Sixth Edition
Fundamentals of
Electric Circuits
Charles K. Alexander | Matthew N.O. Sadiku
sixth edition
Fundamentals of
Electric Circuits
Charles K. Alexander
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Matthew N. O. Sadiku
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Dedicated to our wives, Kikelomo and Hannah, whose understanding and
support have truly made this book possible.
Matthew
and
Chuck
Contents
Chapter 3
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PART 1
DC Circuits
Chapter 1
Basic Concepts
1.1
1.2
1.3
1.4
1.5
1.6
1.7
3.1
3.2
3.3
3.4
3.5
3.6
2
3
Introduction 4
Systems of Units 5
Charge and Current 6
Voltage 9
Power and Energy 10
Circuit Elements 14
Applications 16
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1.8
1.9
Problem Solving
Summary 22
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Basic Laws
29
Introduction 30
Ohm’s Law 30
Nodes, Branches, and Loops 35
Kirchhoff’s Laws 37
Series Resistors and Voltage Division 43
Parallel Resistors and Current Division 44
Wye-Delta Transformations 51
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFN 4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
4.10
Applications
Circuit Theorems
125
Introduction 126
Linearity Property 126
Superposition 128
Source Transformation 133
Thevenin’s Theorem 137
Norton’s Theorem 143
Derivations of Thevenin’s
and Norton’s Theorems 147
Maximum Power Transfer 148
Verifying Circuit Theorems
with PSpice 150
Applications 153
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4.11
Summary
158
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2.8
79
Introduction 80
Nodal Analysis 80
Nodal Analysis with Voltage Sources 86
Mesh Analysis 91
Mesh Analysis with Current Sources 96
Nodal and Mesh Analyses
by Inspection 98
Nodal Versus Mesh Analysis 102
Circuit Analysis with PSpice 103
Applications: DC Transistor Circuits 105
Summary 110
Chapter 4
19
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Chapter 2
3.7
3.8
3.9
3.10
Methods of Analysis
57
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2.9
Summary
63
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Chapter 5
5.1
5.2
Operational Amplifiers
Introduction 174
Operational Amplifiers
173
174
v
vi
Contents
5.3
5.4
5.5
5.6
5.7
5.8
5.9
5.10
Ideal Op Amp 178
Inverting Amplifier 179
Noninverting Amplifier 181
Summing Amplifier 183
Difference Amplifier 185
Cascaded Op Amp Circuits 189
Op Amp Circuit Analysis with PSpice 192
Applications 194
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Instrumentation Amplifiers
5.11
Summary
6.1
6.2
6.3
6.4
6.5
6.6
Capacitors and Inductors
Introduction 214
Capacitors 214
Series and Parallel Capacitors 220
Inductors 224
Series and Parallel Inductors 228
Applications 231
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6.7
Summary
238
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Chapter 7
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
First-Order Circuits 251
Introduction 252
The Source-Free RC Circuit 253
The Source-Free RL Circuit 257
Singularity Functions 263
Step Response of an RC Circuit 271
Step Response of an RL Circuit 278
First-Order Op Amp Circuits 282
Transient Analysis with PSpice 287
Applications 291
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7.10
Summary
8.1
8.2
8.3
8.4
8.5
8.6
197
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Chapter 8
8.7
8.8
8.9
8.10
8.11
Introduction 312
Finding Initial and Final Values 313
The Source-Free Series
RLC Circuit 317
The Source-Free Parallel
RLC Circuit 324
Step Response of a Series RLC
Circuit 329
Step Response of a Parallel RLC
Circuit 334
General Second-Order Circuits 337
Second-Order Op Amp Circuits 342
PSpice Analysis of RLC Circuits 344
Duality 348
Applications 351
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8.12
Summary
354
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AC Circuits 366
Chapter 9
Sinusoids and Phasors
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
Introduction 368
Sinusoids 369
Phasors 374
Phasor Relationships for
Circuit Elements 383
Impedance and Admittance 385
Kirchhoff’s Laws in the Frequency
Domain 387
Impedance Combinations 388
Applications 394
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9.9
Summary
400
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Chapter 10
297
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Second-Order Circuits 311
10.1
10.2
10.3
Sinusoidal Steady-State
Analysis 411
Introduction 412
Nodal Analysis 412
Mesh Analysis 415
367
vii
Contents
10.4
10.5
10.6
10.7
10.8
10.9
Superposition Theorem 419
Source Transformation 422
Thevenin and Norton
Equivalent Circuits 424
Op Amp AC Circuits 429
AC Analysis Using PSpice 431
Applications 435
12.11 Summary
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10.10 Summary
Chapter 13
439
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
3FWJFX2VFTUJPOT 1SPCMFNT Chapter 11
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
AC Power Analysis
541
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT 455
Introduction 456
Instantaneous and Average Power
Maximum Average Power Transfer
Effective or RMS Value 465
Apparent Power and
Power Factor 468
Complex Power 471
Conservation of AC Power 475
Power Factor Correction 479
Applications 481
456
462
13.9
Magnetically Coupled
Circuits 553
Introduction 554
Mutual Inductance 555
Energy in a Coupled Circuit 562
Linear Transformers 565
Ideal Transformers 571
Ideal Autotransformers 579
Three-Phase Transformers 582
PSpice Analysis of Magnetically
Coupled Circuits 584
Applications 589
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13.10 Summary
595
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11.10 Summary
486
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Chapter 12
Three-Phase Circuits
Chapter 14
501
12.1
12.2
12.3
12.4
12.5
Introduction 502
Balanced Three-Phase Voltages 503
Balanced Wye-Wye Connection 507
Balanced Wye-Delta Connection 510
Balanced Delta-Delta
Connection 512
12.6 Balanced Delta-Wye Connection 514
12.7 Power in a Balanced System 517
12.8 Unbalanced Three-Phase
Systems 523
12.9 PSpice for Three-Phase Circuits 527
12.10 Applications 532
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14.1
14.2
14.3
14.4
14.5
14.6
14.7
Frequency Response
611
Introduction 612
Transfer Function 612
The Decibel Scale 615
Bode Plots 617
Series Resonance 627
Parallel Resonance 632
Passive Filters 635
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Active Filters
640
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Scaling
646
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viii
Contents
14.10 Frequency Response Using
PSpice 650
14.11 Computation Using MATLAB 653
14.12 Applications 655
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14.13 Summary
Chapter 17
17.1
17.2
17.3
The Fourier Series
757
Introduction 758
Trigonometric Fourier Series 759
Symmetry Considerations 766
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17.4
17.5
17.6
17.7
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Average Power and RMS Values 780
Exponential Fourier Series 783
Fourier Analysis with PSpice 789
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PART 3
17.8
Advanced Circuit
Analysis 672
15.1
15.2
15.3
15.4
Chapter 18
688
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15.6
15.7
18.1
18.2
18.3
The Convolution Integral 695
Application to Integrodifferential
Equations 703
Summary 706
18.4
18.5
18.6
3FWJFX2VFTUJPOT 1SPCMFNT 18.7
16.1
16.2
16.3
16.4
16.5
16.6
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Summary
743
3FWJFX2VFTUJPOT 1SPCMFNT $PNQSFIFOTJWF1SPCMFNT Fourier Transform
811
Introduction 812
Definition of the Fourier Transform
Properties of the Fourier
Transform 818
Circuit Applications 831
Parseval’s Theorem 834
Comparing the Fourier and
Laplace Transforms 837
Applications 838
Summary
841
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Transform 713
Introduction 714
Circuit Element Models 715
Circuit Analysis 720
Transfer Functions 724
State Variables 728
Applications 735
798
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Chapter 16
Summary
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Transform 673
Introduction 674
Definition of the Laplace
Transform 675
Properties of the Laplace
Transform 677
The Inverse Laplace Transform
795
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17.9
Chapter 15
Applications
Chapter 19
19.1
19.2
19.3
19.4
19.5
19.6
Two-Port Networks
Introduction 852
Impedance Parameters 853
Admittance Parameters 857
Hybrid Parameters 860
Transmission Parameters 865
Relationships Between
Parameters 870
851
812
19.7
19.8
19.9
Contents
ix
Interconnection of Networks 873
Computing Two-Port Parameters
Using PSpice 879
Applications 882
Appendix A
Simultaneous Equations and Matrix
Inversion A
Appendix B
Complex Numbers
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Appendix C
Mathematical Formulas
Appendix D
Answers to Odd-Numbered
Problems A-21
19.10 Summary
891
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Preface
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eventually became the first man-made object to leave our solar system.
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in this textbook. Finally, space is vast so that Voyager 1 will fly past
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Features
New to This Edition
We have added learning objectives to each chapter to reflect what we
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Retained from Previous Editions
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of the solution process and the confidence to solve problems them
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the practice problem without flipping pages or looking at the end of
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Preface
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the interested instructor can go from solutions of first-order circuits
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that will lead to a significantly enhanced career as an engineer. Be
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Preface
The material in the three parts is more than sufficient for a two-semester
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briefly, or assigned as homework. They can be omitted without loss of
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design. More difficult problems are marked with an asterisk (*).
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Prerequisites
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Acknowledgments
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tion a significantly more relevant book. He has checked all the new and
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The sixth edition has benefited greatly from the many outstanding
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and careful corrections and suggestions for clarification which have
xv
xvi
Preface
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Supplements
Instructor and Student Resources
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xvii
®
Required=Results
McGraw-Hill Connect®
Learn Without Limits
Connect is a teaching and learning platform
that is proven to deliver better results for
students and instructors.
Connect empowers students by continually
adapting to deliver precisely what they
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efficient and effective.
Mobile
Connect’s new, intuitive mobile interface gives students
and instructors flexible and convenient, anytime–anywhere
access to all components of the Connect platform.
Students can view
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Adaptive
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About the Authors
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xxi
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About the Authors
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"TTPDJBUJPOGPS$PNQVUJOH.BDIJOFSZ "$. Fundamentals of
Electric Circuits
P A R T
O N E
DCCi rcuits
1
Basic Concepts
2
Basic Laws
3
Methods of Analysis
4
Circuit Theorems
5
Operational Amplifiers
6
Capacitors and Inductors
7
First-Order Circuits
8
Second-Order Circuits
NASA
OUTLINE
c h a p t e r
1
Basic Concepts
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Enhancing Your Skills and Your Career
ABET EC 2000 criteria (3.a), “an ability to apply knowledge
of mathematics, science, and engineering.”
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3
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Chapter 1
Basic Concepts
Learning Objectives
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1.1
Introduction
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JOFMFDUSJDBMFOHJOFFSJOHFEVDBUJPO$JSDVJUUIFPSZJTBMTPW BMVBCMFUP
TUVEFOUTTQFDJBMJ[JOHJOPUIFSCSBODIFTPGUIFQI ZTJDBMTDJFODFTCFDBVTF
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An electric circuit is an interconnection of electrical elements.
Current
–
+
Battery
Figure 1.1
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1.2
5
Systems of Units
+ 9 V (DC)
Antenna
C4
R2
R1
R4
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microphone
R3
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Figure 1.2
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8e commence our study by defining some basic concepts. 5IFTF
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1.2
Systems of Units
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TABLE 1.1
Six basic SI units and one derived unit relevant to this text.
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TABLE 1.2
The SI prefixes.
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Chapter 1
1.3
Basic Concepts
Charge and Current
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Charge is an electrical property of the atomic particles of which matter
consists, measured in coulombs (C).
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A convention is a standard way of
describing something so that others in
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we mean. We will be using IEEE
conventions throughout this book.
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that current is the net floXPGQPTJUJWFDIBSHFT5IVT
Electric current is the time rate of change of charge, measured in
amperes (A).
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1.3
7
Charge and Current
Historical
Andre-Marie Ampere m B'SFODINBUIFNBUJDJBOBOE
physicist, laid the foundation of electrodynamics. He defined the electric
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prolific writer. He formulated the laws of electromagnetics. He inWFOUFE
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DVSSFOUThe best definition is that there are twPXBZTUIBUDVSSFOUDBO
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or time varying.
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An alternating current (ac) is a current that changes direction with
respect to time.
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Chapter 1
–5 A
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(a)
(b)
Figure 1.5
$POWentional current floX B QPTJUJWF
current floX C OFHBUJWe current floX
Example 1.1
Basic Concepts
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0ODFwe define DVSSFOUBTUIFNPWFNFOUPGDIBSHF XFFYQFDUDVS
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rection of current floXJTDPOWFOUJPOBMMZUBLFOBTUIFEJSFDUJPOPGQPTJUJWF
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SFTFOUFEQPTJUJWFMZPSOFHBUJWFMZBTTIPXOJO'JH*OPUIFSX PSET B
OFHBUJWFDVSSFOUPG−A floXJOHJOPOFEJSFDUJPOBTTIPwn in Fig. C JTUIFTBNFBTBDVSSFOUPG+A floXJOHJOUIFPQQPTJUFEJSFDUJPO
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Solution:
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have −×−$FMFDUSPO× FMFDUSPOT=−×−$
Practice Problem 1.1
$BMDVMBUFUIFBNPVOUPGDIBSHFSFQSFTFOUFECZCJMMJPOQSPUPOT
Answer:×−$
Example 1.2
5IFUPUBMDIBS HFFOUFSJOHBUFSNJOBMJTHJ WFOCZ R = UTJO πUN$
$BMDVMBUFUIFDVSSFOUBUU=T
Solution:
ER @@
E UTJOπU N$T= TJOπU+πUDPTπU N"
@@@
J=
=
EU EU
"UU=
J=TJOπ+πDPTπ=+π=N"
Practice Problem 1.2
*GJO&YBNQMF R= −F−U) mC, find the current at U=T
Answer:N"
Example 1.3
%FUFSNJOF UIF UPUBM DIBSHF FOUFSJOH B UFSNJOBM CFUXFFO U = T BOE
U=TJGUIFDVSSFOUQBTTJOHUIFUFSNJOBMJTJ= U−U "
Solution:
Ł
Ł
2= JEU= U− U EU
U=
(
)∣
(
)
U = − − 1 − @@
=$
= U− @@
1.4
9
Voltage
Practice Problem 1.3
The current flowing through an element is
{
"
< U<
J=
U"
U>
$BMDVMBUFUIFDIBSHFFOUFSJOHUIFFMFNFOUGSPNU=UPU=T
Answer:$
1.4
Voltage
"TFxplained briefly in the preWJPVTTFDUJPO UPNP WFUIFFMFDUSPOJOB
DPOEVDUPSJOBQBSUJDVMBSEJSFDUJPOSFRVJSFTTPNFX PSLPSFOFSHZUSBOT
GFS 5IJTX PSLJTQFSGPSNFECZBOF YUFSOBMFMFDUSPNPUJWFGPSDF FNG UZQJDBMMZSFQSFTFOUFECZUIFCBUUFSZJO'JH 5IJTFNGJTBMTPLOP XO
BTWPMUBHFPSQPUFOUJBMEJGGFSFODF5IFWPMUBHFvBCCFUXFFOUXPQPJOUTB
BOECJOBOFMFDUSJDDJSDVJUJTUIFFOFSHZ PSXPSL OFFEFEUPNPWFBVOJU
DIBSHFGSPNCUPBNBUIFNBUJDBMMZ
Δ @@@
EX
vBC=
ER
XIFSFXJTFOFSHZJOKPVMFT + BOERJTDIBSHFJODPVMPNCT $ 5IFWPMU
BHFvBCPSTJNQMZvJTNFBTVSFEJOWPMUT 7 OBNFEJOIPOPSPGUIF*UBMJBO
QIZTJDJTU"MFTTBOESP"OUPOJP7PMUB m XIPJOWented the first
WPMUBJDCBUUFSZ'SPN&R JUJTFWJEFOUUIBU
WPMU=KPVMFDPVMPNC=OFXUPONFUFSDPVMPNC
5IVT
Voltage (or potential difference) is the energy required to move a
unit charge from a reference point (−) to another point (+), measured in
volts (V).
'JHVSFTIP XTUIFW PMUBHFBDSPTTBOFMFNFOU SFQSFTFOUFECZ
BSFDUBOHVMBSCMPDL DPOOFDUFEUPQPJOUT BBOE C 5IFQMVT + BOE
minus −) signs are used to define reference direction or vPMUBHFQPMBSJUZ
5IFvBCDBOCFJOUFSQSFUFEJOUXPXBZT 1PJOUBJTBUBQPUFOUJBMPGvBC
WPMUTIJHIFSUIBOQPJOU C PS UIFQPUFOUJBMBUQPJOU BXJUISFTQFDUUP
QPJOUCJTvBC*UGPMMPXTMPHJDBMMZUIBUJOHFOFSBM
vBC=−vCB
+
vab
–
b
Figure 1.6
1PMBSJUZPGWPMUBHFvBC
'PSFYBNQMF JO'JH XFIB WFUXPSFQSFTFOUBUJPOTPGUIFTBNFW PMU
BHF*O'JH B QPJOU BJT +7BCPWFQPJOU CJO'JH C QPJOU
CJT−7BCPWFQPJOUB8FNBZTBZUIBUJO'JH B UIFSFJTB7
WPMUBHFESPQGSPNBUPCPSFRVJWBMFOUMZB7WPMUBHFSJTFGSPNCUPB
*OPUIFSXPSET BWPMUBHFESPQGSPNBUPCJTFRVJWBMFOUUPBWPMUBHFSJTF
GSPNCUPB
$VSSFOUBOEWPMUBHFBSFUIFUX PCBTJDWBSJBCMFTJOFMFDUSJDDJSDVJUT
5IFDPNNPOUFSNTJHOBMJTVTFEGPSBOFMFDUSJDRVBOUJUZTVDIBTBDVSSFOU
PSBWPMUBHF PSFWFOFMFDUSPNBHOFUJDXBWF XIFOJUJTVTFEGPSDPOWFZJOH
a
+
a
(a)
Figure 1.7
a
–9 V
9V
–
–
+
b
b
(b)
5XPFRVJWBMFOUSFQSFTFOUBUJPOTPGUIF
TBNFWPMUBHFvBC B 1PJOUBJT7BCPWF
QPJOUC C QPJOUCJT−7BCPWFQPJOUB
10
Chapter 1
Basic Concepts
Historical
Alessandro Antonio Volta m BO*UBMJBOQIZTJDJTU JO vented the electric battery—which provided the first continuous flow of
FMFDUSJDJUZ‡BOEUIFDBQBDJUPS
#PSOJOUPBOPCMFGBNJMZJO$PNP *UBMZ 7PMUBXBTQFSGPSNJOH
FMFDUSJDBM FYQFSJNFOUT BUBHF )JT JOWFOUJPOPG UIFCBUUFSZ JO
SFWPMVUJPOJ[FEUIFVTFPGFMFDUSJDJUZ5IFQVCMJDBUJPOPGIJTXPSLJO
NBSLFEUIFCFHJOOJOHPGFMFDUSJDDJSDVJUUIFPSZ7PMUBSFDFJWFENBOZ
IPOPSTEVSJOHIJTMJGFUJNF5IFVOJUPGWPMUBHFPSQPUFOUJBMEJGGFSFODF UIFWPMU XBTOBNFEJOIJTIPOPS
¥6OJWFSTBM*NBHFT(SPVQ
(FUUZ*NBHFT
Keep in mind that electric current is
always through an element and that
electric voltage is always across the
element or between two points.
JOGPSNBUJPO&OHJOFFSTQSFGFSUPDBMMTVDIW BSJBCMFTTJHOBMTSBUIFSUIBO
NBUIFNBUJDBMGVODUJPOTPGUJNFCFDBVTFPGUIFJSJNQPSUBODFJODPNNV OJDBUJPOTBOEPUIFSEJTDJQMJOFT-JL FFMFDUSJDDVSSFOU BDPOTUBOUW PMUBHF
JTDBMMFEB EDWPMUBHFBOEJT SFQSFTFOUFECZ 7 XIFSFBTBTJOVTPJEBMMZ
UJNFWBSZJOHWPMUBHFJTDBMMFEBOBDWPMUBHFBOEJTSFQSFTFOUFECZv"ED
WPMUBHFJTDPNNPOMZQSPEVDFECZBCBUUFSZBDW PMUBHFJTQSPEVDFECZ
BOFMFDUSJDHFOFSBUPS
1.5
Power and Energy
"MUIPVHIDVSSFOUBOEW PMUBHFBSFUIFUX PCBTJDW BSJBCMFTJOBOFMFDUSJD
DJSDVJU UIF ZBSFOPUTVG ficient by themselvFT' PSQSBDUJDBMQVSQPTFT XFOFFEUPLOP XIPXNVDI QPXFSBOFMFDUSJDEF WJDFDBOIBOEMF 8F
BMMLOPXGSPNF YQFSJFODFUIBUBX BUUCVMCHJWFTNPSFMJHIUUIBOB
XBUUCVMC8FBMTP LOPXUIBUXIFO XFQBZ PVSCJMMT UPUIFFMFDUSJD
VUJMJUZDPNQBOJFT XFBSFQBZJOHGPSUIFFMFDUSJDFOFSHZDPOTVNFEPWFS
BDFSUBJOQFSJPEPGUJNF5IVT QPXFSBOEFOFSHZDBMDVMBUJPOTBSFJNQPS
UBOUJODJSDVJUBOBMZTJT
5PSFMBUFQP XFSBOEFOFS HZUPW PMUBHFBOEDVSSFOU XFSFDBMMGSPN
QIZTJDTUIBU
Power is the time rate of expending or absorbing energy, measured in
watts (W).
8FXSJUFUIJTSFMBUJPOTIJQBT
EX
Q= @@@
EU
Δ
1.5
11
Power and Energy
XIFSFQJTQPXFSJOXBUUT 8 XJTFOFSHZJOKPVMFT + BOEUJTUJNFJO
TFDPOET T 'SPN&RT BOE JUGPMMPXTUIBU
ER
EX
EX
@@@
@@@
@@
Q =
=
⋅
= vJ
EU
ER EU
i
+
+
v
v
–
–
p = +vi
p = –vi
(a)
(b)
PS
Q=vJ
i
Figure 1.8
5IFQPXFSQJO&R JTBUJNFWBSZJOHRVBOUJUZBOEJTDBMMFEUIFJOTUBO 3FGFSFODFQPMBSJUJFTGPSQPXFSVTJOHUIF
UBOFPVTQPXFS5IVT UIFQPXFSBCTPSCFEPSTVQQMJFECZBOFMFNFOUJTUIF QBTTJWFTJHODPOWFOUJPO B BCTPSCJOH
QSPEVDUPGUIFWPMUBHFBDSPTTUIFFMFNFOUBOEUIFDVSSFOUUISPVHIJU*GUIF QPXFS C TVQQMZJOHQPXFS
QPXFSIBTB+TJHO QPXFSJTCFJOHEFMJWFSFEUPPSBCTPSCFECZUIFFMFNFOU*G When the voltage and current directions
POUIFPUIFSIBOE UIFQPXFSIBTB−TJHO QPXFSJTCFJOHTVQQMJFECZUIFFMF
conform to Fig. 1.8(b), we have the
NFOU#VUIPXEPXFLOPXXIFOUIFQPXFSIBTBOFHBUJWFPSBQPTJUJWFTJHO
active sign convention and p = +vi.
$VSSFOUEJSFDUJPOBOEWPMUBHFQPMBSJUZQMBZBNBKPSSPMFJOEFUFSNJO
JOHUIFTJHOPGQP XFS*UJTUIFSFGPSFJNQPSUBOUUIBUXFQBZBUUFOUJPOUP
UIFSFMBUJPOTIJQCFUXFFODVSSFOUJBOEWPMUBHFvJO'JH B 5IFWPMU
BHFQPMBSJUZBOEDVSSFOUEJSFDUJPONVTUDPOGPSNXJUIUIPTFTIP XOJO
Fig. B JOPSEFSGPSUIFQP XFSUPIBWFBQPTJUJWFTJHO5IJTJTLOPXO
BTUIFQBTTJWFTJHODPOWFOUJPO#ZUIFQBTTJWFTJHODPOWFOUJPO DVSSFOU
FOUFSTUISPVHIUIFQPTJUJWFQPMBSJUZPGUIFWPMUBHF*OUIJTDBTF Q=+vJPS
vJ>JNQMJFTUIBUUIFFMFNFOUJTBCTPSCJOHQP XFS)PXFWFS JGQ=−vJ
PSvJ< BTJO'JH C UIFFMFNFOUJTSFMFBTJOHPSTVQQMZJOHQPXFS
Passive sign convention is satisfied when the current enters through the
positive terminal of an element and p =+vi. If the current enters through
the negative terminal, p =−vi.
6OMFTTPUIFSXJTFTUBUFE XFXJMMGPMMPXUIFQBTTJWFTJHODPOWFOUJPO
UISPVHIPVUUIJTUFYU'PSFxample, the element in both circuits of Fig. IBTBOBCTPSCJOHQPXFSPG+8CFDBVTFBQPTJUJWFDVSSFOUFOUFSTUIF
QPTJUJWFUFSNJOBMJOCPUIDBTFT*O'JH IP XFWFS UIFFMFNFOUJT
TVQQMZJOHQPXFSPG +8CFDBVTF BQPTJUJWFDVSSFOUFOUFST UIFOFHB
UJWFUFSNJOBM0GDPVSTF BOBCTPSCJOHQPXFSPG−8JTFRVJWBMFOUUPB
TVQQMZJOHQPXFSPG+8*OHFOFSBM
+1PXFSBCTPSCFE=−1PXFSTVQQMJFE
*O GBDU UIF MBX PG DPOTFSWBUJPO PGFOFSHZ NVTU CF PCFZFE JO BOZ
FMFDUSJDDJSDVJU'PSUIJTSFBTPO UIFBMHFCSBJDTVNPGQP XFSJOBDJSDVJU BUBOZJOTUBOUPGUJNF NVTUCF[FSP
∑Q= 5IJTBHain confirms the fBDUUIBUUIFUPUBMQP XFSTVQQMJFEUPUIFDJSDVJU
NVTUCBMBODFUIFUPUBMQPXFSBCTPSCFE
'SPN&R UIFFOFSHZBCTPSCFEPSTVQQMJFECZBOFMFNFOUGSPN
UJNFUUPUJNFUJT
U
Ł
Ł
U
X= QEU
= vJEU
U
U
3A
3A
+
–
4V
4V
–
+
(a)
(b)
Figure 1.9
5XPDBTFTPGBOFMFNFOUXJUIBOBCTPSC
JOHQPXFSPG8 B Q=×=8 C Q=×=8
3A
3A
+
–
4V
4V
–
+
(a)
Figure 1.10
(b)
5XPDBTFTPGBOFMFNFOUXJUIBTVQQMZJOH
QPXFSPG8 B Q=−×=−8 C Q=−×=−8
12
Chapter 1
Basic Concepts
Energy is the capacity to do work, measured in joules (J).
5IFFMFDUSJDQP XFSVUJMJUZDPNQBOJFTNFBTVSFFOFS
8I XIFSF
HZJOX BUUIPVST
8I= +
Example 1.4
"OFOFSHZTPVSDFGPSDFTBDPOTUBOUDVSSFOUPGA for 10 s to floXUISPVHI
BMJHIUC VMC*GL+JTHJ WFOPGGJOUIFGPSNPGMJHIUBOEIFBUFOFS HZ DBMDVMBUFUIFWPMUBHFESPQBDSPTTUIFCVMC
Solution:
5IFUPUBMDIBSHFJT
∆R=J∆U=×=$
5IFWPMUBHFESPQJT
× ∆X
@@@
v=
= @@@@@@@
=7
∆R
Practice Problem 1.4
5PNPWFDIBSHF RGSPNQPJOU CUPQPJOU BSFRVJSFT+'JOEUIFWPMU BHFESPQvBC UIFWPMUBHFBU BQPTJUJWFXJUISFTQFDU UPC JG B R=$ C R=−$
Answer: B 7 C −7
Example 1.5
'JOEUIFQPXFSEFMJWFSFEUPBOFMFNFOUBUU=NTJGUIFDVSSFOUFOUFSJOH
JUTQPTJUJWFUFSNJOBMJT
J=DPTπU"
BOEUIFWPMUBHFJT B v=J C v=EJ∕EU
Solution:
B 5IFWPMUBHFJTv=J=DPTπUIFODF UIFQPXFSJT
Q=vJ=DPTπU8
"UU=NT
Q=DPT π××− =DPTπ=8
(b) We find the voltage and the power as
EJ=
@@
v=
−π TJOπU=−πTJOπU7
EU
Q=vJ=−πTJOπUDPTπU8
"UU=NT
Q=−πTJOπDPTπ8
=−TJO°DPT°=−L8
1.5
13
Power and Energy
Historical
1884 Exhibition *OUIF6OJUFE4UBUFT OPUIJOHQSPNPUFEUIFGVUVSFPG
FMFDUSJDJUZMJLFUIF*OUFSOBUJPOBM&MFDUSJDBM&YIJCJUJPO+VTUJNBHJOF
BXPSMEXJUIPVUFMFDUSJDJUZ BXPSMEJMMVNJOBUFECZDBOEMFTBOEHBTMJHIUT BXPSMEXIFSFUIFNPTUDPNNPOUSBOTQPSUBUJPOXBTCZXBMLJOHBOESJE
JOHPOIPSTFCBDLPSCZIPSTFESBXODBSSJBHF*OUPUIJTXPSMEBOFYIJCJ UJPOXBTcreated that highlighted Thomas Edison and reflected his highly
EFWFMPQFE BCJMJUZUPQSPNPUFIJTJOWFOUJPOTBOEQSPEVDUT)JTFYIJCJU
GFBUVSFETQFDUBDVMBSMJHIUJOHEJTQMBZTQPXFSFECZBOJNQSFTTJWFL8
i+VNCPuHFOFSBUPS
&EXBSE8FTUPOTEZOBNPTBOEMBNQTXFSFGFBUVSFEJOUIF6OJUFE
4UBUFT&MFDUSJD-JHIUJOH$PNQBOZTEJTQMBZ8FTUPOTXFMMLOPXODPM lection of scientific instruments was also shown.
0UIFSQSPNJOFOUFYIJCJUPSTJODMVEFE'SBOL4QSBHVF &MJIV5IPNQTPO BOEUIF#SVTI&MFDUSJD$PNQBOZPG$MFWFMBOE5IF"NFSJDBO*OTUJUVUFPG
Electrical Engineers (AIEE) held its first technical meeting on October m
BUUIF'SBOLMJO*OTUJUVUFEVSJOHUIFFYIJCJU"*&&NFSHFEXJUIUIF*OTUJUVUF
PG3BEJP&OHJOFFST *3& JOUPGPSNUIF*OTUJUVUFPG&MFDUSJDBMBOE
&MFDUSPOJDT&OHJOFFST *&&& 4PVSDF*&&&)JTUPSZ$FOUFS
'JOEUIFQPXFSEFMJWFSFEUPUIFFMFNFOUJO&YBNQMFBUU= 5 ms if UIF
DVSSFOUSFNBJOTUIFTBNFCVUUIFWPMUBHFJT B v=J7
( Ł )
U
C v =+ JEU7
Answer: B 8 C 8
Practice Problem 1.5
14
Chapter 1
Example 1.6
Basic Concepts
)PXNVDIFOFSHZEPFTB8FMFDUSJDCVMCDPOTVNFJOUXPIPVST
Solution:
X=QU= 8 × I × NJOI × TNJO
= +=L+
5IJTJTUIFTBNFBT
X=QU=8×I=8I
Practice Problem 1.6
"IPNFFMFDUSJDIFBUFSESB XT "XIFODPOOFDUFEUPB 7PVUMFU
)PXNVDIFOFSHZJTDPOTVNFECZUIFIFBUFSPWFSBQFSJPEPGIPVST
Answer:LXBUUIPVST
1.6
Circuit Elements
"TXFEJTDVTTFEJO4FDUJPO BOFMFNFOUJTUIFCBTJDCVJMEJOHCMPDLPG
BDJSDVJU"OFMFDUSJDDJSDVJUJTTJNQMZBOJOUFSDPOOFDUJPOPGUIFFMFNFOUT
$JSDVJUBOBMZTJTJTUIFQSPDFTTPGEFUFSNJOJOHWPMUBHFTBDSPTT PSUIFDVS
SFOUTUISPVHI UIFFMFNFOUTPGUIFDJSDVJU
5IFSFBSFUX PUZQFTPGFMFNFOUTGPVOEJOFMFDUSJDDJSDVJUT QBT
TJWFFMFNFOUTBOE BDUJWFFMFNFOUT "OBDUJ WFFMFNFOUJTDBQBCMFPG HFOFSBUJOHFOFSHZXIJMFBQBTTJWFFMFNFOUJTOPU&YBNQMFTPGQBTTJWF
FMFNFOUTBSFSFTJTUPST DBQBDJUPST BOEJOEVDUPST 5ZQJDBMBDUJWFFMF ments include generators, batteries, and operational amplifiers. Our
BJNJOUIJTTFDUJPOJTUPH BJOGBNJMJBSJUZXJUITPNFJNQPSUBOUBDUJ WF
FMFNFOUT
5IFNPTUJNQPSUBOUBDUJ WFFMFNFOUTBSFW PMUBHFPSDVSSFOUTPVSDFT
UIBUHFOFSBMMZEFMJWFSQPwer to the circuit connected to them. 5IFSFBSF
UXPLJOETPGTPVSDFTJOEFQFOEFOUBOEEFQFOEFOUTPVSDFT
An ideal independent source is an active element that provides a
specified voltage or current that is completely independent of other
circuit elements.
v
+
V
–
+
–
(a)
Figure 1.11
(b)
4ZNCPMTGPSJOEFQFOEFOUWPMUBHFTPVSDFT
B VTFEGPSDPOTUBOUPSUJNFWBSZJOHWPMU
BHF C VTFEGPSDPOTUBOUWPMUBHF ED *OPUIFSXPSET BOJEFBMJOEFQFOEFOUWPMUBHFTPVSDFEFMJWFSTUPUIFDJSDVJU
XIBUFWFSDVSSFOUJTOFDFTTBSZUPNBJOUBJOJUTUFSNJOBMW PMUBHF1IZTJDBM
TPVSDFTTVDIBTCBUUFSJFTBOEHFOFSBUPSTNBZCFSFHBSEFEBTBQQSPYJNB
UJPOTUPJEFBMW PMUBHFTPVSDFT'JHVSFTIP XTUIFTZNCPMTGPSJOEF QFOEFOUWPMUBHFTPVSDFT/PUJDFUIBUCPUITZNCPMTJO'JH B BOE
C DBOCFVTFEUPSFQSFTFOUBEDW PMUBHFTPVSDF C VUPOMZUIFTZNCPM
in Fig. B DBOCFVTFEGPSBUJNFW BSZJOHWPMUBHFTPVSDF4JNJMBS MZ BO JEFBM JOEFQFOEFOU DVSSFOU TPVSDF JT BO BDUJWF FMFNFOU UIBU QSP
vides a specified current completely independent of the vPMUBHFBDSPTT
UIFTPVSDF 5IBUJT UIFDVSSFOUTPVSDFEFMJ WFSTUPUIFDJSDVJUXIBUF WFS
1.6
15
Circuit Elements
WPMUBHFJTOFDFTTBSZUPNBJOUBJOUIFEFTJHOBUFEDVSSFOU5IFTZNCPMGPS
BOJOEFQFOEFOUDVSSFOUTPVSDFJTEJTQMBZFEJO'JH XIFSFUIFBSSPX
JOEJDBUFTUIFEJSFDUJPOPGDVSSFOUJ
An ideal dependent (or controlled) source is an active element in
which the source quantity is controlled by another voltage or current.
i
Figure 1.12
4ZNCPMGPSJOEFQFOEFOUDVSSFOUTPVSDF
%FQFOEFOUTPVSDFTBSFVTVBMMZEFTJHOBUFECZEJBNPOETIBQFETZN
CPMT BTTIPXOJO'JH4JODFUIFDPOUSPMPGUIFEFQFOEFOUTPVSDFJT
BDIJFWFECZBWPMUBHFPSDVSSFOUPGTPNFPUIFSFMFNFOUJOUIFDJSDVJU BOE
UIFTPVSDFDBOCFWPMUBHFPSDVSSFOU JUGPMMPXTUIBUUIFSFBSFGPVSQPTTJCMF
UZQFTPGEFQFOEFOUTPVSDFT OBNFMZ
"WPMUBHFDPOUSPMMFEWPMUBHFTPVSDF 7$74 "DVSSFOUDPOUSPMMFEWPMUBHFTPVSDF $$74 "WPMUBHFDPOUSPMMFEDVSSFOUTPVSDF 7$$4 "DVSSFOUDPOUSPMMFEDVSSFOUTPVSDF $$$4 %FQFOEFOUTPVSDFTBSFVTFGVMJONPEFMJOHFMFNFOUTTVDIBTUSBOTJTUPST operational amplifiers, and inteHSBUFEDJSDVJUT"OFYBNQMFPGBDVSSFOU
DPOUSPMMFEWPMUBHFTPVSDFJTTIP XOPOUIFSJHIUIBOETJEFPG'JH XIFSF UIFW PMUBHF JPGUIFW PMUBHFTPVSDF EFQFOETPOUIFDVSSFOU J
UISPVHIFMFNFOU$. Students might be surprised that the Walue of the EF
QFOEFOUWPMUBHFTPVSDFJT J7 BOEOPU J" CFDBVTFJUJTBW PMUBHF
TPVSDF5IFLFZJEFBUPLeep in mind is that a WPMUBHFTPVSDFDPNFTXJUI
QPMBSJUJFT +−) in its symbol, while BDVSSFOUTPVSDFDPNFTXJUIBOBS SPX JSSFTQFDUJWe of what it depends on.
*UTIPVMECFOPUFEUIBUBOJEFBMW PMUBHFTPVSDF EFQFOEFOUPSJO EFQFOEFOU XJMMQSPEVDFBO ZDVSSFOUSFRVJSFEUPFOTVSFUIBUUIFUFSNJ OBMWPMUBHFJTBTTUBUFE XIFSFBTBOJEFBMDVSSFOUTPVSDFXJMMQSPEVDF UIFOFDFTTBSZW oltage to ensure the stated current floX5IVT BOJEFBM
source could in theory supply an infinite amount of enerHZ*UTIPVME
BMTPCFOPUFEUIBUOPUPOMZEPTPVSDFTTVQQMZQP XFSUPBDJSDVJU UIF Z
DBOBCTPSCQPXFSGSPNBDJSDVJUUPP'PSBWPMUBHFTPVSDF XFLOPXUIF
WPMUBHFCVUOPUUIFDVSSFOUTVQQMJFEPSESBXOCZJU#ZUIFTBNFUPLFO XFLOPXUIFDVSSFOUTVQQMJFECZBDVSSFOUTPVSDFC VUOPUUIFW PMUBHF
BDSPTTJU
v
+
–
i
(a)
(b)
Figure 1.13
4ZNCPMTGPS B EFQFOEFOUWPMUBHF
TPVSDF C EFQFOEFOUDVSSFOUTPVSDF
B
A
i
+
5V
−
Q= − =−8
4VQQMJFEQPXFS
'PSQBOEQ, the current floXTJOUPUIFQPTJUJWFUFSNJOBMPGUIFFMFNFOU
JOFBDIDBTF
Q= =8
"CTPSCFEQPXFS
Q= =8
"CTPSCFEQPXFS
10i
Figure 1.14
5IFTPVSDFPOUIFSJHIUIBOETJEFJTB
DVSSFOUDPOUSPMMFEWPMUBHFTPVSDF
Example 1.7
$BMDVMBUFUIFQPXFSTVQQMJFEPSBCTPSCFECZFBDIFMFNFOUJO'JH
Solution:
8FBQQMZUIFTJHODPOWFOUJPOGPSQPXFSTIPXOJO'JHTBOE'PS
Q UIF"DVSSFOUJTPVUPGUIFQPTJUJWFUFSNJOBM PSJOUPUIFOFHBUJWF
UFSNJOBM IFODF
+
–
C
I=5A
+
20 V +
–
p2
–
12 V
p1
Figure 1.15
'PS&YBNQMF
p3
6A
+
8 V p4
–
0.2 I
16
Chapter 1
Basic Concepts
'PSQ XFTIPVMEOPUFUIBUUIFWPMUBHFJT7 QPTJUJWFBUUIFUPQ UIF
TBNFBTUIFWPMUBHFGPSQTJODFCPUIUIFQBTTJWFFMFNFOUBOEUIFEFQFO
EFOUTPVSDFBSFDPOOFDUFEUPUIFTBNFUFSNJOBMT 3FNFNCFSUIBUW PMU
BHFJTBMXBZTNFBTVSFEBDSPTTBOFMFNFOUJOBDJSDVJU 4JODFUIFDVSSFOU
floXTPVUPGUIFQPTJUJWFUFSNJOBM
Q= −* = −× =−8
4VQQMJFEQPXFS
8FTIPVMEPCTFSW FUIBUUIF7JOEFQFOEFOUW
PMUBHFTPVSDFBOE * EFQFOEFOUDVSSFOUTPVSDFBSFTVQQMZJOHQPXFSUPUIFSFTUPGUIFOFU
XPSL XIJMFUIFUXPQBTTJWFFMFNFOUTBSFBCTPSCJOHQPXFS"MTP
Q+Q+Q+Q=−++−=
*OBHSFFNFOUXJUI&R UIFUPUBMQP XFSTVQQMJFEFRVBMTUIFUPUBM
QPXFSBCTPSCFE
Practice Problem 1.7
I=5A
9A 2V
Answer:Q=−8 Q=8 Q=8 Q=8
+–
4A
Figure 1.16
p3
+
-
p1
+
–
'PS1SBDUJDF1SPC
0.6I
p4
3V
–
p2
+
+
5V
–
$PNQVUFUIFQPXFSBCTPSCFEPSTVQQMJFECZFBDIDPNQPOFOUPGUIFDJS
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1.7
Applications2
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UVCFBOEUIFPUIFSXJUIIPXFMFDUSJDVUJMJUJFTEFUFSNJOFZPVSFMFDUSJDCJMM
1.7.1
TV Picture Tube
0OFJNQPSUBOUBQQMJDBUJPOPGUIFNPUJPOPGFMFDUSPOTJTGPVOEJOCPUI
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57DBNFSBSFEVDFTBTDFOFGSPNBOPQUJDBMJNBHFUPBOFMFDUSJDBMTJHOBM
4DBOOJOHJTBDDPNQMJTIFEXJUIBUIJOCFBNPGFMFDUSPOTJOBOJDPOPTDPQF
DBNFSBUVCF
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Fig. 6OMJLFUIFJDPOPTDPQFUVCF XIJDIQSPEVDFTBOFMFDUSPOCFBN
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JODPNJOHTJHOBMThe electron gun, maintained at a high potential, fires
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cal and horizontal deflections so that the spot on the screen where the
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CFBNTUSJLes the fluorescent screen, it giWFTPGGMJHIUBUUIBUTQPU 5IVT UIFCFBNDBOCFNBEFUPiQBJOUuBQJDUVSFPOUIF57TDSFFO
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.PEFSO57UVCFTVTFBEJGGFSFOUUFDIOPMPHZ
1.7
(A)
Plates for
horizontal
horizont
tal deflection
Electron gun
–
Heated filament
(source of electrons)
s)
Cathode
de Anode
(–)
(+)
+
Electron
beam
–
17
Applications
(B)
Plates for
vertical deflection
+
Conductive coating
Fluorescent
screen
Figure 1.17
$BUIPEFSBZUVCF
Historical
Karl Ferdinand Braun and Vladimir K. Zworykin
Karl Ferdinand Braun m PGUIF6OJWFSTJUZPG4USBTCPVSH JOWFOUFEUIF#SBVODBUIPEFSBZUVCF JO5IJTUIFOCFDBNF UIFCB
TJTGPSUIFQJDUVSFUVCFVTFEGPSTPNBOZZFBSTGPSUFMFWJTJPOT*UJTTUJMM
the most economical device today, although the price of flat-screen sys
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m UPEFWFMPQUIFJDPOPTDPQFTPUIBUUIFNPEFSOUFMFWJTJPO
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Solution:
5IFDIBSHFPOBOFMFDUSPOJT
F=−×−$
Example 1.8
18
Chapter 1
Basic Concepts
*GUIFOVNCFSPGFMFDUSPOTJTO UIFOR=OFBOE
i
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−
−
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@@
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J=@@=F
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Figure 1.18
A simplified diagram of the cathode-ray
UVCFGPS&YBNQMF
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site to electron flow as shown in Fig. 1.18, which is a simplified dia
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Practice Problem 1.8
*GBOFMFDUSPOCFBNJOB57QJDUVSFUVCFDBSSJFT FMFDUSPOTTFDPOE
BOEJTQBTTJOHUISPVHIQMBUFTNBJOUBJOFEBUBQPUFOUJBMEJGGFSFODFPG
30 kV, calculate the power in the beam.
Answer:N8
1.7.2
Electricity Bills
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TABLE 1.3
Typical average monthly consumption of household
appliances.
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1.8
19
Problem Solving
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CJMMGPSUIFNPOUIVTJOHUIFGPMMPXJOHSFTJEFOUJBMSBUFTDIFEVMF
Example 1.9
#BTFNPOUIMZDIBSHFPG
'JSTUL8IQFSNPOUIBUDFOUTL8I
/FYUL8IQFSNPOUIBUDFOUTL8I
0WFSL8IQFSNPOUIBUDFOUTL8I
Solution:
8FDBMDVMBUFUIFFMFDUSJDJUZCJMMBTGPMMPXT
#BTFNPOUIMZDIBSHF=
'JSTUL8I!L8I=
/FYUL8I!L8I=
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++
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BWFSBHFDPTUQFSL8IJGPOMZL8IBSFDPOTVNFEJO+VMZXIFOUIF
GBNJMZJTPOWBDBUJPONPTUPGUIFUJNF
Answer:DFOUTL8I
1.8
Problem Solving
"MUIPVHIUIFQSPCMFNTUPCFTPMWFEEVSJOHPOFTDBSFFSXJMMWBSZJODPN
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8FXJMMMJTUUIFTUFQTTJNQMZBOEUIFOFMBCPSBUFPOUIFN
$BSFGVMMZdefineUIFQSPCMFN
1SFTFOUFWFSZUIJOHZPVLOPXBCPVUUIFQSPCMFN
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Practice Problem 1.9
20
Chapter 1
Basic Concepts
somewhat JODPNQMFUF :PVNVTU EPBMMZPV DBOUP NBLFTVSF ZPV
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process well since you will want to come back to it if the first approach
EPFTOPUXPSL
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problem definition, and the process continues. Following this process
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1.8
21
Problem Solving
Example 1.10
4PMWe for the current floXJOHUISPVHIUIFΩSFTJTUPSJO'JH
Solution:
$BSFGVMMZdefineUIFQSPCMFN5IJTJTPOMZBTJNQMFFYBNQMF CVUXF
DBOBMSFBEZTFFUIBUXFEPOPULOPXUIFQPMBSJUZPOUIF7TPVSDF8F
IBWFUIFGPMMPXJOHPQUJPOT8FDBOBTLUIFQSPGFTTPSXIBUUIFQPMBSJUZ
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UP EPOFYU*G XF IBWF UJNF UPXPSL UIF QSPCMFNCPUI XBZT XF DBO
TPMWFGPSUIFDVSSFOUXIFOUIF7TPVSDFJTQMVTPOUPQBOEUIFOQMVT POUIFCPUUPN*GXFEPOPUIBWFUIFUJNFUPXPSLJUCPUIXBZT BTTVNF
B QPMBSJUZBOEUIFODBSFGVMMZ EPDVNFOU ZPVSEFDJTJPO -FU VT BTTVNF
UIBUUIFQSPGFTTPSUFMMTVTUIBUUIFTPVSDFJTQMVTPOUIFCPUUPNBTTIPXO
in Fig. 1.20.
1SFTFOUFWFSZUIJOHZPVLOPXBCPVUUIFQSPCMFN1SFTFOUJOHBMMUIBU
XFLOPXBCPVUUIFQSPCMFNJOWPMWFTMBCFMJOHUIFDJSDVJUDMFBSMZTPUIBU
we define what we seek.
(JWFOUIFDJSDVJUTIPXOJO'JH TPMWFGPSJΩ
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&TUBCMJTIBTFUPG BMUFSOBUJWFTPMVUJPOTBOEEFUFSNJOFUIFPOFUIBU
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0INTMBX OPEBMBOBMZTJT BOENFTIBOBMZTJT
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UJPO CVUJUXJMMMJLFMZUBLFNPSFXPSLUIBOFJUIFSOPEBMPSNFTIBOBMZTJT
5PTPMWFGPSJΩVTJOHNFTIBOBMZTJTXJMMSFRVJSFXSJUJOHUXo TJNVMUBOF
ous equations to find the twPMPPQDVSSFOUTJOEJDBUFEJO'JH6TJOH
OPEBMBOBMZTJTSFRVJSFTTPMWJOHGPSPOMZPOFVOLOPXO5IJTJTUIFFBTJFTU
BQQSPBDI
i1
2Ω
+v
2Ω
5V +
–
v1
–
Loop 1
i3
i2
+
v8Ω
–
8Ω
4Ω
+ v4Ω –
Loop 2
–
+ 3V
Figure 1.21
6TJOHOPEBMBOBMZTJT
5IFSFGPSF XFXJMMTPMWFGPSJΩVTJOHOPEBMBOBMZTJT
"UUFNQUBQSPCMFNTPMVUJPO. We first write down all of the equations
we will need in order to find JΩ
J
Ω=J
v
@@
J
= v
@@
J
Ω= v− @@@@
v− @@@@
v+
@@@@
+ + =
2Ω
5V +
–
4Ω
8Ω
3V
Figure 1.19
*MMVTUSBUJWFFYBNQMF
2Ω
4Ω
i8 Ω
5V +
–
Figure 1.20
Problem definition.
8Ω
–
+ 3V
22
Chapter 1
Basic Concepts
/PXXFDBOTPMWFGPSv
[
]
v− @@@@@@
v − @@@@@@
v +
@@@@@@
+ + =
MFBETUP v− + v + v+ =
v=+ v=+7 v @@
@@
J
= ="
Ω= &WBMVBUFUIFTPMVUJPOBOEDIFDLGPSBDDVSBDZ 8FDBOOPXVTF
,JSDIIPGGTWPMUBHFMBX ,7- UPDIFDLUIFSFTVMUT
v− −
=−"
J=@@@@
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@@
=
=−
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v+ @@@
="
J=@@@@
+
=
=@@
J+J+ J=−++=
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−+vΩ+vΩ=−+ −J× + J× =−+<− − >+ × =−++=
$IFDLT "QQMZJOH,7-UPMPPQ
−vΩ+vΩ−=− J× + J× −
=− × + × −
=−+−=
$IFDLT 4PXFOP XIBWFBW FSZIJHIEF gree of confidence in the accuracZ
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5IJTQSPCMFNIBTCFFOTPMWFETBUJTGBDUPSJMZ
The current through the 8-Ω resistor is 0.25 A flowing down through the
8-Ω resistor.
Practice Problem 1.10
Try applying this process to some of the more difficult problems at the
FOEPGUIFDIBQUFS
1.9
Summary
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SFTVMUT'SPNUIFTF WFOQSJODJQBMVOJUT UIFVOJUTPGPUIFSQI ZTJDBM
RVBOUJUJFTDBOCFEFSJWFE
23
Review Questions
$VSSFOUJTUIFSBUFPGDIBS ge floXQBTUBHJ WFOQPJOUJOBHJ WFO
EJSFDUJPO
ER
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vBC=
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EX
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= vJ
Q=
EU
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current source produces a specific current through its terminals
SFHBSEMFTTPGXIBUJTDPOOFDUFEUPJU
7PMUBHFBOEDVSSFOUTPVSDFTDBOCFEFQFOEFOUPSJOEFQFOEFOU "
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WBSJBCMF
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Review Questions
0OFNJMMJWPMUJTPOFNJMMJPOUIPGBWPMU
B 5SVF
C 'BMTF
C D .7
E (7
C 'BMTF
C BNQFSF
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A charge of 2 C flowing past a given point each
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B C BNQFSFT
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io
vs +
–
6io
Figure 1.22
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24
Chapter 1
Basic Concepts
Problems
Section 1.3
Charge and Current
q (C)
10
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26
Chapter 1
Section 1.7
Basic Concepts
Applications
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27
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Basic Laws
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30
Chapter 2
Basic Laws
Learning Objectives
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31
Ohm’s Law
TABLE 2.1
Resistivities of common materials.
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Chapter 2
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A resistor is either fixFEPSW ariable. Most resistors are of the fixFE
UZQF NFBOJOHUIFJS SFTJTUBODFSFNBJOTDPOTUBOU5IFUXPDPNNPOUZQFT
of fixFESFTJTUPST XJSFXPVOEBOEDPNQPTJUJPO BSFTIPwn in Fig. 5IF
DPNQPTJUJPOSFTJTUPSTBSFVTFEXIFOMBSHFSFTJTUBODFJTOFFEFEThe DJSDVJU
symbol in Fig. 2.1(b) is for a fixFESFTJTUPS7BSJBCMFSFTJTUPSTIBWFBEKVTU
BCMFSFTJTUBODF5IFTZNCPMGPSBWBSJBCMFSFTJTUPSJTTIPXOJO'JH B "DPNNPOWBSJBCMFSFTJTUPSJTLOPXOBTBQPUFOUJPNFUFSPSQPUGPSTIPSU XJUIUIFTZNCPMTIPXOJO'JH C 5IFQPUJTBUISFFUFSNJOBMFMFNFOU
XJUIBTMJEJOHDPOUBDUPSXJQFS #ZTMJEJOHUIFXJQFS UIFSFTJTUBODFTCF tween the wiper terminal and the fixFEUFSNJOBMTW BSZ-JLe fixFESFTJT
UPST WBSJBCMFSFTJTUPSTDBOCFPGFJUIFSXJSF XPVOEPSDPNQPTJUJPOUZQF BTTIPXOJO'JH"MUIPVHISFTJTUPSTMJLFUIPTFJO'JHTBOEBSF
VTFEJODJSDVJUEFTJHOT UPEBZNPTUDJSDVJUDPNQPOFOUTJODMVEJOHSFTJTUPST
BSFFJUIFSTVSGBDFNPVOUFEPSJOUFHSBUFE BTUZQJDBMMZTIPXOJO'JH
2.2
(a)
33
Ohm’s Law
(b)
Figure 2.5
*UTIPVMECFQPJOUFEPVUUIBUOPUBMMSFTJTUPSTPCF Z0INTMB X"
SFTJTUPSUIBUPCFZT0INTMBXJTLOPXOBTBMJOFBSSFTJTUPS*UIBTBDPO
TUBOUSFTJTUBODFBOEUIVTJUTDVSSFOUWPMUBHFDIBSBDUFSJTUJDJTBTJMMVTUSBUFE
JO'JH B *UT JvHSBQIJTBTUSBJHIUMJOFQBTTJOHUISPVHIUIFPSJ HJO"OPOMJOFBSSFTJTUPSEPFTOPUPCF Z0INTMBX*UTSFTJTUBODFWBSJFT
XJUIDVSSFOUBOEJUT JvDIBSBDUFSJTUJDJTUZQJDBMMZTIP wn in Fig. C &YBNQMFTPGEFWJDFTXJUIOPOMJOFBSSFTJTUBODFBSFUIFMJHIUCVMCBOEUIF
EJPEF"MUIPVHIBMMQSBDUJDBMSFTJTUPSTNBZF YIJCJUOPOMJOFBSCFIB WJPS
VOEFSDFSUBJODPOEJUJPOT XFXJMMBTTVNFJOUIJTCPPLUIBUBMMFMFNFOUT
BDUVBMMZEFTJHOBUFEBTSFTJTUPSTBSFMJOFBS
"VTFGVMRVBOUJUZJODJSDVJUBOBMZTJTJTUIFSFDJQSPDBMPGSFTJTUBODF3 LOPXOBTDPOEVDUBODFBOEEFOPUFECZ(
vJ
( = @@
= @@
3
4=℧="7
Figure 2.6
3FTJTUPSTJOBOJOUFHSBUFEDJSDVJUCPBSE
v
5IFDPOEVDUBODFJTBNFBTVSFPGIPXXFMMBOFMFNFOUXJMMDPOEVDU
FMFDUSJDDVSSFOU5IFVOJUPGDPOEVDUBODFJTUIFNIP PINTQFMMFECBDL
XBSE PSSFDJQSPDBMPIN XJUITZNCPM ℧ UIFJOWFSUFEPNFHB"MUIPVHI
FOHJOFFSTPGUFOVTFUIFNIP JOUIJTCPPLXFQSFGFSUPVTFUIFTJFNFOT
4 UIF4*VOJUPGDPOEVDUBODF
5IVT
¥&SJD5PNFZ"MBNZ3'
7BSJBCMFSFTJTUPST B DPNQPTJUJPOUZQF C TMJEFSQPU
¥.D(SBX)JMM&EVDBUJPO.BSL%JFSLFS QIPUPHSBQIFS
Slope = R
i
(a)
v
Slope = R
Conductance is the ability of an element to conduct electric current; it is
measured in mhos (℧) or siemens (S).
i
(b)
5IFTBNFSFTJTUBODFDBOCFF YQSFTTFEJOPINTPSTJFNFOT'
FYBNQMF ΩJTUIFTBNFBT4'SPN&R XFNBZXSJUF
J(v
PS
5IFQPXFSEJTTJQBUFECZBSFTJTUPSDBOCFF YQSFTTFEJOUFSNTPG 3
6TJOH&RT BOE Q =vJ=J3=@@
v 3
Figure 2.7
5IFJvDIBSBDUFSJTUJDPG B BMJOFBS
SFTJTUPS C BOPOMJOFBSSFTJTUPS
34
Chapter 2
Basic Laws
5IFQPXFSEJTTJQBUFECZBSFTJTUPSNBZBMTPCFF YQSFTTFEJOUFSNTPG
(BT
Q=vJ=v(=@@
J (
8FTIPVMEOPUFUXPUIJOHTGSPN&RT BOE 5IFQPXFSEJTTJQBUFEJOBSFTJTUPSJTBOPOMJOFBSGVODUJPOPGFJUIFS
DVSSFOUPSWPMUBHF
4JODF3BOE(BSFQPTJUJWFRVBOUJUJFT UIFQPXFSEJTTJQBUFEJOBSFTJT
UPSJTBM XBZTQPTJUJWF5IVT BSFTJTUPSBM XBZTBCTPSCTQP XFSGSPN
UIFDJSDVJUThis confirms the idea that a resistor is a passiWFFMFNFOU JODBQBCMFPGHFOFSBUJOHFOFSHZ
Example 2.1
"OFMFDUSJDJSPOESBXT"BU7'JOEJUTSFTJTUBODF
Solution:
'SPN0INTMBX
=Ω
3=@v =@@@
J
Practice Problem 2.1
5IFFTTFOUJBMDPNQPOFOUPGBUPBTUFSJTBOFMFDUSJDBMFMFNFOU BSFTJT UPS UIBUDPOWFSUTFMFDUSJDBMFOFSHZUPIFBUFOFSHZ)PXNVDIDVSSFOUJT
ESBXOCZBUPBTUFSXJUISFTJTUBODFΩBU7
Answer:"
Example 2.2
*OUIFDJSDVJUTIPXOJO'JH DBMDVMBUFUIFDVSSFOU J UIFDPOEVDUBODF
( BOEUIFQPXFSQ
i
30 V +
–
5 kΩ
+
v
–
Solution:
5IFWPMUBHFBDSPTTUIFSFTJTUPSJTUIFTBNFBTUIFTPVSDFWPMUBHF 7 CFDBVTFUIFSFTJTUPSBOEUIFWPMUBHFTPVSDFBSFDPOOFDUFEUPUIFTBNF
QBJSPGUFSNJOBMT)FODF UIFDVSSFOUJT
v=@@@@@
J=@@
=N"
3 ×
Figure 2.8
'PS&YBNQMF
5IFDPOEVDUBODFJT
(=@@
=@@@@@
=.N4
3 ×
8FDBODBMDVMBUFUIFQPXFSJOWBSJPVTXBZTVTJOHFJUIFS&RT PS Q=vJ= ×− =N8
PS
Q=J3= ×− ×=N8
PS
Q=v(= .×−=N8
2.3
35
Nodes, Branches, and Loops
'PSUIFDJSDVJUTIPXOJO'JH DBMDVMBUFUIFWPMUBHFv UIFDPOEVDUBODF
( BOEUIFQPXFSQ
Practice Problem 2.2
i
Answer:7 µ4 N8
3 mA
10 kΩ
+
v
–
Figure 2.9
'PS1SBDUJDF1SPC
Example 2.3
"WPMUBHFTPVSDFPGTJOπU7JTDPOOFDUFEBDSPTTBLΩSFTJTUPS'JOE
UIFDVSSFOUUISPVHIUIFSFTJTUPSBOEUIFQPXFSEJTTJQBUFE
Solution:
v=@@@@@@@
J=@@
TJOπU
=TJOπUN"
3 ×
)FODF
Q=vJ=TJOπUN8
"SFTJTUPSBCTPSCTBOJOTUBOUBOFPVTQPXFSPGDPT UN8XIFODPO OFDUFEUPBWPMUBHFTPVSDFW=DPTU7'JOEJBOE3
Practice Problem 2.3
Answer:DPTUN" LΩ
2.3
Nodes, Branches, and Loops
4JODFUIFFMFNFOUTPGBOFMFDUSJDDJSDVJUDBOCFJOUFSDPOOFDUFEJOTF WFSBM
XBZT XFOFFEUPVOEFSTUBOETPNFCBTJDDPODFQUTPGOFUXPSLUPQPMPHZ5P
EJGGFSFOUJBUFCFUXFFOBDJSDVJUBOEBOFUXPSL XFNBZSFHBSEBOFUXPSLBT
BOJOUFSDPOOFDUJPOPGFMFNFOUTPSEFWJDFT XIFSFBTBDJSDVJUJTBOFUXPSL
QSPWJEJOHPOFPSNPSFDMPTFEQBUIT 5IFDPOWFOUJPO XIFOBEESFTTJOH
OFUXPSLUPQPMPHZ JTUPVTFUIFX PSEOFUXPSLSBUIFSUIBODJSDVJU 8FEP
UIJTFWFOUIPVHIUIFXPSEOFUXPSLBOEDJSDVJUNFBOUIFTBNFUIJOHXIFO
VTFEJOUIJTDPOUFYU*OOFUXPSLUPQPMPHZ XFTUVEZUIFQSPQFSUJFTSFMBUJOH
UPUIFQMBDFNFOUPGFMFNFOUTJOUIFOFUXork and the geometric configura
UJPOPGUIFOFUXPSL4VDIFMFNFOUTJODMVEFCSBODIFT OPEFT BOEMPPQT
A branch represents a single element such as a voltage source or a resistor.
*OPUIFSX PSET BCSBODISFQSFTFOUTBO ZUXPUFSNJOBMFMFNFOU 5IFDJS
cuit in Fig. 2.10 has fivFCSBODIFT OBNFMZ UIF7WPMUBHFTPVSDF UIF
"DVSSFOUTPVSDF BOEUIFUISFFSFTJTUPST
A node is the point of connection between two or more branches.
"OPEFJTVTVBMMZJOEJDBUFECZBEPUJOBDJSDVJU*GBTIPSUDJSDVJU
BDPOOFDUJOHXJSF DPOOFDUTUX POPEFT UIFUXPOPEFTDPOTUJUVUFBTJO HMFOPEF5IFDJSDVJUJO'JHIBTUISFFOPEFTB C BOED/PUJDFUIBU
a
5Ω
10 V +
–
b
2Ω
3Ω
c
Figure 2.10
/PEFT CSBODIFT BOEMPPQT
2A
36
Chapter 2
b
5Ω
2Ω
a
3Ω
+
–
10 V
2A
UIFUISFFQPJOUTUIBUGPSNOPEF CBSFDPOOFDUFECZQFSGFDUMZDPOEVDUJOH
XJSFTBOEUIFSFGPSFDPOTUJUVUFBTJOHMFQPJOU5IFTBNFJTUSVFPGUIFGPVS
QPJOUTGPSNJOHOPEF D8FEFNPOTUSBUFUIBU UIFDJSDVJUJO'JHIBT
POMZUISFFOPEFTCZSFESBXJOHUIFDJSDVJUJO'JH5IFUXPDJSDVJUTJO
'JHTBOEBSFJEFOUJDBM)PXFWFS GPSUIFTBLFPGDMBSJUZ OPEFT
CBOEDBSFTQSFBEPVUXJUIQFSGFDUDPOEVDUPSTBTJO'JH
A loop is any closed path in a circuit.
c
Figure 2.11
5IFUISFFOPEFDJSDVJUPG'JHJT
SFESBXO
Basic Laws
"MPPQJTBDMPTFEQBUIGPSNFECZTUBSUJOHBUBOPEF QBTTJOHUISPVHIB
TFUPGOPEFT BOESFUVSOJOHUPUIFTUBSUJOHOPEFXJUIPVUQBTTJOHUISPVHI
BOZOPEFNPSFUIBOPODF"MPPQJTTBJEUPCFJOEFQFOEFOUJGJUDPOUBJOT
BUMFBTUPOFCSBODIXIJDIJTOPUBQBSUPGBO ZPUIFSJOEFQFOEFOUMPPQ
*OEFQFOEFOUMPPQTPSQBUITSFTVMUJOJOEFQFOEFOUTFUTPGFRVBUJPOT
*UJTQPTTJCMFUPGPSNBOJOEFQFOEFOUTFUPGMPPQTXIFSFPOFPGUIF MPPQTEPFTOPUDPOUBJOTVDIBCSBODI*O'JH BCDBXJUIUIF ΩSF
TJTUPSJTJOEFQFOEFOU"TFDPOEMPPQXJUIUIF ΩSFTJTUPSBOEUIFDVSSFOU TPVSDFJTJOEFQFOEFOU5IFUIJSEMPPQDPVMECFUIFPOFXJUIUIFΩSFTJTUPS
JOQBSBMMFMXJUIUIFΩSFTJTUPS5IJTEPFTGPSNBOJOEFQFOEFOUTFUPGMPPQT
"OFUXPSLXJUI CCSBODIFT OOPEFT BOE MJOEFQFOEFOUMPPQTXJMM
TBUJTGZUIFGVOEBNFOUBMUIFPSFNPGOFUXPSLUPQPMPHZ
C=M+O−
"TUIFOFYUUXo definitions shoX DJSDVJUUPQPMPHZJTPGHSFBUWBMVFUP
UIFTUVEZPGWPMUBHFTBOEDVSSFOUTJOBOFMFDUSJDDJSDVJU
Two or more elements are in series if they exclusively share a single node
and consequently carry the same current.
Two or more elements are in parallel if they are connected to the same
two nodes and consequently have the same voltage across them.
&MFNFOUTBSFJOTFSJFTXIFOUIF ZBSFDIBJODPOOFDUFEPSDPOOFDUFETF RVFOUJBMMZ FOEUPFOE' PSFYBNQMF UXPFMFNFOUT BSFJOTFSJFT JGUIFZ
TIBSFPOFDPNNPOOPEFBOEOPPUIFSFMFNFOUJTDPOOFDUFEUPUIBU
DPNNPO OPEF &MFNFOUT JO QBSBMMFM BSF DPOOFDUFE UP UIF TBNF QBJS PG
UFSNJOBMT&MFNFOUTNBZCFDPOOFDUFEJOBX BZUIBUUIFZBSFOFJUIFSJO
TFSJFTOPSJOQBSBMMFM*OUIFDJSDVJUTIPXOJO'JH UIFWPMUBHFTPVSDF
BOEUIF Ω resistor are in series because the same current will floX
UISPVHIUIFN5IFΩSFTJTUPS UIFΩSFTJTUPS BOEUIFDVSSFOUTPVSDF
BSFJOQBSBMMFMCFDBVTFUIFZBSFDPOOFDUFEUPUIFTBNFUXPOPEFTCBOED
BOEDPOTFRVFOUMZIBWFUIFTBNFWPMUBHFBDSPTTUIFN5IFΩBOEΩ
SFTJTUPSTBSFOFJUIFSJOTFSJFTOPSJOQBSBMMFMXJUIFBDIPUIFS
Example 2.4
%FUFSNJOF UIFOVNCFS PGCSBODIFTBOE OPEFTJO UIFDJSDVJU TIPXOJO
'JH*EFOUJGZXIJDIFMFNFOUTBSFJOTFSJFTBOEXIJDIBSFJOQ BSBMMFM
Solution:
4JODFUIFSFBSFGPVSFMFNFOUTJOUIFDJSDVJU UIFDJSDVJUIBTGPVSCSBODIFT
7 Ω Ω,BOE"5IFcircuit has three nodes as identified in FJH
5IFΩSFTJTUPSJTJOTFSJFTXJUIUIF7WPMUBHFTPVSDFCFDBVTFUIFTBNF
current would flow in both. The 6-Ω resistor is in parallel with the 2-A cur
SFOUTPVSDFCFDBVTFCPUIBSFDPOOFDUFEUPUIFTBNFOPEFTBOE
2.4
5Ω
10 V
1
+
–
6Ω
10 V
2A
5Ω
2
+
–
Figure 2.12
37
Kirchhoff’s Laws
2A
6Ω
3
'PS&YBNQMF
Figure 2.13
5IFUISFFOPEFTJOUIFDJSDVJUPG
Fig. 2.12.
Practice Problem 2.4
)PXNBOZCSBODIFTBOEOPEFTEPFTUIFDJSDVJUJO'JHIBWF *EFOUJGZ
UIFFMFNFOUTUIBUBSFJOTFSJFTBOEJOQBSBMMFM
Answer:'JWe branches and three nodes are identified in Fig. 2.15. 5IF
ΩBOEΩSFTJTUPSTBSFJOQBSBMMFM 5IFΩSFTJTUPSBOE7TPVSDF
BSFBMTPJOQBSBMMFM
5Ω
1Ω
2Ω
5Ω
1
+ 10 V
–
Figure 2.14
4Ω
1Ω
2
+ 10 V
–
2Ω
3
'PS1SBDUJDF1SPC
Figure 2.15
"OTXFSGPS1SBDUJDF1SPC
2.4
Kirchhoff’s Laws
0INTMBXCZJUTFMGJTOPUTVGficient to analyze circuits. HoXFWFS XIFOJU
JTDPVQMFEXJUI,JSDIIPGGTUXPMBXT XFIBWFBTVGficient, poXFSGVMTFU
PGUPPMTGPSBOBMZ[JOHBMBSHFWBSJFUZPGFMFDUSJDDJSDVJUT,JSDIIPGGTMBXT
were first introduced in 1847 by the German phZTJDJTU(VTUB W3PCFSU
,JSDIIPGG m 5IFTFMBXTBSFGPSNBMMZLOP XOBT,JSDIIPGGT
DVSSFOUMBX ,$- BOE,JSDIIPGGTWPMUBHFMBX ,7- ,JSDIIPGGs first laXJTCBTFEPOUIFMB XPGDPOTFSWBUJPOPGDIBSHF XIJDISFRVJSFTUIBUUIFBMHFCSBJDTVNPGDIBSHFTXJUIJOBTZTUFNDBOOPU
DIBOHF
Kirchhoff’s current law (KCL) states that the algebraic sum of currents
entering a node (or a closed boundary) is zero.
.BUIFNBUJDBMMZ ,$-JNQMJFTUIBU
/
∑JO = O=
XIFSF/JTUIFOVNCFSPGCSBODIFTDPOOFDUFEUPUIFOPEFBOEJOJTthe OUI
DVSSFOUFOUFSJOH PSMFB WJOH UIFOPEF#ZUIJTMB X DVSSFOUTFOUFSJOHB
OPEFNBZCFSF HBSEFEBTQPTJUJWF XIJMFDVSSFOUTMFB WJOHUIFOPEFNBZ
CFUBLFOBTOFHBUJWFPSWJDFWFSTB
4Ω
38
Chapter 2
Basic Laws
Historical
¥1JYUBMBHF'PUPTUPDL3'
Gustav Robert Kirchhoff m B(FSNBOQIZTJDJTU TUBUFE
UXPCBTJDMBXTJODPODFSOJOHUIFSFMBUJPOTIJQCFUXFFOUIFDVS
SFOUTBOEWPMUBHFTJOBOFMFDUSJDBMOFUXPSL,JSDIIPGGTMBXT BMPOHXJUI
0INTMBX GPSNUIFCBTJTPGDJSDVJUUIFPSZ
#PSOUIFTPOPGBMBXZFSJO,POJHTCFSH &BTU1SVTTJB ,JSDIIPGG
FOUFSFEUIF6OJWFSTJUZPG,POJHTCFSHBUBHFBOEMBUFSCFDBNFBMFDUVS
FSJO#FSMJO)JTDPMMBCPSBUJWFXPSLJOTQFDUSPTDPQZXJUI(FSNBODIFN
JTU3PCFSU#VOTFOMFEUPUIFEJTDPWFSZPGDFTJVNJOBOESVCJEJVN
JO,JSDIIPGGXBTBMTPDSFEJUFEXJUIUIF,JSDIIPGGMBXPGSBEJBUJPO
5IVT ,JSDIIPGGJTGBNPVTBNPOHFOHJOFFST DIFNJTUT BOEQIZTJDJTUT
5PQSPWF,$- BTTVNFBTFUPGDVSSFOUTJL U L= … floXJOUP
BOPEF5IFBMHFCSBJDTVNPGDVSSFOUTBUUIFOPEFJT
J5 U =J U +J U +J U +⋯ *OUFHSBUJOHCPUITJEFTPG&R HJWFT
i5
i1
i2
R5 U =R U +R U +R U +⋯ i4
i3
Figure 2.16
$VSSFOUTBUBOPEFJMMVTUSBUJOH,$-
XIFSFRL U =∫JL U EUBOER5 U =∫J5 U EU#VUUIFMBXPGDPOTFSWB
UJPOPGFMFDUSJDDIBSHFSFRVJSFTUIBUUIFBMHFCSBJDTVNPGFMFDUSJDDIBSHFT
BUUIFOPEFNVTUOPUDIBOHFUIBUJT UIFOPEFTUPSFTOPOFUDIBSHF5IVT R5 U = → J5 U = confirming the vBMJEJUZPG,$-
$POTJEFSUIFOPEFJO'JH"QQMZJOH,$-HJWFT
J + −J +J +J + −J =
Closed boundary
TJODFDVSSFOUTJ J BOEJBSFFOUFSJOHUIFOPEF XIJMFDVSSFOUT JBOEJ
BSFMFBWJOHJU#ZSFBSSBOHJOHUIFUFSNT XFHFU
J +J +J = J +J &RVBUJPO JTBOBMUFSOBUJWFGPSNPG,$-
The sum of the currents entering a node is equal to the sum of the currents leaving the node.
Figure 2.17
"QQMZJOH,$-UPBDMPTFECPVOEBSZ
Two sources (or circuits in general) are
said to be equivalent if they have the
same i-v relationship at a pair of
terminals.
/PUFUIBU,$-BMTPBQQMJFTUPBDMPTFECPVOEBSZ 5IJTNBZCFSF HBSEFEBTBHFOFSBMJ[FEDBTF CFDBVTFBOPEFNBZCFSFHBSEFEBTBDMPTFE
TVSGBDFTISVOLUPBQPJOU*OUX PEJNFOTJPOT BDMPTFECPVOEBSZJTUIF
TBNFBTBDMPTFEQBUI"TUZQJDBMMZJMMVTUSBUFEJOUIFDJSDVJUPG'JH UIFUPUBMDVSSFOUFOUFSJOHUIFDMPTFETVSGBDFJTFRVBMUPUIFUPUBMDVSSFOU
MFBWJOHUIFTVSGBDF
"TJNQMFBQQMJDBUJPOPG,$-JTDPNCJOJOHDVSSFOUTPVSDFTJOQBSBM
MFM5IFDPNCJOFEDVSSFOUJTUIFBMHFCSBJDTVNPGUIFDVSSFOUTVQQMJFE
CZUIFJOEJ WJEVBMTPVSDFT' PSF YBNQMF UIFDVSSFOUTPVSDFTTIP XOJO
2.4
39
Kirchhoff’s Laws
'JH B DBOCFDPNCJOFEBTJO'JH C 5IFDPNCJOFEPSFRVJWB
MFOUDVSSFOUTPVSDFDBOCFGPVOECZBQQMZJOH,$-UPOPEFB
IT
a
*5+ *= *+ *
I2
I1
PS
I3
b
*5 =* −* +* (a)
IT
"DJSDVJUDBOOPUDPOUBJOUXPEJGGFSFOUDVSSFOUT *BOE* JOTFSJFT VOMFTT
*=*PUIFSXJTF,$-XJMMCFWJPMBUFE
,JSDIIPGGTTFDPOEMBXJTCBTFEPOUIFQSJODJQMFPGDPOTFSW BUJPOPG
FOFSHZ
a
IT = I1 – I2 + I3
b
Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages
around a closed path (or loop) is zero.
&YQSFTTFENBUIFNBUJDBMMZ ,7-TUBUFTUIBU
(b)
Figure 2.18
$VSSFOUTPVSDFTJOQBSBMMFM B PSJHJOBM
DJSDVJU C FRVJWBMFOUDJSDVJU
.
XIFSF.JTUIFOVNCFSPGWPMUBHFTJOUIFMPPQ PSUIFOVNCFSPGCSBODIFT
JOUIFMPPQ BOEWNJTUIFNUIWPMUBHF
5PJMMVTUSBUF,7- DPOTJEFSUIFDJSDVJUJO'JH 5IFTJHOPO
FBDIWoltage is the polarity of the terminal encountered first as we traWFM
BSPVOEUIFMPPQ 8FDBOTUBSUXJUIBO ZCSBODIBOEHPBSPVOEUIFMPPQ
FJUIFSDMPDLXJTFPSDPVOUFSDMPDLXJTF4VQQPTFXFTUBSUXJUIUIFW PMU
BHFTPVSDFBOEHPDMPDLXJTFBSPVOEUIFMPPQBTTIP XOUIFOW PMUBHFT
XPVMECF−W +W +W +W BOE−W JOUIBUPSEFS'PSFYBNQMF BTXF
SFBDICSBODI UIFQPTJUJWe terminal is met first; hence, we haWF+W'PS
CSBODI XFSFBDIUIFOF HBUJWe terminal first; hence, −W5IVT ,7-
ZJFMET
XIJDINBZCFJOUFSQSFUFEBT
Sum of voltage drops = Sum of voltage rises
Figure 2.19
5IJTJTBOBMUFSOBUJWFGPSNPG,7-/PUJDFUIBUJGXFIBEUSBWFMFEDPVO
UFSDMPDLXJTF UIFSFTVMUX PVMEIBWFCFFO +W −W +W −W BOE −W XIJDIJTUIFTBNFBTCFGPSFF YDFQUUIBUUIFTJHOTBSFSF WFSTFE)FODF &RT BOE SFNBJOUIFTBNF
8IFOWPMUBHFTPVSDFTBSFDPOOFDUFEJOTFSJFT ,7-DBOCFBQQMJFE
UPPCUBJOUIFUPUBMW PMUBHF5IFDPNCJOFEW PMUBHFJTUIFBMHFCSBJDTVN
PGUIFW PMUBHFTPGUIFJOEJ WJEVBMTPVSDFT' PSFYBNQMF GPSUIFW PMUBHF
TPVSDFTTIPXOJO'JH B UIFDPNCJOFEPSFRVJWBMFOUWPMUBHFTPVSDF
JO'JH C JTPCUBJOFECZBQQMZJOH,7-
−7B C+7 +7 −7 =
–
+ v4
v1 +
–
3FBSSBOHJOHUFSNTHJWFT
W+W+W = W+W
+ v3 –
+ v2 –
–
−W+W+W−W+W= KVL can be applied in two ways: by
taking either a clockwise or a counterclockwise trip around the loop. Either
way, the algebraic sum of voltages
around the loop is zero.
v5
+
∑
WN= N=
"TJOHMFMPPQDJSDVJUJMMVTUSBUJOH,7-
40
Chapter 2
Basic Laws
PS
7B C=7 +7 −7 5PBWPJEWJPMBUJOH,7- BDJSDVJUDBOOPUDPOUBJOUX PEJGGFSFOUWPMUBHFT
7BOE7JOQBSBMMFMVOMFTT7=7
a
+
+ V
1
–
+ V
2
–
Vab
a
–
+ V =V +V –V
S
1
2
3
–
Vab
–
+ V3
b
+
b
–
(a)
(b)
Figure 2.20
7PMUBHFTPVSDFTJOTFSJFT B PSJHJOBMDJSDVJU C FRVJWBMFOUDJSDVJU
'or the circuit in Fig. 2.21(a), find vPMUBHFTvBOEv
2Ω
2Ω
+ v1 –
+ v1 –
v2
20 V +
–
3Ω
i
+
(a)
v2
+
–
20 V +
–
–
Example 2.5
3Ω
(b)
Figure 2.21
'PS&YBNQMF
Solution:
To find vBOE vXFBQQMZ0INTMBXBOE,JSDIIPGGTWPMUBHFMBX "TTVNFUIBUDVSSFOU J flows through the loop as shown in Fig. 2.21(b).
