j 3j j
2 j 0
Therefore, matrix above is skewHermitian.
AT* –j – 1 + j –2
1 + j –3j –j
2 –j 0
A

A
A
a11 a12 a13  a1n
a21 a22 a23  a2n
a31 a32 a33  a3n
    
an1 an2 an3  ann
=
A detA
detA = a11a22a33ann + a12a23a34an1 + a13a24a35an2 +

an1a22a13– –an2a23a14 – an3a24a15 –
A 2
A
a11 a12
a21 a22
=
detA = a11a22 – a21a12
A B
A 1 2
= B 2 –1
3 4
2 0
=
detA detB
detA = 1  4 – 3  2 = 4 – 6 = –2
detB = 2  0 – 2  –1 = 0 – –2 = 2
E.4 Determinants
Let matrix be defined as the square matrix
(E.11)
then, the determinant of , denoted as , is defined as
(E.12)
The determinant of a square matrix of order n is referred to as determinant of order n.
Let be a determinant of order , that is,
(E.13)
Then,
(E.14)
Example E.4
Matrices and are defined as
and
Compute and .
Solution:
Check with MATLAB:
A=[1 2; 3 4]; B=[2 1; 2 0]; % Define matrices A and B
det(A), det(B) % Compute the determinants of A and B

Determinants
ans =
-2
ans =
2
Let be a matrix of order , that is,
(E.15)
A 3
then, is found from
(E.16)
A
a11 a12 a13
a21 a22 a23
a31 a32 a33
=
detA
detA = a11a22a33 + a12a23a31 + a11a22a33
–a11a22a33 – a11a22a33 – a11a22a33
A convenient method to evaluate the determinant of order 3
, is to write the first two columns to
the right of the 3  3
matrix, and add the products formed by the diagonals from upper left to
lower right; then subtract the products formed by the diagonals from lower left to upper right as
shown on the diagram of the next page. When this is done properly, we obtain (E.16) above.
a11 a12 a13
a21 a22 a23
a31 a32 a33
a11 a12
a21 a22
a31 a32 +

This method works only with second and third order determinants. To evaluate higher order
determinants, we must first compute the cofactors; these will be defined shortly.
Example E.5
Compute and if matrices and are defined as
and
detA detB A B
A
2 3 5
1 0 1
2 1 0
= B
2 –3 –4
1 0 –2
0 –5 –6
=

detA
2 3 5 2 3
1 0 1 1 0
2 1 0 2 1
=
detA 2  0  0 + 3  1  1 +
5  1  1
– 2  0  5 – 1  1  2 – 0  1  3
= 11 – 2 = 9
=
detB
2 –3 –4 2 –3
1 0 –2 1 –2
0 –5 –6 2 –6
=
detB 2  0  –6 + –3  –2  0 +
–4  1  –5
– 0  0  –4 – –5  –2  2 – –6  1  –3
= 20 – 38 = –18
=
A n
A
a11 a12 a13  a1n
a21 a22 a23  a2n
a31 a32 a33  a3n
    
an1 an2 an3  ann
=
Solution:
or
Likewise,
or
Check with MATLAB:
A=[2 3 5; 1 0 1; 2 1 0]; det(A) % Define matrix A and compute detA
ans =
9
B=[2 3 4; 1 0 2; 0 5 6];det(B) % Define matrix B and compute detB
ans =
-18
E.5 Minors and Cofactors
Let matrix be defined as the square matrix of order as shown below.
(E.17)
If we remove the elements of its ith row, and jth column, the remaining n – 1
square matrix is
called the minor of , and it is denoted as .
A Mij

Minors and Cofactors
–1i + j Mij aij ij
The signed minor is called the cofactor of and it is denoted as .
Example E.6
Matrix is defined as
(E.18)
A
A
a11 a12 a13
a21 a22 a23
a31 a32 a33
=
Compute the minors , , and the cofactors , and .
Solution:
and
M11
= M12
11 –11 + 1 M11 M11 12 –11 + 2 M12 M12 13 M13 –11 + 3 M13 = = = = – = =
The remaining minors
and cofactors
are defined similarly.
M11 M12 M13 11 12 13
a22 a23
a32 a33
= M11
M21 M22 M23 M31 M32 M33     
21 22 23 31 32 and 33
Example E.7
Compute the cofactors of matrix defined as
(E.19)
Solution:
(E.20)
a21 a23
a31 a33
a21 a22
a31 a32
=
A
A
1 2 –3
2 –4 2
–1 2 –6
=
11 –11 + 1 –4 2
= = 20 12 –11 + 2 2 2
2 –6
= = 10
–1 –6

(E.21)
(E.22)
(E.23)
(E.24)
13 –11 + 3 2 –4
= = = = 6
–1 2
0 21 –12 + 1 2 –3
= = –9 23 –12 + 3 1 2
= = –4
= = –8 32 –13 + 2 1 –3
 = = –8
= = –8
It is useful to remember that the signs of the cofactors follow the pattern below
that is, the cofactors on the diagonals have the same sign as their minors.
Let be a square matrix of any size; the value of the determinant of is the sum of the products
obtained by multiplying each element of any row or any column by its cofactor.
= 2 2 2
Example E.8
(E.25)
Compute the determinant of using the elements of the first row.
Solution:
2 –6
22 –12 + 2 1 –3
–1 –6
–1 2
31 –13 + 1 2 –3
–4 2
2 2
33 –13 + 3 1 2
2 –4
+  +  +
 +  + 
+  +  +
 +  + 
+  +  +
A A
A
A
1 2 –3
2 –4 2
–1 2 –6
=
A
detA 1 –4 2
2 –6
–1 –6
3 2 –4
–1 2
– – = 1  20 – 2  –10 – 3  0 = 40

