Circuitos Eléctricos por James S. Kang.pdf

Electric Circuits
James S. Kang
California State Polytechnic University, Pomona
Australia ● Brazil ● Mexico ● Singapore ● United Kingdom ● United States

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Electric Circuits, First Edition
James S. Kang
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iii
Contents
Preface x
About the Author xvi
Chapter 1
Voltage, Current, Power,
and SourCeS 1
1.1 Introduction 1
1.2 International System of Units 1
1.3 Charge,Voltage, Current, and Power 4
1.3.1 Electric Charge 4
1.3.2 Electric Field 4
1.3.3 Voltage 5
1.3.4 Current 7
1.3.5 Power 9
1.4 Independent Sources 10
1.4.1 Direct Current Sources and Alternating
Current Sources 11
1.5 Dependent Sources 15
1.5.1 Voltage-Controlled Voltage Source (VCVS) 16
1.5.2 Voltage-Controlled Current Source (VCCS) 16
1.5.3 Current-Controlled Voltage Source (CCVS) 16
1.5.4 Current-Controlled Current Source (CCCS) 16
1.6 Elementary Signals 17
1.6.1 Dirac Delta Function 17
1.6.2 Step Function 19
1.6.3 Ramp Function 21
1.6.4 Exponential Decay 23
1.6.5 Rectangular Pulse and Triangular Pulse 24
Summary 27
PrOBLEmS 27
Chapter 2
CirCuit lawS 31
2.1 Introduction 31
2.2 Circuit 31
2.3 Resistor 33
2.4 Ohm’s Law 35
2.5 Kirchhoff’s Current Law (KCL) 38
2.6 Kirchhoff’s Voltage Law (KVL) 46
2.7 Series and Parallel Connection
of Resistors 53
2.7.1 Series Connection of Resistors 53
2.7.2 Parallel Connection of Resistors 58
2.8 Voltage Divider Rule 74
2.8.1 Wheatstone Bridge 80
2.9 Current Divider Rule 82
2.10 Delta-Wye (D-Y) Transformation and Wye-Delta
(Y-D) Transformation 91
2.11 PSpice and Simulink 100
2.11.1 Simulink 104
Summary 104
PrOBLEmS 105
Chapter 3
CirCuit analySiS MethodS 117
3.1 Introduction 117
3.2 Nodal Analysis 118
3.3 Supernode 142
3.4 Mesh Analysis 153
3.5 Supermesh 175
3.6.1 PSpice 190
3.6.2 VCVS 190
3.6.3 VCCS 191
3.6.4 CCVS 192
3.6.5 CCCS 193
3.6.6 Simulink 193
Summary 194
PrOBLEmS 194
Chapter 4
CirCuit theoreMS 208
4.2 Superposition Principle 209
4.3 Source Transformations 221

iv Contents
4.4 Thévenin’s Theorem 234
4.4.1 Finding the thévenin equivalent Voltage Vth 235
4.4.2 Finding the thévenin equivalent Resistance Rth 235
4.5 Norton’s Theorem 263
4.5.1 Finding the norton equivalent Current In 264
4.5.2 Finding the norton equivalent Resistance Rn 264
4.5.3 Relation Between the thévenin equivalent
Circuit and the norton equivalent Circuit 264
4.6 Maximum Power Transfer 284
4.7 PSpice 296
4.7.1 simulink 299
Summary 300
PrOBLEmS 301
Chapter 5
OperatiOnal amplifier CirCuits 314
5.2 Ideal Op Amp 315
5.2.1 Voltage Follower 322
5.3 Sum and Difference 333
5.3.1 summing Amplifier (Inverting
Configuration) 333
5.3.2 summing Amplifier (noninverting
Configuration) 336
5.3.3 Alternative summing Amplifier (noninverting
Configuration) 341
5.3.4 Difference Amplifier 343
5.4 Instrumentation Amplifier 346
5.5 Current Amplifier 347
5.5.1 Current to Voltage Converter (transresistance
Amplifier) 348
5.5.2 negative Resistance Circuit 349
5.5.3 Voltage-to-Current Converter (transconductance
Amplifier) 350
5.6 Analysis of Inverting Configuration 351
5.6.1 Input Resistance 354
5.6.2 output Resistance 354
5.7 Analysis of Noninverting Configuration 358
5.7.1 Input Resistance 360
5.7.2 output Resistance 360
Summary 370
PrOBLEmS 371
Chapter 6
CapaCitOrs and induCtOrs 379
6.2 Capacitors 380
6.2.1 sinusoidal Input to Capacitor 389
6.3 Series and Parallel Connection of Capacitors 390
6.3.1 series Connection of Capacitors 390
6.3.2 Parallel Connection of Capacitors 392
6.4 Op Amp Integrator and Op Amp
Differentiator 395
6.4.1 op Amp Integrator 395
6.4.2 op Amp Differentiator 397
6.5 Inductors 397
6.5.1 sinusoidal Input to Inductor 407
6.6 Series and Parallel Connection of Inductors 408
6.6.1 series Connection of Inductors 408
6.6.2 Parallel Connection of Inductors 409
Summary 416
PrOBLEmS 416
Chapter 7
rC and rl CirCuits 424
7.2 Natural Response of RC Circuit 424
7.2.1 time Constant 428
7.3 Step Response of RC Circuit 435
7.3.1 Initial Value 438
7.3.2 Final Value 438
7.3.4 solution to General First-order Differential
equation with Constant Coefficient and
Constant Input 440
7.4 Natural Response of RL Circuit 448
7.5 Step Response of RL Circuit 459
7.5.1 Initial Value 462
7.5.2 Final Value 462
7.5.4 solution to General First-order Differential
equation with Constant Coefficient and
Constant Input 464
7.6 Solving General First-Order Differential
Equations 476
Summary 494
PrOBLEmS 495
Chapter 8
rlC CirCuits 505
8.2 Zero Input Response of Second-Order
Differential Equations 505
8.2.1 Case 1: overdamped (a . v0 or a1 . 2Ïa0
or z . 1) 507

ConTEnTS v
8.2.2 Case 2: Critically Damped (a 5 v0 or a1 5 2Ïa0
or z 5 1) 509
8.2.3 Case 3: Underdamped (a , v0 or a1 , 2Ïa0
or z , 1) 510
8.3 Zero Input Response of Series RLC Circuit 511
or z . 1) 513
or z 5 1) 513
or z , 1) 513
8.4 Zero Input Response of Parallel RLC Circuit 530
or z . 1) 532
or z 5 1) 532
or z , 1) 532
8.5 Solution of the Second-Order Differential
Equations to Constant Input 545
8.5.1 Particular Solution 545
or z . 1) 546
or z 5 1) 547
or z , 1) 548
8.6 Step Response of a Series RLC Circuit 549
8.6.1 Case 1: overdamped (a . v0 or a12 . Ïa0
or z . 1) 550
or z 5 1) 552
or z , 1) 553
8.7 Step Response of a Parallel RLC Circuit 566
or z . 1) 567
or z 5 1) 569
or z , 1) 570
8.8 General Second-Order Circuits 580
8.9.1 Solving Differential Equations Using Simulink 600
8.9.2 Solving Differential Equations Using PSpice 601
Summary 603
PrOBLEmS 604
Chapter 9
PhaSorS and iMPedanCeS 615
9.2 Sinusoidal Signals 615
9.2.1 Cosine Wave 615
9.2.2 Sine Wave 618
9.3 RMS Value 620
9.4 Phasors 624
9.4.1 Representing Sinusoids in Phasor 627
9.4.2 Conversion Between Cartesian Coordinate
System (Rectangular Coordinate System) and
Polar Coordinate System 629
9.4.3 Phasor Arithmetic 635
9.5 Impedance and Admittance 638
9.5.1 Resistor 639
9.5.2 Capacitor 640
9.5.3 Inductor 642
9.6 Phasor-Transformed Circuit 644
9.7 Kirchhoff’s Current Law and Kirchhoff’s
Voltage Law for Phasors 649
9.8 Series and Parallel Connection of
Impedances 652
9.9 Delta-Wye (D-Y) and Wye-Delta (Y-D)
Transformation 656
Summary 664
PrOBLEmS 664
Chapter 10
analySiS of PhaSor-tranSforMed
CirCuitS 668
10.2 Phasor-Transformed Circuits 669
10.3 Voltage Divider Rule 669
10.4 Current Divider Rule 672
10.7 Superposition Principle 681
10.8 Source Transformation 683
10.9 Thévenin Equivalent Circuit 686
10.9.1 Finding the Thévenin Equivalent
Voltage Vth 687
10.9.2 Finding the Thévenin Equivalent
Impedance Zth 687
10.10 Norton Equivalent Circuit 689
10.11 Transfer Function 692
10.11.1 Series RLC Circuits 701
10.11.2 Parallel RLC Circuits 707
Summary 721
PrOBLEmS 722

vi ConTEnTS
Chapter 11
aC Power 733
11.2 Instantaneous Power,Average Power, Reactive
Power,Apparent Power 733
11.3 Complex Power 739
11.4 Conservation of AC Power 749
11.5 Maximum Power Transfer 752
11.5.1 Maximum Power Transfer for norton
Equivalent Circuit 756
11.6 Power Factor Correction (PFC) 756
Summary 770
PrOBLEmS 770
Chapter 12
three-PhaSe SySteMS 778
12.2 Three-Phase Sources 778
12.2.1 negative Phase Sequence 781
12.3 Balanced Y-Y Circuit 782
12.3.1 Balanced Y-Y Circuit with Wire Impedance 786
12.4 Balanced Y-D Circuit 792
12.4.1 Balanced Y-D Circuit with Wire Impedance 796
12.5 Balanced D-D Circuit 801
12.5.1 Balanced D-D Circuit with Wire Impedance 805
12.6 Balanced D-Y Circuit 813
12.6.1 Balanced D-Y Circuit with Wire Impedance 816
Summary 825
PrOBLEmS 825
Chapter 13
MagnetiCally CouPled
CirCuitS 829
13.2 Mutual Inductance 829
13.2.1 Faraday’s Law 830
13.2.2 Mutual Inductance 831
13.2.3 Mutual Inductance of a Second Coil Wrapped
Around a Solenoid 833
13.3 Dot Convention and Induced Voltage 835
13.3.1 Combined Mutual and Self-Induction
Voltage 838
13.4 Equivalent Circuits 848
13.5 Energy of Coupled Coils 853
13.6 Linear Transformer 855
13.7 Ideal Transformer 865
13.7.1 Autotransformer 874
Summary 881
PrOBLEmS 881
Chapter 14
the laPlaCe tranSforM 886
14.2 Definition of the Laplace Transform 887
14.3 Properties of the Laplace Transform 891
14.3.1 Linearity Property (Superposition Principle) 893
14.3.2 Time-Shifting Property 894
14.3.3 Frequency Translation Property 895
14.3.4 Multiplication by cos(v0t) 898
14.3.5 Multiplication by sin(v0t) 899
14.3.6 Time Differentiation Property 900
14.3.7 Integral Property 902
14.3.8 Frequency Differentiation Property 904
14.3.9 Frequency Integration Property 907
14.3.10 Time-Scaling Property 908
14.3.11 Initial Value Theorem and Final Value
Theorem 910
14.3.12 Initial Value Theorem 910
14.3.13 Final Value Theorem 912
14.4 Inverse Laplace Transform 914
14.4.1 Partial Fraction Expansion 923
14.4.2 Simple Real Poles 925
14.4.3 Complex Poles 928
14.4.4 Repeated Poles 934
14.5 Solving Differential Equations Using the
Laplace Transform 942
Summary 950
PrOBLEmS 951
Chapter 15
CirCuit analySiS in the s-doMain 954
15.2 Laplace-Transformed Circuit Elements 955
15.2.1 Resistor 955
15.2.2 Capacitor 956
15.2.3 Inductor 957
15.3 Laplace-Transformed Circuit 958
15.3.1 Voltage Divider Rule 958
15.3.2 Current Divider Rule 961
15.6 Thévenin Equivalent Circuit in the s-Domain 980