'SPN0INTMBX
v=J v=−J
−+v−v=
"QQMZJOH,7-BSPVOEUIFMPPQHJWFT
4VCTUJUVUJOH&R JOUP&R XFPCUBJO
−+J+J=
PS
J=
4VCTUJUVUJOHJ in Eq. (2.5.1) finally gives
W=7 W=−7
⇒
J="
2.4
41
Kirchhoff’s Laws
Practice Problem 2.5
'JOEWBOEWJOUIFDJSDVJUPG'JH
4Ω
+ v1 –
32 V +
–
Figure 2.22
+ v2 –
+
–
Answer:7 −7
–8 V
2Ω
'PS1SBDUJDF1SPC
Example 2.6
%FUFSNJOFWPBOEJJOUIFDJSDVJUTIPXOJO'JH B i
4Ω
4Ω
2vo
+ –
–
4V +
12 V +
–
2vo
+ –
i
12 V +
–
–
4V +
6Ω
6Ω
+ vo –
+ vo –
(a)
(b)
Figure 2.23
'PS&YBNQMF
Solution:
8FBQQMZ,7-BSPVOEUIFMPPQBTTIPXOJO'JH C 5IFSFTVMUJT
−+J+vP−+J=
"QQMZJOH0INTMBXUPUIFΩSFTJTUPSHJWFT
vP=−J
4VCTUJUVUJOH&R JOUP&R ZJFMET
−+J−J=
⇒
J=−"
BOEWP=7
'JOEWYBOEWPJOUIFDJSDVJUPG'JH
Practice Problem 2.6
Answer:7 −7
10 Ω
70 V +
–
+ vx –
5Ω
+ vo –
Figure 2.24
'PS1SBDUJDF1SPC
+
–
2vx
42
Chapter 2
Example 2.7
'JOEDVSSFOUJPBOEWPMUBHFWPJOUIFDJSDVJUTIPXOJO'JH
Solution:
"QQMZJOH,$-UPOPEFB XFPCUBJO
a
io
+
vo
–
0.5io
Basic Laws
+.JP =JP 4Ω
3A
⇒
JP ="
'PSUIFΩSFTJTUPS 0INTMBXHJWFT
WP=JP =7
Figure 2.25
'PS&YBNQMF
Practice Problem 2.7
15 A
2Ω
'JOEWPBOEJPJOUIFDJSDVJUPG'JH
Answer:7 "
io
3 i
12 Ω
o
3
+
vo
–
Figure 2.26
'PS1SBDUJDF1SPC
Example 2.8
'JOEDVSSFOUTBOEWPMUBHFTJOUIFDJSDVJUTIPXOJO'JH B 8Ω
i1
+ v1 –
30 V +
–
+
v2
–
a
i3
8Ω
i2
+ v1 –
+
v3
–
3Ω
6Ω
30 V +
–
Loop 1
(a)
i1
a
i3
i2
+
v2
–
3 Ω Loop 2
+
v3
–
6Ω
(b)
Figure 2.27
'PS&YBNQMF
Solution:
8FBQQMZ0INTMBXBOE,JSDIIPGGTMBXT#Z0INTMBX
W=J W=J W=J 4JODFUIFWPMUBHFBOEDVSSFOUPGFBDISFTJTUPSBSFSFMBUFECZ0IN T
MBXBTTIPXO XFBSFSFBMMZMPPLJOHGPSUISFFUIJOHTW W W PS J J J "UOPEFB ,$-HJWFT
J −J −J = "QQMZJOH,7-UPMPPQBTJO'JH C
−+W+W = 2.5
43
Series Resistors and Voltage Division
8FFYQSFTTUIJTJOUFSNTPGJBOEJBTJO&R UPPCUBJO
−+J +J =
PS
−J J =@@@@@@
"QQMZJOH,7-UPMPPQ
−W+W=
⇒
W=W
BTFYQFDUFETJODFUIFUXPSFTJTUPSTBSFJOQBSBMMFM8FFYQSFTTWBOEWJO
UFSNTPGJBOEJBTJO&R &RVBUJPO CFDPNFT
J =J J J =@@
⇒
4VCTUJUVUJOH&RT BOE JOUP HJWFT
−J J @@@@@
=
−J −@@
PSJ = "'SPNUIFW BMVFPG J XFOP XVTF&RT UP UP
PCUBJO
J = " J = " W = 7 W = 7 W = 7
Practice Problem 2.8
'JOEUIFDVSSFOUTBOEWPMUBHFTJOUIFDJSDVJUTIPXOJO'JH
Answer:W=7 W=7 W=7 J=" J=N" J="
2.5
Series Resistors and Voltage Division
2Ω
+ v1 –
10 V +
–
i1
i3
4Ω
i2 + v3 –
+
v2
–
–
+
8Ω
6V
5IFOFFEUPDPNCJOFSFTJTUPSTJOTFSJFTPSJOQBSBMMFMPDDVSTTPGSFRVFOUMZ
Figure 2.28
UIBUJUXBSSBOUTTQFDJBMBUUFOUJPO5IFQSPDFTTPGDPNCJOJOHUIFSFTJTUPST 'PS1SBDUJDF1SPC
JTGBDJMJUBUFECZDPNCJOJOHUXPPGUIFNBUBUJNF8JUIUIJTJONJOE DPO
TJEFSUIFTJOHMFMPPQDJSDVJUPG'JH 5IFUXPSFTJTUPSTBSFJOTFSJFT TJODFUIFTBNFDVSSFOU J floXTJOCPUIPGUIFN "QQMZJOH0INTMBXUP
FBDIPGUIFSFTJTUPST XFPCUBJO
W=J3 W=J3 *GXFBQQMZ,7-UPUIFMPPQ NP WJOHJOUIFDMPDLXJTFEJSFDUJPO XF
IBWF
−W+W+W = v
R1
R2
+ v1 –
+ v2 –
+
–
b
PS
J=@@@@@
W 3 3 a
$PNCJOJOH&RT BOE XFHFU
W=W+W=J 3 +3 i
Figure 2.29
"TJOHMFMPPQDJSDVJUXJUIUXPSFTJTUPST
JOTFSJFT
44
Chapter 2
Basic Laws
/PUJDFUIBU&R DBOCFXSJUUFOBT
W=J3FR i
Req
a
JNQMZJOHUIBUUIFUXPSFTJTUPSTDBOCFSFQMBDFECZBOFRVJWBMFOUSFTJTUPS
3FRUIBUJT
+ v –
3FR=3 +3 +
–
v
b
Figure 2.30
&RVJWBMFOUDJSDVJUPGUIF'JHDJSDVJU
5IVT 'JHDBOCFSFQMBDFECZUIFFRVJWBMFOUDJSDVJUJO'JH5IF
UXPDJSDVJUTJO'JHTBOEBSFFRVJWBMFOUCFDBVTFUIFZFYIJCJUUIF
TBNFW PMUBHFDVSSFOUSFMBUJPOTIJQTBUUIFUFSNJOBMT BC"O FRVJWBMFOU
DJSDVJUTVDIBTUIFPOFJO'JHJTVTFGVMJOTJNQMJGZJOHUIFBOBMZTJT
PGBDJSDVJU*OHFOFSBM
The equivalent resistance of any number of resistors connected in series
is the sum of the individual resistances.
Resistors in series behave as a single
resistor whose resistance is equal to
the sum of the resistances of the
individual resistors.
'PS/SFTJTUPSTJOTFSJFTUIFO
/
3FR=3 +3 +⋯+3/ =∑3O O=
5PEFUFSNJOFUIFWPMUBHFBDSPTTFBDISFTJTUPSJO'JH XFTVCTUJ
UVUF&R JOUP&R BOEPCUBJO
3 3 W=@@@@@
W W =@@@@@
W 3 +3 3 +3 /PUJDFUIBUUIFTPVSDFW PMUBHFWJTEJWJEFEBNPOHUIFSFTJTUPSTJOEJSFDU
QSPQPSUJPOUPUIFJSSFTJTUBODFT UIFMBSHFSUIFSFTJTUBODF UIFMBSHFSUIF
WPMUBHFESPQ5IJTJTDBMMFEUIFQSJODJQMFPGWPMUBHFEJWJTJPO BOEUIFDJS
DVJUJO'JHJTDBMMFEBWPMUBHFEJWJEFS*OHFOFSBM JGBWPMUBHFEJWJEFS
IBT/SFTJTUPST 3 3 w 3/ JOTFSJFTXJUIUIFTPVSDFWPMUBHFW UIFOUI
SFTJTUPS 3O XJMMIBWFBWPMUBHFESPQPG
3O WO=@@@@@@@@@@
W
3 +3 +⋯+3/ 2.6
i
v
+
–
i2
R1
Parallel Resistors and Current Division
$POTJEFSUIFDJSDVJUJO'JH XIFSFUX PSFTJTUPSTBSFDPOOFDUFE
in QBSBMMFMBOEUIFSFGPSFIB WFUIFTBNFW PMUBHFBDSPTTUIFN'SPN 0INTMBX
Node a
i1
W=J 3 =J 3 R2
PS
Node b
Figure 2.31
5XPSFTJTUPSTJOQBSBMMFM
W J =@@
W
J =@@
3 3 "QQMZJOH,$-BUOPEFBHJWFTUIFUPUBMDVSSFOUJBT
J=J +J 4VCTUJUVUJOH&R JOUP&R XFHFU
W+@@
J=@@
+@@
W W=W(@@
=@@@
3 3 3 3 ) 3FR
2.6
45
Parallel Resistors and Current Division
XIFSF3FRJTUIFFRVJWBMFOUSFTJTUBODFPGUIFSFTJTUPSTJOQBSBMMFM
=@@
@@@
+@@
3FR 3 3 PS
3+3
= @@@@@@
@@@
3FR
33
PS
33
3FR=@@@@@
3+ 3
5IVT
The equivalent resistance of two parallel resistors is equal to the product of their resistances divided by their sum.
*UNVTUCFFNQIBTJ[FEUIBUUIJTBQQMJFTPOMZUPUX PSFTJTUPSTJOQBSBMMFM
'SPN&R JG3=3UIFO3FR=33
8FDBOFYUFOEUIFSFTVMUJO&R UPUIFHFOFSBMDBTFPGBDJSDVJU
XJUI/SFTJTUPSTJOQBSBMMFM5IFFRVJWBMFOUSFTJTUBODFJT
=@@
@@@
+@@
+⋯+@@@
3FR 3 3 3/ /PUFUIBU3FRJTBMXBZTTNBMMFSUIBOUIFSFTJTUBODFPGUIFTNBMMFTUSFTJTUPS
JOUIFQBSBMMFMDPNCJOBUJPO*G3=3=⋯ = 3/=3 UIFO
3FR=@@
3
/
'PSF YBNQMF JGGPVS ΩSFTJTUPSTBSFDPOOFDUFEJOQBSBMMFM UIFJS
FRVJWBMFOUSFTJTUBODFJTΩ.
*UJTPGUFONPSFDPOWFOJFOUUPVTFDPOEVDUBODFSBUIFSUIBOSFTJTUBODF
XIFOEFBMJOHXJUISFTJTUPSTJOQBSBMMFM'SPN&R UIFFRVJ WBMFOU
DPOEVDUBODFGPS/SFTJTUPSTJOQBSBMMFMJT
(FR = ( +( +( +⋯+(/ Conductances in parallel behave as a
single conductance whose value is
equal to the sum of the individual
conductances.
XIFSF (FR = 3FR ( = 3 ( = 3 ( 3 w (/ 3/
&RVBUJPO TUBUFT
The equivalent conductance of resistors connected in parallel is the
sum of their individual conductances.
5IJTNFBOTUIBUXFNBZSFQMBDFUIFDJSDVJUJO'JHXJUIUIBUJO
'JH/PUJDFUIFTJNJMBSJUZCFUXFFO&RT BOE 5IF
FRVJWBMFOUDPOEVDUBODFPGQBSBMMFMSFTJTUPSTJTPCUBJOFEUIFTBNFX BZ
BTUIFFRVJ WBMFOUSFTJTUBODFPGTFSJFTSFTJTUPST*OUIFTBNFNBOOFS UIF
FRVJWalent conductance of resistors in series is obtained just the TBNF
i
v
+
–
a
v
Req or Geq
b
Figure 2.32
&RVJWBMFOUDJSDVJUUP'JH
46
Chapter 2
Basic Laws
XBZBTUIFSFTJTUBODFPGSFTJTUPSTJOQBSBMMFM 5IVT UIF F RVJWBMFOUDPO
EVDUBODF(FRPG/SFTJTUPSTJOTFSJFT TVDIBTTIPXOJO'JH JT
i
i1 = 0
R1
i2 = i
R2 = 0
=@@
+@@
+@@
+⋯+@@@
@@@
(FR ( ( ( (/ (JWFOUIFUPUBMDVSSFOU JFOUFSJOHOPEF BJO'JH IP XEPXF
PCUBJODVSSFOU JBOE J 8FLOP XUIBUUIFFRVJ WBMFOUSFTJTUPSIBTUIF
TBNFWPMUBHF PS
J3 3 W=J3FR=@@@@@
3 +3 $PNCJOJOH&RT BOE SFTVMUTJO
(a)
i
i1 = i
R1
i2 = 0
R2 = ∞
(b)
Figure 2.33
B "TIPSUFEDJSDVJU C BOPQFODJSDVJU
3 J
J =@@@@@
3 +3 3 J
J =@@@@@
3 +3 XIJDITIPXTUIBUUIFUPUBMDVSSFOU JJTTIBSFECZUIFSFTJTUPSTJOJO WFSTFQSPQPSUJPOUPUIFJSSFTJTUBODFT 5IJTJTLOP XOBTUIF QSJODJQMF
PGDVSSFOUEJWJTJPO BOEUIFDJSDVJUJO'JHJTLOPXOBTBDVSSFOU
EJWJEFS/PUJDFUIBUUIFMBS ger current floXTUISPVHIUIFTNBMMFSSF TJTUBODF
"TBOFYUSFNFDBTF TVQQPTFPOFPGUIFSFTJTUPSTJO'JHJT[FSP TBZ3=UIBUJT 3JTBTIPSUDJSDVJU BTTIP wn in Fig. B 'SPN
&R 3=JNQMJFTUIBU J= J=JThis means that UIFFOUJSF
DVSSFOUJCZQBTTFT3 and floXTUISPVHIUIFTIPSUDJSDVJU3= UIFQBUI
PGMFBTUSFTJTUBODF 5IVTXIFOBDJSDVJUJTTIPSU DJSDVJUFE BTTIP XOJO
'JH B UXPUIJOHTTIPVMECFLFQUJONJOE
5IFFRVJWBMFOUSFTJTUBODF3FR=<4FFXIBU IBQQFOTXIFO3=
JO&R >
The entire current floXTUISPVHIUIFTIPSUDJSDVJU
"TBOPUIFSF YUSFNFDBTF TVQQPTF 3 = ∞,UIBUJT 3 JT BO PQFO
DJSDVJU BTTIPXOJO'JH C The current still floXTUISPVHIUIFQBUI
PGMFBTUSFTJTUBODF 3#ZUBLJOHUIFMJNJUPG&R BT 3 →∞ XF
PCUBJO3FR=3JOUIJTDBTF
*GXFEJWJEFCPUIUIFOVNFSBUPSBOEEFOPNJOBUPSCZ 33 &R CFDPNFT
( J =@@@@@
J
( +( B
( J =@@@@@
J
( +( C
5IVT JOHFOFSBM JGBDVSSFOUEJWJEFSIBT/DPOEVDUPST ( ( w (/ JO
QBSBMMFMXJUIUIFTPVSDFDVSSFOUJ UIFOUIDPOEVDUPS (O XJMMIBWFDVSSFOU
(O JO =@@@@@@@@@@
J
( +( +⋯+(/ 2.6
47
Parallel Resistors and Current Division
*OHFOFSBM JUJTPGUFODPOWFOJFOUBOEQPTTJCMFUPDPNCJOFSFTJTUPSTJO
TFSJFTBOEQBSBMMFMBOESFEVDFBSFTJTUJ WFOFUXPSLUPBTJOHMF FRVJWBMFOU
SFTJTUBODF3FR4VDIBOFRVJ WBMFOUSFTJTUBODFJTUIFSFTJTUBODFCFUXFFO
UIFEFTJHOBUFEUFSNJOBMTPGUIFOFUX PSLBOENVTUF YIJCJUUIFTBNF JW
DIBSBDUFSJTUJDTBTUIFPSJHJOBMOFUXPSLBUUIFUFSNJOBMT
Example 2.9
'JOE3FRGPSUIFDJSDVJUTIPXOJO'JH
Solution:
5PHFU 3FR XFDPNCJOFSFTJTUPSTJOTFSJFTBOEJOQBSBMMFM5IF ΩBOE
ΩSFTJTUPSTBSFJOQBSBMMFM TPUIFJSFRVJWBMFOUSFTJTUBODFJT
4Ω
1Ω
2Ω
Req
5Ω
×
Ω∥Ω=@@@@
=Ω
+
5IFTZNCPM∥JTVTFEUPJOEJDBUFBQBSBMMFMDPNCJOBUJPO "MTP UIFΩ
BOEΩSFTJTUPSTBSFJOTFSJFTIFODFUIFJSFRVJWBMFOUSFTJTUBODFJT
Figure 2.34
4Ω
5IVTUIFDJSDVJUJO'JHJTSFEVDFEUPUIBUJO'JH B *O
Fig. B XFOPUJDFUIBUUIFUX P ΩSFTJTUPSTBSFJOTFSJFT TPUIF
FRVJWBMFOUSFTJTUBODFJT
2Ω
Req
6Ω
(a)
5IJTΩSFTJTUPSJTOPXJOQBSBMMFMXJUIUIFΩSFTJTUPSJO'JH B UIFJSFRVJWBMFOUSFTJTUBODFJT
4Ω
Ω∥Ω=@@@@
×
=.Ω
+
3FR=Ω+.Ω+Ω=.Ω
2Ω
8Ω
Ω+Ω=Ω
3Ω
'PS&YBNQMF
Ω+Ω = Ω
5IFDJSDVJUJO'JH B JTOPXSFQMBDFEXJUIUIBUJO'JH C *O
'JH C UIFUISFFSFTJTUPSTBSFJOTFSJFT)FODF UIFFRVJWBMFOUSFTJT
UBODFGPSUIFDJSDVJUJT
6Ω
8Ω
Req
2.4 Ω
8Ω
(b)
Figure 2.35
&RVJWBMFOUDJSDVJUTGPS&YBNQMF
Practice Problem 2.9
By combining the resistors in Fig. 2.36, find 3FR
Answer:Ω.
4Ω
Req
4Ω
6Ω
3Ω
Figure 2.36
'PS1SBDUJDF1SPC
8Ω
12 Ω
6Ω
10 Ω
48
Chapter 2
Example 2.10
Basic Laws
$BMDVMBUFUIFFRVJWBMFOUSFTJTUBODF3BCJOUIFDJSDVJUJO'JH
a
Rab
10 Ω
c
1Ω
1Ω
d
6Ω
4Ω
3Ω
5Ω
12 Ω
b
b
b
Figure 2.37
'PS&YBNQMF
a
10 Ω
c 1Ω d
2Ω
b
Solution:
5IFΩBOEΩSFTJTUPSTBSFJOQBSBMMFMCFDBVTFUIFZBSFDPOOFDUFEUP
UIFTBNFUXPOPEFTDBOEC5IFJSDPNCJOFESFTJTUBODFJT
Ω∥Ω=@@@@
×
=Ω
+
3Ω
b
6Ω
b
b
a
10 Ω
Ω∥Ω=@@@@
×
=Ω
+
3Ω
b
Ω+Ω=Ω
(b)
&RVJWBMFOUDJSDVJUTGPS&YBNQMF
Ω=@@@@
×
Ω∥
=.Ω
+
5IJTΩSFTJTUPSJTJOTFSJFTXJUIUIFΩSFTJTUPS TPUIBU
Practice Problem 2.10
20 Ω
Rab
18 Ω
20 Ω
9Ω
b
Figure 2.39
'PS1SBDUJDF1SPC
2Ω
1Ω
3B C=+.=.Ω
'JOE3BCGPSUIFDJSDVJUJO'JH
Answer:Ω.
5Ω
16 Ω
8JUIUIFTFUISFFDPNCJOBUJPOT XFDBOSFQMBDFUIFDJSDVJUJO'JHXJUI
UIBUJO'JH B *O'JH B ΩJOQBSBMMFMXJUIΩHJWFTΩ,BT
DBMDVMBUFEJO&R 5IJTΩFRVJWBMFOUSFTJTUBODFJTOPXJOTFSJFT
XJUIUIFΩSFTJTUBODFUPHJWFBDPNCJOFESFTJTUBODFPGΩ+ 2 Ω = 3 Ω.
5IVT XFSFQMBDFUIFDJSDVJUJO 'JH B XJUIUIBU JO'JH C *O
'JH C XFDPNCJOFUIFΩBOEΩSFTJTUPSTJOQBSBMMFMUPHFU
b
Figure 2.38
a
"MTPUIFΩBOEΩSFTJTUPSTBSFJOTFSJFTIFODF UIFJSFRVJWBMFOUSF TJTUBODFJT
c
2Ω
b
4JNJMBSMZ UIFΩBOEΩSFTJTUPSTBSFJOQBSBMMFMTJODFUIF ZBSFDPO
OFDUFEUPUIFTBNFUXPOPEFTEBOEC)FODF
(a)
2.6
49
Parallel Resistors and Current Division
Example 2.11
'JOEUIFFRVJWBMFOUDPOEVDUBODF(FRGPSUIFDJSDVJUJO'JH B 5S
Solution:
5IF4BOE4SFTJTUPSTBSFJOQBSBMMFM TPUIFJSDPOEVDUBODFJT
Geq
8S
6S
4+4=4
5IJT4SFTJTUPSJTOPXJOTFSJFTXJUI4BTTIPXOJO'JH C TP
UIBUUIFDPNCJOFEDPOEVDUBODFJT
(a)
5S
×
@@@@
=4
+
Geq
6S
5IJTJTJOQBSBMMFMXJUIUIF4SFTJTUPS)FODF
(FR=+=4
∥
∥
∥
∥
3FR=@@
(@@
+@@
@@@
1 =@@
@@
+@@@
@@
=@@
12 ) ( ) 1
5Ω
Req
1
6Ω
@×@
Ω
= @@@
= @@@@
@+ @
20 S
(b)
8FTIPVMEOPUFUIBUUIFDJSDVJUJO'JH B JTUIFTBNFBTUIBUJO
'JH D 8IJMFUIFSFTJTUPSTJO'JH B BSFFYQSFTTFEJOTJFNFOT UIPTFJO'JH D BSFFYQSFTTFEJOPINT5PTIPXUIBUUIFDJSDVJUTBSF
the same, we find 3FRGPSUIFDJSDVJUJO'JH D 12 S
=4
(FR= @@@
3FR
1
8Ω
1
12 Ω
(c)
Figure 2.40
'PS&YBNQMF B PSJHJOBMDJSDVJU C JUTFRVJWBMFOUDJSDVJU D TBNFDJSDVJUBT
JO B CVUSFTJTUPSTBSFFYQSFTTFEJOPINT
5IJTJTUIFTBNFBTXFPCUBJOFEQSFWJPVTMZ
$BMDVMBUF(FRJOUIFDJSDVJUPG'JH
Practice Problem 2.11
Answer:S.
7S
5S
16 S
6S
Geq
Figure 2.41
'PS1SBDUJDF1SPC
'JOEJPBOE WPJOUIFDJSDVJUTIP XOJO'JH B $BMDVMBUFUIFQP XFS
EJTTJQBUFEJOUIFΩSFTJTUPS
Solution:
5IFΩBOEΩSFTJTUPSTBSFJOQBSBMMFM TPUIFJSDPNCJOFESFTJTUBODFJT
Ω∥Ω=@@@@
×
=Ω
+
5IVT PVSDJSDVJUSFEVDFTUPUIBUTIPXOJO'JH C /PUJDFUIBU WPJT
OPUBGGFDUFECZUIFDPNCJOBUJPOPGUIFSFTJTUPSTCFDBVTFUIFSFTJTUPSTBSF
Example 2.12
50
Chapter 2
i
4Ω
12 V +
–
JOQBSBMMFMBOEUIFSFGPSFIBWFUIFTBNFWPMUBHFWP'SPN'JH C XF
DBOPCUBJOWPJOUXPXBZT0OFXBZJTUPBQQMZ0INTMBXUPHFU
io
a
+
vo
–
6Ω
3Ω
(a)
4Ω
a
+
vo
–
12 V +
–
="
J=@@@@@
+
BOEIFODF WP=J=×=7"OPUIFSXBZJTUPBQQMZWPMUBHFEJWJ
TJPO TJODFUIF7JO'JH C JTEJWJEFECFUXFFOUIFΩBOEΩ
SFTJTUPST)FODF
b
i
Basic Laws
vP=@@@@@
7 =7
+
4JNJMBSMZ JPDBOCFPCUBJOFEJOUXPXBZT0OFBQQSPBDIJTUPBQQMZ
0INTMBXUPUIFΩSFTJTUPSJO'JH B OPXUIBUXFLOPXWPUIVT
2Ω
b
vP=JP =
⇒
JP =@@
"
"OPUIFSBQQSPBDIJTUPBQQMZDVSSFOUEJWJTJPOUPUIFDJSDVJUJO'JH B OPXUIBUXFLOPXJ CZXSJUJOH
(b)
Figure 2.42
'PS&YBNQMF B PSJHJOBMDJSDVJU (b) its equiWBMFOUDJSDVJU
JP =@@@@@
"
J=@@
" = @@
+
5IFQPXFSEJTTJQBUFEJOUIFΩSFTJTUPSJT
Practice Problem 2.12
i1
12 Ω
QP=WPJP=(@@
)=.8
'JOEWBOEWJOUIFDJSDVJUTIP XOJO'JH "MTPDBMDVMBUFJBOEJ
BOEUIFQPXFSEJTTJQBUFEJOUIFΩBOEΩSFTJTUPST
Answer:W=7 JN" Q8 W=7 J=N" Q=8
+ v1 –
6Ω
30 V +
–
10 Ω
+
v2
–
i2
40 Ω
Figure 2.43
'PS1SBDUJDF1SPC
Example 2.13
'PSUIFDJSDVJUTIP XOJO'JH B EFUFSNJOF B UIFW PMUBHF WP (b) UIFQPXFSTVQQMJFECZUIFDVSSFOUTPVSDF D UIFQP XFSBCTPSCFECZ
FBDISFTJTUPS
Solution:
B 5IFL ΩBOEL ΩSFTJTUPSTBSFJOTFSJFTTPUIBUUIFJSDPNCJOFE
WBMVFJT+=LΩ5IVTUIFDJSDVJUJO'JH B SFEVDFTUPUIBU
TIPXOJO'JH C 8FOPXBQQMZUIFDVSSFOUEJWJTJPOUFDIOJRVFUP
find JBOEJ
J = @@@@@@@@@@@@@
N" =N"
+ J =@@@@@@@@@@@@@
N" =N"
+ 2.7
51
Wye-Delta Transformations
/PUJDFUIBUUIFWPMUBHFBDSPTTUIFLΩBOELΩSFTJTUPSTJTUIFTBNF BOEWP= J= J=7 BTFYQFDUFE
C 1PXFSTVQQMJFECZUIFTPVSDFJT
30 mA
QP=WPJP = N8=.8
6 kΩ
+
vo
–
D 1PXFSBCTPSCFECZUIFLΩSFTJTUPSJT
(a)
Q =JW=J J 3 =J3= × ¢ = .8
Q=J3= ×− =.8
i2
io
1PXFSBCTPSCFECZUIFLΩSFTJTUPSJT
12 kΩ
9 kΩ
30 mA
+
vo
–
i1
9 kΩ
18 kΩ
1PXFSBCTPSCFECZUIFLΩSFTJTUPSJT
(b)
W Q=@@
P=@@@@@
=.8
3
Figure 2.44
'PS&YBNQMF B PSJHJOBMDJSDVJU (b) its equiWBMFOUDJSDVJU
PS
Q=WPJ= N8=.8
/PUJDFUIBUUIFQPXFSTVQQMJFE 8 FRVBMTUIFQPXFSBCTPSCFE +
+=8 5IJTJTPOFXBZPGDIFDLJOHSFTVMUT
'PSUIFDJSDVJUTIPXOJO'JH2.45, find: (a) WBOEW C UIFQPXFSEJT
TJQBUFEJOUIFL ΩBOEL ΩSFTJTUPST BOE D UIFQPXFSTVQQMJFECZ
UIFDVSSFOUTPVSDF
Practice Problem 2.13
3 kΩ
9 kΩ
+
v1
–
30 mA
15 kΩ
+
v2
–
60 kΩ
Figure 2.45
'PS1SBDUJDF1SPC
Answer: B 7 7 C 8 N8 D 8
2.7
Wye-Delta Transformations
4JUVBUJPOTPGUFOBSJTFJODJSDVJUBOBMZTJTXIFOUIFSFTJTUPSTBSFOFJUIFS
JO QBSBMMFM OPS JO TFSJFT 'PS FYBNQMF DPOTJEFS UIF CSJEHF DJSDVJU JO
Fig. )PXEPXFDPNCJOFSFTJTUPST3UISPVHI3XIFOUIFSFTJTUPST
BSFOFJUIFSJOTFSJFTOPSJOQBSBMMFM .BO ZDJSDVJUTPGUIFUZQFTIP XOJO
Fig. 2.46 can be simplified by using three-terminal equiWBMFOUOFUXPSLT
5IFTFBSFUIFXZF : PSUFF 5 OFUX PSLTIPXOJO'JHBOEUIF
EFMUB Δ PSQJ Π OFUXPSLTIPXOJO'JH5IFTFOFUXPSLTPDDVSCZ
UIFNTFMWFTPSBTQBSUPGBMBS HFSOFUXPSL5IFZBSFVTFEJOUISFFQIBTF
OFUXorks, electrical filters, and matching netwPSLT0VSNBJOJOUFSFTU
R1
R2
vs +
–
R5
Figure 2.46
R3
R4
5IFCSJEHFOFUXPSL
R6
52
Chapter 2
Basic Laws
3
1
R1
R2
R1
R2
3
1
R3
R3
2
4
2
(a)
4
(b)
Figure 2.47
5XPGPSNTPGUIFTBNFOFUXPSL B : C 5
Rc
3
1
Rb
Delta to Wye Conversion
Ra
2
4
(a)
Rc
3
1
Rb
Ra
2
IFSFJTJOIPXUPJEFOUJGZUIFNXIFOUIFZPDDVSBTQBSUPGBOFUXPSLBOE
IPXUPBQQMZXZFEFMUBUSBOTGPSNBUJPOJOUIFBOBMZTJTPGUIBUOFUXPSL
4VQQPTFJUJTNPSFDPOWFOJFOUUPXPSLXJUIBXZFOFUXPSLJOBQMBDF
where the circuit contains a delta configuration. 8FTVQFSJNQPTFBXZF
OFUXPSLPOUIFFYJTUJOHEFMUBOFUXork and find the equiWBMFOUSFTJTUBODFT
JOUIFXZFOFUXPSL5PPCUBJOUIFFRVJWBMFOUSFTJTUBODFTJOUIFXZFOFU
XPSL XFDPNQBSFUIFUX POFUXPSLTBOENBL FTVSFUIBUUIFSFTJTUBODF
CFUXFFOFBDIQBJSPGOPEFTJOUIF Δ PS Π OFUXPSLJTUIFTBNFBTUIF
SFTJTUBODFCFUXFFOUIFTBNFQBJSPGOPEFTJOUIF : PS5 OFUXPSL'PS
UFSNJOBMTBOEJO'JHTBOE GPSFYBNQMF
3 : =3 +3 4
3 Δ) = 3C∥(3B + 3D)
(b)
Figure 2.48
5XPGPSNTPGUIFTBNFOFUXPSL B Δ C Π
4FUUJOH3 : =3 Δ)HJWFT
3 3 +3D 3 =3 +3 =@@@@@@@@@
C B
3B +3C +3D
B
3 3 +3C D B
3 =3 +3 =@@@@@@@@@
3B +3C +3D
C
4JNJMBSMZ
3 3 +3D B C
3 =3 +3 =@@@@@@@@@
3B +3C +3D
4VCUSBDUJOH&R D GSPN&R B XFHFU
3 3 −3B D C
3 −3 =@@@@@@@@@
3B +3C +3D
"EEJOH&RT C BOE HJWFT
D
3C 3
3 =@@@@@@@@
3B +3C +3D
D
BOETVCUSBDUJOH&R GSPN&R C ZJFMET
B 3D3
3 =@@@@@@@@
3B +3C +3D 4VCUSBDUJOH&R GSPN&R B XFPCUBJO
C 3B 3
3 =@@@@@@@@
3B +3C +3D
2.7
53
Wye-Delta Transformations
8FEPOPUOFFEUPNFNPSJ[F&RT UP 5PUSBOTGPSNBĴ
OFUXPSLUP: XFDSFBUFBOFYUSBOPEFOBTTIPXOJO'JHBOEGPMMPX
UIJTDPOWFSTJPOSVMF
Each resistor in the Y network is the product of the resistors in the two
adjacent Δ branches, divided by the sum of the three Δ resistors.
Rc
a
b
R2
R1
n
Rb
Ra
0OFDBOGPMMPXUIJTSVMFBOEPCUBJO&RT UP GSPN'JH
R3
Wye to Delta Conversion
5PPCUBJOUIFDPOWFSTJPOGPSNVMBTGPSUSBOTGPSNJOHBXZFOFUXPSLUPBO
FRVJWBMFOUEFMUBOFUXPSL XFOPUFGSPN&RT UP UIBU
3 3 3 3 +3C +3D 3 3 +3 3 +3 3 =@@@@@@@@@@@@@@@@@
B C D B
3B +3C +3D 3B 3C 3D
@@@@@@@
=
3B +3C +3D
%JWJEJOH&R CZFBDIPG&RT UP MFBETUPUIFGPMMP X
JOHFRVBUJPOT
3 3 +33+33
3B =@@@@@@@@@@@@@@@@@
3
3 3 +33+33 3C =@@@@@@@@@@@@@@@@@
3
3 3 +33+33 3D=@@@@@@@@@@@@@@@@@
3
'SPN&RT UP BOE'JH UIFDPO WFSTJPOSVMFGPS:UPΔ
JTBTGPMMPXT
Each resistor in the Δ network is the sum of all possible products of Y
resistors taken two at a time, divided by the opposite Y resistor.
5IF:BOEΔOFUXPSLTBSFTBJEUPCFCBMBODFEXIFO
3 =3 =3 =3: 3B =3C =3D=3Ĵ 6OEFSUIFTFDPOEJUJPOT DPOWFSTJPOGPSNVMBTCFDPNF
3 3: =@@@
Δ
PS
3Δ =3: 0OFNBZX POEFSXI Z 3:JTMFTTUIBO 3Δ 8FMM XFOPUJDFUIBUUIF :DPOOFDUJPOJTMJLFBiTFSJFTuDPOOFDUJPOXIJMFUIFΔDPOOFDUJPOJTMJLF
BiQBSBMMFMuDPOOFDUJPO
/PUFUIBUJONBLJOHUIFUSBOTGPSNBUJPO XFEPOPUUBLFBOZUIJOHPVU
PGUIFDJSDVJUPSQVUJOBOZUIJOHOFX8FBSFNFSFMZTVCTUJUVUJOHEJGGFSFOU
CVUNBUIFNBUJDBMMZFRVJWBMFOUUISFFUFSNJOBMOFUXPSLQBUUFSOTUPDSFBUF
BDJSDVJUJOXIJDISFTJTUPSTBSFFJUIFSJOTFSJFTPSJOQBSBMMFM BMMP XJOHVT
UPDBMDVMBUF3FRJGOFDFTTBSZ
c
Figure 2.49
4VQFSQPTJUJPOPG:BOEΔOFUXPSLTBTBO
BJEJOUSBOTGPSNJOHPOFUPUIFPUIFS
54
Chapter 2
Example 2.14
Basic Laws
$POWFSUUIFΔOFUXPSLJO'JH B UPBOFRVJWBMFOU:OFUXPSL
Rc
a
b
a
b
25 Ω
R1
Rb
10 Ω
15 Ω
5Ω
7.5 Ω
R3
3Ω
R2
Ra
c
c
(a)
(b)
Figure 2.50
'PS&YBNQMF B PSJHJOBMΔOFUXPSL C :FRVJWBMFOUOFUXPSL
Solution:
6TJOH&RT UP XFPCUBJO
3C 3D
3 =@@@@@@@
=@@@
=Ω
=@@@@@@@@
×
3B +3C +3D ++
B 3D3
3 =@@@@@@@
=.Ω
=@@@@@
×
3B +3C +3D
3B 3C 3 =@@@@@@@
=Ω
=@@@@@
×
3B +3C +3D
5IFFRVJWBMFOU:OFUXPSLJTTIPXOJO'JH C Practice Problem 2.14
a
R1
R2
10 Ω
20 Ω
b
5SBOTGPSNUIFXZFOFUXPSLJO'JHUPBEFMUBOFUXPSL
Answer:3B=Ω 3C=Ω 3D=Ω.
40 Ω
R3
c
Figure 2.51
'PS1SBDUJDF1SPC
Example 2.15
0CUBJOUIFFRVJWBMFOUSFTJTUBODF 3BCGPSUIFDJSDVJUJO'JHBOEVTF
it to find current J
Solution:
Define.The problem is clearly defined. Please note, this part nor
NBMMZXJMMEFTFSWFEMZUBLFNVDINPSFUJNF
1SFTFOU$MFBSMZ XIFOXFSFNPWFUIFWoltage source, we end up XJUI
BQVSFMZSFTJTUJWFDJSDVJU4JODFJUJTDPNQPTFEPGEFMUBTBOEXZFT XF
IBWFBNPSFDPNQMF YQSPDFTTPGDPNCJOJOHUIFFMFNFOUTUPHFUIFS 2.7
8FDBOVTFXZFEFMUBUSBOTGPSNBUJPOTBTPOFBQQSPBDIUPfind BTP
MVUJPO*UJT VTFGVMUPMPDBUFUIFXZFT UIFSFBSFUX PPGUIFN POFBU
OBOEUIFPUIFSBUD BOEUIFEFMUBT UIFSFBSFUISFFDBO BCO DOC "MUFSOBUJWF5IFSFBSFEJGGFSFOUBQQSPBDIFTUIBUDBOCFVTFEUPTPMWF
UIJTQSPCMFN4JODFUIFGPDVTPG4FDJTUIFXZFEFMUBUSBOTGPS NBUJPO UIJTTIPVMECFUIFUFDIOJRVFUPVTF"OPUIFSBQQSPBDIXPVME
CFUPTPMW FGPSUIFFRVJ WBMFOUSFTJTUBODFCZJOKFDUJOHPOFBNQJOUP
the circuit and finding the vPMUBHFCFUXFFO BBOE CXFXJMMMFBSO
BCPVUUIJTBQQSPBDIJO$IBQ
5IFBQQSPBDIXFDBOBQQMZIFSFBTBDIFDLX PVMECFUPVTF
a wye-delta transformation as the first solution to the problem.
-BUFSXFDBODIFDLUIFTPMVUJPOCZTUBSUJOHXJUIBEFMUBXZF
USBOTGPSNBUJPO
"UUFNQU*OUIJTDJSDVJU UIFSFBSFUXP:OFUXPSLTBOEUISFFΔOFUXPSLT
5SBOTGPSNJOHKVTUPOFPGUIFTFXJMMTJNQMJGZUIFDJSDVJU*GXFDPO
WFSUUIF
:OFUXPSLDPNQSJTJOHUIFΩ,Ω,BOEΩSFTJTUPST XFNBZTFMFDU
3 =Ω 3 =Ω 55
Wye-Delta Transformations
i
a
a
10 Ω
12.5 Ω
120 V +
–
5Ω
c
n
30 Ω
20 Ω
15 Ω
b
b
Figure 2.52
'PS&YBNQMF
3 =Ω
5IVTGSPN&RT UP XFIBWF
3 3 +3 3 +3 3 @@@@@@@@@@@@@@
3B =@@@@@@@@@@@@@
=×+×+×
3 =Ω
@@@
=
3 3 +3 3 +3 3 @@@
3C =@@@@@@@@@@@@@
==.Ω
3 3 3 +3 3 +3 3 @@@
3D=@@@@@@@@@@@@@
= = Ω
3 8JUIUIF:DPOWFSUFEUPΔ,UIFFRVJWBMFOUDJSDVJU XJUIUIFWPMU
BHFTPVSDFSFNPWFEGPSOPX JTTIPXOJO'JH B $PNCJOJOH
UIFUISFFQBJSTPGSFTJTUPSTJOQBSBMMFM XFPCUBJO
×
∥=@@@@@
=Ω
+
.∥=@@@@@@@@
×
=Ω
+
×
∥=@@@@@
=Ω
+
a
4.545 Ω
a
d
12.5 Ω
17.5 Ω
a
70 Ω
15 Ω
7.292 Ω
35 Ω
21 Ω
c
n
20 Ω
b
b
(a)
1.8182 Ω
15 Ω
10.5 Ω
b
Figure 2.53
30 Ω
2.273 Ω
(b)
&RVJWBMFOUDJSDVJUTUP'JH XJUIUIFWPMUBHFTPVSDFSFNPWFE
(c)
30 Ω
56
Chapter 2
Basic Laws
TPUIBUUIFFRVJWBMFOUDJSDVJUJTTIPwn in Fig. 2.53(b). Hence, we find
3B C= + ∥=@@@@@@@@
×
=Ω
.+
5IFO
WT @@@@
= J=@@@
="
3B C .
8FPCTFSWFUIBUXFIBWFTVDDFTTGVMMZTPMWFEUIFQSPCMFN/PX
XFNVTUFWBMVBUFUIFTPMVUJPO
&WBMVBUF/PXXFNVTUEFUFSNJOFJGUIFBOTXFSJTDPSSFDUBOEUIFO
FWaluate the final solution.
*UJTSFMBUJWFMZFBTZUPDIFDLUIFBOTXFSXFEPUIJTCZTPMWJOH
UIFQSPCMFNTUBSUJOHXJUIBEFMUBXZFUSBOTGPSNBUJPO-FUVTUSBOT GPSNUIFEFMUB DBO JOUPBXZF
-FU3D=Ω 3B=Ω,BOE3O=Ω.5IJTXJMMMFBEUP MFU
ESFQSFTFOUUIFNJEEMFPGUIFXZF 3D3O 3B E=@@@@@@@
=.Ω
=@@@@@@@@
×.
3B +3D+3O ++ .
3 3 @@@@@@
B O
3D E=@@@@
=×.
=.Ω
.
.
3 3 @@@@
B D
3O E=@@@@
=× =.Ω
.
.
5IJTOPXMFBETUPUIFDJSDVJUTIP XOJO'JHVSF D -PPLJOHBU
UIFSFTJTUBODFCFUXFFO EBOEC XFIBWFUXPTFSJFTDPNCJOBUJPOJO
QBSBMMFM HJWJOHVT
.+ .+ @@@@
3E C=@@@@@@@@@@@@@@@@
=.=.Ω
.++.+ .
5IJTJTJOTFSJFTXJUIUIFΩSFTJTUPS CPUIPGXIJDIBSFJOQBS
BMMFMXJUIUIFΩSFTJTUPS5IJTUIFOHJWFTVTUIFFRVJWBMFOUSFTJT
UBODFPGUIFDJSDVJU
.+. @@@@
3B C=@@@@@@@@@@@@
=.=Ω
.+.+ .
5IJTOPXMFBETUP
WT @@@@
= J@@@
="
3B C .
8FOPUFUIBUVTJOHUX PWBSJBUJPOTPOUIFXZFEFMUBUSBOTGPSNBUJPO
MFBETUPUIFTBNFSFTVMUT5IJTSFQSFTFOUTBWFSZHPPEDIFDL
4BUJTGBDUPSZ 4JODFXFIB WFGPVOEUIFEFTJSFEBOTXFSCZEFUFS NJOJOHUIFFRVJ Walent resistance of the circuit first and the an
TXFS DIFDLT UIFO XF DMFBSMZIB WF B TBUJTGBDUPSZ TPMVUJPO 5IJT
SFQSFTFOUTXIBUDBOCFQSFTFOUFEUPUIFJOEJ WJEVBMBTTJHOJOHUIF QSPCMFN
2.8
57
Applications
'PSUIFCSJEHFOFUXork in Fig. 2.54, find 3BCBOEJ
Practice Problem 2.15
i
Answer:Ω "
a
6Ω
48 Ω
240 V +
–
2.8
60 Ω
Applications
3FTJTUPSTBSFPGUFOVTFEUPNPEFMEF WJDFTUIBUDPO WFSUFMFDUSJDBMFOFS HZJOUPIFBUPSPUIFSGPSNTPGFOFSHZ4VDIEFWJDFTJODMVEFDPOEVDUJOH
XJSF MJHIUCVMCT FMFDUSJDIFBUFST TUPWFT PWFOT BOEMPVETQFBLFST*OUIJT
TFDUJPO XFXJMMDPOTJEFSUXPSFBMMJGFQSPCMFNTUIBUBQQMZUIFDPODFQUT
EFWFMPQFEJOUIJTDIBQUFSFMFDUSJDBMMJHIUJOHTZTUFNTBOEEFTJHOPGED
NFUFST
2.8.1
40 Ω
20 Ω
100 Ω
b
Figure 2.54
'PS1SBDUJDF1SPC
Lighting Systems
-JHIUJOHTZTUFNT TVDIBTJOBIPVTFPSPOB$ISJTUNBTUSFF PGUFODPO
TJTUPG /MBNQTDPOOFDUFEFJUIFSJOQBSBMMFMPSJOTFSJFT BTTIP XOJO
Fig. &BDIMBNQJTNPEFMFEBTBSFTJTUPS "TTVNJOHUIBUBMMUIF MBNQTBSFJEFOUJDBMBOEVPJTUIFQPXFSMJOFWPMUBHF UIFWPMUBHFBDSPTT
FBDIMBNQJTVPGPSUIFQBSBMMFMDPOOFDUJPOBOEVP/GPSUIFTFSJFTDPO
OFDUJPO 5IFTFSJFTDPOOFDUJPOJTFBTZUPNBOVGB DUVSFC VUJTTFMEPN VTFEJOQSBDUJDF GPSBUMFBTUUXPSFBTPOT'JSTU JUJTMFTTSFMJBCMFXIFO
BMBNQGB JMT BMMUIFMBNQTHPPVU4FDPOE JUJTIBSEFSUPNBJOUBJO
XIFOBMBNQJTCBE POFNVTUUFTUBMMUIFMBNQTPOFCZPOFUPEFUFDU
UIFGBVMUZPOF
So far, we have assumed that connecting wires are perfect conductors (i.e.,
conductors of zero resistance). In real
physical systems, however, the
resistance of the connecting wire may
be appreciably large, and the
modeling of the system must include
that resistance.
Historical
Thomas Alva Edison m XBTQFSIBQTUIFHSFBUFTU"NFSJDBO
JOWFOUPS)FQBUFOUFE JOWFOUJPOT JODMVEJOHTVDI I JTUPSZNBLJOH
JOWFOUJPOTBTUIFJODBOEFTDFOUFMFDUSJDCVMC UIFQIPOPHSBQI BOEUIF
first commercial motion pictures.
#PSO JO .JMBO 0IJP UIF ZPVOHFTU PG TFWFO DIJMESFO &EJTPO SF
DFJWFEPOMZUISFFNPOUITPGGPSNBMFEVDBUJPOCFDBVTFIFIBUFETDIPPM
)FXBTIPNFTDIPPMFECZIJTNPUIFSBOERVJDLMZCFHBOUPSFBEPO
IJTPXO*O &EJTPOSFBEPOFPG'BSBEBZTCPPLTBOEGPVOEIJT
DBMMJOH)FNPWFEUP.FOMP1BSL /FX+FSTFZ JO XIFSFIFNBO BHFEB XFMMTUBGGFE SFTFBSDI MBCPSBUPSZ.PTUPGIJT JOWFOUJPOTDBNF
PVUPGUIJT laboratory. His laboratory served as a model for modern SF
search PSHBOJ[BUJPOT #FDBVTF PG IJT EJWFSTF JOUFSFTUT BOE UIFPWFS XIFMNJOHOVNCFSPGIJTJOWFOUJPOTBOEQBUFOUT &EJTPOCFHBOUPFTUBC MJTINBOVGBDUVSJOHDPNQBOJFTGPSNBLJOHUIFEFWJDFTIFJOWFOUFE)F
designed the first electric power station to supply electric light. Formal
FMFDUSJDBMFOHJOFFSJOHFEVDBUJPOCFHBOJOUIF NJETXJUI&EJTPOBT
BSPMFNPEFMBOEMFBEFS
-JCSBSZPG$POHSFTT
58
Chapter 2
Basic Laws
1
1
+
Vo
–
Power
plug
2
3
2
+
Vo
–
N
3
N
Lamp
(a)
(b)
Figure 2.55
B 1BSBMMFMDPOOFDUJPOPGMJHIUCVMCT C TFSJFTDPOOFDUJPOPGMJHIUCVMCT
Example 2.16
5ISFFMJHIUCVMCTBSFDPOOFDUFEUPB7CBUUFSZBTTIP XOJO'JH B $BMDVMBUF B UIFUPUBMDVSSFOUTVQQMJFECZUIFCBUUFSZ C UIFDVSSFOU UISPVHIFBDICVMC D UIFSFTJTUBODFPGFBDICVMC
I
9V
15 W
20 W
9V
10 W
(a)
I1
I2
+
V2
–
R2
+
V3
–
R3
+
V1
–
R1
(b)
Figure 2.56
B -JHIUJOHTZTUFNXJUIUISFFCVMCT C SFTJTUJWFDJSDVJUFRVJWBMFOUNPEFM
Solution:
B 5IFUPUBMQP XFSTVQQMJFECZUIFCBUUFSZJTFRVBMUPUIFUPUBMQP XFS
BCTPSCFECZUIFCVMCTUIBUJT
Q=++=8
4JODFQ=7* UIFOUIFUPUBMDVSSFOUTVQQMJFECZUIFCBUUFSZJT
Q *=@@
=@@
="
7
C 5IFCVMCTDBOCFNPEFMFEBTSFTJTUPSTBTTIPXOJO'JH C 4JODF
3 8CVMC JTJOQBSBMMFMXJUIUIFCBUUFSZBTXFMMBTUIFTFSJFTDPN CJOBUJPOPG3BOE3
7 =7 +7 =7
5IFDVSSFOUUISPVHI3JT
Q * =@@
=@@
=."
7 2.8
59
Applications
#Z,$- UIFDVSSFOUUISPVHIUIFTFSJFTDPNCJOBUJPOPG3BOE3JT
* =*−* =−.=."
D 4JODFQ=*3
Q
3 =@@@
=@@@@@
=.Ω
*
. Q
=@@@@@
3 =@@@
=.Ω
*
. Q
=@@@@@
3 =@@@
=.Ω
*
. 3FGFSUP'JHBOEBTTVNFUIFSFBSFTJYMJHIUC VMCTUIBUDBOCFDPO OFDUFEJOQBSBMMFMBOETJYEJG GFSFOUMJHIUC VMCTUIBUDBOCFDPOOFDUFE JO
TFSJFT*OFJUIFSDBTF FBDIMJHIUCVMCJTUPPQFSBUFBU8*GUIFWPMUBHF
BUUIFQMVHJT7GPSUIFQBSBMMFMBOETFSJFTDPOOFDUJPOT DBMDVMBUFUIF
DVSSFOUUISPVHIBOEUIFWPMUBHFBDSPTTFBDICVMCGPSCPUIDBTFT
Practice Problem 2.16
Answer:7BOEN" QBSBMMFM 7BOE" TFSJFT 2.8.2
a
Design of DC Meters
By their nature, resistors are used to control the floXPGDVSSFOU8FUBLF
BEWBOUBHFPGUIJTQSPQFSUZJOTF WFSBMBQQMJDBUJPOT TVDIBTJOBQPUFO UJPNFUFS 'JH 5IFX PSE QPUFOUJPNFUFS EFSJWFEGSPNUIFX PSET
QPUFOUJBMBOE NFUFS JNQMJFTUIBUQPUFOUJBMDBOCFNFUFSFEPVU 5IFQP
UFOUJPNFUFS PSQPUGPSTIPSU JTBUISFFUFSNJOBMEF WJDFUIBUPQFSBUFTPO
UIFQSJODJQMFPGW PMUBHFEJWJTJPO*UJTFTTFOUJBMMZBOBEKVTUBCMFW PMUBHF
EJWJEFS"TBWPMUBHFSFHVMBUPS JUJTVTFEBTBWPMVNFPSMFWFMDPOUSPMPO
SBEJPT 57T BOEPUIFSEFWJDFT*O'JH
3 7P VU=7C D=@@@
CD7
3B D JO
Max
b
Vin +
–
+
Vout
Min –
c
Figure 2.57
5IFQPUFOUJPNFUFSDPOUSPMMJOHQPUFOUJBM
MFWFMT
XIFSF 3BD = 3BC + 3CD5IVT 7PVU EFDSFBTFT PS JODSFBTFT BT UIF TMJEJOH
DPOUBDUPGUIFQPUNPWFTUPXBSEDPSB SFTQFDUJWFMZ
Another application where resistors are used to control current floX
JTJOUIFBOBMPHEDNFUFST‡UIFBNNFUFSWPMUNFUFS BOEPINNFUFS XIJDI
NFBTVSFDVSSFOU WPMUBHF BOESFTJTUBODF SFTQFDUJWFMZ&BDIPGUIFTFNF
UFSTFNQMPZTUIFE"STPOWBMNFUFSNP WFNFOU TIPwn in Fig. 5IF
NPWFNFOUDPOTJTUTFTTFOUJBMMZ PGBNP WBCMFJSPODPSFDPJM NPVOUFEPO
BQJWPUCFUXFFOUIFQPMFTPGBQFSNBOFOUNBHOFU When current floXT
through the coil, it creates a torque which causes the pointer to deflect.
The amount of current through the coil determines the deflection of the
QPJOUFS XIJDIJTSFHJTUFSFEPOBTDBMF attached to UIFNFUFSNPWFNFOU
'PSFYBNQMF JGUIFNFUFSNP WFNFOUJTSBUFEN" ĵ JUX PVMEUBLF
N"UPDBVTF a full-scale deflection of the NFUFSNPWFNFOU#ZJOUSP
EVDJOHBEEJUJPOBMDJSDVJUSZUPUIF E "STPOWBMNFUFSNP WFNFOU BOBN NFUFS WPMUNFUFS PSPINNFUFSDBOCFDPOTUSVDUFE
$POTJEFS'JH XIFSFBOBOBMPHW PMUNFUFSBOEBNNFUFSBSFDPO OFDUFEUPBOFMFNFOU5IFWPMUNFUFSNFBTVSFTUIFWPMUBHFBDSPTTBMPBEBOE
An instrument capable of measuring
voltage, current, and resistance is
called a multimeter or a volt-ohm
meter (VOM).
A load is a component that is receiving
energy (an energy sink), as opposed
to a generator supplying energy (an
energy source). More about loading
will be discussed in Section 4.9.1.
60
Chapter 2
Basic Laws
scale
spring
pointer
S
permanent magnet
N
rotating coil
spring
stationary iron core
Figure 2.58
"E"STPOWBMNFUFSNPWFNFOU
Ammeter
I
A
+
Voltmeter V V
–
Figure 2.59
Element
$POOFDUJPOPGBWPMUNFUFSBOEBO
BNNFUFSUPBOFMFNFOU
JTUIFSFGPSFDPOOFDUFEJOQBSBMMFMXJUIUIFFMFNFOUAs TIPXOJO'JH B UIFWPMUNFUFSDPOTJTUTPGBE"STPO WBMNPWFNFOUJOTFSJFTXJUIBSFTJTUPS XIPTFSFTJTUBODF3NJTEFMJCFSBUFMZNBEFWFSZMBSHF UIFPSFUJDBMMZ infinite),
UPNJOJNJ[FUIFDVSSFOUESBXOGSPNUIFDJSDVJU5PFYUFOEUIFSBOHFPGWPMU
BHFUIBUUIFNFUFSDBONFBTVSF TFSJFTNVMUJQMJFSSFTJTUPSTBSFPGUFODPOOFDUFE
XJUIUIFWPMUNFUFST BTTIPXOJO'JH C 5IFNVMUJQMFSBOHFWPMUNFUFS
JO'JH C DBONFBTVSFW PMUBHFGSPNUP 7 UP7 PSUP 7 EFQFOEJOHPOXIFUIFSUIFTXJUDIJTDPOOFDUFEUP3 3 PS3 SFTQFDUJWFMZ
-FUVTDBMDVMBUFUIFNVMUJQMJFSSFTJTUPS3OGPSUIFTJOHMFSBOHFWPMUNFUFS
JO'JH B PS 3O = 3 3 PS 3GPSUIFNVMUJQMFSBOHF WPMUNFUFSJO
'JH C 8FOFFEUPEFUFSNJOFUIFWBMVFPG3OUPCFDPOOFDUFEJOTFSJFT
XJUIUIFJOUFSOBMSFTJTUBODF3NPGUIFWPMUNFUFS*OBOZEFTJHO XFDPOTJEFS
UIFXPSTUDBTFDPOEJUJPO*OUIJTDBTF UIFXPSTUDBTFPDDVSTXIFOUIFGVMM
TDBMFDVSSFOU*GT=*NfloXTUISPVHIUIFNFUFS5IJTTIPVMEBMTPDPSSFTQPOE
Meter
Multiplier
Rn
+
Probes
Im
Rm
V
–
(a)
R1
1V
R2
10 V
+
Probes V
Meter
Switch
100 V
R3
Im
Rm
–
(b)
Figure 2.60
7PMUNFUFST B TJOHMFSBOHFUZQF C NVMUJQMFSBOHFUZQF
2.8
61
Applications
UPUIFNBYJNVNW PMUBHFSFBEJOHPSUIFGVMMTDBMFW PMUBHF 7GT4JODFUIF NVMUJQMJFSSFTJTUBODF3OJTJOTFSJFTXJUIUIFJOUFSOBMSFTJTUBODF3N
7GT=*GT 3O + 3N Rn
In
Meter
Im
'SPNUIJT XFPCUBJO
Rm
7 3O =@@@
GT−3N *GT
4JNJMBSMZ UIFBNNFUFSNFBTVSFTUIFDVSSFOUUISPVHIUIFMPBEBOEJT DPOOFDUFEJOTFSJFTXJUIJU"TTIPXOJO'JH B UIFBNNFUFSDPOTJTUT
PGBE"STPOWBMNPWFNFOUJOQBSBMMFMXJUIBSFTJTUPSXIPTFSFTJTUBODF 3N
JTEFMJCFSBUFMZNBEFWFSZTNBMM UIFPSFUJDBMMZ [FSP UPNJOJNJ[FUIFW PMU
BHFESPQBDSPTTJU5PBMMPXNVMUJQMFSBOHFT TIVOUSFTJTUPSTBSFPGUFODPO
OFDUFEJOQBSBMMFMXJUI 3NBTTIP wn in Fig. C 5IFTIVOUSFTJTUPST BMMPXUIFNFUFSUPNFBTVSFJOUIFSBOHFmN" mN" PSm" EFQFOEJOHPOXIFUIFSUIFTXJUDIJTDPOOFDUFEUP3 3 PS3 SFTQFDUJWFMZ
/PXPVSPCKFDUJWFJTUPPCUBJOUIFNVMUJQMJFSTIVOU 3OGPSUIFTJOHMF
SBOHFBNNFUFSJO'JH B PS3O= 3 3 PS3GPSUIFNVMUJQMFSBOHF
BNNFUFSJO'JH C 8FOPUJDFUIBU3NBOE3OBSFJOQBSBMMFMBOEUIBU
BUGVMMTDBMFSFBEJOH *=*GT=*N *O XIFSF*OJTUIFDVSSFOUUISPVHIUIF
TIVOUSFTJTUPS3O"QQMZJOHUIFDVSSFOUEJWJTJPOQSJODJQMFZJFMET
I
Probes
(a)
R1
10 mA
R2
100 mA
Switch
1A
R3
Meter
Im
3O *N =@@@@@
* 3O +3N GT
Rm
I
PS
Probes
* 3 =@@@@@
N 3 * GT−*N N
5IFSFTJTUBODF3YPGBMJOFBSSFTJTUPSDBOCFNFBTVSFEJOUX PXBZT
"OJOEJSFDUXBZJTUPNFBTVSFUIFDVSSFOU* that floXTUISPVHIJUCZDPO
OFDUJOHBOBNNFUFSJOTFSJFTXJUIJUBOEUIFW PMUBHF7BDSPTTJUCZDPO OFDUJOHBWPMUNFUFSJOQBSBMMFMXJUIJU BTTIPXOJO'JH B 5IFO
(b)
O
Figure 2.61
"NNFUFST B TJOHMFSBOHFUZQF C NVMUJQMFSBOHFUZQF
3Y=@@
7
*
5IFEJSFDUNFUIPEPGNFBTVSJOHSFTJTUBODFJTUPVTFBOPINNFUFS "OPINNFUFSDPOTJTUTCBTJDBMMZPGBE"STPO WBMNPWFNFOU BW BSJBCMF
SFTJTUPSPSQPUFOUJPNFUFS BOEBCBUUFSZ BTTIPXOJO'JH C "QQMZ
JOH,7-UPUIFDJSDVJUJO'JH C HJWFT
A
I
Rx
+
V
–
V
&= 3+3N +3Y *N PS
(a)
3Y=@@
&− 3+ 3N *N 5IFSFTJTUPS3JTTFMFDUFETVDIUIBUUIFNFUFSHJWes a full-scale deflection;
UIBUJT *N=*GTXIFO3Y=5IJTJNQMJFTUIBU
Ohmmeter
Im
E
&= 3+3N *GT
4VCTUJUVUJOH&R JOUP&R MFBETUP
*GT
3Y= @@
− 3+3N ( *N
)
"TNFOUJPOFE UIFUZQFTPGNFUFSTXFIB WFEJTDVTTFEBSFLOP XOBT
BOBMPHNFUFSTBOEBSFCBTFEPOUIFE"STPOWBMNFUFSNPWFNFOU"OPUIFS
UZQFPGNFUFS DBMMFEB EJHJUBMNFUFS JTCBTFEPOBDUJ WFDJSDVJUFMFNFOUT R
Rm
Rx
(b)
Figure 2.62
5XPXBZTPGNFBTVSJOHSFTJTUBODF
B VTJOHBOBNNFUFSBOEBWPMUNFUFS C VTJOHBOPINNFUFS
62
Chapter 2
Basic Laws
Historical
Samuel F. B. Morse m BO"NFSJDBOQBJOUFS JOWFOUFE
the telegraph, the first practical, commercialized application of
FMFDUSJDJUZ
.PSTFXBTCPSOJO$IBSMFTUPXO .BTTBDIVTFUUT BOETUVEJFEBU:BMF
BOEUIF3PZBM"DBEFNZPG"SUTJO-POEPOUPCFDPNFBOBSUJTU*OUIF
T IFCFDBNFJOUSJHVFEXJUIEFWFMPQJOHBUFMFHSBQI)FIBEBXPSLJOH
NPEFMCZBOEBQQMJFEGPSBQBUFOUJO5IF644FOBUFBQQSP QSJBUFEGVOETGPS.PSTFUPDPOTUSVDUBUFMFHSBQIMJOFCFUXFFO#BMUJNPSF and Washington, D.C. On May 24, 1844, he sent the famous first mes
TBHFi8IBUIBUI(PEXSPVHIUu.PSTFBMTPEFWFMPQFEBDPEFPGEPUTBOE
EBTIFTGPSMFUUFSTBOEOVNCFST GPSTFOEJOHNFTTBHFTPOUIFUFMFHSBQI5IF
EFWFMPQNFOUPGUIFUFMFHSBQIMFEUPUIFJOWFOUJPOPGUIFUFMFQIPOF
-JCSBSZPG$POHSFTT
TVDIBTPQBNQT'PSFYBNQMF BEJHJUBMNVMUJNFUFSEJTQMBZTNFBTVSFNFOUT
PGEDPSBDWPMUBHF DVSSFOU BOESFTJTUBODFBTEJTDSFUFOVNCFST JOTUFBEPG
using a pointer deflection on a continuous scale as in an BOBMPHNVMUJNFUFS
%JHJUBMNFUFSTBSFXIBUZPVXPVMENPTUMJLFMZVTFJOBNPEFSOMBC)PX
FWFS UIFEFTJHOPGEJHJUBMNFUFSTJTCFZPOEUIFTDPQFPGUIJTCPPL
Example 2.17
'PMMPXJOHUIFW PMUNFUFSTFUVQPG'JH EFTJHOBW PMUNFUFSGPSUIF
GPMMPXJOHNVMUJQMFSBOHFT
B m7 C m7 D m7 E m7
"TTVNFUIBUUIFJOUFSOBMSFTJTUBODF3N=LΩBOEUIFGVMMTDBMFDVSSFOU
*GT= 100 μA.
Solution:
8FBQQMZ&R BOEBTTVNFUIBU3 3 3 BOE3DPSSFTQPOEXJUI
SBOHFTm7 m7 m7 BOEm7 SFTQFDUJWFMZ
B 'PSSBOHFm7
−= −=LΩ
3 =@@@@@@@
×¢
C 'PSSBOHFm7
3 = @@@@@@@
−= −=LΩ
×¢
D 'PSSBOHFm7
3 = @@@@@@@
−= −=LΩ
×−
E 'PSSBOHFm7
3 = @@@@@@@
−= −=LΩ
7
×−
/PUFUIBUUIFSBUJPPGUIFUPUBMSFTJTUBODF 3O+3N UPUIFGVMMTDBMFWPMU
BHF7GTJTDPOTUBOUBOEFRVBMUP*GTGPSUIFGPVSSBOHFT5IJTSBUJP HJWFO
JOPINTQFSW PMU PS Ω 7 JTLOP XOBTUIF TFOTJUJWJUZPGUIFW PMUNFUFS
5IFMBSHFSUIFTFOTJUJWJUZ UIFCFUUFSUIFWPMUNFUFS
2.9
'PMMPXJOHUIFBNNFUFSTFUVQPG'JH EFTJHOBOBNNFUFSGPSUIF
GPMMPXJOHNVMUJQMFSBOHFT
B m" C mN" D mN"
5BLFUIFGVMMTDBMFNFUFSDVSSFOUBT*N=N"BOEUIFJOUFSOBMSFTJTUBODF
PGUIFBNNFUFSBT3N=Ω.
Answer:4IVOUSFTJTUPSTNΩ NΩ Ω
2.9
Summary
"SFTJTUPSJTBQBTTJ WFFMFNFOUJOXIJDIUIFW PMUBHF WBDSPTTJUJT
EJSFDUMZQSPQPSUJPOBMUPUIFDVSSFOUJUISPVHIJU5IBUJT BSFTJTUPSJT
BEFWJDFUIBUPCFZT0INTMBX
v=J3
XIFSF3JTUIFSFTJTUBODFPGUIFSFTJTUPS
"TIPSUDJSDVJUJTBSFTJTUPS BQFSGFDUMZ DPOEVDUJOHXJSF XJUI[FSP
SFTJTUBODF 3= "OPQFODJSDVJUJTBSFTJTUPSXJUIinfinite SFTJT
UBODF 3=ñ 5IFDPOEVDUBODF(PGBSFTJTUPSJTUIFSFDJQSPDBMPGJUTSFTJTUBODF
(=@@
3
"CSBODIJTBTJOHMFUX PUFSNJOBMFMFNFOUJOBOFMFDUSJDDJSDVJU "
OPEFJTUIFQPJOUPGDPOOFDUJPOCFUXFFOUX PPSNPSFCSBODIFT "
MPPQJT BDMPTFEQBUI JOBDJSDVJU 5IF OVNCFSPG CSBODIFT C UIF
OVNCFSPGOPEFTO BOEUIFOVNCFSPGJOEFQFOEFOUMPPQTMJOBOFU
XPSLBSFSFMBUFEBT
C=M+O−
,JSDIIPGGTDVSSFOUMB X ,$- TUBUFTUIBUUIFDVSSFOUTBUBO ZOPEF
BMHFCSBJDBMMZTVNUP[FSP*OPUIFSX PSET UIFTVNPGUIFDVSSFOUT
FOUFSJOHBOPEFFRVBMTUIFTVNPGDVSSFOUTMFBWJOHUIFOPEF
,JSDIIPGGTW PMUBHFMB X ,7- TUBUFTUIBUUIFW PMUBHFTBSPVOEB
DMPTFEQBUIBMHFCSBJDBMMZTVNUP[FSP*OPUIFSX PSET UIFTVNPG
WPMUBHFSJTFTFRVBMTUIFTVNPGWPMUBHFESPQT
5XPFMFNFOUTBSFJOTFSJFTXIFOUIF ZBSFDPOOFDUFETFRVFOUJBMMZ FOE UP FOE 8IFO FMFNFOUT BSF JO TFSJFT UIF TBNF DVSSFOU floXT
UISPVHIUIFN J=J 5IFZBSFJOQBSBMMFMJGUIF ZBSFDPOOFDUFEUP
UIFTBNFUXPOPEFT&MFNFOUTJOQBSBMMFMBMXBZTIBWFUIFTBNFWPMU
BHFBDSPTTUIFN W=W 8IFOUXPSFTJTUPST3 = 1( BOE3 = 1( BSFJOTFSJFT UIFJS
FRVJWBMFOUSFTJTUBODF3FRBOEFRVJWBMFOUDPOEVDUBODF(FRBSF
3FR=3 +3 ( ( (FR=@@@@@
( +( 8IFOUXPSFTJTUPST3 = 1( BOE3 = 1( BSFJOQBSBMMFM UIFJS
FRVJWBMFOUSFTJTUBODF3FRBOEFRVJWBMFOUDPOEVDUBODF(FRBSF
3 3 3FR=@@@@@
3 +3 (FR=( +( 63
Summary
Practice Problem 2.17
64
Chapter 2
Basic Laws
5IFWPMUBHFEJWJTJPOQSJODJQMFGPSUXPSFTJTUPSTJOTFSJFTJT
3 W=@@@@@
W 3 +3 3 W=@@@@@
W
3 +3 5IFDVSSFOUEJWJTJPOQSJODJQMFGPSUXPSFTJTUPSTJOQBSBMMFMJT
3 3 J =@@@@@
J J =@@@@@
J
3 +3 3 +3 5IFGPSNVMBTGPSBEFMUBUPXZFUSBOTGPSNBUJPOBSF
D
3C 3
3D3B 3 =@@@@@@@
3 =@@@@@@@
3B +3C +3D
3B +3C +3D
3B 3C 3 =@@@@@@@
3B +3C +3D
5IFGPSNVMBTGPSBXZFUPEFMUBUSBOTGPSNBUJPOBSF
3 3 +3 3 +3 3 3 3 +3 3 +3 3 3B =@@@@@@@@@@@@@
3C =@@@@@@@@@@@@@
3 3 +3 3
+3 3
3 3
3D=@@@@@@@@@@@@@
3 5IFCBTJDMBXTDPWFSFEJOUIJTDIBQUFSDBOCFBQQMJFEUPUIFQSPC
MFNTPGFMFDUSJDBMMJHIUJOHBOEEFTJHOPGEDNFUFST
Review Questions
5IFSFDJQSPDBMPGSFTJTUBODFJT
B WPMUBHF
D DPOEVDUBODF
C Ω
E Ω
6Ω
'PS3FWJFX2VFTUJPO
5IFDVSSFOU*PPG'JHJT
B −"
C −"
D "
5IFNBYJNVNDVSSFOUUIBUB8 LΩSFTJTUPSDBO
TBGFMZDPOEVDUJT
10 A
C L"
E μ"
C D E Io
5IFDVSSFOU*JOUIFDJSDVJUPG'JHJT
B −"
D "
C −"
E "
4A
2A
"OFUXPSLIBTCSBODIFTBOEJOEFQFOEFOUMPPQT
)PXNBOZOPEFTBSFUIFSFJOUIF
OFUXPSL
B + 5V
–
Figure 2.63
C 7
E 7
B L"
D N"
3V +
–
5IFWPMUBHFESPQBDSPTTBL8UPBTUFSUIBUESBXT
"PGDVSSFOUJT
B L7
D 7
C DVSSFOU
E DPVMPNCT
"OFMFDUSJDIFBUFSESBXT"GSPNB7MJOF
5IFSFTJTUBODFPGUIFIFBUFSJT
B Ω
D Ω
I
4Ω
Figure 2.64
'PS3FWJFX2VFTUJPO
E "
65
Problems
*OUIFDJSDVJUJO'JH 7JT
B 7
C 7
D 7
E 7
*OUIFDJSDVJUPG'JH BEFDSFBTFJO3MFBETUPB
EFDSFBTFPG TFMFDUBMMUIBUBQQMZ
B DVSSFOUUISPVHI3
+
C WPMUBHFBDSPTT3
10 V
–
D WPMUBHFBDSPTT3
E QPXFSEJTTJQBUFEJO3
12 V +
–
+ 8V
–
F OPOFPGUIFBCPWF
R1
+
V
–
Vs +
–
Figure 2.65
'PS3FWJFX2VFTUJPO
R2
R3
Figure 2.67
'PS3FWJFX2VFTUJPO
8IJDIPGUIFDJSDVJUTJO'JHXJMMHJWFZPV
7BC= 7 7
5V
+–
5V
a
3V +
–
–+
a
+–
b
"OTXFSTD D C D D C B E E C E
3V +
–
+–
b
1V
1V
(a)
(b)
5V
5V
+–
a
3V +
–
–+
a
–+
b
3V +
–
–+
b
1V
1V
(c)
(d)
Figure 2.66
'PS3FWJFX2VFTUJPO
Problems
Section 2.2
Ohm’s Law
%FTJHOBQSPCMFN DPNQMFUFXJUIBTPMVUJPO UPIFMQ
TUVEFOUTUPCFUUFSVOEFSTUBOE0INTMBX6TFBU
MFBTUUXPSFTJTUPSTBOEPOFWPMUBHFTPVSDF)JOU ZPV
DPVMEVTFCPUISFTJTUPSTBUPODFPSPOFBUBUJNF JUJT
VQUPZPV#FDSFBUJWF
Find the hot resistance of a light bulb rated 60 W,
120 V.
"CBSPGTJMJDPOJTDNMPOHXJUIBDJSDVMBSDSPTT
TFDUJPO*GUIFSFTJTUBODFPGUIFCBSJTΩBUSPPN
UFNQFSBUVSF XIBUJTUIFDSPTTTFDUJPOBMSBEJVTPG
UIFCBS
66
Chapter 2
Basic Laws
B $BMDVMBUFDVSSFOUJJO'JHXIFOUIFTXJUDI
JTJOQPTJUJPO
C 'JOEUIFDVSSFOUXIFOUIFTXJUDIJTJOQPTJUJPO
1
100 Ω
Section 2.4
2
i
+ 40 V
–
250 Ω
Kirchhoff’s Laws
%FTJHOBQSPCMFN DPNQMFUFXJUIBTPMVUJPO UPIFMQ
PUIFSTUVEFOUTCFUUFSVOEFSTUBOE,JSDIIPGGT$VSSFOU
-BX%FTJHOUIFQSPCMFNCZTQFDJGZJOHWBMVFTPG
JB JC BOEJD TIPXOJO'JH BOEBTLJOHUIFNUP
TPMWFGPSWBMVFTPGJ J BOEJ#FDBSFGVMUPTQFDJGZ
SFBMJTUJDDVSSFOUT
ib
Figure 2.68
Section 2.3
i1
i2
'PS1SPC
ia
i3
ic
Nodes, Branches, and Loops
For the network graph in Fig. 2.69, find the number
PGOPEFT CSBODIFT BOEMPPQT
Figure 2.72
'PS1SPC
'JOEJ J BOEJJO'JH
–4 A
i2
1A
–3 A
A
i3
B
–6 A
i1
Figure 2.69
2A
C
–2 A
'PS1SPC
Figure 2.73
*OUIFOFUXPSLHSBQITIPXOJO'JH EFUFSNJOF
UIFOVNCFSPGCSBODIFTBOEOPEFT
'PS1SPC
%FUFSNJOFJBOEJJOUIFDJSDVJUPG'JH
–8 A
4A
i2
i1
–6 A
Figure 2.70
'PS1SPC
Figure 2.74
'PS1SPC
%FUFSNJOFUIFOVNCFSPGCSBODIFTBOEOPEFTJOUIF
DJSDVJUPG'JH
1Ω
12 V +
–
Figure 2.71
'PS1SPC
*OUIFDJSDVJUPG'JH DBMDVMBUF7BOE7
4Ω
8Ω
+
5Ω
2A
+
V1
–
Figure 2.75
'PS1SPC
1V
–
+
+
5V
–
2V
–
+
V2
–
67
Problems
%FUFSNJOF7PJOUIFDJSDVJUJO'JH
*OUIFDJSDVJUJO'JH PCUBJOW W BOEW
+ 30 V –
– 50 V +
+ 20 V –
+
40 V
–
+
10 V +
–
+
v3
–
+
v1
–
14 Ω
16 Ω
+ v2 –
+ 25 V
–
Vo
–
Figure 2.80
'PS1SPC
Figure 2.76
'PS1SPC
For the circuit in Fig. 2.77, use KCL to find the
CSBODIDVSSFOUT*UP*
0CUBJOWUISPVHIWJOUIFDJSDVJUPG'JH
+ v1 –
2A
I2
I4
7A
+
v3
–
v2 –
+
24 V +
–
+ 10 V
–
–+
3A
I1
4A
I3
12 V
Figure 2.81
'PS1SPC
Figure 2.77
'PS1SPC
'JOE*BOE7JOUIFDJSDVJUPG'JH
Given the circuit in Fig. 2.78, use KVL to find the
CSBODIWPMUBHFT7UP7
I
+
+
V1
–
+
3V
–
+
–
4V
+
V3
–
3A
–
+2 V –
10 Ω
4A
20 Ω
Figure 2.78
Figure 2.82
'PS1SPC
From the circuit in Fig. 2.83, find * UIFQPXFSEJT
TJQBUFECZUIFSFTJTUPS BOEUIFQPXFSTVQQMJFECZ
FBDITPVSDF
'PS1SPC
$BMDVMBUFWBOEJYJOUIFDJSDVJUPG'JH
10 V +
–
12 Ω
+ 16 V –
+v–
ix
+
4V
–
10 V
+–
12 V +
–
+ 3i
x
–
'PS1SPC
+–
Figure 2.83
'PS1SPC
I
3Ω
–8 V
Figure 2.79
–2 A
V2
+
+
5V
–
+
V4
–
20 Ω
V
–
68
Chapter 2
Basic Laws
%FUFSNJOFJPJOUIFDJSDVJUPG'JH
io
For the circuit in Fig. 2.88, find 7P7TJOUFSNTPGα,
3 3 3 BOE3*G3=3=3=3 XIBUWBMVF
PGαXJMMQSPEVDF]7P7T]= 10?