Minors and Cofactors
Check with MATLAB:
A=[1 2 3; 2 4 2; 1 2 6]; det(A) % Define matrix A and compute detA
ans =
40
We must use the above procedure to find the determinant of a matrix of order or higher.
Thus, a fourth-order determinant can first be expressed as the sum of the products of the ele-ments
of its first row by its cofactor as shown below.
(E.26)
A
a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44
a11
a22 a23 a24
a32 a33 a34
a42 a43 a44
a21
a12 a13 a14
a32 a33 a34
a42 a43 a44
–
+a31
a12 a13 a14
a22 a23 a24
a42 a43 a44
a41
a12 a13 a14
a22 a23 a24
a32 a33 a34
–
= =
Determinants of order five or higher can be evaluated similarly.
Example E.9
Compute the value of the determinant of the matrix defined as
(E.27)
A 4
A
A
2 –1 0 –3
–1 1 0 –1
4 0 3 –2
–3 0 0 1
=
Solution:
Using the above procedure, we will multiply each element of the first column by its cofactor.
Then,
A=2
1 0 –1
0 3 –2
0 0 1
a
–1
–1 0 –3
0 3 –2
0 0 1
–




b
+4
–1 0 –3
1 0 –1
0 0 1
c
–3
–1 0 –3
1 0 –1
0 3 –2
–
d

Next, using the procedure of Example E.5 or Example E.8, we find
a = 6 b = –3 c = 0 d = –36
detA = a + b + c + d = 6 – 3 + 0 – 36 = –33
, , ,
and thus
We can verify our answer with MATLAB as follows:
A=[ 2 1 0 3; 1 1 0 1; 4 0 3 2; 3 0 0 1]; delta = det(A)
delta =
-33
Some useful properties of determinants are given below.
Property 1: If all elements of one row or one column are zero, the determinant is zero. An exam-ple
of this is the determinant of the cofactor above.
Property 2: If all the elements of one row or column are m times the corresponding elements of
another row or column, the determinant is zero. For example, if
(E.28)
then,
(E.29)
Here, is zero because the second column in is times the first column.
Check with MATLAB:
A=[2 4 1; 3 6 1; 1 2 1]; det(A)
ans =
0
Property 3: If two rows or two columns of a matrix are identical, the determinant is zero. This
follows from Property 2 with .
E.6 Cramer’s Rule
Let us consider the systems of the three equations below:
c
A
2 4 1
3 6 1
1 2 1
=
detA
2 4 1
3 6 1
1 2 1
2 4
3 6
1 2
= = 12 + 4 + 6 – 6 – 4 – 12 = 0
detA A 2
m = 1

Cramer’s Rule
(E.30)
and let
a11x + a12y + a13z = A
a21x + a22y + a23z = B
a31x + a32y + a33z = C

a11 a12 a13
a21 a22 a23
a31 a32 a33
D1
A a11 a13
B a21 a23
C a31 a33
D2
a11 A a13
a21 B a23
a31 C a33
D3
a11 a12 A
a21 a22 B
a31 a32 C
= = = =
Cramer’s rule states that the unknowns x, y, and z can be found from the relations
(E.31)
x
D1
 = ------ y
D2
 = ------ z
D3
 = ------
provided that the determinant  (delta) is not zero.
We observe that the numerators of (E.31) are determinants that are formed from  by the substi-tution
of the known values , , and , for the coefficients of the desired unknown.
A B C
Cramer’s rule applies to systems of two or more equations.
If (E.30) is a homogeneous set of equations, that is, if , then,
are all zero as we found in Property 1 above. Then, also.
Example E.10
Use Cramer’s rule to find , , and if
A = B = C = 0 D1 D2 and D3
(E.32)
x = y = z = 0
v1 v2 v3
2v1 – 5 – v2 + 3v3 = 0
–2v3 – 3v2 – 4v1 = 8
v2 + 3v1 – 4 – v3 = 0
and verify your answers with MATLAB.
Solution:
Rearranging the unknowns , and transferring known values to the right side, we obtain
(E.33)
By Cramer’s rule,
v
2v1 – v2 + 3v3 = 5
–4v1 – 3v2 – 2v3 = 8
3v1 + v2 – v3 = 4


2 –1 3
–4 –3 –2
3 1 –1
2 –1
–4 –3
3 1
= = 6 + 6 – 12 + 27 + 4 + 4 = 35
D1
5 –1 3
8 –3 –2
4 1 –1
5 –1
8 –3
4 1
= = 15 + 8 + 24 + 36 + 10 – 8 = 85
D2
2 5 3
–4 8 –2
3 4 –1
2 5
–4 8
3 4
= = – 16 – 30 – 48 – 72 + 16 – 20 = –170
D3
2 –1 5
–4 –3 8
3 1 4
2 –1
–4 –3
3 1
= = – 24 – 24 – 20 + 45 – 16 – 16 = –55
x1
D1
 ------ 85
----- 17
= = = ----- x2
35
7
D2
 ------ 170
–-------- 34
= = = –----- x3
35
7
–----- 11
= = = –-----
Using relation (E.31) we obtain
(E.34)
We will verify with MATLAB as follows:
% The following script will compute and display the values of v1, v2 and v3.
format rat % Express answers in ratio form
B=[2 1 3; 4 3 2; 3 1 1]; % The elements of the determinant D of matrix B
delta=det(B); % Compute the determinant D of matrix B
d1=[5 1 3; 8 3 2; 4 1 1]; % The elements of D1
detd1=det(d1); % Compute the determinant of D1
d2=[2 5 3; 4 8 2; 3 4 1]; % The elements of D2
detd2=det(d2); % Compute the determinant of D2
d3=[2 1 5; 4 3 8; 3 1 4]; % The elements of D3
detd3=det(d3); % Compute he determinant of D3
v1=detd1/delta; % Compute the value of v1
%
disp('v1=');disp(v1); % Display the value of v1
D3
------ 55
 35
7