ConTEnTS vii
15.7 Norton Equivalent Circuit in the
s-Domain 990
15.8 Transfer Function 997
15.8.1 Sinusoidal Input 998
15.8.2 Poles and Zeros 999
15.9 Convolution 1020
15.9.1 Commutative Property 1021
15.9.2 Associative Property 1021
15.9.3 Distributive Property 1021
15.10 Linear,Time-Invariant (LTI) System 1037
15.10.1 Impulse Response 1038
15.10.2 output of Linear Time-Invariant System 1038
15.10.3 Step Response of LTI System 1039
15.11 Bode Diagram 1040
15.11.1 Linear Scale 1040
15.11.2 dB Scale 1041
15.11.3 Bode Diagram of Constant Term 1044
15.11.4 Bode Diagram of H(s) 5 s 1 1000 1044
15.11.5 Bode Diagram of H(s) 5 100ys 1045
15.11.6 Bode Diagram of H(s) 5 sy1000 1046
15.11.7 Bode Diagram of H(s) 5 104
y(s 1 100)2
1047
15.11.8 Complex Poles and Zeros 1059
15.12 Simulink 1062
Summary 1064
PrOBLEmS 1064
Chapter 16
firSt- and SeCond-order
analog filterS 1074
16.2 Magnitude Scaling and Frequency
Scaling 1075
16.2.1 Magnitude Scaling 1075
16.2.2 Frequency Scaling 1076
16.2.3 Magnitude and Frequency Scaling 1078
16.3 First-Order LPF 1079
16.4 First-Order HPF 1081
16.5 Second-Order LPF 1084
16.5.1 Frequency Response 1085
16.5.2 Magnitude Response 1085
16.5.3 Phase Response 1086
16.5.4 Series RLC LPF 1087
16.5.5 Parallel RLC LPF 1088
16.5.6 Sallen-Key Circuit for the Second-order LPF 1090
16.5.7 Equal R, Equal C Method 1092
16.5.8 normalized Filter 1093
16.5.9 Unity Gain Method 1098
16.6 Second-Order HPF Design 1100
16.6.4 Series RLC HPF 1102
16.6.5 Parallel RLC HPF 1104
16.6.6 Sallen-Key Circuit for the
Second-order HPF 1105
16.6.7 Equal R and Equal C Method 1108
16.6.8 normalization 1109
16.6.9 Unity Gain Method 1110
16.7 Second-Order Bandpass Filter Design 1113
16.7.4 Series RLC Bandpass Filter 1116
16.7.5 Parallel RLC Bandpass Filter 1118
16.7.6 Sallen-Key Circuit for the Second-order
Bandpass Filter 1120
16.7.7 Equal R, Equal C Method 1122
16.7.9 Delyiannis-Friend Circuit 1125
16.8 Second-Order Bandstop Filter Design 1129
16.8.4 Series RLC Bandstop Filter 1132
16.8.5 Parallel RLC Bandstop Filter 1134
16.8.6 Sallen-Key Circuit for the Second-order
Bandstop Filter 1136
16.9 Simulink 1147
Summary 1148
PrOBLEmS 1155
Chapter 17
analog filter deSign 1166
17.2 Analog Butterworth LPF Design 1167
17.2.1 Backward Transformation 1168
17.2.2 Finding the order of the normalized LPF 1168
17.2.3 Finding the Pole Locations 1171
17.3 Analog Butterworth HPF Design 1182
17.4 Analog Butterworth Bandpass Filter
Design 1191
17.5 Analog Butterworth Bandstop Filter
Design 1202
17.6 Analog Chebyshev Type 1 LPF Design 1214
17.7 Analog Chebyshev Type 2 LPF Design 1226
17.8 MATLAB 1242
Summary 1245
PrOBLEmS 1245

viii ConTEnTS
Chapter 18
fourier SerieS 1259
18.2 Signal Representation Using Orthogonal
Functions 1259
18.2.1 orthogonal Functions 1259
18.2.2 Representation of an Arbitrary Signal by
orthogonal Functions 1270
18.2.3 Trigonometric Fourier Series 1278
18.2.4 Proof of orthogonality 1279
18.2.5 Exponential Fourier Series 1282
18.2.6 Proof of orthogonality 1283
18.3 Trigonometric Fourier Series 1283
18.3.1 Trigonometric Fourier Series Using
Cosines only 1286
18.3.2 one-Sided Magnitude Spectrum and one-Sided
Phase Spectrum 1287
18.3.3 DC Level 1296
18.3.4 Time Shifting 1298
18.3.5 Triangular Pulse Train 1302
18.3.6 Sawtooth Pulse Train 1306
18.3.7 Rectified Cosine 1309
18.3.8 Rectified Sine 1313
18.3.9 Average Power of Periodic Signals 1317
18.3.10 Half-Wave Symmetry 1320
18.4 Solving Circuit Problems Using Trigonometric
Fourier Series 1324
18.5 Exponential Fourier Series 1333
18.5.1 Conversion of Fourier Coefficients 1336
18.5.2 Two-Sided Magnitude Spectrum and Two-Sided
Phase Spectrum 1337
18.5.3 Triangular Pulse Train 1343
18.5.4 Sawtooth Pulse Train 1348
18.5.5 Rectified Cosine 1350
18.5.6 Rectified Sine 1353
18.5.7 Average Power of Periodic Signals 1356
18.6 Properties of Exponential Fourier
Coefficients 1357
18.6.1 DC Level 1357
18.6.2 Linearity Property (Superposition
Principle) 1358
18.6.4 Time Reversal Property 1364
18.6.5 Time Differentiation Property 1365
18.6.6 Convolution Property 1365
18.7 Solving Circuit Problems Using Exponential
Fourier Series 1365
Summary 1377
PrOBLEmS 1384
Chapter 19
fourier tranSforM 1399
19.2 Definition of Fourier Transform 1399
19.2.1 Symmetries 1403
19.2.2 Finding Fourier Transform from Fourier
Coefficients 1407
19.3 Properties of Fourier Transform 1408
19.3.1 Linearity Property (Superposition
Principle) 1411
19.3.3 Time-Scaling Property 1414
19.3.4 Symmetry Property (Duality Property) 1416
19.3.5 Time-Reversal Property 1420
19.3.6 Frequency-Shifting Property 1422
19.3.7 Modulation Property 1425
19.3.8 Time-Differentiation Property 1428
19.3.9 Frequency-Differentiation Property 1431
19.3.10 Conjugate Property 1432
19.3.11 Integration Property 1433
19.3.12 Convolution Property 1434
19.3.13 Multiplication Property 1437
19.4 Fourier Transform of Periodic Signals 1439
19.4.1 Fourier Series and Fourier Transform
of Impulse Train 1440
19.5 Parseval’s Theorem 1443
19.6 Simulink 1449
Summary 1452
PrOBLEmS 1452
Chapter 20
two-Port CirCuitS 1457
20.2 Two-Port Circuit 1458
20.2.1 z-Parameters (Impedance Parameters) 1458
20.2.2 y-Parameters (Admittance Parameters) 1464
20.2.3 h-Parameters (Hybrid Parameters) 1470
20.2.4 g-Parameters (Inverse Hybrid Parameters) 1473
20.2.5 ABCD-Parameters (Transmission Parameters,
a-Parameters) 1477
20.2.6 Inverse Transmission Parameters
(b-Parameters) 1485
20.3 Conversion of Parameters 1489
20.3.1 Conversion of z-Parameters to All the other
Parameters 1489
20.3.2 Conversion of z-Parameters to
y-Parameters 1489
20.3.3 Conversion of z-Parameters to ABCD
Parameters 1490

ConTEnTS ix
b-Parameters 1491
h-Parameters 1491
g-Parameters 1492
20.3.7 Conversion of y-Parameters to All the other
Parameters 1493
20.3.8 Conversion of h-Parameters to All the other
Parameters 1494
20.3.9 Conversion of g-Parameters to All the other
Parameters 1494
20.3.10 Conversion of ABCD Parameters to All
the other Parameters 1495
20.3.11 Conversion of b-Parameters to All the other
Parameters 1496
20.4 Interconnection of Two-Port Circuits 1500
20.4.1 Cascade Connection 1500
20.4.2 Series Connection 1502
20.4.3 Parallel Connection 1505
20.4.4 Series-Parallel Connection 1507
20.4.5 Parallel-Series Connection 1508
20.4.6 Cascade Connection for b-Parameters 1508
Summary 1512
PrOBLEmS 1513
Answers to Odd-Numbered Questions 1517
Index 1548

x
Preface
This book is intended to be an introductory text on the subject of electric circuits. It
provides simple explanations of the basic concepts, followed by simple examples and exer-
cises. When necessary, detailed derivations for the main topics and examples are given to
help readers understand the main ideas. MATLAB is a tool that can be used effectively
in Electric Circuits courses. In this text, MATLAB is integrated into selected examples to
illustrate its use in solving circuit problems. MATLAB can be used to check the answers or
solve more complex circuit problems. This text is written for a two-semester sequence or a
three-quarters sequence on electric circuits.
Suggested Course Outlines
The following is a list of topics covered in a typical Electric Circuits courses, with suggested
course outlines.
one-SeMeSter or -Quarter CourSe
If Electric Circuits is offered as a one-semester or one-quarter course, Chapters 1 through
12 can be taught without covering, or only lightly covering, sections 1.6, 2.10, 2.11, 3.6, 4.7,
5.6, 5.7, 5.8, 6.7, 7.6, 7.7, 8.8, 8.9, 9.9, 9.10, 10.12, 11.7, 12.5, 12.6, and 12.7.
two-SeMeSter or -Quarter CourSeS
For two-semester Electric Circuit courses, Chapters 1 through 8, which cover dc circuits,
op amps, and the responses of first-order and second-order circuits, can be taught in the
first semester. Chapters 9 through 20, which cover alternating current (ac) circuits, Laplace
transforms, circuit analysis in the s-domain, two-port circuits, analog filter design and imple-
mentation, Fourier series, and Fourier transform, can then be taught in the second semester.
three-Quarter CourSeS
For three-quarter Electric Circuit courses, Chapters 1 through 5, which cover dc circuits and
op amps,can be taught in the first quarter;Chapters 6 through 13,which cover the responses
of first-order and second-order circuits and ac circuits, can be taught in the second quarter,
and Chapters 14 through 20, which cover Laplace transforms, circuit analysis in the s-do-
main, two-port circuits, analog filter design and implementation, Fourier series, and Fourier
transform, can be taught in the third quarter.
Depending on the catalog description and the course outlines, instructors can pick
and choose the topics covered in the courses that they teach. Several features of this text
are listed next.
Features
After a topic is presented, examples and exercises follow. Examples are chosen to expand
and elaborate the main concept of the topic. In a step-by-step approach, details are worked
out to help students understand the main ideas.