22 Ω
Io
+ 5i
o
–
54 V +
–
R1
Vs +
–
Figure 2.84
𝛼Io
R2
R3
R4
+
Vo
–
Figure 2.88
'PS1SPC
'PS1SPC
'JOE7YJOUIFDJSDVJUPG'JH
For the network in Fig. 2.89, find the current, voltBHF BOEQPXFSBTTPDJBUFEXJUIUIFLΩSFTJTUPS
2 Vx
+
–
1Ω
15 V +
–
5Ω
+
Vx
–
10 kΩ
5 mA
+
Vo
–
0.01Vo
5 kΩ
20 kΩ
Figure 2.89
'PS1SPC
2Ω
Figure 2.85
'PS1SPC
Sections 2.5 and 2.6
'JOE7PJOUIFDJSDVJUJO'JHBOEUIFQPXFS
BCTPSCFECZUIFEFQFOEFOUTPVSDF
10 Ω
10 Ω
25 A
20 Ω
5Ω
40 Ω
'PS1SPC
*OUIFDJSDVJUTIPXOJO'JH EFUFSNJOF7YBOE
UIFQPXFSBCTPSCFECZUIFΩSFTJTUPS
5Ω
+ v
x
$BMDVMBUF*PJOUIFDJSDVJUPG'JH
6Ω
8Ω
–
20 Ω
40 Ω
10 Ω
15 Ω
'PS1SPC
10 Ω
Figure 2.90
'PS1SPC
Figure 2.87
io
25 Ω
2 Vo
Figure 2.86
60 A
'PSUIFDJSDVJUJO'JH JP="$BMDVMBUFJYBOE
UIFUPUBMQPXFSBCTPSCFECZUIFFOUJSFDJSDVJU
ix
+ Vo –
Series and Parallel Resistors
60 Ω
10 V +
–
30 Ω
Figure 2.91
'PS1SPC
Io
3Ω
6Ω
69
Problems
%FTJHOBQSPCMFN VTJOH'JH UPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETFSJFTBOEQBSBMMFMDJSDVJUT
'JOEJUISPVHIJJOUIFDJSDVJUJO'JH
60 Ω
i4
i2 200 Ω
R1
40 Ω
+ v1 –
+
v2
–
Vs +
–
+
v3
–
R2
50 Ω
i1
i3
R3
Figure 2.92
16 A
Figure 2.96
'PS1SPC
'PS1SPC
"MMSFTJTUPST 3 JO'JHBSFΩFBDI'JOE3FR
R
Req
0CUBJOWBOEJJOUIFDJSDVJUPG'JH
i
R
R
R
R
R
9A
+
v
–
4S
1S
6S
2S
3S
R
Figure 2.97
Figure 2.93
'PS1SPC
'PS1SPC
'JOE3FRGPSUIFDJSDVJUJO'JH
25 Ω
Using series/parallel resistance combination, find the
FRVJWBMFOUSFTJTUBODFTFFOCZUIFTPVSDFJOUIFDJSDVJU
PG'JH'JOEUIFPWFSBMMBCTPSCFEQPXFSCZUIF
SFTJTUPSOFUXPSL
180 Ω
60 Ω
60 Ω
Req
600 V +
–
50 Ω
70 Ω
150 Ω
400 Ω
400 Ω
200 Ω
130 Ω
50 Ω
Figure 2.94
'PS1SPC
Figure 2.98
'PS1SPC
'PSUIFDJSDVJUJO'JH EFUFSNJOFJUPJ
3Ω
i3
i2
200 V +
–
Figure 2.95
'PS1SPC
4Ω
$BMDVMBUF7PBOE*PJOUIFDJSDVJUPG'JH
i1
1Ω
i4 2 Ω
70 Ω
i5
200 V +
–
20 Ω
Figure 2.99
'PS1SPC
30 Ω
Io
+
Vo
–
5Ω
70
Chapter 2
Basic Laws
'JOEJBOE7PJOUIFDJSDVJUPG'JH
i
80 Ω
24 Ω
&WBMVBUF3FRMPPLJOHJOUPFBDITFUPGUFSNJOBMTGPS
each of the circuits shown in Fig. 2.103.
50 Ω
3 kΩ
6Ω
25 Ω
3Ω
20 V +
–
20 Ω
60 Ω
30 Ω
2 kΩ
6 kΩ
+
Vo
–
6Ω
3Ω
6 kΩ
20 Ω
(a)
(b)
Figure 2.103
Figure 2.100
'PS1SPC
'PS1SPC
For the ladder network in Fig. 2.104, find * and 3FR
(JWFOUIFDJSDVJUJO'JHBOEUIBUUIFSFTJT
UBODF 3FR MPPLJOHJOUPUIFDJSDVJUGSPNUIFMFGUJT
FRVBMUPΩ EFUFSNJOFUIFWBMVFPG3
I
15 V
R1
2Ω
8Ω
+
–
4Ω
1Ω
6Ω
2Ω
Req
R1
Figure 2.104
'PS1SPC
R1
R1
*G3FR=Ω in the circuit of Fig. 2.105, find 3
Figure 2.101
'PS1SPC
30 Ω
Req
10 Ω
R
12 Ω
60 Ω
'JOE3FRBOEJPJOUIFDJSDVJUPG'JH
12 Ω
12 Ω
Figure 2.105
'PS1SPC
60 Ω
12 Ω
io
2.5 Ω
3FEVDFFBDIPGUIFDJSDVJUTJO'JHUPBTJOHMF
SFTJTUPSBUUFSNJOBMTBC
6Ω
5Ω
80 Ω
35 V +
–
15 Ω
a
20 Ω
8Ω
b
20 Ω
Req
30 Ω
Figure 2.102
'PS1SPC
(a)
a
2Ω
4Ω
5Ω
3Ω
5Ω
10 Ω
b
30 Ω
71
Problems
(a)
2Ω
a
4Ω
5Ω
5Ω
3Ω
8Ω
b
'JOEUIFFRVJWBMFOUSFTJTUBODFBUUFSNJOBMTBCPG
FBDIDJSDVJUJO'JH
10 Ω
10 Ω
4Ω
40 Ω
20 Ω
(b)
Figure 2.106
a
'PS1SPC
30 Ω
50 Ω
b
$BMDVMBUFUIFFRVJWBMFOUSFTJTUBODF3BCBUUFSNJOBMT
BC for each of the circuits in Fig. 2.107.
5Ω
(a)
30 Ω
5Ω
a
20 Ω
10 Ω
12 Ω
40 Ω
20 Ω
5Ω
b
(a)
60 Ω
25 Ω
a
10 Ω
15 Ω
10 Ω
(b)
80 Ω
Figure 2.109
60 Ω
20 Ω
30 Ω
'PS1SPC
b
(b)
Figure 2.107
'PS1SPC
'PSUIFDJSDVJUTJO'JH PCUBJOUIFFRVJWBMFOU
SFTJTUBODFBUUFSNJOBMTBC
'JOE*JOUIFDJSDVJUPG'JH
20 Ω
30 Ω
a
140 V +
–
30 Ω
'PS1SPC
80 Ω
24 Ω
20 Ω
8Ω
36 Ω
14 Ω
36 Ω
36 Ω
b
Figure 2.108
72 Ω
I
Figure 2.110
'PS1SPC
10 Ω
72
Chapter 2
Basic Laws
'JOEUIFFRVJWBMFOUSFTJTUBODF3BCJOUIFDJSDVJUPG
'JH
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOEXZFEFMUBUSBOTGPSNBUJPOTVTJOH
'JH
c
d
5Ω
6Ω
10 Ω
8Ω
a
b
R
9 mA
e
R
3Ω
20 Ω
R
R
R
Figure 2.114
'PS1SPC
f
0CUBJOUIFFRVJWBMFOUSFTJTUBODFBUUIFUFSNJOBMTBC
for each of the circuits in Fig. 2.115.
Figure 2.111
'PS1SPC
a
10 Ω
Section 2.7
30 Ω
Wye-Delta Transformations
10 Ω
10 Ω
$POWFSUUIFDJSDVJUTJO'JHGSPN:UPΔ
20 Ω
20 Ω
b
a
10 Ω
10 Ω
b
a
30 Ω
(a)
20 Ω
b
30 Ω
50 Ω
10 Ω
a
c
c
(a)
(b)
25 Ω
10 Ω
5Ω
Figure 2.112
20 Ω
15 Ω
b
'PS1SPC
(b)
Figure 2.115
'PS1SPC
5SBOTGPSNUIFDJSDVJUTJO'JHGSPNΔUP:
For the circuit shown in Fig. 2.116, find the equivaMFOUSFTJTUBODF"MMSFTJTUPSTBSFΩ
a
60 Ω
60 Ω
b
60 Ω
a
75 Ω
150 Ω
25 Ω
c
c
(a)
(b)
Figure 2.113
'PS1SPC
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
b
Req
Figure 2.116
'PS1SPC
73
Problems
0CUBJOUIFFRVJWBMFOUSFTJTUBODF3BCJOFBDIPGUIF
DJSDVJUTPG'JH*O C BMMSFTJTUPSTIBWFB
WBMVFPGΩ.
30 Ω
16 Ω
40 Ω
30 Ω
20 Ω
a
%FUFSNJOF7JOUIFDJSDVJUPG'JH
60 Ω
+
V
–
100 V +
–
10 Ω
80 Ω
50 Ω
b
15 Ω
35 Ω
10 Ω
20 Ω
12 Ω
Figure 2.120
'PS1SPC
'JOE3FRBOE*JOUIFDJSDVJUPG'JH
(a)
I
a
30 Ω
10 Ω
25 Ω
5Ω
10 Ω
10 Ω
b
50 V +
–
(b)
Figure 2.117
10 Ω
20 Ω
80 Ω
'PS1SPC
15 Ω
10 Ω
$POTJEFSUIFDJSDVJUJO'JH'JOEUIFFRVJWB
MFOUSFTJTUBODFBUUFSNJOBMT B BC, (b) DE
450 Ω
10 Ω
a
300 Ω
b
450 Ω
c
300 Ω
60 Ω
Figure 2.118
'PS1SPC
25 Ω
Req
Figure 2.121
'PS1SPC
Section 2.8
d
Applications
5IF8MJHIUCVMCJO'JHJTSBUFEBU
WPMUT$BMDVMBUFUIFWBMVFPG7TUPNBLFUIFMJHIU
CVMCPQFSBUFBUJUTSBUFEDPOEJUJPOT
150 Watt
$BMDVMBUF*PJOUIFDJSDVJUPG'JH
Vs +
–
Io
100 Ω
50 Ω
Figure 2.122
400 Ω
100 V +
–
200 Ω
800 Ω
400 Ω
Figure 2.119
'PS1SPC
700 Ω
1.7 kΩ
'PS1SPC
"OFOUFSQSJTJOHZPVOHNBOUSBWFMTUP&VSPQF
DBSSZJOHUISFFMJHIUCVMCTIFIBEQVSDIBTFEJO/PSUI
"NFSJDB5IFMJHIUCVMCTIFIBTBSFB8MJHIU
CVMC B8MJHIUCVMC BOEB8MJHIUCVMC&BDI
MJHIUCVMCJTSBUFEBU7)FXJTIFTUPDPOOFDU
UIFTFUPB7TZTUFNUIBUJTGPVOEJO&VSPQF'PS
SFBTPOTXFBSFOPUTVSFPG IFDPOOFDUTUIF8
74
Chapter 2
Basic Laws
MJHIUCVMCJOTFSJFTXJUIBQBSBMMFMDPNCJOBUJPOPGUIF
8MJHIUCVMCBOEUIF8MJHIUCVMCBTTIPXO
JO'JH)PXNVDIQPXFSJTBDUVBMMZCFJOH
EFMJWFSFEUPFBDIMJHIUCVMC 8IBUEPFTIFTFFXIFO
he first turns on the light bulbs?
*TUIFSFBCFUUFSXBZUPDPOOFDUUIFTFMJHIUCVMCTJO
PSEFSUPIBWFUIFNXPSLNPSFFGGFDUJWFMZ
40 W
100 W
110 V
+
–
A
110 V
+
–
B
Figure 2.125
60 W
Figure 2.123
*GBOBNNFUFSXJUIBOJOUFSOBMSFTJTUBODFPGΩ
BOEBDVSSFOUDBQBDJUZPGN"JTUPNFBTVSF" EFUFSNJOFUIFWBMVFPGUIFSFTJTUBODFOFFEFE$BMDV
late the power dissipated in the shunt resistor.
'PS1SPC
*GUIFUISFFCVMCTPG1SPCBSFDPOOFDUFEJO
QBSBMMFMUPUIF7TPVSDF DBMDVMBUFUIFDVSSFOU
UISPVHIFBDICVMC
"TBEFTJHOFOHJOFFS ZPVBSFBTLFEUPEFTJHOB
MJHIUJOHTZTUFNDPOTJTUJOHPGB8QPXFSTVQQMZ
BOEUXPMJHIUCVMCTBTTIPXOJO'JH:PV
NVTUTFMFDUUIFUXPCVMCTGSPNUIFGPMMPXJOHUISFF
BWBJMBCMFCVMCT
3=Ω DPTU= TUBOEBSETJ[F
3=Ω DPTU= TUBOEBSETJ[F
3=Ω DPTU= OPOTUBOEBSETJ[F
5IFTZTUFNTIPVMECFEFTJHOFEGPSNJOJNVNDPTU
TVDIUIBU*MJFTXJUIJOUIFSBOHF*="±QFSDFOU
I
+
70-W
Power
Supply
BOEDFOUTL8I DBMDVMBUFUIFBOOVBMFOFSHZDPTUPG
UIFTZTUFN
Rx
Ry
–
Figure 2.124
'PS1SPC
"UISFFXJSFTZTUFNTVQQMJFTUXPMPBET"BOE#BT
TIPXOJO'JH-PBE"DPOTJTUTPGBNP
UPSESBXJOHBDVSSFOUPG" XIJMFMPBE#JTB1$
ESBXJOH""TTVNJOHIEBZPGVTFGPSEBZT
5IFQPUFOUJPNFUFS BEKVTUBCMFSFTJTUPS 3YJO
'JHJTUPCFEFTJHOFEUPBEKVTUDVSSFOUJYGSPN
10 mA to 1 A. Calculate the values of 3 and 3YUP
BDIJFWFUIJT
ix
110 V
+
–
R
Rx
ix
Figure 2.126
'PS1SPC
%FTJHOBDJSDVJUUIBUVTFTBE"STPOWBMNFUFS XJUI
BOJOUFSOBMSFTJTUBODFPGLΩUIBUSFRVJSFTBDVSSFOU
of 5 mA to cause the meter to deflect full scale) to
CVJMEBWPMUNFUFSUPSFBEWBMVFTPGWPMUBHFTVQUP
WPMUT
"LΩ7WPMUNFUFSSFBET7GVMMTDBMF
B 8IBUTFSJFTSFTJTUBODFJTSFRVJSFEUPNBLFUIF
NFUFSSFBE7GVMMTDBMF
C 8IBUQPXFSXJMMUIFTFSJFTSFTJTUPSEJTTJQBUF
XIFOUIFNFUFSSFBETGVMMTDBMF
B 0CUBJOUIFWPMUBHF7PJOUIFDJSDVJUPG
'JH B C %FUFSNJOFUIFWPMUBHF7ʹPNFBTVSFEXIFOBWPMU
NFUFSXJUILΩJOUFSOBMSFTJTUBODFJTDPOOFDUFE
BTTIPXOJO'JH C 75
Problems
D The finite resistance of the meter introduces an
FSSPSJOUPUIFNFBTVSFNFOU$BMDVMBUFUIFQFSDFOU
FSSPSBT
] ]
7 −7ʹ @@@@@@@
P P
×
7P
E 'JOEUIFQFSDFOUFSSPSJGUIFJOUFSOBMSFTJTUBODF
XFSFLΩ.
"WPMUNFUFSJTVTFEUPNFBTVSF7PJOUIFDJSDVJUJO
'JH5IFWPMUNFUFSNPEFMDPOTJTUTPGBOJEFBM
WPMUNFUFSJOQBSBMMFMXJUIBLΩSFTJTUPS-FU
7T=7 3T=LΩ,BOE3=LΩ.$BMDVMBUF
7PXJUIBOEXJUIPVUUIFWPMUNFUFSXIFO
B 3=LΩ
D 3=LΩ
1 kΩ
2 mA
Rs
+
Vo
–
5 kΩ 4 kΩ
R1
Vs +
–
(a)
R2
1 kΩ
2 mA
5 kΩ
+
Vo
–
250 kΩ
V
Figure 2.129
+
Vo
–
4 kΩ
C 3=LΩ
Voltmeter
'PS1SPC
B $POTJEFSUIF8IFBUTUPOFCSJEHFTIPXOJO
'JH$BMDVMBUFWB WC BOEWBC
(b)
Figure 2.127
'PS1SPC
(b) 3FXPSLQBSU B JGUIFHSPVOEJTQMBDFEBU
B instead of P
B 'JOEUIFDVSSFOU*JOUIFDJSDVJUPG'JH B C "OBNNFUFSXJUIBOJOUFSOBMSFTJTUBODFPGΩJT
JOTFSUFEJOUIFOFUXPSLUPNFBTVSF*ʹ BTTIPXOJO
'JH C 8IBUJT*ʹ?
D $BMDVMBUFUIFQFSDFOUFSSPSJOUSPEVDFECZUIF
NFUFSBT
_____
*−*ʹ
× 100%
*
]
]
8 kΩ
25 V +
–
a
16 Ω
b
12 kΩ
o
I
15 kΩ
10 kΩ
Figure 2.130
'PS1SPC
4V +
–
40 Ω
60 Ω
'JHVSFSFQSFTFOUTBNPEFMPGBTPMBS
QIPUPWPMUBJDQBOFM(JWFOUIBU7T=7 3=Ω BOEJ-=2 A, find 3-
(a)
I'
Ammeter
16 Ω
R1
4V +
–
40 Ω
60 Ω
iL
Vs +
–
(b)
Figure 2.128
'PS1SPC
Figure 2.131
'PS1SPC
RL
76
Chapter 2
Basic Laws
'JOE7PJOUIFUXPXBZQPXFSEJWJEFSDJSDVJUJO
'JH
1Ω
1Ω
2Ω
Vo
10 V +
–
1Ω
5IFDJSDVJUJO'JHJTUPDPOUSPMUIFTQFFEPGB
NPUPSTVDIUIBUUIFNPUPSESBXTDVSSFOUT" " BOE"XIFOUIFTXJUDIJTBUIJHI NFEJVN BOEMPX
QPTJUJPOT SFTQFDUJWFMZ5IFNPUPSDBOCFNPEFMFE
BTBMPBESFTJTUBODFPGNΩ.%FUFSNJOFUIFTFSJFT
ESPQQJOHSFTJTUBODFT3 3 BOE3
1Ω
Low
1Ω
R1
10-A, 0.01-Ω fuse
Medium
Figure 2.132
High
'PS1SPC
R2
6V
"OBNNFUFSNPEFMDPOTJTUTPGBOJEFBMBNNF
UFSJOTFSJFTXJUIBΩSFTJTUPS*UJTDPOOFDUFE
XJUIBDVSSFOUTPVSDFBOEBOVOLOPXOSFTJTUPS
3YBTTIPXOJO'JH5IFBNNFUFSSFBEJOH
JTOPUFE8IFOBQPUFOUJPNFUFS3JTBEEFEBOE
BEKVTUFEVOUJMUIFBNNFUFSSFBEJOHESPQTUPPOF
IBMGJUTQSFWJPVTSFBEJOH UIFO3=Ω.8IBUJT
UIFWBMVFPG3Y
R3
Motor
Figure 2.134
'PS1SPC
'JOE3BCJOUIFGPVSXBZQPXFSEJWJEFSDJSDVJUJO
'JH"TTVNFFBDI3=Ω
R
R
20 Ω
A
I
Ammeter
model
R
R
R
a
R
R
R
R
R
R
R
Rx
R
R
b
Figure 2.133
'PS1SPC
Figure 2.135
'PS1SPC
R
77
Comprehensive Problems
Comprehensive Problems
3FQFBU1SPCGPSUIFFJHIUXBZEJWJEFSTIPXOJO
'JH
1
1
"OFMFDUSJDQFODJMTIBSQFOFSSBUFEN8 7JT
connected to a 9-V battery as shown in Fig. 2.138.
$BMDVMBUFUIFWBMVFPGUIFTFSJFTESPQQJOHSFTJTUPS3Y
OFFEFEUPQPXFSUIFTIBSQFOFS
1
1
1
1
1
1
1
Switch
1
1
a
Rx
1
1
1
1
1
9V
1
1
Figure 2.138
1
'PS1SPC
1
1
1
1
1
1
1
1
1
1
1
b
Figure 2.136
'PS1SPC
A loudspeaker is connected to an amplifier as shown
JO'JH*GBΩMPVETQFBLFSESBXTUIFNBYJ
mum power of 12 W from the amplifier, determine
UIFNBYJNVNQPXFSBΩMPVETQFBLFSXJMMESBX
4VQQPTFZPVSDJSDVJUMBCPSBUPSZIBTUIFGPMMPXJOH
TUBOEBSEDPNNFSDJBMMZBWBJMBCMFSFTJTUPSTJOMBSHF
RVBOUJUJFT
Ω
Ω
Ω
LΩ
6TJOHTFSJFTBOEQBSBMMFMDPNCJOBUJPOTBOEBNJOJ
NVNOVNCFSPGBWBJMBCMFSFTJTUPST IPXXPVMEZPV
PCUBJOUIFGPMMPXJOHSFTJTUBODFTGPSBOFMFDUSPOJD
DJSDVJUEFTJHO
B Ω
D LΩ
Amplifier
LΩ
Loudspeaker
Figure 2.139
'PS1SPC
C Ω
E LΩ
*OUIFDJSDVJUJO'JH UIFXJQFSEJWJEFTUIF
QPUFOUJPNFUFSSFTJTUBODFCFUXFFOα3BOE −α)3 ≤α≤ 1. 'JOEWPWT
For a specific application, the circuit shown in
'JHXBTEFTJHOFETPUIBU*-=N"BOE
UIBU3JO=LƗ8IBUBSFUIFWBMVFTPG3BOE3
R
vs +
–
R
vo
𝛼R
–
Figure 2.137
'PS1SPC
IL
R1
+
1A
5 kΩ
Figure 2.140
'PS1SPC
Rin
10 kΩ
R2
10 kΩ
78
Chapter 2
Basic Laws
5IFQJOEJBHSBNPGBSFTJTUBODFBSSBZJTTIPXOJO
'JH'JOEUIFFRVJWBMFOUSFTJTUBODFCFUXFFO
UIFGPMMPXJOH
5XPEFMJDBUFEFWJDFTBSFSBUFEBTTIPXOJO'JH
'JOEUIFWBMVFTPGUIFSFTJTUPST3BOE3
OFFEFEUPQPXFSUIFEFWJDFTVTJOHB7CBUUFSZ
B BOE
C BOE
D BOE
60-mA, 2-Ω fuse
4
3
40 Ω
40 Ω
1
Figure 2.142
'PS1SPC
30 Ω
75 Ω
'PS1SPC
30 Ω
2
36 V, 720 mW
Device 2
36 V
R2
30 Ω
Figure 2.141
R1
Device 1
12 V, 75 mW
c h a p t e r
3
Methods of
Analysis
No great work is ever done in a hurry. To develop a great scientific dis
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Enhancing Your Career
Career in Electronics
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DVSSFOUMFWFMT5IJTEJTUJODUJPOOPMPOHFSIPMET BTQPXFSTFNJDPOEVDUPS
EFWJDFTPQFSBUFBUIJHIMF WFMTPGDVSSFOU 5PEBZ FMFDUSPOJDT JTSFHBSEFE
BTUIFTDJFODFPGUIFNPUJPOPGDIBS HFTJOBH BT WBDVVN PSTFNJDPO EVDUPS.PEFSOFMFDUSPOJDTJO WPMWFTUSBOTJTUPSTBOEUSBOTJTUPSDJSDVJUT
5IFFBSMJFSFMFDUSPOJDDJSDVJUTXFSFBTTFNCMFEGSPNDPNQPOFOUT.BO Z
FMFDUSPOJDDJSDVJUTBSFOPXQSPEVDFEBTJOUFHSBUFEDJSDVJUT GBCSJDBUFEJO
BTFNJDPOEVDUPSTVCTUSBUFPSDIJQ
Electronic circuits find applications in manZBSFBT TVDIBTBVUPNB
UJPO CSPBEDBTUJOH DPNQVUFST BOEJOTUSVNFOUBUJPO5IFSBOHFPGEFWJDFT
UIBUVTFFMFDUSPOJDDJSDVJUTJTFOPSNPVTBOEJTMJNJUFEPOMZCZPVSJNBHJ
OBUJPO3BEJP UFMFWJTJPO DPNQVUFST BOETUFSFPTZTUFNTBSFCVUBGFX
"OFMFDUSJDBMFOHJOFFSVTVBMMZQFSGPSNTEJ WFSTFGVODUJPOTBOEJT MJLFMZUPVTF EFTJHO PSDPOTUSVDUTZTUFNTUIBUJODPSQPSBUFTPNFGPSNPG FMFDUSPOJDDJSDVJUT5IFSFGPSF BOVOEFSTUBOEJOHPGUIFPQFSBUJPOBOEBOBM
ZTJTPGFMFDUSPOJDTJTFTTFOUJBMUPUIFFMFDUSJDBMFOHJOFFS &MFDUSPOJDTIBT CFDPNFBTQFDJBMUZEJTUJODUGSPNPUIFSEJTDJQMJOFTXJUIJOFMFDUSJDBMFOHJ neering. Because the field of electronics is eWFSBEWBODJOH BOFMFDUSPOJDT
FOHJOFFSNVTUVQEBUFIJTIFSLOPXMFEHFGSPNUJNFUPUJNF5IFCFTUXBZ
UPEPUIJTJTCZCFJOHBNFNCFSPGBQSPGFTTJPOBMPSHBOJ[BUJPOTVDIBTUIF
*OTUJUVUFPG&MFDUSJDBMBOE&MFDUSPOJDT&OHJOFFST *&&& 8JUIBNFNCFS
TIJQPGPWFS UIF*&&&JTUIFMBSHFTUQSPGFTTJPOBMPSHBOJ[BUJPOJO
UIFXorld. Members benefit immensely from the numerous magB[JOFT KPVSOBMT USBOTBDUJPOT BOEDPOGFSFODFTZNQPTJVNQSPDFFEJOHTQVCMJTIFE
ZFBSMZCZ*&&&:PVTIPVMEDPOTJEFSCFDPNJOHBO*&&&NFNCFS
5SPVCMFTIPPUJOHBOFMFDUSPOJDDJSDVJU
CPBSE
¥#SBOE91JDUVSFT1VODI4UPDL3'
79
80
Chapter 3
Methods of Analysis
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
6OEFSTUBOE,JSDIIPGGTDVSSFOUMBX
6OEFSTUBOE,JSDIIPGGTWPMUBHFMBX
%FWFMPQBOVOEFSTUBOEJOHPGIPXUPVTF,JSDIIPGGTDVSSFOU
MBXUPXSJUFOPEBMFRVBUJPOTBOEUIFOUPTPMWFGPSVOLOPXO
OPEFWPMUBHFT
%FWFMPQBOVOEFSTUBOEJOHPGIPXUPVTF,JSDIIPGGTWPMUBHF
MBXUPXSJUFNFTIFRVBUJPOTBOEUIFOUPTPMWFGPSVOLOPXO
MPPQDVSSFOUT
&YQMBJOIPXUPVTF14QJDFUPTPMWFGPSVOLOPXOOPEFWPMUBHFT
BOEDVSSFOUT
3.1
Introduction
)BWJOHVOEFSTUPPEUIFGVOEBNFOUBMMBXTPGDJSDVJUUIFPSZ 0INTMBXBOE
,JSDIIPGGTMBXT XFBSFOPXQSFQBSFEUPBQQMZUIFTFMBXTUPEFWFMPQUXP
QPXFSGVMUFDIOJRVFTGPSDJSDVJUBOBMZTJTOPEBMBOBMZTJT XIJDIJTCBTFE
POBTZTUFNBUJDBQQMJDBUJPOPG,JSDIIPG GTDVSSFOUMBX ,$- BOENFTI
BOBMZTJT XIJDIJTCBTFEPOBTZTUFNBUJDBQQMJDBUJPOPG,JSDIIPGGTWPMUBHF
MBX ,7- 5IFUXPUFDIOJRVFTBSFTPJNQPSUBOUUIBUUIJTDIBQUFSTIPVME
CFSF HBSEFEBTUIFNPTUJNQPSUBOUJOUIFCPPL4UVEFOUTBSFUIFSFGPSF
FODPVSBHFEUPQBZDBSFGVMBUUFOUJPO
8JUIUIFUXPUFDIOJRVFTUPCFEFWFMPQFEJOUIJTDIBQUFS XFDBOBOB
MZ[FBOZMJOFBSDJSDVJUCZPCUBJOJOHBTFUPGTJNVMUBOFPVTFRVBUJPOTUIBUBSF
UIFOTPMWFEUPPCUBJOUIFSFRVJSFEWBMVFTPGDVSSFOUPSWPMUBHF0OFNFUIPE
PGTPMWJOHTJNVMUBOFPVTFRVBUJPOTJOWPMWFT$SBNFSTSVMF XIJDIBMMPXTVT
UPDBMDVMBUFDJSDVJUWBSJBCMFTBTBRVPUJFOUPGEFUFSNJOBOUT5IFFYBNQMFT
JOUIFDIBQUFSXJMMJMMVTUSBUFUIJTNFUIPE"QQFOEJYA also briefly summa
SJ[FTUIFFTTFOUJBMTUIFSFBEFSOFFETUPLOP XGPSBQQMZJOH$SBNFS TSVMF
"OPUIFSNFUIPEPGTPMWJOHTJNVMUBOFPVTFRVBUJPOTJTUPVTF ."5-"# B
DPNQVUFSTPGUXBSFEJTDVTTFEJO"QQFOEJY&
"MTPJOUIJTDIBQUFS XFJOUSPEVDFUIFVTFPG14QJDFGPS8JOEPXT B
DJSDVJUTJNVMBUJPODPNQVUFSTPGUXBSFQSPHSBNUIBUXFXJMMVTFUISPVHI PVUUIFUF YU'JOBMMZ XFBQQMZUIFUFDIOJRVFTMFBSOFEJOUIJTDIBQUFSUP
BOBMZ[FUSBOTJTUPSDJSDVJUT
3.2
Nodal analysis is also known as the
node-voltage method.
Nodal Analysis
/PEBMBOBMZTJTQSPWJEFTBHFOFSBMQSPDFEVSFGPSBOBMZ[JOHDJSDVJUTVTJOH
OPEFWPMUBHFTBTUIFDJSDVJUW BSJBCMFT$IPPTJOHOPEFW PMUBHFTJOTUFBE
PG FMFNFOU WPMUBHFT BT DJSDVJU WBSJBCMFT JT DPOWFOJFOU BOE SFEVDFT UIF
OVNCFSPGFRVBUJPOTPOFNVTUTPMWFTJNVMUBOFPVTMZ
5PTJNQMJGZNBUUFST XFTIBMMBTTVNFJOUIJTTFDUJPOUIBUDJSDVJUTEP
OPUDPOUBJOWPMUBHFTPVSDFT$JSDVJUTUIBUDPOUBJOWPMUBHFTPVSDFTXJMMCF
BOBMZ[FEJOUIFOFYUTFDUJPO
*O OPEBM BOBMZTJT XF BSF JOUFSFTUFE JO finding UIF OPEF WPMUBHFT
(JWFOBDJSDVJUXJUIOOPEFTXJUIPVUWPMUBHFTPVSDFT UIFOPEBMBOBMZTJT
PGUIFDJSDVJUJOWPMWFTUBLJOHUIFGPMMPXJOHUISFFTUFQT
3.2
81
Nodal Analysis
Steps to Determine Node Voltages:
4FMFDUBOPEFBTUIFSFGFSFODFOPEF"TTJHOWPMUBHFTv v vO − UP UIF SFNBJOJOH O − OPEFT 5IF WPMUBHFT BSF SFGFS
FODFEXJUISFTQFDUUPUIFSFGFSFODFOPEF
"QQMZ,$-UPFBDIPGUIF O −OPOSFGFSFODFOPEFT6TF
0INTMBXUPFYQSFTTUIFCSBODIDVSSFOUTJOUFSNT PGOPEF
WPMUBHFT
4PMWFUIFSFTVMUJOHTJNVMUBOFPVTFRVBUJPOTUPPCUBJOUIFVO LOPXOOPEFWPMUBHFT
8FTIBMMOPXFYQMBJOBOEBQQMZUIFTFUISFFTUFQT
The first step in nodal analysis is selecting a node as the SFGFSFODFPS
EBUVNOPEF5IFSFGFSFODFOPEFJTDPNNPOMZDBMMFEUIFHSPVOETJODFJU
JTBTTVNFEUPIB WF[FSPQPUFOUJBM"SFGFSFODFOPEFJTJOEJDBUFEC ZBOZ
PGUIFUISFFTZNCPMTJO'JH5IFUZQFPGHSPVOEJO'JH D JTDBMMFE
B DIBTTJTHSPVOEBOEJTVTFEJOEF WJDFTXIFSFUIFDBTF FODMPTVSF PS
DIBTTJTBDUTBTBSFGFSFODFQPJOUGPSBMMDJSDVJUT8IFOUIFQPUFOUJBMPGUIF
FBSUIJTVTFEBTSFGFSFODF XFVTFUIF FBSUIHSPVOEJO'JH B PS C 8FTIBMMBMXBZTVTFUIFTZNCPMJO'JH C 0ODFXFIB WFTFMFDUFEBSFGFSFODFOPEF XFBTTJHOW PMUBHFEFT JHOBUJPOTUPOPOSFGFSFODFOPEFT$POTJEFS GPSF YBNQMF UIFDJSDVJUJO
Fig. B /PEFJTUIFSFGFSFODFOPEF v= XIJMFOPEFTBOE
BSFBTTJHOFEWPMUBHFTvBOEv SFTQFDUJWFMZ,FFQJONJOEUIBUUIFOPEF
WPMUBHFTare defined XJUISFTQFDUUPUIFSFGFSFODFOPEF"TJMMVTUSBUFEJO
'JH B FBDIOPEFWPMUBHFJTUIFWPMUBHFSJTFGSPNUIFSFGFSFODFOPEF
UPUIFDPSSFTQPOEJOHOPOSFGFSFODFOPEFPSTJNQMZUIFW PMUBHFPGUIBU
OPEFXJUISFTQFDUUPUIFSFGFSFODFOPEF
"TUIFTFDPOETUFQ XFBQQMZ,$-UPFBDIOPOSFGFSFODFOPEFJOUIF
DJSDVJU5PBWPJEQVUUJOHUPPNVDIJOGPSNBUJPOPOUIFTBNFDJSDVJU UIF
DJSDVJUJO'JH B JTSFESB XOJO'JH C XIFSFXFOP XBEEJ J BOEJBTUIFDVSSFOUTUISPVHISFTJTUPST 3 3 BOE 3 SFTQFDUJWFMZ"U
OPEF BQQMZJOH,$-HJWFT
*=*+J+J
(a)
(c)
(b)
Figure 3.1
$PNNPOTZNCPMTGPSJOEJDBUJOHB
SFGFSFODFOPEF B DPNNPOHSPVOE (b) ground, (c) chassis ground.
The number of nonreference nodes is
equal to the number of independent
equations that we will derive.
I2
R2
1
I1
+
v1
–
+
v2
–
R1
0
R3
(a)
I2
"UOPEF
2
*+J=J
8FOPXBQQMZ0INTMBXUPFYQSFTTUIFVOLOPXODVSSFOUTJ J BOEJJO
UFSNTPGOPEFWPMUBHFT5IFLFZJEFBUPCFBSJONJOEJTUIBU TJODFSFTJT
UBODFJTBQBTTJWFFMFNFOU CZUIFQBTTJWFTJHODPOWFOUJPO DVSSFOUNVTU
BMXays floXGSPNBIJHIFSQPUFOUJBMUPBMPXFSQPUFOUJBM
I1
i2
R2
i2
v2
i1
i3
R1
R3
(b)
Current flows from a higher potential to a lower potential in a resistor.
Figure 3.2
5ZQJDBMDJSDVJUGPSOPEBMBOBMZTJT
8FDBOFYQSFTTUIJTQSJODJQMFBT
vIJHIFS−vMPXFS
@@@@@@@@@@@
J=
3
v1
82
Chapter 3
Methods of Analysis
/PUFUIBUUIJTQSJODJQMFJTJOBHSFFNFOUXJUIUIFX ay we defined resis
UBODFJO$IBQUFS TFF'JH 8JUIUIJTJONJOE XFPCUBJOGSPN
Fig. 3.2(b),
v−
@@@@@@
J=
PS
J=(v
3
v−v
J=@@@@@@
PS
J=( v−v 3
v−
J=@@@@@@
PS
J=(v
3
4VCTUJUVUJOH&R JO&RT BOE SFTVMUT SFTQFDUJWFMZ JO
v @@@@@@
v −v
@@@
*=*+
+ 3
3
v−v @@@
v
*+@@@@@@
= 3
3
*OUFSNTPGUIFDPOEVDUBODFT &RT BOE CFDPNF
Appendix A discusses how to use
Cramer’s rule.
*=*+(v+( v−v *+( v−v =(v
5IFUIJSETUFQJOOPEBMBOBMZTJTJTUPTPMWFGPSUIFOPEFWPMUBHFT*G
XFBQQMZ,$-UP O−OPOSFGFSFODFOPEFT XFPCUBJO O−TJNVMUB OFPVTFRVBUJPOTTVDIBT&RT BOE PS BOE ' PSUIF
DJSDVJUPG'JH XFTPMWF&RT BOE PS BOE UPPC
UBJOUIFOPEFWPMUBHFTvBOEvVTJOHBOZTUBOEBSENFUIPE TVDIBTUIF
TVCTUJUVUJPONFUIPE UIFFMJNJOBUJPONFUIPE $SBNFS TSVMF PSNBUSJY
JOWFSTJPO5PVTFFJUIFSPGUIFMBTUUXPNFUIPET POFNVTUDBTUUIFTJNVM
UBOFPVTFRVBUJPOTJONBUSJYGPSN'PSFYBNQMF &RT BOE DBO
CFDBTUJONBUSJYGPSNBT
][ ] [
[
]
v
(+(
−(
* −*
= −( (+( v * XIJDIDBOCFTPMWFEUPHFUvBOEv&RVBUJPOXJMMCFHFOFSBMJ[FEJO
4FDUJPO 5IFTJNVMUBOFPVTFRVBUJPOTNBZBMTPCFTPMW FEVTJOHDBM DVMBUPSTPSXJUITPGUX BSFQBDLBHFTTVDIBT ."5-"# .BUIDBE .BQMF BOE2VBUUSP1SP
Example 3.1
$BMDVMBUFUIFOPEFWPMUBHFTJOUIFDJSDVJUTIPXOJO'JH B Solution:
$POTJEFS'JH C XIFSFUIFDJSDVJUJO'JH B IBTCFFOQSFQBSFEGPS
OPEBMBOBMZTJT/PUJDFIPXUIFDVSSFOUTBSFTFMFDUFEGPSUIFBQQMJDBUJPO
PG,$-&YDFQUGPSUIFCSBODIFTXJUIDVSSFOUTPVSDFT UIFMBCFMJOHPGUIF
DVSSFOUTJTBSCJUSBSZCVUDPOTJTUFOU #ZDPOTJTUFOU XFNFBOUIBUJG GPS FYBNQMF XFBTTVNFUIBUJFOUFSTUIFΩSFTJTUPSGSPNUIFMFGUIBOETJEF JNVTUMFBWFUIFSFTJTUPSGSPNUIFSJHIUIBOETJEF 5IFSFGFSFODFOPEFJT
TFMFDUFE BOEUIFOPEFWPMUBHFTvBOEvBSFOPXUPCFEFUFSNJOFE
"UOPEF BQQMZJOH,$-BOE0INTMBXHJWFT
J=J+J
⇒
v−v @@@@@@
v −
@@@@@@
=
+ .VMUJQMZJOHFBDIUFSNJOUIFMBTUFRVBUJPOCZ XFPCUBJO
=v− v+v
3.2
83
Nodal Analysis
PS
5A
v− v=
"UOPEF XFEPUIFTBNFUIJOHBOEHFU
v −v
v−
@@@@@@
+=+@@@@@@
⇒
J+J=J+J
1
4Ω
2Ω
.VMUJQMZJOHFBDIUFSNCZSFTVMUTJO
2
6Ω
10 A
v−v+=+v
PS
−v+v=
(a)
5A
/PXXFIBWF UXPTJNVMUBOFPVT&RT BOE 8FDBO TPMWF
UIFFRVBUJPOTVTJOHBOZNFUIPEBOEPCUBJOUIFWBMVFTPGvBOEv
■ METHOD 1 6TJOHUIFFMJNJOBUJPOUFDIOJRVF XFBEE&RT BOE v=
⇒
v=7
4VCTUJUVUJOHv=JO&R HJWFT
v−=
■ METHOD 2 5PVTF$SBNFSTSVMF XFOFFEUPQVU&RT BOE
JONBUSJYGPSNBT
− v
v = −
5IFEFUFSNJOBOUPGUIFNBUSJYJT
−
Δ= =−=
−
][ ] [ ]
[
i2
v1
i3
i1 = 5
4Ω
v2
i4 = 10
i2 i
5
6Ω
2Ω
10 A
=7
v=@@@
⇒
[
i1 = 5
(b)
Figure 3.3
'PS&YBNQMF B PSJHJOBMDJSDVJU (b) circuit for analysis.
]
8FOPXPCUBJOvBOEvBT
−
Δ
@@@@@@@@
+
@@@@@@@@
@@@=
v=
=
=7
Δ
Δ
Δ @@@@@@@@
− @@@@@@@@
+ @@@
=
v= =
= 7
Δ
Δ
ǀ
ǀ
ǀ
ǀ
HJWJOHVTUIFTBNFSFTVMUBTEJEUIFFMJNJOBUJPONFUIPE
*GXFOFFEUIFDVSSFOUT XFDBOFBTJMZDBMDVMBUFUIFNGSPNUIFW
BMVFT
PGUIFOPEBMWPMUBHFT
v−v
v
J=@@@@@@
J= " = −" J=@@= "
v
@@
J= = "
J= " 5IFGBDUUIBU J is negative shows that the current flows in the direction
PQQPTJUFUPUIFPOFBTTVNFE
0CUBJOUIFOPEFWPMUBHFTJOUIFDJSDVJUPG'JH
Answer: v=307 v=−7
Practice Problem 3.1
84
Chapter 3
Example 3.2
14 A
%FUFSNJOFUIFWPMUBHFTBUUIFOPEFTJO'JH B 5Ω
1
4Ω
Methods of Analysis
Solution:
5IFDJSDVJUJOUIJTFYBNQMFIBTUISFFOPOSFGFSFODFOPEFT VOMJLFUIFQSF
WJPVTFYBNQMFXIJDIIBTUXPOPOSFGFSFODFOPEFT8FBTTJHOWPMUBHFTUP
UIFUISFFOPEFTBTTIPXOJO'JH C BOEMBCFMUIFDVSSFOUT
"UOPEF
v−v @@@@@@
v −v
= J+ JY
⇒
=@@@@@@
+ .VMUJQMZJOHCZBOESFBSSBOHJOHUFSNT XFHFU
2
5Ω
7A
Figure 3.4
⇒
JY= J+ J
v−v @@@@@@
v −v
v −
@@@@@@
=@@@@@@
+ 1
2Ω
8Ω
2
i1
v1
3
3A
4Ω
3A
4Ω
4Ω
ix
v−v−v= "UOPEF
'PS1SBDUJDF1SPC
2ix
2Ω
8Ω
i2
i1
v3
i3
ix
ix
i2
v2
2ix
4Ω
3A
0
(a)
(b)
Figure 3.5
'PS&YBNQMF B PSJHJOBMDJSDVJU C DJSDVJUGPSBOBMZTJT
.VMUJQMZJOHCZBOESFBSSBOHJOHUFSNT XFHFU
−v+ v−v= "UOPEF
J+ J= JY
⇒
v −v
v−v @@@@@@@@@
v −v
@@@@@@
+@@@@@@
= .VMUJQMZJOHCZ SFBSSBOHJOHUFSNT BOEEJWJEJOHCZ XFHFU
v−v+ v= 8FIBWFUISFFTJNVMUBOFPVTFRVBUJPOTUPTPMWFUPHFUUIFOPEFWPMUBHFT
v v BOEv8FTIBMMTPMWFUIFFRVBUJPOTJOUISFFXBZT
■ METHOD 1
BOE 6TJOHUIFFMJNJOBUJPOUFDIOJRVF XFBEE&RT v−v= PS
= v−v=@@@
"EEJOH&RT BOE HJWFT
−v+ v= ⇒
v= v
3.2
Nodal Analysis
4VCTUJUVUJOH&R JOUP&R ZJFMET
v−v= ⇒
v= v= v= 7
'SPN&R XFHFU
v= v−v= v−v= −v= −7
5IVT
v= 7 ■ METHOD 2
JONBUSJYGPSN
v= 7 v= −7
5PVTF$SBNFSTSVMF XFQVU&RT UP [
][ ] [ ]
− −
v
−
−
v = −
v 'SPNUIJT XFPCUBJO
Δ
v=@@@
Δ
Δ
v=@@@
Δ
Δ
v=@@@
Δ
XIFSFΔ Δ Δ BOEΔBSFUIFEFUFSNJOBOUTUPCFDBMDVMBUFEBTGPMMPXT
"TFYQMBJOFE JO"QQFOEJY" UPDBMDVMBUF UIFEFUFSNJOBOU PGBCZ 3 matrix, we repeat the first two rows and cross multiply.
ǀ
ǀ
−
− Δ=
− −=
− −
−
−
ǀ ǀ
− −
−
−
− − +
− − − +
+
= −+ + −−= ǀ ǀ
ǀ ǀ
ǀ ǀ
4JNJMBSMZ XFPCUBJO
− −
−
Δ=
− − − − +
− − +
+
−
−
−
−
Δ= − +
− − − − +
+
−
= + + −−−= = + −−−+ = − −
Δ= = + + −−−= −
− +
−
− − − +
+
−
85
86
Chapter 3
Methods of Analysis
Thus, we find
Δ
Δ
= 7 = 7
@@@
@@@
v=@@@=
v=@@@=
Δ Δ Δ
− @@@@
v=@@@=
= −7
Δ
BTXFPCUBJOFEXJUI.FUIPE
■ METHOD 3 8F OPX VTF ."5-"# UP TPMWF UIF NBUSJY &RVB
tion (3.2.6) can be written as
"7= #
⇒
7= "−#
XIFSF"JTUIFCZTRVBSFNBUSJY #JTUIFDPMVNOWFDUPS BOE 7JTB
DPMVNOWFDUPSDPNQSJTFEPGv v BOEvUIBUXFXBOUUPEFUFSNJOF8F
VTF."5-"#UPEFUFSNJOF7BTGPMMPXT
>>"=< − − − − − >
>>#=< >ʹ
>>7=JOW " #
7=
−
5IVT v= 7 v= 7 BOEv= −7 BTPCUBJOFEQSFWJPVTMZ
Practice Problem 3.2
Find the voltages at the three nonreference nodes in the circuit of Fig. Answer:v= 7 v= −7 v= 7
2Ω
1
3Ω
4ix
2
4A
ix
4Ω
3
6Ω
3.3
Nodal Analysis with Voltage Sources
8FOPXDPOTJEFSIP XWPMUBHFTPVSDFTBG GFDUOPEBMBOBMZTJT8 FVTFUIF DJSDVJUJO'JHGPSJMMVTUSBUJPO$POTJEFSUIFGPMMPXJOHUXPQPTTJCJMJUJFT
Figure 3.6
'PS1SBDUJDF1SPC
■ CASE 1 *GBWPMUBHFTPVSDFJTDPOOFDUFECFUXFFOUIFSFGFSFODF
OPEFBOEBOPOSFGFSFODFOPEF XFTJNQMZTFUUIFWPMUBHFBUUIFOPO
SFGFSFODFOPEFFRVBMUPUIFWPMUBHFPGUIFWPMUBHFTPVSDF*O'JH GPS
FYBNQMF
v= 7
5IVT PVSBOBMZTJTJTTPNFwhat simplified by this knoXMFEHFPGUIFWPMU
BHFBUUIJTOPEF
■ CASE 2 *GUIFWPMUBHFTPVSDF EFQFOEFOUPSJOEFQFOEFOU JTDPO OFDUFECFUXFFOUXPOPOSFGFSFODFOPEFT UIFUXPOPOSFGFSFODFOPEFT
3.3
87
Nodal Analysis with Voltage Sources
4Ω
Supernode
i4
2Ω
v1
i1
5V
v2
v3
+–
i2
10 V +
–
i3
8Ω
6Ω
Figure 3.7
"DJSDVJUXJUIBTVQFSOPEF
GPSNBHFOFSBMJ[FEOPEFPSTVQFSOPEFXFBQQMZCPUI,$-BOE,7-UP
EFUFSNJOFUIFOPEFWPMUBHFT
A supernode may be regarded as a
closed surface enclosing the voltage
source and its two nodes.
A supernode is formed by enclosing a (dependent or independent)
voltage source connected between two nonreference nodes and any
elements connected in parallel with it.
*O'JH OPEFTBOEGPSNBTVQFSOPEF 8 FDPVMEIBWFNPSFUIBO
UXPOPEFTGPSNJOHBTJOHMFTVQFSOPEF' PSFYBNQMF TFFUIFDJSDVJUJO
'JH 8FBOBMZ[FBDJSDVJUXJUITVQFSOPEFTVTJOHUIFTBNFUISFF
TUFQTNFOUJPOFEJOUIFQSF WJPVTTFDUJPOF YDFQUUIBUUIFTVQFSOPEFTBSF
USFBUFEEJG GFSFOUMZ 8IZ #FDBVTFBOFTTFOUJBMDPNQPOFOUPGOPEBM
BOBMZTJTJTBQQMZJOH,$- XIJDISFRVJSFTLOP XJOHUIFDVSSFOUUISPVHI
FBDIFMFNFOU5IFSFJTOPXBZPGLOPXJOHUIFDVSSFOUUISPVHIBWPMUBHF
TPVSDFJOBEWBODF)PXFWFS, KCL must be satisfied at a supernode likF
BOZPUIFSOPEF)FODF BUUIFTVQFSOPEFJO'JH
J+ J= J+ J
B
v −v
v −
v−v @@@@@@
v− @@@@@@
@@@@@@
+ =@@@@@@
+ C
PS
5PBQQMZ,JSDIIPGGTWPMUBHFMBXUPUIFTVQFSOPEFJO'JH XFSFESBX
UIFDJSDVJUBTTIPXOJO'JH(PJOHBSPVOEUIFMPPQJOUIFDMPDLXJTF
EJSFDUJPOHJWFT
−v+ + v= ⇒
v−v= 5V
+
'SPN&RT C BOE XFPCUBJOUIFOPEFWPMUBHFT
/PUFUIFGPMMPXJOHQSPQFSUJFTPGBTVQFSOPEF
5IFW PMUBHFTPVSDFJOTJEFUIFTVQFSOPEFQSP WJEFTBDPOTUSBJOU FRVBUJPOOFFEFEUPTPMWFGPSUIFOPEFWPMUBHFT
"TVQFSOPEFIBTOPWPMUBHFPGJUTPXO
"TVQFSOPEFSFRVJSFTUIFBQQMJDBUJPOPGCPUI,$-BOE,7-
Figure 3.8
+–
+
v2
v3
–
–
"QQMZJOH,7-UPBTVQFSOPEF
88
Chapter 3
Example 3.3
'PSUIFDJSDVJUTIPwn in Fig. 3.9, find the node vPMUBHFT
10 Ω
2V
v1
Solution:
5IFTVQFSOPEFDPOUBJOTUIF7TPVSDF OPEFTBOE BOEUIF Ω
SFTJTUPS"QQMZJOH,$-UPUIFTVQFSOPEFBTTIPXOJO'JH B HJWFT
v2
+–
2Ω
2A
Methods of Analysis
= J+ J+ 4Ω
7A
&YQSFTTJOHJBOEJJOUFSNTPGUIFOPEFWPMUBHFT
v− @@@@@@
v −
=@@@@@@
+ + ⇒
= v+ v+ PS
Figure 3.9
'PS&YBNQMF
v= −−v
5PHFUUIFSFMBUJPOTIJQCFUXFFOvBOEv XFBQQMZ,7-UPUIFDJSDVJUJO
'JH C (PJOHBSPVOEUIFMPPQ XFPCUBJO
−v−+ v= ⇒
'SPN&RT BOE XFXSJUF
v= v+ v= v+ = −−v
PS
v= −
⇒
v= −7
BOEv= v+ = −7/PUFUIBUUIF ΩSFTJTUPSEPFTOPUNBLF
BOZEJGGFSFODFCFDBVTFJUJTDPOOFDUFEBDSPTTUIFTVQFSOPEF
2 v2
v1
i2 7 A
i1
2A
2Ω
2A
4Ω
2V
1
+
7A
+–
1
v1
v2
–
–
(b)
(a)
Figure 3.10
"QQMZJOH B ,$-UPUIFTVQFSOPEF C ,7-UPUIFMPPQ
Practice Problem 3.3
3Ω
Figure 3.11
+–
14 V +
–
'JOEvBOEJJOUIFDJSDVJUPG'JH
6V
4Ω
+
v
–
'PS1SBDUJDF1SPC
2Ω
Answer:−N7 "
i
6Ω
2
+
3.3
89
Nodal Analysis with Voltage Sources
Example 3.4
'JOEUIFOPEFWPMUBHFTJOUIFDJSDVJUPG'JH
3Ω
+ vx –
20 V
+–
1
2Ω
6Ω
2
3vx
3
+–
4
4Ω
10 A
1Ω
Figure 3.12
'PS&YBNQMF
Solution:
/PEFTBOEGPSNBTVQFSOPEFTPEPOPEFTBOE8FBQQMZ,$-UP
UIFUXPTVQFSOPEFTBTJO'JH B "UTVQFSOPEF
J+ = J+ J
&YQSFTTJOHUIJTJOUFSNTPGUIFOPEFWPMUBHFT
v−v
v−v @@
v
@@@@@@
@@@@@@
=
+
+
PS
v+ v−v−v= "UTVQFSOPEF
⇒
J= J+ J+ J
v−v @@
v
v −v
v @@
@@@@@@
@@@@@@
=
+
+
PS
v+ v−v−v= 3Ω
3Ω
i2
2Ω
i1
6Ω
v2
v1
+ vx –
+ vx –
i1
i3
v3
i3
10 A
Loop 3
v4
i5
i4
4Ω
1Ω
+
v1
+–
Loop 1
–
(a)
Figure 3.13
"QQMZJOH B ,$-UPUIFUXPTVQFSOPEFT C ,7-UPUIFMPPQT
3𝜐x
i3
20 V
+
6Ω
+
v2
v3
–
–
(b)
+ –
Loop 2
+
v4
–
90
Chapter 3
Methods of Analysis
8FOPXBQQMZ,7-UPUIFCSBODIFTJOWPMWJOHUIFWPMUBHFTPVSDFTBT
TIPXOJO'JH C 'PSMPPQ
−v+ + v= ⇒
v− v= 'PSMPPQ
−v+ vY+ v= #VUvY= v−vTPUIBU
v−v−v= 'PSMPPQ
vY−vY+ J−= #VUJ= v−vBOEvY= v−v)FODF
−v−v+ v+ v= 8FOFFEGPVSOPEFW PMUBHFT v v v BOE v BOEJUSFRVJSFTPOMZ
four out of the five Eqs. (3.4.1) to (3.4.5) to find them. Although the fiGUI
FRVBUJPOJTSFEVOEBOU JUDBOCFVTFEUPDIFDLSFTVMUT8FDBOTPMWF&RT
UP EJSFDUMZVTJOH ."5-"#8FDBOFMJNJOBUFPOFOPEF
WPMUBHFTPUIBUXFTPMW FUISFFTJNVMUBOFPVTFRVBUJPOTJOTUFBEPGGPVS 'SPN&R v= v− 4VCTUJUVUJOHUIJT JOUP&RT BOE
SFTQFDUJWFMZ HJWFT
v−v−v= v−v−v= BOE
&RVBUJPOT BOE DBOCFDBTUJONBUSJYGPSNBT
[
][ ] [ ]
−
−
v
−
−
v = − − v 6TJOH$SBNFSTSVMFHJWFT
ǀ
ǀ
−
−
Δ= −
= − −
− −
ǀ
ǀ
ǀ
ǀ
−
−
Δ=
−
= −
−
− −
−
Δ= −= − −
5IVT XFBSSJWFBUUIFOPEFWPMUBHFTBT
ǀ
ǀ
− Δ= = − − Δ
Δ
−
−
@@@@@
@@@@@@
v=@@@=
7
= 7 v=@@@=
=
Δ
Δ
−
−
Δ
@@@@
v=@@@=
= −7
Δ −
BOEv= v−= 78FIBWFOPUVTFE&R JUDBOCFVTFE
UPDSPTTDIFDLSFTVMUT
3.4
91
Mesh Analysis
Practice Problem 3.4
'JOEv v BOEvJOUIFDJSDVJUPG'JHVTJOHOPEBMBOBMZTJT
6Ω
25 V
v1
+–
i
3.4
2Ω
Mesh Analysis
.FTIBOBMZTJTQSPWJEFTBOPUIFSHFOFSBMQSPDFEVSFGPSBOBMZ[JOHDJSDVJUT VTJOHNFTIDVSSFOUTBTUIFDJSDVJUWBSJBCMFT6TJOHNFTIDVSSFOUTJOTUFBE
PGFMFNFOUDVSSFOUTBTDJSDVJUWBSJBCMFTJTDPOWFOJFOUBOESFEVDFTUIF
OVNCFSPGFRVBUJPOTUIBUNVTUCFTPMWFETJNVMUBOFPVTMZ3FDBMMUIBUB
MPPQJTBDMPTFEQBUIXJUIOPOPEFQBTTFENPSFUIBOPODF"NFTIJTB
MPPQUIBUEPFTOPUDPOUBJOBOZPUIFSMPPQXJUIJOJU
Nodal analysis applies KCL to find unknoXOWPMUBHFTJOBHJ WFO
circuit, while mesh analysis applies KVL to find unknoXODVSSFOUT
.FTIBOBMZTJTJTOPURVJUFBTHFOFSBMBTOPEBMBOBMZTJTCFDBVTFJUJT POMZBQQMJDBCMFUPBDJSDVJUUIBUJT QMBOBS"QMBOBS DJSDVJUJTPOFUIBU DBOCFESBXOJOBQMBOFXJUIOPCSBODIFTDSPTTJOHPOFBOPUIFSPUIFS
XJTFJUJTOPOQMBOBS"DJSDVJUNBZIBWFDSPTTJOHCSBODIFTBOETUJMMCF
QMBOBSJGJUDBOCFSFESB XOTVDIUIBUJUIBTOPDSPTTJOHCSBODIFT' PS
FYBNQMF UIFDJSDVJUJO'JH B IBTUX PDSPTTJOHCSBODIFT C VUJU
DBOCFSFESBXOBTJO'JH C )FODF UIFDJSDVJUJO'JH B JT QMBOBS)PXFWFS UIFDJSDVJUJO'JHJTOPOQMBOBS CFDBVTFUIFSF JTOPX BZUPSFESB XJUBOEB WPJEUIFCSBODIFTDSPTTJOH/POQMBOBS
DJSDVJUTDBOCFIBOEMFEVTJOHOPEBMBOBMZTJT CVUUIFZXJMMOPUCFDPO
TJEFSFEJOUIJTUFYU
5i
v2
+ –
Answer:v= 7 v= −7 v= 7
v3
4Ω
3Ω
Figure 3.14
'PS1SBDUJDF1SPC
Mesh analysis is also known as loop
analysis or the mesh-current method.
1A
2Ω
5Ω 6Ω
1Ω
3Ω
4Ω
7Ω
8Ω
1Ω
(a)
5Ω
4Ω
6Ω
7Ω
1A
2Ω
3Ω
2Ω
13 Ω
5A
12 Ω
11 Ω
9Ω
8Ω
10 Ω
1Ω
5Ω
8Ω
Figure 3.16
"OPOQMBOBSDJSDVJU
5o understand mesh analysis, we should first eYQMBJONPSFBCPVU
XIBUXFNFBOCZBNFTI
A mesh is a loop that does not contain any other loops within it.
4Ω
3Ω
6Ω
7Ω
(b)
Figure 3.15
B "QMBOBSDJSDVJUXJUIDSPTTJOHCSBODIFT C UIFTBNFDJSDVJUSFESBXOXJUIOPDSPTT
JOHCSBODIFT
92
Chapter 3
Methods of Analysis
a
I1
R1
b
I2
R2
c
I3
V1 +
–
i2
i1
R3
e
f
+ V
2
–
d
Figure 3.17
"DJSDVJUXJUIUXPNFTIFT
Although path abcdefa is a loop and
not a mesh, KVL still holds. This is the
reason for loosely using the terms
loop analysis and mesh analysis to
mean the same thing.
*O'JH GPSFYBNQMF QBUIT BCFGBBOE CDEFCBSFNFTIFT CVUQBUI
BCDEFGBJTOPUBNFTI5IFDVSSFOUUISPVHIBNFTIJTLOPXOBT NFTI
DVSSFOU. In mesh analysis, we are interested in applying KVL to find the
NFTIDVSSFOUTJOBHJWFODJSDVJU
*OUIJTTFDUJPO XFXJMMBQQMZNFTIBOBMZTJTUPQMBOBSDJSDVJUTUIBUEP
OPUDPOUBJODVSSFOUTPVSDFT*OUIFOFYUTFDUJPO XFXJMMDPOTJEFSDJSDVJUT
XJUIDVSSFOUTPVSDFT*OUIFNFTIBOBMZTJTPGBDJSDVJUXJUIONFTIFT XF
UBLFUIFGPMMPXJOHUISFFTUFQT
Steps to Determine Mesh Currents:
"TTJHONFTIDVSSFOUTJ J JOUPUIFONFTIFT
"QQMZ,7-UPFBDI PG UIF ONFTIFT 6TF 0INTMBX UP FYQSFTTUIFWPMUBHFTJOUFSNTPGUIFNFTIDVSSFOUT
4PMWFUIFSFTVMUJOH OTJNVMUBOFPVTFRVBUJPOTUPHFUUIFNFTI
DVSSFOUT
The direction of the mesh current is
arbitrary—(clockwise or counterclockwise)—and does not affect the validity
of the solution.
5PJMMVTUSBUFUIFTUFQT DPOTJEFSUIFDJSDVJUJO'JH The first
TUFQSFRVJSFTUIBUNFTIDVSSFOUT JBOE JBSFBTTJHOFEUPNFTIFTBOE
"MUIPVHIBNFTIDVSSFOUNBZCFBTTJHOFEUPFBDINFTIJOBOBSCJ USBSZEJSFDUJPO JUJTDPOWFOUJPOBMUPBTTVNFUIBUFBDINFTIDVSSFOUfloXT
DMPDLXJTF
"TUIFTFDPOETUFQ XFBQQMZ,7-UPFBDINFTI"QQMZJOH,7-UP
NFTI XFPCUBJO
−7+ 3J+ 3 J−J = PS
3+ 3 J−3J= 7
'PSNFTI BQQMZJOH,7-HJWFT
The shortcut way will not apply if one
mesh current is assumed clockwise
and the other assumed counterclockwise, although this is permissible.
3J+ 7+ 3 J−J = PS
−3J+ 3+ 3 J= −7
Note in Eq. (3.13) that the coefficient of JJTUIFTVNPGUIFSFTJTUBODFT
JOthe first NFTI XIJMF UIFcoefficient PGJJTUIF OFHBUJWFPGUIF SFTJT
UBODFDPNNPOUPNFTIFTBOE/PXPCTFSWFUIBUUIFTBNFJTUSVFJO
Eq. 5IJTDBOTFSWFBTBTIPSUDVUXBZPGXSJUJOHUIFNFTIFRVB UJPOT8FXJMMFYQMPJUUIJTJEFBJO4FDUJPO
3.4
93
Mesh Analysis
5IFUIJSETUFQJTUPTPMWFGPSUIFNFTIDVSSFOUT1VUUJOH&RT BOE JONBUSJYGPSNZJFMET
[
][ ] [ ]
*= J−J
J
7
3+3
−3
= J
−3 3+3 −7 XIJDIDBOCFTPMWFEUPPCUBJOUIFNFTIDVSSFOUT JBOE J8FBSFBU MJCFSUZUPVTFBOZUFDIOJRVFGPSTPMWJOHUIFTJNVMUBOFPVTFRVBUJPOT
"DDPSEJOHUP&R JGBDJSDVJUIBT OOPEFT CCSBODIFT BOEMJO
EFQFOEFOUMPPQTPSNFTIFT UIFO M= C−O+ )FODF MJOEFQFOEFOU
TJNVMUBOFPVT FRVBUJPOT BSF SFRVJSFE UP TPMWF UIF DJSDVJU VTJOH NFTI
BOBMZTJT
/PUJDFUIBUUIFCSBODIDVSSFOUTBSFEJGGFSFOUGSPNUIFNFTIDVSSFOUT
VOMFTTUIFNFTIJTJTPMBUFE5PEJTUJOHVJTICFUXFFOUIFUXPUZQFTPGDVS
SFOUT XFVTFJGPSBNFTIDVSSFOUBOE*GPSBCSBODIDVSSFOU5IFDVSSFOU
FMFNFOUT* * BOE*BSFBMHFCSBJDTVNTPGUIFNFTIDVSSFOUT*UJTF WJ
EFOUGSPN'JHUIBU
*= J *= J Example 3.5
'or the circuit in Fig. 3.18, find the branch currents * * BOE *VTJOH
NFTIBOBMZTJT
I1
Solution:
We first obtain the mesh currents using KVL. For mesh 1,
5Ω
10 Ω
15 V +
–
PS
i1
Figure 3.18
J+ J+ J−J −= 'PS&YBNQMF
J= J−
■ METHOD 1 6TJOHUIFTVCTUJUVUJPONFUIPE XFTVCTUJUVUF
Eq. (3.5.2) into Eq. (3.5.1), and write
J−−J= ⇒
J= "
'SPN&R J= J−= −= "5IVT
*= J= " *= J= " *= J−J= ■ METHOD 2 5PVTF$SBNFSTSVMF XFDBTU&RT BOE JONBUSJYGPSNBT
[
][ ] [ ]
− J
=
−
J
i2
+ 10 V
–
'PSNFTI
PS
6Ω
I3
−+ J+ J−J + = J−J= I2
4Ω
94
Chapter 3
Methods of Analysis
8FPCUBJOUIFEFUFSNJOBOUT
ǀ
ǀ
ǀ
−
Δ= = + = 5IVT
ǀ
¢
Δ= = −= ¢
Δ
J=@@@= " Δ
ǀ
ǀ
Δ=
= + = ¢ Δ
J=@@@= "
Δ
BTCFGPSF
Practice Problem 3.5
$BMDVMBUFUIFNFTIDVSSFOUTJBOEJPGUIFDJSDVJUPG'JH
Answer:J= " J= 200N"
15 Ω
6Ω
90 V +
–
10 Ω
i1
i2
4Ω
+ 40 V
–
5Ω
Figure 3.19
'PS1SBDUJDF1SPC
Example 3.6
Use mesh analysis to find the current *PJOUIFDJSDVJUPG'JH
Solution:
8FBQQMZ,7-UPUIFUISFFNFTIFTJOUVSO'PSNFTI
−+ J−J + J−J = i1
A
i2
PS
Io
24 V +
–
24 Ω
4Ω
i1
12 Ω
Figure 3.20
i2
10 Ω
'PS&YBNQMF
i3
J−J−J= 'PSNFTI
J+ J−J + J−J = +
–
4Io
PS
−J+ J−J= 'PSNFTI
*P+ J−J + J−J = 3.4
Mesh Analysis
#VUBUOPEF" *P= J−J TPUIBU
J−J + J− J + J−J = PS
−J− J+ J= *ONBUSJYGPSN &RT UP CFDPNF
[
][ ] [ ]
J
− −
J = − −
− −
J ǀ ǀ
ǀ ǀ
ǀ ǀ
ǀ ǀ
8FPCUBJOUIFEFUFSNJOBOUTBT
Δ=
− −
−
−
− − − +
− − − − − +
−
+
= −−−−−= −
−
− −
− −
−
−
− +
− +
+
−
Δ=
− − − − −
−
−
− +
− +
+
Δ=
Δ=
= −= = + = −
−
= + = −
−
− +
−
− − +
−
+
8FDBMDVMBUFUIFNFTIDVSSFOUTVTJOH$SBNFSTSVMFBT
Δ
@@@@
J=@@@=
= " Δ Δ
= "
@@@@
J=@@@=
Δ Δ
= "
@@@@
J=@@@=
Δ 5IVT *P= J−J= "
95
96
Chapter 3
Practice Problem 3.6
Using mesh analysis, find *PJOUIFDJSDVJUPG'JH
6Ω
Io
16 V +
–
4Ω
i1
Answer:−"
i3
8Ω
2Ω
–
+
i2
3.5
10io
'PS1SBDUJDF1SPC
4Ω
i1
■ CASE 1 8IFOBDVSSFOUTPVSDFFYJTUTPOMZJOPOFNFTI$POTJEFS
UIFDJSDVJUJO'JH GPSFYBNQMF8FTFUJ= −"BOEXSJUFBNFTI
FRVBUJPOGPSUIFPUIFSNFTIJOUIFVTVBMXBZUIBUJT
3Ω
6Ω
5A
i2
Mesh Analysis with Current Sources
"QQMZJOHNFTIBOBMZTJTUPDJSDVJUTDPOUBJOJOHDVSSFOUTPVSDFT EFQFOEFOU PSJOEFQFOEFOU NBZBQQFBSDPNQMJDBUFE#VUJUJTBDUVBMMZNVDIFBTJFS UIBOXIBUXFFODPVOUFSFEJOUIFQSFWJPVTTFDUJPO CFDBVTFUIFQSFTFODF PGUIFDVSSFOUTPVSDFTSFEVDFTUIFOVNCFSPGFRVBUJPOT$POTJEFSUIFGPM
MPXJOHUXPQPTTJCMFDBTFT
Figure 3.21
10 V +
–
Methods of Analysis
−+ J+ J−J = Figure 3.22
J= −"
⇒
■ CASE 2 8IFOBDVSSFOUTPVSDFFYJTUTCFUXFFOUXPNFTIFT$PO
TJEFSUIFDJSDVJUJO'JH B GPSFYBNQMF8FDSFBUFB TVQFSNFTICZ
FYDMVEJOHUIFDVSSFOUTPVSDFBOEBOZFMFNFOUTDPOOFDUFEJOTFSJFTXJUI
JU BTTIPXOJO'JH C 5IVT
"DJSDVJUXJUIBDVSSFOUTPVSDF
A supermesh results when two meshes have a (dependent or independent) current source in common.
6Ω
10 Ω
6Ω
10 Ω
i1
i2
2Ω
20 V +
–
i1
i2
4Ω
6A
i1
0
(a)
i2
Exclude these
elements
20 V +
–
4Ω
(b)
Figure 3.23
B 5XPNFTIFTIBWJOHBDVSSFOUTPVSDFJODPNNPO C BTVQFSNFTI DSFBUFECZFYDMVEJOHUIFDVSSFOU
TPVSDF
"TTIPXOJO'JH C XFDSFBUFBTVQFSNFTIBTUIFQFSJQIFSZPGUIFUXP
NFTIFTBOEUSFBUJUEJGGFSFOUMZ *GBDJSDVJUIBTUXPPSNPSFTVQFSNFTIFT UIBUJOUFSTFDU UIFZTIPVMECFDPNCJOFEUPGPSNBMBSHFSTVQFSNFTI 8IZ
USFBUUIFTVQFSNFTIEJGGFSFOUMZ #FDBVTFNFTIBOBMZTJTBQQMJFT,7-‡
XIJDI SFRVJSFT UIBUXFLOPXUIF WPMUBHFBDSPTTFBDICSBODI‡BOEXF EPOPULOPXUIFWPMUBHFBDSPTTBDVSSFOUTPVSDFJOBEWBODF)PXFWFS B TVQFSNFTINVTUTBUJTGZ,7-MJLFBOZPUIFSNFTI5IFSFGPSF BQQMZJOH
,7-UPUIFTVQFSNFTIJO'JH C HJWFT
−+ J+ J+ J= 3.5
97
Mesh Analysis with Current Sources
PS
J+ J= 8FBQQMZ,$-UPBOPEFJOUIFCSBODIXIFSFUIFUXPNFTIFTJOUFSTFDU
"QQMZJOH,$-UPOPEFJO'JH B HJWFT
J= J+ 4PMWJOH&RT BOE XFHFU
J= −" J= "
/PUFUIFGPMMPXJOHQSPQFSUJFTPGBTVQFSNFTI
5IFDVSSFOUTPVSDFJOUIFTVQFSNFTIQSP
WJEFTUIFDPOTUSBJOU FRVBUJPOOFDFTTBSZUPTPMWFGPSUIFNFTIDVSSFOUT
"TVQFSNFTIIBTOPDVSSFOUPGJUTPXO
"TVQFSNFTISFRVJSFTUIFBQQMJDBUJPOPGCPUI,7-BOE,$-
Example 3.7
'or the circuit in Fig. 3.24, find JUPJVTJOHNFTIBOBMZTJT
2Ω
i1
i1
4Ω
P
i2
5A
6Ω
i2
2Ω
Io
3Io
i2
Q
i3
8Ω
i4
+ 10 V
–
i3
Figure 3.24
'PS&YBNQMF
Solution:
/PUFUIBUNFTIFTBOEGPSNBTVQFSNFTICFDBVTFUIFZIBWFBOJOEF
QFOEFOUDVSSFOUTPVSDFJODPNNPO"MTP NFTIFTBOEGPSNBOPUIFS
TVQFSNFTICFDBVTFUIFZIBWFBEFQFOEFOUDVSSFOUTPVSDFJODPNNPO
5IFUXPTVQFSNFTIFTJOUFSTFDUBOEGPSNBMBSHFSTVQFSNFTIBTTIPXO
"QQMZJOH,7-UPUIFMBSHFSTVQFSNFTI
J+ J+ J−J + J= PS
J+ J+ J−J= 'PSUIFJOEFQFOEFOUDVSSFOUTPVSDF XFBQQMZ,$-UPOPEF1
J= J+ 'PSUIFEFQFOEFOUDVSSFOUTPVSDF XFBQQMZ,$-UPOPEF2
J= J+ *P
98
Chapter 3
Methods of Analysis
#VU*P= −J IFODF
J= J−J
"QQMZJOH,7-JONFTI
J+ J−J + = PS
J−J= −
'SPN&RT UP J= −" Practice Problem 3.7
i3
5Ω
24 V +
–
i1
20 Ω
Figure 3.25
I2
G2
v2
G1
G3
R1
R2
R3
i3
3.6
[
+ V2
–
B 5IFDJSDVJUJO'JH C UIFDJSDVJU
JO'JH
][ ] [
]
v
(+(
−(
* −*
= −( (+( v * 0CTFSWFUIBUFBDIPGUIFEJBHPOBMUFSNTJTUIFTVNPGUIFDPOEVDUBODFT
DPOOFDUFEEJSFDUMZUPOPEF PS XIJMF UIFPGGEJBHPOBMUFSNTBSF UIF
OFHBUJWFTPGUIFDPOEVDUBODFTDPOOFDUFECFUXFFOUIFOPEFT"MTP FBDI
UFSNPOUIFSJHIUIBOETJEFPG&R JTUIFBMHFCSBJDTVNPGUIFDVS
SFOUTFOUFSJOHUIFOPEF
*OHFOFSBM JGBDJSDVJUXJUIJOEFQFOEFOUDVSSFOUTPVSDFTIBT /OPO
SFGFSFODFOPEFT UIFOPEFW PMUBHFFRVBUJPOTDBOCFXSJUUFOJOUFSNTPG
UIFDPOEVDUBODFTBT
(b)
Figure 3.26
Nodal and Mesh Analyses
by Inspection
(a)
i1
Answer: J=" J=N" J="
5IJTTFDUJPOQSFTFOUTBHFOFSBMJ[FEQSPDFEVSFGPSOPEBMPSNFTIBOBMZTJT
*UJTBTIPSUDVUBQQSPBDICBTFEPONFSFJOTQFDUJPOPGBDJSDVJU
8IFOBMMTPVSDFTJOBDJSDVJUBSFJOEFQFOEFOUDVSSFOUTPVSDFT XF EPOPUOFFEUPBQQMZ,$-UPFBDIOPEFUPPCUBJOUIFOPEFW
PMUBHF
FRVBUJPOTBTXFEJEJO4FDUJPO 8FDBOPCUBJOUIFFRVBUJPOTCZ
NFSFJOTQFDUJPOPGUIFDJSDVJU "TBOF YBNQMF MFUVTSFF YBNJOFUIF DJSDVJUJO'JH TIPXOBHBJOJO'JH B GPSDPOWFOJFODF5IF
DJSDVJUIBTUXPOPOSFGFSFODFOPEFTBOEUIFOPEFFRVBUJPOTXFSFEF
SJWFEJO4FDUJPOBT
'PS1SBDUJDF1SPC
V1 +
–
J= "
6TFNFTIBOBMZTJTUPEFUFSNJOFJ J BOEJJO'JH
5Ω
i2
3Ω
I1
J= " 10 Ω
4A
v1
J= −" [
][ ] [ ]
J
( ( ⋯ (/
v
J
( ( ⋯ (/
v
=
⋮
⋮
⋮
⋮
⋮
⋮
(/ (/ ⋯ (// v/ J/ 3.6
Nodal and Mesh Analyses by Inspection
PSTJNQMZ
(W=J
XIFSF
(LL=4VNPGUIFDPOEVDUBODFTDPOOFDUFEUPOPEFL
(LK=(KL=/FHBUJWFPGUIFTVNPGUIFDPOEVDUBODFTEJSFDUMZ
DPOOFDUJOHOPEFTLBOEK L≠K
vL=6OLOPXOWPMUBHFBUOPEFL
JL=4VNPGBMMJOEFQFOEFOUDVSSFOUTPVSDFTEJSFDUMZDPOOFDUFEUP
OPEFL XJUIDVSSFOUTFOUFSJOHUIFOPEFUSFBUFEBTQPTJUJWF
(JTDBMMFEUIF DPOEVDUBODFNBUSJYWJTUIFPVUQVUW FDUPSBOE JJTUIF
JOQVUWFDUPS&RVBUJPO DBOCFTPMWFEUPPCUBJOUIFVOLOPXOOPEF
WPMUBHFT,FFQJONJOEUIBUUIJTJTW BMJEGPSDJSDVJUTXJUIPOMZJOEFQFO EFOUDVSSFOUTPVSDFTBOEMJOFBSSFTJTUPST
4JNJMBSMZ XFDBOPCUBJONFTIDVSSFOUFRVBUJPOTCZJOTQFDUJPOXIFO
BMJOFBSSFTJTUJWFDJSDVJUIBTPOMZJOEFQFOEFOUWPMUBHFTPVSDFT$POTJEFS
UIFDJSDVJUJO'JH TIPXOBHBJOJO'JH C GPSDPOWFOJFODF5IF
DJSDVJUIBTUXPOPOSFGFSFODFOPEFTBOEUIFOPEFFRVBUJPOTXFSFEFSJWFE
JO4FDUJPOBT
[
][ ] [ ]
J
3+3
−3
v
=
J
−3 3+3 −v 8FOPUJDFUIBUFBDIPGUIFEJBHPOBMUFSNTJTUIFTVNPGUIFSFTJTUBODFTJO
UIFSFMBUFENFTI XIJMFFBDIPGUIFPGGEJBHPOBMUFSNTJTUIFOFHBUJWFPG
UIFSFTJTUBODFDPNNPOUPNFTIFTBOE&BDIUFSNPOUIF SJHIUIBOE
TJEFPG&R JTUIFBMHFCSBJDTVNUBLFODMPDLXJTFPGBMMJOEFQFOEFOU
WPMUBHFTPVSDFTJOUIFSFMBUFENFTI
*OHFOFSBM JGUIFDJSDVJUIBT /NFTIFT UIFNFTIDVSSFOUFRVBUJPOT
DBOCFFYQSFTTFEJOUFSNTPGUIFSFTJTUBODFTBT
[
][ ] [ ]
J
3 3 ⋯ 3/
v
J
3 3 ⋯ 3/
v
=
⋮
⋮
⋮
⋮
⋮
⋮
3/ 3/ ⋯ 3// J/ v/ 3J=W
PSTJNQMZ
XIFSF
3LL=4VNPGUIFSFTJTUBODFTJONFTIL
3LK=3KL=/FHBUJWFPGUIFTVNPGUIFSFTJTUBODFTJODPNNPOXJUI
NFTIFTLBOEK L≠K
JL=6OLOPXONFTIDVSSFOUGPSNFTILJOUIFDMPDLXJTFEJSFDUJPO
WL=4VNUBLFODMPDLXJTFPGBMMJOEFQFOEFOUWPMUBHFTPVSDFTJO
NFTIL XJUIWPMUBHFSJTFUSFBUFEBTQPTJUJWF
3JTDBMMFEUIF SFTJTUBODFNBUSJY JJTUIFPVUQVUW FDUPSBOE WJTUIF
JOQVUW FDUPS 8FDBOTPMW F&R UPPCUBJOUIFVOLOP XONFTI
DVSSFOUT
99
100
Example 3.8
Chapter 3
Methods of Analysis
8SJUFUIFOPEFW PMUBHFNBUSJYFRVBUJPOTGPSUIFDJSDVJUJO'JHCZ
JOTQFDUJPO
2A
1Ω
3A
8Ω
5 Ω v2
v1
10 Ω
8Ω
v3
1A
4Ω
v4
4A
2Ω
Figure 3.27
'PS&YBNQMF
Solution:
5IFDJSDVJUJO'JHIBTGPVSOPOSFGFSFODFOPEFT TPXFOFFEGPVS
OPEFFRVBUJPOT5IJTJNQMJFTUIBUUIFTJ[FPGUIFDPOEVDUBODFNBUSJY ( JTCZ5IFEJBHPOBMUFSNTPG( JOTJFNFOT BSF
(=@@
+@@@
= +@@
(=@@
+@@
=
+@@
(=@@
+@@
= +@@
(=@@
+@@
=
5IFPGGEJBHPOBMUFSNTBSF
=− (=−@@
(=(=
=− =−
(=− (=−@@
(=−@@
=−
(=− (=−@@
(= (= (=− (=−
5IFJOQVUDVSSFOUWFDUPSJIBTUIFGPMMPXJOHUFSNT JOBNQFSFT
J= J=−−=− J= J=+=
5IVT UIFOPEFWPMUBHFFRVBUJPOTBSF
−
v
−
−
− −
v
=
−
−
v
−
− v [
][ ] [ ]
XIJDIDBOCFTPMWFE VTJOH."5-"#UPPCUBJOUIFOPEFWPMUBHFT v v v BOEv
3.6
#ZJOTQFDUJPO PCUBJOUIF OPEFWPMUBHF FRVBUJPOTGPSUIF DJSDVJU JO
Fig. 3.28.