Gaussian Elimination Method
v1=
17/7
v2=
-34/7
v3=
-11/7
These are the same values as in (E.34)
E.7 Gaussian Elimination Method
We can find the unknowns in a system of two or more equations also by the Gaussian elimina-tion
method. With this method, the objective is to eliminate one unknown at a time. This can be
done by multiplying the terms of any of the equations of the system by a number such that we
can add (or subtract) this equation to another equation in the system so that one of the
unknowns will be eliminated. Then, by substitution to another equation with two unknowns, we
can find the second unknown. Subsequently, substitution of the two values found can be made
into an equation with three unknowns from which we can find the value of the third unknown.
This procedure is repeated until all unknowns are found. This method is best illustrated with the
following example which consists of the same equations as the previous example.
Example E.11
Use the Gaussian elimination method to find , , and of the system of equations
(E.35)
v1 v2 v3
2v1 – v2 + 3v3 = 5
–4v1 – 3v2 – 2v3 = 8
3v1 + v2 – v3 = 4
Solution:
As a first step, we add the first equation of (E.35) with the third to eliminate the unknown v2 and
we obtain the equation
(E.36)
5v1 + 2v3 = 9
Next, we multiply the third equation of (E.35) by 3, and we add it with the second to eliminate
, and we obtain the equation
(E.37)
v2
5v1 – 5v3 = 20
Subtraction of (E.37) from (E.36) yields

(E.38)
7v3 11 or v3
Now, we can find the unknown from either (E.36) or (E.37). By substitution of (D.38) into
(E.36) we obtain
(E.39)
+    9 or v1
–-----
Finally, we can find the last unknown from any of the three equations of (E.35). By substitu-tion
----- 33
– ----- 35
– ----- 34
= = = –-----
into the first equation we obtain
(E.40)
These are the same values as those we found in Example E.10.
The Gaussian elimination method works well if the coefficients of the unknowns are small inte-gers,
as in Example E.11. However, it becomes impractical if the coefficients are large or fractional
numbers.
E.8 The Adjoint of a Matrix
Let us assume that is an n square matrix and is the cofactor of . Then the adjoint of ,
denoted as , is defined as the n square matrix below.
(E.41)
We observe that the cofactors of the elements of the ith row (column) of are the elements of
the ith column (row) of .
Example E.12
Compute if Matrix is defined as
11
7
= – = –-----
v1
5v1 2 11
7
17
7
= = -----
v2
v2 2v1 + 3v3 – 5 34
7
7
7
7
A ij aij A
adjA
adjA
11 21 31  n1
12 22 32  n2
13 23 33  n3
    
1n 2n 3n  nn
=
A
adjA
adjA A

Singular and NonSingular Matrices
(E.42)
Solution:
A
1 2 3
1 3 4
1 4 3
=
adjA
3 4
4 3
– 2 3
2 3
4 3
3 4
– 1 3
1 4
1 3
= =
1 3
2 3
3 4
–
1 3
1 4
– 1 2
1 2
1 4
1 3
–7 6 –1
1 0 –1
1 –2 1
E.9 Singular and NonSingular Matrices
An square matrix is called singular if ; if , is called nonsingular.
n A detA = 0 detA  0 A
Example E.13
(E.43)
A
A
1 2 3
2 3 4
3 5 7
=
Determine whether this matrix is singular or nonsingular.
Solution:
detA
1 2 3
2 3 4
3 5 7
= = 21 + 24 + 30 – 27 – 20 – 28 = 0
Therefore, matrix is singular.
1 2
2 3
3 5
A

E.10 The Inverse of a Matrix
If A and B are n square matrices such that AB = BA = I , where I is the identity matrix, B
is
called the inverse of , denoted as , and likewise, is called the inverse of , that is,
A B A–1 = A B
If a matrix is non-singular, we can compute its inverse from the relation
(E.44)
= ------------adjA
------------adjA 1
-----
Example E.14
(E.45)
Compute its inverse, that is, find
Solution:
Here, , and since this is a non-zero value, it is possible to com-pute
the inverse of using (E.44).
From Example E.12,
Then,
(E.46)
Check with MATLAB:
A=[1 2 3; 1 3 4; 1 4 3], invA=inv(A) % Define matrix A and compute its inverse
A =
1 2 3
1 3 4
1 4 3
A B–1 =
A A–1
A–1 1
detA
A
A
1 2 3
1 3 4
1 4 3
=
A–1
detA = 9 + 8 + 12 – 9 – 16 – 6 = –2
A
adjA
–7 6 –1
1 0 –1
1 –2 1
=
A–1 1
detA
–2
–7 6 –1
1 0 –1
1 –2 1
3.5 –3 0.5
–0.5 0 0.5
–0.5 1 –0.5
= = =