PREFACE xi
In addition to analyzing RC, RL, and RLC circuits connected in series or parallel in
the time domain and the frequency domain, analyses of circuits different from RC, RL, and
RLC circuits and connected other than in series and parallel are provided. Also, general
input signals that are different from unit step functions are included in the analyses.
In the analog filter design, the specifications of the filter are translated into its trans-
fer function in cascade form. From the transfer function, each section can be designed with
appropriate op amp circuits. The normalized component values for each section are found
by adopting a simplification method (equal R equal C or unity gain). Then, magnitude
scaling and frequency scaling are used to find the final component values.The entire design
procedure, from the specifications to the circuit design, is detailed, including the PSpice
simulation used to verify the design.
Before the discussion of Fourier series, orthogonal functions and the representation
of square integrable functions as a linear combination of a set of orthogonal functions are
introduced. The set of orthogonal functions for Fourier series representation consists of
cosines and sines. The Fourier coefficients for the square pulse train, triangular pulse train,
sawtooth pulse train, and rectified sines and cosines are derived.The Fourier coefficients of
any variation of these waveforms can be found by applying the time-shifting property and
finding the dc component.
MATLAB can be an effective tool in solving problems in electric circuits. Simple
functions such as calculating the equivalent resistance or impedance of parallel connec-
tion of resistors, capacitors, and inductors; conversion from Cartesian coordinates to polar
coordinates; conversion from polar coordinates to Cartesian coordinates; conversion from
the wye configuration to delta configuration; and conversion from delta configuration to
wye configuration provide accurate answers in less time.These simple functions can be part
of scripts that enable us to find solutions to typical circuit problems.
The complexity of taking the inverse Laplace transforms increases as the order
increases. MATLAB can be used to solve equations and to find integrals, transforms,
inverse transforms, and transfer functions. The application of MATLAB to circuit analysis
is demonstrated throughout the text when appropriate. For example, after finding inverse
Laplace transforms by hand using partial fraction expansion, answers from MATLAB are
provided as a comparison.
Examples of circuit simulation using OrCAD PSpice and Simulink are given at the
end of each chapter. Simulink is a tool that can be used to perform circuit simulations. In
Simulink, physical signals can be converted to Simulink signals and vice versa. Simscapes
include many blocks that are related to electric circuits. Simulink can be used in computer
assignments or laboratory experiments.
The Instructor’s Solution Manual for the exercises and end-of-chapter problems is
available for instructors.This manual includes MATLAB scripts for selected problems as a
check on the accuracy of the solutions by hand.
Overview of Chapters
In Chapter 1, definitions of voltage, current, power, and energy are given.Also, independent
voltage source and current source are introduced, along with dependent voltage sources and
current sources.
In Chapter 2, nodes, branches, meshes, and loops are introduced. Ohm’s law is explained.
Kirchhoff’s current law (KCL), Kirchhoff’s voltage law (KVL), the voltage divider rule,
and the current divider rule are explained with examples.
In Chapter 3, nodal analysis and mesh analysis are discussed in depth.The nodal analysis
and mesh analysis are used extensively in the rest of the text.
Chapter 4 introduces circuit theorems that are useful in analyzing electric circuits and
electronic circuits.The circuit theorems discussed in this chapter are the superposition

xii PREFACE
principle, source transformations,Thévenin’s theorem, Norton’s theorem, and maximum
power transfer.
Chapter 5 introduces op amp circuits. Op amp is a versatile integrated circuit (IC) chip
that has wide-ranging applications in circuit design.The concept of the ideal op amp model
is explained, along with applications in sum and difference, instrumentation amplifier,
and current amplifier. Detailed analysis of inverting configuration and noninverting
configuration is provided.
In Chapter 6, the energy storage elements called capacitors and inductors are discussed.
The current voltage relation of capacitors and inductors are derived.The energy stored on
the capacitors and inductors are presented.
In Chapter 7, the transformation of RC and RL circuits to differential equations and
solutions of the first-order differential equations to get the responses of the circuits
are presented. In the general first-order circuits, the input signal can be dc, ramp signal,
exponential signal, or sinusoidal signal.
In Chapter 8, the transformation of series RLC and parallel RLC circuits to the second-
order differential equations, as well as solving the second-order differential equations to
get the responses of the circuits are presented. In the general second-order circuits, the
input signal can be dc, ramp signal, exponential signal, or sinusoidal signal.
Chapter 9 introduces sinusoidal signals, phasors, impedances, and admittances.Also,
transforming ac circuits to phasor-transformed circuits is presented, along with analyzing
phasor transformed circuits using KCL, KVL, equivalent impedances, delta-wye
transformation, and wye-delta transformation.
The analysis of phasor-transformed circuits is continued in Chapter 10 with the
introduction of the voltage divider rule, current divider rule, nodal analysis, mesh analysis,
superposition principle, source transformation,Thévenin equivalent circuit, Norton
equivalent circuit, and transfer function.This analysis is similar to the one for resistive
circuits with the use of impedances.
Chapter 11 presents information on ac power.The definitions of instantaneous power,
average power, reactive power, complex power, apparent power, and power factor are also
given, and power factor correction is explained with examples.
As an extension of ac power, the three-phase system is presented in Chapter 12.The
connection of balanced sources (wye-connected or delta-connected) to balanced loads
(wye-connected or delta connected) are presented, both with and without wire impedances.
Magnetically coupled circuits, which are related to ac power, are discussed in Chapter 13.
Mutual inductance, induced voltage, dot convention, linear transformers, and ideal
transformers are introduced.
The Laplace transform is introduced in Chapter 14.The definition of the transform, region
of convergence, transform, and inverse transform are explained with examples.Various
properties of Laplace transform are also presented with examples.
The discussion on Laplace transform is continued in Chapter 15. Electric circuits can
be transformed into an s-domain by replacing voltage sources and current sources to
the s-domain and replacing capacitors and inductors to impedances.The circuit laws
and theorems that apply to resistive circuits also apply to s-domain circuits.The time
domain signal can be obtained by taking the inverse Laplace transform of the s-domain
representation.The differential equations in the time domain are transformed to algebraic
equations in the s-domain.The transfer function in the s-domain is defined as the ratio

PREFACE xiii
of the output signal in the s-domain to the input signal in the s-domain.The concept of
convolution is introduced with a number of examples.Also, finding the convolution using
Laplace transforms are illustrated in the same examples. Plotting the magnitude response
and phase response of a circuit or a system using the Bode diagram is introduced.
The first-order and the second-order analog filters that are building blocks for the
higher-order filters are presented in Chapter 16.The filters can be implemented by
interconnecting passive elements consisting of resistors, capacitors, and inductors.
Alternatively, filters can be implemented utilizing op amp circuits. Sallen and Key circuits
for implementing second-order filters are discussed as well, along with design examples.
The discussion on analog filter design is extended in Chapter 17.A filter is designed to
meet the specifications of the filter.The transfer function that satisfies the specification
is found. From the transfer function, the corner frequency and Q value can be found.
Then, the normalized component values and scaled component values are found. PSpice
simulations can be used to verify the design.
Orthogonal functions and the representation of signals as a linear combination of a set
of orthogonal functions are introduced in Chapter 18. If the set of orthogonal functions
consists of harmonically related sinusoids or exponential functions, the representation is
called the Fourier series. Fourier series representation of common signals, including the
square pulse train, triangular pulse train, sawtooth waveform, and rectified cosine and sine,
are presented in detail, with examples.The derivation and application of the time-shifting
property of Fourier coefficients are provided. In addition, the application of the Fourier
series representation in solving circuit problems are presented, along with examples.
As the period of a periodic signal is increased to infinity, the signal becomes nonperiodic,
the discrete line spectrums become a continuous spectrum, and multiplying the Fourier
coefficients by the period produces the Fourier transform, as explained in Chapter 19.
Important properties of the Fourier transform, including time shifting, frequency shifting,
symmetry, modulation, convolution, and multiplication, are introduced, along with
interpretation and examples.
Two-port circuits are defined and analyzed in Chapter 20. Depending on which of the
parameters are selected as independent variables, there are six different representations
for two-port circuits.The coefficients of the representations are called parameters.The six
parameters (z, y, h, g, ABCD, b) for two-port circuits are presented along with examples.
The conversion between the parameters and the interconnection of parameters are
provided in this chapter.
Instructor resources
Cengage Learning’s secure, password-protected Instructor Resource Center contains help-
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Microsoft PowerPoint slides, test banks, and an Instructor’s Solution Manual, with detailed
solutions to all the problems from the text.The Instructor Resource Center can be accessed
at https://login.cengage.com.
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xiv PREFACE
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adjust their course to maximize student success.

PREFACE xv
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acknowledgments
I wish to acknowledge and thank the Global Engineering team at Cengage Learning for
their dedication to this new book: Timothy Anderson, Product Director; Ashley Kaupert,
Associate Media Content Developer; Kim Kusnerak, Senior Content Project Manager;
Kristin Stine, Marketing Manager; Elizabeth Brown and Brittany Burden, Learning Solu-
tions Specialists; and Alexander Sham, Product Assistant.They have skillfully guided every
aspect of this text’s development and production to successful completion. I also would
like to express my appreciation to the following reviewers, whose helpful comments and
suggestions improved the manuscript:
Elizabeth Brauer, Northern Arizona University
Mario Edgardo Magana, Oregon State University
Malik Elbuluk,The University of Akron
Timothy A. Little, Dalhousie University
Ahmad Nafisi, California Polytechnic State University—San Luis Obispo
Scott Norr, University of Minnesota—Duluth
Nadipuram Prasad, New Mexico State University
Vignesh Rajamani, Oklahoma State University
Pradeepa Yahampath, University of Manitoba

xvi
About the Author
Dr. James S. Kang is a professor of electrical and computer engineering at the California
State Polytechnic University, Pomona, commonly known as Cal Poly Pomona. Cal Poly
Pomona is famous for its laboratory-oriented, hands-on approach to engineering education.
Most of the electrical and computer engineering courses offered there include a compan-
ion laboratory course. Students design, build, and test practical circuits in the laboratory
based on the theory that they learned in the lecture course. This book, Electric Circuits,
incorporates this philosophy.