Answer:
101
Nodal and Mesh Analyses by Inspection
[
][ ] [ ]
− −
v
−
v
=
−
−
v
−
− v Practice Problem 3.8
1Ω
4Ω
v3
v4
3A
v1
5Ω
20 Ω
v2
1Ω
2A
2A
Figure 3.28
'PS1SBDUJDF1SPC
#ZJOTQFDUJPO XSJUFUIFNFTIDVSSFOUFRVBUJPOTGPSUIFDJSDVJUJO'JH
5Ω
i1
4V
2Ω
2Ω
10 V +
–
4Ω
i2
1Ω
i4
2Ω
+–
4Ω
3Ω
1Ω
3Ω
i5
i3
+ 6V
–
+ 12 V
–
Figure 3.29
'PS&YBNQMF
Solution:
We have five meshes, so the resistance matrix is 5 by 5. The diagonal
UFSNT JOPINT BSF
3=++++=
3=++= 3=++= 3=++= 3=+=
5IFPGGEJBHPOBMUFSNTBSF
3=− 3=− 3==3
3=− 3=− 3=− 3=−
3=− 3=− 3==3
3= 3=− 3= 3=−
3= 3=− 3= 3=−
5IFJOQVUWPMUBHFWFDUPSWIBTUIFGPMMPXJOHUFSNTJOWPMUT
v= v=−=
v=−+=− v= v=−
Example 3.9
102
Chapter 3
Methods of Analysis
5IVT UIFNFTIDVSSFOUFRVBUJPOTBSF
−
−
[
−
−
−
−
−
−
][ ] [ ]
−
−
J
−
J
−
J
=
− J
J −
'SPN UIJT XF DBO VTF ."5-"# UP PCUBJO NFTI DVSSFOUT J J J J BOEJ
Practice Problem 3.9
#ZJOTQFDUJPO PCUBJOUIFNFTIDVSSFOUFRVBUJPOTGPSUIFDJSDVJUJO
'JH
50 Ω
20 Ω
i2
30 V +
–
15 Ω
i1
+ 12 V
–
30 Ω
i3
20 Ω
i4
i5
–
+ 20 V
80 Ω
60 Ω
Figure 3.30
'PS1SBDUJDF1SPC
Answer:
−
−
[
3.7
−
−
−
−
−
−
−
J
J
J
−
−
=
J
−
J ][ ] [
]
Nodal Versus Mesh Analysis
#PUIOPEBMBOENFTIBOBMZTFTQSPWJEFBTZTUFNBUJDXBZPGBOBMZ[JOHB
DPNQMFYOFUXPSL4PNFPOFNBZBTL(J WFOBOFUX PSLUPCFBOBMZ[FE IPXEPXFLOPXXIJDINFUIPEJTCFUUFSPSNPSFFGficient? 5IFDIPJDFPG
UIFCFUUFSNFUIPEJTEJDUBUFECZUXPGBDUPST
3.8
103
Circuit Analysis with PSpice
The first fBDUPSJTUIFOBUVSFPGUIFQBSUJDVMBSOFUXPSL/FUXPSLTUIBU
DPOUBJONBOZTFSJFTDPOOFDUFEFMFNFOUT WPMUBHFTPVSDFT PSTVQFSNFTIFT
BSF NPSF TVJUBCMF GPS NFTI BOBMZTJT XIFSFBT OFUXPSLT XJUI QBSBMMFM
DPOOFDUFEFMFNFOUT DVSSFOUTPVSDFT PSTVQFSOPEFTBSFNPSFTVJUBCMF
GPSOPEBMBOBMZTJT"MTP BDJSDVJUXJUIGFXFSOPEFTUIBONFTIFTJTCFUUFS
BOBMZ[FEVTJOHOPEBMBOBMZTJT XIJMFBDJSDVJUXJUIGF XFSNFTIFTUIBO
OPEFTJTCFUUFSBOBMZ[FEVTJOHNFTIBOBMZTJT 5IFLFZJTUPTFMFDUUIF
NFUIPEUIBUSFTVMUTJOUIFTNBMMFSOVNCFSPGFRVBUJPOT
5IFTFDPOEGBDUPSJTUIFJOGPSNBUJPOSFRVJSFE*GOPEFW PMUBHFTBSF
SFRVJSFE JUNBZCFFYQFEJFOUUPBQQMZOPEBMBOBMZTJT*GCSBODIPSNFTI
DVSSFOUTBSFSFRVJSFE JUNBZCFCFUUFSUPVTFNFTIBOBMZTJT
*UJTIFMQGVMUPCFGBNJMJBSXJUICPUINFUIPETPGBOBMZTJT GPSBUMFBTU
UXPSFBTPOT'JSTU POFNFUIPEDBOCFVTFEUPDIFDLUIFSFTVMUTGSPNUIF
PUIFSNFUIPE JGQPTTJCMF4FDPOE TJODFFBDINFUIPEIBTJUTMJNJUBUJPOT POMZPOFNFUIPENBZCFTVJUBCMFGPSBQBSUJDVMBSQSPCMFN'PSFYBNQMF NFTIBOBMZTJTJTUIFPOMZNFUIPEUPVTFJOBOBMZ[JOHUSBOTJTUPSDJSDVJUT BTXFTIBMMTFFJO4FDUJPO#VUNFTIBOBMZTJTDBOOPUFBTJMZCFVTFEUP
TPMWFBOPQBNQDJSDVJU BTXFTIBMMTFFJO$IBQUFS CFDBVTFUIFSFJTOP
EJSFDUXBZUPPCUBJOUIFW PMUBHFBDSPTTUIFPQBNQJUTFMG' PSOPOQMBOBS
OFUXPSLT OPEBMBOBMZTJTJTUIFPOMZPQUJPO CFDBVTFNFTIBOBMZTJTPOMZ
BQQMJFTUPQMBOBSOFUX PSLT"MTP OPEBMBOBMZTJTJTNPSFBNFOBCMFUP
TPMVUJPOCZDPNQVUFS BTJUJTFBTZUPQSPHSBN5IJTBMMPXTPOFUPBOBMZ[F
DPNQMJDBUFEDJSDVJUTUIBUEFGZIBOEDBMDVMBUJPO "DPNQVUFSTPGUX BSF
QBDLBHFCBTFEPOOPEBMBOBMZTJTJTJOUSPEVDFEOFYU
3.8
Circuit Analysis with PSpice
14QJDF JT B DPNQVUFS TPGUXBSF DJSDVJU BOBMZTJT QSPHSBN UIBU XF XJMM
HSBEVBMMZ MFBSO UP VTF UISPVHIPVU UIF DPVSTF PG UIJT UFYU 5IJT TFDUJPO
JMMVTUSBUFTIPXUPVTF 14QJDFGPS8JOEPXTUPBOBMZ[FUIFEDDJSDVJUTXF
IBWFTUVEJFETPGBS
5IF SFBEFS JT FYQFDUFE UP SFWJFX 4FDUJPOT % UISPVHI % PG
"QQFOEJY%CFGPSFQSPDFFEJOHJOUIJTTFDUJPO*UTIPVMECFOPUFEUIBU
14QJDFJTPOMZIFMQGVMJOEFUFSNJOJOHCSBODIWPMUBHFTBOEDVSSFOUTXIFO
UIFOVNFSJDBMWBMVFTPGBMMUIFDJSDVJUDPNQPOFOUTBSFLOPXO
6TF14QJDF to find the node vPMUBHFTJOUIFDJSDVJUPG'JH
Appendix D provides a tutorial on
using PSpice for Windows.
Example 3.10
Solution:
1 20 Ω 2
The first step is to draw the given circuit using Schematics. If one fol
MPXTUIFJOTUSVDUJPOTHJWFOJO"QQFOEJYTFDUJPOT%BOE% UIFTDIF +
30 Ω
NBUJD JO'JH JTQSPEVDFE #FDBVTF UIJTJT BEDBOBMZTJT XF VTF 120 V –
WPMUBHFTPVSDF7%$BOEDVSSFOUTPVSDF*%$5IFQTFVEPDPNQPOFOU
7*&810*/54BSFBEEFEUPEJTQMBZUIFSFRVJSFEOPEFWPMUBHFT0ODFUIF
0
DJSDVJUJTESBXOBOETBWFEBTFYBNTDI XFSVO14QJDFCZTFMFDUJOH Figure 3.31
"OBMZTJT4JNVMBUF5IFDJSDVJUJTTJNVMBUFEBOEUIFSFTVMUTBSFEJTQMBZFE 'PS&YBNQMF
10 Ω
40 Ω
3
3A
104
Chapter 3
Methods of Analysis
120.0000
1
81.2900
R1
2
20
+
120 V
–
R3
89.0320
3
10
IDC
V1
R2
R4
30
40
3A
I1
0
Figure 3.32
'PS&YBNQMFUIFTDIFNBUJDPGUIFDJSDVJUJO'JH
on VIEWPOINTS and also saved in output file FYBNPVU5IFPVUQVU
file includes the following:
/0%& 70-5"(& /0%& 70-5"(& /0%& 70-5"(&
JOEJDBUJOHUIBU7=7 7=7 7=7
Practice Problem 3.10
'PSUIFDJSDVJUJO'JH VTF14QJDF to find the node voltages.
500 mA
1
60 Ω
30 Ω
100 Ω
2
50 Ω
3
+ 50 V
–
25 Ω
0
Figure 3.33
'PS1SBDUJDF1SPC
Answer: 7=−7 7=7 7=7
*OUIFDJSDVJUPG'JH EFUFSNJOFUIFDVSSFOUTJ J BOEJ
1Ω
4Ω
2Ω
i1
24 V +
–
2Ω
Figure 3.34
'PS&YBNQMF
3vo
+–
Example 3.11
8Ω
i2
4Ω
i3
+
vo
–
3.9
105
Applications: DC Transistor Circuits
Solution:
5IFTDIFNBUJDJTTIPXOJO'JH 5IFTDIFNBUJDJO'JH
JODMVEFTUIFPVUQVUSFTVMUT JNQMZJOHUIBUJUJTUIFTDIFNBUJDEJTQMBZFE
POUIFTDSFFO BGUFSUIFTJNVMBUJPO /PUJDFUIBUUIFWPMUBHFDPOUSPMMFE
WPMUBHFTPVSDF&JO'JHJTDPOOFDUFETPUIBUJUTJOQVUJTUIFWPMU BHFBDSPTTUIF ΩSFTJTUPSJUTHBJOJTTFUFRVBMUP*OPSEFSUPEJT QMBZUIFSFRVJSFEDVSSFOUT XFJOTFSUQTFVEPDPNQPOFOU*130#&4JO
UIFBQQSPQSJBUFCSBODIFT5IFTDIFNBUJDJTTBWFEBT FYBNTDIBOE
TJNVMBUFECZTFMFDUJOH"OBMZTJT4JNVMBUF5IFSFTVMUTBSFEJTQMBZFEPO
IPROBES as shown in Fig. 3.35 and saved in output file FYBNPVU
From the output file or the IPROBES, we obtain J=J="BOE
J="
E
– +
2
E1
–+
R5
R1
1
R6
4
24 V
+
–
R2
2
R3
8
R4
4
V1
1.333E + 00
1.333E + 00
2.667E + 00
0
Figure 3.35
5IFTDIFNBUJDPGUIFDJSDVJUJO'JH
6TF14QJDFUPEFUFSNJOFDVSSFOUTJ J BOEJJOUIFDJSDVJUPG'JH
Practice Problem 3.11
Answer:J=−N" J=" J="
i1
4Ω
2A
3.9
†
Applications: DC Transistor Circuits
.PTUPGVTEFBMXJUIFMFDUSPOJDQSPEVDUTPOBSPVUJOFCBTJTBOEIB WF
TPNFF YQFSJFODFXJUIQFSTPOBMDPNQVUFST "CBTJDDPNQPOFOUGPS
UIFJOUFHSBUFEDJSDVJUTGPVOEJOUIFTFFMFDUSPOJDTBOEDPNQVUFSTJTUIF BDUJWF UISFFUFSNJOBMEFWJDFLOPXOBTUIFUSBOTJTUPS6OEFSTUBOEJOHUIF
USBOTJTUPSJTFTTFOUJBMCFGPSFBOFOHJOFFSDBOTUBSUBOFMFDUSPOJDDJSDVJU
EFTJHO
'JHVSFEFQJDUTWBSJPVTLJOETPG USBOTJTUPSTDPNNFSDJBMMZBWBJM
BCMF5IFSFBSFUX PCBTJDUZQFTPGUSBOTJTUPST CJQPMBSKVODUJPOUS BOTJT
UPST #+5T BOE field-efGFDUUS BOTJTUPST '&5T )FSF XFDPOTJEFSPOMZ
the BJTs, which were the first of the twPBOEBSFTUJMMVTFEUPEBZ 0VS
PCKFDUJWFJTUPQSFTFOUFOPVHIEFUBJMBCPVUUIF#+5UPFOBCMFVTUPBQQMZ
UIFUFDIOJRVFTEFWFMPQFEJOUIJTDIBQUFSUPBOBMZ[FEDUSBOTJTUPSDJSDVJUT
2Ω
10 V +
–
Figure 3.36
i2
1Ω
'PS1SBDUJDF1SPC
i3
i1
2Ω
106
Chapter 3
Methods of Analysis
Historical
William Schockley m +PIO#BSEFFO m BOE
8BMUFS#SBUUBJO m DPJOWFOUFEUIFUSBOTJTUPS
/PUIJOHIBTIBEBHSFBUFSJNQBDUPOUIFUSBOTJUJPOGSPNUIFi*OEVT
USJBM"HFuUPUIFi"HFPGUIF&OHJOFFSuUIBOUIFUSBOTJTUPS*BNTVSFUIBU
%S4IPDLMFZ %S#BSEFFO BOE%S#SBUUBJOIBEOPJEFBUIFZXPVMEIBWF
UIJTJODSFEJCMFFGGFDUPOPVSIJTUPSZ8IJMFXPSLJOHBU#FMM-BCPSBUPSJFT UIFZTVDDFTTGVMMZEFNPOTUSBUFEUIFQPJOUDPOUBDUUSBOTJTUPS JOWFOUFECZ
#BSEFFOBOE#SBUUBJOJO BOEUIFKVODUJPOUSBOTJTUPS XIJDI4IPDL
MFZDPODFJWFEJOBOETVDDFTTGVMMZQSPEVDFEJO
It is interesting to note that the idea of the field-effect transistor, the
most commonly used one today, was first conceived in 1925–1928 by
+&-JMJFOGFME B(FSNBOJNNJHSBOUUPUIF6OJUFE4UBUFT5IJTJTFWJEFOU
from his patents of what appears to be a field-effect transistor. Unfortu
OBUFMZ UIFUFDIOPMPHZUPSFBMJ[FUIJTEFWJDFIBEUPXBJUVOUJMXIFO
Shockley’s field-effect transistor became a reality. Just think what today
XPVMECFMJLFJGXFIBEUIJTUSBOTJTUPSZFBSTFBSMJFS
'PSUIFJSDPOUSJCVUJPOTUPUIFDSFBUJPOPGUIFUSBOTJTUPS %S4IPDLMFZ %S#BSEFFO BOE%S#SBUUBJOSFDFJWFE JO UIF/PCFM1SJ[FJOQIZT
JDT*UTIPVMECFOPUFEUIBU%S#BSEFFOJTUIFPOMZJOEJWJEVBMUPXJOUXP
/PCFMQSJ[FTJOQIZTJDTUIFTFDPOEDBNFMBUFSGPSXPSLJOTVQFSDPOEVD
UJWJUZBUUIF6OJWFSTJUZPG*MMJOPJT
$PVSUFTZPG-VDFOU
5FDIOPMPHJFT#FMM-BCT
Collector
C
n
p
Base
B
n
E
Emitter
(a)
Collector
Figure 3.37
C
7BSJPVTUZQFTPGUSBOTJTUPST
¥.D(SBX)JMM&EVDBUJPO.BSL%JFSLFS QIPUPHSBQIFS
p
Base
B
n
p
E
Emitter
(b)
Figure 3.38
5XPUZQFTPG#+5TBOEUIFJSDJSDVJU
TZNCPMT B OQO C QOQ
5IFSFBSFUX PUZQFTPG#+5T OQOBOE QOQ XJUIUIFJSDJSDVJUTZN CPMTBTTIPXOJO'JH&BDIUZQFIBTUISFFUFSNJOBMT EFTJHOBUFEBT
FNJUUFS & CBTF # BOEDPMMFDUPS $ 'PSUIF OQOUSBOTJTUPS UIF DVS
SFOUTBOEWoltages of the transistor are specified as in Fig. 3.39. "QQMZJOH
,$-UP'JH B HJWFT
*&=*#+*$
3.9
107
Applications: DC Transistor Circuits
XIFSF*& *$ BOE*#BSFFNJUUFS DPMMFDUPS BOECBTFDVSSFOUT SFTQFDUJWFMZ
4JNJMBSMZ BQQMZJOH,7-UP'JH C HJWFT
IC
IB
7$&+7&#+7#$=
C
B
XIFSF 7$& 7&# BOE 7#$BSFDPMMFDUPS FNJUUFS FNJUUFSCBTF BOECBTF
DPMMFDUPSWPMUBHFT5IF#+5DBOPQFSBUFJOPOFPGUISFFNPEFTBDUJ WF DVUPGG BOETBUVSBUJPO8IFOUSBOTJTUPSTPQFSBUFJOUIFBDUJWFNPEF UZQJ
DBMMZ7#& ≃ 7
E
(a)
*$=α*&
IE
C
XIFSF αJTDBMMFEUIF DPNNPOCBTFDVSS FOU HBJO*O&R αEFOPUFTUIFGSBDUJPOPGFMFDUSPOTJOKFDUFECZUIFFNJUUFSUIBUBSFDPM MFDUFECZUIFDPMMFDUPS"MTP
*$=β*#
XIFSFβJTLOPXOBTUIFDPNNPOFNJUUFSDVSSFOUHBJO5IFαBOEβBSF
DIBSBDUFSJTUJDQSPQFSUJFTPGBHJWFOUSBOTJTUPSBOEBTTVNFDPOTUBOUWBMVFT
GPSUIBUUSBOTJTUPS5ZQJDBMMZ αUBLFTWBMVFTJOUIFSBOHFPGUP XIJMFβUBLFTWBMVFTJOUIFSBOHFPGUP'SPN&RT UP JUJTFWJEFOUUIBU
*&= +β *#
β=@@@@@
α −α
+
+
VCB
B
–
VCE
+
VBE
–
–
E
(b)
Figure 3.39
5IFUFSNJOBMWBSJBCMFTPGBOOQOUSBOTJTUPS
B DVSSFOUT C WPMUBHFT
BOE
5IFTFFRVBUJPOTTIPXUIBU JOUIFBDUJWFNPEF UIF#+5DBOCFNPEFMFE
BTBEFQFOEFOUDVSSFOUDPOUSPMMFEDVSSFOUTPVSDF 5IVT JODJSDVJUBOBMZ
TJT UIFEDFRVJ WBMFOUNPEFMJO'JH C NBZCFVTFEUPSFQMBDFUIF
OQOUSBOTJTUPSJO'JH B 4JODFβJO&R JTMBSHF BTNBMMCBTF
DVSSFOUDPOUSPMTMBS HFDVSSFOUTJOUIFPVUQVUDJSDVJU$POTFRVFOUMZ UIF
CJQPMBSUSBOTJTUPSDBOTFSWe as an amplifier QSPEVDJOHCPUIDVSSFOUH BJO
BOEWPMUBHFHain. Such amplifiers can be used to furnish a DPOTJEFSBCMF
BNPVOUPGQPXFSUPUSBOTEVDFSTTVDIBTMPVETQFBLFSTPSDPOUSPMNPUPST
C
+
IB
B
+
VCE
VBE
–
B
IB
IC
+
VBE
–
+
C
βIB
VCE
–
E
(a)
–
E
(b)
Figure 3.40
B "OOQOUSBOTJTUPS C JUTEDFRVJWBMFOUNPEFM
*UTIPVMECFPCTFSW FEJOUIFGPMMP XJOHF YBNQMFTUIBUPOFDBOOPU
EJSFDUMZBOBMZ[FUSBOTJTUPSDJSDVJUTVTJOHOPEBMBOBMZTJTCFDBVTFPGUIFQP
UFOUJBMEJGGFSFODFCFUXFFOUIFUFSNJOBMTPGUIFUSBOTJTUPS 0OMZXIFOUIF USBOTJTUPSJTSFQMBDFECZJUTFRVJWBMFOUNPEFMDBOXFBQQMZOPEBMBOBMZTJT
In fact, transistor circuits provide motivation to study dependent sources.
108
Chapter 3
Example 3.12
Methods of Analysis
'JOE*# *$ BOEvPJOUIFUSBOTJTUPSDJSDVJUPG'JH "TTVNFUIBUUIF
USBOTJTUPSPQFSBUFTJOUIFBDUJWFNPEFBOEUIBUβ=
IC
200 kΩ
+
4V
–
IB
+
Input
loop
VBE
–
100 Ω
+
vo
–
Output
loop
+
6V
–
Figure 3.41
'PS&YBNQMF
Solution:
'PSUIFJOQVUMPPQ ,7-HJWFT
−+*# ¤ +7#&=
4JODF7#&=7JOUIFBDUJWFNPEF
*#=@@@@@@@
−
=•"
¤
#VU
*$=β*#=¤•"=N"
'PSUIFPVUQVUMPPQ ,7-HJWFT
−vP−*$+=
PS
vP=−*$=−=7
/PUFUIBUvP=7$&JOUIJTDBTF
Practice Problem 3.12
1 kΩ
+
15 kΩ
+
+
6V
–
VCE
VBE
–
150 Ω
–
+
vo
–
Figure 3.42
'PS1SBDUJDF1SPC
'PSUIFUSBOTJTUPSDJSDVJUJO'JH MFU
%FUFSNJOFWPBOE7$&
Answer: 7 N7
+
18 V
–
β =BOE 7#& =7
3.9
109
Applications: DC Transistor Circuits
Example 3.13
'PSUIF#+5DJSDVJUJO'JH β=BOE7#&=7'JOEvP
Solution:
1 kΩ
Define.The circuit is clearly defined and the problem is clearly stat
FE5IFSFBQQFBSUPCFOPBEEJUJPOBMRVFTUJPOTUIBUOFFEUPCFBTLFE
1SFTFOU8FBSFUPEFUFSNJOFUIFPVUQVUWPMUBHFPGUIFDJSDVJUTIPXO
JO'JH5IFDJSDVJUDPOUBJOTBOJEFBMUSBOTJTUPSXJUIβ=
BOE7#&=7
"MUFSOBUJWF8FDBOVTFNFTIBOBMZTJTUPTPMW FGPS vP8FDBOSF QMBDFUIFUSBOTJTUPSXJUIJUTFRVJWBMFOUDJSDVJUBOEVTFOPEBMBOBMZTJT
8FDBOUSZCPUIBQQSPBDIFTBOEVTFUIFNUPDIFDLFBDIPUIFS "T
BUIJSEDIFDL XFDBOVTFUIFFRVJ WBMFOUDJSDVJUBOETPMW FJUVTJOH
14QJDF
"UUFNQU
■ METHOD 1
Working with Fig. 3.44(a), we start with the first loop.
−+L*+L *−* =
*− *=¤−
PS
1 kΩ
+
100 kΩ
+
2V
–
vo
200 kΩ
I1
–
I2
(a)
100 kΩ
+
2V
–
V1
IB
0.7 V
200 kΩ
1 kΩ
150IB
+
+
16 V
–
I3
+
–
–
+
16 V
–
𝜈o
(b)
R1
700.00mV
14.58 V
100k
+
2V
–
R3
1k
R2
200k
0.7 V
+
F1
–
F
(c)
Figure 3.44
4PMVUJPOPGUIFQSPCMFNJO&YBNQMF B .FUIPE C .FUIPE D .FUIPE
+
16 V
–
+
100 kΩ
+
2V
–
Figure 3.43
vo
200 kΩ
'PS&YBNQMF
–
+
16 V
–
110
Chapter 3
Methods of Analysis
/PXGPSMPPQ
L *− * +7#&=
PS
−*+*=−¤−
4JODFXFIBWFUXPFRVBUJPOTBOEUXPVOLOPXOT XFDBOTPMWFGPS
BOE*"EEJOH&R UP XFHFU
*=¤−"
BOE
*
*= −+ −∕=•"
4JODF*=−*=−N" XFDBOOPXTPMWFGPSvPVTJOHMPPQ
−vP+L*+=
PS
vP=−+=7
■ METHOD 2 3FQMBDJOHUIFUSBOTJTUPSXJUIJUTFRVJWBMFOUDJSDVJU QSPEVDFTUIFDJSDVJUTIPXOJO'JH C 8FDBOOPXVTFOPEBM
BOBMZTJTUPTPMWFGPSvP
"UOPEFOVNCFS7=7
− ∕L+∕L+*#=
PS
*#=•"
"UOPEFOVNCFSXFIBWF
*#+ vP− ∕L=
PS
vP=−¤¤¤−=7
&WBMVBUF 5IFBOTXFSTDIFDL C VUUPGVSUIFSDIFDLXFDBO
VTF 14QJDF .FUIPE XIJDIHJ WFTVTUIFTPMVUJPOTIP XOJO
'JH D 4BUJTGBDUPSZ $MFBSMZ XFIBWFPCUBJOFEUIFEFTJSFEBOTXFSXJUIB
Wery high confidence leWFM8FDBOOPXQSFTFOUPVSXPSLBTBTPMV
UJPOUPUIFQSPCMFN
Practice Problem 3.13
Answer: 7 •"
10 kΩ
Io +
120 kΩ
+
1V
–
+
10 kΩ
vo
VBE
–
Figure 3.45
'PS1SBDUJDF1SPC
–
5IFUSBOTJTUPSDJSDVJUJO'JHIBT β=BOE 7#&=7'JOE vP
BOE*P
+
20 V
–
3.10
Summary
/PEBMBOBMZTJTJT UIFBQQMJDBUJPOPG ,JSDIIPGGTDVSSFOUMBX BUUIF
OPOSFGFSFODFOPEFT *UJTBQQMJDBCMFUPCPUIQMBOBSBOEOPOQMBOBS
DJSDVJUT 8FFYQSFTTUIFSFTVMUJOUFSNTPGUIFOPEFWPMUBHFT4PMW JOHUIFTJNVMUBOFPVTFRVBUJPOTZJFMETUIFOPEFWPMUBHFT
"TVQFSOPEFDPOTJTUTPGUXPOPOSFGFSFODFOPEFTDPOOFDUFECZB EFQFOEFOUPSJOEFQFOEFOU WPMUBHFTPVSDF
.FTIBOBMZTJTJTUIFBQQMJDBUJPOPG,JSDIIPGGTWPMUBHFMBXBSPVOE
NFTIFTJOBQMBOBSDJSDVJU8FFYQSFTTUIFSFTVMUJOUFSNTPGNFTI
DVSSFOUT4PMWJOHUIFTJNVMUBOFPVTFRVBUJPOTZJFMETUIFNFTI
DVSSFOUT
111
Review Questions
"TVQFSNFTIDPOTJTUTPGUXPNFTIFTUIBUIBWFB EFQFOEFOUPSJOEF
QFOEFOU DVSSFOUTPVSDFJODPNNPO
/PEBMBOBMZTJTJTOPSNBMMZVTFEXIFOBDJSDVJUIBTGFXFSOPEFFRVB
UJPOTUIBONFTIFRVBUJPOT.FTIBOBMZTJTJTOPSNBMMZVTFEXIFOB
DJSDVJUIBTGFXFSNFTIFRVBUJPOTUIBOOPEFFRVBUJPOT
$JSDVJUBOBMZTJTDBOCFDBSSJFEPVUVTJOH14QJDF
%$USBOTJTUPSDJSDVJUTDBOCFBOBMZ[FEVTJOHUIFUFDIOJRVFTDPWFSFE
JOUIJTDIBQUFS
Review Questions
"UOPEFJOUIFDJSDVJUPG'JH BQQMZJOH,$-
HJWFT
'PSUIFDJSDVJUJO'JH vBOEvBSFSFMBUFEBT
C v=J − +v
E v=−J − +v
B v=J++v
D v=−J++v
− W @@
v −v
v
B +@@@@@@@
= +@@@@@@
v −v
v − @@
v
C +@@@@@@@
= +@@@@@@
v − v
− v @@@@@@
−v
+@@@@@@
D +@@@@@@@
= 12 V
8V
6Ω
v1
+–
i
+
–
v2
4Ω
v − v
v − @@@@@@
−v
+@@@@@@
E +@@@@@@@
= Figure 3.47
'PS3FWJFX2VFTUJPOTBOE
8Ω
2A
3Ω
v1
1
12 V
+
–
6Ω
4Ω
2
v2
*OUIFDJSDVJUPG'JH UIFWPMUBHFvJT
B −7
D 7
C −7
E 7
5IFDVSSFOUJJOUIFDJSDVJUPG'JHJT
B −"
D "
6Ω
C −"
E "
4Ω
Figure 3.46
'PS3FWJFX2VFTUJPOTBOE
10 V
*OUIFDJSDVJUPG'JH BQQMZJOH,$-BUOPEF
HJWFT
v −v
v
v
B @@@@@@
+@@=@@
v −v
v
v
C @@@@@@
+@@=@@
v −v
− v @@
v
D @@@@@@
+@@@@@@@
= v −v
v − @@
v
E @@@@@@
+@@@@@@@
= +
–
i
+ 6V
–
2Ω
Figure 3.48
'PS3FWJFX2VFTUJPOTBOE
5IFMPPQFRVBUJPOGPSUIFDJSDVJUJO'JHJT
B −+J++J=
C +J++J=
D +J − +J=
E −+J − +J=
112
Chapter 3
Methods of Analysis
*OUIFDJSDVJUPG'JH DVSSFOUJJT
B "
C "
D "
E "
5IF14QJDFQBSUOBNFGPSBDVSSFOUDPOUSPMMFEWPMU
BHFTPVSDFJT
B &9
2Ω
20 V +
–
+
v
–
3Ω
4Ω
Figure 3.49
'PS3FWJFX2VFTUJPOTBOE
5IFWPMUBHFWBDSPTTUIFDVSSFOUTPVSDFJOUIFDJSDVJU
PG'JHJT
B 7
C 7
D 7
E (9
B *UNVTUCFDPOOFDUFEJOTFSJFT
C *UQMPUTUIFCSBODIDVSSFOU
D *UEJTQMBZTUIFDVSSFOUUISPVHIUIFCSBODIJO
XIJDIJUJTDPOOFDUFE
E *UDBOCFVTFEUPEJTQMBZWPMUBHFCZDPOOFDUJOHJU
JOQBSBMMFM
F *UJTVTFEPOMZGPSEDBOBMZTJT
G *UEPFTOPUDPSSFTQPOEUPBQBSUJDVMBSDJSDVJU
FMFNFOU
i2
2A
D )9
8IJDIPGUIFGPMMPXJOHTUBUFNFOUTBSFOPUUSVFPGUIF
QTFVEPDPNQPOFOU*130#&
1Ω
i1
C '9
E 7
"OTXFSTB D B D D B E C D C E
Problems
Sections 3.2 and 3.3
Nodal Analysis
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEOPEBMBOBMZTJT
'JOEUIFDVSSFOUT*UISPVHI*BOEUIFWPMUBHFvPJO
UIFDJSDVJUPG'JH
vo
R1
I1
R2
Ix
12 V +
–
8A
20 Ω
I3
30 Ω
20 A
I4
60 Ω
+ 9V
–
R3
Figure 3.50
Figure 3.52
'PS1SPCBOE1SPC
10 Ω
I2
'PS1SPC
'PSUIFDJSDVJUJO'JH PCUBJOvBOEv
2Ω
6A
v1
10 Ω
Figure 3.51
'PS1SPC
5Ω
(JWFOUIFDJSDVJUJO'JH DBMDVMBUFUIFDVSSFOUT
JUISPVHIJ
v2
4Ω
3A
3A
6A
i1
i2
20 Ω
10 Ω
Figure 3.53
'PS1SPC
i3
40 Ω
40 Ω
i4
2A
113
Problems
0CUBJOvPJOUIFDJSDVJUPG'JH
%FUFSNJOF*CJOUIFDJSDVJUJO'JHVTJOHOPEBM
BOBMZTJT
120 V
+–
60 V +
–
120 kΩ
Ib
120 kΩ
30 kΩ
+
vo
–
Figure 3.54
24 V +
–
50 Ω
4PMWFGPS7JOUIFDJSDVJUPG'JHVTJOHOPEBM
BOBMZTJT
5Ω
10 V +
–
'JOE*PJOUIFDJSDVJUPG'JH
1Ω
4Ω
+
V1
–
2 Io
4A
+ 20 V
–
10 Ω
Figure 3.55
Io
8Ω
'PS1SPC
"QQMZOPEBMBOBMZTJTUPTPMWFGPS7YJOUIFDJSDVJUPG
'JH
'JOE7PBOEUIFQPXFSEJTTJQBUFEJOBMMUIFSFTJTUPST
JOUIFDJSDVJUPG'JH
12 Ω
+
Vx
60 Ω
–
30 Ω
0.05Vx
Figure 3.56
6Ω
vo
–
Figure 3.57
60 V +
–
–
+
12 Ω
24 V
'PS1SPC
Using nodal analysis, find vPJOUIFDJSDVJUPG'JH
+
6Ω
Vo
Figure 3.60
'PS1SPC
4Ω
2Ω
Figure 3.59
'PS1SPC
2A
150 Ω
'PS1SPC
10 Ω
+ –
Figure 3.58
'PS1SPC
60Ib
250 Ω
4Ω
20 Ω
20 Ω
60 V +
–
'PS1SPCBOE1SPC
6TJOHOPEBMBOBMZTJT EFUFSNJOF7PJOUIFDJSDVJUJO
'JH
20 Ω
+
–
5vo
40 V +
–
Figure 3.61
'PS1SPC
10 Ω
Ix
20 Ω
4 Ix
10 Ω
+
Vo
–
114
Chapter 3
Methods of Analysis
$BMDVMBUFvBOEvJOUIFDJSDVJUPG'JHVTJOH
OPEBMBOBMZTJT
v1
Using nodal analysis, find current JPJOUIFDJSDVJUPG
'JH
600 V v
2
+–
10 Ω
io
30 Ω
50 Ω
15 A
4Ω
Figure 3.62
2Ω
10 Ω
8Ω
60 V +
–
'PS1SPC
Using nodal analysis, find vPJOUIFDJSDVJUPG'JH
3io
Figure 3.66
'PS1SPC
12.5 A
%FUFSNJOFUIFOPEFWPMUBHFTJOUIFDJSDVJUJO'JH
VTJOHOPEBMBOBMZTJT
8Ω
2Ω
1Ω
+
–
–
+ 50 V
4Ω
300 V
+–
100 V
+
vo
–
Figure 3.63
'PS1SPC
40 Ω
1
Apply nodal analysis to find JPBOEUIFQPXFSEJT
TJQBUFEJOFBDISFTJTUPSJOUIFDJSDVJUPG'JH
40 Ω
2
20 Ω
3
40 Ω
15 A
2A
Figure 3.67
10 V
+–
io
'PS1SPC
3S
6S
5S
4A
Use nodal analysis to find v v BOEvJOUIFDJSDVJU
PG'JH
Figure 3.64
'PS1SPC
3A
%FUFSNJOFWPMUBHFTvUISPVHIvJOUIFDJSDVJUPG
'JHVTJOHOPEBMBOBMZTJT
2Ω
v1
2S
2vo
v1
2A
Figure 3.65
'PS1SPC
+–
1S
v2
+
vo
–
8S
4S
v3
5A
+ 13 V
–
8Ω
4Ω
v2
4Ω
v3
8Ω
2Ω
+
–
Figure 3.68
'PS1SPC
12 V
115
Problems
For the circuit in Fig. 3.69, find v v BOEvVTJOH
OPEBMBOBMZTJT
6TFOPEBMBOBMZTJTBOE."5-"# to find 7PJOUIF
DJSDVJUPG'JH
240 V
+–
20i
v1
+–
8Ω
v3
20 Ω
v2
i
40 Ω
10 Ω
4A
40 Ω
+ Vo –
4Ω
2A
2Ω
2Ω
1Ω
Figure 3.69
1Ω
Figure 3.73
'PS1SPC
For the circuit in Fig. 3.70, find vBOEvVTJOHOPEBM
BOBMZTJT
Io
5Ω
6TFOPEBMBOBMZTJTBMPOHXJUI."5-"#UPEFUFSNJOF
UIFOPEFWPMUBHFTJO'JH
2Io
20 Ω
v1
'PS1SPC
v2
20 Ω
v4
5Ω
10 Ω
+–
1Ω
v1
v2
10 Ω
10 Ω
v3
30 V
Figure 3.70
8Ω
4A
'PS1SPC
30 Ω
20 Ω
%FUFSNJOFvBOEvJOUIFDJSDVJUPG'JH
Figure 3.74
10 Ω
10 Ω
v1
'PS1SPC
3A
v2
+ vo –
10 V
$BMDVMBUFUIFOPEFWPMUBHFTv v BOEvJOUIF
DJSDVJUPG'JH
10 Ω
+
–
20 Ω
–
+
5vo
3A
Figure 3.71
'PS1SPC
Use nodal analysis to find 7PJOUIFDJSDVJUPG'JH
60 Ω
30 V +
–
Figure 3.72
'PS1SPC
30 Ω
+
30 Ω
Vo
–
5Ω
v1
3Vo
io
v2
5Ω
20 Ω
+–
60 Ω
10 Ω
3A
15 V +
–
Figure 3.75
'PS1SPC
+
–
v3
5Ω
15 Ω
4io
–
+ 10 V
116
Chapter 3
Methods of Analysis
6TFOPEBMBOBMZTJTUPEFUFSNJOFWPMUBHFTv v BOE
vJOUIFDJSDVJUPG'JH
Using nodal analysis, find vPBOEJPJOUIFDJSDVJUPG
'JH
4S
3io
v1
10 Ω io
v2
1S
96 V
40 Ω
20 Ω
80 V +
–
v3
1S
–+
+
–
4vo
2io 80 Ω
+
vo
–
io
2A
2S
4S
2S
4A
Figure 3.79
'PS1SPC
Figure 3.76
'PS1SPC
6TF."5-"# to find the voltages at nodes B C D BOEEJOUIFDJSDVJUPG'JH
'JOEUIFOPEFWPMUBHFTGPSUIFDJSDVJUJO'JH
c
1Ω
5Ω
10 Ω
20 Ω
d
8Ω
4Ω
+ vo –
4Ω
v1
b
8Ω
16 Ω
–
60 V +
4Io
–+
4Ω
1A
2vo
v2
v3 2 Ω
Io
1Ω
+ 10 V
–
4Ω
+ 90 V
–
a
Figure 3.80
Figure 3.77
'PS1SPC
'PS1SPC
6TF."5-"#UPTPMWFGPSUIFOPEFWPMUBHFTJOUIF
DJSDVJUPG'JH
V4
2A 3S
1S
1S
V1
5A
2S
1S
4S
V2
0CUBJOUIFOPEFWPMUBHFTv v BOEvJOUIFDJSDVJU
PG'JH
2S
5 kΩ
V3
50 V
6A
v1
6A
Figure 3.78
'PS1SPC
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
Figure 3.81
'PS1SPC
–+
v2
75 V
+–
+ 120 V
–
v3
10 kΩ
117
Problems
Sections 3.4 and 3.5
8Ω
Mesh Analysis
4Ω
5Ω
8IJDIPGUIFDJSDVJUTJO'JHJTQMBOBS 'PSUIF
QMBOBSDJSDVJU SFESBXUIFDJSDVJUTXJUIOPDSPTTJOH
CSBODIFT
1Ω
6Ω 3Ω
7Ω
1Ω
3Ω
4Ω
2Ω
5Ω
4A
2Ω
(b)
Figure 3.83
6Ω
'PS1SPC
3FXPSL1SPCVTJOHNFTIBOBMZTJT
2A
6TFNFTIBOBMZTJTUPPCUBJOJB JC BOEJDJOUIFDJSDVJU
JO'JH
(a)
20 Ω
3Ω
ia
12 V
ic
10 Ω
5Ω
+
–
ib
2Ω
30 V +
–
15 Ω
45 V
+
–
4Ω
5Ω
Figure 3.84
1Ω
'PS1SPC
(b)
4PMWF1SPCVTJOHNFTIBOBMZTJT
Figure 3.82
'PS1SPC
"QQMZNFTIBOBMZTJTUPUIFDJSDVJUJO'JHBOE
PCUBJO*P
4Ω
3Ω
%FUFSNJOFXIJDIPGUIFDJSDVJUTJO'JHJTQMBOBS
BOESFESBXJUXJUIOPDSPTTJOHCSBODIFT
60 V +
–
10 A
2Ω
1Ω
+ 22.5 V
–
1Ω
5Ω
7Ω
10 V
2Ω
Io
2Ω
1Ω
1Ω
3Ω
+
–
6Ω
4Ω
(a)
5A
4Ω
Figure 3.85
'PS1SPC
6TJOH'JHGSPN1SPC EFTJHOBQSPCMFNUP
IFMQPUIFSTUVEFOUTCFUUFSVOEFSTUBOENFTIBOBMZTJT
118
Chapter 3
Methods of Analysis
For the bridge network in Fig. 3.86, find JPVTJOH
NFTIBOBMZTJT
io
6TFNFTIBOBMZTJTUPPCUBJOJPJOUIFDJSDVJUPG
'JH
2 kΩ
6 kΩ
6 kΩ
4A
2 kΩ
56 V +
–
400 V +
–
100 Ω
io
50 Ω
4 kΩ
4 kΩ
50 Ω
100 Ω
+–
Figure 3.86
50 V
'PS1SPC
Figure 3.90
Apply mesh analysis to find JJO'JH
'PS1SPC
10 Ω
i1
2Ω
6V
i
4Ω
1Ω
i2
'JOEDVSSFOUJJOUIFDJSDVJUPG'JH
+–
5Ω
i3
4Ω
+ 8V
–
8Ω
4A
Figure 3.87
2Ω
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQTUVEFOUT
CFUUFSVOEFSTUBOENFTIBOBMZTJTVTJOHNBUSJDFT
20 Ω
30 Ω
10 Ω
i
30 V +
–
6Ω
3Ω
1Ω
Figure 3.91
'PS1SPC
V1
+
–
i1
40 Ω
30 Ω
i2
i3
–
+ V3
+–
$BMDVMBUFUIFNFTIDVSSFOUTJBOEJJO'JH
V2
Figure 3.88
'PS1SPC
10 Ω
Use mesh analysis to find vBCBOEJPJOUIFDJSDVJUPG
'JH
80 V +
–
80 V +
–
Figure 3.89
'PS1SPC
20 Ω
i1
10 Ω i2
+
–
io
30 Ω
30 Ω
20 Ω
+ vo –
40 V +
–
20 Ω
10 Ω
+
vab
–
Figure 3.92
'PS1SPC
30 Ω
3FXPSL1SPCVTJOHNFTIBOBMZTJT
2vo
119
Problems
%FUFSNJOFUIFDVSSFOUUISPVHIUIFLΩSFTJTUPSJO
UIFDJSDVJUPG'JHVTJOHNFTIBOBMZTJT
Apply mesh analysis to find vPJOUIFDJSDVJUPG
'JH
vo
3 kΩ
400 V
4 kΩ
2 kΩ
+
–
10 Ω
10 Ω
10 Ω
5 kΩ
+–
100 V
1 kΩ
6V +
–
10 kΩ
+ 4V
–
10 Ω
–
+ 3V
10 A
Figure 3.96
Figure 3.93
'PS1SPC
'PS1SPC
Use mesh analysis to find J J BOEJJOUIFDJSDVJU
PG'JH
'JOEvPBOEJPJOUIFDJSDVJUPG'JH
+
vo
–
3Ω
1Ω
vo
12 V +
–
2Ω
2Ω
i1
4Ω
i3
+ 54 V
–
2io
8Ω
3A
io
2Ω
i2
+
–
2vo
Figure 3.97
'PS1SPC
Figure 3.94
'PS1SPC
'JOEUIFNFTIDVSSFOUTJOUIFDJSDVJUPG'JH
VTJOH."5-"#
2 kΩ
Use mesh analysis to find the current JPJOUIFDJSDVJU
PG'JH
I5
6 kΩ
io
4Ω
52 V +
–
Figure 3.95
'PS1SPC
I3
10 Ω
8 kΩ
1 kΩ
2Ω
8Ω
3io
8 kΩ
12 V +
–
Figure 3.98
'PS1SPC
I1
I4
4 kΩ
3 kΩ
I2
3 mA
120
Chapter 3
Methods of Analysis
'JOEUIFNFTIDVSSFOUTJ J BOEJJOUIFDJSDVJUJO
'JH
1 kΩ
1 kΩ
'JOEJ J BOEJJOUIFDJSDVJUPG'JH
30 Ω
1 kΩ
i2
10 Ω
1 kΩ
+
12 V –
i1
1Ω
i2
+
10 V –
10 Ω
–
+ 12 V
i3
i1
i3
+ 120 V
–
30 Ω
30 Ω
Figure 3.99
'PS1SPC
*OUIFDJSDVJUPG'JH TPMWFGPS* * BOE*
10 V
6Ω
I1
1A
I3
12 Ω
'PS1SPC
3FXPSL1SPCVTJOHNFTIBOBMZTJT
+–
4A
Figure 3.103
I2
$BMDVMBUFUIFQPXFSEJTTJQBUFEJOFBDISFTJTUPSJOUIF
DJSDVJUPG'JH
2Ω
4Ω
0.5io
+–
8V
io
'PS1SPC
20 Ω
+ 100 V
–
10 Ω
10 Ω
'PS1SPC
+
v1
–
10 Ω
$BMDVMBUFUIFDVSSFOUHBJOJP∕JTJOUIFDJSDVJUPG
'JH
50 Ω
20 Ω
+
vo
–
is
Figure 3.101
'PS1SPC
In the circuit of Fig. 3.102, find the values of 3 7 BOE7HJWFOUIBUJP=N"
10 kΩ
Io
6 kΩ
60 mA
10 kΩ
6 kΩ
Figure 3.102
'PS1SPC
2Ω
Figure 3.104
10 Ω
v2
–
+ 280 V
–
1Ω
%FUFSNJOFvBOEvJOUIFDJSDVJUPG'JH
+
8Ω
4Ω
Figure 3.100
+
V1
–
R
60 Ω
io
–
+
75 Ω
90 Ω
9vo
Figure 3.105
'PS1SPC
'JOEUIFNFTIDVSSFOUTJ J BOEJJOUIFOFUXPSLPG
'JH
4 kΩ
+
V2
–
100 V
+
–
i1
Figure 3.106
'PS1SPC
8 kΩ
4 mA
i2
2 kΩ
2i1
i3
+ 40 V
–
121
Problems
'JOEvYBOEJYJOUIFDJSDVJUTIPXOJO'JH
io
+
vx
–
0.8io
5Ω
10 Ω
2Ω
5A
10 Ω
3A
0.25 vx
8SJUFBTFUPGNFTIFRVBUJPOTGPSUIFDJSDVJUJO
'JH6TF."5-"#UPEFUFSNJOFUIFNFTI
DVSSFOUT
8Ω
Figure 3.107
8Ω
'PS1SPC
+
30 V –
4Ω
i1
+
– 6Ω
12 V
10 Ω
+
– 24 V
2Ω
4Ω
i3
8Ω
i2
2Ω
+
– 40 V
6Ω
4Ω
i4
i5
8Ω
+
– 32 V
Figure 3.110
'PS1SPC
Section 3.6
'JOEWPBOEJPJOUIFDJSDVJUPG'JH
50 Ω
10 Ω
io
+ vo –
0CUBJOUIFOPEFWPMUBHFFRVBUJPOTGPSUIFDJSDVJUJO
'JHCZJOTQFDUJPO5IFOTPMWFGPS7P
+
–
10 Ω
Nodal and Mesh Analyses
by Inspection
5A
4io
250 V +
–
40 Ω
4Ω
2Ω
5A
0.2vo
+ Vo –
3Vo
Figure 3.108
10 Ω
5Ω
10 A
'PS1SPC
Figure 3.111
'PS1SPC
6TF."5-"#UPTPMWFGPSUIFNFTIDVSSFOUTJOUIF
DJSDVJUPG'JH
6V
3Ω
–+
1Ω
5Ω
12 V
+
–
i1
Figure 3.109
'PS1SPC
i4
2Ω
1Ω
6Ω
i2
I2
10 V
4Ω
1Ω
6TJOH'JH EFTJHOBQSPCMFN UPTPMWFGPS7P UPIFMQPUIFSTUVEFOUTCFUUFSVOEFSTUBOEOPEBM
BOBMZTJT5SZZPVSCFTUUPDPNFVQXJUIWBMVFTUP
NBLFUIFDBMDVMBUJPOTFBTJFS
–+
i5
R2
1Ω
8Ω
i3
6Ω
R3
+
I1
–
+ 9V
Figure 3.112
'PS1SPC
R1
Vo
–
R4
+ V
1
–
122
Chapter 3
Methods of Analysis
'PSUIFDJSDVJUTIPXOJO'JH XSJUFUIFOPEF
WPMUBHFFRVBUJPOTCZJOTQFDUJPO
#ZJOTQFDUJPO XSJUFUIFNFTIDVSSFOUFRVBUJPOTGPS
UIFDJSDVJUJO'JH
20 Ω
20 kΩ
40 V
50 mA
v2
5 kΩ
+–
100 mA
10 kΩ
5 kΩ
i3
10 Ω
30 Ω
+ 20 V
–
i4
20 Ω
Figure 3.116
40 mA
10 kΩ
i1
20 Ω
v3
+–
v1
i2
15 V
'PS1SPC
8SJUFUIFNFTIDVSSFOUFRVBUJPOTGPSUIFDJSDVJUJO
'JH
Figure 3.113
'PS1SPC
10 Ω
10 Ω
+–
10 V
15 V +
–
8SJUFUIFOPEFWPMUBHFFRVBUJPOTCZJOTQFDUJPOBOE
UIFOEFUFSNJOFWBMVFTPG7BOE7JOUIFDJSDVJUPG
'JH
i1
10 Ω
10 Ω
4ix
V1
1S
2S
5S
10 Ω
10 Ω
10 V
+–
10 Ω
i4
10 Ω
i3
15 V +
–
V2
ix
20 A
i2
10 Ω
10 Ω
+–
10 V
7A
Figure 3.117
'PS1SPC
#ZJOTQFDUJPO PCUBJOUIFNFTIDVSSFOUFRVBUJPOTGPS
UIFDJSDVJUJO'JH
Figure 3.114
'PS1SPC
R1
8SJUFUIFNFTIDVSSFOUFRVBUJPOTGPSUIFDJSDVJUJO
'JH/FYU EFUFSNJOFUIFWBMVFTPGJ J BOEJ
V1
+
–
R2
i1
30 V +
–
V2
R6
'PS1SPC
3Ω
1Ω
i1
R7
+
–
i4
+ V
4
–
R8
+–
V3
Figure 3.118
2Ω
4Ω
Figure 3.115
i3
R5
i2
R4
i3
5Ω
R3
i2
+ 15 V
–
'PS1SPC
Section 3.8
Circuit Analysis with PSpice or
MultiSim
6TF14QJDFPS.VMUJ4JNUPTPMWF1SPC
6TF14QJDFPS.VMUJ4JNUPTPMWF1SPC
123
Problems
4PMWFGPS7BOE7JOUIFDJSDVJUPG'JHVTJOH
14QJDFPS.VMUJ4JN
5IFGPMMPXJOHQSPHSBNJTUIF4DIFNBUJDT/FUMJTUPGB
QBSUJDVMBSDJSDVJU%SBXUIFDJSDVJUBOEEFUFSNJOFUIF
WPMUBHFBUOPEF
2ix
5Ω
V1
V2
2Ω
5A
1Ω
2A
Section 3.9
$BMDVMBUFvPBOE*PJOUIFDJSDVJUPG'JH
ix
Io
Figure 3.119
'PS1SPC
4PMWF1SPCVTJOH14QJDFPS.VMUJ4JN
3FXPSL1SPCVTJOH14QJDFPS.VMUJ4JN
'JOEUIFOPEBMWPMUBHFTvUISPVHIvJOUIFDJSDVJUPG
'JHVTJOH14QJDFPS.VMUJ4JN
+–
10 Ω
12 Ω
vo
100
15 mV +
–
+
+
–
50Io
vo
20 kΩ
–
Figure 3.121
'PS1SPC
v3
For the simplified transistor circuit of Fig. 3.122,
DBMDVMBUFUIFWPMUBHFvP
4Ω
8A
2Ω
v4
1Ω
v2
4 kΩ
An audio amplifier with a resistance of 9 ΩTVQQMJFT
QPXFSUPBTQFBLFS8IBUTIPVMECFUIFSFTJTUBODFPG
UIFTQFBLFSGPSNBYJNVNQPXFSUPCFEFMJWFSFE
6Io
v1
Applications
Io
1 kΩ
+ 20 V
–
I
400I
47 mV +
–
5 kΩ
2 kΩ
Figure 3.120
'PS1SPC
+
vo
–
Figure 3.122
'PS1SPC
6TF14QJDFPS.VMUJ4JNUPTPMWFUIFQSPCMFNJO
&YBNQMF
*GUIF4DIFNBUJDT/FUMJTUGPSBOFUXPSLJTBTGPMMPXT ESBXUIFOFUXPSL
For the circuit in Fig. 3.123, find the gain vP∕vT
500 Ω
3.5 kΩ
vs +
–
Figure 3.123
'PS1SPC
+
v1
–
500 Ω
–
+
80 v1 500 Ω
+
vo
–
124
Chapter 3
Methods of Analysis
%FUFSNJOFUIFHBJOWP∕WT of the transistor amplifier
DJSDVJUJO'JH
2 kΩ
vs +
–
Io
4 kΩ
4 kΩ
For the transistor circuit of Fig. 3.127, find *# 7$& BOEvP5BLFβ= 7#&=7
–vo
1000
5 kΩ
+
–
+
vo
–
50Io
10 kΩ
+
IB
6 kΩ
VCE
+
9V
–
–
Figure 3.124
3V
'PS1SPC
2 kΩ
400 Ω
For the transistor circuit shown in Fig. 3.125, find *#
BOE7$&-FUβ= BOE7#&=7
+
vo
–
Figure 3.127
0.7 V 100 kΩ
– +
'PS1SPC
+ 15 V –
2.25 V
+
–
1 kΩ
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUSBOTJTUPST.BLFTVSFZPV
VTFSFBTPOBCMFOVNCFST
Figure 3.125
VCC
'PS1SPC
R3
$BMDVMBUFvTGPSUIFUSBOTJTUPSJO'JHHJWFOUIBU
vP=7 β= 7#&=7
R2
vs
1 kΩ
10 kΩ
vs
500 Ω
+
vo
–
R4
+
18 V
–
Figure 3.126
'PS1SPC
Comprehensive Problem
3FXPSL&YBNQMFXJUIIBOEDBMDVMBUJPO
R1
Figure 3.128
'PS1SPC
c h a p t e r
Circuit Theorems
4
:PVSTVDDFTTBTBOFOHJOFFSXJMMCFEJSFDUMZQSPQPSUJPOBMUPZPVSBCJMJUZ
UPDPNNVOJDBUF
‡$IBSMFT,"MFYBOEFS
Enhancing Your Skills and Your Career
Enhancing Your Communication Skills
5BLJOHBDPVSTFJODJSDVJUBOBMZTJTJTPOFTUFQJOQSFQBSJOHZPVSTFMGGPS
BDBSFFSJOFMFDUSJDBMFOHJOFFSJOH&OIBODJOHZPVSDPNNVOJDBUJPOTLJMMT
XIJMFJOTDIPPMTIPVMEBMTPCFQBSUPGUIBUQSFQBSBUJPO BTBMBSHFQBSUPG
ZPVSUJNFXJMMCFTQFOUDPNNVOJDBUJOH
1FPQMFJOJOEVTUSZIBWFDPNQMBJOFEBHBJOBOEBHBJOUIBUHSBEVBUJOH
FOHJOFFSTBSFJMMQSFQBSFEJOXSJUUFOBOEPSBMDPNNVOJDBUJPO "OFOHJ
OFFSXIPDPNNVOJDBUFTFGGFDUJWFMZCFDPNFTBWBMVBCMFBTTFU
:PVDBOQSPCBCMZTQFBLPSXSJUFFBTJMZBOERVJDLMZ#VUIPXFGGFD
UJWFMZEPZPVDPNNVOJDBUF 5IFBSUPGFGGFDUJWFDPNNVOJDBUJPOJTPGUIF
VUNPTUJNQPSUBODFUPZPVSTVDDFTTBTBOFOHJOFFS
'PSFOHJOFFSTJOJOEVTUSZ DPNNVOJDBUJPOJTL FZUPQSPNPUBCJMJUZ $POTJEFSUIFSFTVMUPGBTVSWFZPG64DPSQPSBUJPOTUIBUBTLFEXIBUGBD
tors influence managerial promotion. 5IFTVSW FZJODMVEFTBMJTUJOHPG
QFSTPOBMRVBMJUJFTBOEUIFJSJNQPSUBODFJOBEW BODFNFOU:PVNBZCF
TVSQSJTFEUPOPUFUIBUiUFDIOJDBMTLJMMCBTFEPOFYQFSJFODFuQMBDFEGPVSUI
GSPNUIFCPUUPN"UUSJCutes such as self-confidence, ambition, fleYJCJMJUZ NBUVSJUZ BCJMJUZUPNBL FTPVOEEFDJTJPOT HFUUJOHUIJOHTEPOFXJUIBOE
UISPVHIQFPQMF BOEDBQBDJUZGPSIBSEXPSLBMMSBOLFEIJHIFS"UUIFUPQ
PGUIFMJTUX BTiBCJMJUZUPDPNNVOJDBUF u5IFIJHIFSZPVSQSPGFTTJPOBM
DBSFFSQSPHSFTTFT UIFNPSFZPVXJMMOFFEUPDPNNVOJDBUF 5IFSFGPSF ZPVTIPVMESFHBSEFGGFDUJWFDPNNVOJDBUJPOBTBOJNQPSUBOUUPPMJOZPVS
FOHJOFFSJOHUPPMDIFTU
-FBSOJOHUPDPNNVOJDBUFFG GFDUJWFMZJTBMJGFMPOHUBTLZPVTIPVME
BMXBZTXPSLUPXBSE5IFCFTUUJNFUPCFHJOJTXIJMFTUJMMJOTDIPPM$PO
UJOVBMMZMPPLGPSPQQPSUVOJUJFTUPEF WFMPQBOETUSFOHUIFOZPVSSFBEJOH XSJUJOH MJTUFOJOH BOETQFBLJOHTLJMMT:PVDBOEPUIJTUISPVHIDMBTTSPPN
QSFTFOUBUJPOT UFBNQSPKFDUT BDUJ WFQBSUJDJQBUJPOJOTUVEFOUPS HBOJ[B
UJPOT BOEFOSPMMNFOUJODPNNVOJDBUJPODPVSTFT5IFSJTLTBSFMFTTOPX
UIBOMBUFSJOUIFXPSLQMBDF
"CJMJUZUPDPNNVOJDBUFFGGFDUJWFMZJTSF
HBSEFECZNBOZBTUIFNPTUJNQPSUBOU
TUFQUPBOFYFDVUJWFQSPNPUJPO
¥*54UPDL1VODI4UPDL3'
125
126
Chapter 4
Circuit Theorems
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
%FWFMPQBOEFOIBODFZPVSTLJMMTJOVTJOHOPEBMBOBMZTJTBOE
NFTIBOBMZTJTUPBOBMZ[FCBTJDDJSDVJUT
6OEFSTUBOEIPXMJOFBSJUZXPSLTXJUICBTJDDJSDVJUT
&YQMBJOUIFQSJODJQMFPGTVQFSQPTJUJPOBOEIPXJUDBOCFVTFE
UPIFMQBOBMZ[FDJSDVJUT
6OEFSTUBOEUIFWBMVFPGTPVSDFUSBOTGPSNBUJPOBOEIPXJUDBO
CFVTFEUPTJNQMJGZDJSDVJUT
3FDPHOJ[F5IFWFOJOTBOE/PSUPOTUIFPSFNTBOELOPXIPX
UIFy can lead to greatly simplified circuits.
&YQMBJOUIFNBYJNVNQPXFSUSBOTGFSDPODFQU
4.1
Introduction
"NBKPSBEW BOUBHFPGBOBMZ[JOHDJSDVJUTVTJOH,JSDIIPG GTMBXTBTXF
EJEJO$IBQUFSJTUIBUXFDBOBOBMZ[FBDJSDVJUXJUIPVUUBNQFSJOHXJUI
JUTPSJHJOBMconfiguration. "NBKPSEJTBEWBOUBHFPGUIJTBQQSPBDIJTUIBU GPSBMBSHF DPNQMFYDJSDVJU UFEJPVTDPNQVUBUJPOJTJOWPMWFE
5IFHSPXUIJOBSFBTPGBQQMJDBUJPOPGFMFDUSJDDJSDVJUTIBTMFEUPBO
FWPMVUJPOGSPNTJNQMFUPDPNQMFYDJSDVJUT5PIBOEMFUIFDPNQMFYJUZ FO
HJOFFSTPWFSUIFZFBSTIBWFEFWFMPQFETPNFUIFPSFNTUPTJNQMJGZDJSDVJU
BOBMZTJT4VDIUIFPSFNTJODMVEF5IFWFOJOTBOE/PSUPOTUIFPSFNT4JODF
UIFTFUIFPSFNTBSFBQQMJDBCMFUPMJOFBS circuits, we first discuss the DPODFQU
PGDJSDVJUMJOFBSJUZ*OBEEJUJPOUPDJSDVJUUIFPSFNT XFEJTDVTTUIFDPODFQUT
PGTVQFSQPTJUJPO TPVSDFUSBOTGPSNBUJPO BOENBYJNVNQPXFSUSBOTGFSJO
UIJTDIBQUFS5IFDPODFQUTXFEF WFMPQBSFBQQMJFEJOUIFMBTUTFDUJPOUP
TPVSDFNPEFMJOHBOESFTJTUBODFNFBTVSFNFOU
4.2
Linearity Property
-JOFBSJUZJTUIFQSPQFSUZPG BOFMFNFOUEFTDSJCJOHBMJOFBSSFMBUJPOTIJQ
CFUXFFODBVTFBOEFGGFDU"MUIPVHIUIFQSPQFSUZBQQMJFTUPNBOZDJSDVJU
FMFNFOUT XFTIBMMMJNJUJUTBQQMJDBCJMJUZUPSFTJTUPSTJOUIJTDIBQUFS5IF
QSPQFSUZJTBDPNCJOBUJPOPGCPUIUIFIPNPHFOFJUZ TDBMJOH QSPQFSUZ
BOEUIFBEEJUJWJUZQSPQFSUZ
5IFIPNPHFOFJUZQSPQFSUZSFRVJSFTUIBUJGUIFJOQVU BMTPDBMMFEUIF
FYDJUBUJPO JTNVMUJQMJFECZBDPOTUBOU UIFOUIFPVUQVU BMTPDBMMFEUIF
SFTQPOTF JTNVMUJQMJFECZUIFTBNFDPOTUBOU'PSBSFTJTUPS GPSFYBNQMF 0INTMBXSFMBUFTUIFJOQVUJUPUIFPVUQVUv
v = J3
*GUIFDVSSFOUJTJODSFBTFECZBDPOTUBOUL UIFOUIFWPMUBHFJODSFBTFTDPS
SFTQPOEJOHMZCZLUIBUJT
LJ3=Lv
4.2
127
Linearity Property
5IFBEEJUJWJUZQSPQFSUZSFRVJSFTUIBUUIFSFTQPOTFUPBTVNPGJOQVUT
JTUIFTVNPGUIFSFTQPOTFTUPFBDIJOQVUBQQMJFETFQBSBUFMZ 6TJOHUIF
WPMUBHFDVSSFOUSFMBUJPOTIJQPGBSFTJTUPS JG
v=J3
B
BOE
v=J3
C
UIFOBQQMZJOH J+J HJWFT
v= J+J 3=J3+J3=W+W
8FTBZ UIBUB SFTJTUPSJT BMJOFBS FMFNFOUCFDBVTF UIF WPMUBHFDVSSFOU
relationship satisfies both the homogeneity and the additiWJUZQSPQFSUJFT
*OHFOFSBM BDJSDVJUJTMJOFBSJGJUJTCPUIBEEJUJWFBOEIPNPHFOFPVT"
MJOFBSDJSDVJUDPOTJTUTPGPOMZMJOFBSFMFNFOUT MJOFBSEFQFOEFOUTPVSDFT BOEJOEFQFOEFOUTPVSDFT
A linear circuit is one whose output is linearly related (or directly proportional) to its input.
For example, when current i1 flows
through resistor R, the power is p1 = Ri12,
and when current i2 flows through R, the
power is p2 = Ri22. If current i1 + i2 flows
through R, the power absorbed is p3 =
R (i1 + i2)2 = Ri12 + Ri22 + 2Ri1i2 ≠ p1 + p2.
Thus, the power relation is nonlinear.
5ISPVHIPVUUIJTCPPLXFDPOTJEFSPOMZMJOFBSDJSDVJUT/PUFUIBUTJODF
Q=J3=v∕3 NBLJOHJUBRVBESBUJDGVODUJPOSBUIFSUIBOBMJOFBSPOF UIF
SFMBUJPOTIJQCFUXFFOQPXFSBOEWPMUBHF PSDVSSFOU JTOPOMJOFBS 5IFSF
GPSF UIFUIFPSFNTDPWFSFEJOUIJTDIBQUFSBSFOPUBQQMJDBCMFUPQPXFS
5PJMMVTUSBUFUIFMJOFBSJUZQSJODJQMF DPOTJEFSUIFMJOFBSDJSDVJUTIPXO
JO'JH 5IFMJOFBSDJSDVJUIBTOPJOEFQFOEFOUTPVSDFTJOTJEFJU*UJT
FYDJUFECZBW PMUBHFTPVSDF vT XIJDITFSW FTBTUIFJOQVU 5IFDJSDVJU
vs +
–
JTUFSNJOBUFECZBMPBE 38FNBZUBL FUIFDVSSFOU JUISPVHI 3BTUIF
PVUQVU4VQQPTFvT=7HJWFTJ=""DDPSEJOHUPUIFMJOFBSJUZQSJO
DJQMF vT=7XJMMHJWFJ="#ZUIFTBNFUPLFO J=N"NVTUCF
EVFUPvT=N7
Figure 4.1
i
Linear circuit
R
"MJOFBSDJSDVJUXJUIJOQVUvTBOEPVUQVUJ
Example 4.1
'or the circuit in Fig. 4.2, find *PXIFOvT=7BOEvT=7
2Ω
Solution:
"QQMZJOH,7-UPUIFUXPMPPQT XFPCUBJO
+ vx –
J− J+ vT=
−J+ J− vY− vT=
6Ω
i1
"EEJOH&RT BOE ZJFMET
Figure 4.2
'PS&YBNQMF
⇒
J=−J
4VCTUJUVUJOHUIJTJO&R XFHFU
−J+ vT=
⇒
vT
@@@
J=
4Ω
i2
vs
−J+ J− vT=
Io
4Ω
#VUvY=J&RVBUJPO CFDPNFT
J+ J=
8Ω
+
–
–
+
3vx
128
Chapter 4
Circuit Theorems
8IFOvT=7
"
@@@
*P=J=
8IFOvT=7
"
*P=J=@@@
TIPXJOHUIBUXIFOUIFTPVSDFWBMVFJTEPVCMFE *PEPVCMFT
Practice Problem 4.1
12 Ω
is
4Ω
8Ω
'or the circuit in Fig. 4.3, find vPXIFOJT=BOEJT="
Answer:7 7
+
vo
–
Figure 4.3
'PS1SBDUJDF1SPC
Example 4.2
"TTVNF*P= A and use linearity to find the actual vBMVFPG *PJOUIF
DJSDVJUPG'JH
I4
6Ω
2 V
2
I2
2Ω
1
I3
I s = 15 A
7Ω
3Ω
1 V
Io
I1
4Ω
5Ω
Figure 4.4
'PS&YBNQMF
Solution:
*G*P=" UIFO 7= + *P=7BOE *=7∕=""QQMZJOH
,$-BUOPEFHJWFT
*=*+ *P="
7
7=7+ *=+ =7 *=@@@
="
"QQMZJOH,$-BUOPEFHJWFT
*=*+*="
5IFSFGPSF *T="5IJTTIPXTUIBUBTTVNJOH*P=HJWFT*T=" UIF
BDUVBMTPVSDFDVSSFOUPG"XJMMHJWF*P="BTUIFBDUVBMWBMVF
Practice Problem 4.2
12 Ω
40 V +
–
Figure 4.5
'PS1SBDUJDF1SPC
5Ω
8Ω
+
Vo
–
"TTVNFUIBU7P=7BOEVTFMJOFBSJUZUPDBMDVMBUFUIFBDUVBMWBMVFPG7P
JOUIFDJSDVJUPG'JH
Answer:7
4.3
4.3
129
Superposition
Superposition
*GBDJSDVJUIBTUX PPSNPSFJOEFQFOEFOUTPVSDFT POFX BZUPEFUFSNJOF
UIFWBMVFPGB specific vBSJBCMF WPMUBHFPSDVSSFOU JTUPVTFOPEBMPS
NFTIBOBMZTJTBTJO$IBQUFS"OPUIFSXBZJTUPEFUFSNJOFUIFDPOUSJCV
UJPOPGFBDIJOEFQFOEFOUTPVSDFUPUIFW BSJBCMFBOEUIFOBEEUIFNVQ
5IFMBUUFSBQQSPBDIJTLOPXOBTUIFTVQFSQPTJUJPOQSJODJQMF
5IFJEFBPGTVQFSQPTJUJPOSFTUTPOUIFMJOFBSJUZQSPQFSUZ
The superposition principle states that the voltage across (or current
through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent
source acting alone.
Superposition is not limited to circuit
analysis but is applicable in many
fields where cause and effect bear a
linear relationship to one another.
5IFQSJODJQMFPGTVQFSQPTJUJPOIFMQTVTUPBOBMZ[FBMJOFBSDJSDVJUXJUI
NPSFUIBOPOFJOEFQFOEFOUTPVSDFCZDBMDVMBUJOHUIFDPOUSJC VUJPOPG
FBDIJOEFQFOEFOUTPVSDFTFQBSBUFMZ)PXFWFS UPBQQMZUIFTVQFSQPTJUJPO
QSJODJQMF XFNVTULFFQUXPUIJOHTJONJOE
8FDPOTJEFSPOFJOEFQFOEFOUTPVSDFBUBUJNFXIJMFBMMPUIFSJOEF
QFOEFOUTPVSDFTBSF UVSOFEPGG5IJTJNQMJFTUIBUXFSFQMBDFF WFSZ
WPMUBHFTPVSDFCZ 7 PSBTIPSUDJSDVJU BOEF WFSZDVSSFOUTPVSDF
CZ" PSBOPQFODJSDVJU 5IJTXBZXFPCUBJOBTJNQMFSBOENPSF
NBOBHFBCMFDJSDVJU
%FQFOEFOUTPVSDFTBSFMFGUJOUBDUCFDBVTFUIF ZBSFDPOUSPMMFECZ
DJSDVJUWBSJBCMFT
8JUIUIFTFJONJOE XFBQQMZUIFTVQFSQPTJUJPOQSJODJQMFJOUISFFTUFQT
Steps to Apply Superposition Principle:
5VSOPGGBMMJOEFQFOEFOUTPVSDFTF YDFQUPOFTPVSDF'JOEUIF
PVUQVU WPMUBHFPSDVSSFOU EVFUPUIBUBDUJ WFTPVSDFVTJOHUIF
UFDIOJRVFTDPWFSFEJO$IBQUFSTBOE
3FQFBUTUFQGPSFBDIPGUIFPUIFSJOEFQFOEFOUTPVSDFT
'JOE UIF UPUBM DPOUSJCVUJPO CZBEEJOH BMHFCSBJDBMMZ BMM UIF DPOUSJCVUJPOTEVFUPUIFJOEFQFOEFOUTPVSDFT
"OBMZ[JOHBDJSDVJUVTJOHTVQFSQPTJUJPOIBTPOFNBKPSEJTBEWBOUBHF
*UNBZWFSZMJLFMZJOWPMWFNPSFXPSL*GUIFDJSDVJUIBTUISFFJOEFQFOEFOU
TPVSDFT XFNBZIB WFUPBOBMZ[FUISFFTJNQMFSDJSDVJUTFBDIQSP WJEJOH
UIFDPOUSJCVUJPOEVFUPUIFSFTQFDUJWFJOEJWJEVBMTPVSDF)PXFWFS TVQFS
QPTJUJPOEPFTIFMQSFEVDFBDPNQMF YDJSDVJUUPTJNQMFSDJSDVJUTUISPVHI
SFQMBDFNFOUPGWPMUBHFTPVSDFTCZTIPSU DJSDVJUTBOEPGDVSSFOUTPVSDFT
CZPQFODJSDVJUT
,FFQJONJOEUIBUTVQFSQPTJUJPOJTCBTFEPOMJOFBSJUZ 'PSUIJTSFB TPO JUJTOPUBQQMJDBCMFUPUIFFG GFDUPOQP XFSEVFUPFBDITPVSDF CF DBVTFUIFQP XFSBCTPSCFECZBSFTJTUPSEFQFOETPOUIFTRVBSFPGUIF
WPMUBHFPSDVSSFOU*GUIFQPXFSWBMVFJTOFFEFE UIFDVSSFOUUISPVHI PS
Woltage across) the element must be calculated first using superposition.
Other terms such as killed, made
inactive, deadened, or set equal to
zero are often used to convey the
same idea.