Solution of Simultaneous Equations with Matrices
invA =
3.5000 -3.0000 0.5000
-0.5000 0 0.5000
-0.5000 1.0000 -0.5000
A A–1 I
Multiplication of a matrix by its inverse produces the identity matrix , that is,
(E.47)
AA–1 = I or A–1A = I
Example E.15
Prove the validity of (E.47) for the Matrix defined as
Proof:
Then,
and
A
A 4 3
2 2
=
detA 8 – 6 2 and adjA 2 –3
–2 4
= = =
------------adjA 12
A–1 1
= = =
detA
-- 2 –3
–2 4
1 –3  2
–1 2
AA–1 4 3
= = = = I
2 2
1 –3  2
–1 2
4 – 3 – 6 + 6
2 – 2 – 3 + 4
1 0
0 1
E.11 Solution of Simultaneous Equations with Matrices
Consider the relation
(E.48)
AX = B
where A and B are matrices whose elements are known, and X
is a matrix (a column vector)
whose elements are the unknowns. We assume that and are conformable for multiplica-tion.
A–1
Multiplication of both sides of (E.48) by yields:
(E.49)
or
A X
A–1AX = A–1B = IX = A–1B

(E.50)
Therefore, we can use (E.50) to solve any set of simultaneous equations that have solutions. We
will refer to this method as the inverse matrix method of solution of simultaneous equations.
 
 
 
Example E.16
For the system of the equations
(E.51)
compute the unknowns using the inverse matrix method.
Solution:
In matrix form, the given set of equations is where
(E.52)
Then,
(E.53)
or
(E.54)
Next, we find the determinant , and the adjoint .
Therefore,
X=A–1B
2x1 + 3x2 + x3 = 9
x1 + 2x2 + 3x3 = 6
 
3x1 + x2 + 2x3 = 8
x1 x2 and x3
AX = B
A
2 3 1
1 2 3
3 1 2
= X
x1
x2
x3
= B
9
6
8
  =
X = A–1B
x1
x2
x3
2 3 1
1 2 3
3 1 2
–1
9
6
8
=
detA adjA
detA = 18 and adjA
1 –5 7
7 1 –5
–5 7 1
=

A–1 1
------------ adjA 1
= =
detA
-----
18
1 –5 7
7 1 –5
–5 7 1
and with relation (E.53) we obtain the solution as follows:
(E.55)
X
x1
x2
x3
1
18
-----
1 –5 7
7 1 –5
–5 7 1
9
6
8
1
18
-----
35
29
5
35  18
29  18
5  18
1.94
1.61
0.28
= = = = =
A–1 B
To verify our results, we could use the MATLAB’s inv(A) function, and then multiply by .
However, it is easier to use the matrix left division operation ; this is MATLAB’s solu-tion
X = A B
A–1B A  X = B X B
of for the matrix equation , where matrix is the same size as matrix .
For this example,
A=[2 3 1; 1 2 3; 3 1 2]; B=[9 6 8]'; X=A B
X =
1.9444
1.6111
0.2778
Example E.17
For the electric circuit of Figure E.1,
+
V = 100 v
1  2  2 
9  9  4 
I1 I2 I3
Figure E.1. Electric circuit for Example E.17
the loop equations are
(E.56)
10I1 – 9I2 = 100
–9I1 + 20I2 – 9I3 = 0
–9I2 + 15I3 = 0

Use the inverse matrix method to compute the values of the currents , , and
Solution:
For this example, the matrix equation is or , where
I1 I2 I3
RI = V I = R–1V
R
10 –9 0
–9 20 –9
0 –9 15
 = =
= V
100
0
0
and I
I1
I2
I3
R–1
R–1 1
= ------------ adjR
detR
R
 =
detR = 975 adjR
219 135 81
135 150 90
81 90 119
------------adjR 1
R–1 1
= =
detR
--------
975
219 135 81
135 150 90
81 90 119
I
I1
I2
I3
1
975
--------
219 135 81
135 150 90
81 90 119
100
0
0
100
--------
975
219
135
81
22.46
13.85
8.31
= = = =
The next step is to find . It is found from the relation
(E.57)
Therefore, we must find the determinant and the adjoint of . For this example, we find that
(E.58)
Then,
and
Check with MATLAB:
R=[10 9 0; 9 20 9; 0 9 15]; V=[100 0 0]'; I=RV; fprintf(' n');...
fprintf('I1 = %4.2f t', I(1)); fprintf('I2 = %4.2f t', I(2)); fprintf('I3 = %4.2f t', I(3)); fprintf(' n')
I1 = 22.46 I2 = 13.85 I3 = 8.31
We can also use subscripts to address the individual elements of the matrix. Accordingly, the
MATLAB script above could also have been written as:
R(1,1)=10; R(1,2)=9; % No need to make entry for A(1,3) since it is zero.
R(2,1)=9; R(2,2)=20; R(2,3)=9; R(3,2)=9; R(3,3)=15; V=[100 0 0]'; I=RV; fprintf(' n');...
fprintf('I1 = %4.2f t', I(1)); fprintf('I2 = %4.2f t', I(2)); fprintf('I3 = %4.2f t', I(3)); fprintf(' n')