1
Chapter 1
Voltage, Current,
Power, and Sources
1.1 Introduction
The seven base units of the International System of Units (SI), along with derived units rel-
evant to electrical and computer engineering, are presented in this chapter.The definitions
of the terms voltage, current, and power are given as well.
A voltage source with voltage Vs provides a constant potential difference to the circuit
connected between the positive terminal and the negative terminal. A current source with
current Is provides a constant current of Is amperes to the circuit connected to the two termin-
als. If the voltage from the voltage source is constant with time, the voltage source is called
the direct current (dc) source. Likewise, if the current from the current source is constant
with time, the current source is called the dc source. If the voltage from the voltage source is
a sinusoid, the voltage source is called alternating current (ac) voltage source. Likewise, if the
current from the current source is a sinusoid,the current source is called the ac current source.
The voltage or current on the dependent sources depends solely on the controlling
voltage or controlling current.Dependent sources are introduced along with circuit symbols.
The elementary signals that are useful throughout the text are introduced next. The
elementary signals are Dirac delta function, step function, ramp function, rectangular pulse,
triangular pulse, and exponential decay.
1.2 International System of Units
The International System of Units (SI) is the modern form of the metric system derived
from the meter-kilogram-second (MKS) system. The SI system is founded on seven base
units for the seven quantities assumed to be mutually independent. Tables 1.1–1.6, which

2 Chapter 1 Voltage, Current, Power, and SourCeS
give information on the SI system, come from the NIST Reference on Constants, Units,
and Uncertainty (http://physics.nist.gov/cuu/Units/units.html), the official reference of the
National Institute of Standards and Technology.
A meter is defined as the length of a path traveled by light in a vacuum during a time
interval of 1299,792,458 [(≈ 1(3 3 108
)] of a second.
A kilogram is equal to the mass of the international prototype of the kilogram.
TABLe 1.1
SI Base Units.
Base Quantity Name Symbol
Length meter m
Mass kilogram kg
Time second s
Electric current ampere A
Thermodynamic temperature kelvin K
Amount of a substance mole mol
Luminous intensity candela cd
TABLe 1.2
Examples of SI
Derived Units.
Derived Quantity Name Symbol
Area square meter m2
Volume cubic meter m3
Speed, velocity meter per second ms
Acceleration meter per second squared ms2
Wave number reciprocal meter m21
Mass density kilogram per cubic meter kgm3
Specific volume cubic meter per kilogram m3
kg
Current density ampere per square meter Am2
Magnetic field strength ampere per meter Am
Luminance candela per square meter cdm2
TABLe 1.3
SI Derived Units
with Special
Names and
Symbols.
Expression in terms
of other SI units
Plane angle radian rad —
Solid angle steradian sr —
Frequency hertz Hz —
Force newton N —
Pressure, stress pascal Pa Nm2
Energy, work, quantity of heat joule J N ? m
Power, radiant flux watt W Js
Electric charge, quantity
of electricity
coulomb C —
Electric potential difference, volt V W/A
electromotive force
Capacitance farad F CV
Electric resistance ohm V VA
Electric conductance siemens S AV
Magnetic flux weber Wb V ? s
Magnetic flux density tesla T Wbm2
Inductance henry H WbA
Celsius temperature degrees Celsius 8C —
Luminous flux lumen lm cd ? sr
Illuminance lux lx lmm2

1.2 International System of units 3
TABLe 1.4
Examples of SI
Derived Units
with Names
and Symbols
(Including
Special Names
and Symbols.)
Dynamic viscosity Pascal second Pa?s
Moment of force newton meter N?m
Surface tension newton per meter Nm
Angular velocity radian per second rads
Angular acceleration radian per second squared rads2
Heat flux density, irradiance watt per square meter Wm2
Thermal conductivity watt per meter kelvin W(m?K)
Energy density joule per cubic meter Jm3
Electric field strength volt per meter Vm
Electric charge density coulomb per cubic meter Cm3
Electric flux density coulomb per square meter Cm2
Permittivity farad per meter Fm
Permeability henry per meter Hm
Exposure (X- and -rays) coulomb per kilogram Ckg
TABLe 1.5
Metric Prefixes.
Prefix Symbol Magnitude
yocto y 10224
zepto z 10221
atto a 10218
femto f 10215
pico p 10212
nano n 1029
micro 1026
milli m 1023
centi c 1022
deci d 1021
deka da 101
hecto h 102
kilo k 103
mega M 106
giga G 109
tera T 1012
peta P 1015
exa E 1018
zetta Z 1021
yotta Y 1024
TABLe 1.6
Units Outside
the SI That Are
Accepted for
Use with the SI
System.
Name Symbol Value in SI Units
Minute (time) min 1 min 5 60 s
Hour h 1 h 5 60 min 5 3600 s
Day d 1 d 5 24 h 5 86,400 s
Degree (angle) ° 1° 5 (/180) rad
Minute (angle) 9 19 5 (160)° 5 (10,800) rad
Second (angle) 0 10 5 (160)9 5 (648,000) rad
Liter L 1 L 5 1 dm3
5 1023
m3
Metric ton t 1 t 5 1000 kg
Neper Np 1 Np 5 20 log10(e) dB 5 20ln(10) dB
Bel B 1 B 5 (12) ln(10) Np, 1 dB 5 0.1 B
Electronvolt eV 1 eV 5 1.60218 3 10219
J
Unified atomic mass unit u 1 u 5 1.66054 3 10227
kg
Astronomical unit ua 1 ua 5 1.49598 3 1011
m

A second is the duration of 9,192,631,770 periods of the radiation corresponding to
the transition between the two hyperfine levels of the ground state of the cesium 133 atom.
An ampere is the constant current which, if maintained in two straight parallel con-
ductors of infinite length, of negligible circular cross section, and placed 1 meter apart in
vacuum, would produce between these conductors a force equal to 2 3 1027
newtons per
meter of length.
A kelvin, is 1273.16 of the thermodynamic temperature of the triple point of water.
A mole is the amount of substance of a system that contains as many elementary
entities as there are atoms in 0.012 kilogram of carbon 12; its symbol is mol.When the mole
is used, the elementary entities must be specified; they may be atoms, molecules, ion, elec-
trons, other particles, or specified groups of such particles.
The candela is the luminous intensity,in a given direction,of a source that emits mono-
chromatic radiation of frequency 540 3 1012
hertz (Hz) and that has the radiant intensity in
that direction of 1683 watt per steradian.
1.3 Charge, Voltage, Current, and power
1.3.1 ElEctric chargE
Atoms are the basic building blocks of matter.The nucleus of atoms consists of protons and
neutrons. Electrons orbit around the nucleus. Protons are positively charged, and electrons
are negatively charged, while neutrons are electrically neutral.The amount of charge on the
proton is given by
e 5 1.60217662 3 10219
C
Here, the unit for charge is in coulombs (C).
2e 5 21.60217662 3 10219
C
Notice that the charge is quantized as the integral multiple of e. Since there are equal
numbers of protons and electrons in an atom, it is electrically neutral. When a plastic is
rubbed by fur, some electrons from the fur are transferred to the plastic. Since the fur lost
electrons and the plastic gained them, the former is positively charged and the latter neg-
atively charged. When the fur and the plastic are placed close together, they attract each
other. Opposite charges attract, and like charges repel. However, since the electrons and
protons are not destroyed, the total amount of charge remains the same. This is called the
conservation of charge.
1.3.2 ElEctric FiEld
According to Coulomb’s law, the magnitude of force between two charged bodies is pro-
portional to the charges Q and q and inversely proportional to the distance squared; that is,
F 5
1
4«
Qq
r2
(1.1)
Here, « is permittivity of the medium.The permittivity of free space, «0, is given by
«0 5
1
4c2
1027
(Fym) 5 8.8541878176 3 10212
(Fym) (1.2)
Here, c is the speed of light in the vacuum, given by c 5 299,792,458 ms ≈ 3 3 108
ms.
The unit for permittivity is farads per meter (Fm).The direction of the force coincides with
the line connecting the two bodies. If the charges have the same polarity, the two bodies
F 5
1
4«
Qq
r2
(1.1)
«0 5
1
4c2
1027
(Fy
(Fy
(F m) 5 8.8541878176 3 10212
(Fy
(Fy
(F m) (1.2)

1.3 Charge, Voltage, Current, and Power 5
repel each other. On the other hand, if the charges have the opposite polarity, they attract
each other.
If a positive test charge with magnitude q is brought close to a positive point charge
with magnitude Q, the test charge will have a repulsive force. The magnitude of the force is
inversely proportional to the distance squared between the point charge and the test charge.
The presence of the point charge creates a field around it,where charged particles experience
force. This is called an electric field, which is defined as the force on a test charge q as the
charge q decreases to zero; that is,
E 5 lim
qS0
F
q
(Vm) (1.3)
The electric field is a force per unit charge. The electric field E is a vector quantity whose
direction is the same as that of the force. Figure 1.1 shows the electric field for a positive
point charge and charged parallel plates.
E 5 lim
qS0
F
q
(Vm) (1.3)
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
d
E
E
B
A
rA rB
Q
2Q
Q
(a) (b)
Figure 1.1
Electric field for
(a) a point charge and
(b) parallel plates.
If an object with charge q is placed in the presence of electric field E, the object will
experience a force as follows:
F 5 qE (1.4)
For a positive point charge Q, the electric field is given by
E 5
1
4«
Q
r2
ar (1.5)
where ar is a unit vector in the radial direction from the positive point charge Q. For paral-
lel plates with area S per plate, distance d between the plates, the electric field is constant
within the plates and the magnitude of the electric field is given by
E 5
Q
«S
(1.6)
The direction of the field is from the plate with positive charges to the plate with neg-
ative charges, as shown in Figure 1.1(b).
1.3.3 VoltagE
If a positive test charge dq is moved against the electric field created by a positive charge,
an external agent must apply work to the test charge. Let dwAB be the amount of the work
F 5 qE (1.4)
E 5
1
4«
Q
r2
ar (1.5)
E 5
Q
«S
(1.6)

needed to move the test charge from B (initial) to A (final). Here, dwAB is the potential
energy in joules. Then, the potential difference between points A and B is defined as the
work done per unit charge against the force; that is,
vAB 5 vA 2 vB 5
dwAB
dq
(JC) (1.7)
The unit for the potential difference is joules per coulomb, which is also called a
volt (V):
1 V 5 1 JC
The potential difference between A and B is called voltage. The potential difference
between points A and B is given by
vAB 5 vA 2 vB 5 2#
A
B
E ? d/ (1.8)
The negative sign implies that moving against the electric field increases the potential.
For a positive point charge Q at origin with an electric field given by Equation (1.5), the
potential difference between two points A and B with distances rA and rB, respectively, from
Q is given by
vAB 5 vA 2 vB 5 2#
rA
rB
1
4«
Q
r2
dr 5 2
Q
4« 1
21
r 2*
rA
rB
5
Q
4« 1
1
rA
2
1
rB
2 V (1.9)
Notice that the integral of 1/r2
is 21/r. If rB is infinity, the potential difference is
vAB 5 vA 2 vB 5 vA 5
Q
4«rA
V (1.10)
The potential is zero at infinity. This is a reference potential. For the parallel plates
shown in Figure 1.1(b), the potential difference between A and B is
v 5 Ed 5
Q
«S
d (1.11)
If the potential at B is set at zero (vB 5 0), the potential at point A is given by
vA 5
dwA
dq
(J/C) (1.12)
or simply
v 5
dw
dq
(J/C) (1.13)
The potential difference v is called voltage. A battery is a device that converts chem-
ical energy to electrical energy.When a positive charge is moved from the negative terminal
to the positive terminal through the 12-V battery,the battery does 12 joules of work on each
unit charge.The potential energy of the charge increases by 12 joules.The battery provides
energy to the rest of the circuit.
vAB 5 vA 2 vB 5
dwAB
dq
(JC) (1.7)
vAB 5 vA 2 vB 5 2#
A
B
E ? d/ (1.8)
vAB 5 vA 2 vB 5 2#
rA
rA
r
rB
1
4«
Q
r2
dr 5 2
Q
4« 1
21
r 2*
rA
rA
r
rB
5
Q
4« 1
1
rA
rA
r
2
1
rB
rB
r 2 V (1.9)
vAB 5 vA 2 vB 5 vA 5
Q
4«rA
rA
r
V (1.10)
v 5 Ed 5
Q
«S
d (1.11)
vA 5
dwA
dq
(J/C) (1.12)
v 5
dw
dq
(J/C) (1.13)