130
Chapter 4
Example 4.3
Use the superposition theorem to find vJOUIFDJSDVJUPG'JH
8Ω
6V +
–
Solution:
4JODFUIFSFBSFUXPTPVSDFT MFU
+
v
–
4Ω
v=v+v
XIFSFvBOEvBSFUIFDPOUSJCVUJPOTEVFUPUIF7W PMUBHFTPVSDFBOE
UIF"DVSSFOUTPVSDF SFTQFDUJ WFMZ 5PPCUBJO v XFTFUUIFDVSSFOU
TPVSDFUP[FSP BTTIP XOJO'JH B "QQMZJOH,7-UPUIFMPPQJO 'JH B HJWFT
3A
Figure 4.6
'PS&YBNQMF
8Ω
6V +
–
i1
4Ω
5IVT
+
v1
–
i2
i3
+
v2
–
4Ω
3A
v=v+v=+=7
Practice Problem 4.3
Using the superposition theorem, find vPJOUIFDJSDVJUPG'JH
5Ω
2Ω
v=J=7
5PHFUv XFTFUUIFWPMUBHFTPVSDFUP[FSP BTJO'JH C 6TJOHDVS
SFOUEJWJTJPO
="
J=@@@@@
+ )FODF
'PS&YBNQMF B DBMDVMBUJOHv C DBMDVMBUJOHv
vo
–
J="
And we find
Figure 4.7
+
⇒
v=J=7
(b)
3Ω
J− =
8FNBZBMTPVTFWPMUBHFEJWJTJPOUPHFUvCZXSJUJOH
=7
v=@@@@@
+ (a)
8Ω
Circuit Theorems
8A
Answer:7
+ 40 V
–
Figure 4.8
'PS1SBDUJDF1SPC
Example 4.4
'JOEJPJOUIFDJSDVJUPG'JHVTJOHTVQFSQPTJUJPO
2Ω
3Ω
5io
1Ω
4A
+ –
+–
XIFSF J′P BOE J″P BSF EVF UP UIF " DVSSFOU TPVSDF BOE 7 WPMUBHF
TPVSDFSFTQFDUJWFMZ5PPCUBJOJ′P XFUVSOPGGUIF7TPVSDFTPUIBUXF
IBWFUIFDJSDVJUJO'JH B 8FBQQMZNFTIBOBMZTJTJOPSEFSUPPCUBJO
J′P'PSMPPQ
20 V
io
4Ω
5Ω
Figure 4.9
'PS&YBNQMF
Solution:
5IFDJSDVJUJO'JHJO WPMWFTBEFQFOEFOUTPVSDF XIJDINVTUCFMFGU
JOUBDU8FMFU
JP=J′P+J″P
J="
−J+J− J− J′P=
'PSMPPQ
4.3
131
Superposition
2Ω
2Ω
i1
5io'
1Ω
4A
i1
io'
4Ω
i5
5Ω
i3
+–
20 V
(b)
(a)
Figure 4.10
'PS&YBNQMF"QQMZJOHTVQFSQPTJUJPOUP B PCUBJOJP′ C PCUBJOJ″P
'PSMPPQ
−J−J+ J+ J′P=
J=J− J′P=− J′P
#VUBUOPEF
4VCTUJUVUJOH&RT BOE JOUP&RT BOE HJ WFT
UXPTJNVMUBOFPVTFRVBUJPOT
J− J′P=
J+J′P=
XIJDIDBOCFTPMWFEUPHFU
"
@@@
J′P=
5PPCUBJO J″
P XFUVSOPG GUIF"DVSSFOUTPVSDFTPUIBUUIFDJSDVJU
CFDPNFTUIBUTIPXOJO'JH C 'PSMPPQ ,7-HJWFT
J−J− J″P=
−J+J− + J″P=
BOEGPSMPPQ
#VUJ = −J″
P4VCTUJUVUJOHUIJTJO&RT BOE HJWFT
J−J″P=
J+J″P=−
"
J″P=−@@@
XIJDIXFTPMWFUPHFU
+ –
io''
0
5io''
1Ω
+ –
i3
5Ω
i4
3Ω
i2
3Ω
/PXTVCTUJUVUJOH&RT BOE JOUP&R HJWFT
=−"
JP=−@@@
4Ω
132
Chapter 4
Practice Problem 4.4
Use superposition to find vYJOUIFDJSDVJUPG'JH
20 Ω vx
25 V +
–
5A
Circuit Theorems
Answer:vY=7
4Ω
0.1vx
Figure 4.11
'PS1SBDUJDF1SPC
Example 4.5
24 V
+–
'or the circuit in Fig. 4.12, use the superposition principle to find J
8Ω
Solution:
*OUIJTDBTF XFIBWFUISFFTPVSDFT-FU
4Ω
4Ω
J=J+J+J
i
12 V +
–
Figure 4.12
'PS&YBNQMF
3Ω
3A
XIFSFJ J BOE JBSFEVFUPUIF7 7 BOE"TPVSDFTSFTQFD UJWFMZ5PHFUJ DPOTJEFSUIFDJSDVJU JO'JH B $PNCJOJOH Ω PO
UIFSJHIUIBOETJEF JOTFSJFTXJUI ΩHJWFTΩ5IFΩJOQBSBMMFM
XJUIΩHJWFT×∕=Ω5IVT
="
J=@@@
5PHFU J DPOTJEFSUIFDJSDVJUJO'JH C "QQMZJOHNFTIBOBMZTJT
HJWFT
JB−JC+ =
⇒
JB−JC= −
JC−JB=
⇒
J JB= @@
C
4VCTUJUVUJOH&R JOUP&R HJWFT
J=JC=−
5PHFUJ DPOTJEFSUIFDJSDVJUJO'JH D 6TJOHOPEBMBOBMZTJTHJWFT
v @@@@@@
v −v
= @@@
+ ⇒
v− v @@
v
v
@@@@@@
⇒
=+ @@
=v− v
v v=@@@
4VCTUJUVUJOH&R JOUP&R MFBETUPv=BOE
v
J=@@="
5IVT
J=J+J+J=−+="
4.4
Source Transformation
133
8Ω
4Ω
3Ω
4Ω
i1
i1
12 V +
–
12 V +
–
3Ω
3Ω
(a)
24 V
8Ω
+–
4Ω
ia
ib
8Ω
4Ω
4Ω
4Ω
v1
i2
i3
3Ω
3Ω
(b)
v2
3A
(c)
Figure 4.13
'PS&YBNQMF
'JOE*JOUIFDJSDVJUPG'JHVTJOHUIFTVQFSQPTJUJPOQSJODJQMF
6Ω
8V +
–
2Ω
I
8Ω
2A
+ 6V
–
Figure 4.14
'PS1SBDUJDF1SPC
Answer:N"
4.4
Source Transformation
8FIBWFOPUJDFEUIBUTFSJFTQBSBMMFMDPNCJOBUJPOBOEXZFEFMUBUSBOT
GPSNBUJPOIFMQTJNQMJGZDJSDVJUT 4PVSDFUSBOTGPSNBUJPOJTBOPUIFSUPPM
GPSTJNQMJGZJOHDJSDVJUT#BTJDUPUIFTFUPPMTJTUIFDPODFQUPG FRVJWB
MFODF8FSFDBMMUIBUBOFRVJWBMFOUDJSDVJUJTPOFXIPTFvJDIBSBDUFSJTUJDT
BSFJEFOUJDBMXJUIUIFPSJHJOBMDJSDVJU
*O4FDUJPO XFTB XUIBUOPEFW PMUBHF PSNFTIDVSSFOU FRVB UJPOTDBOCFPCUBJOFECZNFSFJOTQFDUJPOPGBDJSDVJUXIFOUIFTPVSDFT
BSFBMMJOEFQFOEFOUDVSSFOU PSBMMJOEFQFOEFOUW PMUBHF TPVSDFT*UJT
UIFSFGPSFFYQFEJFOUJODJSDVJUBOBMZTJTUPCFBCMFUPTVCTUJUVUFBW PMUBHF
TPVSDFJOTFSJFTXJUIBSFTJTUPSGPSBDVSSFOUTPVSDFJOQBSBMMFMXJUIB
Practice Problem 4.5
134
Chapter 4
Circuit Theorems
SFTJTUPS PSWJDFWFSTB BTTIPXOJO'JH&JUIFSTVCTUJUVUJPOJTLOPXO
BTBTPVSDFUSBOTGPSNBUJPO
R
vs
a
a
+
–
is
R
b
b
Figure 4.15
5SBOTGPSNBUJPOPGJOEFQFOEFOUTPVSDFT
A source transformation is the process of replacing a voltage source vs
in series with a resistor R by a current source is in parallel with a resistor
R, or vice versa.
5IFUX PDJSDVJUTJO'JHBSFFRVJ WBMFOU‡QSPWJEFEUIF ZIB WFUIF
TBNFW PMUBHFDVSSFOUSFMBUJPOBUUFSNJOBMT BC*UJTFBTZUPTIP XUIBU
UIFZBSFJOEFFEFRVJ WBMFOU*GUIFTPVSDFT BSFUVSOFEPG G UIFFRVJ WBMFOU
SFTJTUBODFBUUFSNJOBMT BCJOCPUIDJSDVJUTJT 3"MTP XIFOUFSNJOBMT
BC are short-circuited, the short-circuit current floXJOHGSPN BUP CJT
JTD=vT∕3JOUIF DJSDVJUPOUIF MFGUIBOETJEFBOEJTD=JTGPSUIF DJSDVJU
POUIFSJHIUIBOETJEF5IVT vT∕3=JTJOPSEFSGPSUIFUXPDJSDVJUTUPCF
FRVJWBMFOU)FODF TPVSDFUSBOTGPSNBUJPOSFRVJSFTUIBU
v
vT=JT3
PS
JT=@@
T
3
4PVSDFUSBOTGPSNBUJPOBMTPBQQMJFTUPEFQFOEFOUTPVSDFT QSP WJEFE
XFDBSFGVMMZIBOEMFUIFEFQFOEFOUW BSJBCMF"TTIP XOJO'JH B
EFQFOEFOUWPMUBHFTPVSDFJOTFSJFTXJUIBSFTJTUPSDBOCFUSBOTGPSNFEUPB
EFQFOEFOUDVSSFOUTPVSDFJOQBSBMMFMXJUIUIFSFTJTUPSPSWJDFWFSTBXIFSF
XFNBLe sure that Eq. (4.5) is satisfied.
R
vs
a
+
–
a
is
b
R
b
Figure 4.16
5SBOTGPSNBUJPOPGEFQFOEFOUTPVSDFT
-JLFUIFXZFEFMUBUSBOTGPSNBUJPOXFTUVEJFEJO$IBQUFS BTPVSDF
USBOTGPSNBUJPOEPFTOPUBG GFDUUIFSFNBJOJOHQBSUPGUIFDJSDVJU 8IFO
BQQMJDBCMF TPVSDFUSBOTGPSNBUJPOJTBQP XFSGVMUPPMUIBUBMMP XTDJSDVJU
NBOJQVMBUJPOTUPFBTFDJSDVJUBOBMZTJT)PXFWFS XFTIPVMELFFQUIFGPM
MPXJOHQPJOUTJONJOEXIFOEFBMJOHXJUITPVSDFUSBOTGPSNBUJPO
/PUFGSPN'JH PS'JH UIBUUIFBSSP XPGUIFDVSSFOUTPVSDF
JTEJSFDUFEUPXBSEUIFQPTJUJWFUFSNJOBMPGUIFWPMUBHFTPVSDF
/PUFGSPN&R UIBUTPVSDFUSBOTGPSNBUJPOJTOPUQPTTJCMFXIFO
3= XIJDIJTUIFDBTFXJUIBOJEFBMWPMUBHFTPVSDF)PXFWFS GPSB
QSBDUJDBM OPOJEFBMWPMUBHFTPVSDF 3≠4JNJMBSMZ BOJEFBMDVSSFOU
TPVSDFXJUI 3=∞DBOOPUCFSFQMBDFECZBfinite WPMUBHFTPVSDF
.PSFXJMMCFTBJEPOJEFBMBOEOPOJEFBMTPVSDFTJO4FDUJPO
4.4
135
Source Transformation
Use source transformation to find vPJOUIFDJSDVJUPG'JH
Example 4.6
Solution:
8e first transform the current and vPMUBHFTPVSDFTUPPCUBJOUIFDJS DVJUJO'JH B $PNCJOJOHUIFΩBOEΩSFTJTUPSTJOTFSJFTBOE
USBOTGPSNJOHUIF7W PMUBHFTPVSDFHJ WFTVT'JH C 8FOPX
DPNCJOFUIFΩBOEΩSFTJTUPSTJOQBSBMMFMUPHFUΩ8FBMTPDPN
CJOFUIF"BOE"DVSSFOUTPVSDFTUPHFUB"TPVSDF 5IVT CZ
SFQFBUFEMZBQQMZJOHTPVSDFUSBOTGPSNBUJPOT XFPCUBJOUIFDJSDVJUJO
'JH D 2Ω
4Ω
3A
8Ω
3Ω
+
vo
–
+ 12 V
–
Figure 4.17
'PS&YBNQMF
2Ω
4Ω
–
12 V +
+
vo
–
8Ω
3Ω
4A
(a)
6Ω
2A
i
+
vo
–
8Ω
3Ω
4A
8Ω
(b)
+
vo
–
2Ω
2A
(c)
Figure 4.18
'PS&YBNQMF
BOE
8FVTFDVSSFOUEJWJTJPOJO'JH D UPHFU
="
J=@@@@@
+
vP=J= =7
"MUFSOBUJWFMZ TJODFUIF ΩBOEΩSFTJTUPSTJO'JH D BSFJO
QBSBMMFM UIFZIBWFUIFTBNFWPMUBHFvPBDSPTTUIFN)FODF
×
@@@@@
vP= ∥ " =
=7
'JOEJPJOUIFDJSDVJUPG'JHVTJOHTPVSDFUSBOTGPSNBUJPO
5V
1Ω
–+
6Ω
5A
Figure 4.19
3Ω
'PS1SBDUJDF1SPC
Answer:"
io
7Ω
3A
4Ω
Practice Problem 4.6
136
Chapter 4
Example 4.7
'JOEvYJO'JHVTJOHTPVSDFUSBOTGPSNBUJPO
4Ω
0.25vx
2Ω
6V
+
–
+
vx
–
2Ω
Circuit Theorems
+ 18 V
–
Figure 4.20
'PS&YBNQMF
Solution:
5IFDJSDVJUJO'JHJO WPMWFTBW PMUBHFDPOUSPMMFEEFQFOEFOUDVS SFOUTPVSDF 8FUSBOTGPSNUIJTEFQFOEFOUDVSSFOUTPVSDFBTXFMMBTUIF
7JOEFQFOEFOUW PMUBHFTPVSDFBTTIP XOJO'JH B 5IF7
WPMUBHFTPVSDFJTOPUUSBOTGPSNFECFDBVTFJUJTOPUDPOOFDUFEJOTFSJFT
XJUIBOZSFTJTUPS5IFUX P ΩSFTJTUPSTJOQBSBMMFMDPNCJOFUPHJ WFB
ΩSFTJTUPS XIJDIJTJOQBSBMMFMXJUIUIF"DVSSFOUTPVSDF 5IFDVS
SFOUTPVSDFJTUSBOTGPSNFEUPBW PMUBHFTPVSDFBTTIPXOJO'JH C /PUJDFUIBUUIFUFSNJOBMTGPSvYBSFJOUBDU"QQMZJOH,7-BSPVOEUIFMPPQ
in Fig. 4.21(b) giWFT
−+J+vY+=
vx
4Ω
3A
2Ω
2Ω
1Ω
+–
+
vx
–
vx
4Ω
+–
+
+ 18 V
–
3V +
–
vx
i
+ 18 V
–
–
(a)
(b)
Figure 4.21
'PS&YBNQMF"QQMZJOHTPVSDFUSBOTGPSNBUJPOUPUIFDJSDVJUJO'JH
"QQMZJOH,7-UPUIFMPPQDPOUBJOJOHPOMZUIF7W PMUBHFTPVSDF UIF
ΩSFTJTUPS BOEvYZJFMET
−+J+vY=
⇒
vY=−J
4VCTUJUVUJOHUIJTJOUP&R XFPCUBJO
+J+−J=
⇒
J=−"
"MUFSOBUJWFMZ XFNBZBQQMZ,7-UPUIFMPPQDPOUBJOJOH vY UIFΩSF
TJTUPS UIFW PMUBHFDPOUSPMMFEEFQFOEFOUW PMUBHFTPVSDF BOEUIF7
WPMUBHFTPVSDFJO'JH C 8FPCUBJO
−vY+J+vY+=
⇒
J=−"
5IVT vY=−J=7
Practice Problem 4.7
Use source transformation to find JYJOUIFDJSDVJUTIPXOJO'JH
5Ω
ix
24 mA
Figure 4.22
10 Ω
'PS1SBDUJDF1SPC
Answer:N"
–
+
2ix
4.5
4.5
I
Thevenin’s Theorem
*UPGUFOPDDVSTJOQSBDUJDFUIBUBQBSUJDVMBSFMFNFOUJOBDJSDVJUJTWBSJBCMF
VTVBMMZDBMMFEUIF MPBE) while other elements are fixFE"TBUZQJDBM
FYBNQMF BIPVTFIPMEPVUMFUUFSNJOBMNBZCFDPOOFDUFEUPEJG GFSFOUBQ
QMJBODFTDPOTUJUVUJOHBW BSJBCMFMPBE&BDIUJNFUIFW BSJBCMFFMFNFOUJT
DIBOHFE UIFFOUJSFDJSDVJUIBTUPCFBOBMZ[FEBMMPWFSBHBJO5PBWPJEUIJT
QSPCMFN 5IFWFOJOTUIFPSFNQSP vides a technique by which the fixFE
QBSUPGUIFDJSDVJUJTSFQMBDFECZBOFRVJWBMFOUDJSDVJU
"DDPSEJOHUP 5IFWFOJOTUIFPSFN UIFMJOFBSDJSDVJUJO'JH B DBOCFSFQMBDFECZUIBUJO'JH C 5IFMPBEJO'JHNBZCFB
TJOHMFSFTJTUPSPSBOPUIFSDJSDVJU 5IFDJSDVJUUPUIFMFGUPGUIFUFSNJOBMT
BCJO'JH C JTLOP XOBTUIF 5IFWFOJOFRVJWBMFOUDJS DVJUJUX BT
EFWFMPQFE JO CZ . -FPO5IFWFOJO m B 'SFODI UFMF
HSBQIFOHJOFFS
Thevenin’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a voltage source VTh in series
with a resistor RTh, where VTh is the open-circuit voltage at the terminals
and RTh is the input or equivalent resistance at the terminals when the
independent sources are turned off.
5IFQSPPGPGUIFUIFPSFNXJMMCFHJ WFOMBUFS JO4FDUJPO0VS
NBKPSDPODFSOSJHIUOPXJTIPw to find the 5IFWFOJOFRVJWBMFOUWPMUBHF
75IBOESFTJTUBODF 35I5PEPTP TVQQPTFUIFUX PDJSDVJUTJO 'JH
BSFFRVJWBMFOU 5XPDJSDVJUTBSFTBJEUPCF FRVJWBMFOUJGUIF ZIBWFUIF
TBNFWoltage-current relation at their terminals. Let us find out what will
NBLFUIFUX PDJSDVJUTJO'JHFRVJ WBMFOU*GUIFUFSNJOBMT BCBSF
NBEFPQFODJSDVJUFE CZSFNP ving the load), no current floXT TPUIBU
UIFPQFODJSDVJUWPMUBHFBDSPTTUIFUFSNJOBMTBCJO'JH B NVTUCF
FRVBMUPUIFWPMUBHFTPVSDF75IJO'JH C TJODFUIFUXPDJSDVJUTBSF
FRVJWBMFOU5IVT 75IJTUIFPQFODJSDVJUW PMUBHFBDSPTTUIFUFSNJOBMTBT
TIPXOJO'JH B UIBUJT
75I=vPD
Linear
two-terminal
circuit
a
+
voc
–
b
Linear circuit with
all independent
sources set equal
to zero
V Th = voc
RTh = R in
(a)
(b)
a
R in
b
Figure 4.24
'JOEJOH75IBOE35I
"HBJO XJUIUIFMPBEEJTDPOOFDUFEBOEUFSNJOBMT
BCPQFO
DJSDVJUFE XF UVSOPG GBMM JOEFQFOEFOUTPVSDFT 5IFJOQVUSFTJTUBODF
PSFRVJ WBMFOU SFTJTUBODF PGUIF EFBEDJSDVJU BUUIF UFSNJOBMT BC JO
'JH B NVTUCFFRVBMUP35IJO'JH C CFDBVTFUIFUXPDJSDVJUT
BSF FRVJWBMFOU5IVT 35IJTUIFJOQVUSFTJTUBODFBUUIFUFSNJOBMTXIFO
UIFJOEFQFOEFOUTPVSDFTBSFUVSOFEPGG BTTIPXOJO'JH C UIBUJT
35I=3JO
137
Thevenin’s Theorem
a
+
V
–
Linear
two-terminal
circuit
Load
b
(a)
R Th
I
a
+
V
–
VTh +
–
Load
b
(b)
Figure 4.23
3FQMBDJOHBMJOFBSUXPUFSNJOBMDJSDVJU
CZJUT5IFWFOJOFRVJWBMFOU B PSJHJOBM
DJSDVJU C UIF5IFWFOJOFRVJWBMFOU
DJSDVJU
138
Chapter 4
Circuit with
all independent
sources set equal
to zero
RTh =
vo
io
5o apply this idea in finding the 5IFWFOJOSFTJTUBODF35I XFOFFEUP
DPOTJEFSUXPDBTFT
io
a
+ v
– o
■ CASE 1 *GUIFOFUXPSLIBTOPEFQFOEFOUTPVSDFT XFUVSOPG GBMM
JOEFQFOEFOUTPVSDFT 35IJTUIFJOQVUSFTJTUBODFPGUIFOFUX PSLMPPLJOH
CFUXFFOUFSNJOBMTaBOEb BTTIPXOJO'JH C b
■ CASE 2 *GUIFOFUXPSLIBTEFQFOEFOUTPVSDFT XFUVSOPG GBMMJO
EFQFOEFOUTPVSDFT"TXJUITVQFSQPTJUJPO EFQFOEFOUTPVSDFTBSFOPUUP
CFUVSOFEPGGCFDBVTFUIFZBSFDPOUSPMMFECZDJSDVJUWBSJBCMFT8FBQQMZB
WPMUBHFTPVSDFvPBUUFSNJOBMTaBOEbBOEEFUFSNJOFUIFSFTVMUJOHDVSSFOU
JP5IFO35I=vP∕JP BTTIPXOJO'JH B "MUFSOBUJWFMZ XFNBZJO
TFSUBDVSSFOUTPVSDFJPBUUFSNJOBMTBCBTTIPwn in Fig. 4.25(b) and find
UIFUFSNJOBMWPMUBHFvP"HBJO35I=vP∕JP&JUIFSPGUIFUXPBQQSPBDIFT
XJMMHJWFUIFTBNFSFTVMU*OFJUIFSBQQSPBDIXFNBZBTTVNFBO ZWBMVF
PGvPBOEJP'PSFYBNQMF XFNBZVTFvP=7PSJP=" PSFWFOVTF
unspecified vBMVFTPGvPPSJP
(a)
a
Circuit with
all independent
sources set equal
to zero
RTh =
vo
io
+
vo
–
io
b
(b)
Figure 4.25
'JOEJOH35IXIFODJSDVJUIBTEFQFOEFOU
TPVSDFT
Later we will see that an alternative way
of finding RTh is RTh = voc ∕isc.
a
IL
Linear
circuit
RL
b
(a)
R Th
a
IL
VTh
+
–
RL
(b)
b
Figure 4.26
"DJSDVJUXJUIBMPBE B PSJHJOBMDJSDVJU C 5IFWFOJOFRVJWBMFOU
Example 4.8
32 V +
–
4Ω
1Ω
12 Ω
2A
'PS&YBNQMF
*UPGUFOPDDVSTUIBU 35IUBL FTBOF HBUJWFW BMVF*OUIJTDBTF UIF OFHBUJWFSFTJTUBODF v=−J3 JNQMJFTUIBUUIFDJSDVJUJTTVQQMZJOHQPXFS
5IJTJTQPTTJCMFJOBDJSDVJUXJUIEFQFOEFOUTPVSDFT&YBNQMFXJMM
JMMVTUSBUFUIJT
5IFWFOJOTUIFPSFNJTW FSZJNQPSUBOUJODJSDVJUBOBMZTJT*UIFMQT
TJNQMJGZBDJSDVJU "MBSHFDJSDVJUNBZCFSFQMBDFECZBTJOHMFJOEFQFO EFOUWPMUBHFTPVSDFBOEBTJOHMFSFTJTUPS5IJTSFQMBDFNFOUUFDIOJRVFJTB
QPXFSGVMUPPMJODJSDVJUEFTJHO
"TNFOUJPOFEFBSMJFS BMJOFBSDJSDVJUXJUIBWBSJBCMFMPBEDBOCFSF
QMBDFECZUIF5IFWFOJOFRVJWBMFOU FYDMVTJWFPGUIFMPBE5IFFRVJWBMFOU
OFUXPSLCFIBWFTUIFTBNFX BZFYUFSOBMMZBTUIFPSJHJOBMDJSDVJU$PO TJEFSBMJOFBSDJSDVJUUFSNJOBUFECZBMPBE 3- BTTIPXOJO'JH B 5IFDVSSFOU*-UISPVHIUIFMPBEBOEUIFWPMUBHF7-BDSPTTUIFMPBEBSFFBT
JMZEFUFSNJOFEPODFUIF 5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUBUUIFMPBE T
UFSNJOBMTJTPCUBJOFE BTTIP XOJO'JH C 'SPN 'JH C XF
PCUBJO
75I
@@@@@@@@
*-=
B
35I+33-
7
7-=3-*-=@@@@@@@@
C
35I+3- 5I
/PUFGSPN'JH C UIBUUIF 5IFWFOJOFRVJWBMFOUJTBTJNQMFW PMUBHF
EJWJEFS ZJFMEJOH7-CZNFSFJOTQFDUJPO
'JOEUIF5IFWFOJOFRVJWBMFOUDJSDVJUPGUIFDJSDVJUTIPXOJO'JH UP
UIFMFGUPGUIFUFSNJOBMT BCThen find the current through 3-= BOEΩ
a
RL
b
Figure 4.27
Circuit Theorems
Solution:
8e find 35ICZUVSOJOHPG GUIF7W PMUBHFTPVSDF SFQMBDJOHJU
XJUIBTIPSUDJSDVJU BOEUIF"DVSSFOUTPVSDF SFQMBDJOHJUXJUIBO
4.5
139
Thevenin’s Theorem
PQFODJSDVJU 5IFDJSDVJUCFDPNFTXIBUJTTIP XOJO'JH B 5IVT
×
@@@@@@
35I=∥+=
+ =Ω
4Ω
1Ω
4Ω
a
+
R Th
12 Ω
1Ω
VTh
32 V +
–
i1
12 Ω
i2
2A
VTh
–
b
(a)
a
b
(b)
Figure 4.28
'or Example 4.8: (a) finding 35I, (b) finding 75I
5o find 75I DPOTJEFSUIFDJSDVJUJO'JH C "QQMZJOHNFTIBOBM
ZTJTUPUIFUXPMPPQT XFPCUBJO
−+J+ J−J = J=−"
4PMWJOHGPSJ XFHFUJ="5IVT
75I= J−J = + =7
"MUFSOBUJWFMZ JUJTF WFOFBTJFSUPVTFOPEBMBOBMZTJT 8FJHOPSFUIF Ω
resistor since no current floXTUISPVHIJU"UUIFUPQOPEF ,$-HJWFT
−75I
75I
@@@@@@@@
@@@
+=
PS
−75I+=75I
⇒
75I=7
BTPCUBJOFECFGPSF8e could also use source transformation to find 75I
5IF5IFWFOJOFRVJWBMFOUDJSDVJUJTTIP XOJO'JH 5IFDVSSFOU
UISPVHI3-JT
7
=@@@@@@
*-=@@@@@@@@
5I 35I+3- +3-
4Ω
a
IL
30 V +
–
8IFO3-=
RL
b
="
*-=@@@
8IFO3-=
="
@@@
*-=
8IFO3-=
="
@@@
*-=
Figure 4.29
5IF5IFWFOJOFRVJWBMFOUDJSDVJUGPS
&YBNQMF
140
Chapter 4
Practice Problem 4.8
90 Ω
90 Ω
a
2A
60 Ω
6TJOH5IFWFOJOs theorem, find the equiWBMFOUDJSDVJUUPUIFMFGUPGUIF
UFSNJOBMTJOUIFDJSDVJUPG'JHThen find *
I
180 V
+
–
Circuit Theorems
Answer:75I=7 35I=Ω *="
15 Ω
b
Figure 4.30
'PS1SBDUJDF1SPC
Example 4.9
'JOEUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO'JHBUUFSNJOBMTBC
2vx
Solution:
5IJTDJSDVJUDPOUBJOTBEFQFOEFOUTPVSDF VOMJL FUIFDJSDVJUJOUIFQSF WJPVTF YBNQMF 5o find 35I XFTFUUIFJOEFQFOEFOUTPVSDFFRVBMUP [FSPC VUMFB WFUIFEFQFOEFOUTPVSDFBMPOF#FDBVTFPGUIFQSFTFODF PGUIFEFQFOEFOUTPVSDF IP XFWFS XFF YDJUFUIFOFUX PSLXJUIBW PMU
BHF TPVSDF vP DPOOFDUFE UPUIFUFSNJOBMT BT JOEJDBUFE JO 'JH B 8FNBZTFU WP = 7UPFBTFDBMDVMBUJPO TJODFUIFDJSDVJUJTMJOFBS Our goal is to find the current JPUISPVHIUIFUFSNJOBMT BOEUIFOPCUBJO
35I=∕JP "MUFSOBUJWFMZ, we may insert a 1-A current source, find the
DPSSFTQPOEJOHWPMUBHFvP BOEPCUBJO35I=vP∕
–+
2Ω
2Ω
5A
+
vx
–
4Ω
a
6Ω
b
Figure 4.31
'PS&YBNQMF
2vx
2vx
–+
–+
i1
i3
2Ω
4Ω
+
vx
–
2Ω
io
i2
6Ω
+ v =1V
– o
i3
2Ω
2Ω
a
5A
i1
4Ω
+
vx
–
+
6Ω
i2
voc
–
b
(a)
a
b
(b)
Figure 4.32
'JOEJOH35IBOE75IGPS&YBNQMF
"QQMZJOHNFTIBOBMZTJTUPMPPQJOUIFDJSDVJUPG'JH B
SFTVMUTJO
−vY+ J−J =
PS
vY=J−J
#VU−J=vY=J−JIFODF
J=−J
'PSMPPQTBOE BQQMZJOH,7-QSPEVDFT
J+ J−J + J−J =
J−J +J+=
4.5
141
Thevenin’s Theorem
4PMWJOHUIFTFFRVBUJPOTHJWFT
"
J=−@@
#VUJP=−J=∕")FODF
7=Ω
@@@@
35I=
JP
5PHFU75I, we find vPDJOUIFDJSDVJUPG'JH C "QQMZJOHNFTI
BOBMZTJT XFHFU
J=
−vY+ J−J =
⇒
vY=J−J
J−J + J−J +J=
PS
6Ω
J−J−J=
20 V
#VU J−J =vY4PMWJOHUIFTFFRVBUJPOTMFBETUPJ=∕)FODF
75I=vPD=J=7
5IF5IFWFOJOFRVJWBMFOUJTBTTIPXOJO'JH
'JOEUIF5IFWFOJOFRVJWBMFOUDJSDVJUPGUIFDJSDVJUJO'JHUPUIFMFGU
PGUIFUFSNJOBMT
a
+
–
b
Figure 4.33
5IF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO
'JH
Practice Problem 4.9
5Ω
Answer:75I=7 35I=NΩ
6V +
–
Ix
3Ω
1.5Ix
a
4Ω
b
Figure 4.34
'PS1SBDUJDF1SPC
%FUFSNJOFUIF 5IFWFOJOFRVJ WBMFOUPGUIFDJSDVJUJO'JH B BU
UFSNJOBMTBC
Solution:
Define.The problem is clearly defined; we are to determine the
5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUTIPXOJO'JH B 1SFTFOU5IFDJSDVJUDPOUBJOTBΩSFTJTUPSJOQBSBMMFMXJUIB
ΩSFTJTUPS5IFTFBSF JOUVSO JOQBSBMMFMXJUIBEFQFOEFOUDVS
SFOUTPVSDF*UJTJNQPSUBOUUPOPUFUIBUUIFSFBSFOPJOEFQFOEFOU
TPVSDFT
"MUFSOBUJWFThe first thing to consider is that, since we haWFOP
JOEFQFOEFOUTPVSDFTJOUIJTDJSDVJU XFNVTUFYDJUFUIFDJSDVJUFYUFS
OBMMZ*OBEEJUJPO XIFOZPVIBWFOPJOEFQFOEFOUTPVSDFTUIFWBMVF
GPS75IXJMMCFFRVBMUP[FSPTPZPVXJMMPOMZIBWe to find 35I
Example 4.10
142
Chapter 4
5IFTJNQMFTUBQQSPBDIJTUPFYDJUFUIFDJSDVJUXJUIFJUIFSB7
WPMUBHFTPVSDFPSB"DVSSFOUTPVSDF#FDBVTFXFXJMMFOEVQ
XJUIBOFRVJWBMFOUSFTJTUBODF FJUIFSQPTJUJWFPSOFHBUJWF *QSFGFS
UPVTFUIFDVSSFOUTPVSDFBOEOPEBMBOBMZTJTXIJDIXJMMZJFMEB
WPMUBHFBUUIFPVUQVUUFSNJOBMTFRVBMUPUIFSFTJTUBODF XJUI"
floXJOHJO vPJTFRVBMUPUJNFTUIFFRVJWBMFOUSFTJTUBODF "TBOBMUFSOBUJWF UIFDJSDVJUDPVMEBMTPCFFYDJUFECZB7
Woltage source and mesh analysis could be used to find the equiWB
MFOUSFTJTUBODF
"UUFNQU8FTUBSUCZXSJUJOHUIFOPEBMFRVBUJPOBUaJO'JH C BTTVNJOHJP="
a
ix
4Ω
2ix
2Ω
b
(a)
vo
a
ix
4Ω
2ix
2Ω
io
b
8ix
–
+
a
9Ω
ix
2Ω
i1
i2
+ 10 V
–
(c)
a
b
(d)
Figure 4.35
9Ω
+ 10 V
–
i
'PS&YBNQMF
JY= − vP ∕=−vP∕
4VCTUJUVUJOH&R JOUP&R ZJFMET
−vP∕ + vP − ∕+ vP − ∕+ − =
= − + @+ @ vP − b
–4 Ω
JY+ vP− ∕+ vP− ∕+ − =
(JWFOUIBUXFIBWFUXPVOLOPXOTBOEPOMZPOFFRVBUJPO XFXJMM
OFFEBDPOTUSBJOUFRVBUJPO
(b)
4Ω
Circuit Theorems
PS
vP=−7
4JODFvP=× 35I UIFO35I=vP∕= −Ω
5IFOFHBUJWFWBMVFPGUIFSFTJTUBODFUFMMTVTUIBU BDDPSE
JOHUPUIFQBTTJWFTJHODPOWFOUJPO UIFDJSDVJUJO'JH B JT
TVQQMZJOHQPXFS0GDPVSTF UIFSFTJTUPSTJO'JH B DBOOPU
TVQQMZQPXFS UIFZBCTPSCQPXFS JUJTUIFEFQFOEFOUTPVSDFUIBU
TVQQMJFTUIFQPXFS5IJTJTBOFYBNQMFPGIPXBEFQFOEFOUTPVSDF
BOESFTJTUPSTDPVMECFVTFEUPTJNVMBUFOFHBUJWFSFTJTUBODF
&WBMVBUF'JSTUPGBMM XFOPUFUIBUUIFBOTXFSIBTBOFHBUJWFWBMVF
8FLOPXUIJTJTOPUQPTTJCMFJOBQBTTJWFDJSDVJU CVUJOUIJTDJSDVJU
XFEPIBWFBOBDUJWFEFWJDF UIFEFQFOEFOUDVSSFOUTPVSDF 5IVT UIFFRVJWBMFOUDJSDVJUJTFTTFOUJBMMZBOBDUJWFDJSDVJUUIBUDBOTVQQMZ
QPXFS
/PXXFNVTUFWBMVBUFUIFTPMVUJPO5IFCFTUXBZUPEPUIJT
JTUPQFSGPSNBDIFDL VTJOHBEJGGFSFOUBQQSPBDI BOETFFJGXF
PCUBJOUIFTBNFTPMVUJPO-FUVTUSZDPOOFDUJOHBΩSFTJTUPSJO
TFSJFTXJUIB7WPMUBHFTPVSDFBDSPTTUIFPVUQVUUFSNJOBMTPG
UIFPSJHJOBMDJSDVJUBOEUIFOUIF5IFWFOJOFRVJWBMFOU5PNBLFUIF
DJSDVJUFBTJFSUPTPMWF XFDBOUBLFBOEDIBOHFUIFQBSBMMFMDVSSFOU
TPVSDFBOEΩSFTJTUPSUPBTFSJFTWPMUBHFTPVSDFBOEΩ SFTJTUPS
CZVTJOHTPVSDFUSBOTGPSNBUJPO5IJT XJUIUIFOFXMPBE HJWFTVT
UIFDJSDVJUTIPXOJO'JH D 8FDBOOPXXSJUFUXPNFTIFRVBUJPOT
JY + J + J − J =
J − J + J + =
/PUF XFPOMZIBWFUXPFRVBUJPOTCVUIBWFUISFFVOLOPXOT TPXF
OFFEBDPOTUSBJOUFRVBUJPO8FDBOVTF
JY=J − J
4.6
143
Norton’s Theorem
5IJTMFBETUPBOFXFRVBUJPOGPSMPPQ4JNQMJGZJOHMFBETUP
+ − J + − + J=
PS
−J + J=
PS
J=J
−J + J=−
Substituting the first equation into the second giWFT
−J + J=−
PS
J=−∕=−"
6TJOHUIF5IFWFOJOFRVJWBMFOUJTRVJUFFBTZTJODFXFIBWFPOMZPOF
MPPQ BTTIPXOJO'JH E −J + J + =
PS
J=−∕=−"
4BUJTGBDUPSZ $MFBSMZXFIBWFGPVOEUIFWBMVFPGUIFFRVJWBMFOU
DJSDVJUBTSFRVJSFECZUIFQSPCMFNTUBUFNFOU$IFDLJOHEPFTWBMJ
EBUFUIBUTPMVUJPO XFDPNQBSFEUIFBOTXFSXFPCUBJOFECZVTJOH
UIFFRVJWBMFOUDJSDVJUXJUIPOFPCUBJOFECZVTJOHUIFMPBEXJUI
UIFPSJHJOBMDJSDVJU 8FDBOQSFTFOUBMMUIJTBTBTPMVUJPOUPUIF
QSPCMFN
Practice Problem 4.10
0CUBJOUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO'JH
Answer:75I=7 35I=−Ω
4.6
10 Ω
+
vx
–
Norton’s Theorem
5Ω
15 Ω
Figure 4.36
'PS1SBDUJDF1SPC
Norton’s theorem states that a linear two-terminal circuit can be replaced by an equivalent circuit consisting of a current source IN in parallel
with a resistor RN, where IN is the short-circuit current through the terminals and RN is the input or equivalent resistance at the terminals when the
independent sources are turned off.
5IVT UIFDJSDVJUJO'JH B DBOCFSFQMBDFECZUIFPOFJO'JH C 5IFQSPPGPG/PSUPO TUIFPSFNXJMMCFHJ WFOJOUIFOF YUTFDUJPO
'PSOPX XFBSFNBJOMZDPODFSOFEXJUIIP XUPHFU3/BOE*/8Ffind
3/JOUIFTBNFX ay we find 35I*OG BDU GSPNXIBUXFLOP XBCPVU TPVSDFUSBOTGPSNBUJPO UIF5IFWFOJOBOE/PSUPOSFTJTUBODFTBSFFRVBM
UIBUJT
3/=35I
a
b
*O BCPVUZFBSTBGUFS5IFWFOJOQVCMJTIFEIJTUIFPSFN &-/PSUPO BO"NFSJDBOFOHJOFFSBU#FMM 5FMFQIPOF-BCPSBUPSJFT QSPQPTFEBTJNJMBS
UIFPSFN
4vx
+–
Linear
two-terminal
circuit
b
(a)
a
IN
RN
b
5o find the Norton current */ XFEFUFSNJOFUIFTIPSUDJSDVJUDVSSFOU
floXJOHGSPNUFSNJOBM aUP bJOCPUIDJSDVJUTJO'JH*UJTF WJEFOU
a
(b)
Figure 4.37
B 0SJHJOBMDJSDVJU C /PSUPOFRVJWBMFOU
DJSDVJU
144
Chapter 4
UIBUUIFTIPSUDJSDVJUDVSSFOUJO'JH C JT */5IJTNVTUCFUIFTBNF
TIPSUDJSDVJUDVSSFOUGSPNUFSNJOBM aUP bJO'JH B TJODFUIFUX P
DJSDVJUTBSFFRVJWBMFOU5IVT
a
Linear
two-terminal
circuit
Circuit Theorems
isc = IN
b
Figure 4.38
*/=JTD
TIPXOJO'JH %FQFOEFOUBOEJOEFQFOEFOUTPVSDFTBSFUSFBUFEUIF
TBNFXBZBTJO5IFWFOJOTUIFPSFN
0CTFSWFUIFDMPTFSFMBUJPOTIJQCFUXFFO/PSUPO TBOE 5IFWFOJOT
UIFPSFNT3/=35IBTJO&R BOE
'JOEJOH/PSUPODVSSFOU*/
75I
@@@@
*/=
35I
5IJTJTFTTFOUJBMMZTPVSDFUSBOTGPSNBUJPO' PSUIJTSFBTPO TPVSDFUSBOT GPSNBUJPOJTPGUFODBMMFE5IFWFOJO/PSUPOUSBOTGPSNBUJPO
4JODF75I */ BOE35IBSFSFMBUFEBDDPSEJOHUP&R UPEFUFS NJOFUIF5IFWFOJOPS/PSUPOFRVJWalent circuit requires that we find:
The Thevenin and Norton equivalent
circuits are related by a source
transformation.
r 5IFPQFODJSDVJUWPMUBHFvPDBDSPTTUFSNJOBMTBBOEC
r 5IFTIPSUDJSDVJUDVSSFOUJTDBUUFSNJOBMTBBOEC
r 5IFFRVJWBMFOUPSJOQVUSFTJTUBODF3JOBUUFSNJOBMTaBOEbXIFOBMM
JOEFQFOEFOUTPVSDFTBSFUVSOFEPGG
8FDBODBMDVMBUFBO ZUXPPGUIFUISFFVTJOHUIFNFUIPEUIBUUBL FTUIF
MFBTUFGGPSUBOEVTFUIFNUPHFUUIFUIJSEVTJOH0IN TMBX&YBNQMF
XJMMJMMVTUSBUFUIJT"MTP TJODF
75I=WPD
B
*/=JTD
C
W@@@
35I= PD=3/
D
JTD
UIFPQFODJSDVJUBOETIPSUDJSDVJUUFTUTBSFTVGficient to find anZ5IFWFOJO
PS/PSUPOFRVJWBMFOU PGBDJSDVJUXIJDIDPOUBJOTBUMFBTUPOFJOEFQFOEFOU
TPVSDF
Example 4.11
'JOEUIF/PSUPOFRVJ
UFSNJOBMTBC
8Ω
a
4Ω
2A
+ 12 V
–
8Ω
Figure 4.39
'PS&YBNQMF
5Ω
b
WBMFOUDJSDVJUPGUIFDJSDVJUJO'JHBU
Solution:
8e find 3/JOUIFTBNFX ay we find 35IJOUIF5IFWFOJOFRVJWBMFOUDJS
DVJU4FUUIFJOEFQFOEFOUTPVSDFTFRVBMUP[FSP5IJTMFBETUPUIFDJSDVJU
in Fig. 4.40(a), from which we find 3/5IVT
×
@@@@@@
3/=∥ + + =∥ =
=Ω
5o find */ XFTIPSUDJSDVJUUFSNJOBMT aBOEb BTTIPXOJO'JH C 8FJHOPSFUIFΩ SFTJTUPSCFDBVTFJUIBTCFFOTIPSUDJSDVJUFE"QQMZJOH
NFTIBOBMZTJT XFPCUBJO
J=" J − J − =
'SPNUIFTFFRVBUJPOT XFPCUBJO
J="=JTD=*/
4.6
8Ω
8Ω
a
i1
RN
5Ω
4Ω
145
Norton’s Theorem
8Ω
a
isc = IN
i2
4Ω
2A
5Ω
+ 12 V
–
8Ω
b
b
(a)
(b)
8Ω
i4
4Ω
i3
2A
+ 12 V
–
8Ω
5Ω
+
a
VTh = voc
–
b
(c)
Figure 4.40
'or Example 4.11; finding: (a) RN C IN=isc D V5I=voc
"MUFSOBUJWFMZ XFNBZEFUFSNJOF*/GSPN75I∕35I8FPCUBJO75IBT
UIFPQFODJSDVJUWPMUBHFBDSPTTUFSNJOBMTaBOEbJO'JH D 6TJOH
NFTIBOBMZTJT XFPCUBJO
J="
J − J − =
⇒
J="
BOE
WPD=75I=J=7
)FODF
75I @@
* =@@@
=="
/
35I
BTPCUBJOFEQSF WJPVTMZ 5IJTBMTPTFSW es to confirm Eq. (4.12c) that
35I = vPD∕JTD = ∕ = Ω 5IVT UIF/PSUPOFRVJ WBMFOUDJSDVJUJTBT
TIPXOJO'JH
'JOEUIF/PSUPOFRVJ WBMFOUDJSDVJUGPSUIFDJSDVJUJO'JH BU
UFSNJOBMTab
a
4Ω
1A
b
Figure 4.41
/PSUPOFRVJWBMFOUPGUIFDJSDVJUJO'JH
Practice Problem 4.11
Answer:3/=Ω */="
90 Ω
450 V +
–
4A
90 Ω
a
180 Ω
b
Figure 4.42
'PS1SBDUJDF1SPC
146
Chapter 4
Example 4.12
Circuit Theorems
6TJOH /PSUPOT UIFPSFN find 3/ BOE */ PG UIF DJSDVJU JO'JH BU
UFSNJOBMTBC
2 ix
5Ω
ix
Solution:
5o find 3/ XFTFUUIFJOEFQFOEFOUWPMUBHFTPVSDFFRVBMUP[FSPBOEDPO
OFDUBWPMUBHFTPVSDFPGvP=7 PSBOy unspecified vPMUBHFvP UPUIF
UFSNJOBMT8FPCUBJOUIFDJSDVJUJO'JH B 8FJHOPSFUIFΩSFTJT
UPSCFDBVTFJUJTTIPSUDJSDVJUFE "MTPEVFUPUIFTIPSUDJSDVJU UIF Ω
SFTJTUPS UIFWPMUBHFTPVSDF BOEUIFEFQFOEFOUDVSSFOUTPVSDFBSFBMMJO
7
QBSBMMFM)FODF JY="UOPEFa JP=@@@@
Ω
=" BOE
a
+ 10 V
–
4Ω
b
Figure 4.43
v
=Ω
@@@
3/=@@
P=
JP 'PS&YBNQMF
5o find */ XFTIPSUDJSDVJUUFSNJOBMTBBOEC and find the current JTD as indicated in Fig. 4.44(b). Note from this figure that the 4 Ω resistor UIF
7WPMUBHFTPVSDF UIFΩSFTJTUPS BOEUIFEFQFOEFOUDVSSFOUTPVSDF
BSFBMMJOQBSBMMFM)FODF
="
JY= @@@
"UOPEFa ,$-HJWFT
+ J = + ="
JTD=@@@
Y
5IVT
*/="
2ix
2ix
5Ω
ix
a
io
+ v =1V
– o
4Ω
5Ω
ix
4Ω
a
isc = IN
+ 10 V
–
b
b
(a)
(b)
Figure 4.44
'or Example 4.12: (a) finding 3/, (b) finding */
Practice Problem 4.12
'JOEUIF/PSUPOFRVJ
UFSNJOBMTab
2vx
+–
6Ω
10 A
2Ω
+
vx
–
a
b
Figure 4.45
'PS1SBDUJDF1SPC
WBMFOUDJSDVJUPGUIFDJSDVJUJO'JHBU
Answer:3/=Ω, */="
4.7
4.7
a
Derivations of Thevenin’s and
Norton’s Theorems
v="J + "WT + "vT + "JT + "JT
+
v
–
i
*OUIJTTFDUJPO XFXJMMQSP WF5IFWFOJOTBOE/PSUPO TUIFPSFNTVTJOH
UIFTVQFSQPTJUJPOQSJODJQMF
$POTJEFSUIFMJOFBSDJSDVJUJO'JH B *UJTBTTVNFEUIBUUIFDJS
DVJUDPOUBJOTSFTJTUPSTBOEEFQFOEFOUBOEJOEFQFOEFOUTPVSDFT8FIBWF
BDDFTTUPUIFDJSDVJUWJBUFSNJOBMT BBOE C UISPVHIXIJDIDVSSFOUGSPN
BOFYUFSOBMTPVSDFJTBQQMJFE0VSPCKFDUJWFJTUPFOTVSFUIBUUIFWPMUBHF
DVSSFOUSFMBUJPOBUUFSNJOBMT aBOEbJTJEFOUJDBMUPUIBUPGUIF 5IFWFOJO
FRVJWBMFOUJO'JH C ' PSUIFTBL FPGTJNQMJDJUZ TVQQPTFUIFMJO FBSDJSDVJUJO'JH B DPOUBJOTUX PJOEFQFOEFOUW PMUBHFTPVSDFT vT
BOEvTBOEUX PJOEFQFOEFOUDVSSFOUTPVSDFT JTBOE JT8FNBZPCUBJO
BOZDJSDVJUW BSJBCMF TVDIBTUIFUFSNJOBMW PMUBHFv CZBQQMZJOHTVQFS QPTJUJPO5IBUJT XFDPOTJEFSUIFDPOUSJC VUJPOEVFUPFBDIJOEFQFOEFOU
TPVSDFJODMVEJOHUIFF YUFSOBMTPVSDF J#ZTVQFSQPTJUJPO UIFUFSNJOBM
WPMUBHFvJT
147
Derivations of Thevenin’s and Norton’s Theorems
Linear
circuit
b
(a)
R Th
a
+
+ V
Th
–
v
i
–
b
(b)
Figure 4.46
%FSJWBUJPOPG5IFWFOJOFRVJWBMFOU B B
DVSSFOUESJWFODJSDVJU C JUT5IFWFOJO
FRVJWBMFOU
XIFSF" " " " BOE"BSFDPOTUBOUT&BDIUFSNPOUIFSJHIUIBOE
TJEFPG&R JTUIFDPOUSJC VUJPOPGUIFSFMBUFEJOEFQFOEFOUTPVSDF
UIBUJT "JJTUIFDPOUSJC VUJPOUP vEVFUPUIFF YUFSOBMDVSSFOUTPVSDF J "vTJTUIFDPOUSJCVUJPOEVFUPUIFWPMUBHFTPVSDFvT BOETPPO8FNBZ
DPMMFDUUFSNTGPSUIFJOUFSOBMJOEFQFOEFOUTPVSDFTUPHFUIFSBT # TPUIBU
&R CFDPNFT
v="J + #
XIFSF#="vT + "vT + "JT + "JT8FOPXXBOUUPFWBMVBUFUIF
WBMVFTPGDPOTUBOUT "BOE #8IFOUIFUFSNJOBMT aBOE bBSFPQFO
DJSDVJUFE J=BOEv=#5IVT #JTUIFPQFODJSDVJUWPMUBHFvPD XIJDI
JTUIFTBNFBT75I TP
#=75I
8IFOBMMUIFJOUFSOBMTPVSDFTBSFUVSOFEPGG #=5IFDJSDVJUDBOUIFO
CFSFQMBDFECZBOFRVJWBMFOUSFTJTUBODF3FR XIJDIJTUIFTBNFBT35I BOE
&R CFDPNFT
v="J=35IJ
⇒
"=35I
4VCTUJUVUJOHUIFWBMVFTPG"BOE#JO&R HJWFT
v=35IJ + 75I
J=$v + %
a
Linear
circuit
v +
–
b
XIJDIFYQSFTTFTUIFWPMUBHFDVSSFOUSFMBUJPOBUUFSNJOBMTaBOEbPGUIF
DJSDVJUJO'JH C 5IVT UIFUXPDJSDVJUTJO'JH B BOE C BSFFRVJWBMFOU
8IFOUIFTBNFMJOFBSDJSDVJUJTESJ WFOCZBW PMUBHFTPVSDF vBT
TIPwn in Fig. 4.47(a), the current floXJOHJOUPUIFDJSDVJUDBOCFPCUBJOFE
CZTVQFSQPTJUJPOBT
i
(a)
i
a
v +
–
RN
b
XIFSF$vJTUIFDPOUSJC VUJPOUP JEVFUPUIFF YUFSOBMWPMUBHFTPVSDF v
BOE %DPOUBJOTUIFDPOUSJC VUJPOTUP iEVFUPBMMJOUFSOBMJOEFQFOEFOU
TPVSDFT 8IFOUIFUFSNJOBMT BCBSFTIPSUDJSDVJUFE v =TPUIBU IN
(b)
Figure 4.47
%FSJWBUJPOPG/PSUPOFRVJWBMFOU B B
WPMUBHFESJWFODJSDVJU C JUT/PSUPO
FRVJWBMFOU
148
Chapter 4
Circuit Theorems
J=%=−JTD XIFSFJTD is the short-circuit current floXJOHPVUPGUFSNJOBM
a XIJDIJTUIFTBNFBTUIF/PSUPODVSSFOU*/ JF
%=−*/
8IFOBMMUIFJOUFSOBMJOEFQFOEFOUTPVSDFTBSFUVSOFEPG G % =BOE
UIFDJSDVJUDBOCFSFQMBDFECZBOFRVJWBMFOUSFTJTUBODF3FR PSBOFRVJWB
MFOUDPOEVDUBODF (FR=∕3FR XIJDIJTUIFTBNFBT 35IPS 3/5IVT &R CFDPNFT
v − * @@@
J=
/
35I
5IJTFYQSFTTFTUIFW PMUBHFDVSSFOUSFMBUJPOBUUFSNJOBMT abPGUIFDJS cuit in Fig. 4.47(b), confirming that the twPDJSDVJUTJO'JH B BOE
C BSFFRVJWBMFOU
4.8
RTh
*ONBOZQSBDUJDBMTJUVBUJPOT BDJSDVJUJTEFTJHOFEUPQSPWJEFQPXFSUPB
MPBE5IFSFBSFBQQMJDBUJPOTJOBSFBTTVDIBTDPNNVOJDBUJPOTXIFSFJU
JTEFTJSBCMFUPNBYJNJ[FUIFQPXFSEFMJWFSFEUPBMPBE8FOPXBEESFTT
UIFQSPCMFNPGEFMJ WFSJOHUIFNBYJNVNQP XFSUPBMPBEXIFOHJ WFOB
TZTUFNXJUILOPXOJOUFSOBMMPTTFT*UTIPVMECFOPUFEUIBUUIJTXJMMSFTVMU
in significant internal losses greater than or equal to the poXFSEFMJWFSFE
UPUIFMPBE
5IF5IFWFOJOFRVJWalent is useful in finding the maximum poXFSB
MJOFBSDJSDVJUDBOEFMJWFSUPBMPBE8FBTTVNFUIBUXFDBOBEKVTUUIFMPBE
SFTJTUBODF3-*GUIFFOUJSFDJSDVJUJTSFQMBDFE CZJUT5IFWFOJOFRVJWBMFOU
FYDFQU GPSUIF MPBE BT TIPXOJO 'JH UIFQPXFS EFMJWFSFE UPUIF
MPBEJT
75I
Q=J3-= @@@@@@@@
3-
( 35I + 3- )
a
i
VTh +
–
Maximum Power Transfer
RL
b
Figure 4.48
5IFDJSDVJUVTFEGPSNBYJNVNQPXFS
USBOTGFS
'PSBHJWFODJSDVJU 75IBOE35I are fixFE#ZWBSZJOHUIFMPBESFTJT
UBODF3- UIFQPXFSEFMJWFSFEUPUIFMPBEWBSJFTBTTLFUDIFEJO'JH
8FOPUJDFGSPN'JHUIBUUIFQPXFSJTTNBMMGPSTNBMMPSMBSHFWBMVFT
PG3-CVUNBYJNVNGPSTPNFWBMVFPG3-CFUXFFOBOE∞8FOPXXBOU
UPTIPXUIBUUIJTNBYJNVNQPXFSPDDVSTXIFO3-JTFRVBMUP35I5IJTJT
LOPXOBTUIFNBYJNVNQPXFSUIFPSFN
p
pmax
0
Figure 4.49
RTh
RL
1PXFSEFMJWFSFEUPUIFMPBEBTBGVODUJPO
PG3-
Maximum power is transferred to the load when the load resistance
equals the Thevenin resistance as seen from the load (RL = RTh).
5PQSPWFUIFNBYJNVNQPXFSUSBOTGFSUIFPSFN XFEJGGFSFOUJBUFpJO
&R XJUISFTQFDUUP3-BOETFUUIFSFTVMUFRVBMUP[FSP8FPCUBJO
[
]
EQ
3 + 3- − 3- 35I + 3@@@@=V
5I @@@@@@@@@@@@@@@@@@@@@@@@
5I
E335I + 3- [
]
3 + 3- − 3=V5I @@@@@@@@@@@@@@@
5I
=
35I + 3- 4.8
149
Maximum Power Transfer
5IJTJNQMJFTUIBU
= 35I + 3- − 3- = 35I − 3- XIJDIZJFMET
3-=35I
TIPXJOH UIBU UIFNBYJNVN QPXFS USBOTGFS UBLFT QMBDF XIFO UIF MPBE
SFTJTUBODF3-FRVBMTUIF5IFWFOJOSFTJTUBODF35I8e can readily confirm
UIBU&R HJWFTUIFNBYJNVNQPXFSCZTIPXJOHUIBUEQ∕E3-<
5IFNBYJNVNQP XFSUSBOTGFSSFEJTPCUBJOFECZTVCTUJUVUJOH
&R JOUP&R GPS
75I
QNBY=@@@@
35I
The source and load are said to be
matched when RL = RTh.
&RVBUJPO BQQMJFTPOMZXIFO 3-=35I8IFO3-≠ 35I XFDPN
QVUFUIFQPXFSEFMJWFSFEUPUIFMPBEVTJOH&R Example 4.13
'JOEUIFW BMVFPG 3-GPSNBYJNVNQP XFSUSBOTGFSJOUIFDJSDVJUPG 'JH'JOEUIFNBYJNVNQPXFS
6Ω
3Ω
12 V +
–
2Ω
12 Ω
a
RL
2A
b
Figure 4.50
'PS&YBNQMF
Solution:
8e need to find the 5IFWFOJOSFTJTUBODF 35IBOEUIF 5IFWFOJOWPMUBHF
75IBDSPTTUIFUFSNJOBMTBC5PHFU35I XFVTFUIFDJSDVJUJO'JH B BOEPCUBJO
×
35I= + + ∥= + @@@@@@
=Ω
6Ω
3Ω
12 Ω
6Ω
2Ω
RTh
12 V +
–
3Ω
2Ω
+
i1
12 Ω
i2
2A
VTh
–
(a)
Figure 4.51
'or Example 4.13: (a) finding 35I, (b) finding 75I
(b)
150
Chapter 4
Circuit Theorems
5PHFU75I XFDPOTJEFSUIFDJSDVJUJO'JH C "QQMZJOHNFTIBOBMZ
TJTHJWFT
− + J − J= J=−"
4PMWJOHGPSJ XFHFUJ=−∕"QQMZJOH,7-BSPVOEUIFPVUFSMPPQUP
HFU75IBDSPTTUFSNJOBMTBC XFPCUBJO
− + J + J + + 75I=
⇒
75I=7
'PSNBYJNVNQPXFSUSBOTGFS
3-=35I=Ω
BOEUIFNBYJNVNQPXFSJT
7 5I
=8
QNBY=@@@@
=@@@@@
3- × Practice Problem 4.13
60 Ω
120 Ω
Answer:Ω N8
+ vx –
9V +
–
Figure 4.52
%FUFSNJOFUIFWBMVFPG3-UIBUXJMMESBXUIFNBYJNVNQPXFSGSPNUIF
SFTUPGUIFDJSDVJUJO'JH$BMDVMBUFUIFNBYJNVNQPXFS
30 Ω
+
–
'PS1SBDUJDF1SPC
RL
3vx
4.9
Verifying Circuit Theorems with PSpice
*OUIJTTFDUJPO XFMFBSOIPXUPVTF14QJDFUPWFSJGZUIFUIFPSFNTDPWFSFE
JOUIJTDIBQUFS. Specifically XFXJMMDPOTJEFSVTJOH%$4XFFQBOBMZTJTUP
find the 5IFWFOJOPS/PSUPOFRVJ WBMFOUBUBO ZQBJSPGOPEFTJOBDJSDVJU BOEUIFNBYJNVNQP XFSUSBOTGFSUPBMPBE 5IFSFBEFSJTBEWJTFEUPSFBE Section D.3 of "QQFOEJY%JOQSFQBSBUJPOGPSUIJTTFDUJPO
5Pfind the 5IFWFOJOFRVJWBMFOUPGBDJSDVJUBUBQBJSPGPQFOUFSNJ
OBMTVTJOH 14QJDF XFVTFUIFTDIFNBUJDFEJUPSUPESB XUIFDJSDVJUBOE
JOTFSUBOJOEFQFOEFOUQSPCJOHDVSSFOUTPVSDF TBZ *Q BUUIFUFSNJOBMT
5IFQSPCJOHDVSSFOU TPVSDFNVTUIBWFBQBSUOBNF *43$8FUIFO QFS
GPSNB%$4XFFQPO*Q BTEJTDVTTFEJO4FDUJPO%5ZQJDBMMZ XFNBZ
MFUUIFDVSSFOUUISPVHI*QW BSZGSPNUP"JO"JODSFNFOUT"GUFS
TBWJOHBOETJNVMBUJOHUIFDJSDVJU XFVTF1SPCFUPEJTQMBZBQMPUPGUIF
WPMUBHFBDSPTT*QWFSTVTUIFDVSSFOUUISPVHI*Q5IF[FSPJOUFSDFQUPGUIF
QMPUHJWFTVTUIF5IFWFOJOFRVJWBMFOUWPMUBHF XIJMFUIFTMPQFPGUIFQMPU
JTFRVBMUPUIF5IFWFOJOSFTJTUBODF
5o find the Norton equiWBMFOUJOWPMWFTTJNJMBSTUFQTF YDFQUUIBUXF
JOTFSUBQSPCJOHJOEFQFOEFOUW PMUBHFTPVSDF XJUIBQBSUOBNF 743$ TBZ 7Q BUUIFUFSNJOBMT 8FQFSGPSNB%$4XFFQPO 7QBOEMFU 7Q
WBSZGSPNUP7JO7JODSFNFOUT"QMPUPGUIFDVSSFOUUISPVHI
7QWFSTVTUIFWPMUBHFBDSPTT7QJTPCUBJOFEVTJOHUIF1SPCFNFOVBGUFS
TJNVMBUJPO5IF[FSPJOUFSDFQUJTFRVBMUPUIF/PSUPODVSSFOU XIJMFUIF
TMPQFPGUIFQMPUJTFRVBMUPUIF/PSUPODPOEVDUBODF
5o find the maximum poXFSUSBOTGFSUPBMPBEVTJOH 14QJDFJO WPMWFTQFSGPSNJOHB%$QBSBNFUSJD 4XFFQPOUIFDPNQPOFOUW BM
VFPG 3-JO'JHBOEQMPUUJOHUIFQP XFSEFMJ WFSFEUPUIFMPBEBT
BGVODUJPOPG 3-"DDPSEJOHUP'JH UIFNBYJNVNQP XFSPDDVST
4.9
151
Verifying Circuit Theorems with PSpice
XIFO 3- = 35I 5IJT JT CFTU JMMVTUSBUFE XJUI BO FYBNQMF BOE &Y
ample 4.15 proWJEFTPOF
8FVTF743$BOE*43$BTQBSUOBNFTGPSUIFJOEFQFOEFOUWPMUBHF
BOEDVSSFOUTPVSDFT SFTQFDUJWFMZ
Example 4.14
$POTJEFSUIFDJSDVJUJO'JH TFF&YBNQMF 6TF 14QJDF to find
UIF5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUT
Solution:
B 5o find the 5IFWFOJOSFTJTUBODF35IBOE5IFWFOJOWPMUBHF75IBUUIF
UFSNJOBMTBC in the circuit in Fig. 4.31, we first use Schematics to draX
UIFDJSDVJUBTTIPXOJO'JH B /PUJDFUIBUBQSPCJOHDVSSFOUTPVSDF
*JTJOTFSUFEBUUIFUFSNJOBMT6OEFS "OBMZTJT∕4FUQVU XFTFMFDU%$
4XFFQ*OUIF%$4XFFQEJBMPHCPY XFTFMFDU-JOFBSGPSUIF
4XFFQ5ZQF
BOE$VSSFOU4PVSDFGPSUIF4XFFQ7BS5ZQF8FFOUFS*VOEFSUIF/BNF
CPY BT4UBSU7BMVF BT&OE7BMVF BOEBT*ODSFNFOU"GUFSTJNVMB
UJPO XFBEEUSBDF7 *m GSPNUIF14QJDF"∕%XJOEPXBOEPCUBJOUIF
QMPUTIPXOJO'JH C 'SPNUIFQMPU XFPCUBJO
75I=;FSPJOUFSDFQU=7 − @@@@@@@
35I=4MPQF=
=Ω
5IFTFBHSFFXJUIXIBUXFHPUBOBMZUJDBMMZJO&YBNQMF
26 V
I1
R4
4
E1
++
–
–
GAIN=2
R2
R4
2
2
R3
6
24 V
I2
0
22 V
20 V
0A
0.2 A
= V(I2:_)
(a)
Figure 4.53
'or Example 4.14: (a) schematic and (b) plot for finding 35IBOE75I
C 5o find the Norton equiWBMFOU XFNPEJGZUIFTDIFNBUJDJO'JH B CZSFQMBZJOHUIFQSPCJOHDVSSFOUTPVSDFXJUIBQSPCJOHW PMUBHFTPVSDF
75IFSFTVMUJTUIFTDIFNBUJDJO'JH B "HBJO JOUIF%$4XFFQ
EJBMPHCPY XFTFMFDU -JOFBS GPSUIF 4XFFQ 5ZQFBOE 7PMUBHF4PVSDF
GPSUIF4XFFQ7BS5ZQF8FFOUFS7VOEFS/BNFCPY BT4UBSU7BMVF BT &OE7BMVF BOEBT *ODSFNFOU6OEFSUIF 14QJDF"∕%8JOEPX XFBEEUSBDF* 7 BOEPCUBJOUIFQMPUJO'JH C 'SPNUIFQMPU XFPCUBJO
*/=;FSPJOUFSDFQU="
− @@@@@@@@@@@@
(/=4MPQF=
=4
0.4 A
(b)
0.6 A
0.8 A
1.0 A
152
Chapter 4
Circuit Theorems
3.4 A
I1
R4
4
R2
R1
2
2
E1
++
–
–
GAIN=2
R3
6
3.3 A
V1 +
–
3.2 A
3.1 A
0V
0
0.2 V
I(V1)
0.4 V
0.6 V
V_V1
0.8 V
1.0 V
(b)
(a)
Figure 4.54
'or Example 4.14: (a) schematic and (b) plot for finding (NBOE*/
Practice Problem 4.14
3FXPSL1SBDUJDF1SPCVTJOH14QJDF
Answer:75I=7 35I=NΩ
Example 4.15
3FGFSUPUIFDJSDVJUJO'JH6TF14QJDFto find UIFNBYJNVNQPXFS
USBOTGFSUP3-
1 kΩ
1V +
–
RL
Figure 4.55
'PS&YBNQMF
%$-*$,-UIFWBMVFLPG3 SFQSFTFOUJOH3- UPPQFOVQUIF4FU
"UUSJCVUF7BMVFEJBMPHCPY
3FQMBDFLXJUI\3-^BOEDMJDL0,UPBDDFQUUIFDIBOHF
PARAMETERS:
RL
2k
R1
DC=1 V
+
–
V1
1k
R2
{RL}
0
Figure 4.56
Solution:
8FOFFEUPQFSGPSNB%$4XFFQPO 3-UPEFUFSNJOFXIFOUIFQP XFS
BDSPTTJUJTNBYJNVN 8e first draXUIFDJSDVJUVTJOH4DIFNBUJDTBT
TIPXOJO'JH0ODFUIFDJSDVJUJTESB XO XFUBL FUIFGPMMP XJOH
UISFFTUFQTUPGVSUIFSQSFQBSFUIFDJSDVJUGPSB%$4XFFQ
The first step inWPMWes defining the vBMVFPG3-BTBQBSBNFUFS TJODF
XFXBOUUPWBSZJU5PEPUIJT
4DIFNBUJDGPSUIFDJSDVJUJO'JH
/PUFUIBUUIFDVSMZCSBDLFUTBSFOFDFTTBSZ
The second step is to define parameter5PBDIJFWFUIJT
4FMFDU%SBX∕(FU/FX1BSU∕-JCSBSJFT⋯∕TQFDJBMTMC
5ZQF1"3".JOUIF1BSU/BNFCPYBOEDMJDL0,
%3"(UIFCPYUPBOZQPTJUJPOOFBSUIFDJSDVJU
$-*$,-UPFOEQMBDFNFOUNPEF
%$-*$,-UPPQFOVQUIF1BSU/BNF1"3".EJBMPHCPY
$-*$,-PO/".&=BOEFOUFS3- XJUIOPDVSMZCSBDLFUT JOUIF
7BMVFCPY BOE$-*$,-4BWF"UUSUPBDDFQUDIBOHF
$-*$,-PO7"-6&=BOEFOUFSLJOUIF7BMVFCPY BOE$-*$,-
4BWF"UUSUPBDDFQUDIBOHF
$MJDL0,
4.10
153
Applications
5IFWBMVFLJOJUFNJTOFDFTTBSZGPSBCJBTQPJOUDBMDVMBUJPOJU
DBOOPUCFMFGUCMBOL
5IFUIJSETUFQJTUPTFUVQUIF%$4XFFQUPTXFFQUIFQBSBNFUFS
5PEPUIJT
4FMFDU"OBMZTJT∕4FUQVUUPCSJOHVQUIF%$4XFFQEJBMPHCPY
'PSUIF4XFFQ5ZQF TFMFDU-JOFBS PS0DUBWFGPSBXJEFSBOHFPG3- 'PSUIF4XFFQ7BS5ZQF TFMFDU(MPCBM1BSBNFUFS
6OEFSUIF/BNFCPY FOUFS3-
*OUIF4UBSU7BMVFCPY FOUFS
*OUIF&OE7BMVFCPY FOUFSL
*OUIF*ODSFNFOUCPY FOUFS
$MJDL0,BOE$MPTFUPBDDFQUUIFQBSBNFUFST
"GUFSUBLJOHUIFTFTUFQTBOETB WJOHUIFDJSDVJU XFBSFSFBEZUP
TJNVMBUF4FMFDU "OBMZTJT4JNVMBUF*GUIFSFBSFOPFSSPST XFTFMFDU
"EE5SBDFJOUIF 14QJDF "%XJOEP XBOEUZQF −7 3 ∗* 3 JO
UIF 5SBDF$PNNBOE CPY<5IFOF HBUJWFTJHOJTOFFEFETJODF* 3 JT OFHBUJWF>5IJTHJWFTUIFQMPUPGUIFQPXFSEFMJWFSFEUP3-BT3-WBSJFT
GSPN ΩUPL Ω8FDBO BMTPPCUBJO UIFQPXFSBCTPSCFECZ 3-CZ
UZQJOH7 3 ∗7 3 3-JOUIF5SBDF$PNNBOECPY&JUIFSXBZ XF
PCUBJOUIFQMPUJO'JH*UJTFWJEFOUGSPNUIFQMPUUIBUUIFNBYJNVN
QPXFSJTμ8/PUJDFUIBUUIFNBYJNVNPDDVSTXIFO 3-=LΩ BT
FYQFDUFEBOBMZUJDBMMZ
'JOEUIF NBYJNVNQP XFS USBOTGFSSFE UP 3-JG UIFL ΩSFTJTUPS JO
'JHJTSFQMBDFECZBLΩSFTJTUPS
250 uW
200 uW
150 uW
100 uW
50 uW
0
2.0 K
4.0 K
–V(R2:2)*I(R2)
RL
6.0 K
Figure 4.57
'PS&YBNQMFUIFQMPUPGQPXFS
BDSPTT3-
Practice Problem 4.15
Answer:μ8
Rs
4.10
Applications
vs +
–
*OUIJTTFDUJPOXFXJMMEJTDVTTUX PJNQPSUBOUQSBDUJDBMBQQMJDBUJPOTPG
UIFDPODFQUTDP WFSFEJOUIJTDIBQUFSTPVSDFNPEFMJOHBOESFTJTUBODF
NFBTVSFNFOU
4.10.1
(a)
Source Modeling
4PVSDFNPEFMJOHQSPWJEFTBOFYBNQMFPGUIFVTFGVMOFTTPGUIF5IFWFOJO
PSUIF/PSUPOFRVJWBMFOU"OBDUJWFTPVSDFTVDIBTBCBUUFSZJTPGUFODIBS
BDUFSJ[FECZJUT 5IFWFOJOPS/PSUPOFRVJ WBMFOUDJSDVJU"OJEFBMWPMUBHF
TPVSDFQSPWJEFTBDPOTUBOUW PMUBHFJSSFTQFDUJWFPGUIFDVSSFOUESB XOCZ
UIFMPBE XIJMFBOJEFBMDVSSFOUTPVSDFTVQQMJFTBDPOTUBOUDVSSFOUSFHBSE
MFTTPGUIFMPBEWPMUBHF"T'JHTIPXT QSBDUJDBMWPMUBHFBOEDVSSFOU
TPVSDFTBSFOPUJEFBM EVFUPUIFJS JOUFSOBMSFTJTUBODFTPS TPVSDFSFTJT
UBODFT3TBOE3Q5IFZCFDPNF JEFBMBT 3T→BOE 3Q→∞5PTIPX
UIBUUIJTJTUIFDBTF DPOTJEFSUIFFGGFDUPGUIFMPBEPOWPMUBHFTPVSDFT Rp
is
(b)
Figure 4.58
B 1SBDUJDBMWPMUBHFTPVSDF C QSBDUJDBM
DVSSFOUTPVSDF
154
Chapter 4
Circuit Theorems
BTTIPXOJO'JH B #ZUIFWPMUBHFEJWJTJPOQSJODJQMF UIFMPBEWPMU
age is
3
v-= @@@@@@@
- v
3T+ 3- T
"T 3-JODSFBTFT UIFMPBEW PMUBHFBQQSPBDIFTBTPVSDFW PMUBHF WT BT
JMMVTUSBUFEJO'JH C 'SPN&R XFTIPVMEOPUFUIBU
5IFMPBEW PMUBHFXJMMCFDPOTUBOUJGUIFJOUFSOBMSFTJTUBODF 3T
PGUIFTPVSDFJT[FSPPS BUMFBTU 3T ≪ 3-*OPUIFSX PSET UIF
TNBMMFS3TJTDPNQBSFEXJUI 3- UIFDMPTFSUIFW PMUBHFTPVSDFJT
UPCFJOHJEFBM
vL
Rs
vs
+
–
vL
Ideal source
vs
+
Practical source
RL
–
0
(a)
(b)
RL
Figure 4.59
B 1SBDUJDBMWPMUBHFTPVSDFDPOOFDUFEUPBMPBE3- C MPBEWPMU
BHFEFDSFBTFTBT3-EFDSFBTFT
8IFOUIFMPBEJTEJTDPOOFDUFE JF UIFTPVSDFJTPQFODJSDVJUFETP
UIBU3- →∞ vPD=vT5IVT vTNBZCFSF HBSEFEBTUIF VOMPBEFE
TPVSDFWPMUBHF5IFDPOOFDUJPOPGUIFMPBEDBVTFTUIFUFSNJOBMWPMU
BHFUPESPQJONBHOJUVEFUIJTJTLOPXOBTUIFMPBEJOHFGGFDU
IL
Rp
is
RL
(a)
IL
Ideal source
is
Practical source
0
RL
(b)
Figure 4.60
B 1SBDUJDBMDVSSFOUTPVSDFDPOOFDUFEUP
BMPBE3- C MPBEDVSSFOUEFDSFBTFTBT3-
JODSFBTFT
5IF TBNF BSHVNFOU DBO CF NBEF GPSB QSBDUJDBM DVSSFOU TPVSDF XIFO
DPOOFDUFEUPBMPBEBTTIP XOJO'JH B #ZUIFDVSSFOUEJ WJTJPO
QSJODJQMF
3Q
J-=@@@@@@@
J
3Q + 3- T
'JHVSF C TIPXTUIFWBSJBUJPOJOUIFMPBEDVSSFOUBTUIFMPBESFTJT UBODFJODSFBTFT"HBJO XFOPUJDFBESPQJODVSSFOUEVFUPUIFMPBE MPBE
JOHFGGFDU BOEMPBEDVSSFOUJTDPOTUBOU JEFBMDVSSFOUTPVSDF XIFOUIF
JOUFSOBMSFTJTUBODFJTWFSZMBSHF JF 3Q→∞PS BUMFBTU 3Q≫3- 4PNFUJNFT XFOFFEUPLOPXUIFVOMPBEFETPVSDFWPMUBHFvTBOEUIF
JOUFSOBMSFTJTUBODF 3TPGBW PMUBHFTPVSDF 5o find vTBOE 3T XFGPMMP X
UIFQSPDFEVSFJMMVTUSBUFEJO'JH'JSTU XFNFBTVSFUIFPQFODJSDVJU
WPMUBHFvPDBTJO'JH B BOETFU
vT=vPD
5IFO XFDPOOFDUBW BSJBCMFMPBE 3-BDSPTTUIFUFSNJOBMTBTJO 'JH C 8FBEKVTUUIFSFTJTUBODF3-VOUJMXFNFBTVSFBMPBEWPMU
BHF PG FYBDUMZ POFIBMG PG UIF PQFODJSDVJU WPMUBHF v- = vPD∕ CF DBVTFOPX3-=35I=3T"UUIBUQPJOU XFEJTDPOOFDU3-BOENFBTVSF
JU8FTFU
3T=3-
'PSFYBNQMF BDBSCBUUFSZNBZIBWFvT=7BOE3T=Ω
4.10
+
Signal
source
+
vL
Signal
source
voc
–
155
Applications
RL
–
(b)
(a)
Figure 4.61
B .FBTVSJOHvPD C NFBTVSJOHv-
Example 4.16
5IFUFSNJOBMW PMUBHFPGBW PMUBHFTPVSDFJT 7XIFODPOOFDUFEUPB
8MPBE 8IFOUIFMPBEJTEJTDPOOFDUFE UIFUFSNJOBMW PMUBHFSJTFTUP
7 B $BMDVMBUFUIFTPVSDFW PMUBHF vTBOEJOUFSOBMSFTJTUBODF 3T
C %FUFSNJOFUIFWPMUBHFXIFOBOΩMPBEJTDPOOFDUFEUPUIFTPVSDF
Solution:
B 8FSFQMBDFUIFTPVSDFCZJUT5IFWFOJOFRVJWBMFOU5IFUFSNJOBMWPMU
BHFXIFOUIFMPBEJTEJTDPOOFDUFEJTUIFPQFODJSDVJUWPMUBHF
vT=vPD=7
8IFOUIF MPBEJTDPOOFDUFE BTTIPXO JO'JH B v- = 7BOE
p-=8)FODF
v-
@@@
Q-= 3-
⇒
v-
@@
Rs
=Ω
3-=Q =@@@
vs
+
+
–
vL
(a)
5IFWPMUBHFBDSPTT3TJTUIFEJGGFSFODFCFUXFFOUIFTPVSDFWPMUBHFvTBOE
UIFMPBEWPMUBHFv- PS
− ==3T J- =Ω
3T=@@@
*-
2.4 Ω
+ 5IFNFBTVSFEPQFODJSDVJUWoltage across a certain amplifier is 16 75IF
WPMUBHFESPQTUP7XIFOBΩMPVETQFBLFSJTDPOOFDUFEUPUIFBNQMJ
fier$BMDVMBUFUIFWPMUBHFXIFOBΩMPVETQFBLFSJTVTFEJOTUFBE
Answer:7
+
12.4 V +
–
C /PXUIBUXFIBWFUIF5IFWFOJOFRVJWBMFOUPGUIFTPVSDF XFDPOOFDU
UIF Ɨ MPBE BDSPTT UIF5IFWFOJO FRVJWBMFOU BT TIPXO JO 'JH C 6TJOHWPMUBHFEJWJTJPO XFPCUBJO
@@@@@@@
v=
=7
RL
–
5IFMPBEDVSSFOUJT
v
=
"
@@
-=@@@
J-=@@@
3- iL
v
8Ω
–
(b)
Figure 4.62
'PS&YBNQMF
Practice Problem 4.16
156
Historical note: The bridge was
invented by Charles Wheatstone
(1802–1875), a British professor who
also invented the telegraph, as Samuel
Morse did independently in the
United States.