I1 = 22.46 I2 = 13.85 I3 = 8.31
Spreadsheets also have the capability of solving simultaneous equations with real coefficients
using the inverse matrix method. For instance, we can use Microsoft Excel’s MINVERSE (Matrix
Inversion) and MMULT (Matrix Multiplication) functions, to obtain the values of the three cur-rents
in Example E.17.
The procedure is as follows:
1.We begin with a blank spreadsheet and in a block of cells, say B3:D5, we enter the elements of
matrix R as shown in Figure D.2. Then, we enter the elements of matrix V
in G3:G5.
2. Next, we compute and display the inverse of R , that is, R–1
. We choose B7:D9 for the ele-ments
of this inverted matrix. We format this block for number display with three decimal
places. With this range highlighted and making sure that the cell marker is in B7, we type the
formula
=MININVERSE(B3:D5)
and we press the Crtl-Shift-Enter keys simultaneously. We observe that R–1
appears in these
cells.
3. Now, we choose the block of cells G7:G9 for the values of the current I
. As before, we high-light
them, and with the cell marker positioned in G7, we type the formula
=MMULT(B7:D9,G3:G5)
and we press the Crtl-Shift-Enter keys simultaneously. The values of then appear in G7:G9.
I
A B C D E F G H
10 -9 0 100
R= -9 20 -9 V= 0
0 -9 15 0
0.225 0.138 0.083 22.462
R-1= 0.138 0.154 0.092 I= 13.846
0.083 0.092 0.122 8.3077
Figure E.2. Solution of Example E.17 with a spreadsheet
1234567
89
10
Example E.18
For the phasor circuit of Figure E.18

C
R3 = 100 
Figure E.3. Circuit for Example E.18
--------------------------------
--------------------------------
the current can be found from the relation
(E.59)
and the voltages and can be computed from the nodal equations
(E.60)
and
(E.61)
Compute, and express the current in both rectangular and polar forms by first simplifying like
terms, collecting, and then writing the above relations in matrix form as , where
, , and
Solution:
The matrix elements are the coefficients of and . Simplifying and rearranging the nodal
equations of (E.60) and (E.61), we obtain
(E.62)
Next, we write (E.62) in matrix form as
(E.63)
+

85  R1
R2 50 
L
IX
VS
j100 
j200 
170
V1 V2
IX
IX
V1 – V2
R3
= -------------------
V1 V2
V1 – 1700
85
V1 – V2
100
-------------------
V1 – 0
j200
+ + --------------- = 0
V2 – 1700
–j100
V2 – V1
100
-------------------
V2 – 0
50
+ + --------------- = 0
Ix
YV = I
Y = Admit tance V = Voltage I = Current
Y V1 V2
0.0218 – j0.005V1 – 0.01V2 = 2
–0.01V1 + 0.03 + j0.01V2 = j1.7
0.0218 – j0.005 –0.01
–0.01 0.03 + j0.01
Y
V1
V2
V
2
j1.7
I
=




where the matrices Y , V , and I
are as indicated.
We will use MATLAB to compute the voltages and , and to do all other computations.
The script is shown below.
Y=[0.02180.005j 0.01; 0.01 0.03+0.01j]; I=[2; 1.7j]; V=YI; % Define Y, I, and find V
fprintf('n'); % Insert a line
disp('V1 = '); disp(V(1)); disp('V2 = '); disp(V(2)); % Display values of V1 and V2
V1 =
1.0490e+002 + 4.9448e+001i
V2 =
53.4162 + 55.3439i
V1 V2
Next, we find from
R3=100; IX=(V(1)V(2))/R3 % Compute the value of IX
IX =
IX
0.5149 - 0.0590i
This is the rectangular form of . For the polar form we use the MATLAB script
magIX=abs(IX), thetaIX=angle(IX)*180/pi % Compute the magnitude and the angle in
degrees
magIX =
0.5183
thetaIX =
-6.5326
Therefore, in polar form,
IX
IX = 0.518–6.53
Spreadsheets have limited capabilities with complex numbers, and thus we cannot use them to
compute matrices that include complex numbers in their elements as in Example E.18.

E.12 Exercises
For Exercises 1, 2, and 3 below, the matrices , , , and are defined as:
1. Perform the following computations, if possible. Verify your answers with MATLAB.
a. b. c. d.
e. f. g. h.
a. b. c. d.
e. f. g. h.
a. b. c. d. e. f.
4. Solve the following systems of equations using Cramer’s rule. Verify your answers with MAT-LAB.
a. b.
5. Repeat Exercise 4 using the Gaussian elimination method.
6. Solve the following systems of equations using the inverse matrix method. Verify your answers
with MATLAB.
a. b.
A B C D
A
1 –1 –4
5 7 –2
3 –5 6
= B
5 9 –3
–2 8 2
7 –4 6
= C=
4 6
–3 8
5 –2
D 1 –2 3
–3 6 –4
=
A + B A + C B + D C + D
A – B A – C B – D C – D
A  B A  C B  D C  D
B A  C A  D A  D ·
 C
detA detB detC detD detA  B detA  C
x1 – 2x2 + x3 = –4
–2x1 + 3x2 + x3 = 9
3x1 + 4x2 – 5x3 = 0
– x1 + 2x2 – 3x3 + 5x4 = 14
x1 + 3x2 + 2x3 – x4 = 9
3x1–3x2 + 2x3 + 4x4 = 19
4x1 + 2x2 + 5x3 + x4 = 27
1 3 4
3 1 –2
2 3 5
x1
x2
x3