1.3.4 currEnt
In the absence of an electric field, the free electrons in the conduction band of conduct-
ors such as copper wire make random movements. The number of electrons crossing a
cross-sectional area of the copper wire from left to right will equal the number of elec-
trons crossing the same cross-sectional area from right to left.The net number of electrons
crossing this area will be zero. When an electric field is applied along the copper wire, the
negatively charged electrons will move toward the direction of higher potential. The cur-
rent is defined as the total amount of charge q passing through a cross-sectional area in
t seconds; that is,
I 5
q
t
(1.14)
The unit for the current is coulombs per second (C/s) or amperes (A). If the amount
of charge crossing the area changes with time, the current is defined as
i(t) 5
dq(t)
dt
(1.15)
The direction of current is defined as the direction of positive charges. Since the
charge carriers inside the conductors are electrons, the direction of electrons is opposite to
the direction of the current. Figure 1.2 shows the directions of the electric field, current, and
electron inside a conductor.
i(t) 5
dq(t)
dt
(1.15)
E
I
e
Figure 1.2
The directions of
E, I, and e.
The charge transferred between time t1 and t2 can be obtained by integrating the cur-
rent from t1 and t2; that is,
q 5 #
t2
t1
i()d (1.16)
q 5 #
t2
t2
t
t1
i()d
d
d (1.16)
The charge flowing into a circuit element for t $ 0 is given by
q(t) 5 2 3 1023
(1 2 e21000t
) coulomb
Find the current flowing into the element for t $ 0.
i(t) 5
dq(t)
dt
5 2 3 1023
3 1000e21000t
A 5 2e21000t
A for t $ 0
exAmpLe 1.1
I 5
q
t
(1.14)

The current flowing into a circuit element is given by
i(t) 5 5 sin(210t) mA
for t $ 0. Find the charge flowing into the device for t $ 0.Also, find the total charge
entered into the device at t 5 0.05 s.
q(t) 5 #
t
0
i()d 5
5 3 1023
210
f1 2 cos(210t)g
5 7.9577 3 1025
f1 2 cos(210t)g coulomb
At t 5 0.05 s, we have
q(0.05) 5 1.5915 3 1024
f1 2 cos(210 3 0.05)g 5 1.5915 3 1024
coulombs
Exercise 1.1
The charge flowing into a circuit element for t $ 0 is given by
q(t) 5 4 3 1023
e22000t
coulomb
Find the current flowing into the element for t $ 0.
Answer:
i(t) 5
dq(t)
dt
5 28e22000t
A for t $ 0
ExAmplE 1.2
Exercise 1.2
The current flowing into a circuit element is given by
i(t) 5 5 cos(210t) mA
for t $ 0. Find the charge flowing into the device for t $ 0.Also, find the total charge
entered into the device at t 5 0.0125 s.
Answer:
q(t) 5 #
t
0
i()d 5
5 3 1023
210
sin(210t) 5 7.9577 3 1025
sin(210t) coulombs
q(0.0125) 5 7.9577 3 1025
sin(210 3 0.0125) 5 5.6270 3 1025
coulombs

1.3.5 PowEr
The battery provides a constant potential difference (voltage) of v volts from the negative
terminal to the positive terminal. When a positive charge dq is moved from the negative
terminal to the positive terminal through the battery, the potential energy is increased by
dq v 5 dw.When the positive charge dq moves through the rest of the circuit from the posi-
tive terminal to the negative terminal,the potential energy is decreased by the same amount
(dq v).The rate of potential energy loss is given by
p 5
dw
dt
5
dqv
dt
5 iv (1.17)
The rate of energy loss is defined as power. Equation (1.17) can be rewritten as
dw 5 dqv 5 pdt (1.18)
The energy is the product of power and time. If Equation (1.18) is integrated as a
function of time, we get
w(t) 5 #
t
2`
p()d (1.19)
According to Equation (1.19), the energy is the integral of power.As shown in Equa-
tion (1.17), power is the derivative of energy. Taking the derivative of Equation (1.19),
we obtain
p(t) 5
dw(t)
dt
(1.20)
If the voltage and the current are time-varying, the power is also time-varying. If the
voltage and current are expressed as a function of time, Equation (1.17) can be written as
p(t) 5 i(t)v(t) (1.21)
The power given by Equation (1.21) is called instantaneous power. According to
Equation (1.21), instantaneous power is the product of current and voltage as a function
of time. In the passive sign convention, if the direction of current is from the positive ter-
minal of a device, through the device, and to the negative terminal of the device [as shown
in Figure 1.3(a)], the power is positive. On the other hand, if the current leaves the positive
terminal of a device, flows through the rest of the circuit, and enters the negative terminal
of the device [as shown in Figure 1.3(b)], the power is negative.
If power is positive [i.e., p(t) . 0], the element is absorbing power. On the other
hand, if power is negative, the element is delivering (supplying) power. In a given circuit,
the total absorbed power equals the total delivered or supplied power. This is called con-
servation of power.
p 5
dw
dt
5
dqv
dt
5 iv (1.17)
dw 5 dqv 5 pdt (1.18)
w(t) 5 #
t
2`
p()d
d
d (1.19)
p(t) 5 i(t)v(t) (1.21)
i(t)
1
v(t)
2
(a)
i(t)
1
v(t)
2
(b)
Figure 1.3
(a) Power is positive.
(b) Power is negative.
Let the voltage across an element be v(t) 5 100 cos(260t) V, and the current though the element from positive
terminal to negative terminal be i(t) 5 5 cos(2π60t) A for t $ 0. Find the instantaneous power p(t) and plot p(t).
exAmpLe 1.3
continued
p(t) 5
dw(t)
dt
(1.20)

p(t) 5 i(t) v(t) 5 5 cos(260t) 3 100 cos(260t) 5 500 cos2
(260t)
5 250 1 250 cos(2 3 120t) W
The power p(t) is shown in Figure 1.4. Since p(t) $ 0 for all t, the element is not
delivering power any time. On average, the element absorbs 250 W of power.
20.015 20.01 20.005 0 0.005 0.01 0.015
0
100
200
300
400
500
t (s)
p(t)
(W)
Power
Figure 1.4
Plot of p(t).
exercise 1.3
Let the voltage across an element be v(t) 5 100 cos(2p60t) V and the current though the element from positive
terminal to negative terminal be i(t) 5 6 sin(2p60t) A for t $ 0. Find the instantaneous power p(t) and plot p(t).
p(t) 5 i(t) v(t) 5 6 sin(260t) 3 100 cos(260t) 5 300 sin(2120t) W.
The power p(t) is shown in Figure 1.5. Since p(t) . 0 half of the time and p(t) , 0 the
other half of the time, the element absorbs power for 1240 s, then delivers power for
the next 1240 s, and then repeats the cycle. On average, the element does not absorb
any power.
20.015 20.01 20.005 0 0.005 0.01 0.015
2300
2200
2100
0
100
200
300
t (s)
p(t)
(W)
Power
Figure 1.5
Power p(t).
Example 1.3 continued
1.4 Independent Sources
A voltage source with voltage Vs provides a constant potential difference to the circuit con-
nected between the positive terminal and the negative terminal. The circuit notations for
the voltage source are shown in Figure 1.6.
If a positive charge Dq is moved from the negative terminal to the positive terminal
through the voltage source, the potential energy of the charge is increased by DqVs. If a
negative charge with magnitude Dq is moved from the positive terminal to the negative
terminal through the voltage source,the potential energy of the charge is increased by DqVs.
A battery is an example of a voltage source.
2
1
1
2
Vs Vs
(a) (b)
Figure 1.6
Circuit symbols for
voltage sources.

A current source with current Is provides a constant current of Is amperes to the
circuit connected to the two terminals. The circuit notation for the current source is shown
in Figure 1.7.
1.4.1 dirEct currEnt SourcES and altErnating
currEnt SourcES
If the voltage from the voltage source is constant with time, the voltage source is called the
direct current (dc) source. Likewise, if the current from the current source is constant with
time, the current source is called the direct current (dc) source.
If the voltage from the voltage source is a sinusoid, as shown in Figure 1.8, the
voltage source is called alternating current (ac) voltage source. Likewise, if the current
from the current source is a sinusoid, the current source is called alternating current (ac)
current source. A detailed discussion of ac signals is given in Chapter 9. The circuit nota-
tion for an ac voltage source and ac current source are shown in Figure 1.9. The phase is
given in degrees.The circuit notation for dc voltage shown in Figure 1.6(a) and the circuit
notation for dc current shown in Figure 1.7 are also used for ac voltage and ac current,
respectively.
22 T 21.5 T 2T 20.5 T 0 0.5 T T 1.5 T 2 T
2Vm
0
Vm
t (s)
v(t)
(V)
Figure 1.8
Plot of a cosine
wave with period T,
amplitude Vm, and
phase zero.
Is
Figure 1.7
A circuit symbol for
the current source.
When dc voltage sources are connected in series, they can be com-
bined into a single equivalent dc voltage source, as shown in Figure 1.10,
where V3 5 V1 1 V2 5 4.5 V 1 7.5 V 5 12 V. If there are other com-
ponents, such as the resistors between V1 and V2 in the circuit shown
in Figure 1.10, the voltage sources can be combined, so long as all the
components are connected in series. Resistors are discussed further in
Chapter 2.
When dc current sources are connected in parallel, they can
be combined into a single equivalent dc current source, as shown in
Figure 1.11, where I3 5 I1 1 I2 5 3 A 1 5 A 5 8 A. If other com-
ponents such as resistors are connected in parallel to I1 and I2 in the
circuit shown in Figure 1.11, the current sources can be combined,
so long as all the components are connected in parallel between the
same points.
Figure 1.10
An equivalent voltage source.
4.5 Vdc
7.5 Vdc
2
1 V1
12 Vdc
2
1 V3
2
1 V2
1
2
Vs
100 Vac
0 Vdc AC PHASE 5 120
(a)
1
2
Is
10 Aac
0 Adc AC PHASE 5 30
(b)
Figure 1.9
Circuit symbols
for (a) ac voltage
source; (b) ac
current source.

I1
3 Adc
I2
5 Adc
I3
8 Adc
Figure 1.11
An equivalent
current source.
Redraw the circuit shown in Figure 1.12 with one voltage source and one current source, without affecting the
voltages across and currents through the resistors in the circuit.
example 1.4
I1
3 mA
2 mA
I2
1
2
1
2
5 V
3 V
V1
R1
3 kV
R2
4 kV
R3
6 kV
V2
0
Figure 1.12
Circuit for
EXAMPLE 1.4.
Since V1 and V2 are part of a single wire, they can be combined into the single voltage
source V3. Since V2 has the same polarity as V1, the value of V3 is given by
V3 5 V1 1 V2 5 5 V 1 3 V 5 8 V
Since I1 and I2 are connected between the same points in the circuit, they can be
combined into the single current source I3. Since I2 has the same polarity as I1, the value
of I3 is given by
I3 5 I1 1 I2 5 3 mA 1 2 mA 5 5 mA
The equivalent circuit, with one voltage source and one current source, is shown in
Figure 1.13.
continued

I1
5 mA
1
2
8 V
V1
R1
3 kV
R2
4 kV
R3
6 kV
0
Figure 1.13
A circuit with one
current source and
one voltage source.
exercise 1.4
Redraw the circuit shown in Figure 1.14 with one voltage source and one current source, without affecting the
voltages across and currents through the resistors in the circuit.
I1
3 mA
I2
2 mA
1 2
7 V
V1
R1
2 kV
R2
2 kV
R3
4 kV
2 V
V2
1
2
0
Figure 1.14
Circuit for
EXERCISE 1.4.
I3
1 mA
1
2
5 V
V3
R1
2 kV
R2
2 kV
R3
4 kV
0
Figure 1.15
A circuit with one
current source and
one voltage source.
Answer:
The equivalent circuit with one voltage source and one current source is shown in
Figure 1.15.