R1
v +
–
R2
R3
Galvanometer
+
v1
–
Figure 4.63
+
v2
–
5IF8IFBUTUPOFCSJEHF3YJTUIF
SFTJTUBODFUPCFNFBTVSFE
Rx
Chapter 4
Circuit Theorems
4.10.2
Resistance Measurement
"MUIPVHIUIFPINNFUFSNFUIPEQSP WJEFTUIFTJNQMFTUX BZUPNFB TVSFSFTJTUBODF NPSFBDDVSBUFNFBTVSFNFOUNBZCFPCUBJOFEVTJOHUIF
8IFBUTUPOF CSJEHF 8IJMF PINNFUFST BSF EFTJHOFE UP NFBTVSF SFTJT
UBODFJOMPX NJE PSIJHISBOHF B8IFBUTUPOFCSJEHFJTVTFEUPNFBTVSF
SFTJTUBODFJOUIFNJESBOHF TBZ CFUXFFOΩBOE.Ω7FSZMPXWBMVFT
PGSFTJTUBODFTBSFNFBTVSFEXJUIBmilliohmmeter,XIJMFWFSZIJHIWBMVFT
BSFNFBTVSFEXJUIB.FHHFSUFTUFS
5IF 8IFBUTUPOF CSJEHF PS SFTJTUBODF CSJEHF DJSDVJU JT VTFE JOB
OVNCFSPGBQQMJDBUJPOT)FSFXFXJMMVTFJUUPNFBTVSFBOVOLOP
XO
SFTJTUBODF 5IFVOLOP XOSFTJTUBODF 3YJTDPOOFDUFEUPUIFCSJEHFBT
TIPXOJO'JH 5IFWBSJBCMFSFTJTUBODFJTBEKVTUFEVOUJMOPDVSSFOU
floXTUISPVHIUIFHBMWBOPNFUFS XIJDIJTFTTFOUJBMMZBE"STPOWBMNPWF
NFOUPQFSBUJOHBTBTFOTJUJ WFDVSSFOUJOEJDBUJOHEFWJDFMJLFBOBNNFUFS
JOUIFNJDSPBNQSBOHF6OEFSUIJTDPOEJUJPO v=v BOEUIFCSJEHFJT
TBJEUPCFCBMBODFE4JODFOPDVSSFOUfloXTUISPVHIUIFHBMWBOPNFUFS 3
BOE3CFIBWFBTUIPVHIUIF ZXFSFJOTFSJFTTPEP 3BOE 3Y5IFGBDU
that no current floXTUISPVHIUIFHBMWBOPNFUFSBMTPJNQMJFTUIBUv=W
"QQMZJOHUIFWPMUBHFEJWJTJPOQSJODJQMF
3
3
v=@@@@@@@
v=v=@@@@@@@
Y v
3 + 3
3 + 3Y
Hence, no current floXTUISPVHIUIFHBMWBOPNFUFSXIFO
3
3
@@@@@@@
=@@@@@@@
Y ⇒
33=33Y
3 + 3 3 + 3Y
PS
3
@@@
3Y=
3 3 *G3=3 BOE 3JTBEKVTUFE VOUJMOPDVSSFOU floXTUISPVHIUIF HBMWB
OPNFUFS UIFO3Y=3
)Pw do we find the current through the gBMWBOPNFUFSXIFOUIF
8IFBUTUPOFCSJEHFJTVOCBMBODFE 8e find the 5IFWFOJOFRVJWBMFOU 75I
BOE35I XJUISFTQFDUUPUIFH BMWBOPNFUFSUFSNJOBMT*G 3NJTUIFSFTJT UBODFPGUIFH BMWBOPNFUFS UIFDVSSFOUUISPVHIJUVOEFSUIFVOCBMBODFE
DPOEJUJPOJT
75I
*=@@@@@@@@
35I + 3N
&YBNQMFXJMMJMMVTUSBUFUIJT
Example 4.17
*O'JH 3=ΩBOE3=Ω5IFCSJEHFJTCBMBODFEXIFO3
JTBEKVTUFEUPCFΩ%FUFSNJOFUIFVOLOPXOSFTJTUBODF3Y
Solution:
6TJOH&R HJWFT
3
@@@
@@@@
3Y=
3 =
=Ω
3 4.10
Applications
"8IFBUTUPOFCSJEHFIBT3=3=LΩ3JTBEKVTUFEVOUJMOPDVSSFOU
floXTUISPVHIUIFHBMWBOPNFUFS"UUIBUQPJOU 3=LΩ8IBUJTUIF
WBMVFPGUIFVOLOPXOSFTJTUBODF
157
Practice Problem 4.17
Answer:LΩ
5IFDJSDVJUJO'JHSFQSFTFOUTBOVOCBMBODFECSJEHF*GUIFH
BMWBOPN
FUFSIBTBSFTJTUBODFPGΩ, find the current through the HBMWBOPNFUFS
400 Ω
3 kΩ
40 Ω
a
220 V +
–
G
b
600 Ω
1 kΩ
Figure 4.64
6OCBMBODFECSJEHFPG&YBNQMF
Solution:
8e first need to replace the circuit by its 5IFWFOJOFRVJ WBMFOUBUUFS NJOBMT aBOE b 5IF 5IFWFOJOSFTJTUBODFJTGPVOEVTJOHUIFDJSDVJUJO
'JH a /PUJDFUIBUUIFLΩBOE,ΩSFTJTUPSTBSFJOQBSBMMFM TP
BSFUIFBOE ΩSFTJTUPST 5IFUXPQBSBMMFMDPNCJOBUJPOTGPSNB
TFSJFTDPNCJOBUJPOXJUISFTQFDUUPUFSNJOBMTBBOEC)FODF
35I=∥ + ∥
×
× =@@@@@@@@@@@
= + =Ω
+ @@@@@@@@@
+ + 5o find the 5IFWFOJOW PMUBHF XFDPOTJEFSUIFDJSDVJUJO'JH C 6TJOHUIFWPMUBHFEJWJTJPOQSJODJQMFHJWFT
v=@@@@@@@@@@@
=7 + =7
v=@@@@@@@@@
+ "QQMZJOH,7-BSPVOEMPPQabHJWFT
−v + 75I + v=
PS
75I=v − v= − =−7
)BWJOHEFUFSNJOFEUIF5IFWFOJOFRVJWBMFOU XFfind the current UISPVHI
UIFHBMWBOPNFUFSVTJOH'JH D V5I
− =−N"
*(=@@@@@@@@
=@@@@@@@@
35I + 3N + 5IFOFHBUJWe sign indicates that the current floXTJOUIFEJSFDUJPOPQQP TJUFUPUIFPOFBTTVNFE UIBUJT GSPNUFSNJOBMCUPUFSNJOBMB
Example 4.18
158
Chapter 4
Circuit Theorems
400 Ω
3 kΩ
a
RTh
+
220 V +
–
b
600 Ω
1 kΩ
400 Ω
3 kΩ
1 kΩ
+
v1
–
(a)
a
VTh
–
b
+
v2
–
600 Ω
(b)
RTh
a
IG
40 Ω
VTh +
–
G
b
(c)
Figure 4.65
'PS&YBNQMF B 'JOEJOH35I, (b) finding 75I D EFUFSNJOJOHUIFDVSSFOUUISPVHIUIFHBMWBOPNFUFS
Practice Problem 4.18
20 Ω
30 Ω
0CUBJOUIFDVSSFOUUISPVHIUIFH BMWBOPNFUFS IB WJOHBSFTJTUBODFPG Ω JOUIF8IFBUTUPOFCSJEHFTIPXOJO'JH
Answer:N"
G
14 Ω
40 Ω
60 Ω
16 V
Figure 4.66
'PS1SBDUJDF1SPC
4.11
Summary
"MJOFBSOFUX PSLDPOTJTUTPGMJOFBSFMFNFOUT MJOFBSEFQFOEFOU
TPVSDFT BOEMJOFBSJOEFQFOEFOUTPVSDFT
/FUXPSLUIFPSFNTBSFVTFEUPSFEVDFBDPNQMFYDJSDVJUUPBTJNQMFS
POF UIFSFCZNBLJOHDJSDVJUBOBMZTJTNVDITJNQMFS
5IFTVQFSQPTJUJPOQSJODJQMFTUBUFTUIBUGPSBDJSDVJUIB WJOHNVMUJQMF
JOEFQFOEFOUTPVSDFT UIFWPMUBHFBDSPTT PSDVSSFOUUISPVHI BOFMF
NFOUJTFRVBMUPUIFBMHFCSBJDTVNPGBMMUIFJOEJWJEVBMWPMUBHFT PS
DVSSFOUT EVFUPFBDIJOEFQFOEFOUTPVSDFBDUJOHPOFBUBUJNF
4PVSDFUSBOTGPSNBUJPOJTBQSPDFEVSFGPSUSBOTGPSNJOHBW
PMUBHF
TPVSDFJOTFSJFTXJUIBSFTJTUPSUPBDVSSFOUTPVSDFJOQBSBMMFMXJUIB
SFTJTUPS PSWJDFWFSTB
5IFWFOJOTBOE/PSUPOTUIFPSFNTBMMPXVTUPJTPMBUFBQPSUJPOPGB
OFUXPSLXIJMFUIFSFNBJOJOHQPSUJPOPGUIFOFUX PSLJTSFQMBDFECZ
BOFRVJWBMFOUOFUXPSL5IF5IFWFOJOFRVJWBMFOUDPOTJTUTPGBW PMU
BHFTPVSDF75IJOTFSJFTXJUIBSFTJTUPS35I XIJMFUIF/PSUPOFRVJWB
MFOUDPOTJTUTPGBDVSSFOUTPVSDF*/JOQBSBMMFMXJUIBSFTJTUPS3/5IF
UXPUIFPSFNTBSFSFMBUFECZTPVSDFUSBOTGPSNBUJPO
3/=35I 75I
*/=@@@
35I
159
Review Questions
'PSBHJ WFO5IFWFOJOFRVJWBMFOUDJSDVJU NBYJNVNQP XFSUSBOTGFS
PDDVSTXIFO3-=35IUIBUJT XIFOUIFMPBESFTJTUBODFJTFRVBMUP
UIF5IFWFOJOSFTJTUBODF
5IFNBYJNVNQP XFSUSBOTGFSUIFPSFNTUBUFTUIBUUIFNBYJNVN
QPXFSJTEFMJWFSFECZBTPVSDFUPUIFMPBE3-XIFO3-JTFRVBMUP
35I UIF5IFWFOJOSFTJTUBODFBUUIFUFSNJOBMTPGUIFMPBE
14QJDFDBOCFVTFEUPW FSJGZUIFDJSDVJUUIFPSFNTDP WFSFEJOUIJT
DIBQUFS
4PVSDFNPEFMJOHBOESFTJTUBODFNFBTVSFNFOUVTJOHUIF8IFBUTUPOF
CSJEHFQSPWJEFBQQMJDBUJPOTGPS5IFWFOJOTUIFPSFN
Review Questions
5IFDVSSFOUUISPVHIBCSBODIJOBMJOFBSOFUXPSLJT
"XIFOUIFJOQVUTPVSDFWPMUBHFJT7*GUIFWPMU
BHFJTSFEVDFEUP7BOEUIFQPMBSJUZJTSFWFSTFE UIF
DVSSFOUUISPVHIUIFCSBODIJT
B ¢"
C ¢"
E "
F "
D "
B 5SVF
'PSTVQFSQPTJUJPO JUJTOPUSFRVJSFEUIBUPOMZPOF
JOEFQFOEFOUTPVSDFCFDPOTJEFSFEBUBUJNFBOZ
OVNCFSPGJOEFQFOEFOUTPVSDFTNBZCFDPOTJEFSFE
TJNVMUBOFPVTMZ
B 5SVF
8IJDIQBJSPGDJSDVJUTJO'JHBSFFRVJWBMFOU
B BBOEC
C CBOEE
D BBOED
E DBOEE
20 V +
–
5IFTVQFSQPTJUJPOQSJODJQMFBQQMJFTUPQPXFS
DBMDVMBUJPO
C 'BMTF
4A
(a)
C 'BMTF
3FGFSUP'JH5IF5IFWFOJOSFTJTUBODFBU
UFSNJOBMTBBOECJT
B Ω
C Ω
D Ω
E Ω
5Ω
5Ω
C 'BMTF
B 5SVF
5IF/PSUPOSFTJTUBODF3/JTFYBDUMZFRVBMUPUIF
5IFWFOJOSFTJTUBODF35I
5Ω
4A
(b)
20 V +
–
(c)
5Ω
5Ω
(d)
Figure 4.68
'PS3FWJFX2VFTUJPO
50 V +
–
a
b
20 Ω
Figure 4.67
'PS3FWJFX2VFTUJPOTUP
5IF5IFWFOJOWPMUBHFBDSPTTUFSNJOBMTBBOECPGUIF
DJSDVJUJO'JHJT
B 7
C 7
D 7
E 7
5IF/PSUPODVSSFOUBUUFSNJOBMTBBOECPGUIFDJSDVJU
JO'JHJT
B "
C "
D "
E "
"MPBEJTDPOOFDUFEUPBOFUXPSL"UUIFUFSNJOBMTUP
XIJDIUIFMPBEJTDPOOFDUFE 35I=ΩBOE75I=
75IFNBYJNVNQPTTJCMFQPXFSTVQQMJFEUPUIF
MPBEJT
B 8
C 8
D 8
E 8
5IFTPVSDFJTTVQQMZJOHUIFNBYJNVNQPXFSUP
UIFMPBEXIFOUIFMPBESFTJTUBODFFRVBMTUIFTPVSDF
SFTJTUBODF
B 5SVF
C 'BMTF
"OTXFSTC B C E C B B D D B
160
Chapter 4
Circuit Theorems
Problems
Section 4.2
Linearity Property
$BMDVMBUFUIFDVSSFOUJPJOUIFDJSDVJUPG'JH
8IBUWBMVFPGJOQVUWPMUBHFJTOFDFTTBSZUPNBLFJP
FRVBMUPBNQT
5Ω
'PSUIFDJSDVJUJO'JH BTTVNFvP=7 BOEVTF
linearity to find the actual vBMVFPGvP
2Ω
15 V +
–
25 Ω
vo 2 Ω
3Ω
6Ω
6Ω
4Ω
io
30 V +
–
40 Ω
15 Ω
Figure 4.73
'PS1SPC
Figure 4.69
'PS1SPC
'PSUIFMJOFBSDJSDVJUTIPXOJO'JH VTFMJOFBSJUZ
UPDPNQMFUFUIFGPMMPXJOHUBCMF
&YQFSJNFOU
7T
7P
7
7
7
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEMJOFBSJUZ
R2
I
R1
R4
R3
7
¢7
+
vo
–
R5
Figure 4.70
+
Vo
–
Linear
circuit
Vs +
–
'PS1SPC
B *OUIFDJSDVJUPG'JH DBMDVMBUFvPBOEJP
XIFOvT= 7
C 'JOEvPBOEJPXIFOvT=7
D 8IBUBSFvPBOEJPXIFOFBDIPGUIFΩSFTJTUPST
JTSFQMBDFECZBΩ SFTJTUPSBOEvT=7
Figure 4.74
'PS1SPC
6TFMJOFBSJUZBOEUIFBTTVNQUJPOUIBU7P=7UP
find the actual vBMVFPG7PJO'JH
10 Ω
+ Vx –
1Ω
1Ω
vs
30 Ω
1Ω
+
–
1Ω
+
vo
–
io
1Ω
'PS1SPC
3Ω
6Ω
4Ω
5Ω
4Ω
Superposition
Using superposition, find 7PJOUIFDJSDVJUPG'JH
$IFDLXJUI14QJDFPS.VMUJ4JN
2Ω
io
'PS1SPC
'PS1SPC
6TFMJOFBSJUZUPEFUFSNJOFJPJOUIFDJSDVJUPG'JH
Figure 4.72
Figure 4.75
Section 4.3
Figure 4.71
40 Ω
1A
9A
Figure 4.76
'PS1SPC
Vo
1Ω
3Ω
+ 9V
–
+ 3V
–
161
Problems
(JWFOUIBU*=BNQTXIFO7T=WPMUTBOE
*T=−BNQTBOE*=BNQXIFO7T=WPMUTBOE
*T= VTFTVQFSQPTJUJPOBOEMJOFBSJUZUPEFUFSNJOFUIF
WBMVFPG*XIFO7T=WPMUTBOE*T=BNQT
Use superposition to find vPJOUIFDJSDVJUPG'JH
4A
8Ω
Is
I
12 V
10 Ω
2A
Figure 4.77
–+
Vs +
–
5Ω
+
vo
–
Figure 4.81
'PS1SPC
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETVQFSQPTJUJPO/PUF UIF
MFUUFSLJTBHBJOZPVDBOTQFDJGZUPNBLFUIFQSPCMFN
FBTJFSUPTPMWFCVUNVTUOPUCF[FSP
Apply the superposition principle to find vPJOUIF
DJSDVJUPG'JH
6Ω
2A
R
kVab
+ –
V +
–
+
I
a
4Ω
Vab
–
20 V +
–
b
Figure 4.78
+
vo
–
1A
3Ω
Figure 4.82
'PS1SPC
'PS1SPC
Use the superposition principle to find JPBOEvPJO
UIFDJSDVJUPG'JH
io 10 Ω
40 Ω
'or the circuit in Fig. 4.83, use superposition to find
J$BMDVMBUFUIFQPXFSEFMJWFSFEUPUIFΩSFTJTUPS
20 Ω
+ vo –
6A
2Ω
4io
–
+ 30 V
Figure 4.79
1Ω
20 V +
–
2A
i
2Ω
4Ω
–
+ 16 V
3Ω
Figure 4.83
'PS1SPC
'PS1SPCTBOE
%FUFSNJOFvPJOUIFDJSDVJUPG'JHVTJOHUIF
TVQFSQPTJUJPOQSJODJQMF
(JWFOUIFDJSDVJUJO'JH VTFTVQFSQPTJUJPOUP
PCUBJOJP
2A
36 A
6Ω
5Ω
io
4Ω
4Ω
3Ω
2Ω
10 Ω
5Ω
+ v –
o
12 V +
–
Figure 4.80
'PS1SPC
3Ω
12 Ω
+ 19 V
–
108 V
+
–
Figure 4.84
'PS1SPC
18 A
162
Chapter 4
Circuit Theorems
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETPVSDFUSBOTGPSNBUJPO
6TFTVQFSQPTJUJPOUPPCUBJOvYJOUIFDJSDVJUPG
'JH$IFDLZPVSSFTVMUVTJOH14QJDFPS
.VMUJ4JN
30 Ω
10 Ω
io
20 Ω
+ vx –
90 V +
–
60 Ω
30 Ω
6A
+
–
R1
V +
–
40 V
R2
+
vo
–
I
Figure 4.89
Figure 4.85
'PS1SPC
'PS1SPC
Use superposition to find 7PJOUIFDJSDVJUPG'JH
'PSUIFDJSDVJUJO'JH VTFTPVSDF
transformation to find J
10 Ω
– +
2 Vo
5Ω
+
Vo
–
3A
10 Ω
i
10 Ω
5Ω
2A
4A
4Ω
+ 20 V
–
Figure 4.90
Figure 4.86
'PS1SPC
'PS1SPC
6TFTVQFSQPTJUJPOUPTPMWFGPSvYJOUIFDJSDVJUPG
'JH
ix
2Ω
8Ω
10 A –4 A
3FGFSSJOHUP'JH VTFTPVSDFUSBOTGPSNBUJPOUP
EFUFSNJOFUIFDVSSFOUBOEQPXFSBCTPSCFECZUIF
Ω SFTJTUPS
8Ω
+
vx
–
3A
– +
10 Ω
3Ω
6Ω
+ 15 V
–
4ix
Figure 4.87
Figure 4.91
'PS1SPC
Section 4.4
'PS1SPC
Source Transformation
Use source transformation to find the vPMUBHF7YJO
UIFDJSDVJUPG'JH
6TFTPVSDFUSBOTGPSNBUJPOUPSFEVDFUIFDJSDVJU
CFUXFFOUFSNJOBMTaBOEbTIPXOJO'JHUPB
TJOHMFWPMUBHFTPVSDFJOTFSJFTXJUIBTJOHMFSFTJTUPS
20 Ω
20 V +
–
10 Ω
20 V +
–
3A
a
20 Ω
1A
8Ω
20 Ω
40 V +
–
+ 30 V
–
b
Figure 4.88
'PS1SPC
Figure 4.92
'PS1SPC
10 Ω
+ Vx –
10 Ω
2Vx
163
Problems
0CUBJOvPJOUIFDJSDVJUPG'JHVTJOHTPVSDF
USBOTGPSNBUJPO$IFDLZPVSSFTVMUVTJOH14QJDFPS
.VMUJ4JN
Use source transformation to find vPJOUIFDJSDVJUPG
'JH
2A
4 kΩ
4Ω
3A
5Ω
+ vo –
– +
1 kΩ
3 mA
6A
+–
2Ω
3vo
2 kΩ
9Ω
+
vo
–
Figure 4.97
'PS1SPC
30 V
Figure 4.93
'PS1SPC
Use source transformation to find JPJOUIFDJSDVJUPG
'JH
6TFTPVSDFUSBOTGPSNBUJPOPOUIFDJSDVJUTIPXOJO
Fig 4.98 to find JY
ix
24 Ω
60 Ω
5Ω
3A
io
+
–
2Ω
6A
12 V
4Ω
20 V
Figure 4.94
+
–
30 Ω
10 Ω
0.7ix
Figure 4.98
'PS1SPC
%FUFSNJOFvYJOUIFDJSDVJUPG'JHVTJOHTPVSDF
USBOTGPSNBUJPO
'PS1SPC
Apply source transformation to find vYJOUIFDJSDVJU
PG'JH
10 Ω
a
12 Ω
b
20 Ω
40 Ω
8A
+
–
40 V
6Ω
+ vx –
30 V +
–
+ vx –
50 V +
–
3Ω
+
–
8Ω
2vx
Figure 4.99
'PS1SPC
Figure 4.95
Use source transformation to find JYJOUIFDJSDVJUPG
'JH
'PS1SPCTBOE
Use source transformation to find *PJO'JH
1Ω
Io
10 Ω
4Ω
ix
+ Vo –
8V +
–
Figure 4.96
'PS1SPC
3Ω
1
V
3 o
60 V +
–
Figure 4.100
'PS1SPC
15 Ω
0.5ix
50 Ω
40 Ω
164
Chapter 4
Sections 4.5 and 4.6
Circuit Theorems
Thevenin’s and Norton’s
Theorems
'JOEUIF/PSUPOFRVJWBMFOUXJUISFTQFDUUPUFSNJOBMT
BCJOUIFDJSDVJUTIPXOJO'JH
%FUFSNJOFUIF5IFWFOJOFRVJWBMFOUDJSDVJU TIPXOJO
'JH BTTFFOCZUIFPINSFTJTUPS
Then calculate the current floXJOHUISPVHIUIFPIN
SFTJTUPS
+–
a
10 kΩ
6 mA
13 Ω
30 V
20 kΩ
6 kΩ
60 Ω
b
7Ω
Figure 4.104
+ 240 V
–
30 Ω
'PS1SPC
"QQMZ5IFWFOJOs theorem to find 7PJOUIFDJSDVJUPG
'JH
Figure 4.101
'PS1SPC
1Ω
4Ω
6TJOH'JH EFTJHOBQSPCMFNUIBUXJMMIFMQ
PUIFSTUVEFOUTCFUUFSVOEFSTUBOE5IFWFOJOFRVJWBMFOU
DJSDVJUT
5Ω
16 Ω
2.5 A
+
–
10 Ω
50 V
+
Vo
–
Figure 4.105
'PS1SPC
I
R1
V +
–
R3
0CUBJOUIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTBCPG
UIFDJSDVJUTIPXOJO'JH
a
5A
R2
10 Ω
b
Figure 4.102
16 Ω
a
10 Ω
'PS1SPCTBOE
5Ω
30 V +
–
b
6TF5IFWFOJOs theorem to find vPJO1SPC
4PMWFGPSUIFDVSSFOUiJOUIFDJSDVJUPG'JH
VTJOH5IFWFOJOTUIFPSFN )JOU'JOEUIF5IFWFOJO
FRVJWBMFOUTFFOCZUIFΩSFTJTUPS
Figure 4.106
'PS1SPC
'JOEUIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTBCPGUIF
DJSDVJUJO'JH
+V –
o
i
10 Ω
150 V +
–
Figure 4.103
'PS1SPC
12 Ω
+ 60 V
–
10 kΩ
40 Ω
70 V
+
–
Figure 4.107
'PS1SPC
20 kΩ
a
b
+
–
4Vo
165
Problems
'JOEUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUTBU
UFSNJOBMTBCPGUIFDJSDVJUTIPXOJO'JH
14 V
14 Ω
–+
1A
6Ω
a
3A
6Ω
'JOEUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO
'JHBTTFFOCZMPPLJOHJOUPUFSNJOBMTBBOEC
6Ω
10 A
5Ω
a
4Ω
b
b
Figure 4.112
'PS1SPC
Figure 4.108
'PS1SPC
'or the circuit in Fig. 4.109, find the 5IFWFOJO
FRVJWBMFOUCFUXFFOUFSNJOBMTBBOEC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOE/PSUPOFRVJWBMFOUDJSDVJUT
20 Ω
–
+ 20 V
20 Ω
10 Ω
a
R2
b
I
a
R1
R3
10 Ω
b
10 Ω
5A
10 Ω
30 V +
–
Figure 4.113
'PS1SPC
0CUBJOUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUT
PGUIFDJSDVJUJO'JHXJUISFTQFDUUPUFSNJOBMT
BBOEC
Figure 4.109
'PS1SPC
'JOEUIF5IFWFOJOFRVJWBMFOUMPPLJOHJOUPUFSNJOBMT
BCPGUIFDJSDVJUJO'JHBOETPMWFGPSJY
10 Ω
20 V +
–
6Ω
a
10 Ω
a
b
Iv
ix
20 Ω
1A
5Ω
2A
20 Ω
+
–
20Iv
b
Figure 4.114
Figure 4.110
'PS1SPC
'PS1SPC
'PSUIFDJSDVJUJO'JH PCUBJOUIF5IFWFOJO
FRVJWBMFOUBTTFFOGSPNUFSNJOBMT
B BC
C CD
3Ω
+
24 V –
10io
1Ω
a
io
4Ω
b
2Ω
%FUFSNJOFUIF/PSUPOFRVJWBMFOUBUUFSNJOBMTBCGPS
UIFDJSDVJUJO'JH
5Ω
'PS1SPC
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
2Ω
a
4Ω
b
2A
c
Figure 4.111
5A
+ –
Figure 4.115
'PS1SPC
'JOEUIF/PSUPOFRVJWBMFOUMPPLJOHJOUPUFSNJOBMT
BCPGUIFDJSDVJUJO'JH-FU7=7 *=" 3=Ω 3= Ω BOE3= Ω
166
Chapter 4
Circuit Theorems
0CUBJOUIF/PSUPOFRVJWBMFOUPGUIFDJSDVJUJO
'JHUPUIFMFGUPGUFSNJOBMTBC6TFUIFSFTVMU
to find current J
6Ω
12 V
1 kΩ
a
+–
i
5Ω
4Ω
2A
'JOEUIF5IFWFOJOFRVJWBMFOUCFUXFFOUFSNJOBMTBC
PGUIFDJSDVJUJO'JH
4A
3V +
–
Io
a
+
–
2Vx
40Io +
Vx
–
50 Ω
b
Figure 4.120
b
'PS1SPC
Figure 4.116
'PS1SPC
(JWFOUIFDJSDVJUJO'JH PCUBJOUIF/PSUPO
FRVJWBMFOUBTWJFXFEGSPNUFSNJOBMT
B Bb
0CUBJOUIF/PSUPOFRVJWBMFOUBUUFSNJOBMTBCPGUIF
DJSDVJUJO'JH
C DE
a
6Ω
120 V +
–
b
3Ω
I
8 kΩ
4Ω
c
6A
2Ω
2V +
–
0.001Vab
+
–
80I
50 kΩ
a
+
Vab
–
b
d
Figure 4.121
'PS1SPC
Figure 4.117
'PS1SPC
'PSUIFUSBOTJTUPSNPEFMJO'JH PCUBJOUIF
5IFWFOJOFRVJWBMFOUBUUFSNJOBMTBC
3 kΩ
a
io
+
21 V –
6TF/PSUPOs theorem to find 7PJOUIFDJSDVJUPG
'JH
20io
12 kΩ
2 kΩ
10 kΩ
2 kΩ
b
+
120 V +
–
24 kΩ
10 mA 20 kΩ
Figure 4.118
'PS1SPC
Vo
–
Figure 4.122
'JOEUIF/PSUPOFRVJWBMFOUBUUFSNJOBMTBCPGUIF
DJSDVJUJO'JH
+ Vo –
10 Ω
a
'PS1SPC
0CUBJOUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUT
BUUFSNJOBMTBCGPSUIFDJSDVJUJO'JH
20 Ω
2 Vo
+
–
50 V +
–
6Ω
b
Figure 4.119
'PS1SPC
2Ω
3Ω
3A
+
vx
–
0.5vx
a
10 Ω
b
Figure 4.123
'PS1SPCTBOE
167
Problems
5IFOFUXPSLJO'JHNPEFMTBCJQPMBSUSBOTJTUPS
common-emitter amplifier connected to a load. Find
UIF5IFWFOJOSFTJTUBODFTFFOCZUIFMPBE
ib
vs
0.1io
bib
R1
+
–
'JOEUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO
'JH
R2
10 Ω
RL
a
+
vo
–
io
Figure 4.124
40 Ω
'PS1SPC
20 Ω
+–
%FUFSNJOFUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUTBU
UFSNJOBMTBCPGUIFDJSDVJUJO'JH
b
2vo
Figure 4.128
'PS1SPC
20 Ω
10 Ω
'JOEUIF/PSUPOFRVJWBMFOUGPSUIFDJSDVJUJO
'JH
a b
8A
50 Ω
40 Ω
10 Ω
Figure 4.125
'PS1SPCTBOE
+
vo
–
'or the circuit in Fig. 4.126, find the 5IFWFOJOBOE
/PSUPOFRVJWBMFOUDJSDVJUTBUUFSNJOBMTBC
2A
20 Ω
0.5vo
Figure 4.129
'PS1SPC
18 V
+–
a
4Ω
6Ω
3A
0CUBJOUIF5IFWFOJOFRVJWBMFOUTFFOBUUFSNJOBMTBC
PGUIFDJSDVJUJO'JH
b
4Ω
5Ω
1Ω
10ix
10 V
Figure 4.126
+
–
2Ω
'PS1SPCTBOE
b
0CUBJOUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUT
BUUFSNJOBMTBCPGUIFDJSDVJUJO'JH
2Ω
12 V +
–
6Ω
2Ω
a
6Ω
6Ω
–
+ 12 V
Figure 4.130
'PS1SPC
'PSUIFDJSDVJUTIPXOJO'JH EFUFSNJOFUIF
SFMBUJPOTIJQCFUXFFO7PBOE*o
+ 12 V
–
4Ω
2Ω
32 V +
–
b
Figure 4.127
'PS1SPC
a
ix
+–
Figure 4.131
'PS1SPC
2Ω
12 Ω
Io
+
Vo
–
168
Chapter 4
Section 4.8
Circuit Theorems
%FUFSNJOFUIFNBYJNVNQPXFSEFMJWFSFEUPUIFWBSJ
BCMFSFTJTUPS3TIPXOJOUIFDJSDVJUPG'JH
Maximum Power Transfer
'JOEUIFNBYJNVNQPXFSUIBUDBOCFEFMJWFSFEUP
UIFSFTJTUPS3JOUIFDJSDVJUPG'JH
30 V
2Ω
3Ω
–+
R
2 Vx
60 V +
–
5Ω
+
Figure 4.132
Vx
'PS1SPC
5IFWBSJBCMFSFTJTUPS3JO'JHJTBEKVTUFEVOUJM
JUBCTPSCTUIFNBYJNVNQPXFSGSPNUIFDJSDVJU
B $BMDVMBUFUIFWBMVFPG3GPSNBYJNVNQPXFS
C %FUFSNJOFUIFNBYJNVNQPXFSBCTPSCFECZ3
80 kΩ
10 Ω
– +
3A
18 A
–
20 Ω
10 Ω
20 Ω
R
Figure 4.136
'PS1SPC
20 kΩ
100 V
R
+–
10 kΩ
'PSUIFDJSDVJUJO'JH XIBUSFTJTUPSDPOOFDUFE
BDSPTTUFSNJOBMTBCXJMMBCTPSCNBYJNVNQPXFS
GSPNUIFDJSDVJU 8IBUJTUIBUQPXFS
90 kΩ
Figure 4.133
'PS1SPC
$POTJEFSUIFΩSFTJTUPSJO'JH'JSTUDPNQVUF
UIF5IFWFOJOFRVJWBMFOUDJSDVJUBTTFFOCZUIFƗ
SFTJTUPS$PNQVUFUIFWBMVFPGRUIBUSFTVMUTJO5IFWFOJO
FRVJWBMFOUSFTJTUBODFFRVBMUPUIFΩSFTJTUBODFBOE
UIFODBMDVMBUFQPXFSEFMJWFSFEUPUIFΩSFTJTUPS
/PXMFU3=Ɨ Ω BOE∞ DBMDVMBUFUIFQPXFS
EFMJWFSFEUPUIFΩSFTJTUPSJOFBDIDBTF8IBUDBO
ZPVTBZBCPVUUIFWBMVFPG3UIBUXJMMSFTVMUJOUIFNBYJ
NVNQPXFSUIBUDBOCFEFMJWFSFEUPUIFΩSFTJTUPS
–+
3A
R
30 V
10 Ω
30 Ω
3 kΩ
8V +
–
10 kΩ
+
vo
–
1 kΩ
–
+
120vo
a
40 kΩ
b
Figure 4.137
'PS1SPC
B 'PSUIFDJSDVJUJO'JH PCUBJOUIF5IFWFOJO
FRVJWBMFOUBUUFSNJOBMTBC
C $BMDVMBUFUIFDVSSFOUJO3-=Ω
D 'JOE3-GPSNBYJNVNQPXFSEFMJWFSBCMFUP3-
E %FUFSNJOFUIBUNBYJNVNQPXFS
60 Ω
Figure 4.134
'PS1SPC
'JOEUIFNBYJNVNQPXFSUSBOTGFSSFEUPSFTJTUPS3JO
UIFDJSDVJUPG'JH
10 kΩ
100 V +
–
+
𝜐o
–
2A
4Ω
22 kΩ
4A
40 kΩ 0.006𝜐o
30 kΩ
R
6Ω
RL
2Ω
+–
20 V
Figure 4.135
'PS1SPC
Figure 4.138
'PS1SPC
a
b
169
Problems
%FUFSNJOFUIFNBYJNVNQPXFSUIBUDBOCFEFMJWFSFE
UPUIFWBSJBCMFSFTJTUPS3JOUIFDJSDVJUPG'JH
10 Ω
60 V
25 Ω
20 Ω
'PSUIFDJSDVJUJO'JH VTF14QJDFPS.VMUJ4JN
to find the 5IFWFOJOFRVJWBMFOUBUUFSNJOBMTBC
Section 4.10
R
+
–
6TF14QJDFPS.VMUJ4JN to find the 5IFWFOJO
FRVJWBMFOUDJSDVJUBUUFSNJOBMTBCPGUIFDJSDVJU
JO'JH
5Ω
Figure 4.139
'PS1SPC
'PSUIFCSJEHFDJSDVJUTIPwn in Fig. 4.140, find the
MPBE3-GPSNBYJNVNQPXFSUSBOTGFSBOEUIFNBYJ
NVNQPXFSBCTPSCFECZUIFMPBE
Applications
"OBVUPNPCJMFCBUUFSZIBTBOPQFODJSDVJUWPMUBHF
PG7XIJDIESPQTUP7XIFODPOOFDUFEUP
UXP8IFBEMJHIUT8IBUJTUIFSFTJTUBODFPGUIF
IFBEMJHIUTBOEUIFWBMVFPGUIFJOUFSOBMSFTJTUBODFPG
UIFCBUUFSZ
5IFGPMMPXJOHSFTVMUTXFSFPCUBJOFEGSPN
NFBTVSFNFOUTUBLFOCFUXFFOUIFUXPUFSNJOBMTPGB
SFTJTUJWFOFUXPSL
5FSNJOBM7PMUBHF
5FSNJOBM$VSSFOU
7
"
7
"
'JOEUIF5IFWFOJOFRVJWBMFOUPGUIFOFUXPSL
R1
vs
RL
+
–
R3
R4
R2
Figure 4.140
'PS1SPC
'PSUIFDJSDVJUJO'JH EFUFSNJOFUIFWBMVFPG
3TVDIUIBUUIFNBYJNVNQPXFSEFMJWFSFEUPUIF
MPBEJTN8
8IFODPOOFDUFEUPBΩSFTJTUPS BCBUUFSZIBTB
UFSNJOBMWPMUBHFPG7CVUQSPEVDFT7PO
BOPQFODJSDVJU%FUFSNJOFUIF5IFWFOJOFRVJWBMFOU
DJSDVJUGPSUIFCBUUFSZ
5IF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTabPGUIF
MJOFBSOFUXPSLTIPXOJO'JHJTUPCFEFUFS
NJOFECZNFBTVSFNFOU8IFOBLΩSFTJTUPSJT
DPOOFDUFEUPUFSNJOBMTa-b UIFWPMUBHF7BCJTNFB
TVSFEBT78IFOBLΩSFTJTUPSJTDPOOFDUFEUP
UIFUFSNJOBMT 7BCJTNFBTVSFEBT7%FUFSNJOF
B UIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTBC C 7BC
XIFOBLΩSFTJTUPSJTDPOOFDUFEUPUFSNJOBMTBC
R
a
Linear
R
network
R
1V +
–
+ 2V + 3V
–
–
b
RL
Figure 4.141
'PS1SPC
Section 4.9
Verifying Circuit Theorems
with PSpice
Figure 4.142
'PS1SPC
"CMBDLCPYXJUIBDJSDVJUJOJUJTDPOOFDUFEUPBWBSJ
BCMFSFTJTUPS"OJEFBMBNNFUFS XJUI[FSPSFTJTUBODF BOEBOJEFBMWoltmeter (with infinite resistance) are
VTFEUPNFBTVSFDVSSFOUBOEWPMUBHFBTTIPXOJO
'JH5IFSFTVMUTBSFTIPXOJOUIFUBCMFPO
UIFOFYUQBHF
i
4PMWF1SPCVTJOH14QJDFPS.VMUJ4JN-FU
7= 7 *=" 3= Ω 3=Ω BOE
3=Ω
A
Black
box
6TF14QJDFPS.VMUJ4JNUPTPMWF1SPC
6TF14QJDFPS.VMUJ4JNUPTPMWF1SPC
0CUBJOUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO
'JHVTJOH14QJDFPS.VMUJ4JN
Figure 4.143
'PS1SPC
V
R
170
Chapter 4
Circuit Theorems
B 'JOEiXIFO3=Ω
C %FUFSNJOFUIFNBYJNVNQPXFSGSPNUIFCPY
3Ω
77
J"
"USBOTEVDFSJTNPEFMFEXJUIBDVSSFOUTPVSDF*TBOE
BQBSBMMFMSFTJTUBODF3T5IFDVSSFOUBUUIFUFSNJOBMT
PGUIFTPVSDFJTNFBTVSFEUPCFN"XIFOBO
BNNFUFSXJUIBOJOUFSOBMSFTJTUBODFPGΩJTVTFE
B *GBEEJOHBLΩSFTJTUPSBDSPTTUIFTPVSDF
UFSNJOBMTDBVTFTUIFBNNFUFSSFBEJOHUPGBMMUP
N" DBMDVMBUF*TBOE3T
C 8IBUXJMMUIFBNNFUFSSFBEJOHCFJGUIF
SFTJTUBODFCFUXFFOUIFTPVSDFUFSNJOBMTJT
DIBOHFEUPLΩ
$POTJEFSUIFDJSDVJUJO'JH"OBNNFUFSXJUI
JOUFSOBMSFTJTUBODF3JJTJOTFSUFECFUXFFO"BOE#UP
NFBTVSF*P%FUFSNJOFUIFSFBEJOHPGUIFBNNFUFSJG
B 3J=Ω C 3J=Ω )JOU'JOEUIF
5IFWFOJOFRVJWBMFOUDJSDVJUBUUFSNJOBMTBC
5IF8IFBUTUPOFCSJEHFDJSDVJUTIPXOJO'JHJT
VTFEUPNFBTVSFUIFSFTJTUBODFPGBTUSBJOHBVHF5IF
BEKVTUBCMFSFTJTUPSIBTBMJOFBSUBQFSXJUIBNBYJNVN
WBMVFPGΩ*GUIFSFTJTUBODFPGUIFTUSBJOHBVHF
JTGPVOEUPCFΩ XIBUGSBDUJPOPGUIFGVMMTMJEFS
USBWFMJTUIFTMJEFSXIFOUIFCSJEHFJTCBMBODFE
Rs
2 kΩ
vs +
–
4 kΩ
G
100 Ω
Rx
Figure 4.146
'PS1SPC
B *OUIF8IFBUTUPOFCSJEHFDJSDVJUPG'JH
TFMFDUUIFWBMVFTPG3BBOE3CTVDIUIBUUIFCSJEHF
DBONFBTVSF3YJOUIFSBOHFPGmΩ
C 3FQFBUGPSUIFSBOHFPGmΩ
+–
a
2 kΩ
Io
30 kΩ
4 mA
V
5 kΩ
b
Ra
150 Ω
20 kΩ
+ 60 V
–
G
Rx
Rb
10 kΩ
Figure 4.147
Figure 4.144
'PS1SPC
'PS1SPC
$POTJEFSUIFDJSDVJUJO'JH B 3FQMBDFUIF
SFTJTUPS3-CZB[FSPSFTJTUBODFBNNFUFSBOEEFUFS
NJOFUIFBNNFUFSSFBEJOH C 5PWFSJGZUIFSFDJQSPD
JUZUIFPSFN JOUFSDIBOHFUIFBNNFUFSBOEUIF7
TPVSDFBOEEFUFSNJOFUIFBNNFUFSSFBEJOHBHBJO
$POTJEFSUIFCSJEHFDJSDVJUPG'JH*TUIF
CSJEHFCBMBODFE *GUIFLΩSFTJTUPSJTSFQMBDFECZ
BOLΩSFTJTUPS XIBUSFTJTUPSDPOOFDUFECFUXFFO
UFSNJOBMTabBCTPSCTUIFNBYJNVNQPXFS 8IBUJT
UIJTQPXFS
2 kΩ
6 kΩ
3 kΩ
10 kΩ
12 kΩ
'PS1SPC
220 V +
–
RL
12 V +
–
Figure 4.145
20 kΩ
a
5 kΩ
15 kΩ
Figure 4.148
'PS1SPC
b
10 kΩ
171
Comprehensive Problems
Comprehensive Problems
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vs +
–
R
βix
Ro
+ V –
o
3
Figure 4.149
10 Ω
'PS1SPC
60 Ω
"OBUUFOVBUPSJTBOJOUFSGBDFDJSDVJUUIBUSFEVDFTUIF
WPMUBHFMFWFMXJUIPVUDIBOHJOHUIFPVUQVUSFTJTUBODF
B #ZTQFDJGZJOH3TBOE3QPGUIFJOUFSGBDFDJSDVJUJO
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7
@@@
P= 7H
Rp
Attenuator
Figure 4.150
+
Vo
–
RL
4
10 Ω
40 Ω
1
+ 9V –
Figure 4.151
'PS1SPC
RL
6 kΩ
+
15 V
–
B
Load
Req
3 kΩ
'PS1SPC
E
"EDWPMUNFUFSXJUIBTFOTJUJWJUZPGLΩ∕7JTVTFE
to find the 5IFWFOJOFRVJWBMFOUPGBMJOFBSOFUXPSL
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A common-emitter amplifier circuit is shoXOJO
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Vg +
–
8Ω
3FR=35I=3H=Ω
C 6TJOHUIFJOUFSGBDFEFTJHOFEJOQBSU B DBMDVMBUF
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7H=7
Rg
8Ω
10 Ω
C m7TDBMF7
0CUBJOUIF5IFWFOJOWPMUBHFBOEUIF5IFWFOJO
SFTJTUBODFPGUIFOFUXPSL
Rc
Figure 4.152
'PS1SPC
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5
Operational
Amplifiers
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173
174
Chapter 5
Operational Amplifiers
Learning Objectives
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5.1
Introduction
The term operational amplifier was
introduced in 1947 by John Ragazzini
and his colleagues, in their work on
analog computers for the National
Defense Research Council after World
War II. The first op amps used vacuum
tubes rather than transistors.
)BWJOHMFBSOFEUIFCBTJDMB XTBOEUIFPSFNTGPSDJSDVJUBOBMZTJT XFBSF
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UIFPQFSational amplifier PSPQBNQGPSTIPSU5IFPQBNQJTBWFSTBUJMF
DJSDVJUCVJMEJOHCMPDL
An op amp may also be regarded as a
voltage amplifier with very high gain.
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EJGGFSFOUJBUFJU5IFBCJMJUZPGUIFPQBNQUPQFSGPSNUIFTFNBUIFNBUJDBM
PQFSBUJPOTJTUIFSFBTPOJUJTDBMMFEBOPQFSBUJPOBMamplifier*UJTBMTPUIF
SFBTPOGPSUIFXJEFTQSFBEVTFPGPQBNQTJOBOBMPHEFTJHO0QBNQTBSF
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TJWF FBTZUPVTF BOEGVOUPXPSLXJUI
8FCFHJOCZEJTDVTTJOHUIFJEFBMPQBNQBOEMBUFSDPOTJEFSUIFOPO
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DJSDVJUT TVDIBTUIFJO WFSUFS WPMUBHFGPMMPXFS TVNNFS BOEEJG GFSFODF
amplifier8FXJMMBMTPBOBMZ[FPQBNQDJSDVJUTXJUI14QJDF'JOBMMZ XF
MFBSOIPXBO PQBNQJT VTFEJO EJHJUBMUPBOBMPHDPOWFSUFSTBOEJOTUSV
mentation amplifiers.
The op amp is an electronic unit that behaves like a voltage-controlled
voltage source.
5.2
Operational Amplifiers
An operational amplifier is designed so that it performs some mathemati
DBMPQFSBUJPOTXIFOF YUFSOBMDPNQPOFOUT TVDIBTSFTJTUPSTBOEDBQBDJ UPST BSFDPOOFDUFEUPJUTUFSNJOBMT5IVT An op amp is an active circuit element designed to perform mathematical operations of addition, subtraction, multiplication, division, differentiation, and integration.
5IFPQBNQJTBOFMFDUSPOJDEFWJDFDPOTJTUJOHPGBDPNQMFYBSSBOHF
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5.2
175
Operational Amplifiers
UPUSFBUUIFPQBNQBTBDJSDVJUC VJMEJOHCMPDLBOETJNQMZTUVEZXIBU
UBLFTQMBDFBUJUTUFSNJOBMT
0QBNQTBSFDPNNFSDJBMMZB WBJMBCMFJOJOUFHSBUFEDJSDVJUQBDLBHFT
JOTFWFSBMGPSNT'JHVSFTIP XTBUZQJDBMPQBNQQBDLBHF "UZQJDBM
POFJTUIFFJHIUQJOEVBMJOMJOFQBDLBHF PS%*1 TIPXOJO'JH B 1JOPSUFSNJOBMJTVOVTFE BOEUFSNJOBMTBOEBSFPGMJUUMFDPODFSOUP
VTThe fivFJNQPSUBOUUFSNJOBMTBSF
5IFJOWFSUJOHJOQVU QJO
5IFOPOJOWFSUJOHJOQVU QJO
5IFPVUQVU QJO
5IFQPTJUJWFQPXFSTVQQMZ7 + QJO
5IFOFHBUJWFQPXFSTVQQMZ7 − QJO
Figure 5.1
A typical operational amplifier
¥.D(SBX)JMM&EVDBUJPO.BSL%JFSLFS QIPUPHSBQIFS
The pin diagram in Fig. 5.2(a)
corresponds to the 741 generalpurpose op amp made by Fairchild
Semiconductor.
5IFDJSDVJUTZNCPMGPSUIFPQBNQJTUIFUSJBOHMFJO'JH C BTTIP
XO UIFPQBNQIBTUX PJOQVUTBOEPOFPVUQVU 5IFJOQVUTBSFNBSL FEXJUI
NJOVT − BOEQMVT + UPTQFDJGZJOWFSUJOHBOEOPOJOWFSUJOHJOQVUT SF
TQFDUJWFMZ"OJOQVUBQQMJFEUPUIFOPOJOWFSUJOHUFSNJOBMXJMMBQQFBSXJUI
UIFTBNFQPMBSJUZBUUIFPVUQVU XIJMFBOJOQVUBQQMJFEUPUIFJO WFSUJOH
UFSNJOBMXJMMBQQFBSJOWFSUFEBUUIFPVUQVU
"TBOBDUJWFFMFNFOU UIFPQBNQNVTUCFQPXFSFECZBWPMUBHFTVQ
QMZBTUZQJDBMMZTIPXOJO'JH"MUIPVHIUIFQPXFSTVQQMJFTBSFPGUFO
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TVQQMZDVSSFOUTNVTUOPUCFPWFSMPPLFE#Z,$
JP=J+J+J++J−
5IFFRVJWBMFOUDJSDVJUNPEFMPGBOPQBNQJTTIPXOJO'JH5IF
PVUQVUTFDUJPODPOTJTUTPGBW PMUBHFDPOUSPMMFETPVSDFJOTFSJFTXJUIUIF
V+
Balance
1
8
No connection
+
Inverting input
2
7
V
Noninverting input
3
6
Output
V–
4
5
Balance
Inverting input 2
–
Noninverting input 3
+
7
6 Output
415
V–
Offset Null
(b)
(a)
Figure 5.2
A typical op amp: (a) pin configuration, (b) circuit symbol.
v1
i1
2
7
i+
io
6
3
i2
4
i–
Figure 5.3
1PXFSJOHUIFPQBNQ
–
vd
+
+
VCC
–
+
VCC
–
Ri
Ro
+
–
vo
Avd
v2
Figure 5.4
5IFFRVJWalent circuit of the nonideal op amp.
176
Chapter 5
Operational Amplifiers
PVUQVUSFTJTUBODF3P*UJTFWJEFOUGSPN'JHUIBUUIFJOQVUSFTJTUBODF
3JJTUIF5IFWFOJOFRVJWBMFOUSFTJTUBODFTFFOBUUIFJOQVUUFSNJOBMT XIJMF
UIFPVUQVUSFTJTUBODF3PJTUIF5IFWFOJOFRVJWBMFOUSFTJTUBODFTFFOBUUIF
PVUQVU5IFEJGGFSFOUJBMJOQVUWPMUBHFvEJTHJWFOCZ
vE=v−v
XIFSFvJTUIFWPMUBHFCFUXFFOUIFJOWFSUJOHUFSNJOBMBOEHSPVOEBOEv
JTUIFWPMUBHFCFUXFFOUIFOPOJOWFSUJOHUFSNJOBMBOEHSPVOE5IFPQBNQ
TFOTFTUIFEJGGFSFODFCFUXFFOUIFUXPJOQVUT NVMUJQMJFTJUCZUIFHBJO" BOEDBVTFTUIFSFTVMUJOHWPMUBHFUPBQQFBSBUUIFPVUQVU5IVT UIFPVUQVU
vPJTHJWFOCZ
Sometimes, voltage gain is expressed
in decibels (dB), as discussed in
Chapter 14.
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WP="WE=" W−W
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5IFDPODFQUPGGFFECBDLJTDSVDJBMUPPVSVOEFSTUBOEJOHPGPQBNQ
DJSDVJUT"OFHBUJWFGFFECBDLJTBDIJFWFEXIFOUIFPVUQVUJTGFECBDLUP
UIFJOWFSUJOHUFSNJOBMPGUIFPQBNQ"T&YBNQMFTIPXT XIFOUIFSF
JTBGFFECBDLQBUIGSPNPVUQVUUPJOQVU UIFSBUJPPGUIFPVUQVUWPMUBHFUP
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VTFEJODJSDVJUTXJUIGFFECBDLQBUIT
"QSBDUJDBMMJNJUBUJPOPGUIFPQBNQJTUIBUUIFNBHOJUVEFPGJUT
PVUQVUWPMUBHFDBOOPUF YDFFE]7$$ ]*OPUIFSX PSET UIFPVUQVUW PMUBHF
JTEFQFOEFOUPOBOEJTMJNJUFECZUIFQP XFSTVQQMZWPMUBHF'JHVSF
JMMVTUSBUFTUIBUUIFPQBNQDBOPQFSBUFJOUISFFNPEFT EFQFOEJOHPOUIF
EJGGFSFOUJBMJOQVUWPMUBHFvE 1PTJUJWFTBUVSBUJPO vP=7$$
-JOFBSSFHJPO −7$$≤vP="vE≤7$$
/FHBUJWFTBUVSBUJPO vP=−7$$
*GXFBUUFNQUUPJODSFBTFvECFZPOEUIFMJOFBSSBOHF UIFPQBNQCFDPNFT
TBUVSBUFEBOEZJFMET vP=7$$PS vP=−7$$5ISPVHIPVUUIJTCPPL XF
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UIFPVUQVUWPMUBHFJTSFTUSJDUFECZ
vo
Positive saturation
VCC
−7$$≤vP≤7$$
TABLE 5.1
0
Negative saturation
Figure 5.5
vd
–VCC
0QBNQPVUQVUWPMUBHFvPBTBGVODUJPOPG
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Typical ranges for op amp parameters.
1BSBNFUFS
0QFOMPPQHBJO "
*OQVUSFTJTUBODF 3J
0VUQVUSFTJTUBODF 3P
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UP
UPΩ
UPΩ
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`
`Ω
Ω
Operational Amplifiers
177
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QPTTJCJMJUZPGTBUVSBUJPONVTUCFCPSOFJONJOEXIFOPOFEFTJHOTXJUI
PQBNQT UPB WPJEEFTJHOJOHPQBNQDJSDVJUTUIBUXJMMOPUX PSLJOUIF
MBCPSBUPSZ
Throughout this book, we assume that
an op amp operates in the linear range.
Keep in mind the voltage constraint on
the op amp in this mode.
5.2
Example 5.1
"PQBNQIBTBOPQFOMPPQWPMUBHFHBJOPG× JOQVUSFTJTUBODF
PG.Ω BOEPVUQVUSFTJTUBODFPGΩ5IFPQBNQJTVTFEJOUIFDJSDVJU
PG'JH B 'JOEUIFDMPTFEMPPQHBJOvP∕vT%FUFSNJOFDVSSFOUJXIFO
vT= 7
20 kΩ
20 kΩ
10 kΩ
1
–
741
+
vs +
–
10 kΩ
i
1
O
+
vo
–
vs
+
–
–
vd
+
(a)
o
Ri = 2 MΩ
(b)
Figure 5.6
'PS&YBNQMF B PSJHJOBMDJSDVJU C UIFFRVJWBMFOUDJSDVJU
Solution:
6TJOHUIFPQBNQNPEFMJO'JH XFPCUBJOUIFFRVJWBMFOUDJSDVJUPG
Fig. 5.6(a) as shown in Fig. 5.6(b). We now solve the circuit in Fig. C CZVTJOHOPEBMBOBMZTJT"UOPEF ,$-HJWFT
vT−v
v
v −vP
=@@@@@@@@@@
+@@@@@@@
@@@@@@@@
× × ×
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v +v
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v1
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= N"
×
+
–
i
Avd
O
178
Chapter 5
Operational Amplifiers
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Practice Problem 5.1
+
741
–
vs
+
–
io
40 kΩ
5 kΩ
20 kΩ
If the same 741 op amp in Example 5.1 is used in the circuit of Fig. DBMDVMBUFUIFDMPTFEMPPQHBJOvP∕vT'JOEJPXIFOvT=7
Answer: μ"
+
vo
–
5.3
Ideal Op Amp
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PQBNQT"OPQBNQJTJEFBMJGJUIBTUIFGPMMPXJOHDIBSBDUFSJTUJDT
Figure 5.7
'PS1SBDUJDF1SPC
Infinite openMPPQHBJO "≃∞
Infinite input resistance, 3J≃∞
;FSPPVUQVUSFTJTUBODF 3P≃
An ideal op amp is an amplifier with infinite open-loop gain, infinite
input resistance, and zero output resistance.
i1 = 0
+
v1
–
i2 = 0
–
vd
+
+
v2 = v1
–
–
+
+
"MUIPVHIBTTVNJOHBOJEFBMPQBNQQSPWJEFTPOMZBOBQQSPYJNBUF
analysis, most modern amplifiers haWFTVDIMBSHFHBJOTBOEJOQVUJN QFEBODFTUIBUUIFBQQSPYJNBUFBOBMZTJTJTBHPPEPOF6OMFTTTUBUFE
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'PSDJSDVJUBOBMZTJT UIFJEFBMPQBNQJTJMMVTUSBUFEJO'JH XIJDI
JTEFSJWFEGSPNUIFOPOJEFBMNPEFMJO'JH5XPJNQPSUBOUQSPQFSUJFT
PGUIFJEFBMPQBNQBSF
5IFDVSSFOUTJOUPCPUIJOQVUUFSNJOBMTBSF[FSP
vo
–
Figure 5.8
J= J=
This is due to infinite input resistance. An infinite resistance be
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BOEDVSSFOUDBOOPUFOUFSUIFPQBNQ#VUUIFPVUQVUDVSSFOUJTOPU
OFDFTTBSJMZ[FSPBDDPSEJOHUP&R 5IFWPMUBHFBDSPTTUIFJOQVUUFSNJOBMTJTFRVBMUP[FSPJF
*EFBMPQBNQNPEFM
vE=v−v=
PS
The two characteristics can be
exploited by noting that for voltage
calculations the input port behaves as
a short circuit, while for current calculations the input port behaves as an
open circuit.
v=v
5IVT BOJEFBMPQBNQIBT[FSPDVSSFOUJOUPJUTUX PJOQVUUFSNJOBMT
BOEUIFW PMUBHFCFUXFFOUIFUX PJOQVUUFSNJOBMTJTFRVBMUP[FSP
&RVBUJPOT BOE BSFF YUSFNFMZJNQPSUBOUBOETIPVMECF
SFHBSEFEBTUIFLFZIBOEMFTUPBOBMZ[JOHPQBNQDJSDVJUT
5.4
179
Inverting Amplifier
Example 5.2
3FXPSL1SBDUJDF1SPCVTJOHUIFJEFBMPQBNQNPEFM
Solution:
8FNBZSFQMBDFUIFPQBNQJO'JHCZJUTFRVJWBMFOU
model in Fig. BTXFEJEJO&YBNQMF#VUXFEPOPUSFBMMZOFFEUPEPUIJT8FKVTU
OFFEUPLFFQ&RT BOE JONJOEBTXFBOBMZ[FUIFDJSDVJUJO
'JH5IVT UIF'JHDJSDVJUJTQSFTFOUFEBTJO'JH/PUJDFUIBU
v=vT
4JODFJ= UIF BOE LΩSFTJTUPSTBSFJOTFSJFTUIFTBNFDVSSFOU
flows through them. vJTUIFWPMUBHFBDSPTTUIF LΩSFTJTUPS)FODF VTJOHUIFWPMUBHFEJWJTJPOQSJODJQMF
v
v=@@@@@
v =@@
P
+ P "DDPSEJOHUP&R v=v
i2 = 0
v2
+
v1
–
i1 = 0
vs +
–
i0
O
+
40 kΩ
5 kΩ
vo
–
20 kΩ
Figure 5.9
'PS&YBNQMF
4VCTUJUVUJOH&RT BOE JOUP&R ZJFMETUIFDMPTFE MPPQHBJO
v
v
vT=@@
P ⇒ @@
vP =
T
XIJDIJTWFSZDMPTFUPUIFWBMVFPGPCUBJOFEXJUIUIFOPOJEFBM
NPEFMJO1SBDUJDF1SPC5IJTTIPXTUIBUOFHMJHJCMZTNBMMFSSPSSF TVMUTGSPNBTTVNJOHJEFBMPQBNQDIBSBDUFSJTUJDT
"UOPEF0
v
v
JP=@@@@@@
+@@@
PN"
P + 'SPN&R XIFOvT=7 vP=74VCTUJUVUJOHGPSvP=9 7JO
&R QSPEVDFT
JP=+=N"
5IJT BHBJO JT DMPTF UP UIF WBMVFPG N" PCUBJOFE JO 1SBDUJDF
Prob. 5.1 with the nonideal model.
Practice Problem 5.2
3FQFBU&YBNQMFVTJOHUIFJEFBMPQBNQNPEFM
Answer:− μ"
i2
i1
5.4
Inverting Amplifier
*OUIJTBOEUIFGPMMPXJOHTFDUJPOT XFDPOTJEFSTPNFVTFGVMPQBNQDJS DVJUTUIBUPGUFOTFSW FBTNPEVMFTGPSEFTJHOJOHNPSFDPNQMF YDJSDVJUT
The first of such op amp circuits is the inWerting amplifier shoXOJO
Fig. *OUIJTDJSDVJU UIFOPOJO WFSUJOHJOQVUJTHSPVOEFE vJJTDPO OFDUFEUPUIFJO WFSUJOHJOQVUUISPVHI 3 BOEUIFGFFECBDLSFTJTUPS 3G JT
DPOOFDUFECFUXFFOUIFJOWFSUJOHJOQVUBOEPVUQVU0VSHPBMJTUPPCUBJO
R1
v1
Rf
0A
– –
0V
v2 +
+
1
vi
+
–
Figure 5.10
5IFJOWerting amplifier
+
vo
–
180
Chapter 5
Operational Amplifiers
UIFSFMBUJPOTIJQCFUXFFOUIFJOQVUW PMUBHFvJBOEUIFPVUQVUW PMUBHFvP
"QQMZJOH,$-BUOPEF
vJ−v @@@@@@
v −v
J=J ⇒ @@@@@@
= P
3M
3G
A key feature of the inverting amplifier
is that both the input signal and the
feedback are applied at the inverting
terminal of the op amp.
#VUv=v=GPSBOJEFBMPQBNQ TJODFUIFOPOJO WFSUJOHUFSNJOBMJT
HSPVOEFE)FODF
v
v
@@J=− @@P
3M
3G
PS
3G
vP=− @@vJ
3M
Note there are two types of gains: The
one here is the closed-loop voltage
gain Av , while the op amp itself has an
open-loop voltage gain A.
+
vi
R1
–
–
+
Rf
v
R1 i
–
3FGFSUPUIFPQBNQJO'JH*G vJ=7 DBMDVMBUF B UIFPVUQVU
WPMUBHFvP BOE C UIFDVSSFOUJOUIFLΩSFTJTUPS
25 kΩ
–
+
Solution:
B 6TJOH&R 3G
vP
@@
@@
@@
vJ=−3M =− =−
+
vo
–
vP=−vJ=− =−7
Figure 5.12
C 5IFDVSSFOUUISPVHIUIFLΩSFTJTUPSJT
'PS&YBNQMF
vJ− @@@@@@@@
J=@@@@@
= −
=μ"
3M
×
Practice Problem 5.3
'JOEUIFPVUQVUPGUIFPQBNQDJSDVJUTIPXOJO'JH$BMDVMBUFUIF
DVSSFOUUISPVHIUIFGFFECBDLSFTJTUPS
280 kΩ
4 kΩ
45 mV +
–
Figure 5.13
'PS1SBDUJDF1SPC
5IFWPMUBHFHBJOJT"v=vP∕vJ=−3G∕35IFEFTJHOBUJPOPGUIFDJSDVJUJO
'JHBTBOJOWFSUFSBSJTFTGSPNUIFOFHBUJWFTJHO5IVT
/PUJDFUIBUUIFH BJOJTUIFGFFECBDLSFTJTUBODFEJ WJEFECZUIFJO QVUSFTJTUBODFXIJDINFBOTUIBUUIFHBJOEFQFOETPOMZPOUIFFYUFSOBM
FMFNFOUTDPOOFDUFEUPUIFPQBNQ*OWJFXPG&R BOFRVJWBMFOU
DJSDVJUGPSUIFJOWFrting amplifier JTTIPXOJO'JH5IFJOWFSUJOH
amplifier is used, for eYBNQMF JOBDVSSFOUUPWPMUBHFDPOWFSUFS
Example 5.3
+
–
vo
"OFRVJWBMFOUDJSDVJUGPSUIFJOWFSUFSJO
'JH
vi
An inverting amplifier reverses the polarity of the input signal while amplifying it.
+
Figure 5.11
10 kΩ
–
+
Answer:−7 μ"
+
vo
–
5.5
181
Noninverting Amplifier
Example 5.4
%FUFSNJOFvPJOUIFPQBNQDJSDVJUTIPXOJO'JH
40 kΩ
Solution:
"QQMZJOH,$-BUOPEFB
20 kΩ
v −vP @@@@@@
−vB
@@@@@@
B
=
LΩ LΩ
a
b
6V +
–
vB−vP=−vB ⇒ vP=vB−
#VUvB=vC=7GPSBOJEFBMPQBNQ CFDBVTFPGUIF[FSPWPMUBHFESPQ
BDSPTTUIFJOQVUUFSNJOBMTPGUIFPQBNQ)FODF
vP=−=−7
–
+
2V +
–
+
vo
–
Figure 5.14
'PS&YBNQMF
/PUJDFUIBUJGvC==vB UIFOvP=− BTFYQFDUFEGSPN&R 5XPLJOETPGDVSSFOU UPWPMUBHFDPOWFSUFST BMTPLOPXOBT USBOTSFTJT
tance amplifiers BSFTIPXOJO'JH
Practice Problem 5.4
B 4IPXUIBUGPSUIFDPOWFSUFSJO'JH B
v
@@
P=−3
JT
C 4IPXUIBUGPSUIFDPOWFSUFSJO'JH C
3
3
v
@@
P=−3M +@@@
+@@@ 3M 3 )
JT
(
Answer:1SPPG
R
–
+
is
R2
R1
+
vo
–
is
(a)
R3
–
+
+
vo
–
(b)
Figure 5.15
'PS1SBDUJDF1SPC
i2
i1
R1
v1
v2
5.5
Noninverting Amplifier
"OPUIFS JNQPSUBOUBQQMJDBUJPO PGUIF PQBNQJT UIFOPOJO WFSUJOH
amplifier shoXOJO'JH*OUIJTDBTF UIFJOQVUW PMUBHFvJJTBQQMJFE
EJSFDUMZBUUIFOPOJO WFSUJOHJOQVUUFSNJOBM BOESFTJTUPS 3JTDPOOFDUFE
vi
+
+
-–
Rf
–
+
+
vo
–
Figure 5.16
5IFOPOJOWerting amplifier
182
Chapter 5
Operational Amplifiers
CFUXFFOUIFHSPVOEBOEUIFJO WFSUJOHUFSNJOBM8FBSFJOUFSFTUFEJOUIF
PVUQVUWPMUBHFBOEUIFWPMUBHFHBJO"QQMJDBUJPOPG,$-BUUIFJOWFSUJOH
UFSNJOBMHJWFT
v−vP
−v
J=J ⇒ @@@@@@
=@@@@@@
3M
3G
#VUv=v=vJ&RVBUJPO CFDPNFT
¢v vJ−vP
@@@
J=@@@@@@
3M
3G
PS
vP=(+@@
3 )vJ
3G
M
5IFWPMUBHFHBJOJT"v=vP∕vJ=+3G ∕3 XIJDIEPFTOPUIBWFBOFHB
UJWFTJHO5IVT UIFPVUQVUIBTUIFTBNFQPMBSJUZBTUIFJOQVU
–
+
vi
A noninverting amplifier is an op amp circuit designed to provide a
positive voltage gain.
+
+
–
vo = vi
"HBJOXFOPUJDFUIBUUIFHBJOEFQFOETPOMZPOUIFFYUFSOBMSFTJTUPST
/PUJDFUIBUJGGFFECBDLSFTJTUPS3G= TIPSUDJSDVJU PS3=∞ PQFO
DJSDVJU PSCPUI UIFHBJOCFDPNFT6OEFSUIFTFDPOEJUJPOT 3G = BOE
3=∞ UIFDJSDVJUJO'JHCFDPNFTUIBUTIPXOJO'JH XIJDI
JTDBMMFEB WPMUBHFGPMMPXFS PSVOJUZHBJOamplifier CFDBVTFUIFPVUQVU
GPMMPXTUIFJOQVU5IVT GPSBWPMUBHFGPMMPXFS
–
Figure 5.17
5IFWPMUBHFGPMMPXFS
First
stage
+
vi
–
–
+
Figure 5.18
+
vo
–
Second
stage
"WPMUBHFGPMMPXFSVTFEUPJTPMBUFUXP
DBTDBEFETUBHFTPGBDJSDVJU
Example 5.3
vP=vJ
4VDIBDJSDVJUIBTBWF SZIJHIJOQVUJNQFEBODFBOEJTUIFSFGPSFVTF GVMBTBOJOUFSNFEJBUFTUBHF PSCVGfer) amplifier to isolate one circuit
GSPNBOPUIFS BTQPSUSBZFEJO'JH 5IFW PMUBHFGPMMP XFSNJOJ NJ[FTJOUFSBDUJPOCFUXFFOUIFUX PTUBHFTBOEFMJNJOBUFTJOUFSTUBHF MPBEJOH
'PSUIFPQBNQDJSDVJUJO'JH DBMDVMBUFUIFPVUQVUWPMUBHFvP
Solution:
8FNBZTPMWFUIJTJOUXPXBZTVTJOHTVQFSQPTJUJPOBOEVTJOHOPEBM
BOBMZTJT
■ METHOD 1 6TJOHTVQFSQPTJUJPO XFMFU
vP=vP+vP
5.6
183
Summing Amplifier
XIFSFvPJTEVFUPUIF7WPMUBHFTPVSDF BOEvPJTEVFUPUIF7JOQVU
5PHFUvP XFTFUUIF7TPVSDFFRVBMUP[FSP6OEFSUIJTDPOEJUJPO UIF
DJSDVJUCFDPNFTBOJOWFSUFS)FODF&R HJWFT
=−7
vP=−@@
5PHFU vP XFTFUUIF 7TPVSDFFRVBMUP[FSP5IFDJSDVJUCFDPNFTB
noninverting amplifier so that Eq. (5.11) applies.
vP=(
+@@
)=7
10 kΩ
4 kΩ
a
–
+
b
6V +
–
+
4V +
–
vo
–
Figure 5.19
'PS&YBNQMF
5IVT
vP=vP+vP=−+=−7
■ METHOD 2 "QQMZJOH,$-BUOPEFB
−v
vB−vP
@@@@@@
B
=@@@@@@
#VUvB=vC= BOETP
−vP
@@@@@
−
=@@@@@
⇒ =−vP
PSvP=−7 BTCFGPSF
Practice Problem 5.5
$BMDVMBUFvPJOUIFDJSDVJUPG'JH
Answer:7
4 kΩ
9V +
–
8 kΩ
+
–
+
5 kΩ
vo
2 kΩ
5.6
–
Summing Amplifier
#FTJEFTamplification, the PQBNQDBOQFSGPSNBEEJUJPOBOETVCUSBDUJPO
The addition is performed by the summing amplifier coWFSFEJOUIJTTFD
UJPOUIFTVCUSBDUJPOJTQFSGPSNFECZUIFEJG ference amplifier coWFSFEJO
UIFOFYUTFDUJPO
Figure 5.20
'PS1SBDUJDF1SPC
v1
A summing amplifier is an op amp circuit that combines several inputs
and produces an output that is the weighted sum of the inputs.
v2
v3
R1
R2
R3
i1
i2
i3
Rf
i
i
0
–
a
+
0
+
vo
–
The summing amplifier TIP XOJO'JH JTBW BSJBUJPOPGUIF
JOWerting amplifier*UUBLFTBEWBOUBHFPGUIFG BDUUIBUUIFJOWFSUJOHDPO
figuration can handle manZJOQVUTBUUIFTBNFUJNF 8FLFFQJONJOE
Figure 5.21
The summing amplifier
184
Chapter 5
Operational Amplifiers
UIBUUIFDVSSFOUFOUFSJOHFBDIPQBNQJOQVUJT[FSP "QQMZJOH,$-BU
OPEFBHJWFT
J=J+J+J
v −v
v −v
J=@@@@@@
B
J=@@@@@@
B
3
3
v−vB
v −v
J=@@@@@@
J=@@@@@@
B P
3
3G
#VU
8FOPUFUIBUvB=BOETVCTUJUVUF&R JOUP&R 8FHFU
3G
3G
3G
vP=− @@@
v+@@@
v+@@@
v 3
3 )
( 3
JOEJDBUJOHUIBUUIFPVUQVUW PMUBHFJTBXFJHIUFETVNPGUIFJOQVUT' PS
UIJTSFBTPO UIFDJSDVJUJO'JHJTDBMMFEB TVNNFS/FFEMFTTUPTBZ UIFTVNNFSDBOIBWFNPSFUIBOUISFFJOQVUT
Example 5.6
$BMDVMBUFvPBOEJPJOUIFPQBNQDJSDVJUJO'JH
5 kΩ
10 kΩ
a
2.5 kΩ
2V +
–
+
–
1V
b
–
+
io
2 kΩ
+
vo
–
Figure 5.22
'PS&YBNQMF
Solution:
5IJTJTBTVNNFSXJUIUXPJOQVUT6TJOH&R HJWFT
vP=−[@@
+@@@
]=− + =−7
5IFDVSSFOUJPJTUIFTVNPGUIFDVSSFOUTUISPVHIUIFBOELΩSFTJT
UPST#PUIPGUIFTFSFTJTUPSTIBWFWPMUBHF vP=−7BDSPTTUIFN TJODF
vB=vC=)FODF
v − @@@@@@
v − JP=@@@@@@
P + P N"=−−=−N"
5.7
Difference Amplifier
Practice Problem 5.6
'JOEvPBOEJPJOUIFPQBNQDJSDVJUTIPXOJO'JH
20 kΩ
8 kΩ
10 kΩ
1.5 V
+
–
–
+
6 kΩ
2V +
–
185
io
+
vo
–
4 kΩ
+ 1.2 V
–
Figure 5.23
'PS1SBDUJDF1SPC
Answer:−7 −N"
5.7
Difference Amplifier
%JGGFSFODF PSEJG ferential) amplifiers are used in vBSJPVTBQQMJDBUJPOT
XIFSFUIFSFJTBOFFEUPBNQMJGZUIFEJG GFSFODFCFUXFFOUX PJOQVUTJH OBMT 5IFy are first cousins of the instrumentation amplifier UIFNPTU
useful and popular amplifier XIJDIXFXJMMEJTDVTTJO4FDUJPO
A difference amplifier is a device that amplifies the difference between
two inputs but rejects any signals common to the two inputs.
$POTJEFSUIFPQBNQDJSDVJUTIP XOJO'JH, FFQJONJOEUIBU
[FSPDVSSFOUTFOUFSUIFPQBNQUFSNJOBMT"QQMZJOH,$-UPOPEFB
v−vB @@@@@@
v −v
@@@@@@
= B P
3
3
PS
3
3
vP= @@@
+ vB−@@@
v 3
( 3
)
R2
R1
R3
v1
+
–
+ v
– 2
Figure 5.24
0
va
%JGference amplifier
0
vb
–
+
R4
+
vo
–
The difference amplifier is also known
as the subtractor, for reasons to be
shown later.
186
Chapter 5
Operational Amplifiers
"QQMZJOH,$-UPOPEFC
v −v
v −
@@@@@@
C
=@@@@@
C 3
3
PS
3
vC=@@@@@@@
v
3+ 3 #VUvB=vC4VCTUJUVUJOH&R JOUP&R ZJFMET
3
3
3
@@@
vP= @@@+ @@@@@@@
v
− v
3
( 3
) 3+3
PS
3 +3∕3
3
vP=@@@@@@@@@@@@
v−@@@
v 3
3 +3∕3
4JODFBEJGference amplifier must reject a signal common to the twPJO
puts, the amplifier must haWFUIFQSPQFSUZUIBUvP=XIFOv=v5IJT
QSPQFSUZFYJTUTXIFO
3
3
@@@
=@@@
3 3
5IVT XIFOUIFPQBNQDJSDVJUJTBEJG
CFDPNFT
ference amplifier &R 3
vP=@@@
v−v 3
*G3=3BOE 3=3 UIFEJG ference amplifier becomes a TVCUSBDUPS XJUIUIFPVUQVU
Example 5.7
vP=v−v
%FTJHOBOPQBNQDJSDVJUXJUIJOQVUT
vP=−v+v
vBOE
vTVDIUIBU Solution:
5IFDJSDVJUSFRVJSFTUIBU
vP=v−v
5IJTDJSDVJUDBOCFSFBMJ[FEJOUXPXBZT
Design 1 *GXFEFTJSFUPVTFPOMZPOFPQBNQ XFDBOVTFUIFPQBNQ
DJSDVJUPG'JH$PNQBSJOH&R XJUI&R XFTFF
@@@
3= ⇒ 3=3
3
5.7
187
Difference Amplifier
"MTP
@@
+3∕3
@@@@@@@@@@
@@@@@@@@@
= ⇒ =@@
+3∕3
+3∕3 PS
3
=+@@@ ⇒ 3=3
3
*GXFDIPPTF 3 =L ΩBOE 3 =L Ω UIFO 3 =L ΩBOE
3 = 20 kΩ
Design 2 *GXFEFTJSFUPVTFNPSFUIBOPOFPQBNQ XFNBZDBTDBEF
an inverting amplifier and a twoJOQVUJOWFSUJOHTVNNFS BTTIPXOJO
'JH'PSUIFTVNNFS
vP=−vB−v
vB=−v
3R3
v2
R3
5R1
–
+
va
BOEGPSUIFJOWFSUFS
$PNCJOJOH&RT BOE HJWFT
5R1
v1
–
+
R1
Figure 5.25
'PS&YBNQMF
vP=v−v
XIJDIJTUIFEFTJSFESFTVMU*O'JH XFNBZTFMFDU 3=LΩBOE
3=LΩPS3=3=LΩ
Practice Problem 5.7
%FTJHOBEJGference amplifier with gBJO
Answer:5ZQJDBM3=3=LΩ 3=3=LΩ
"Oinstrumentation amplifier shown in Fig. 5.26 is an amplifier of low
MFWFMTJHOBMTVTFEJOQSPDFTTDPOUSPMPSNFBTVSFNFOUBQQMJDBUJPOTBOE
DPNNFSDJBMMZBWBJMBCMFJOTJOHMFQBDLBHFVOJUT4IPXUIBU
3
3
+@@@@
v−v
vP=@@@
3 (
3 )
Solution:
We recognize that the amplifier " in Fig. 5.26 is a difference amplifier.
5IVT GSPN&R 3
vP=@@@
vP−vP 3
4JODFUIFPQBNQT"BOE"ESBXOPDVSSFOU DVSSFOUJ flows through
UIFUISFFSFTJTUPSTBTUIPVHIUIFZXFSFJOTFSJFT)FODF
vP−vP=J 3+3+3 =J 3+3 Example 5.8
vo
188
Chapter 5
Operational Amplifiers
+
v1 +
–
0
0
–
+
v2
vo1
– A1
R1
R2
R3
va
R4
–
i
+
vb
R3
A2
A3
vo
R1
vo2
R2
+
–
Figure 5.26
Instrumentation amplifier; for Example 5.8.
#VU
v −v
J=@@@@@@
B C
3
BOEvB=v vC=v5IFSFGPSF
v−v
J=@@@@@@
3
*OTFSUJOH&RT BOE JOUP&R HJWFT
3
3
vP=@@@
+@@@@
v−v
3 (
3 )
as required. We will discuss the instrumentation amplifier in detail in
4FDUJPO
Practice Problem 5.8
0CUBJOJP in the instrumentation amplifier circuit of Fig. 5.27.
6.98 V
+
40 kΩ
–
20 kΩ
–
+
–
7V
+
Figure 5.27
io
20 kΩ
40 kΩ
50 kΩ
Instrumentation amplifier; for Practice Prob
Answer:μ"
5.8
5.8
189
Cascaded Op Amp Circuits
Cascaded Op Amp Circuits
"TXFLOPX PQBNQDJSDVJUTBSFNPEVMFTPSCVJMEJOHCMPDLTGPSEFTJHO
JOHDPNQMF YDJSDVJUT*UJTPGUFOOFDFTTBSZJOQSBDUJDBMBQQMJDBUJPOTUP
DPOOFDUPQBNQDJSDVJUTJODBTDBEF JF IFBEUPUBJM UPBDIJF WFBMBSHF
PWFSBMMH BJO*OHFOFSBM UX PDJSDVJUTBSFDBTDBEFEXIFOUIF ZBSFDPO nected in tandem, one behind another in a single file.