–3
–2
0
=
2 4 3 –2
2 –4 1 3
–1 3 –4 2
2 –2 2 1
x1
x2
x3
x4

1
10
–14
7
=

References and Suggestions for Further Study
A. The following publications by The MathWorks, are highly recommended for further study. They
are available from The MathWorks, 3 Apple Hill Drive, Natick, MA, 01760,
www.mathworks.com.
1. Getting Started with MATLAB
2. Using MATLAB
3. Using MATLAB Graphics
4. Using Simulink
5. SimPowerSystems for Use with Simulink
6. FixedPoint Toolbox
7. Simulink FixedPoint
8. RealTime Workshop
9. Signal Processing Toolbox
10. Getting Started with Signal Processing Blockset
10. Signal Processing Blockset
11. Control System Toolbox
12. Stateflow
B. Other references indicated in text pages and footnotes throughout this text, are listed below.
1. Mathematics for Business, Science, and Technology, ISBN 9781934404010
2. Numerical Analysis Using MATLAB and Excel, ISBN 9781934404034
3. Circuit Analysis II with MATLAB Applications, ISBN 0970951159
4. Signals and Systems with MATLAB Computing and Simulink Modeling,
ISBN 9781934404119
5. Electronic Devices and Amplifier Circuits with MATLAB Applications, ISBN 9781934404133
6. Digital Circuit Analysis and Design with Simulink Modeling and Introduction to CPLDs and FPGAs,
ISBN 9781934404058
Circuit Analysis I with MATLAB Computing and Simulink / SimPowerSystems Modeling R1

7. Introduction to Simulink with Engineering Applications, ISBN 9781934404096
8. Introduction to Stateflow with Applications, ISBN 9781934404072
9. Reference Data for Radio Engineers, ISBN 0672212188, Howard W. Sams & Co.
10. Electronic Engineers’ Handbook, ISBN 0070209812, McGrawHill
R2 Circuit Analysis I with MATLAB Computing and Simulink / SimPowerSystems Modeling

Index
Symbols and Numerics complex excitation function 6-3, 6-23 delta function
complex number(s) defined 10-7
% (percent) symbol in MATLAB A-2 addition B-2 sampling property 10-8
3-dB down 4-4 conjugate A-3, B-3 sifting property 10-9
defined A-3, B-2 demos in MATLAB A-2
A division B-4 dependent source(s)
exponential form B-5 current 1-11, 3-38
abs(z) MATLAB function A-23 multiplication B-3 voltage 1-11, 3-38
admittance 6-17 polar form B-5 determinant C-9
ampere 1-2, 1-19 rectangular form B-5 device(s)
ampere capacity of wires 2-30 subtraction B-2 active 1-11, 1-20
amplifier 4-1, 4-32 complex power 8-16 passive 1-11, 1-20
buffer 4-20 conductance 2-2 dielectric 5-16, 5-29
unity gain 4-13, 4-20 conj(A) MATLAB function C-8 differential input amplifier 4-5
analog-to-digital converter 8-28, 8-33 conjugate of a complex number B-3 digital filter 7-21
angle(z) MATLAB function A-23 conv(a,b) MATLAB function A-6 diode(s) 1-12
attenuation 4-13, 4-33 conversion factors 1-16 Dirac function 10-9
attenuator 4-1 conductor sizes for interior wiring 2-33 direct current 1-4
average value 8-2, 8-31 coulomb 1-1, 1-19 discontinuous function 10-1
axis MATLAB command A-16 Cramer’s rule 3-2, C-16, C-17 disp(A) MATLAB function 7-19, A-32
critical frequency 4-13, 4-33 display formats in MATLAB A-31
B current 1-1 division in MATLAB A-18
current division expressions 2-25 dot multiplication operator in MATLAB A-20
bandwidth 4-4 current flow driving functions 6-1
box MATLAB command A-12 conventional 1-2 duality 6-18, 6-25
branch 2-5 electron 1-2
current gain 4-2 E
C current limiting devices 2-2
current ratings for editor window in MATLAB A-1
capacitance 5-1, 5-17 electronic equipment 2-30 editor/debugger in MATLAB A-1
capacitance combinations 5-24 current source effective (RMS) value of sinusoids 8-5
capacitor(s) 1-11, 1-20, 5-16 combinations 2-14 effective values 8-4
in parallel 5-25 ideal 1-11 efficiency 3-44
in series 5-24 independent 1-11 eight-to-three line encoder 8-29
chemical processes 1-17, 1-20 practical 3-21 electric field 5-16, 5-17, 5-29
circuit(s) cutoff frequency electric filters - see filters
defined 1-13, 1-20 band-elimination filter 4-15 energy dissipated in a resistor 2-4
analysis with loop equations 3-8 band-pass filter 4-15 energy stored in a capacitor 5-21
analysis with mesh equations 3-8 high-pass filter 4-14 energy stored in an inductor 5-12
analysis with nodal equations 3-1 low-pass filter 4-13 eps in MATLAB A-22
with non-linear devices 3-42 lower 4-4 Euler’s identities B-4
clc MATLAB command A-2 upper 4-4 excitations 6-1
clear MATLAB command A-2 exit MATLAB command A-2
combined mesh 3-17 D exponential form of complex numbers B-5
combined node 3-6 exponentiation in MATLAB A-18
command screen in MATLAB A-1 data points in MATLAB A-14 eye(n) in MATLAB C-7
command window in MATLAB A-1 DC (Direct Current) 1-4 eye(size(A)) in MATLAB C-7
commas in MATLAB A-8 decibel 4-2, A-13
comment line in MATLAB A-2 deconv(c,d) MATLAB function A-6, A-7 F
comparators 8-29 default color in MATLAB A-15
complementary function 9-1 default in MATLAB A-12 Farad 5-17, 5-29
complete response 10-16 default line in MATLAB A-15 Faraday’s law of
complex conjugate A-4, B-3 default marker in MATLAB A-15 electromagnetic induction 5-2
IN-1