An ac voltage waveform can be represented as
v(t) 5 Vm cos1
2t
T
1 2V (1.22)
Here,Vm is the amplitude (peak value) of the cosine wave,T is the period of the cosine
wave, and f is the phase of the cosine wave.The peak-to-peak amplitude is 2Vm.The cosine
wave repeats itself every T seconds. The number of periods per second, called frequency
and denoted by f, is given by
f 5
1
T
Hz (1.23)
The unit for the frequency is 1/s and is called hertz (Hz). In terms of the frequency in
hertz, the ac voltage waveform can be written as
v(t) 5 Vm cos(2ft 1 ) V (1.24)
Since the angle changes by 2 radians in one period, and there are f periods in
1 second, the changes in angle in 1 second is given by
5 2f 5
2
T
(1.25)
The parameter is called the angular velocity of the cosine wave and has a unit of
radians per second (rads). In terms of radian frequency , the cosine wave becomes
v(t) 5 Vm cos(t 1 ) (1.26)
The ac current waveform can be written as
i(t) 5 Im cos1
2t
T
1 25 Im cos(2ft 1 ) 5 Im cos(t 1 ) A (1.27)
If the cosine wave shown in Figure 1.8 is shifted to the right by T4, we get
v(t) 5 Vm cos321t 2
T
42
T
45 Vm cos1
2
T
t 2

225 Vm sin1
2
T
t2
5 Vm sin(2ft) 5 Vm sin(t)
(1.28)
The sine wave given by Equation (1.28) is shown in Figure 1.16.
v(t) 5 Vm
Vm
V cos1
2t
T
1 2V (1.22)
f 5
1
T
Hz (1.23)
v(t) 5 Vm
Vm
V cos(2ft
ft
1 ) V (1.24)
5 2f
f
5
2
T
(1.25)
v(t) 5 Vm
Vm
V cos(t 1 ) (1.26)
i(t) 5 Im
Im
I cos1
2t
T
1 25 Im
Im
I cos(2ft
ft
1 ) 5 Im
Im
I cos(t 1 ) A (1.27)
22 T 21.5 T 2T 20.5 T 0 0.5 T T 1.5 T 2 T
2Vm
0
Vm
t (s)
v(t)
(V)
Figure 1.16
Plot of a sine wave
with period T
and amplitude Vm.
Notice that if the phase of the cosine is shifted by 290 degrees, the cosine wave
becomes a sine wave.
v(t) 5 Vm
Vm
V cos321t 2
T
42
T
45 Vm
Vm
V cos1
2
T
t 2

225 Vm
Vm
V sin1
2
T
t2
5 Vm
Vm
V sin(2ft
ft
) 5 Vm
Vm
V sin(t)
(1.28)

1.5 dependent Sources 15
1.5 Dependent Sources
The voltage sources and current sources discussed previously are called independent
sources because they are stand-alone sources that provide power to the external circuit
connected to the sources. Usually, the independent sources convert one form of energy to
electrical energy. For example, a battery converts chemical energy into electrical energy.
21 27/8 26/8 25/8 24/8 23/8 22/8 21/8 0 1/8 2/8 3/8 4/8 5/8 6/8 7/8 1
24
22
0
2
4
6
8
10
t (ms)
v(t)
(V)
Figure 1.17
A sinusoid for
EXAMPLE 1.5.
Since the period is T 5 1 ms, the frequency is f 5 1T 5 11ms 5 1000 Hz 5 1 kHz.
The radian frequency is 2f 5 21000 5 6283.1853 rad/s.The difference between the
maximum and minimum is 8 2 (24) 5 12 V, which is the peak-to-peak amplitude.The
peak value of the amplitude is Vm 5 12V2 5 6 V.The average amplitude is (8 2 4)2 5
2 V, which is the dc component.The cosine wave is shifted to the left by T8 ms, which is
4 rad 5 458.Therefore, the equation is given by
v(t) 5 2 1 6 cos(21000t 1 458) V
Find the equation of the sinusoidal signal shown in Figure 1.17.
exAmpLe 1.5
exercise 1.5
Plot v(t) 5 4 1 2 cos(22000t 2 728) V.
Answer:
The signal v(t) is shown in Figure 1.18.
25 24 23 22 21 0 1 2 3 4 5
3 1024
2
3
4
5
6
t (s)
v(t)
(V)
Figure 1.18
The plot of v(t).

A solar cell converts energy from the Sun into electrical energy. A wind turbine converts
wind energy into electrical energy. The amount of energy supplied from the source to the
circuit per unit time is the power of the source. Dependent sources do not have the ability
to convert one form of energy into electrical energy.The voltage or current of the depend-
ent sources depend solely on the controlling voltage or controlling current.The dependent
sources are used to model integrated circuit (IC) devices.
Depending on whether the dependent source is a voltage source or a current source,
and whether the dependent source is controlled by a voltage or a current, there are four
different dependent sources.The four types of dependent sources are:
Voltage-controlled voltage source (VCVS)
Voltage-controlled current source (VCCS)
Current-controlled voltage source (CCVS)
Current-controlled current source (CCCS)
These four types of dependent sources are discussed next.
1.5.1 VoltagE-controllEd VoltagE SourcE (VcVS)
The voltage on the VCVS is proportional to the controlling voltage, which is the voltage in
another part of the circuit. For example, the controlling voltage can be the voltage across a
circuit element in another part of the circuit. Let vd be the controlled voltage and vc be the
controlling voltage.Then, we have
vd 5 kv vc
where kv is the unitless (V/V) proportionality constant. Figure 1.19 shows the circuit symbol
for VCVS.
1.5.2 VoltagE-controllEd currEnt SourcE (VccS)
The current on the VCCS is proportional to the controlling voltage. Let id be the controlled
current and vc be the controlling voltage.Then, we have
id 5 gm vc
where gm is the conductance in siemens (S). Figure 1.20 shows the circuit symbol for VCCS.
1.5.3 currEnt-controllEd VoltagE SourcE (ccVS)
The voltage on the CCVS is proportional to the controlling current, the current in another
part of the circuit. For example, the controlling current can be the current through a circuit
element in another part of the circuit. Let vd be the controlled voltage and ic be the con-
trolling current.Then, we have
vd 5 rm ic
where rm is the resistance in ohms (V). Figure 1.21 shows the circuit symbol for CCVS.
1.5.4 currEnt-controllEd currEnt SourcE (cccS)
The current on the CCCS is proportional to the controlling current. Let id be the controlled
current and ic be the controlling current.Then, we have
id 5 ki ic
where ki is the unitless (A/A) proportionality constant. Figure 1.22 shows the circuit symbol
for CCCS.
kv vc
1
2
Figure 1.19
Circuit symbol
for VCVS.
gm vc
Figure 1.20
Circuit symbol
for VCCS.
rm ic
1
2
Figure 1.21
Circuit symbol
for CCVS.
ki ic
Figure 1.22
Circuit symbol
for CCCS.

1.6 elementary Signals 17
Vs
va
0.2 va
2 V
R3
6 V
R1
2 V
R4
5 V
R2
3 V
1
2
1
2
0
Figure 1.23
Circuit for
EXAMPLE 1.6.
The current through the VCCS in the direction indicated in Figure 1.23 ( T ) is
0.2 va 5 0.2 (AV) 3 0.9851 V 5 0.1970 A
In the circuit shown in Figure 1.23, the controlling voltage, which is the voltage across R2, is va 5 0.9851 V.
Find the controlled current through the VCCS.
exAmpLe 1.6
exercise 1.6
In the circuit shown in Figure 1.24, the controlling current, which is the current through R2, is ia 5 0.5625 A.
Find the controlled voltage across the CCVS.
Vs
ia
2 ia
3 V
R3
R1
2 V 4 V
R4
2 V
R2
3 V
1
2
1
2
0
Figure 1.24
Circuit for
EXERCISE 1.6.
Answer:
2 ia 5 1.125 V.
1.6 elementary Signals
Several elementary signals that will be useful in later chapters are presented in this section.
1.6.1 dirac dElta Function
A rectangular pulse with height 1/ and width is shown in Figure 1.25.The pulse is centered
at 2/2 and the area of the pulse is 1.The rectangular pulse can be written as

f(t) 5
1

rect1t 1

2

2 (1.29)
f(
f(
f t) 5
1

rect1t 1

2

2 (1.29)
If the pulse width is decreased to zero, the height of the pulse is
increased to infinity while maintaining the area at 1.The limiting form of
a rectangular pulse shown in Figure 1.25 as →0 is defined as the Dirac
delta function (or delta function) and is denoted by (t); that is,
(t) 5 lim
S0
1

rect1
t 1

2
2 (1.30)
The mathematical symbol for the Dirac delta function is shown in
Figure 1.26.
(t) 5 lim
S0
1

rect1
t 1

2
2 (1.30)
2
1
0
0
(t)
t (s)
Figure 1.26
Symbol for the Dirac
delta function.
f(t)
t (s)
0
2
1/
Figure 1.25
A rectangular pulse.
5
3
4
2
1
0
0 1 2 3 4 5
f(t)
t (s)
25 24 23 22 21
Figure 1.27
Plot of f(t).
The Dirac delta function is located at t 5 1 and has an area of 4.The signal f(t) is shown
in Figure 1.27.
Plot f(t) 5 4 d(t 2 1).
exAmpLe 1.7
If the rectangular pulse given by Equation (1.29) is shifted to the right by 2,it becomes
g(t) 5
1

rect 1
t
2 (1.31)
The Dirac delta function can also be defined as
(t) 5 lim
S0
1

rect 1
t
2 (1.32)
g(t) 5
1

rect 1
t
2 (1.31)
(t) 5 lim
S0
1

rect 1
t
2 (1.32)

When a continuous signal f(t) is multiplied by (t 2 a) and integrated from 2` to `,
we obtain f(a); that is,
#
`
2`
f(t)(t 2 a)dt 5 f(a) (1.33)
This result is called the sifting property of the delta function because it sifts out a
single value of f(t), f(a), at the location of the delta function (t 5 a). To prove the sifting
property, we replace (t 2 a) with
(t 2 a) 5 lim
S0
1

rect1
t 2 a
2
Then, the integral becomes
#
`
2`
f(t)(t 2 a)dt 5 lim
S0
1
#
`
2`
f(t)rect1
t 2 a
2dt 5 lim
S0
1
#
a1

2
a2

2
f(t)rect1
t 2 a
2dt
As → 0, f(t) → f(a) for (a 2 2) , t , (a 1 2).Thus, the integral becomes
#
`
2`
f(t)(t 2 a)dt 5 lim
S0
1
#
a1

2
a2

2
f(a) 3 1dt 5 lim
S0
1

f(a) 5 f(a)
1.6.2 StEP Function
The unit step function u(t) is the integral of the Dirac delta function (t). If Equation (1.29)
is integrated, we obtain
#
t
2`
f()d 5
1
#
t
2`
rect1
1

2
2d 5
0, t , 2
5t

1 1, 2 # t , 0
1, 0 # t
(1.34)
exercise 1.7
Plot f(t) 5 22 d(t 1 3).
4
2
0
22
24
0 1 2 3 4 5
f(t)
t (s)
25 24 23 22 21
Answer:
f(t) is shown in Figure 1.28.
Figure 1.28
Plot of f(t).
#
`
2`
f(
f(
f t)(t 2 a)dt 5 f(
f(
f a) (1.33)
#
t
2`
f(
f(
f )d
d
d 5
1
#
t
2`
rect1
1