A cascade connection is a head-to-tail arrangement of two or more op
amp circuits such that the output of one is the input of the next.
8IFOPQBNQDJSDVJUTBSFDBTDBEFE FBDIDJSDVJUJOUIFTUSJOHJT
DBMMFEB TUBHFUIFPSJHJOBMJOQVUTJHOBMJTJODSFBTFECZUIFH BJOPGUIF
JOEJWJEVBMTUBHF 0Q BNQDJSDVJUT IBWFUIF BEWBOUBHF UIBUUIFZ DBOCF
DBTDBEFEXJUIPVUDIBOHJOHUIFJSJOQVU PVUQVUSFMBUJPOTIJQT5IJTJTEVFUP
UIFGact that each (ideal) op amp circuit has infinite input resistance and
[FSPPVUQVUSFTJTUBODF'JHVSFEJTQMBZTBCMPDLEJBHSBNSFQSFTFO UBUJPOPGUISFFPQBNQDJSDVJUTJODBTDBEF4JODFUIFPVUQVUPGPOFTUBHF
JTUIFJOQVUUPUIFOFYUTUBHF UIFPWFSBMMHBJOPGUIFDBTDBEFDPOOFDUJPOJT
UIFQSPEVDUPGUIFHBJOTPGUIFJOEJWJEVBMPQBNQDJSDVJUT PS
"="""
"MUIPVHIUIFDBTDBEFDPOOFDUJPOEPFTOPUBGGFDUUIFPQBNQJOQVUPVUQVU
SFMBUJPOTIJQT DBSFNVTUCFF YFSDJTFEJOUIFEFTJHOPGBOBDUVBMPQBNQ
DJSDVJUUPFOTVSFUIBUUIFMPBEEVFUPUIFOF YUTUBHFJOUIFDBTDBEFEPFT
OPUTBUVSBUFUIFPQBNQ
+
v1
–
Stage 1
A1
+
v2 = A1v1
–
Stage 2
A2
+
v3 = A2v2
–
Stage 3
A3
+
vo = A3v3
–
Figure 5.28
"UISFFTUBHFDBTDBEFEDPOOFDUJPO
'JOEvPBOEJPJOUIFDJSDVJUJO'JH
Example 5.9
Solution:
This circuit consists of two noninverting amplifiers cascaded. At the out
put of the first op amp,
+
–
vB=(+@@
) =N7
20 mV +
–
12 kΩ
3 kΩ
"UUIFPVUQVUPGUIFTFDPOEPQBNQ
v −v
JP=@@@@@@
P C
N"
+
–
b
10 kΩ
4 kΩ
io
+
vo
–
vP=(+@@
)vB= + =N7
5IFSFRVJSFEDVSSFOUJPJTUIFDVSSFOUUISPVHIUIFLΩSFTJTUPS
a
Figure 5.29
'PS&YBNQMF
190
Chapter 5
Operational Amplifiers
#VUvC=vB=N7)FODF
− × −3
JP=@@@@@@@@@@@@@@@@
=μ"
×3
Practice Problem 5.9
+
–
5V +
–
%FUFSNJOFvPBOEJPJOUIFPQBNQDJSDVJUJO'JH
Answer:7 μ"
+
–
+
200 kΩ
vo
io
–
50 kΩ
Figure 5.30
'PS1SBDUJDF1SPC
Example 5.10
*Gv=7BOEv= 2 V, find vPJOUIFPQBNQDJSDVJUPG'JH
A
6 kΩ
v1
2 kΩ
–
+
5 kΩ
a
10 kΩ
B
–
+
8 kΩ
v2
4 kΩ
–
+
C
vo
15 kΩ
b
Figure 5.31
'PS&YBNQMF
Solution:
Define. The problem is clearly defined.
1SFTFOU8JUIBOJOQVUPGvPG7BOEPGvPG7 EFUFSNJOF
UIFPVUQVUWPMUBHFPGUIFDJSDVJUTIPXOJO'JHVSF5IFPQBNQ
circuit is actually composed of three circuits. The first DJSDVJUBDUT
as an amplifier of gain − −LΩ∕LΩ GPSvBOEUIFTFDPOE
functions as an amplifier of gain − −LΩ∕LΩ GPSv5IFMBTU
DJSDVJUTFSWFTBTBTVNNFSPGUXPEJGGFSFOUHBJOTGPSUIFPVUQVUPG
UIFPUIFSUXPDJSDVJUT
"MUFSOBUJWF5IFSFBSFEJGGFSFOUXBZTPGXPSLJOHXJUIUIJTDJSDVJU
#FDBVTFJUJOWPMWFTJEFBMPQBNQT UIFOBQVSFMZNBUIFNBUJDBM
5.8
191
Cascaded Op Amp Circuits
BQQSPBDIXJMMXPSLRVJUFFBTJMZ"TFDPOEBQQSPBDIXPVMECFUP
VTF14QJDF as a confirmation of the math.
"UUFNQU Let the output of the first op amp circuit be designated
BTvBOEUIFPVUQVUPGUIFTFDPOEPQBNQDJSDVJUCFEFTJHOBUFEBT
v5IFOXFHFU
v=−v =−×=−7
v=−v =−×=−7
*OUIFUIJSEDJSDVJUXFIBWF
vP=− LΩ∕LΩ v +<− LΩ∕LΩ v>
=− − − ∕ −
=+ = 7
&WBMVBUF5PQSPQFSMZFWBMVBUFPVSTPMVUJPO XFOFFEUPJEFOUJGZ
BSFBTPOBCMFDIFDL)FSFXFDBOFBTJMZVTF14QJDFUPQSPWJEFUIBU
DIFDL
/PXXFDBOTJNVMBUFUIJTJO14QJDF5IFSFTVMUTBSFTIPXOJO
'JH
R4
R6
1V
+ v1
2 kΩ
6 kΩ
OPAMP
–
+
–
–3.000
R2
5 kΩ
U1
R1
8.667 V
10 kΩ
OPAMP
–
R5
R7
+ v2
2V
–
4 kΩ
–4.000
8 kΩ
OPAMP
–
+
+
R3
15 kΩ
U2
Figure 5.32
'PS&YBNQMF
8FPCUBJOUIFTBNFSFTVMUTVTJOHUXPFOUJSFMZEJGGFSFOUUFDIOJRVFT
(the first is to treat the op amp circuits as just gains and a summer
BOEUIFTFDPOEJTUPVTFDJSDVJUBOBMZTJTXJUI14QJDF 5IJTJTB
WFSZHPPENFUIPEPGBTTVSJOHUIBUXFIBWFUIFDPSSFDUBOTXFS
4BUJTGBDUPSZ We are satisfied we have obtained the asked for
SFTVMUT8FDBOOPXQSFTFOUPVSXPSLBTBTPMVUJPOUPUIF
QSPCMFN
U3
192
Chapter 5
Practice Problem 5.10
Operational Amplifiers
*Gv=7BOEv=7, find vPJOUIFPQBNQDJSDVJUPG'JH
60 kΩ
20 kΩ
–
+
v1
–
+
+
–
30 kΩ
50 kΩ
10 kΩ
v2 +
–
vo
–
+
Figure 5.33
'PS1SBDUJDF1SPC
Answer:7
5.9
Op Amp Circuit Analysis with PSpice
14QJDFGPS8JOEPXTEPFTOPUIBWFBNPEFMGPSBOJEFBMPQBNQ BMUIPVHI
POFNBZDSFBUFPOFBTBTVCDJSDVJUVTJOHUIF $SFBUF4VCDJSDVJUMJOFJO
UIF5PPMTNFOV3BUIFSUIBODSFBUJOHBOJEFBMPQBNQ XFXJMMVTFPOF
PGUIFGPVSOPOJEFBM DPNNFSDJBMMZB WBJMBCMFPQBNQTTVQQMJFEJOUIF
14QJDFMJCSBSZFWBMTMC5IFPQBNQNPEFMTIBWFUIFQBSUOBNFT-' -. -. BOEV" BTTIPXOJO'JH&BDIPGUIFNDBOCF
PCUBJOFEGSPN%SBX(FU/FX1BSUMJCSBSJFTFWBMMJCPSCZTJNQMZ
TFMFDUJOH%SBX(FU/FX1BSUBOEUZQJOHUIFQBSUOBNFJOUIF1BSU/BNF
EJBMPHCPY BTVTVBM/PUFUIBUFBDIPGUIFNSFRVJSFTEDTVQQMJFT XJUIPVU
XIJDIUIFPQBNQXJMMOPUXPSL5IFEDTVQQMJFTTIPVMECFDPOOFDUFEBT
TIPXOJO'JH
U2
U4
3
+
7
5
V+ B2
2
6
+
6
V+ BB /S
LF411
V– G
1
–
4
LM111
(a) JFET–input op
amp subcircuit
(b) Op amp
subcircuit
2
–
V–
4
B1
1
Figure 5.34
U3
85
3
7
3
/POJEFBMPQBNQNPEFMBWBJMBCMFJO14QJDF
2
+
–
4 U1A
V+
1
V–
11
LM324
(c) Five–
connection
op amp subcircuit
3
2
+
–
7
5
V+ 052
V–
4
6
051
1
uA741
(d) Five–connection
op amp subcircuit
5.9
Op Amp Circuit Analysis with PSpice
193
Example 5.11
6TF14QJDFUPTPMWFUIFPQBNQDJSDVJUGPS&YBNQMF
Solution:
6TJOH4DIFNBUJDT XFESBXUIFDJSDVJUJO'JH B BTTIPXOJO
Fig.Á5.35. Notice that the positive terminal of the voltage source vTJT
DPOOFDUFEUPUIFJOWFSUJOHUFSNJOBM QJO WJBUIFLΩSFTJTUPS XIJMF
UIFOPOJOWFSUJOHUFSNJOBM QJO JTHSPVOEFEBTSFRVJSFEJO'JH B "MTP OPUJDFIPXUIFPQBNQJTQPXFSFEUIFQPTJUJWFQPXFSTVQQMZ
UFSNJOBM7 + QJO JTDPOOFDUFEUPB 7EDWPMUBHFTPVSDF XIJMF UIFOFHBUJWFQPXFSTVQQMZUFSNJOBM7− QJO JTDPOOFDUFEUP −7
Pins 1 and 5 are left floating because they are used for offset null adjust
NFOU XIJDIEPFTOPUDPODFSOVTJOUIJTDIBQUFS#FTJEFTBEEJOHUIFED
QPXFSTVQQMJFTUPUIFPSJHJOBMDJSDVJUJO'JH B XFIBWFBMTPBEEFE
QTFVEPDPNQPOFOUT7*&810*/5BOE*130#&UPSFTQFDUJWFMZNFBTVSF
UIFPVUQVUWPMUBHF vPBUQJOBOEUIFSFRVJSFEDVSSFOU JUISPVHIUIF LΩSFTJTUPS
0
–
VS +
2V
3
R1
10 K
V2
+
U1
2
+
–
7
5
V+ 052
V–
6
–3.9983
15 V
+
051
4
–
1
–
uA741
15 V
0
V3
R2
1.999E–04
20 K
Figure 5.35
4DIFNBUJDGPS&YBNQMF
"GUFSTB WJOHUIFTDIFNBUJD XFTJNVMBUFUIFDJSDVJUCZTFMFDUJOH
"OBMZTJT4JNVMBUFBOEIB WFUIFSFTVMUTEJTQMBZFEPO 7*&810*/5BOE
*130#&'SPNUIFSFTVMUT UIFDMPTFEMPPQHBJOJT
vP @@@@@@@
− =−
@@
vT = BOEJ=N" JOBHSFFNFOUXJUIUIFSFTVMUTPCUBJOFEBOBMZUJDBMMZ
JO&YBNQMF
3FXPSL1SBDUJDF1SPCVTJOH14QJDF
Answer: μ"
Practice Problem 5.11
194
Chapter 5
5.10
Analog
output
Four-bit
DAC
(a)
V3
V2
R1
†
Applications
5IFPQBNQJTBGVOEBNFOUBMC VJMEJOHCMPDLJONPEFSOFMFDUSPOJD JOTUSVNFOUBUJPO*UJTVTFEF YUFOTJWFMZJONBO ZEF WJDFT BMPOHXJUI
SFTJTUPSTBOEPUIFSQBTTJ WFFMFNFOUT*UTOVNFSPVTQSBDUJDBMBQQMJDB tions include instrumentation amplifiers, digitalUPBOBMPHDPOWFSUFST BOBMPHDPNQVUFST MFWel shifters, filters, calibration circuits, inWFSUFST TVNNFST JOUF HSBUPST EJGGFSFOUJBUPST TVCUSBDUPST MPH BSJUINJDBNQMJ fiers, comparators, gyrators, oscillators, rectifiers, reHVMBUPST WPMUBHF
UPDVSSFOUDPO WFSUFST DVSSFOU UPWPMUBHF DPOWFSUFST BOE DMJQQFST
4PNFPGUIFTFXFIBWFBMSFBEZDPOTJEFSFE8FXJMMDPOTJEFSUXPNPSF
BQQMJDBUJPOTIFSFUIFEJHJUBMUPBOBMPHDPOWFSUFSBOEUIFJOTUSVNFOUB
tion amplifier
Digital
input
(0000 –1111)
V1
Operational Amplifiers
R2
MSB
V4
R3
R4
LSB
Rf
–
+
(b)
Vo
5.10.1 Digital-to-Analog Converter
5IFEJHJUBMUPBOBMPHDPOWFSUFS %"$ USBOTGPSNTEJHJUBMTJHOBMTJOUP
BOBMPHGPSN "UZQJDBMF YBNQMFPGBGPVS CJU% "$JTJMMVTUSBUFEJO
Fig. B 5IFGPVSCJU%"$DBOCFSFBMJ[FEJONBOZXBZT"TJN
QMFSFBMJ[BUJPOJTUIF CJOBSZXFJHIUFEMBEEFS TIPXOJO'JH C 5IFCJUTBSFXFJHIUTBDDPSEJOHUPUIFNBHOJUVEFPGUIFJSQMBDFW BM
VF CZEFTDFOEJOHW BMVFPG 3G∕3OTPUIBUFBDIMFTTFSCJUIBTIBMGUIF XFJHIUPGUIFOF YUIJHIFS 5IJTJTPC WJPVTMZBOJO WFSUJOHTVNNJOH
amplifier5IFPVUQVUJTSFMBUFEUPUIFJOQVUTBTTIP XOJO&R 5IVT
Figure 5.36
'PVSCJU%"$ B CMPDLEJBHSBN C CJOBSZXFJHIUFEMBEEFSUZQF
In practice, the voltage levels may be
typically 0 and ± 5 V.
Example 5.12
3G
3G
3G
3G
−7P=@@
7M+ @@@
7+ @@@
7+ @@@
7
3M
3
3
3
*OQVU7JTDBMMFEUIF most significant bit .4# XIJMFJOQVU 7JTUIF
least significant bit -4# &BDIPGUIFGPVSCJOBSZJOQVUT7 7DBO
BTTVNFPOMZUXPWPMUBHFMFWFMTPS7#ZVTJOHUIFQSPQFSJOQVUBOE
GFFECBDLSFTJTUPSW BMVFT UIF% "$QSPWJEFTBTJOHMFPVUQVUUIBUJTQSP QPSUJPOBMUPUIFJOQVUT
*OUIFPQBNQDJSDVJUPG'JH C MFU
3G =L Ω 3 =L Ω 3=LΩ 3=LΩ BOE3=LΩ0CUBJOUIFBOBMPHPVUQVUGPS
CJOBSZJOQVUT<> <> <> <>
Solution:
4VCTUJUVUJOHUIFHJWFOWBMVFTPGUIFJOQVUBOEGFFECBDLSFTJTUPSTJO
Eq. (5.23) gives
3G
3G
3G
3G
−7P=@@
7M+ @@@
7+ @@@
7+ @@@
7
3M
3
3
3
=7M+ 0.57+ 0.257+ 0.1257
6TJOHUIJTFRVBUJPO BEJHJUBMJOQVU< 7777> =<>QSPEVDFTBO
BOBMPHPVUQVUPG−7P=7<7777>=<>HJWFT−7P=7
5.10
195
Applications
4JNJMBSMZ
<7777>=<>
<7777>=<>
<7777>=<>
⋮
<7777>=<>
⇒ −7P=7
⇒ −7P=+=7
⇒ −7P=7
⇒ −7P=+++
=7
5BCMFTVNNBSJ[FT UIFSFTVMUPGUIFEJHJUBM UPBOBMPHDPOWFSTJPO
/PUFUIBUXFIBWFBTTVNFEUIBUFBDICJUIBTBWBMVFPG75IVT JO
UIJTTZTUFN XFDBOOPUSFQSFTFOUBWPMUBHFCFUXFFOBOE GPS
FYBNQMF5IJTMBDLPGSFTPMVUJPOJTBNBKPSMJNJUBUJPOPGEJHJUBMUPBOBMPH
DPOWFSTJPOT'PSHSFBUFSBDDVSBDZ BXPSESFQSFTFOUBUJPOXJUIBHSFBUFS
OVNCFSPGCJUTJTSFRVJSFE&WFOUIFOBEJHJUBMSFQSFTFOUBUJPOPGBOBOB MPHWPMUBHFJTOFWFSFYBDU*OTQJUFPGUIJTJOFYBDUSFQSFTFOUBUJPO EJHJUBM
SFQSFTFOUBUJPOIBTCFFOVTFEUPBDDPNQMJTISFNBSLBCMFUIJOHTTVDIBT
BVEJP$%TBOEEJHJUBMQIPUPHSBQIZ
TABLE 5.2
Input and output values of the four-bit DAC.
#JOBSZJOQVU
<VVVV>
%FDJNBMWBMVF
0VUQVU
−Vo
Practice Problem 5.11
"UISFFCJU%"$JTTIPXOJO'JH
B %FUFSNJOF]7P]GPS<777>=<>
C 'JOE]7P]JG<777>=<>
D *G]7P]=7JTEFTJSFE XIBUTIPVMECF<777>
E 5PHFU]7P]=7 XIBUTIPVMECF<777>
Answer:7 7 <> <>
v1
v2
v3
10 kΩ
20 kΩ
40 kΩ
Figure 5.37
10 kΩ
–
+
5ISFFCJU%"$GPS1SBDUJDF1SPC
vo
196
Chapter 5
Operational Amplifiers
5.10.2 Instrumentation Amplifiers
0OFPGUIFNPTUVTFGVMBOEW FSTBUJMFPQBNQDJSDVJUTGPSQSFDJTJPONFB TVSFNFOUBOEQSPDFTTDPOUSPMJTUIF instrumentation amplifier *" TP
DBMMFECFDBVTFPGJUTXJEFTQSFBEVTFJONFBTVSFNFOUTZTUFNT 5ZQJDBM
applications of IAs include isolation amplifiers, thermocouple amplifi
FST BOEEBUBBDRVJTJUJPOTZTUFNT
The instrumentation amplifier is an eYUFOTJPOPGUIFEJGGFSFODFBN
plifier in that it amplifies the difGFSFODFCFUXFFOJUTJOQVUTJHOBMT "T
TIPwn in Fig. 5.26 (see Example 5.8), an instrumentation amplifier typi
DBMMZDPOTJTUTPGUISFFPQBNQTBOETFWFOSFTJTUPST'PSDPOWFOJFODF UIF
amplifier is shoXOBHBJOJO'JH B XIFSFUIFSFTJTUPSTBSFNBEFFRVBM
FYDFQUGPSUIFFYUFSOBMHBJOTFUUJOHSFTJTUPS 3( DPOOFDUFECFUXFFOUIF
HBJOTFUUFSNJOBMT'JHVSF C TIPXTJUTTDIFNBUJDTZNCPM&YBNQMF
TIPXFEUIBU
vP="v v−v Inverting input v
1
Gain set
R
+
–1
R
R
RG
R
Gain set
–
+2
Noninverting input v2
–
+3
vo
Output
R
–
R
+
(a)
(b)
Figure 5.38
B The instrumentation amplifier with an eYUFSOBMSFTJTUBODFUPBEKVTUUIFHBJO C TDIFNBUJDEJBHSBN
XIFSFUIFWPMUBHFHBJOJT
"v=1+@@@
23
3(
"TTIP XO JO Fig. 5.39, the instrumentation amplifier amplifies small
EJGGFSFOUJBMTJHOBMW PMUBHFTTVQFSJNQPTFEPOMBS HFSDPNNPO NPEF
–
RG
+
Small differential signals riding on larger
common-mode signals
Figure 5.39
Instrumentation amplifier
5IF*"SFKFDUTDPNNPOWPMUBHFTCut amplifies small signal vPMUBHFT
Amplified differential signal,
no common-mode signal
5.11
Summary
197
WPMUBHFT4JODFUIFDPNNPONPEFWPMUBHFTBSFFRVBM UIFZDBODFMFBDI
PUIFS
5IF*"IBTUISFFNBKPSDIBSBDUFSJTUJDT
5IFWPMUBHFHBJOJTBEKVTUFECZPOFFYUFSOBMSFTJTUPS3(
5IFJOQVUJNQFEBODF PGCPUIJOQVUTJTW FSZIJHIBOE EPFTOPUWBSZ
BTUIFHBJOJTBEKVTUFE
5IFPVUQVU vPEFQFOETPOUIFEJG GFSFODFCFUXFFOUIFJOQVUT vBOE
v OPUPOUIFWPMUBHFDPNNPOUPUIFN DPNNPONPEFWPMUBHF %VFUPUIFXJEFTQSFBEVTFPG*"T NBOVG BDUVSFSTIBWFEFWFMPQFE
UIFTF amplifiers on singleQBDLBHFVOJUT " UZQJDBMF YBNQMF JTUIF
-) EFWFMPQFECZ/BUJPOBM4FNJDPOEVDUPS 5IFH BJODBOCFW BS
JFEGSPNUP CZBOF YUFSOBMSFTJTUPSXIPTFW BMVFNBZWBSZGSPN
UPLΩ
*O'JH MFU 3=LΩ v=7 BOEv=7*G3(JT
BEKVTUFEUP Ω EFUFSNJOF B UIFWPMUBHFHBJO C UIFPVUQVUWPMU BHFvP
Example 5.13
Solution:
B 5IFWPMUBHFHBJOJT
× "v=+ @@@
3= + @@@@@@@@
= 3(
C 5IFPVUQVUWPMUBHFJT
vP="v v− v = − = N7= N7
%FUFSNJOFUIFWBMVFPGUIFFYUFSOBMHBJOTFUUJOHSFTJTUPS3(SFRVJSFEGPS
UIF*"JO'JHUPQSPEVDFBHBJOPGXIFO3=LΩ
Answer:Ω
5.11
Summary
5IFPQBNQJTBIJHI Hain amplifier that has high input resistance
BOEMPXPVUQVUSFTJTUBODF
5BCMFTVNNBSJ[FTUIFPQBNQDJSDVJUTDPOTJEFSFEJOUIJTDIBQUFS
5IFFYQSFTTJPOGPSUIFH ain of each amplifier circuit holds whether
UIFJOQVUTBSFED BD PSUJNFWBSZJOHJOHFOFSBM
Practice Problem 5.13
198
Chapter 5
Operational Amplifiers
TABLE 5.3
Summary of basic op amp circuits.
0QBNQDJSDVJU
/BNFPVUQVUJOQVUSFMBUJPOTIJQ
*OWerting amplifier
3
vP−@@@
vJ
3
R2
vi
R1
–
+
vo
/POJOWerting amplifier
+3
vP@@@@@@
v
3 J
R2
R1
vi
–
+
vo
7PMUBHFGPMMPXFS
vi
v1
v2
v3
v1
–
+
vo
R1
Rf
R2
R1
3G
( M
3G
3G
)
vP− __
v __
v __
v 3
3
3
vo
%JGference amplifier
R2
–
+
v2
4VNNFS
–
+
R3
R1
vPvJ
3
vP__
v−v
3
M
vo
R2
An ideal op amp has an infinite input resistance, a zero output resis
tance, and an infinite gBJO
'PSBOJEFBM PQBNQ UIFDVSSFOUJOUPFBDI PGJUTUXPJOQVUUFSNJ
OBMTJT[FSP BOEUIFW PMUBHFBDSPTTJUTJOQVUUFSNJOBMTJTOF HMJHJCMZ
TNBMM
*OBOJO Werting amplifier UIFPVUQVUW PMUBHFJTBOF HBUJWFNVMUJQMF
PGUIFJOQVU
*OBOPOJOWerting amplifier UIFPVUQVUJTBQPTJUJ WFNVMUJQMFPGUIF
JOQVU
*OBWPMUBHFGPMMPXFS UIFPVUQVUGPMMPXTUIFJOQVU
In a summing amplifier UIFPVUQVUJTUIFXFJHIUFETVNPGUIFJOQVUT
*OBEJGGFSFODFamplifier UIFPVUQVUJTQSPQPSUJPOBMUPUIFEJGGFSFODF
PGUIFUXPJOQVUT
0QBNQDJSDVJUTNBZCFDBTDBEFEXJUIPVUDIBOHJOHUIFJSJOQVU
PVUQVUSFMBUJPOTIJQT
14QJDFDBOCFVTFEUPBOBMZ[FBOPQBNQDJSDVJU
5ZQJDBMBQQMJDBUJPOTPGUIFPQBNQDPOTJEFSFEJOUIJTDIBQUFSJODMVEF UIFEJHJUBMUPBOBMPHDPOWerter and the instrumentation amplifier
199
Review Questions
Review Questions
5IFUXPJOQVUUFSNJOBMTPGBOPQBNQBSFMBCFMFEBT
B IJHIBOEMPX
C QPTJUJWFBOEOFHBUJWF
D JOWFSUJOHBOEOPOJOWFSUJOH
E EJGGFSFOUJBMBOEOPOEJGGFSFOUJBM
'PSBOJEFBMPQBNQ XIJDIPGUIFGPMMPXJOH
TUBUFNFOUTBSFOPUUSVF
B 5IFEJGGFSFOUJBMWPMUBHFBDSPTTUIFJOQVU
UFSNJOBMTJT[FSP
C 5IFDVSSFOUJOUPUIFJOQVUUFSNJOBMTJT[FSP
D 5IFDVSSFOUGSPNUIFPVUQVUUFSNJOBMJT[FSP
E 5IFJOQVUSFTJTUBODFJT[FSP
F 5IFPVUQVUSFTJTUBODFJT[FSP
WPMUBHFJT
B −N7
C −N7
D N7
E N7
3FGFSUP'JH*GvTN7 WPMUBHFvBJT
B −N7
C N7
D ∕N7
E N7
5IFQPXFSBCTPSCFECZUIFLΩSFTJTUPSJO
'JHJT
B N8
C N8
D N8
E N8
'PSUIFDJSDVJUJO'JH WPMUBHFvPJT
B −7
D −7
C −7
E −7
+
–
10 kΩ
2 kΩ
1V
–
+
+
–
6V +
–
ix
+
vo
–
3 kΩ
4 kΩ
+
2 kΩ
vo
–
Figure 5.42
'PS3FWJFX2VFTUJPOT
Figure 5.40
'PS3FWJFX2VFTUJPOTBOE
'PSUIFDJSDVJUJO'JH DVSSFOUJYJT
B OPOJOWFSUFS
B μ"
D μ"
C WPMUBHFGPMMPXFS
C μ"
E ∕μ"
D TVNNFS
*GvTJOUIFDJSDVJUPG'JH DVSSFOUJPJT
B −N"
D ∕N"
Which of these amplifiers is used in a digitalUP
BOBMPHDPOWFSUFS
C −N"
E ∕N"
*GvTN7JOUIFDJSDVJUPG'JH UIFPVUQVU
8 kΩ
4 kΩ
a
10 mV +
–
Figure 5.41
–
+
vs +
–
'PS3FWJFX2VFTUJPOT BOE
2 kΩ
io
+
vo
–
E EJGference amplifier
Difference amplifiers are used in (please check all
that apply):
(a) instrumentation amplifiers
C WPMUBHFGPMMPXFST
D WPMUBHFSFHVMBUPST
E CVGGFST
(e) summing amplifiers
G ) subtracting amplifiers
"OTXFSTD D E C C B D E C D B G
200
Chapter 5
Operational Amplifiers
Problems
Section 5.2
Operational Amplifiers
+
741
–
5IFFRVJWBMFOUNPEFMPGBDFSUBJOPQBNQJTTIPXO
JO'JH%FUFSNJOF
B UIFJOQVUSFTJTUBODF
C UIFPVUQVUSFTJTUBODF
D UIFWPMUBHFHBJOJOE#
vo
+–
1 mV
Figure 5.45
'PS1SPC
60 Ω
–
vd
1.5 MΩ
+
+
–
8 × 104vd
5IFPQBNQJO'JHIBT3J=LΩ 3P=Ω "= 'JOEUIFEJGGFSFOUJBM
WPMUBHFvEBOEUIFPVUQVUWPMUBHFvP
Figure 5.43
'PS1SPC
– +
vd
+
–
5IFPQFOMPPQHBJOPGBOPQBNQJT $BMDVMBUFUIFPVUQVUWPMUBHFXIFOUIFSFBSFJOQVUT
PG+μV on the inverting terminal and +μ7
POUIFOPOJOWFSUJOHUFSNJOBM
%FUFSNJOFUIFWPMUBHFJOQVUUPUIFJOWFSUJOHUFSNJOBM
PGBOPQBNQXIFO¢μ7JTBQQMJFEUPUIFOPO
JOWFSUJOHUFSNJOBMBOEUIFPVUQVUUISPVHIBOPQFO
MPPQHBJOPG JT7
5IFPVUQVUWPMUBHFPGBOPQBNQJT−7XIFOUIF
OPOJOWFSUJOHJOQVUJTN7*GUIFPQFOMPPQHBJO
of the op amp is 2 × XIBUJTUIFJOWFSUJOH
JOQVU
10 kΩ
1 mV
'PS1SPC
Section 5.3
vi
+
–
Ideal Op Amp
0CUBJOvPGPSFBDIPGUIFPQBNQDJSDVJUTJO'JH
10 kΩ
2 kΩ
–
+
+
1 mA
–
2V +
+
vo
–
vo
–
+
1V +
–
2 kΩ
–
(a)
Figure 5.44
'PS1SPC
+
vo
–
+
–
Figure 5.46
'PSUIFPQBNQDJSDVJUPG'JH UIFPQBNQIBT
BOPQFOMPPQHBJOPG BOJOQVUSFTJTUBODFPG
LΩ BOEBOPVUQVUSFTJTUBODFPGΩ'JOEUIF
WPMUBHFHBJOvP∕vJVTJOHUIFOPOJEFBMNPEFMPGUIF
op amp.
–
+
100 kΩ
6TJOHUIFTBNFQBSBNFUFSTGPSUIFPQBNQ
in Example 5.1, find vPJOUIFPQBNQDJSDVJUPG
Fig. 5.45.
(b)
Figure 5.47
'PS1SPC
%FUFSNJOFvPGPSFBDIPGUIFPQBNQDJSDVJUTJO
Fig. 5.48.
+
vo
–
201
Problems
25 kΩ
2 kΩ
5 kΩ
–
+
+
+
– 4V
1 mA
vs
vo
–
+
+
–
+
vo
–
10 kΩ
–
Figure 5.51
'PS1SPC
+
–
+
–
3V +
–
'JOEvPBOEJPJOUIFDJSDVJUPG'JH
1V
2 kΩ
+
vo
–
10 kΩ
1V
Figure 5.48
+
–
io
+
–
90 kΩ
'PS1SPC
+
vo
–
100 kΩ
10 kΩ
50 kΩ
'JOEUIFHBJOvP∕vTPGUIFDJSDVJUJO'JH
Figure 5.52
20 kΩ
'PS1SPC
+
–
+
10 kΩ
vs +
–
%FUFSNJOFUIFPVUQVUWPMUBHFvPJOUIFDJSDVJUPG
Fig. 5.53.
10 kΩ
vo
10 kΩ
–
20 kΩ
Figure 5.49
2 mA
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFSTUV
EFOUTCFUUFSVOEFSTUBOEIPXJEFBMPQBNQTXPSL
R1
–
R3
V +
–
R4
'PS1SPC
R5
+
vo
–
5 kΩ
Inverting Amplifier
B %FUFSNJOFUIFSBUJPvP∕JTJOUIFPQBNQDJSDVJUPG
'JH
io
+
–
+
Figure 5.53
Section 5.4
R2
10 kΩ
+
vo
–
C &WBMVBUFUIFSBUJPGPS3=LΩ 3=LΩ 3=LΩ
R1
R3
R2
Figure 5.50
–
+
'PS1SPC
is
$BMDVMBUFUIFWPMUBHFSBUJPvP∕vTGPSUIFPQBNQ
DJSDVJUPG'JH"TTVNFUIBUUIFPQBNQJT
JEFBM
+
vo
–
Figure 5.54
'PS1SPC
202
Chapter 5
Operational Amplifiers
6TJOH'JH EFTJHOBQSPCMFNUPIFMQTUVEFOUT
CFUUFSVOEFSTUBOEJOWFSUJOHPQBNQT
%FUFSNJOFJPJOUIFDJSDVJUPG'JH
2 kΩ
4 kΩ
10 kΩ
R3
R1
ix
+
–
V
iy
–
+
750 mV +
–
io
–
+
4 kΩ
2 kΩ
Figure 5.58
R4
'PS1SPC
R2
*OUIFDJSDVJUPG'JH DBMDVMBUFvPPGvT=7
8 kΩ
Figure 5.55
'PS1SPC
4 kΩ
$BMDVMBUFUIFHBJOvP∕vJXIFOUIFTXJUDIJO'JH
JTJO
B QPTJUJPO
C QPTJUJPO
80 kΩ
5 kΩ
vi
+
–
–
+
+
vo
vs +
–
–
Figure 5.59
'PS1SPC
1
2 MΩ
9V +
–
D QPTJUJPO
12 kΩ
2 kΩ
4 kΩ
2
$BMDVMBUFvPJOUIFPQBNQDJSDVJUPG'JH
10 kΩ
3
–
+
4 kΩ
10 kΩ
+
vo
–
–
+
4V +
–
+
vo
2V +
–
–
Figure 5.56
'PS1SPC
Figure 5.60
'PS1SPC
'PSUIFDJSDVJUTIPXOJO'JHVSF TPMWFGPSUIF
5IFWFOJOFRVJWBMFOUDJSDVJUMPPLJOHJOUPUFSNJOBMT"
BOE#
For the op amp circuit in Fig. 5.61, find the voltage
HBJOvP∕vT
10 kΩ
10 kΩ
+
9V –
Design an inverting amplifier with a gain of −
Rf
–
+
R1
2.5 Ω
vs +
–
R2
+
vo
–
Figure 5.57
'PS1SPC
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
–
+
Figure 5.61
'PS1SPC
203
Problems
In the circuit shown in Fig. 5.62, find LJOUIFWPMUBHF
USBOTGFSGVODUJPOvP=LvT
'JOEJPJOUIFPQBNQDJSDVJUPG'JH
50 kΩ
Rf
R1
–
+
R2
vs
–+
vo
Figure 5.66
–
'PS1SPC
Figure 5.62
%FUFSNJOFUIFWPMUBHFHBJOvP∕vJPGUIFPQBNQ
DJSDVJUJO'JH
'PS1SPC
Section 5.5
20 kΩ
+
R4
R3
+ 0.4 V
–
10 kΩ
–
+
io
Noninverting Amplifier
R1
+
–
$BMDVMBUFvPJOUIFPQBNQDJSDVJUPG'JH
vi +
–
12 kΩ
–
+
+
15 V –
20 kΩ
+
vo
–
R2
R1
+
vo
R2
–
Figure 5.67
'PS1SPC
In the circuit shown in Fig. 5.68, find JYBOEUIF
QPXFSBCTPSCFECZUIFLΩSFTJTUPS
Figure 5.63
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFSTUV
EFOUTCFUUFSVOEFSTUBOEOPOJOWFSUJOHPQBNQT
+
–
V +
–
io
ix
+
3.6 V –
30 kΩ
20 kΩ
R3
R2
R1
60 kΩ
–
+
Figure 5.68
'PS1SPC
Figure 5.64
For the circuit in Fig. 5.69, find JY
'PS1SPC
12 kΩ
'JOEvPJOUIFPQBNQDJSDVJUPG'JH
16 Ω
7.5 V +
–
Figure 5.65
'PS1SPC
v1
–
+
24 Ω
6 kΩ
v2 8 Ω
ix
12 Ω
+
vo
–
25 mA
3 kΩ
6 kΩ
+
–
+
vo
–
Figure 5.69
'PS1SPC
204
Chapter 5
Operational Amplifiers
$BMDVMBUFJYBOEvPJOUIFDJSDVJUPG'JH'JOEUIF
QPXFSEJTTJQBUFECZUIFLΩSFTJTUPS
ix
+
–
10 mV +
–
+
–
vs +
–
R1
20 kΩ
R2
b
50 kΩ
60 kΩ
30 kΩ
10 kΩ
+
vo
–
Figure 5.73
'PS1SPC
Section 5.6
Summing Amplifier
Determine the output of the summing amplifier in
'JH
Figure 5.70
3V
'PS1SPC
–+
5.4 V
3FGFSUPUIFPQBNQDJSDVJUJO'JH$BMDVMBUFJY
BOEUIFQPXFSBCTPSCFECZUIFLΩSFTJTUPS
–+
–3.5 V
+–
1 kΩ
ix
+
–
4 kΩ
2.5 mA
3 kΩ
2 kΩ
'PS1SPC
V2
+–
(JWFOUIFPQBNQDJSDVJUTIPXOJO'JH FYQSFTT
vPJOUFSNTPGvBOEv
V3
–+
R2
+
–
R3
V4
R4
20 kΩ
–
+
30 kΩ
+
vo
–
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFSTUV
dents better understand summing amplifiers.
V1
vin
30 kΩ
'PS1SPC
–+
R1
10 kΩ
Figure 5.74
Figure 5.71
v1
v2
a
+–
+
vo
–
Figure 5.72
'PS1SPC
R1
+
R2
–
R3
+
vo
–
R5
R4
Figure 5.75
'PS1SPC
'PSUIFPQBNQDJSDVJUJO'JH EFUFSNJOFUIF
WBMVFPGvJOPSEFSUPNBLFvP=−7
10 kΩ
50 kΩ
–3 V
v2
Design a noninverting amplifier with a gain of 7.5.
For the circuit shown in Fig. 5.73, find the Thevenin
FRVJWBMFOUBUUFSNJOBMTBC )JOU To find 35I BQQMZ
BDVSSFOUTPVSDFJPBOEDBMDVMBUFvP
5V
Figure 5.76
'PS1SPC
20 kΩ
50 kΩ
–
+
vo
205
Problems
3FGFSSJOHUPUIFDJSDVJUTIPXOJO'JH EFUFS
NJOF7PJOUFSNTPG7BOE7
100 kΩ
6TJOHPOMZUXPPQBNQT EFTJHOBDJSDVJUUPTPMWF
v− v @@@@@@
v − v
+ −vPVU=@@@@@@
200 kΩ
Section 5.7
100 kΩ
V1 +
–
–
+
V2 +
–
10 Ω
+
40 Ω
Difference Amplifier
The circuit in Fig. 5.79 is for a difference amplifier.
'JOEvPHJWFOUIBUv=7BOEv=7
Vo
–
30 kΩ
2 kΩ
Figure 5.77
'PS1SPC
"Oaveraging amplifierJTBTVNNFSUIBUQSPWJEFT
BOPVUQVUFRVBMUPUIFBWFSBHFPGUIFJOQVUT#Z
VTJOHQSPQFSJOQVUBOEGFFECBDLSFTJTUPSWBMVFT one can get
−vPVU=@@
v+ v+ v+ v
v1 +
–
v2 +
–
+
+
vo
20 kΩ
–
Figure 5.79
'PS1SPC
6TJOHBGFFECBDLSFTJTUPSPGLΩ EFTJHOBO
averaging amplifier with four inputs.
5IFGFFECBDLSFTJTUPSPGBUISFFJOQVUBWFSBHJOH
summing amplifier is 50 kΩ8IBUBSFUIFWBMVFTPG
3 3 BOE3
The circuit in Fig. 5.80 is a differential amplifier
ESJWFOCZBCSJEHF'JOEvP
The feedback resistor of a fiveJOQVUBWFSBHJOHTVN
ming amplifier is 40 kΩ8IBUBSFUIFWBMVFTPG3 3 3 3 BOE3
4IPXUIBUUIFPVUQVUWPMUBHFvPPGUIFDJSDVJUJO
Fig. 5.78 is
3 + 3
vP=@@@@@@@@@
3 v + 3v
3 3+ 3 20 kΩ
10 kΩ
R3
v1
v2
R1
–
–
+
40 kΩ
+
vo
60 kΩ
20 kΩ
vo
80 kΩ
30 kΩ
15.5 mV
R4
80 kΩ
Figure 5.80
'PS1SPC
R2
Figure 5.78
Design a difference amplifier to have a gain of 4 and
BDPNNPONPEFJOQVUSFTJTUBODFPGLΩBUFBDI
JOQVU
'PS1SPC
%FTJHOBOPQBNQDJSDVJUUPQFSGPSNUIFGPMMPXJOH
PQFSBUJPO
vP=v−v
–
2 kΩ
"MMSFTJTUBODFTNVTUCF≤LΩ
%FTJHOBDJSDVJUUPBNQMJGZUIFEJGGFSFODFCFUXFFO
UXPJOQVUTCZ
B 6TFPOMZPOFPQBNQ
C 6TFUXPPQBNQT
206
Chapter 5
Operational Amplifiers
6TJOHUXPPQBNQT EFTJHOBTVCUSBDUPS
R2
2
R1
%FTJHOBOPQBNQDJSDVJUTVDIUIBU
vP=v+ v− v− v
-FUBMMUIFSFTJTUPSTCFJOUIFSBOHFPGUPLΩ
RG
–
+
vi
–
+
+
R1
The ordinary difference amplifier for fixedHBJO
PQFSBUJPOJTTIPXOJO'JH B *UJTTJNQMFBOE
SFMJBCMFVOMFTTHBJOJTNBEFWBSJBCMF0OFXBZPG
QSPWJEJOHHBJOBEKVTUNFOUXJUIPVUMPTJOHTJNQMJDJUZ
BOEBDDVSBDZJTUPVTFUIFDJSDVJUJO'JH C "OPUIFSXBZJTUPVTFUIFDJSDVJUJO'JH D 4IPXUIBU
R2
2
vo
R2
2
R2
2
–
(c)
Figure 5.81
'PS1SPC
B GPSUIFDJSDVJUJO'JH B
Section 5.8 Cascaded Op Amp Circuits
v
3
@@@
vP= @@@@
J
3
%FUFSNJOFUIFWPMUBHFUSBOTGFSSBUJPvP∕vTJOUIFPQ
BNQDJSDVJUPG'JH XIFSF3=LΩ
C GPSUIFDJSDVJUJO'JH C
vP @@@@
3 @@@@@@@
@@@
vJ= 3
3
+@@@@
3(
R
R
D GPSUIFDJSDVJUJO'JH D
R
v
3
3
@@
vP= @@@
+@@@@
J
3 (
3( )
+
R
vo
R2
–
(a)
R1
2
–
vi
+
R2
R1
2
+
R1
2
R2
'PS1SPC
*OBDFSUBJOFMFDUSPOJDEFWJDF BUISFFstage amplifier
JTEFTJSFE XIPTFPWFSBMMWPMUBHFHBJOJTE#5IF
individual voltage gains of the first two stages are
UPCFFRVBM XIJMFUIFHBJOPGUIFUIJSEJTUPCFPOF
fourth of each of the first two. Calculate the voltage
HBJOPGFBDI
+
vo
–
(b)
–
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEDBTDBEFEPQBNQT
–
RG
R1
2
vo
R
Figure 5.82
+
+
R1
+
–
–
–
vi
+
+
–
vs
R2
R1
–
+
R2
R1
+
vi
–
Figure 5.83
'PS1SPC
–
+
R4
R3
–
+
207
Problems
'JOEvPJOUIFPQBNQDJSDVJUPG'JH
vs1
25 kΩ
50 kΩ
100 kΩ
–
+
100 kΩ
+
–
20 kΩ –0.2 V 10 kΩ
vo
1.2 V
–
+
50 kΩ
%FUFSNJOFvPJOUIFDJSDVJUPG'JH
10 kΩ
20 kΩ
–
+
100 kΩ
40 kΩ
–
+
50 kΩ
vs2
vo
Figure 5.88
'PS1SPC
Figure 5.84
'PS1SPC
0CUBJOUIFDMPTFEMPPQWPMUBHFHBJOvP∕vJPGUIF
DJSDVJUJO'JH
$BMDVMBUFJPJOUIFPQBNQDJSDVJUPG'JH
Rf
10 kΩ
1 kΩ
–
+
1.8 V +
–
R2
2 kΩ
–
+
5 kΩ
R1
io
vi
4 kΩ
3 kΩ
R3
–
+
+
–
–
+
+
vo
R4
–
Figure 5.89
Figure 5.85
'PS1SPC
'PS1SPC
*OUIFPQBNQDJSDVJUPG'JH EFUFSNJOFUIF
WPMUBHFHBJOvP∕vT5BLF3=LΩ
2R
R
R3
4R
R
–
+
%FUFSNJOFUIFHBJOvP∕vJPGUIFDJSDVJUJO'JH
R2
R1
–
+
+
vs +
–
–
+
vi +
–
vo
R4
R5
R6
–
+
+
vo
–
–
Figure 5.90
'PS1SPC
Figure 5.86
'PS1SPC
$BMDVMBUFvP∕vJJOUIFPQBNQDJSDVJUPG'JH
For the op amp circuit shown in Fig. 5.91, find
vP∕vT
G4
+
+
–
100 kΩ
40 kΩ
50 kΩ
vi
100 kΩ
–
Figure 5.87
'PS1SPC
G
80 kΩ
G1
–
+
+
vo
–
vs +
–
Figure 5.91
'PS1SPC
–
+
G3
G
G2
–
+
+
vo
–
208
Chapter 5
Operational Amplifiers
'JOEvPJOUIFDJSDVJUPG'JH BTTVNJOHUIBU
3G=∞ PQFODJSDVJU 'JOEvPJOUIFPQBNQDJSDVJUPG'JH
10 kΩ
100 kΩ
+
–
100 kΩ
Rf
100 kΩ
–
+
30 kΩ
70 mV +
–
30 kΩ
–
+
70 kΩ
15 kΩ
+
5 kΩ
vo
–
+
–
Figure 5.92
15 mV
'PS1SPC
For the circuit in Fig. 5.93, find vP
+
–
+
–
+
vo
–
6 kΩ
2 kΩ
1 kΩ
Figure 5.95
'PS1SPCTBOE
25 kΩ
40 kΩ
20 kΩ
6V +
–
20 kΩ
–
+
+
4V –
3FQFBUUIFQSFWJPVTQSPCMFNJG3G=LΩ
100 kΩ
–
+
10 kΩ
+
vo
2V +
–
%FUFSNJOFvPJOUIFPQBNQDJSDVJUPG'JH
–
Figure 5.93
30 kΩ
'PS1SPC
10 kΩ
A
60 kΩ
10 kΩ
80 kΩ
–
+
1.2 V +
–
20 kΩ
–
+
20 kΩ
–
+
2V +
–
vo
20 kΩ
10 kΩ
3V
10 kΩ
–
+
+
–
2.8 V +
–
4V
Figure 5.94
Figure 5.96
'PS1SPC
–
+
+
–
'PS1SPC
–
C
+
10 kΩ
80 kΩ
40 kΩ
20 kΩ
+
1V +
–
0CUBJOUIFPVUQVUvPJOUIFDJSDVJUPG'JH
–
40 kΩ
B
vo
209
Problems
'JOEJPJOUIFPQBNQDJSDVJUPG'JH
%FUFSNJOFvPJOUIFPQBNQDJSDVJUPG'JH
100 kΩ
10 kΩ
20 kΩ
5 kΩ
40 kΩ
–
+
1.5 V +
–
80 kΩ
10 kΩ
10 kΩ
–
+
+
2.25 V –
–
+
+
20 kΩ
vo
–
+ 0.6 V
–
Figure 5.100
'PS1SPC
Section 5.9
50 kΩ
1.6 kΩ
–
+
20 kΩ
+
0.9 V –
+
–
30 kΩ
io
–
+
100 kΩ
32 kΩ
Op Amp Circuit Analysis with
PSpice
3FXPSL&YBNQMFVTJOHUIFOPOJEFBMPQBNQ
-.JOTUFBEPGV"
4PMWF1SPCVTJOH14QJDFPS.VMUJ4JNBOE
PQBNQV"
Figure 5.97
'PS1SPC
4PMWF1SPCVTJOH14QJDFPS.VMUJ4JNBOE
PQBNQ-.
'JOEUIFMPBEWPMUBHFv-JOUIFDJSDVJUPG'JH
6TF14QJDFPS.VMUJ4JNUPPCUBJOvPJOUIFDJSDVJUPG
'JH
10 kΩ
100 kΩ
20 kΩ
5.6 V +
–
30 kΩ
40 kΩ
250 kΩ
–
+
–
+
–
+
20 kΩ
2 kΩ
–
+
1V +
–
+
vL
–
2V +
–
+
vo
–
Figure 5.101
'PS1SPC
Figure 5.98
'PS1SPC
%FUFSNJOFvPJOUIFPQBNQDJSDVJUPG'JH VTJOH14QJDFPS.VMUJ4JN
%FUFSNJOFUIFMPBEWPMUBHFv-JOUIFDJSDVJUPG
Fig. 5.99.
5V +
–
50 kΩ
10 kΩ
5 kΩ
+ 5.4 V
–
Figure 5.99
'PS1SPC
–
+
20 kΩ
10 kΩ
+
–
–
100 kΩ
+
20 kΩ
4 kΩ
+
vL
–
1V
+
–
Figure 5.102
'PS1SPC
10 kΩ
40 kΩ
–
+
+
vo
–
210
Chapter 5
Operational Amplifiers
6TF14QJDFPS.VMUJ4JNUPTPMWF1SPC
6TF14QJDFPS.VMUJ4JNUPWFSJGZUIFSFTVMUTJO
&YBNQMF"TTVNFOPOJEFBMPQBNQT-.
%FTJHOBWPMUBHFDPOUSPMMFEJEFBMDVSSFOUTPVSDF
XJUIJOUIFPQFSBUJOHMJNJUTPGUIFPQBNQ XIFSFUIF
PVUQVUDVSSFOUJTFRVBMUPvT U μ"
Section 5.10 Applications
"GPVSbit DAC covers a voltage range of 0 to 10 V.
$BMDVMBUFUIFSFTPMVUJPOPGUIF%"$JOWPMUTQFS
EJTDSFUFCJOBSZTUFQ
%FTJHOBTJYCJUEJHJUBMUPBOBMPHDPOWFSUFS
B *G]7P]=7JTEFTJSFE XIBUTIPVME
<777777>CF
C $BMDVMBUF]7P]JG<777777>=<>
D 8IBUJTUIFNBYJNVNWBMVF]7P]DBOBTTVNF
'JHVSFEJTQMBZTBUXPPQBNQJOTUSVNFOUB
tion amplifier. Derive an expression for vPJOUFSNT
PGv and v. How can this amplifier be used as a
TVCUSBDUPS
"GPVSCJU33MBEEFS%"$JTQSFTFOUFEJO'JH
v1
B 4IPXUIBUUIFPVUQVUWPMUBHFJTHJWFOCZ
7
7
7
7
@@@
+ @@@
+ @@@
+ @@@
−7P=3G(
3
3
3
3 )
–
R2
C *G3G=LΩBOE3=LΩ, find |7P]GPS
<7777>=<>BOE<7777>=<>
Rf
V1
V2
V3
V4
2R
2R
2R
2R
R4
+
R3
v2
R1
–
vo
+
Figure 5.105
–
+
Vo
'PS1SPC
R
R
R
Figure 5.106 shows an instrumentation amplifier
ESJWFOCZBCSJEHF0CUBJOUIFHBJOvP∕vJPGUIF
amplifier.
2R
Figure 5.103
20 kΩ
'PS1SPC
In the op amp circuit of Fig. 5.104, find the value of
3TPUIBUUIFQPXFSBCTPSCFECZUIFLΩSFTJTUPSJT
N8%FUFSNJOFUIFQPXFSHBJO
vi
40 kΩ
80 kΩ
–
+
+
1 kΩ
25 kΩ
500 kΩ
Figure 5.106
'PS1SPC
500 kΩ
–
+
10 kΩ
–
'PS1SPC
25 kΩ
2 kΩ
20 kΩ
Figure 5.104
+
–
10 kΩ
R
10 V +
–
30 kΩ
vo
211
Comprehensive Problems
Comprehensive Problems
%FTJHOBDJSDVJUUIBUQSPWJEFTBSFMBUJPOTIJQCFUXFFO
PVUQVUWPMUBHFvPBOEJOQVUWPMUBHFvTTVDIUIBU
vP=vT−5XPPQBNQT B7CBUUFSZ BOE
TFWFSBMSFTJTUPSTBSFBWBJMBCMF
3FGFSUPUIFbridge amplifierTIPXOJO'JH
%FUFSNJOFUIFWPMUBHFHBJOvP∕vJ
90 kΩ
30 kΩ
5IFPQBNQDJSDVJUJO'JHJTBDVSSFOUBNQMJ
fier'JOEUIFDVSSFOUHBJOJP∕JT of the amplifier.
40 kΩ
50 kΩ
20 kΩ
–
+
4 kΩ
io
5 kΩ
is
2 kΩ
vi
–
+
+
–
+
vo
–
Figure 5.109
'PS1SPC
R3
'PS1SPC
R1
A noninverting current amplifier is portrayed in
Fig. 5.108. Calculate the gain JP∕JT5BLF3=LΩ
BOE3=LΩ
+
–
+
vi
R1
is
–
+
R4
R2
iL
R2
io
–
R2
Figure 5.110
'PS1SPC
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Figure 5.108
–
+
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c h a p t e r
6
Capacitors and
Inductors
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the man to whom the idea first occurs.
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Enhancing Your Skills and Your Career
ABET EC 2000 criteria (3.c), “an ability to design a system,
component, or process to meet desired needs.”
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Probably the most important part of design is clearly defining what
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213
214
Chapter 6
Capacitors and Inductors
Learning Objectives
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In contrast to a resistor, which spends
or dissipates energy irreversibly, an
inductor or capacitor stores or releases
energy (i.e., has a memory).
Dielectric with permittivity є
Metal plates,
each with area A
6.1
Introduction
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6.2
Capacitors
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A capacitor consists of two conducting plates separated by an insulator
(or dielectric).
d
Figure 6.1
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6.2
215
Capacitors
Historical
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–
+
–
+
–
+
+
v
–q
–
Figure 6.2
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Capacitance is the ratio of the charge on one plate of a capacitor to the
voltage difference between the two plates, measured in farads (F).
Alternatively, capacitance is the amount
of charge stored per plate for a unit
voltage difference in a capacitor.
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Capacitor voltage rating and capacitance are typically inversely rated due
to the relationships in Eqs. (6.1) and
(6.2). Arcing occurs if d is small and V
is high.
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216
Chapter 6
i
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(a)
Capacitors and Inductors
Figure 6.3
Circuit symbols for capacitors: (a) fixFE
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(a)
(b)
(c)
Figure 6.4
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(a)
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in metal or plastic films. Electrolytic capacitors produce vFSZIJHI
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(b)
Figure 6.5
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$PVSUFTZPG+PIBOTPO
According to Eq. (6.4), for a capacitor
to carry current, its voltage must vary
with time. Hence, for constant voltage,
i =0.
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6.2
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∫
v U = @@
t
$ −∞
217
Capacitors
J τ Eτ
i
Slope = C
PS
0
dv/dt
Figure 6.6
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U
∫ J τ Eτ+W U
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$ U
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Ev
EU
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U
vU
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v −∞
−∞
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v −∞
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field.
8FTIPVMEOPUFUIFGPMMPXJOHJNQPSUBOUQSPQFSUJFTPGBDBQBDJUPS
/PUFGSPN&R UIBUXIFOUIFW PMUBHFBDSPTTBDBQBDJUPSJTOPU
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UPSJT[FSP5IVT
A capacitor is an open circuit to dc.
)PXFWFS JGBCBUUFSZ EDWPMUBHF JTDPOOFDUFEBDSPTTBDBQBDJUPS UIFDBQBDJUPSDIBSHFT
An alternative way of looking at this is
using Eq. (6.9), which indicates that
energy is proportional to voltage
squared. Since injecting or extracting
energy can only be done over some
finite time, voltage cannot change
instantaneously across a capacitor.
218
Chapter 6
5IFWPMUBHFPOUIFDBQBDJUPSNVTUCFDPOUJOVPVT
v
v
Capacitors and Inductors
The voltage on a capacitor cannot change abruptly.
t
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t
(a)
(b)
Figure 6.7
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QPTTJCMF
Leakage resistance
Capacitance
Figure 6.8
$JSDVJUNPEFMPGBOPOJEFBMDBQBDJUPS
Example 6.1
B $BMDVMBUFUIFDIBSHFTUPSFEPOBQ'DBQBDJUPSXJUI7BDSPTTJU
C 'JOEUIFFOFSHZTUPSFEJOUIFDBQBDJUPS
Solution:
B 4JODFR=$v
R=×−×=Q$
C 5IFFOFSHZTUPSFEJT
$v=@@
××−×=Q+
X=@@
Practice Problem 6.1
8IBUJTUIFWPMUBHFBDSPTTB μ'DBQBDJUPSJGUIFDIBSHFPOPOFQMBUF
JTN$ )PXNVDIFOFSHZJTTUPSFE
Answer:7 N+
Example 6.2
5IFWPMUBHFBDSPTTBμ'DBQBDJUPSJT
v U =DPTU7
$BMDVMBUFUIFDVSSFOUUISPVHIJU
Solution:
By definition, the current is
J U =$@@@
Ev=×−@@
E DPTU
EU
EU
=−×−××TJOU=−TJOU"
6.2
219
Capacitors
*GBμ'DBQBDJUPSJTDPOOFDUFEUPBWPMUBHFTPVSDFXJUI
Practice Problem 6.2
v U =DPT U 7
EFUFSNJOFUIFDVSSFOUUISPVHIUIFDBQBDJUPS
Answer:−TJO U N"
Example 6.3
%FUFSNJOFUIFWPMUBHFBDSPTTBμ'DBQBDJUPSJGUIFDVSSFOUUISPVHIJUJT
−U
J U =F
N"
"TTVNFUIBUUIFJOJUJBMDBQBDJUPSWPMUBHFJT[FSP
Solution:
U JEτ+v BOEv =
4JODFv=@@
∫
$ U
v=@@@@@@@@
F−τEτ·−
− ∫
×
|
U
×
=@@@@@@@
−F−U 7
F−τ =
−
5IFDVSSFOUUISPVHIBμ'DBQBDJUPSJTJ U =TJOπUN"$BMDV
MBUFUIFWPMUBHFBDSPTTJUBUU=NTBOEU=NT5BLFv =
Practice Problem 6.3
Answer:N7 7
%FUFSNJOF UIFDVSSFOU UISPVHIB μ'DBQBDJUPS XIPTF WPMUBHFJT
TIPXOJO'JH
Solution:
5IFWPMUBHFXBWFGPSNDBOCFEFTDSJCFENBUIFNBUJDBMMZBT
{
U7
−U7
v U =
−+U7 {
<U<
<U<
<U<
PUIFSXJTF
4JODFJ=$Ev∕EUBOE$=μ' XFUBLFUIFEFSJWBUJWFPGvUPPCUBJO
−
J U =×−×
{
N"
−N"
=
N" <U<
<U<
<U< PUIFSXJTF
<U<
<U<
<U< PUIFSXJTF
5IVT UIFDVSSFOUXBWFGPSNJTBTTIPXOJO'JH
Example 6.4
v(t)
50
0
1
2
3
t
4
–50
Figure 6.9
'PS&YBNQMF
i (mA)
10
0
1
–10
Figure 6.10
'PS&YBNQMF
2
3
4
t
220
Chapter 6
Practice Problem 6.4
i (mA)
100
Capacitors and Inductors
An initially uncharged 1-mF capacitor has the current shown in Fig. BDSPTTJU$BMDVMBUFUIFWPMUBHFBDSPTTJUBUU=NTBOEU=NT
Answer:N7 N7
0
2
4
6
t (ms)
Figure 6.11
'PS1SBDUJDF1SPC
Example 6.5
0CUBJOUIFFOFS HZTUPSFEJOFBDIDBQBDJUPSJO'JH B VOEFSED
DPOEJUJPOT
2 mF
Solution:
6OEFSEDDPOEJUJPOT XFSFQMBDFFBDIDBQBDJUPSXJUIBOPQFODJSDVJU BT
TIPXOJO'JH C 5IFDVSSFOUUISPVHIUIFTFSJFTDPNCJOBUJPOPGUIF
LΩBOELΩSFTJTUPSTJTPCUBJOFECZDVSSFOUEJWJTJPOBT
2 kΩ
6 mA
5 kΩ
3 kΩ
4 kΩ
J=@@@@@@@@
N" =N"
++
4 mF
)FODF UIFWPMUBHFTvBOEvBDSPTTUIFDBQBDJUPSTBSF
(a)
v=J=7
+ v1 –
BOEUIFFOFSHJFTTUPSFEJOUIFNBSF
i
2 kΩ
5 kΩ
6 mA
3 kΩ
+
v2
–
v=J=7
4 kΩ
$ v=@@
×− =N+
X=@@
$ v =@@
×− =N+
X=@@
(b)
Figure 6.12
'PS&YBNQMF
Practice Problem 6.5
Under dc conditions, find the energy stored in the capacitors in Fig. 6.13.
Answer:N+ N+
3 kΩ
1 kΩ
50 V +
–
30 μF
20 μF
Figure 6.13
'PS1SBDUJDF1SPC
6 kΩ
6.3
Series and Parallel Capacitors
8FLOPXGSPNSFTJTUJ WFDJSDVJUTUIBUUIFTFSJFTQBSBMMFMDPNCJOBUJPOJTB QPXFSGVMUPPMGPSSFEVDJOHDJSDVJUT 5IJTUFDIOJRVFDBOCFF YUFOEFEUP
TFSJFTQBSBMMFMDPOOFDUJPOTPGDBQBDJUPST XIJDIBSFTPNFUJNFTFODPVO UFSFE8FEFTJSFUPSFQMBDFUIFTFDBQBDJUPSTCZBTJOHMFFRVJWBMFOUDBQBDJ
UPS$FR
*OPSEFSUPPCUBJOUIFFRVJ WBMFOUDBQBDJUPS $FRPG /DBQBDJUPSTJO
QBSBMMFM DPOTJEFSUIFDJSDVJUJO'JH B 5IFFRVJ WBMFOUDJSDVJUJT
6.3
JO'JH C /PUFUIBUUIFDBQBDJUPSTIB WFUIFTBNFW PMUBHFvBDSPTT
UIFN"QQMZJOH,$-UP'JH B
J=J+J+J+ ⋯+J/
221
Series and Parallel Capacitors
i
i1
i2
i3
iN
C1
C2
C3
CN
–
(a)
#VUJL=$LEv∕EU)FODF
J=$@@@
Ev+$@@@
Ev+$@@@
Ev+⋯+$/@@@
Ev
EU
EU
EU
EU
=(
∑$L)
/
L=
@@@
Ev=$FR@@@
Ev
EU
i
$FR=$+$+$+⋯+$/
+
v
–
(b)
Figure 6.14
XIFSF
Ceq
EU B 1BSBMMFMDPOOFDUFE/DBQBDJUPST C FRVJWBMFOUDJSDVJUGPSUIFQBSBMMFM
DBQBDJUPST
The equivalent capacitance of N parallel-connected capacitors is the
sum of the individual capacitances.
8FPCTFSWFUIBUDBQBDJUPST JOQBSBMMFMDPNCJOF JOUIFTBNF NBOOFSBT
SFTJTUPSTJOTFSJFT
8FOP XPCUBJO $FRPG /DBQBDJUPSTDPOOFDUFEJOTFSJFTCZ
DPNQBSJOHUIFDJSDVJUJO'JH B XJUIUIFFRVJ WBMFOUDJSDVJUJO
Fig. C /PUFUIBUUIFTBNFDVSSFOUJfloXT BOEDPOTFRVFOUMZUIF
TBNFDIBS HF UISPVHIUIFDBQBDJUPST "QQMZJOH,7-UPUIFMPPQJO
'JH B
v=v+v+v+⋯+v/
i
v +
–
C1
C2
C3
CN
+ v1 –
+ v2 –
+ v3 –
+ vN –
U
J τ Eτ + v U 5IFSFGPSF
#VU vL = @@@
∫
L $L U
(a)
i
J τ Eτ+v U +@@@
UJ τ Eτ+v U
v=@@@
$ U
$ U
∫
U
∫
v +
–
U
J τ Eτ+v U
+⋯+@@@
∫
/ $/ U
+⋯+@@@
UJ τ Eτ+v U +v U
= @@@
+@@@
∫
$/ ) U
( $ $
U
XIFSF
J τ Eτ+v U
=@@@
∫
$FR U
=@@@
+@@@
+@@@
+⋯+@@@
@@@
$FR $ $ $
$/ Ceq
+
v
–
(b)
Figure 6.15
B 4FSJFTDPOOFDUFE/DBQBDJUPST C FRVJWBMFOUDJSDVJUGPSUIFTFSJFT
DBQBDJUPS
+⋯+v/ U
+
v
222
Chapter 6
Capacitors and Inductors
5IFJOJUJBMWPMUBHFv U BDSPTT$FRJTSFRVJSFECZ,7-UPCFUIFTVNPG
UIFDBQBDJUPSWPMUBHFTBUU0SBDDPSEJOHUP&R v U =v U +v U +⋯+v/ U
5IVT BDDPSEJOHUP&R The equivalent capacitance of series-connected capacitors is the
reciprocal of the sum of the reciprocals of the individual capacitances.
/PUFUIBUDBQBDJUPSTJOTFSJFTDPNCJOFJOUIFTBNFNBOOFSBTSFTJTUPST
JOQBSBMMFM'PS/= JF UXPDBQBDJUPSTJOTFSJFT &R CFDPNFT
+@@@
@@@
=@@@
$FR
$
$
PS
$$
$FR=@@@@@@@
$+$
Example 6.6
'JOEUIFFRVJ WBMFOUDBQBDJUBODFTFFOCFUXFFOUFSNJOBMT BBOE CPGUIF
DJSDVJUJO'JH
5 μF
60 μF
a
20 μF
6 μF
20 μF
Ceq
b
Figure 6.16
'PS&YBNQMF
Solution:
5IF μ'BOE μ'DBQBDJUPSTBSFJOTFSJFTUIFJSFRVJWBMFOUDBQBDJ
tance is
×
@@@@@@
=μ'
+
5IJT μ'DBQBDJUPSJTJOQBSBMMFMXJUIUIF μ'BOE μ'DBQBDJUPST
UIFJSDPNCJOFEDBQBDJUBODFJT
++=μ'
5IJT μ'DBQBDJUPSJTJOTFSJFTXJUIUIF μ'DBQBDJUPS)FODF UIF
FRVJWBMFOUDBQBDJUBODFGPSUIFFOUJSFDJSDVJUJT
×
$FR=@@@@@@@
=μ'
+
6.3
223
Series and Parallel Capacitors
Practice Problem 6.6
'JOEUIFFRVJWBMFOUDBQBDJUBODFTFFOBUUIFUFSNJOBMTPGUIFDJSDVJUJO
'JH
50 μF
60 μF
Answer:μ'
Ceq
70 μF
20 μF
120 μF
Figure 6.17
'PS1SBDUJDF1SPC
Example 6.7
'or the circuit in Fig. 6.18, find the vPMUBHFBDSPTTFBDIDBQBDJUPS
Solution:
We first find the equivalent capacitance $FR TIPXOJO'JH5IFUXP
QBSBMMFMDBQBDJUPSTJO'JHDBOCFDPNCJOFEUPHFU +=N'
5IJTN'DBQBDJUPSJTJOTFSJFTXJUIUIFN'BOEN'DBQBDJUPST
5IVT
5IFUPUBMDIBSHFJT
$FR=@@@@@@@@@@
N'=N'
@@
+@@
+@@
20 mF
30 mF
+ v1 –
+ v2 –
30 V +
–
+
v3
–
40 mF
20 mF
Figure 6.18
'PS&YBNQMF
R=$FRv=×−×=$
This is the charge on the 20-mF and 30-mF capacitors, because they BSF
in series with the 30-V source. (A crude way to see this is UPJNBHJOFUIBU
DIBSHFBDUTMJLFDVSSFOU TJODFJ=ER∕EU 5IFSFGPSF
R
v=@@@=@@@@@@@@@
=7
$ ×−
R
v=@@@=@@@@@@@@@
=7
$ ×−
30 V +
–
Ceq
Figure 6.19
&RVJWBMFOUDJSDVJUGPS'JH
)BWJOHEFUFSNJOFEvBOEv XFOPXVTF,7-UPEFUFSNJOFvCZ
v=−v−v=7
"MUFSOBUJWFMZ TJODFUIFN'BOEN'DBQBDJUPSTBSFJOQBSBMMFM UIFZIBWFUIFTBNFWPMUBHFvBOEUIFJSDPNCJOFEDBQBDJUBODFJT+=
N'5IJTDPNCJOFEDBQBDJUBODFJTJOTFSJFTXJUIUIFN'BOEN'
DBQBDJUPSTBOEDPOTFRVFOUMZIBTUIFTBNFDIBSHFPOJU)FODF
R
v=@@@@@@
=7
=@@@@@@@@@
N' ×−
'JOEUIFWPMUBHFBDSPTTFBDIPGUIFDBQBDJUPSTJO'JH
Answer:v=7 v=7 v=7 v=7
Practice Problem 6.7
40 μF
60 μF
+ v1 –
+
v2
150 V +
–
–
+ v3 –
Figure 6.20
20 μF
'PS1SBDUJDF1SPC
+
v4
–
30 μF
224
Chapter 6
Length, ℓ
Cross-sectional area, A
Core material
Number of turns, N
Figure 6.21
5ZQJDBMGPSNPGBOJOEVDUPS
6.4
Capacitors and Inductors
Inductors
"OJOEVDUPSJTBQBTTJWFFMFNFOUEFTJHOFEUPTUPSFFOFSHZJOJUTNBHOFUJD
field. Inductors find numerous applications in electronic and poXFSTZT
UFNT5IFZBSFVTFEJOQPXFSTVQQMJFT USBOTGPSNFST SBEJPT 57T SBEBST BOEFMFDUSJDNPUPST
"OZDPOEVDUPSPGFMFDUSJDDVSSFOUIBTJOEVDUJWFQSPQFSUJFTBOENBZ
CFSFHBSEFEBTBOJOEVDUPS#VUJOPSEFSUPFOIBODFUIFJOEVDUJ WFFGGFDU BQSBDUJDBMJOEVDUPSJTVTVBMMZGPSNFEJOUPBD ZMJOESJDBMDPJMXJUINBO Z
UVSOTPGDPOEVDUJOHXJSF BTTIPXOJO'JH
An inductor consists of a coil of conducting wire.
*GDVSSFOUJTBMMPXFEUPQBTTUISPVHIBOJOEVDUPS JUJTGPVOEUIBUUIFWPMU
BHFBDSPTTUIFJOEVDUPSJTEJSFDUMZQSPQPSUJPOBMUPUIFUJNFSBUFPGDIBOHF
PGUIFDVSSFOU6TJOHUIFQBTTJWFTJHODPOWFOUJPO
In view of Eq. (6.18), for an inductor
to have voltage across its terminals, its
current must vary with time. Hence,
v = 0 for constant current through
the inductor.
EJ W=- @@
EU
XIFSF-JTUIFDPOTUBOUPGQSPQPSUJPOBMJUZDBMMFEUIF JOEVDUBODFPGUIF
JOEVDUPS5IFVOJUPGJOEVDUBODFJTUIFIFOSZ ) OBNFEJOIPOPSPGUIF
"NFSJDBOJOWentor Joseph Henry (1797–1878). It is clear from Eq. UIBUIFOSZFRVBMTWPMUTFDPOEQFSBNQFSF
Inductance is the property whereby an inductor exhibits opposition to
the change of current flowing through it, measured in henrys (H).
(a)
5IFJOEVDUBODFPGBOJOEVDUPSEFQFOETPOJUTQI ZTJDBMEJNFOTJPO
BOEDPOTUSVDUJPO' PSNVMBTGPSDBMDVMBUJOHUIFJOEVDUBODFPGJOEVDUPST
PGEJGGFSFOUTIBQFTBSFEFSJ WFEGSPNFMFDUSPNBHOFUJDUIFPSZBOEDBOCF
GPVOEJOTUBOEBSEFMFDUSJDBMFOHJOFFSJOHIBOECPPLT'PSFYBNQMF GPSUIF
JOEVDUPS TPMFOPJE TIPXOJO'JH
/ μ"
-=@@@@@
ℓ
(b)
(c)
Figure 6.22
7BSJPVTUZQFTPGJOEVDUPST B TPMFOPJEBM
XPVOEJOEVDUPS C UPSPJEBMJOEVDUPS (c) axial lead inductor
¥.D(SBX)JMM&EVDBUJPO.BSL%JFSLFS QIPUPHSBQIFS
XIFSF/JTUIFOVNCFSPGUVSOT ℓJTUIFMFOHUI "JTUIFDSPTTTFDUJPOBM
BSFB BOE μJTUIFQFSNFBCJMJUZPGUIFDPSF 8FDBOTFFGSPN&R UIBUJOEVDUBODFDBOCFJODSFBTFECZJODSFBTJOHUIFOVNCFSPGUVSOTPG
DPJM VTJOHNBUFSJBMXJUIIJHIFSQFSNFBCJMJUZBTUIFDPSF JODSFBTJOHUIF
DSPTTTFDUJPOBMBSFB PSSFEVDJOHUIFMFOHUIPGUIFDPJM
-JLFDBQBDJUPST DPNNFSDJBMMZB WBJMBCMFJOEVDUPSTDPNFJOEJG GFS
FOUWBMVFTBOEUZQFT 5ZQJDBMQSBDUJDBMJOEVDUPSTIB WFJOEVDUBODFWBMVFT
SBOHJOHGSPNBGFXNJDSPIFOSZT μ) BTJODPNNVOJDBUJPOTZTUFNT UP
UFOTPGIFOSZT ) BTJOQP wer systems. Inductors may be fixFEPSWBSJ
BCMF5IFDPSFNBZCFNBEFPGJSPO TUFFM QMBTUJD PSBJS 5IFUFSNTDPJM
BOEDIPLFBSFBMTPVTFEGPSJOEVDUPST$PNNPOJOEVDUPSTBSFTIP XOJO
'JH5IFDJSDVJUTZNCPMTGPSJOEVDUPSTBSFTIP wn in Fig. GPM
MPXJOHUIFQBTTJWFTJHODPOWFOUJPO
&RVBUJPO JTUIFW PMUBHFDVSSFOUSFMBUJPOTIJQGPSBOJOEVDUPS 'JHVSFTIP XTUIJTSFMBUJPOTIJQHSBQIJDBMMZGPSBOJOEVDUPSXIPTF
6.4
225
Inductors
Historical
Joseph Henry m BO"NFSJDBOQIZTJDJTU EJTDPWFSFEJOEVD
JOEVDUBODFJTJOEFQFOEFOUPGDVSSFOU4VDIBOJOEVDUPSJTLOP XOBTB
MJOFBSJOEVDUPS'PSBOPOMJOFBSJOEVDUPS UIFQMPUPG&R XJMMOPU
CFBTUSBJHIUMJOFCFDBVTFJUTJOEVDUBODFW BSJFTXJUIDVSSFOU 8FXJMM
BTTVNFMJOFBSJOEVDUPSTJOUIJTUFYUCPPLVOMFTTTUBUFEPUIFSXJTF
5IFDVSSFOUWPMUBHFSFMBUJPOTIJQJTPCUBJOFEGSPN&R BT
vEU
EJ=@@
-
i
+
v
–
L
(a)
U v τ Eτ
J=@@
∫
- −∞
PS
i
+
v
–
L
+
v
–
(b)
L
(c)
$JSDVJUTZNCPMTGPSJOEVDUPST B BJSDPSF C JSPODPSF D WBSJBCMFJSPODPSF
v
U
W τ Eτ+J U
J=@@
∫
- U
XIFSFJ U JTUIFUPUBMDVSSFOUGPS −∞<U<UBOEJ −∞ =5IFJEFB
PGNBLJOHJ −∞ =JTQSBDUJDBMBOESFBTPOBCMF CFDBVTFUIFSFNVTUCF
BUJNFJOUIFQBTUXIFOUIFSFXBTOPDVSSFOUJOUIFJOEVDUPS
5IFJOEVDUPSJTEFTJHOFEUPTUPSFFOFS gy in its magnetic field. 5IF
FOFSHZTUPSFEDBOCFPCUBJOFEGSPN&R 5IFQPXFSEFMJWFSFEUP
UIFJOEVDUPSJT
Q=vJ=(-@@
EJ)J
EU
5IFFOFSHZTUPSFEJT
i
Figure 6.23
*OUFHSBUJOHHJWFT
/0""T1FPQMF$PMMFDUJPO
UBODFBOEDPOTUSVDUFEBOFMFDUSJDNPUPS
#PSO JO "MCBOZ /FX :PSL )FOSZ HSBEVBUFE GSPN "MCBOZ "DBE
FNZBOEUBVHIUQIJMPTPQIZBU1SJODFUPO6OJWFSTJUZGSPNUP
He was the first secretary of the Smithsonian Institution. He conduct
FE TFWFSBMFYQFSJNFOUTPOFMFDUSPNBHOFUJTNBOEEFWFMPQFEQPXFSGVM
FMFDUSPNBHOFUTUIBUDPVMEMJGUPCKFDUTXFJHIJOHUIPVTBOETPGQPVOET
*OUFSFTUJOHMZ +PTFQI)FOSZEJTDPWFSFEFMFDUSPNBHOFUJDJOEVDUJPOCFGPSF
Faraday but failed to publish his findings. The unit of inductance, UIF
IFOSZ XBTOBNFEBGUFSIJN
U
U
X=∫ Q τ Eτ=-∫ @@
EJJEτ
−∞
−∞ Eτ
U
-J U − @@
-J −∞
=- JEJ=@@
−∞
∫
Slope = L
0
Figure 6.24
di/dt
7PMUBHFDVSSFOUSFMBUJPOTIJQPGBOJOEVDUPS
226
Chapter 6
Capacitors and Inductors
4JODFJ −∞ =
-J
X=@@
8FTIPVMEOPUFUIFGPMMPXJOHJNQPSUBOUQSPQFSUJFTPGBOJOEVDUPS
/PUFGSPN&R UIBUUIFW PMUBHFBDSPTTBOJOEVDUPSJT[FSP
XIFOUIFDVSSFOUJTDPOTUBOU5IVT
An inductor acts like a short circuit to dc.
. "OJNQPSUBOUQSPQFSUZPGUIFJOEVDUPSJTJUTPQQPTJUJPOUPUIFDIBOHF
in current floXJOHUISPVHIJU
i
i
The current through an inductor cannot change instantaneously.
t
t
(a)
(b)
Figure 6.25
$VSSFOUUISPVHIBOJOEVDUPS B BMMPXFE C OPUBMMPXBCMFBOBCSVQUDIBOHFJTOPU
QPTTJCMF
Since an inductor is often made of a
highly conducting wire, it has a very
small resistance.