feedback 4-4 imaginary M
negative 4-5 axis B-2
positive 4-5 number B-2 magnetic field 5-1, 5-16, 5-29
figure window in MATLAB A-13 impedance 6-14 magnetic flux 5-2, 5-29
filter inductance 5-2 matrix, matrices
active 4-13 inductive adjoint C-20
all-pass 7-22 reactance 6-15, 6-23 cofactor of C-12
analog 7-23 susceptance 6-18, 6-23 conformable for addition C-2
band-elimination 4-15, 4-33, 7-22 inductor(s) conformable for multiplication C-4
band-pass 4-14, 4-33, 7-22 defined 1-11, 1-20, 5-2 congugate of C-8
band-rejection 4-15, 4-33, 7-22 in parallel 5-15 defined C-1
band-stop 4-15, 4-33, 7-22 in series 5-14 diagonal of C-1, C-6
high-pass 4-14, 4-33, 7-22 initial condition 5-3 Hermitian C-9
low-pass 4-13, 4-33, 7-22 initial rate of decay 9-3, 9-11 identity C-6
passive 4-13, 7-23 instantaneous values 2-1 inverse of C-21
phase shift 7-22 int(f,a,b) MATLAB function 1-7 left division in MATLAB C-24
RC high-pass 7-25 International System of Units 1-14 lower triangular C-6
RC low-pass 7-23 minor of C-12
stop-band 4-15, 4-33, 7-22 J multiplication using MATLAB A-20
flash converter 8-28 non-singular C-21
flux linkage 5-2, 5-29 j operator B-1 singular C-21
fmax(f,x1,x2) MATLAB function A-29 scalar C-6
fmin(f,x1,x2) MATLAB function A-29 K skew-Hermitian C-9
forced response 6-4, 10-16, 10-22 skew-symmetric C-9
format command in MATLAB A-31 KCL 2-6 square C-1
format in MATLAB A-31 Kirchhoff’s Current Law 2-6 symmetric C-8
fplot MATLAB command A-27 Kirchhoff’s Voltage Law 2-7 theory 3-2
fplot(fcn,lims) KVL 2-7 trace of C-2
MATLAB command A-27 transpose C-7
fprintf(format,array) L upper triangular C-5
MATLAB command 7-19, A-32 zero C-2
frequency response A-12 left-hand rule 5-1 maximum power
frequency-domain to time-domain lims = MATLAB function A-27 transfer theorem 3-35, 7-35
transformation 6-6, 6-23 linear mechanical forms of energy 1-17, 1-20
full-wave rectification circuit 3-38 mesh
function file in MATLAB A-26 devices 1-11 combined 3-18
fzero(f,x) MATLAB function A-26 factor A-9 defined 2-6
inductor 5-2 equations 2-10, 3-1, 5-25, 7-5
G passive element 3-37 generalized 3-17
linearity 3-37 mesh(x,y,z) MATLAB function A-18
Gaussian elimination method C-19 lines of magnetic flux 5-1, 5-29 meshgrid(x,y) MATLAB function A-18
grid MATLAB command A-12 linspace(values) MATLAB command A-14 metric system 1-14, 1-20
ground ln (natural log) A-13 m-file in MATLAB A-1, A-26
defined 2-1, 2-14 load mho 2-2
virtual 4-17 capacitive 8-15, 8-32 Military Standards 2-27
gtext(‘string’) MATLAB function A-13 inductive 8-15, 8-32 MINVERSE in Excel C-26
lighting 2-33 MMULT in Excel C-26, C-27
H resistive 8-11 multiplication of complex numbers B-3
log (common log) A-13 multiplication in MATLAB A-18
half-power points 4-4 log(x) MATLAB function A-13 multirange ammeter/milliammeter 8-24
half-wave rectification 8-3 log10(x) MATLAB function A-13
Heavyside function 10-9 log2(x) MATLAB function A-13 N
Henry 5-3, 5-29 loglog(x,y) MATLAB function A-13
loop NaN in MATLAB A-26
I defined 2-5 National Electric Code (NEC) 2-30
equations 3-1, 3-13 natural response
imag(z) MATLAB function A-23 circuits with single 2-10 9-1, 9-9, 10-16, 10-22
IN-2