2
2d
d
d 5
0, t , 2
5t

1 1, 2 # t , 0
1, 0 # t
(1.34)

The unit step function is defined as the limiting form of Equation (1.34). In the limit
as → 0, Equation (1.34) becomes
u(t) 5 50, t , 0
1, 0 # t
(1.35)
Notice that at t 5 0, u(t) 5 1. The unit step function defined by Equation (1.35) is
shown in Figure 1.29.
2
1
0
0 1 2 3 4 5
u(t)
t (s)
25 24 23 22 21
Figure 1.29
A unit step function.
2
1
0
0 1 2 3 4 5
f(t)
t (s)
25 24 23 22 21
Figure 1.30
Plot of f(t).
Notice that u(t) 5 1 for t $ 0 and zero for t , 0, and u(t 2 2) 5 1 for t $ 2 and zero
for t , 2.Thus, u(t) 2 u(t 2 2) 5 0 for t $ 2, and u(t) 2 u(t 2 2) 5 1 for 0 # t , 2, and
zero for t , 0.The signal f(t) is shown in Figure 1.30.
Plot f(t) 5 u(t) 2 u(t 2 2).
exAmpLe 1.8
exercise 1.8
Plot f(t) 5 22 u(t 1 3).
25 24 23 22 21 0 1 2 3 4 5
23
22
21
0
1
2
t (s)
f
(
t
)
Answer:
Figure 1.31
Plot of f(t).
u(t) 5 50, t , 0
1, 0 # t
(1.35)

If Equation (1.31) is integrated, we obtain
#
t
2`
g()d 5
1
#
t
2`
rect1

2d 5
5
0, t ,
2
2
t

1
1
2
,
2
2
# t ,

2
1,

2
# t
(1.36)
If u(t) is defined as the limiting form of Equation (1.36) as → 0, we obtain
u(t) 5
5
0, t , 0
1
2
, t 5 0
1, 0 # t
(1.37)
In this text, the definition of u(t) given by Equation (1.35) is used. Since u(0) 5 1, it
does include voltages and currents at t 5 0.
1.6.3 ramP Function
A unit ramp function is defined by
r(t) 5 t u(t) (1.38)
The unit ramp function is shown in Figure 1.32.The unit ramp function is the integral
of the unit step function:
r(t) 5 #
t
2`
u()d (1.39)
The derivative of the unit ramp function is the unit step function.
u(t) 5
dr(t)
dt
(1.40)
#
t
2`
g()d
d
d 5
1
#
t
2`
rect1

2d
d
d 5
5
0, t ,
2
2
t

1
1
2
,
2
2
# t ,

2
1,

2
# t
(1.36)
u(t) 5
5
0, t , 0
1
2
, t 5 0
1, 0 # t
(1.37)
25 24 23 22 21 0 1 2 3 4 5
0
1
2
3
4
5
t (s)
r
(
t
)
Figure 1.32
A unit ramp function.
r(t) 5 t u(t) (1.38)
r(t) 5 #
t
2`
u()d
d
d (1.39)
u(t) 5
dr(t)
dt
(1.40)

For t 0, f(t) 5 0.
For 0 # t 1, f(t) is a linear line with slope of 2.
For 6 # t, f(t) 5 0.
The waveform f(t) is shown in Figure 1.33.
Plot f(t) 5 2tu(t) 2 4(t 2 1)u(t 2 1) 1 4(t 2 3)u(t 2 3) 2 4(t 2 5)u(t 2 5) 1 2(t 2 6)u(t 2 6).
exAmpLe 1.9
t (s)
f(t)
21 0 1 2 3 4 5 6 7
23
22
21
0
1
2
3
Figure 1.33
Waveform f(t).
exercise 1.9
Plot f(t) 5 tu(t) 2 (t 2 1)u(t 2 1) 2 (t 2 3)u(t 2 3) 1 (t 2 4)u(t 2 4).
21 0 1 2 3 4 5
0
1
2
t (s)
f(t)
Figure 1.34
Waveform f(t).
Answer:
The waveform is shown in Figure 1.34.
Find the equation of the waveform shown in Figure 1.35.
exAmpLe 1.10
continued

For t , 0, f(t) 5 0.
For 0 # t , 1, f(t) is a linear line with slope of 3.Thus, f(t) 5 3tu(t).
For 1 # t , 3, f(t) is a linear line with slope of 23.To change the slope from 3 to 23,
we need to add 2 6(t 2 1)u(t 2 1).At this point, we have f(t) 5 3tu(t) 2 6(t 2 1)u(t 2 1).
For 3 # t , 6, f(t) is a linear line with slope of 1.To change the slope from 23 to 1,
we need to add 4(t 2 3)u(t 2 3).At this point, we have f(t) 5 3tu(t) 2 6(t 2 1)u(t 2 1) 1
4(t 2 3)u(t 2 3).
t (s)
f(t)
21 0 1 2 3 4 5 6 7
23
22
21
0
1
2
3
Figure 1.35
Waveform f(t).
For 6 # t, f(t) 5 0.To change the slope from 1 to 0, we need to add 2(t 2 6)u(t 2 6).
Thus, we have the final equation given by
f(t) 5 3tu(t) 2 6(t 2 1)u(t 2 1) 1 4(t 2 3)u(t 2 3) 2 (t 2 6)u(t 2 6).
1.6.4 ExPonEntial dEcay
A signal that decays exponentially can be written as
f(t) 5 e-at
u(t), a . 0 (1.41)
The signal f(t) for a 5 0.5 is shown in Figure 1.37.
f(
f(
f t) 5 e-at
u(t), a . 0 (1.41)
21 0 1 2 3
0
1
t (s)
f(t)
exercise 1.10
Find the equation of the waveform shown in Figure 1.36.
Answer:
f(t) 5 tu(t) 2 2(t 2 1)u(t 2 1) 1 (t 2 2)u(t 2 2).
Figure 1.36
Waveform for
EXERCISE 1.10.

A damped cosine and damped sine, respectively, can be written as
f(t) 5 e2at
cos(bt)u(t), a . 0 (1.42)
f(t) 5 e2at
sin(bt)u(t), a . 0 (1.43)
A damped cosine signal is shown in Figure 1.38 for a 5 0.5 and b 5 4.
f(
f(
f t) 5 e2at
cos(bt)u(t),
),
), a . 0 (1.42)
f(
f(
f t) 5 e2at
sin(bt)u(t),
),
), a . 0 (1.43)
25 24 23 22 21 0 1 2 3 4 5
21
0
1
t (s)
f(t)
Figure 1.38
A damped
cosine signal.
0
f(t)
A
0
t (s)
/2
2/2
Figure 1.39
A rectangular pulse.
1.6.5 rEctangular PulSE and triangular PulSE
A rectangular pulse with amplitude A and pulse width t is shown in Figure 1.39.The center
of the pulse is at t 5 0.
The rectangular pulse shown in Figure 1.39 is denoted by
f(t) 5 Arect1
t
2
Plot f(t) 5 rect1
t 1 1
2 21 3rect1
t 2 1
2 22 2rect1
t 2 3.5
3 2.
exAmpLe 1.11
The first rectangle is centered at t 5 21 and has a height of 1 and width of 2.The
second rectangle is centered at t 5 1 and has a height of 3 and width of 2.The third
rectangle is centered at t 5 3.5 and has a height of 22 and width of 3.The waveform
continued
25 24 23 22 21 0 1 2 3 4 5
0
1
t (s)
f(t)
Figure 1.37
f(t) 5 e-at
u(t), a 5 0.5.

A triangular pulse with amplitude A and base 2 is shown in Figure 1.42. The center
of the pulse is at t 5 0.
23 22 21 0 1 2 3 4 5 6
24
23
22
21
0
1
2
3
4
t (s)
f(t)
Figure 1.40
Waveform f(t).
exercise 1.11
Plot f(t) 5 23rect1
t 1 5
2 21 2 rect 1
t 2 5
4 2.
f(t)
210 29 28 27 26 25 24 23 22 21 0 1 2 3 4 5 6 7 8 9 10
24
23
22
21
0
1
2
3
4
t (s)
Answer:
The waveform f(t) is shown in Figure 1.41.
0
f(t)
A
0
t (s)

2
Figure 1.42
A triangular pulse.
Figure 1.41
Waveform f(t).

26 25 24 23 22 21 0 1 2 3 4 5 6
24
23
22
21
0
1
2
3
4
t (s)
f(t)
Figure 1.43
Waveform f(t).
The first triangle is centered at t 5 22 and has a height of 2 and base of 4.The second tri-
angle is centered at t 5 0 and has a height of 22 and base of 4.The third triangle is centered
at t 5 2 and has a height of 2 and base of 4.The waveform f(t) is shown in Figure 1.43.
Plot f(t) 5 2tri1
t 1 2
2 22 2tri1
t
221 2tri1
t 2 2
2 2.
exAmpLe 1.12
exercise 1.12
Find the equation of the waveform f(t) shown in Figure 1.44.
210 29 28 27 26 25 24 23 22 21 0 1 2 3 4 5 6 7 8 9 10
22
21
0
1
2
3
4
t (s)
f(t)
Figure 1.44
Waveform for
EXERCISE 1.12.
Answer:
f(t) 5 rect 1
t
421 2tri 1
t
22
The triangular pulse shown in Figure 1.42 is denoted by
f(t) 5 A tri1
t
2

Problems 27
In this chapter, the seven base units of the International
System of Units (SI), along with derived units relevant to
electrical and computer engineering, are presented. The
definitions of voltage, current, and power, among other
terms, are given.The potential difference per unit charge
between A and B is called voltage between A and B,
vAB 5 wAB/q,where wAB is the amount of the work needed
to move the test charge from B to A.
Current is defined as the rate of change of charge:
i(t) 5
dq(t)
dt
Power is the product of current and voltage:
p(t) 5 i(t)v(t)
Energy is the integral of power:
w(t) 5 #
t
2`
p()d
The four types of dependent sources are:
Voltage-controlled voltage source (VCVS)
Voltage-controlled current source (VCCS)
Current-controlled voltage source (CCVS)
Current-controlled current source (CCCS)
1.1 Find the current flowing through an element
if the charge flowing through the element is
given by
q(t) 5 50.002t, C t $ 0
0, t , 0
1.2 Find the current flowing through an element if the
charge flowing through the element is given by
q(t) 5 55e20.2t
, C t $ 0
0, t , 0
given by
q(t) 5 58(1 2 e20.003t
), C t $ 0
0, t , 0
given by
q(t) 5 57te20.003t
, C t $ 0
0, t , 0
prOBLemS
SummArY
1.7 Find the total charge passing through an element at
one cross section over the time interval 0 # t # 5 s
if the current through the same cross section is
given by
i(t) 5 5 mA
0 1 2 3 4 5 6 7
21
20.5
0
0.5
1
3 1023
t (s)
q(t)
(C)
Figure p1.6
given by
q(t) 5 58 3 1026
sin(2 3 1000t), C t $ 0
0, t , 0
1.6 The charge entering an element is shown in
Figure P1.6. Plot the current through the
element for 0 # t , 7 s.

1.8 Find the total charge passing through an element
at one cross section over the time interval
0 # t # 5 s if the current through the same
cross section is given by
i(t) 5 55e20.2t
A, t $ 0
0, t , 0
i(t) 5 53(1 2 e20.5t
) A, t $ 0
0, t , 0
i(t) 5 52te23t
A, t $ 0
0, t , 0
i(t) 5 57 sin(ty5) A, t $ 0
0, t , 0
1.12 Find the power in the circuit element shown in
Figure P1.12 and state whether the element is
absorbing power or delivering power.
2 A
5 V
1
2
Figure p1.12
3 A
2 V
1
2
Figure p1.13
4 mA
5 V
2
1
Figure p1.14
10 mA
12 V
2
1
Figure p1.15
1.16 Find the power p(t) on the element when the
current through the element i(t) from positive ter-
minal to negative terminal and voltage v(t) across
the element are given by
i(t) 5 2 mA, v(t) 5 5 V
current through the element i(t) from positive
terminal to negative terminal and voltage v(t)
across the element are given by
i(t) 5 25 cos(21000t) mA,
v(t) 5 5 sin(21000t) V

Problems 29
i(t) 5 60 e20.07t
u(t) mA,
v(t) 5 7 e20.08t
u(t) V
i(t) 5 8 cos(2100t) mA,
v(t) 5 3 cos(2100t) V
i(t) 5 6 sin(2100t) mA,
v(t) 5 2 sin(2100t) V
1.21 Redraw the circuit shown in Figure P1.21 with
one voltage source and one current source,
without affecting the voltages across and
currents through the resistors in the circuit.
I1
2 mA
I3
4 mA
1
2
2 V
V1
R2
2 kV
1
2
4 V
V2
2
1
3 V
V3
R1
1 kV
R3
3 kV
R4
6 kV
R5
3 kV
I2
3 mA
0
Figure p1.22
0
.
1
2
5
0
.
2
5
0
.
3
7
5
0
.
5
0
.
6
2
5
0
.
7
5
0
.
8
7
5
1
10
8
6
4
2
0
22
24
26
28
210
212
t (s)
v(t)
(V)
2
0
.
8
7
5
2
0
.
6
2
5
2
0
.
5
2
0
.
3
7
5
2
0
.
2
5
2
0
.
1
2
5
0
2
0
.
7
5
2
1
Figure p1.24
R1
2 kV
R3
4 kV
R5
5 kV
R2
3 kV
R4
1 kV
1
2
1
2
Vs
3 V va
2
1
0.001 va
0.5 va
0
Figure p1.25
1.23 Plot v(t) 5 22 1 6 cos(25000t 2 908) V.
1.24 Find the equation of the sinusoid shown in
Figure P1.24.
I1
5 mA
R2
V1
4 kV
R3
1 kV
V2
R1
5 kV
I2
3 mA
2 V
5 V
2
1
1
2
0
Figure p1.21
1.25 In the circuit shown in Figure P1.25, the
controlling voltage, which is the voltage across
R2, is va 5 1.2908 V. Find the controlled voltage
across the VCVS and the controlled current
through the VCCS.
1.22 Redraw the circuit shown in Figure P1.22 with one
voltage source and one current source,without
affecting the voltages across and currents through
the resistors in the circuit.

1.26 In the circuit shown in Figure P1.26, the
controlling current, which is the current through
R1, is ia 5 0.8714 mA. Find the controlled
voltage across the CCVS and the controlled
current through the CCCS.
ia
R3
4 kV
R1
1 kV
R2
3 kV
R4
2 kV
1
2
Vs
3 V
0.6 ia
500 ia
1
2
0
Figure p1.26
1.28 Plot
f(t) 5 25 (t 1 2) 1 7 (t 2 6)
1.29 Plot
f(t) 5 2t u(t) 2 4(t 2 1) u(t 2 1) 1 3 (t 2 3)
3 u(t 2 3) 2 (t 2 5) u(t 2 5)
1.30 Plot
f(t) 5 22t u(t) 1 6(t 2 2) u(t 2 2) 2 5(t 2 3)
3 u(t 2 3) 1 (t 2 5) u(t 2 5)
1.31 Plot
f(t) 5 2 rect 1
t 1 3
4 2
1.32 Plot
f(t) 5 2 tri 1
t 1 4
2 2
1.27 Plot
f(t) 5 u(t) 2 3 u(t 2 2) 1 6 u(t 2 5)
2 4 u(t 2 8)

31
Chapter 2
Circuit Laws
2.1 Introduction
Nodes, branches, loops, and meshes are defined in this chapter. The equation of resistance
of a conductor is expressed as a function of conductivity (or resistivity), and the dimension
of the conductor. Ohm’s law is introduced.
Kirchhoff’s current law (KCL) and Kirchhoff’s voltage law (KVL) are presented in this
chapter.ThesetwoKirchhoff’slawsprovidethetheoreticalbasisforthenodalanalysisandmesh
analysis discussed in the next chapter and applied in circuit analysis in the rest of the chapters.
Finding the equivalent resistance of series and parallel connection of resistors are dis-
cussed. Simple circuit rules can be applied to analyze circuits after simplifying the circuits
using equivalent resistances.
The voltage divider rule and the current divider rule are useful tools to analyze cir-
cuits without too much effort.
If a circuit contains resistors in wye (Y) shape, it can be changed to delta (D) shape. On
the other hand,if a circuit contains resistors in delta shape,it can be changed to wye shape.The
transformation from wye to delta and delta to wye may make it easier to simplify the circuit.
2.2 Circuit
A circuit is an interconnection of elements, which can be voltage sources, current sources,
resistors, capacitors, inductors, coupled coils, transformers, op amps, etc.A node is a point in
a circuit where two or more elements are joined.A simple node is a node that connects two
elements.A path in a circuit is a series of connected elements from a node to another node
that does not go to the same node more than once.A branch is a path in a circuit consisting
of a single element. The voltage of a node measured with respect to a reference node is
called node voltage. The ground node where the voltage is at ground level is usually taken
to be the reference node.A loop of a circuit is a closed path starting from a node and return-
ing to the same node. The loop must have minimum of two branches in its closed path.
A mesh is a loop that does not contain another loop inside it.

32 Chapter 2 CirCuit Laws
In the circuit shown in Figure 2.1, there are three nodes: 0, 1,
and 2. Elements B, C, and D are joined at node 0. Node 1
connects elements A and B, and node 2 connects elements A, C,
and D. If node 0 is the ground node, then the potential is set to
zero at node 0.The voltages at node 1 and node 2 are measured
with respect to node 0.
There are four branches in the circuit shown in Figure 2.1:A,
B, C, and D.There are three loops in the circuit shown in Figure 2.1:
0-B-1-A-2-D-0
0-B-1-A-2-C-0
0-C-2-D-0
There are two meshes in the circuit shown in Figure 2.1:
0-B-1-A-2-C-0
0-C-2-D-0
The loop 0-B-1-A-2-D-0 contains two meshes: 0-B-1-A-2-C-0 and 0-C-2-D-0.
Find all the nodes, loops, and meshes for the circuit shown in Figure 2.1.
ExamplE 2.1
D
B
E F
0
1
2
A
C
3
Exercise 2.1
Find all the nodes, meshes, and loops for the circuit shown in Figure 2.2.
FigurE 2.2
Circuit for
EXERCISE 2.1.
answer:
Nodes: 0, 1, 2, 3
Meshes:
1-A-3-C-2-B-1
0-D-1-B-2-E-0
0-E-2-C-3-F-0
Loops: In addition to the meshes, we have
0-D-1-A-3-F-0
0-D-1-B-2-C-3-F-0
0-D-1-A-3-C-2-E-0
0-E-2-B-1-A-3-F-0
B
A
C D
0
1 2
FigurE 2.1
Circuit for EXAMPLE 2.1.

2.3 resistor 33
2.3 resistor
A resistor is a circuit component that regulates the flow of current.The resistance of a resistor
measures its ability to limit the current. When the resistance value is large, the amount of
current flow through the resistor is small. On the other hand, if the resistance value is small,
the amount of current flow through the resistor is large. The resistance value of a resistor is
determined by the resistivity of the material used to make it, as well as its dimensions.
Low-power resistors can be made from carbon composition material made of fine
granulated graphite mixed with clay. For high power, wire-wound resistors can be used.The
wire-wound resistors are constructed by twisting a wire made of nichrome or similar mate-
rial around a ceramic core.The circuit symbol for a resistor is shown in Figure 2.3.
The current density is defined as the amount of current through the unit area. If A is
the cross-sectional area of a conductor that carries a constant current I, the current density
is given by
J 5
I
A
(2.1)
The current is obtained by integrating the current density through the area.Thus,we have
I 5 #J ? dA (2.2)
It can be shown that the current density is proportional to the electric field intensity;
that is,
J 5 E (2.3)
where is the conductivity of the material. The unit for conductivity is siemens per meter
(S/m). Equation (2.3) is called the microscopic Ohm’s law.
Let the length of a cylindrical conductor with the cross-sectional area A be /. Let the
potential difference between the ends of the conductor be V.This potential difference gen-
erates a constant electric field E inside the conductor.The potential difference V is related
to the electric field through
V 5 E/ (2.4)
Thus, we have E 5 Vy/. Substituting this result into Equation (2.3), we get
J 5 E 5 1
V
/ 2 (2.5)
Since J 5 IyA, Equation (2.5) becomes
I
A
5 1
V
/ 2 (2.6)
Solving for V in Equation (2.6), we get
V 5
/
A
I 5 RI (2.7)
where R is defined as the resistance of the conductor.This is called the macroscopic Ohm’s law.
The SI unit for the resistance R is the ohm (V 5 V/A).The resistance of a material is given by
R 5
/
A
(2.8)
R1
1 kV
R2
4.7 kV
FigurE 2.3
Circuit symbol
for a resistor.

34 Chapter 2 CirCuit Laws
The resistance is proportional to the length and inversely proportional to the conduct-
ivity and the cross-sectional area. The resistivity of a material is defined as the reciprocal
of the conductivity .Thus, we have
5
1

(2.9)
The SI unit for resistivity is V ? m. In terms of the resistivity, the resistance R can be
written as
R 5
/
A
(2.10)
The resistance is proportional to the resistivity and length and inversely proportional
to the cross-sectional area. The inverse of resistance is called conductance and is denoted
by G.The unit for conductance is S (siemens). Notice that S 5 V21
5 A/V.
G 5
1
R
(2.11)
With resistivity (r) and cross-sectional area (A) fixed, let R1 be the resistance when
the length is /1, R2 be the resistance when the length is /2, and R be the resistance when the
length is / 5 /1 1 /2.Then, we have
R 5
/
A
5
(/1 1 /2)
A
5
/1
A
1
/2
A
5 R1 1 R2 (2.12)
Equation (2.12) says that when two resistors are connected in series, the equivalent
resistance is the sum of the two resistances. In general, as shown in section 2.7 later in this
chapter, if n resistors, R1, R2, . . ., Rn, are connected in series, the equivalent resistance is
given by
Req 5 R1 1 R2 1 . . . 1 Rn (2.13)
With resistivity (r) and length (/) fixed, let R1 be the resistance when the cross-
sectional area is A1, R2 be the resistance when the cross-sectional area is A2, and R be the
resistance when the cross-sectional area is A 5 A1 1 A2.Then, we have
R 5
/
A
5
/
A1 1 A2
5
1
A1
/
1
A2
/
5
1
1
/
A1
1
1
/
A2
5
1
1
R1
1
1
R2
5
R1R2
R1 1 R2
(2.14)
Equation (2.14) says that when two resistors are connected in parallel, the equival-
ent resistance is given by R1R2/(R11R2). In general, as shown in section 2.7 later in this
chapter, if n resistors, R1, R2, . . ., Rn, are connected in parallel, the equivalent resistance
is given by
Req 5
1
1
R1
1
1
R2
1 Á 1
1
Rn
(2.15)

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