NEC 2-30 complex 8-16, 8-17 series connection 2-8, 2-16, 2-17
negative charge 5-16 gain 4-2 short circuit 2-2
network in a capacitor 5-22 SI Derived Units 1-17
active 1-13, 1-20 in an inductor 5-11 siemens 2-2
passive 1-13, 1-20 in a resistor 2-3, 2-4, 2-28 signal 4-1, 4-32
topology 3-1 instantaneous 8-4 single ended output amplifier 4-5
newton 1-1, 1-19 power factor 8-10 single node-pair parallel circuit 2-14
nodal analysis 2-14, 3-1, 7-1 defined 8-10 slope converter 8-28
node lagging 8-15 solar energy 1-17, 1-20
combined 3-6 leading 8-15 sources of energy 1-17, 1-20
defined 2-5 power factor correction 8-18 standard prefixes 1-15
generalized 3-6 power triangle 8-16 Standards for Electrical and
equations 2-14, 3-2, 5-25, 7-1 prefixes 1-15, 1-16 Electronic Devices 2-26
non-reference 3-1 principle of superposition 3-41 steady-state conditions 5-12
reference 3-1 string in MATLAB A-18
non-linear devices 1-11 Q subplot(m,n,p) MATLAB command A-18
Norton’s theorem 3-33, 7-10 substitution method of solving a system
nuclear energy 1-17, 1-20 quad MATLAB function 1-8 of simultaneous equations 3-2
quad(‘f’,a,b,tol) MATLAB function 1-8 supermesh 3-17
O quad8 MATLAB function 1-8 supernode 3-6
quadratic factors A-9 superposition principle 3-38, 7-6
Ohm 2-1 quit MATLAB command A-2 susceptance
Ohm’s law 2-1 capacitive 6-18, 6-25
Ohm’s law for AC circuits 6-14 R inductive 6-18, 6-25
Ohmmeter 8-26
parallel type 8-26 rational polynomials A-8 T
series type 8-26 reactance
shunt type 8-26 capacitive 6-15, 6-24 temperature scales equivalents 1-16
op amp 4-5 inductive 6-15, 6-24 text(x,y,’string’) MATLAB function A-14
inverting mode 4-6 real text(x,y,z,’string’) MATLAB function A-16
non-inverting mode 4-9 axis B-2 Thevenin’s theorem 3-23, 7-10
open circuit 2-2 number B-2 time constant 9-3, 9-11, 10-18, 10-24
operational amplifier - see op amp real(z) MATLAB function A-23 time-domain to frequency-domain
regulation 3-45 transformation 6-5, 6-23
P resistance 2-1 time-window converter 8-28
input 4-28 title(‘string’) MATLAB command A-12
parallel connection 2-8, 2-17, 2-18 negative 2-3 total response 10-1, 10-14
particular solution 6-4 output 4-28 tracking converter 8-28
passive sign convention 1-9, 1-19 resistive network 8-29 transient response 9-1
periodic functions of time 8-1 resistors 1-11, 2-2 transistors 1-11
phasor analysis in amplifier circuits 7-14 color code 2-27 trivial solution 9-2
phasor diagram 7-17 failure rate 2-27 two-terminal device 1-4, 1-19
plot(x,y) MATLAB command A-10, A-12 shunt (parallel) 8-22
plot3(x,y,z) MATLAB command A-15 tolerance 2-27 U
polar plot in MATLAB A-24 response 6-1, 6-23
polar(theta,r) MATLAB function A-23 right-hand rule 5-1 unit impulse function 10-7
poly(r) MATLAB function A-4 RMS value of sinusoids 8-5 unit ramp function 10-6
polyder(p) MATLAB function A-6 RMS values of sinusoids with unit step function 10-1
polynomial construction from different frequencies 8-7
known roots in MATLAB A-4 roots(p) MATLAB function A-3, A-8 V
polyval(p,x) MATLAB function A-6 round(n) MATLAB function A-24
potential difference 1-4 virtual ground 4-17
power S volt 1-5, 1-19
absorbed 1-8, 1-19 voltage
average 8-9, 8-14 script file in MATLAB A-26 defined 1-4
in capacitive loads 8-11 semicolons in MATLAB A-8 dividers 2-2
in inductive loads 8-11 semilogx(x,y) MATLAB command A-12 division expressions 2-22
in a resistive loads 8-11 semilogy(x,y) MATLAB command A-12 drop 1-5
IN-3

follower 4-20
gain 4-2
instantaneous 1-6
rise 1-5
voltage source
combinations 2-14
ideal 1-11
independent 1-11
practical 3-20
voltmeter 8-24
W
watt 1-8
watt-hour meter 8-28
wattage 2-4, 2-29
wattmeter 8-28
weber 5-1, 5-29
Wheatstone bridge 8-27, 8-32
X
xlabel(‘string’) MATLAB command A-12
Y
ylabel(‘string’) MATLAB command A-12
Z
zero potential 2-14
IN-4

Students and working professionals will find Circuit
Analysis I with MATLAB® Computing and
Simulink®/SimPowerSystems Modeling to be a con-cise
and easy-to-learn text. It provides complete,
clear, and detailed explanations of the principal elec-trical
engineering concepts, and these are illustrated
with numerous practical examples.
This text includes the following chapters and appendices:
• Basic Concepts and Definitions • Analysis of Simple Circuits • Nodal and Mesh Equations -
Circuit Theorems • Introduction to Operational Amplifiers • Inductance and Capacitance
• Sinusoidal Circuit Analysis • Phasor Circuit Analysis • Average and RMS Values, Complex Power,
and Instruments • Natural Response • Forced and Total Response in RL and RC Circuits •
Introduction to MATLAB® • Introduction to Simulink® • Introduction to SimPowerSystems®
• Review of Complex Numbers • Matrices and Determinants
Each chapter and appendix contains numerous practical applications supplemented with detailed
instructions for using MATLAB, Simulink, and SimPowerSystems to obtain quick and accurate
results.
Steven T. Karris is the founder and president of Orchard Publications, has undergraduate and
graduate degrees in electrical engineering, and is a registered professional engineer in California
and Florida. He has more than 35 years of professional engineering experience and more than 30
years of teaching experience as an adjunct professor, most recently at UC Berkeley, California. His
area of interest is in The MathWorks, Inc.™products and the publication of MATLAB® and
Simulink® based texts.
Visit us on the Internet
or email us: info@orchardpublications.com
ISBN-13: 978-1-934404-18-8
ISBN-10: 1-934404-18-7
$70.00 U.S.A.
Circuit Analysis I
Simulink®/SimPowerSystems Modeling

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