L
Rw
Cw
Figure 6.26
$JSDVJUNPEFMGPSBQSBDUJDBMJOEVDUPS
Example 6.8
"DDPSEJOHUP&R BEJTDPOUJOVPVTDIBOHFJOUIFDVSSFOU
UISPVHIBOJOEVDUPSSFRVJSFTBOinfinite WPMUBHF XIJDIJTOPUQIZTJ
DBMMZQPTTJCMF 5IVT BOJOEVDUPSPQQPTFTBOBCSVQUDIBOHFJOUIF
DVSSFOUUISPVHIJU'PSFYBNQMF UIFDVSSFOUUISPVHIBOJOEVDUPSNBZ
UBLFUIFGPSNTIP XOJO'JH B XIFSFBTUIFJOEVDUPSDVSSFOU
DBOOPUUBLFUIFGPSNTIPXOJO'JH C JOSFBMMJGFTJUVBUJPOTEVF
UPUIFEJTDPOUJOVJUJFT)P XFWFS UIFW PMUBHFBDSPTTBOJOEVDUPSDBO
DIBOHFBCSVQUMZ
-JLFUIFJEFBMDBQBDJUPS UIFJEFBMJOEVDUPSEPFTOPUEJTTJQBUF FOFSHZ5IFFOFSHZTUPSFEJOJUDBOCFSFUSJF WFEBUBMBUFSUJNF
5IFJOEVDUPSUBL FTQPXFSGSPNUIFDJSDVJUXIFOTUPSJOHFOFS HZ
BOE EFMJWFSTQP XFS UPUIFDJSDVJU XIFO SFUVSOJOHQSF WJPVTMZ
TUPSFEFOFSHZ
" QSBDUJDBM OPOJEFBM JOEVDUPS IBT B significant SFTJTUJWF DPNQP
OFOU BTTIPXOJO'JH5IJTJTEVFUPUIFG BDUUIBUUIFJOEVDUPS
JTNBEFPGBDPOEVDUJOHNBUFSJBMTVDIBTDPQQFS XIJDIIBTTPNF
SFTJTUBODF5IJTSFTJTUBODFJTDBMMFEUIFXJOEJOHSFTJTUBODF3X BOEJU
BQQFBSTJOTFSJFTXJUIUIFJOEVDUBODFPGUIFJOEVDUPS 5IFQSFTFODF
PG3XNBLFTJUCPUIBOFOFS HZTUPSBHF EFWJDFBOE an FOFSHZEJTTJ
QBUJPOEFWJDF4JODF 3XJTVTVBMMZW FSZTNBMM JUJTJHOPSFEJONPTU
DBTFT5IFOPOJEFBMJOEVDUPSBMTPIBTBXJOEJOHDBQBDJUBODF$XEVF
UPUIFDBQBDJUJWFDPVQMJOHCFUXFFOUIFDPOEVDUJOHDPJMT $XJTWFSZ
TNBMMBOEDBOCFJHOPSFEJONPTUDBTFT FYDFQUBUIJHIGSFRVFODJFT
8FXJMMBTTVNFJEFBMJOEVDUPSTJOUIJTCPPL
5IFDVSSFOUUISPVHIB)JOEVDUPSJTJ U =UF−U"'JOEUIFWPMUBHF
BDSPTTUIFJOEVDUPSBOEUIFFOFSHZTUPSFEJOJU
Solution:
4JODFv=-EJ∕EUBOE-=)
v=@@
E UF−U =F−U+U − F−U=F−U −U 7
EU
6.4
Inductors
227
5IFFOFSHZTUPSFEJT
-J=@@
UF−U=UF−U+
X=@@
*GUIFDVSSFOUUISPVHIBN)JOEVDUPSJTJ U =TJO U) mA, find the
UFSNJOBMWPMUBHFBOEUIFFOFSHZTUPSFE
Practice Problem 6.8
Answer:DPT U N7 TJO U μ+
'JOEUIFDVSSFOUUISPVHIB)JOEVDUPSJGUIFWPMUBHFBDSPTTJUJT
{
v U =
U
Example 6.9
U>
U< Also, find the enerHZTUPSFEBUU=T"TTVNFJ v >
Solution:
Uv τ Eτ+J U BOE-=)
4JODFJ=@@
∫
- U
U =U"
U τEτ+=×@@
J=@@
∫
5IFQPXFSQ=vJ=U BOEUIFFOFSHZTUPSFEJTUIFO
|
U=L+
X=∫ QEU=∫ UEU=@@
"MUFSOBUJWFMZ XFDBOPCUBJOUIFFOFSHZTUPSFEVTJOH&R CZ
XSJUJOH
|
-J −@@
-J =@@
× −=L+
X =@@
BTPCUBJOFECFGPSF
5IFUFSNJOBMWPMUBHFPGB)JOEVDUPSJT v = − U 7'JOEUIF
current flowing through it at U=TBOEUIFFOFSHZTUPSFEJOJUBU U=T
"TTVNFJ ="
Answer:−" +
Practice Problem 6.9
228
Chapter 6
Example 6.10
i
1Ω
Consider the circuit in Fig. 6.27(a). Under dc conditions, find: (a) J v$ BOEJ- C UIFFOFSHZTUPSFEJOUIFDBQBDJUPSBOEJOEVDUPS
5Ω
iL
4Ω
12 V +
–
+
vC
–
1F
1Ω
Solution:
B 6OEFSEDDPOEJUJPOT XFSFQMBDF UIFDBQBDJUPSXJUIBOPQFO DJSDVJU
BOEUIFJOEVDUPSXJUIBTIPSUDJSDVJU BTJO'JH C *UJTFWJEFOUGSPN
'JH C UIBU
2H
="
J=J-=@@@@@
+
5IFWPMUBHFv$JTUIFTBNFBTUIFWPMUBHFBDSPTTUIFΩSFTJTUPS)FODF
(a)
i
Capacitors and Inductors
5Ω
v$=J=7
iL
4Ω
12 V +
–
C 5IFFOFSHZJOUIFDBQBDJUPSJT
+
vC
–
$v =@@
=+
X$=@@
$ (b)
BOEUIBUJOUIFJOEVDUPSJT
Figure 6.27
-J =@@
=+
X-=@@
- 'PS&YBNQMF
Practice Problem 6.10
iL
6H
Answer:7 " + +
+
vC
–
2Ω
6Ω
10 A
4F
6.5
Figure 6.28
'PS1SBDUJDF1SPC
i
+
v
L1
%FUFSNJOFv$ J- BOEUIF FOFSHZTUPSFEJOUIFDBQBDJUPS BOEJOEVDUPSJO
UIFDJSDVJUPG'JHVOEFSEDDPOEJUJPOT
L3
L2
+ v – +v – +v –
1
2
3
LN
...
+v –
N
–
(a)
/PXUIBUUIFJOEVDUPSIBTCFFOBEEFEUPPVSMJTUPGQBTTJWe elements, it JT
OFDFTTBSZUPFYUFOEUIFQPXFSGVMUPPMPGTFSJFTQBSBMMFMDPNCJOBUJPO8F
OFFEUPLOPXIPw to find the equiWBMFOUJOEVDUBODFPGBTFSJFTDPOOFDUFE
PSQBSBMMFMDPOOFDUFETFUPGJOEVDUPSTGPVOEJOQSBDUJDBMDJSDVJUT
$POTJEFSBTFSJFTDPOOFDUJPOPG/JOEVDUPST BTTIPXOJO'JH B XJUIUIFFRVJWBMFOUDJSDVJUTIPXOJO'JH C 5IFJOEVDUPSTIBWFUIF
TBNFDVSSFOUUISPVHIUIFN"QQMZJOH,7-UPUIFMPPQ
i
Series and Parallel Inductors
v=v+v+v+⋯+v/
4VCTUJUVUJOHvL = -LEJ∕EUSFTVMUTJO
+
v
L eq
–
(b)
Figure 6.29
B "TFSJFTDPOOFDUJPOPG/JOEVDUPST C FRVJWBMFOUDJSDVJUGPSUIFTFSJFT
JOEVDUPST
EJ
v=-@@
EJ +-@@
EJ +-@@
EJ +⋯+-/ @@
EU
EU
EU
EU
EJ
= -+-+-+⋯+-/ @@
EU /
EJ =-FR@@
EJ
= ( ∑-L)@@
EU
EU
L=
-FR=-+-+-+⋯+-/
XIFSF
6.5
229
Series and Parallel Inductors
i
5IVT
The equivalent inductance of series-connected inductors is the sum of
the individual inductances.
*OEVDUPST JO TFSJFT BSF DPNCJOFE JO FYBDUMZ UIF TBNF XBZ BT SFTJTUPST
in series.
8FOPXDPOTJEFSBQBSBMMFMDPOOFDUJPOPG /JOEVDUPST BTTIPXOJO
'JH B XJUIUIFFRVJ WBMFOUDJSDVJUJO'JH C 5IFJOEVDUPST
IBWFUIFTBNFWPMUBHFBDSPTTUIFN6TJOH,$
J=J+J+J+⋯+J/
∫
U
∫
U
vEU+J U
+⋯+@@@
∫
/ -/ U
+⋯+@@@
UvEU+J U +J U
= @@
+@@
∫
-/ ) U
( - -
+⋯+J/ U
XIFSF
/
/
UvEU+∑
U
@@@
= ∑@@
JL U = ∫ vEU+J U
∫
-FR U
(L= -L ) U
L=
=@@
+@@
+@@
+⋯+@@@
@@@
-FR - - -
-/
5IFJOJUJBMDVSSFOUJ U UISPVHI-FRBUU = UJTFYQFDUFECZ,$-UPCFUIF
TVNPGUIFJOEVDUPSDVSSFOUTBUU5IVT BDDPSEJOHUP&R J U =J U +J U +⋯+J/ U
"DDPSEJOHUP&R The equivalent inductance of parallel inductors is the reciprocal of the
sum of the reciprocals of the individual inductances.
/PUFUIBUUIFJOEVDUPSTJOQBSBMMFMBSFDPNCJOFEJOUIFTBNFXBZBTSFTJT
UPSTJOQBSBMMFM
'PSUXPJOEVDUPSTJOQBSBMMFM /= &R CFDPNFT
=@@
+@@
@@@
-FR - -
L1
PS
--
-FR=@@@@@@@
-+-
i3
L2
iN
L3
LN
(a)
i
+
v
L eq
–
B "QBSBMMFMDPOOFDUJPOPG/JOEVDUPST C FRVJWBMFOUDJSDVJUGPSUIFQBSBMMFM
JOEVDUPST
U
- U
–
i2
Figure 6.30
vEU+J U +@@
vEU+J U
J=@@
- U
v
i1
(b)
UvEU+J U IFODF
#VUJL=@@
∫
L -L U
+
"TMPOHBTBMMUIFFMFNFOUTBSFPGUIFTBNFUZQF UIF∆:USBOTGPSNBUJPOT
GPSSFTJTUPSTEJTDVTTFEJO4FDUJPODBOCFF YUFOEFEUPDBQBDJUPSTBOE
JOEVDUPST
230
Chapter 6
Capacitors and Inductors
TABLE 6.1
Important characteristics of the basic elements.†
3FMBUJPO
3FTJTUPS 3
$BQBDJUPS $
*OEVDUPS -
WJ
W=J3
UJ τ Eτ+W U
W=@@
∫
$ U
EJ
W=-@@
EU
JW
J=W∕3
J=$@@@
EW
EU
UW τ Eτ+J U
J=@@
∫
- U
QPSX
W
Q=J3=@@
3
-J
X=@@
4FSJFT
3FR=3+3
$W
X=@@
$$
$FR=@@@@@@@
$+$
1BSBMMFM
33
3FR=@@@@@@@
$FR=$+$
3+3
--
-FR=@@@@@@@
-+-
"UED
4BNF
4IPSUDJSDVJU
-FR=-+-
0QFODJSDVJU
$JSDVJUWBSJBCMF
UIBUDBOOPU
DIBOHFBCSVQUMZ /PUBQQMJDBCMF W
J
n
1BTTJWFTJHODPOWFOUJPOJTBTTVNFE
*UJTBQQSPQSJBUFBUUIJTQPJOUUPTVNNBSJ[FUIFNPTUJNQPSUBOUDIBS
BDUFSJTUJDTPGUIFUISFFCBTJDDJSDVJUFMFNFOUTXFIBWFTUVEJFE5IFTVN
NBSZJTHJWFOJO5BCMF
Example 6.11
20 H
4H
L eq
'JOEUIFFRVJWBMFOUJOEVDUBODFPGUIFDJSDVJUTIPXOJO'JH
7H
8H
12 H
Solution:
5IF) ) BOE)JOEVDUPSTBSFJOTFSJFTUIVT DPNCJOJOHUIFN
HJWFTB)JOEVDUBODF5IJT)JOEVDUPSJTJOQBSBMMFMXJUIUIF)
JOEVDUPSTPUIBUUIFZBSFDPNCJOFE UPHJWF
@@@@@@
×
=)
10 H
Figure 6.31
'PS&YBNQMF
+
5IJT)JOEVDUPSJTJOTFSJFTXJUIUIF)BOE)JOEVDUPST)FODF
-FR=++=)
Practice Problem 6.11
$BMDVMBUFUIFFRVJWBMFOUJOEVDUBODFGPSUIFJOEVDUJWFMBEEFSOFUXPSLJO
'JH
L eq
20 mH
100 mH
40 mH
50 mH
40 mH
30 mH
Figure 6.32
'PS1SBDUJDF1SPC
Answer:N)
20 mH
6.6
231
Applications
Example 6.12
'PSUIFDJSDVJUJO'JH J U = −F−U N"*GJ =−1 mA, find:
B J C v U v U BOEv U D J U BOEJ U i
Solution:
+ v1 –
+
B 'SPNJ U = −F
−U
2H
v
N" J = − =N"4JODFJ=J+J
i1
4H
–
J =J −J =− − =N"
i2
+
v2
–
12 H
Figure 6.33
C 5IFFRVJWBMFOUJOEVDUBODFJT
'PS&YBNQMF
-FR=+║=+=)
5IVT
v U =-FR@@
EJ = − − F−UN7=F−UN7
EU
BOE
v U =@@
EJ = − − F−UN7=F−UN7
EU
4JODFv=v+v
v U =v U −v U =F−UN7
D 5IFDVSSFOUJJTPCUBJOFEBT
U
U v EU+J =@@@@
J U =@@
∫ F−UEU+N"
∫
|
U
−U
=−F−U +N"=−F
++=−F−UN"
4JNJMBSMZ
U F−UEU−N"
U v EU+J =@@@@
J U =@@@
∫
∫
|
U
−U
=−F−U −N"=−F
+−=−F−UN"
/PUFUIBUJ U +J U =J U Practice Problem 6.12
*OUIFDJSDVJUPG'JH J U =F−U"*GJ = 7 A, find:
B J C J U BOEJ U D v U v U BOEv U Answer: B " C − + F
−F−U7 −F−U7
−U
−U
" − + F
i2
−U
" D −F 7
i
+ v1
+
v
3H
i1
–
6H
–
6.6
Applications
$JSDVJUFMFNFOUTTVDIBTSFTJTUPSTBOEDBQBDJUPSTBSFDPNNFSDJBMMZ
BWBJMBCMFJOFJUIFSEJTDSFUF GPSNPSJOUF HSBUFEDJSDVJU *$ GPSN6OMJLF
DBQBDJUPSTBOESFTJTUPST JOEVDUPSTXJUIBQQSFDJBCMFJOEVDUBODFBSFEJG ficult to produce on IC substrates. 5IFSFGPSF JOEVDUPST DPJMT VTVBMMZ
Figure 6.34
'PS1SBDUJDF1SPC
+
v2
–
8H
232
Chapter 6
Capacitors and Inductors
DPNFJOEJTDSFUFGPSNBOEUFOEUPCFNPSFC VMLZBOEF YQFOTJWF' PS
UIJTSFBTPO JOEVDUPSTBSFOPUBTWFSTBUJMFBTDBQBDJUPSTBOESFTJTUPST BOE
UIFZBSFNPSFMJNJUFEJOBQQMJDBUJPOT)PXFWFS UIFSFBSFTFWFSBMBQQMJDB
UJPOTJOXIJDIJOEVDUPSTIBWFOPQSBDUJDBMTVCTUJUVUF5IFZBSFSPVUJOFMZ
VTFEJOSFMBZT EFMBZT TFOTJOHEFWJDFT QJDLVQIFBET UFMFQIPOFDJSDVJUT SBEJPBOE 57SFDFJWFST QPXFSTVQQMJFT FMFDUSJDNPUPST NJDSPQIPOFT BOEMPVETQFBLFST UPNFOUJPOBGFX
$BQBDJUPSTBOEJOEVDUPSTQPTTFTTUIFGPMMPXJOHUISFFTQFDJBMQSPQFS
UJFTUIBUNBLFUIFNWFSZVTFGVMJOFMFDUSJDDJSDVJUT
5IFDBQBDJUZUPTUPSFFOFS HZNBLFTUIFNVTFGVMBTUFNQPSBSZW PMU
BHFPSDVSSFOUTPVSDFT5IVT UIFZDBOCFVTFEGPSHFOFSBUJOHBMBSHF
BNPVOUPGDVSSFOUPSWPMUBHFGPSBTIPSUQFSJPEPGUJNF
$BQBDJUPSTPQQPTFBO ZBCSVQUDIBOHFJOW PMUBHF XIJMFJOEVDUPST
PQQPTFBOZBCSVQUDIBOHFJODVSSFOU5IJTQSPQFSUZNBLFTJOEVDUPST
VTFGVMGPSTQBSLPSBSDTVQQSFTTJPOBOEGPSDPOWFSUJOHQVMTBUJOHED
WPMUBHFJOUPSFMBUJWFMZTNPPUIEDWPMUBHF
$BQBDJUPSTBOEJOEVDUPSTBSFGSFRVFOD ZTFOTJUJ WF 5IJTQSPQFSUZ
NBLFTUIFNVTFGVMGPSGSFRVFODZEJTDSJNJOBUJPO
The first twPQSPQFSUJFTBSFQVUUPVTFJOEDDJSDVJUT XIJMFUIFUIJSE
POFJTUBLFOBEWBOUBHFPGJOBDDJSDVJUT 8FXJMMTFFIP XVTFGVMUIFTF
QSPQFSUJFTBSFJOMBUFSDIBQUFST' PSOPX DPOTJEFSUISFFBQQMJDBUJPOT JOWPMWJOHDBQBDJUPSTBOEPQBNQTJOUFHSBUPS EJGGFSFOUJBUPS BOEBOBMPH
DPNQVUFS
6.6.1
i1
R1
v1
*NQPSUBOUPQ BNQDJSDVJUTUIBUVTF FOFSHZTUPSBHFFMFNFOUTJODMVEF JOUFHSBUPSTBOEEJG GFSFOUJBUPST 5IFTFPQBNQDJSDVJUTPGUFOJO WPMWF
SFTJTUPSTBOEDBQBDJUPSTJOEVDUPST DPJMT UFOEUPCFNPSFC VMLZBOE
FYQFOTJWF
5IFPQBNQJOUFHSBUPSJTVTFEJOOVNFSPVTBQQMJDBUJPOT FTQFDJBMMZ
JOBOBMPHDPNQVUFST UPCFEJTDVTTFEJO4FDUJPO
Rf
i2
0A
–
–
0V
v2 + +
1
+
vi
+
vo
–
Integrator
An integrator is an op amp circuit whose output is proportional to the
integral of the input signal.
–
*GUIFGFFECBDLSFTJTUPS 3GJOUIFG BNJMJBSJO Werting amplifier of
Fig. B JTSFQMBDFECZBDBQBDJUPS XFPCUBJOBOJEFBMJOUF HSBUPS BT
TIPXOJO'JH C *UJTJOUFSFTUJOHUIBUXFDBOPCUBJOBNBUIFNBUJDBM
SFQSFTFOUBUJPOPGJOUFHSBUJPOUIJTXBZ"UOPEFBJO'JH C
(a)
C
iC
iR
R
+
vi
a
+
+
vo
#VU
v
J3=@@J 3
–
–
J3=J$
–
Ev
P
J$=−$@@@
EU
4VCTUJUVUJOHUIFTFJO&R XFPCUBJO
(b)
Figure 6.35
3FQMBDJOHUIFGFFECBDLSFTJTUPSJOUIF
JOWerting amplifier in (a) produces an
JOUFHSBUPSJO C v
Ev
@@
J =−$@@@P
B
v EU
EvP=−@@@
3$ J
C
3
EU
6.6
233
Applications
*OUFHSBUJOHCPUITJEFTHJWFT
U
v τ Eτ
vP U −vP =−@@@
∫
3$ J
5PFOTVSFUIBUvP = JUJTBMXBZTOFDFTTBSZUPEJTDIBSHFUIFJOUFHSBUPST
DBQBDJUPSQSJPSUPUIFBQQMJDBUJPOPGBTJHOBM"TTVNJOHvP =
U
W τ Eτ
WP=−@@@
∫
3$ J
XIJDITIPXTUIBUUIFDJSDVJUJO'JH C QSP WJEFTBOPVUQVUW PMUBHF
QSPQPSUJPOBMUPUIFJOUFHSBMPGUIFJOQVU*OQSBDUJDF UIFPQBNQJOUF HSB
UPSSFRVJSFTBGFFECBDLSFTJTUPSUPSFEVDFEDHBJOBOEQSFWFOUTBUVSBUJPO
$BSFNVTUCFUBL FOUIBUUIFPQBNQPQFSBUFTXJUIJOUIFMJOFBSSBOHFTP
UIBUJUEPFTOPUTBUVSBUF
Example 6.13
*Gv=DPT UN7BOE v=UN7, find vPJOUIFPQBNQDJSDVJUJO 'JH"TTVNFUIBUUIFWPMUBHFBDSPTTUIFDBQBDJUPSJTJOJUJBMMZ[FSP
Solution:
5IJTJTBTVNNJOHJOUFHSBUPS BOE
∫
∫
v1
v EU−@@@@
v =−@@@@
v EU
P
3$ 3 MΩ
–
+
v2
3$ 2 μF
100 kΩ
U
=−@@@@@@@@@@@@@@@@
∫ DPT τ Eτ
×××− U
−@@@@@@@@@@@@@@@@@@
τEτ
∫
×××− Figure 6.36
'PS&YBNQMF
TJOU−@@@
U @@@
@@@@
=−@@
=−TJOU−U N7
5IFJOUFHSBUPSJO'JH C IBT3=LΩ $=μ'%FUFSNJOFUIF
PVUQVUWPMUBHFXIFOBEDWPMUBHFPGN7JTBQQMJFEBU U="TTVNF
UIBUUIFPQBNQJTJOJUJBMMZOVMMFE
Answer:−UN 7
6.6.2
Differentiator
A differentiator is an op amp circuit whose output is proportional to the
rate of change of the input signal.
*O'JH B JGUIFJOQVUSFTJTUPSJTSFQMBDFECZBDBQBDJUPS UIF
SFTVMUJOHDJSDVJUJTBEJG GFSFOUJBUPS TIPXOJO'JH "QQMZJOH,$-
BUOPEFB
J3=J$
Practice Problem 6.13
vo
234
Chapter 6
Capacitors and Inductors
#VU
v
Ev
J3=−@@P
J
J$=$ @@@
3
EU
4VCTUJUVUJOHUIFTFJO&R ZJFMET
R
iR
–
+
a
+
vi
–
+
vo
–
Figure 6.37
"OPQBNQEJGGFSFOUJBUPS
Example 6.14
TIPXJOHUIBUUIFPVUQVUJTUIFEFSJ WBUJWFPGUIFJOQVU%JG GFSFOUJBUPSDJS
DVJUTBSFFMFDUSPOJDBMMZVOTUBCMFCFDBVTFBO ZFMFDUSJDBMOPJTFXJUIJOUIF
DJSDVJUJTFYBHHFSBUFECZUIFEJGGFSFOUJBUPS'PSUIJTSFBTPO UIFEJGGFSFO
UJBUPSDJSDVJUJO'JHJTOPUBTVTFGVMBOEQPQVMBSBTUIFJOUFHSBUPS*U
JTTFMEPNVTFEJOQSBDUJDF
Solution:
5IJTJTBEJGGFSFOUJBUPSXJUI
0.2 μF
–
+
+
–
4LFUDIUIFPVUQVUW PMUBHFGPSUIFDJSDVJUJO'JH B HJWFOUIFJOQVU
WPMUBHFJO'JH C 5BLFvP=BUU=
5 kΩ
vi
EW
J
WP=−3$ @@@
EU
C
iC
+
vo
–
3$=×××−=−T
'PS<U<NT XFDBOFYQSFTTUIFJOQVUWPMUBHFJO'JH C BT
{
vJ=
(a)
U
−U <U<NT
<U<NT
5IJTJTSFQFBUFEGPS <U<NT6TJOH &R UIFPVUQVUJTPC tained as
vo(V)
{
Ev
−7
vP=−3$@@@
J=
EU
7
4
<U<NT
<U<NT
5IVT UIFPVUQVUJTBTTLFUDIFEJO'JH
0
2
4
6
8
t (ms)
(b)
Figure 6.38
'PS&YBNQMF
vo (V)
2
0
2
4
6
8
t (ms)
–2
Figure 6.39
0VUQVUPGUIFDJSDVJUJO'JH B Practice Problem 6.14
5IFEJGGFSFOUJBUPSJO'JHIBT 3=L ΩBOE$ = μ'(JWFO
UIBUvJ=U7 EFUFSNJOFUIFPVUQVUvP
Answer: −N7
6.6
6.6.3
Applications
235
Analog Computer
0QBNQTXFSFJOJUJBMMZEF WFMPQFEGPSFMFDUSPOJDBOBMPHDPNQVUFST
"OBMPHDPNQVUFSTDBOCFQSPHSBNNFEUPTPMW FNBUIFNBUJDBMNPEFMTPG
NFDIBOJDBMPSFMFDUSJDBMTZTUFNT5IFTFNPEFMTBSFVTVBMMZFYQSFTTFEJO
UFSNTPGEJGGFSFOUJBMFRVBUJPOT
5PTPMW FTJNQMFEJG GFSFOUJBMFRVBUJPOTVTJOHUIFBOBMPHDPNQVUFS
SFRVJSFTDBTDBEJOHUISFFUZQFTPGPQBNQDJSDVJUTJOUF HSBUPSDJSDVJUT summing amplifiers, and inWFSUJOHOPOJOWerting amplifiers for neHBUJWF
QPTJUJWFTDBMJOH5IFCFTUXBZUPJMMVTUSBUFIPXBOBOBMPHDPNQVUFSTPMWFT
BEJGGFSFOUJBMFRVBUJPOJTXJUIBOFYBNQMF
4VQQPTFXFEFTJSFUIFTPMVUJPOY U PGUIFFRVBUJPO
B @@@
E Y+C@@@
EY+DY=G U EU
EU
U>
XIFSF B C BOE D BSFDPOTUBOUT BOE G U JTBOBSCJUSBSZGPSDJOHGVOD UJPOThe solution is obtained by first solving the highest-order deriWBUJWF
UFSN4PMWJOHGPSE Y∕EUZJFMET
GU
@@@
E Y=@@@ −@@
C @@@
EY−@@
DY
EU
B
B EU
B
5PPCUBJO EY∕EU UIF E Y∕EUUFSNJTJOUF HSBUFEBOEJOWFSUFE'JOBMMZ UP
PCUBJOY UIFEY∕EUUFSNJTJOUFHSBUFEBOEJOWFSUFE5IFGPSDJOHGVODUJPO
JTJOKFDUFEBUUIFQSPQFSQPJOU 5IVT UIFBOBMPHDPNQVUFSGPSTPMWJOH
Eq. JTJNQMFNFOUFECZDPOOFDUJOHUIFOFDFTTBSZTVNNFST JOWFSU
FST BOEJOUF HSBUPST"QMPUUFSPSPTDJMMPTDPQF NBZCFVTFEUPWJF XUIF
PVUQVUY PSEY∕EU PSEY∕EU EFQFOEJOHPOXIFSFJUJTDPOOFDUFEJOUIF
TZTUFN
"MUIPVHIUIFBCPWFFYBNQMFJTPOBTFDPOEPSEFSEJGGFSFOUJBMFRVB
UJPO BOZEJGGFSFOUJBMFRVBUJPODBOCFTJNVMBUFECZBOBOBMPH DPNQVUFS
DPNQSJTJOHJOUFHSBUPST JOWFSUFST BOEJOWFSUJOHTVNNFST#VUDBSFNVTU
CFF YFSDJTFEJOTFMFDUJOHUIFW BMVFTPGUIFSFTJTUPSTBOEDBQBDJUPST UP
FOTVSFUIBUUIFPQBNQTEPOPUTBUVSBUFEVSJOHUIFTPMVUJPOUJNFJOUFSWBM
5IFBOBMPHDPNQVUFSTXJUIWBDVVNUVCFTXFSFCVJMUJOUIFTBOE
T3FDFOUMZUIFJSVTFIBTEFDMJOFE 5IFZIB WFCFFOTVQFSTFEFECZ
NPEFSOEJHJUBMDPNQVUFST)P XFWFS XFTUJMMTUVEZBOBMPHDPNQVUFSTGPS
UXPSFBTPOT'JSTU UIFB WBJMBCJMJUZPGJOUFHSBUFEPQBNQTIBTNBEFJUQPT TJCMFUPCVJMEBOBMPHDPNQVUFSTFBTJMZBOEDIFBQMZ4FDPOE VOEFSTUBOEJOH
BOBMPHDPNQVUFSTIFMQTXJUIUIFBQQSFDJBUJPOPGUIFEJHJUBMDPNQVUFST
%FTJHOBOBOBMPHDPNQVUFSDJSDVJUUPTPMWFUIFEJGGFSFOUJBMFRVBUJPO
Ev
Ev
@@@@
P
+@@@
P+v =TJOU EU
EU
P
U>
TVCKFDUUP vP = − vP′ = XIFSFUIFQSJNFSFGFSTUPUIFUJNF EFSJWBUJWF
Solution:
Define.8FIBWe a clearly defined problem and eYQFDUFETPMVUJPO
*NJHIUSFNJOEUIFTUVEFOUUIBUNBOZUJNFTUIFQSPCMFNJTOPUTP
well defined and this portion of the problem-solving process could
Example 6.15
236
Chapter 6
Capacitors and Inductors
SFRVJSFNVDINPSFFGGPSU*GUIJTJTTP UIFOZPVTIPVMEBMXBZTLFFQ
JONJOEUIBUUJNFTQFOUIFSFXJMMSFTVMUJONVDIMFTTFGGPSUMBUFSBOE
NPTUMJLFMZTBWFZPVBMPUPGGSVTUSBUJPOJOUIFQSPDFTT
1SFTFOU$MFBSMZ VTJOHUIFEFWJDFTEFWFMPQFEJO4FDUJPOXJMM
BMMPXVTUPDSFBUFUIFEFTJSFEBOBMPHDPNQVUFSDJSDVJU8FXJMM
OFFEUIFJOUFHSBUPSDJSDVJUT QPTTJCMZDPNCJOFEXJUIBTVNNJOH
DBQBCJMJUZ BOEPOFPSNPSFJOWFSUFSDJSDVJUT
"MUFSOBUJWF5IFBQQSPBDIGPSTPMWJOHUIJTQSPCMFNJT
TUSBJHIUGPSXBSE8FXJMMOFFEUPQJDLUIFDPSSFDUWBMVFTPG
SFTJTUBODFTBOEDBQBDJUPSTUPBMMPXVTUPSFBMJ[FUIFFRVBUJPOXFBSF
SFQSFTFOUJOHThe final output of the circuit will giWFUIFEFTJSFE
SFTVMU
"UUFNQUThere are an infinite number of possibilities for picking
UIFSFTJTUPSTBOEDBQBDJUPST NBOZPGXIJDIXJMMSFTVMUJODPSSFDU
TPMVUJPOT&YUSFNFWBMVFTPGSFTJTUPSTBOEDBQBDJUPSTXJMMSFTVMUJO
JODPSSFDUPVUQVUT'PSFYBNQMF MPXWBMVFTPGSFTJTUPSTXJMMPWFSMPBE
UIFFMFDUSPOJDT1JDLJOHWBMVFTPGSFTJTUPSTUIBUBSFUPPMBSHFXJMM
DBVTFUIFPQBNQTUPTUPQGVODUJPOJOHBTJEFBMEFWJDFT5IFMJNJUT
DBOCFEFUFSNJOFEGSPNUIFDIBSBDUFSJTUJDTPGUIFSFBMPQBNQ
We first solve for the second derivative as
EvP
Ev
@@@@
=TJOU−@@@
P−vP
EU
EU
4PMWJOHUIJTSFRVJSFTTPNFNBUIFNBUJDBMPQFSBUJPOT JODMVE
JOHTVNNJOH TDBMJOH BOEJOUFHSBUJPO*OUFHSBUJOHCPUITJEFTPG
Eq. (6.15.1) giWFT
U
Ev
Ev τ
@@@P=−∫ (−TJO τ +@@@@@
P +vP τ )Eτ+vP′ EU
Eτ
XIFSFvP′ =8FJNQMFNFOU&R VTJOHUIFTVNNJOH
JOUFHSBUPSTIPXOJO'JH B 5IFWBMVFTPGUIFSFTJTUPSTBOE
DBQBDJUPSTIBWFCFFODIPTFOTPUIBU3$=GPSUIFUFSN
U
v τ Eτ
−@@@
∫
3$ P
0UIFSUFSNTJOUIFTVNNJOHJOUFHSBUPSPG&R BSF
JNQMFNFOUFEBDDPSEJOHMZ5IFJOJUJBMDPOEJUJPOEvP ∕EU=JT
JNQMFNFOUFECZDPOOFDUJOHB7CBUUFSZXJUIBTXJUDIBDSPTTUIF
DBQBDJUPSBTTIPXOJO'JH B 5IFOFYUTUFQJTUPPCUBJOvPCZJOUFHSBUJOHEvP∕EUBOEJOWFSUJOH
UIFSFTVMU
U
EvP τ
vP=−∫ −@@@@@
Eτ+v Eτ
5IJTJTJNQMFNFOUFEXJUIUIFDJSDVJUJO'JH C XJUIUIFCBUUFSZ
HJWJOHUIFJOJUJBMDPOEJUJPOPG−78FOPXDPNCJOFUIFUXP
DJSDVJUTJO'JH B BOE C UPPCUBJOUIFDPNQMFUFDJSDVJUTIPXO
JO'JH D 8IFOUIFJOQVUTJHOBMTJOUJTBQQMJFE XFPQFO
UIFTXJUDIFTBUU=UPPCUBJOUIFPVUQVUXBWFGPSNvP XIJDINBZ
CFWJFXFEPOBOPTDJMMPTDPQF
6.6
–
–10 sin (4t)
vo
1V
1 MΩ
+
237
Applications
t=0
–
1 μF
4V
+
1 μF
1 MΩ
–
+
dvo 0.5 MΩ
dt
1 MΩ
dvo
dt
1 MΩ
dvo
dt
1 MΩ
+
–
vo
–vo
–
+
vo
(b)
–
1V
+
–
t=0
+
4V
1 V 1 μF
1 MΩ
1 MΩ
–
+
(a)
10 sin (4t)
t=0
–
+
t=0
1 μF
1 MΩ
–
+
0.5 MΩ
dvo
dt
(c)
Figure 6.40
'PS&YBNQMF
&WBMVBUF5IFBOTXFSMPPLTDPSSFDU CVUJTJU *GBOBDUVBMTPMVUJPO
GPSvPJTEFTJSFE UIFOBHPPEDIFDLXould be to first find the
TPMVUJPOCZSFBMJ[JOHUIFDJSDVJUJO14QJDF5IJTSFTVMUDPVMEUIFOCF
DPNQBSFEXJUIBTPMVUJPOVTJOHUIFEJGGFSFOUJBMTPMVUJPODBQBCJMJUZ
PG."5-"#
Since all we need to do is check the circuit and confirm that it
SFQSFTFOUTUIFFRVBUJPO XFIBWFBOFBTJFSUFDIOJRVFUPVTF8FKVTU
HPUISPVHIUIFDJSDVJUBOETFFJGJUHFOFSBUFTUIFEFTJSFEFRVBUJPO
)PXFWFS XFTUJMMIBWFDIPJDFTUPNBLF8FDPVMEHPUISPVHI
UIFDJSDVJUGSPNMFGUUPSJHIUCVUUIBUXPVMEJOWPMWFEJGGFSFOUJBUJOHUIF
SFTVMUUPPCUBJOUIFPSJHJOBMFRVBUJPO"OFBTJFSBQQSPBDIXPVMECF
UPHPGSPNSJHIUUPMFGU5IJTJTUIFBQQSPBDIXFXJMMVTFUPDIFDLUIF
BOTXFS
4UBSUJOHXJUIUIFPVUQVU vP XFTFFUIBUUIFSJHIUIBOEPQBNQ
JTOPUIJOHNPSFUIBOBOJOWFSUFSXJUIBHBJOPGPOF5IJTNFBOTUIBU
UIFPVUQVUPGUIFNJEEMFDJSDVJUJT−vP5IFGPMMPXJOHSFQSFTFOUTUIF
BDUJPOPGUIFNJEEMFDJSDVJU
U Ev
U
−vP=−(∫ @@@
PEτ+vP )=−(vP +v
P )
Eτ
|
=− vP U −vP +vP XIFSFvP =−7JTUIFJOJUJBMWPMUBHFBDSPTTUIFDBQBDJUPS
8FDIFDLUIFDJSDVJUPOUIFMFGUUIFTBNFXBZ
U
Ev
Ev
Ev
@@@P=− ∫ −@@@@
PEτ−vP′ =−(−@@@P+vP′ −vP′ )
EU
EU
Eτ
1 MΩ
1 MΩ
–
+
vo
238
Chapter 6
Capacitors and Inductors
/PX BMMXFOFFEUPWerify is that the input to the first op amp is
−EvP∕EU
-PPLJOHBUUIFJOQVUXFTFFUIBUJUJTFRVBMUP
∕− @@@
Ev
Ev
−TJO U +vP+@@@@@@@
P
P=−TJO U +vP+@@@
EU
.Ω EU
XIJDIEPFTQSPEVDF−EvP∕EUGSPNUIFPSJHJOBMFRVBUJPO
4BUJTGBDUPSZ 5IFTPMVUJPOXFIBWFPCUBJOFEJTTBUJTGBDUPSZ8FDBO
OPXQSFTFOUUIJTXPSLBTBTPMVUJPOUPUIFQSPCMFN
Practice Problem 6.15
%FTJHOBOBOBMPHDPNQVUFSDJSDVJUUPTPMWFUIFEJGGFSFOUJBMFRVBUJPO
Ev
Ev
@@@@
P
+@@@
P+vP=DPTU U>
EU
EU
TVCKFDUUPvP = vP′ =
Answer:4FF'JH XIFSF3$=T
2V
t=0
C
R
d 2vo
dt 2
C
R
–
+
R
R
2
–
+
R
R
vo
R
3
–
+
–
+
d 2vo
dt 2
R
R
cos (10t) +
–
–
+
R
4
Figure 6.41
'PS1SBDUJDF1SPC
6.7
Summary
5IFDVSSFOUUISPVHIBDBQBDJUPSJTEJSFDUMZQSPQPSUJPOBMUPUIFUJNF
SBUFPGDIBOHFPGUIFWPMUBHFBDSPTTJU
J=$ @@@
Ev
EU
5IFDVSSFOUUISPVHIBDBQBDJUPSJT[FSPVOMFTTUIFW PMUBHFJTDIBOH
JOH5IVT BDBQBDJUPSBDUTMJLFBOPQFODJSDVJUUPBEDTPVSDF
239
Review Questions
5IFWPMUBHFBDSPTTBDBQBDJUPSJTEJSFDUMZQSPQPSUJPOBMUPUIFUJNF
JOUFHSBMPGUIFDVSSFOUUISPVHIJU
U JEτ=@@
UJEτ+v U
v=@@
∫
∫
$ −∞
$ U
5IFWPMUBHFBDSPTTBDBQBDJUPSDBOOPUDIBOHFJOTUBOUMZ
$BQBDJUPSTJOTFSJFTBOEJOQBSBMMFMBSFDPNCJOFEJOUIFTBNFX BZ
BTDPOEVDUBODFT
5IFWPMUBHFBDSPTTBOJOEVDUPSJTEJSFDUMZQSPQPSUJPOBMUPUIFUJNF
SBUFPGDIBOHFPGUIFDVSSFOUUISPVHIJU
EJ
v=-@@
EU
5IFWPMUBHFBDSPTTUIFJOEVDUPSJT[FSPVOMFTTUIFDVSSFOUJTDIBOH JOH5IVT BOJOEVDUPSBDUTMJLFBTIPSUDJSDVJUUPBEDTPVSDF
5IFDVSSFOUUISPVHIBOJOEVDUPSJTEJSFDUMZQSPQPSUJPOBMUPUIFUJNF
JOUFHSBMPGUIFWPMUBHFBDSPTTJU
UvEτ+J U
U vEτ=@@
J=@@
∫
∫
- −∞
- U
5IFDVSSFOUUISPVHIBOJOEVDUPSDBOOPUDIBOHFJOTUBOUMZ
*OEVDUPSTJOTFSJFTBOEJOQBSBMMFMBSFDPNCJOFEJOUIFTBNFX BZ
SFTJTUPSTJOTFSJFTBOEJOQBSBMMFMBSFDPNCJOFE
"UBOZHJWFOUJNFU UIFFOFSHZTUPSFEJOBDBQBDJUPSJT @$v XIJMF
UIFFOFSHZTUPSFEJOBOJOEVDUPSJT@-J
5ISFFBQQMJDBUJPODJSDVJUT UIFJOUF HSBUPS UIFEJGGFSFOUJBUPS BOEUIF
BOBMPHDPNQVUFS DBOCFSFBMJ[FEVTJOHSFTJTUPST DBQBDJUPST BOE
op amps.
Review Questions
8IBUDIBSHFJTPOB'DBQBDJUPSXIFOJUJT
DPOOFDUFEBDSPTTB7TPVSDF
B $
D $
0
C KPVMFT
E GBSBET
8IFOUIFUPUBMDIBSHFJOBDBQBDJUPSJTEPVCMFE UIF
FOFSHZTUPSFE
B SFNBJOTUIFTBNF
D JTEPVCMFE
10
C $
E $
$BQBDJUBODFJTNFBTVSFEJO
B DPVMPNCT
D IFOSZT
v(t)
C JTIBMWFE
E JTRVBESVQMFE
$BOUIFWPMUBHFXBWFGPSNJO'JHCFBTTPDJBUFE
XJUIBSFBMDBQBDJUPS
B :FT
C /P
1
2
t
– 10
Figure 6.42
'PS3FWJFX2VFTUJPO
5IFUPUBMDBQBDJUBODFPGUXPN'TFSJFTDPOOFDUFE
DBQBDJUPSTJOQBSBMMFMXJUIBN'DBQBDJUPSJT
B N'
E N'
C N'
F N'
D N'
240
Chapter 6
Capacitors and Inductors
*O'JH JGJ=DPTUBOEv=TJOU UIF
element is:
B BSFTJTUPS
C BDBQBDJUPS
*OEVDUPSTJOQBSBMMFMDBOCFDPNCJOFEKVTUMJLFSFTJT
UPSTJOQBSBMMFM
B 5SVF
D BOJOEVDUPS
C 'BMTF
'PSUIFDJSDVJUJO'JH UIFWPMUBHFEJWJEFS
GPSNVMBJT
-+-
-+-
vT
C v=@@@@@@@
vT
B v=@@@@@@@
-
-
i
v +
–
Element
-
D v=@@@@@@@
v
-+- T
Figure 6.43
'PS3FWJFX2VFTUJPO
L1
+ v –
1
")JOEVDUPSDIBOHFTJUTDVSSFOUCZ"JOT5IF
WPMUBHFQSPEVDFEBUUIFUFSNJOBMTPGUIFJOEVDUPSJT
B 7
D 7
-
E v=@@@@@@@
v
-+- T
C 7
E 7
*GUIFDVSSFOUUISPVHIBN)JOEVDUPSJODSFBTFT
GSPN[FSPUP" IPXNVDIFOFSHZJTTUPSFEJOUIF
JOEVDUPS
B N+
D N+
vs
+
–
+
v2
–
L2
Figure 6.44
'PS3FWJFX2VFTUJPO
"OTXFSTB E E C D C B C B E
C N+
E N+
Problems
Section 6.2
Capacitors
*GUIFWPMUBHFBDSPTTB'DBQBDJUPSJTUF−U7 find the current and the power.
"μ'DBQBDJUPSIBTFOFSHZX U =DPTU+
%FUFSNJOFUIFDVSSFOUUISPVHIUIFDBQBDJUPS
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOEIPXDBQBDJUPSTXPSL
"WPMUBHFBDSPTTBDBQBDJUPSJTFRVBMUP
<mDPT U)] V and the current flowing through
JUJTFRVBMUPTJO U μ"%FUFSNJOFUIFWBMVFPG
UIFDBQBDJUBODF$BMDVMBUFUIFQPXFSCFJOHTUPSFE
CZUIFDBQBDJUPS
5IFWPMUBHFBDSPTTBμ'DBQBDJUPSJTTIPXOJO
Fig. 6.45. Find the current waveform.
v (t) V
10
0
0
–10
Figure 6.45
'PS1SPC
2
4
6
8
2
4
6
8
10
12 t (ms)
–10
Figure 6.46
'PS1SPC
"UU= UIFWPMUBHFBDSPTTBN'DBQBDJUPSJT7
$BMDVMBUFUIFWPMUBHFBDSPTTUIFDBQBDJUPSGPSU>
XIFODVSSFOUU mA flows through it.
"N'DBQBDJUPSIBTUIFUFSNJOBMWPMUBHF
7
U≤
v= −U
−U
"F
+#F
7
U≥
v(t) V
10
5IFWPMUBHFXBWFGPSNJO'JHJTBQQMJFEBDSPTT
Bμ'DBQBDJUPS%SBXUIFDVSSFOUXBWFGPSN
UISPVHIJU
t (ms)
{
*GUIFDBQBDJUPSIBTBOJOJUJBMDVSSFOUPGA, find:
B UIFDPOTUBOUT"BOE#
C UIFFOFSHZTUPSFEJOUIFDBQBDJUPSBUU=
D UIFDBQBDJUPSDVSSFOUGPSU>
241
Problems
5IFDVSSFOUUISPVHIB'DBQBDJUPSJT −F−U "
%FUFSNJOFUIFWPMUBHFBOEQPXFSBUU=T"TTVNF
v =
5IFWPMUBHFBDSPTTBN'DBQBDJUPSJTTIPXOJO
Fig. 6.47. Determine the current through the DBQBDJUPS
v (t) (V)
5XPDBQBDJUPST BOEμ' BSFDPOOFDUFE
UPB7TPVSDF'JOEUIFFOFSHZTUPSFEJOFBDI
DBQBDJUPSJGUIFZBSFDPOOFDUFEJO
B QBSBMMFM
C TFSJFT
5IFFRVJWBMFOUDBQBDJUBODFBUUFSNJOBMTBCJOUIF
DJSDVJUPG'JHJTμF. Calculate the value of $
16
30 μF
C
a
0
1
2
3
10 μF
t (μs)
4
b
Figure 6.47
'PS1SPC
Figure 6.50
"N'DBQBDJUPSIBTUIFDVSSFOUXBWFGPSNTIPXO
JO'JH"TTVNJOHUIBUv =7 TLFUDIUIF
WPMUBHFXBWFGPSNv U 'PS1SPC
%FUFSNJOFUIFFRVJWBMFOUDBQBDJUBODFGPSFBDIPGUIF
DJSDVJUTPG'JH
12 F
4F
i(t) (mA)
15
6F
3F
10
5
0
2
–5
4
6
8
4F
t (s)
(a)
6F
–10
Figure 6.48
5F
'PS1SPC
4F
(b)
"WPMUBHFPGF−U7BQQFBSTBDSPTTBQBSBMMFM
DPNCJOBUJPOPGBN'DBQBDJUPSBOEBΩ
SFTJTUPS$BMDVMBUFUIFQPXFSBCTPSCFECZUIF
QBSBMMFMDPNCJOBUJPO
3F
6F
2F
'JOEUIFWPMUBHFBDSPTTUIFDBQBDJUPSTJOUIFDJSDVJU
PG'JHVOEFSEDDPOEJUJPOT
10 Ω
2F
4F
3F
(c)
50 Ω
Figure 6.51
'PS1SPC
70 Ω
C1
+
v1
–
20 Ω
+ 60 V
–
+
v2
–
C2
'JOE$FRJOUIFDJSDVJUPG'JHJGBMMDBQBDJUPST
BSFμ'
Figure 6.49
'PS1SPC
Section 6.3
Series and Parallel Capacitors
4FSJFTDPOOFDUFEBOEQ'DBQBDJUPSTBSF
QMBDFEJOQBSBMMFMXJUITFSJFTDPOOFDUFEBOE
Q'DBQBDJUPST%FUFSNJOFUIFFRVJWBMFOU
DBQBDJUBODF
Ceq
Figure 6.52
'PS1SPC
242
Chapter 6
Capacitors and Inductors
'JOEUIFFRVJWBMFOUDBQBDJUBODFCFUXFFOUFSNJOBMT
B and CJOUIFDJSDVJUPG'JH"MMDBQBDJUBODFT
BSFJOμ'
40 μF
10 μF
12
5 μF
20 μF
40
15 μF
a
50
20
12
10 μF
35 μF
80
30
15 μF
10
a
b
Figure 6.56
b
'PS1SPC
60
6TJOH'JH EFTJHOBQSPCMFNUIBUXJMMIFMQ
PUIFSTUVEFOUTCFUUFSVOEFSTUBOEIPXDBQBDJUPSTXPSL
UPHFUIFSXIFODPOOFDUFEJOTFSJFTBOEJOQBSBMMFM
Figure 6.53
'PS1SPC
'JOEUIFFRVJWBMFOUDBQBDJUBODFBUUFSNJOBMTBCPG
UIFDJSDVJUJO'JH
C1
V +
–
a
2 μF
3 μF
3 μF
'PS1SPC
*OUIFDJSDVJUTIPXOJO'JHBTTVNFUIBUUIF
DBQBDJUPSTXFSFJOJUJBMMZVODIBSHFEBOEUIBUUIF
DVSSFOUTPVSDFIBTCFFODPOOFDUFEUPUIFDJSDVJUMPOH
FOPVHIGPSBMMUIFDBQBDJUPSTUPSFBDITUFBEZTUBUF OP
current flowing through the capacitors). Determine
UIFWPMUBHFBDSPTTFBDIDBQBDJUPSBOEUIFFOFSHZ
TUPSFEJOFBDI
30 μF
30 μF
2 μF
2 μF
3 μF
C4
Figure 6.57
1 μF
1 μF
C3
C2
3 μF
15 mA
10 μF
10 kΩ
20 μF
18 μF
b
Figure 6.58
Figure 6.54
'PS1SPC
%FUFSNJOFUIFFRVJWBMFOUDBQBDJUBODFBUUFSNJOBMT
BCPGUIFDJSDVJUJO'JH
5 μF
6 μF
'PS1SPC
B 4IPXUIBUUIFWPMUBHFEJWJTJPOSVMFGPSUXP
DBQBDJUPSTJOTFSJFTBTJO'JH B JT
$
$
v v=@@@@@@@
v
v=@@@@@@@
$+$ T
$+$ T
4 μF
C1
a
2 μF
3 μF
BTTVNJOHUIBUUIFJOJUJBMDPOEJUJPOTBSF[FSP
12 μF
b
Figure 6.55
'PS1SPC
vs +
–
+ v1 –
+
v2
–
(a)
0CUBJOUIFFRVJWBMFOUDBQBDJUBODFPGUIFDJSDVJUJO
'JH
Figure 6.59
'PS1SPC
C2
is
(b)
i1
i2
C1
C2
243
Problems
C 'PSUXPDBQBDJUPSTJOQBSBMMFMBTJO'JH C TIPXUIBUUIFDVSSFOUEJWJTJPOSVMFJT
$
J J=@@@@@@@
$+$ T
$
J=@@@@@@@
J
$+$ T
BTTVNJOHUIBUUIFJOJUJBMDPOEJUJPOTBSF[FSP
5ISFFDBQBDJUPST $=μ' $=μ' BOE
$ = 20 μ' BSFDPOOFDUFEJOQBSBMMFMBDSPTTB
7TPVSDF%FUFSNJOF
B UIFUPUBMDBQBDJUBODF
C UIFDIBSHFPOFBDIDBQBDJUPS
D UIFUPUBMFOFSHZTUPSFEJOUIFQBSBMMFM
DPNCJOBUJPO
"TTVNJOHUIBUUIFDBQBDJUPSTBSFJOJUJBMMZVODIBSHFE find vP U JOUIFDJSDVJUPG'JH
is (mA)
6 μF
90
is
+
vo (t)
–
3 μF
0
2 t (s)
1
Figure 6.62
'PS1SPC
(JWFOUIBUGPVSμ'DBQBDJUPSTDBOCFDPOOFDUFE
in series and in parallel, find the minimum and
NBYJNVNWBMVFTUIBUDBOCFPCUBJOFECZTVDI
TFSJFTQBSBMMFMDPNCJOBUJPOT
*Gv = 0, find v U J U BOEJ U JOUIFDJSDVJUPG
'JH
0CUBJOUIFFRVJWBMFOUDBQBDJUBODFPGUIFOFUXPSL
TIPXOJO'JH
is (mA)
30
40 μF
0
50 μF
30 μF
1
2
3
5
4
t
–30
10 μF
20 μF
i1
Figure 6.60
is
'PS1SPC
%FUFSNJOF$FRGPSFBDIDJSDVJUJO'JH
6 μF
i2
+
v
–
4 μF
Figure 6.63
'PS1SPC
C
C eq
C
C
C
C
*OUIFDJSDVJUJO'JH MFUJT=F−UN"BOE
UIFWPMUBHFBDSPTTFBDIDBQBDJUPSJTFRVBMUP[FSPBU
U = 0. Determine vBOEvBOEUIFFOFSHZTUPSFEJO
FBDIDBQBDJUPSGPSBMMU>
(a)
C
36 μF
C
C eq
24 μF
C
(b)
Figure 6.61
C
+
–
v2
is
10 Ω
'PS1SPC
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
v1
Figure 6.64
'PS1SPC
+
–
60 μF
244
Chapter 6
Capacitors and Inductors
0CUBJOUIF5IFWFOJOFRVJWBMFOUBUUIFUFSNJOBMT BC PGUIFDJSDVJUTIPXOJO'JH1MFBTFOPUFUIBU
5IFWFOJOFRVJWBMFOUDJSDVJUTEPOPUHFOFSBMMZFYJTU
GPSDJSDVJUTJOWPMWJOHDBQBDJUPSTBOESFTJTUPST5IJTJT
BTQFDJBMDBTFXIFSFUIF5IFWFOJOFRVJWBMFOUDJSDVJU
EPFTFYJTU
*GUIFWPMUBHFXBWFGPSNJO'JHJTBQQMJFEBDSPTT
UIFUFSNJOBMTPGB)JOEVDUPS DBMDVMBUFUIFDVSSFOU
UISPVHIUIFJOEVDUPS"TTVNFJ =−"
v(t) (V)
5F
45 V +
–
10
a
3F
2F
0
b
Figure 6.65
Inductors
5IFDVSSFOUUISPVHIBN)JOEVDUPSJTF−U∕"
'JOEUIFWPMUBHFBOEUIFQPXFSBUU=T
"OJOEVDUPSIBTBMJOFBSDIBOHFJODVSSFOUGSPN
100 mA to 200 mA in 2 ms and induces a voltage of
N7$BMDVMBUFUIFWBMVFPGUIFJOEVDUPS
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFSVOEFS
TUBOEIPXJOEVDUPSTXPSL
5IFDVSSFOUUISPVHIBN)JOEVDUPSJTTJOU"
π
'JOEUIFWPMUBHF BOEUIFFOFSHZTUPSFEBUU=___
T
{
J U =
"N)JOEVDUPSJTDPOOFDUFEJOQBSBMMFMXJUIB
LΩSFTJTUPS5IFDVSSFOUUISPVHIUIFJOEVDUPSJT
J U =F−UN" B 'JOEUIFWPMUBHFv-BDSPTTUIF
JOEVDUPS C 'JOEUIFWPMUBHFv3BDSPTTUIFSFTJTUPS
D %PFTv3 U +v- U = E $BMDVMBUFUIFFOFSHZ
TUPSFEJOUIFJOEVDUPSBUU=
*GUIFWPMUBHFXBWFGPSNJO'JHJTBQQMJFEUPB
25-mH inductor, find the inductor current J U GPS
<U<TFDPOET"TTVNFJ =
UF−U" v(t)
U<
5
U> 'JOEUIFWPMUBHFv U 0
v U =<F−U+U+>7
GPSU>
%FUFSNJOFUIFDVSSFOUJ U UISPVHIUIFJOEVDUPS
"TTVNFUIBUJ ="
5IFDVSSFOUUISPVHIBN)JOEVDUPSJTTIPXOJO
'JH%FUFSNJOFUIFWPMUBHFBDSPTTUIFJOEVDUPS
BUU= BOENT
1
'PS1SPC
'JOEv$ J- BOEUIFFOFSHZTUPSFEJOUIFDBQBDJUPS
BOEJOEVDUPSJOUIFDJSDVJUPG'JHVOEFSED
DPOEJUJPOT
2Ω
3A
'PS1SPC
t
Figure 6.68
10
Figure 6.66
2
–5
i(A)
0
t
5
4
5IFDVSSFOUJOBN)JOEVDUPSJODSFBTFTGSPN
UPN" TUFBEZTUBUF )PXNVDIFOFSHZJTTUPSFE
JOUIFJOEVDUPS
5IFWPMUBHFBDSPTTBN)JOEVDUPSJTHJWFOCZ
3
2
Figure 6.67
5IFDVSSFOUUISPVHIBN)JOEVDUPSJT
1
'PS1SPC
'PS1SPC
Section 6.4
5IFWPMUBHFBDSPTTB)JOEVDUPSJT −F−U 7
*GUIFJOJUJBMDVSSFOUUISPVHIUIFJOEVDUPSJT" find the current and the energy stored in the inductor
BUU=T
4Ω
+
vC
–
2F
0.5 H
5Ω
2
4
6
t (ms)
Figure 6.69
'PS1SPC
iL
245
Problems
'PSUIFDJSDVJUJO'JH DBMDVMBUFUIFWBMVFPG3UIBU
XJMMNBLFUIFFOFSHZTUPSFEJOUIFDBQBDJUPSUIFTBNF
BTUIBUTUPSFEJOUIFJOEVDUPSVOEFSEDDPOEJUJPOT
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEIPXJOEVDUPSTCFIBWF
XIFODPOOFDUFEJOTFSJFTBOEXIFODPOOFDUFEJO
QBSBMMFM
R
160 μF
2Ω
5A
L4
4 mH
L2
Figure 6.70
Leq
'PS1SPC
Under steady-state dc conditions, findJBOEvJOUIF
DJSDVJUJO'JH
10 Ω
25 V +
–
v
Figure 6.74
i
'JOE-FRBUUIFUFSNJOBMTPGUIFDJSDVJUJO'JH
10 Ω
10 H
Figure 6.71
6 mH
'PS1SPC
Section 6.5
L5
L6
–
20 H
15 Ω
L1
'PS1SPC
10 F
+
L3
8 mH
a
5 mH
Series and Parallel Inductors
'JOEUIFFRVJWBMFOUJOEVDUBODFPGUIFDJSDVJUJO
Fig. 6.72. Assume all inductors are 40 mH.
12 mH
8 mH
6 mH
b
4 mH
8 mH
10 mH
Figure 6.75
'PS1SPC
Figure 6.72
'JOEUIFFRVJWBMFOUJOEVDUBODFMPPLJOHJOUPUIF
UFSNJOBMTPGUIFDJSDVJUJO'JH
'PS1SPC
"OFOFSHZTUPSBHFOFUXPSLDPOTJTUTPGTFSJFT
DPOOFDUFEBOEN)JOEVDUPSTJOQBSBMMFMXJUI
TFSJFTDPOOFDUFEBOEN)JOEVDUPST$BMDVMBUF
UIFFRVJWBMFOUJOEVDUBODF
a
100 mH
%FUFSNJOF-FRBUUFSNJOBMTBC of the circuit in Fig. 10 mH
100 mH
60 mH
60 mH
25 mH
60 mH
20 mH
a
60 mH
150 mH
b
b
30 mH
300 mH
Figure 6.73
'PS1SPC
Figure 6.76
'PS1SPC
30 mH
246
Chapter 6
Capacitors and Inductors
'JOE-FRJOFBDIPGUIFDJSDVJUTJO'JH
The current waveform in Fig. 6.80 flows through a
)JOEVDUPS4LFUDIUIFWPMUBHFBDSPTTUIFJOEVDUPS
PWFSUIFJOUFSWBM<U<T
L
()
L
Leq
L
L
2
L
0
1
2
3
4
5
6
Figure 6.80
(a)
'PS1SPC
L
L
L
L
B 'PSUXPJOEVDUPSTJOTFSJFTBTJO'JH B TIPXUIBUUIFWPMUBHFEJWJTJPOQSJODJQMFJT
L
Leq
(b)
-
v=@@@@@@@
v -+- T
Figure 6.77
'PS1SPC
BTTVNJOHUIBUUIFJOJUJBMDPOEJUJPOTBSF[FSP
C 'PSUXPJOEVDUPSTJOQBSBMMFMBTJO'JH C TIPXUIBUUIFDVSSFOUEJWJTJPOQSJODJQMFJT
'JOE-FRJOUIFDJSDVJUPG'JH
-
J=@@@@@@@
J -+- T
L
L
L
BTTVNJOHUIBUUIFJOJUJBMDPOEJUJPOTBSF[FSP
+ v –
1
L
-
J=@@@@@@@
J
-+- T
L1
L
L
L
-
v=@@@@@@@
v
-+- T
L
vs
+
–
+
v2
–
is
L2
i1
i2
L1
L2
L eq
(a)
Figure 6.78
'PS1SPC
'PS1SPC
%FUFSNJOF-FRUIBUNBZCFVTFEUPSFQSFTFOUUIF
JOEVDUJWFOFUXPSLPG'JHBUUIFUFSNJOBMT
i
a
L eq
2
4H
*OUIFDJSDVJUPG'JH JP ="%FUFSNJOF
JP U BOEvP U GPSU>
di
dt
io (t)
+–
3H
5H
4e–2t A
b
Figure 6.79
'PS1SPC
(b)
Figure 6.81
Figure 6.82
'PS1SPC
3H
5H
+
–
vo
247
Problems
$POTJEFSUIFDJSDVJUJO'JH'JOE B -FR J U BOEJ U JGJT=F−UN" C vP U), (c) FOFSHZTUPSFEJO
UIFN)JOEVDUPSBUU=T
i1
+
vo
–
is
5IFTXJUDIJO'JHIBTCFFOJOQPTJUJPO"GPSB
MPOHUJNF"UU= UIFTXJUDINPWFTGSPNQPTJUJPO
"UP#5IFTXJUDIJTBNBLFCFGPSFCSFBLUZQFTP
UIBUUIFSFJTOPJOUFSSVQUJPOJOUIFJOEVDUPSDVSSFOU
'JOE
B J U GPSU> C vKVTUBGUFSUIFTXJUDIIBTCFFONPWFEUPQPTJUJPO# D v U MPOHBGUFSUIFTXJUDIJTJOQPTJUJPO#
i2
4 mH
20 mH
6 mH
4Ω
B
i
L eq
Figure 6.83
12 V
'PS1SPC
+
–
0.5 H
t=0A
+
v
–
5Ω
6A
Figure 6.86
$POTJEFSUIFDJSDVJUJO'JH(JWFOUIBU
v U =F−UN7GPSU>BOEJ =−N" find: (a) J C J U BOEJ U 'PS1SPC
5IFJOEVDUPSTJO'JHBSFJOJUJBMMZDIBSHFEBOE
BSFDPOOFDUFEUPUIFCMBDLCPYBUU=*GJ =" J =−" BOEv U =F−UN7 U≥ 0, find:
25 mH
+
i1(t)
i2(t)
v(t)
20 mH
60 mH
B UIFFOFSHZJOJUJBMMZTUPSFEJOFBDIJOEVDUPS
C UIFUPUBMFOFSHZEFMJWFSFEUPUIFCMBDLCPYGSPN
U =UPU=∞
D J U BOEJ U U≥
E J U U≥
–
Figure 6.84
i(t)
'PS1SPC
+
Black box v
t=0
i1
i2
5H
20 H
–
*OUIFDJSDVJUPG'JH TLFUDIvP
Figure 6.87
'PS1SPC
+
vo
–
i1 (t)
i2(t)
2H
5IFDVSSFOUJ U UISPVHIBN)JOEVDUPSJTFRVBM JONBHOJUVEF UPUIFWPMUBHFBDSPTTJUGPSBMMWBMVFTPG
UJNF*GJ = 2 A, find J U i2 (t) (A)
4
i1 (t) (A)
3
0
Figure 6.85
'PS1SPC
3
6 t (s)
Section 6.6
0
2
4
6 t (s)
Applications
"OPQBNQJOUFHSBUPSIBT3=LΩBOE$=μ'
*GUIFJOQVUWPMUBHFJTvJ=TJOUN7 PCUBJOUIF
PVUQVUWPMUBHF"TTVNFUIBUBUUFRVBMUP[FSP UIF
PVUQVUJTFRVBMUP[FSP
248
Chapter 6
Capacitors and Inductors
"7EDWPMUBHFJTBQQMJFEUPBOJOUFHSBUPSXJUI
3 =LΩ $=μ'BUU=)PXMPOHXJMM
JUUBLFGPSUIFPQBNQUPTBUVSBUFJGUIFTBUVSBUJPO
WPMUBHFTBSF+7BOE−7 "TTVNFUIBUUIF
JOJUJBMDBQBDJUPSWPMUBHFXBT[FSP
4IPXUIBUUIFDJSDVJUJO'JHJTBOPOJOWFSUJOH
JOUFHSBUPS
R
"OPQBNQJOUFHSBUPSXJUI3=.ΩBOE
$=μ'IBTUIFJOQVUXBWFGPSNTIPXOJO
'JH1MPUUIFPVUQVUXBWFGPSN
R
–
+
R
vi
vi (mV)
+
R
vo
+
–
C
–
20
Figure 6.90
10
0
'PS1SPC
1
2
3
4 5
6
t (ms)
5IFUSJBOHVMBSXBWFGPSNJO'JH B JTBQQMJFEUP
UIFJOQVUPGUIFPQBNQEJGGFSFOUJBUPSJO'JH C 1MPUUIFPVUQVU
–10
–20
Figure 6.88
v (t)
'PS1SPC
10
6TJOHBTJOHMFPQBNQ BDBQBDJUPS BOESFTJTUPSTPG
LΩPSMFTT EFTJHOBDJSDVJUUPJNQMFNFOU
U
vP=− vJ τ Eτ
∫
0
1
2
3
4
t (ms)
–10
"TTVNFvP = BUU = (a)
4IPXIPXZPVXPVMEVTFBTJOHMFPQBNQUPHFOFSBUF
20 kΩ
U
vP=−∫ v+v+v Eτ
0.01 μF
*GUIFJOUF HSBUJOHDBQBDJUPSJT $ = μ' PCUBJOUIF
PUIFSDPNQPOFOUWBMVFT
vi
"UU=NT DBMDVMBUFvPEVFUPUIFDBTDBEFE
JOUFHSBUPSTJO'JH"TTVNFUIBUUIFJOUFHSBUPST
BSFSFTFUUP7BUU=
+
–
–
+
+
vo
–
(b)
Figure 6.91
'PS1SPC
2 μF
10 kΩ
1V +
–
Figure 6.89
'PS1SPC
–
+
0.5 μF
20 kΩ
–
+
+
vo
–
"OPQBNQEJGGFSFOUJBUPSIBT3=LΩBOE
C = 10 μ'5IFJOQVUWPMUBHFJTBSBNQS U =U mV.
'JOEUIFPVUQVUWPMUBHF
"WPMUBHFXBWFGPSNIBTUIFGPMMPXJOHDIBSBDUFSJTUJDT
BQPTJUJWFTMPQFPG7TGPSNTGPMMPXFECZB
OFHBUJWFTMPQFPG7TGPSNT*GUIFXBWFGPSN
is applied to a differentiator with 3=LΩ $= 10 μ' TLFUDIUIFPVUQVUWPMUBHFXBWFGPSN
249
Comprehensive Problems
5IFPVUQVUvPPGUIFPQBNQDJSDVJUJO'JH B is shown in Fig. 6.92(b). Let 3J=3G=.ΩBOE
C =μ'%FUFSNJOFUIFJOQVUWPMUBHFXBWFGPSN
BOETLFUDIJU
%FTJHOBOBOBMPHDPNQVUFSDJSDVJUUPTPMWFGPSv U HJWFOUIFGPMMPXJOHFRVBUJPOBOEBWBMVFGPSG U BOE
UIBUW =7
Ev U ∕EU +vU=GUEU
'JHVSFQSFTFOUTBOBOBMPHDPNQVUFSEFTJHOFE
to solve a differential equation. Assuming G U JT
LOPXO TFUVQUIFFRVBUJPOGPSG U Rf
C
Ri
vi
1 μF
–
+
+
–
+
vo
–
1 MΩ
1 μF
1 MΩ
–
+
1 MΩ
vo(t)
–
+
100 kΩ
(a)
100 kΩ
vo
4
0
500 kΩ
–
+
–
+
200 kΩ
–f (t)
Figure 6.93
1
2
3
4
t (s)
'PS1SPC
–4
(b)
Figure 6.92
%FTJHOBOBOBMPHDPNQVUFSUPTJNVMBUFUIFGPMMPXJOH
FRVBUJPOUPTPMWFGPSv U BTTVNFUIFJOJUJBM
DPOEJUJPOTBSF[FSP Ev U ∕EU + Ev U ∕EU =G U
'PS1SPC
%FTJHOBOPQBNQDJSDVJUTVDIUIBU
%FTJHOBOBOBMPHDPNQVUFSUPTJNVMBUF
Ev
Ev
@@@@
+@@@
P+v =TJOU
P
vP=vT+∫ vT EU
EU
EU
P
XIFSFv = −7BOEv′ = XIFSF vTBOE vPBSFUIFJOQVUW
WPMUBHF SFTQFDUJWFMZ
PMUBHFBOEPVUQVU Comprehensive Problems
:PVSMBCPSBUPSZIBTBWBJMBCMFBMBSHFOVNCFSPG
μ'DBQBDJUPSTSBUFEBU75PEFTJHOBDBQBDJ
UPSCBOLPGμ'SBUFEBU7 IPXNBOZμ'
DBQBDJUPSTBSFOFFEFEBOEIPXXPVMEZPVDPOOFDU
UIFN
"ON)JOEVDUPSJTVTFEJOBGVTJPOQPXFS
FYQFSJNFOU*GUIFDVSSFOUUISPVHIUIFJOEVDUPSJT
J U =DPT π U N" GPSBMMU> 0 s, find the power
CFJOHEFMJWFSFEUPUIFJOEVDUPSBOEUIF
FOFSHZTUPSFEJOJUBUU=T
250
Chapter 6
Capacitors and Inductors
"TRVBSFXBWFHFOFSBUPSQSPEVDFTUIFWPMUBHF
XBWFGPSNTIPXOJO'JH B 8IBULJOEPGB
DJSDVJUDPNQPOFOUJTOFFEFEUPDPOWFSUUIFWPMUBHF
waveform to the triangular current waveform shown
JO'JH C $BMDVMBUFUIFWBMVFPGUIFDPNQP
OFOU BTTVNJOHUIBUJUJTJOJUJBMMZVODIBSHFE
4
0
1
3
2
4
t (ms)
(b)
v (V)
Figure 6.94
5
0
i (A)
'PS1SPC
1
2
3
–5
(a)
4
t (ms)
"OFMFDUSJDNPUPSDBOCFNPEFMFEBTBTFSJFT
DPNCJOBUJPOPGBΩSFTJTUPSBOEN)JOEVDUPS
*GBDVSSFOUJ U =UF−U A flows through the series
combination, find the voltage across the combination.
c h a p t e r
First-Order Circuits
7
8FMJve in deeds, not years; in thoughts, not breaths; in feelings, not JO
figures on a dial. We should count time in heart-throbs. He most livFT
XIPUIJOLTNPTU GFFMTUIFOPCMFTU BDUTUIFCFTU
‡1+#BJMFZ
Enhancing Your Career
Careers in Computer Engineering
&MFDUSJDBMFOHJOFFSJOHFEVDBUJPOIBTHPOFUISPVHIESBTUJDDIBOHFTJOSF
DFOUEFDBEFT.PTUEFQBSUNFOUTIBWFDPNFUPCFLOPXOBT%FQBSUNFOU
PG&MFDUSJDBMBOE$PNQVUFS&OHJOFFSJOH FNQIBTJ[JOHUIFSBQJEDIBOHFT
EVFUPDPNQVUFST$PNQVUFSTPDDVQ ZBQSPNJOFOUQMBDFJONPEFSOTP DJFUZBOEFEVDBUJPO5IFZIBWFCFDPNFDPNNPOQMBDFBOEBSFIFMQJOH
UPDIBOHFUIFG BDFPGSFTFBSDI EF WFMPQNFOU QSPEVDUJPO C VTJOFTT BOE
FOUFSUBJONFOU5IFTDJFOUJTU FOHJOFFS EPDUPS BUUPSOFZ UFBDIFS BJSMJOF
QJMPU CVTJOFTTQFSTPO‡BMNPTUBOyone benefits from a computer’TBCJMJ
UJFTUPTUPSFMBSHFBNPVOUTPGJOGPSNBUJPOBOEUPQSPDFTTUIBUJOGPSNBUJPO
JOWFSZTIPSUQFSJPETPGUJNF 5IFJOUFSOFU BDPNQVUFSDPNNVOJDBUJPO
OFUXPSL JTFTTFOUJBMJOC VTJOFTT FEVDBUJPO BOEMJCSBSZTDJFODF$PN QVUFSVTBHFDPOUJOVFTUPHSPXCZMFBQTBOECPVOET
"OFEVDBUJPOJODPNQVUFSFOHJOFFSJOHTIPVMEQSPWJEFCSFBEUIJOTPGU
XBSF IBSEXBSFEFTJHO BOECBTJDNPEFMJOHUFDIOJRVFT*UTIPVMEJODMVEF DPVSTFTJOEBUBTUSVDUVSFT EJHJUBMTZTUFNT DPNQVUFSBSDIJUFDUVSF NJDSP QSPDFTTPST JOUFSGBDJOH TPGUXBSFFOHJOFFSJOH BOEPQFSBUJOHTZTUFNT
Electrical engineers who specialize in computer engineering find
jobs in computer industries and in numerous fields where computers
BSFCFJOHVTFE$PNQBOJFTUIBUQSPEVDFTPGUXBSFBSFHSPXJOHSBQJEMZJO
OVNCFSBOETJ[FBOEQSPWJEJOHFNQMPZNFOUGPSUIPTFXIPBSFTLJMMFEJO
QSPHSBNNJOH"OFYDFMMFOUXBZUPBEWBODFPOFTLOPXMFEHFPGDPNQVU
FSTJTUPKPJOUIF*&&&$PNQVUFS4PDJFUZ XIJDITQPOTPSTEJWFSTFNBHB
[JOFT KPVSOBMT BOEDPOGFSFODFT
$PNQVUFSEFTJHOPGWFSZMBSHFTDBMF
JOUFHSBUFE 7-4* DJSDVJUT
$PVSUFTZ#SJBO'BTU $MFWFMBOE4UBUF
6OJWFSTJUZ
251
252
Chapter 7
First-Order Circuits
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
Understand solutions to unforced, first-order linear difGFSFOUJBM
FRVBUJPOT
$PNQSFIFOETJOHVMBSJUZFRVBUJPOTBOELOPXUIFJSJNQPSUBODF
JOTPMWJOHMJOFBSEJGGFSFOUJBMFRVBUJPOT
6OEFSTUBOEUIFFGfect of unit step sources on first-order linear
EJGGFSFOUJBMFRVBUJPOT
&YQMBJOIPw dependent sources and op amps influence simple
first-order linear difGFSFOUJBMFRVBUJPOT
6TF14QJDFUPTPMWFTJNQMFUSBOTJFOUDJSDVJUTXJUIBOJOEVDUPS
PSBDBQBDJUPS
7.1
Introduction
/PXUIBUXFIB WFDPOTJEFSFEUIFUISFFQBTTJ WFFMFNFOUT SFTJTUPST DB QBDJUPST BOEJOEVDUPST BOEPOFBDUJWFFMFNFOU UIFPQBNQ JOEJWJEVBMMZ XFBSFQSFQBSFEUPDPOTJEFSDJSDVJUTUIBUDPOUBJOWBSJPVTDPNCJOBUJPOTPG
UXPPSUISFFPGUIFQBTTJ WFFMFNFOUT*OUIJTDIBQUFS XFTIBMMF YBNJOF
UXPUZQFTPGTJNQMFDJSDVJUTBDJSDVJUDPNQSJTJOHBSFTJTUPSBOEDBQBDJUPS
BOEBDJSDVJUDPNQSJTJOHBSFTJTUPSBOEBOJOEVDUPS 5IFTFBSFDBMMFE3$
BOE3- DJSDVJUT SFTQFDUJWFMZ"TTJNQMFBTUIFTFDJSDVJUTBSF UIF y find
DPOUJOVBMBQQMJDBUJPOTJOFMFDUSPOJDT DPNNVOJDBUJPOT BOEDPOUSPMTZT UFNT BTXFTIBMMTFF
8FDBSSZPVUUIFBOBMZTJTPG3$BOE3-DJSDVJUTCZBQQMZJOH,JSDI
IPGGTMBXT BTXFEJEGPSSFTJTUJ WFDJSDVJUT 5IFPOMZEJG GFSFODFJTUIBU
BQQMZJOH,JSDIIPGGTMBXTUPQVSFMZSFTJTUJWFDJSDVJUTSFTVMUTJOBMHFCSBJD
FRVBUJPOT XIJMFBQQMZJOHUIFMB XTUP3$BOE3-DJSDVJUTQSPEVDFTEJG
GFSFOUJBMFRVBUJPOT XIJDIBSFNPSFEJGficult to solvFUIBOBMHFCSBJDFRVB
UJPOT5IFEJGGFSFOUJBMFRVBUJPOTSFTVMUJOHGSPNBOBMZ[JOH3$BOE3-DJS
cuits are of the first order)FODF UIFDJSDVJUTBSFDPMMFDUJWFMZLOPXOBT
firTUPSEFSDJSDVJUT
A first-order circuit is characterized by a first-order differential equation.
*OBEEJUJPOUPUIFSFCFJOHUX o types of first-order circuits 3$BOE
3- UIFSFBSFUX PXBZTUPF YDJUFUIFDJSDVJUT The first wBZJTCZJOJ UJBMDPOEJUJPOTPGUIFTUPSBHFFMFNFOUTJOUIFDJSDVJUT*OUIFTFTPDBMMFE
TPVSDFGSFFDJSDVJUT XFBTTVNFUIBUFOFS HZJTJOJUJBMMZTUPSFEJOUIFDB QBDJUJWFPSJOEVDUJ WFFMFNFOU 5IFFOFSHZDBVTFT current to floXJOUIF
DJSDVJUBOEJTHSBEVBMMZEJTTJQBUFEJOUIFSFTJTUPST "MUIPVHITPVSDFGSFF
circuits are by definition free of independent sources, theZNBZIB WF
EFQFOEFOUTPVSDFT5IFTFDPOEXBZPGFxciting first-order circuits is by
JOEFQFOEFOU TPVSDFT *O UIJTDIBQUFS UIF JOEFQFOEFOU TPVSDFT XF XJMM
DPOTJEFSBSFEDTPVSDFT *OMBUFSDIBQUFST XFTIBMMDPOTJEFSTJOVTPJEBM
BOEFYQPOFOUJBMTPVSDFT 5IFUXo types of first-order circuits and the
UXPXBZTPGFYDJUJOHUIFNBEEVQUPUIFGPVSQPTTJCMFTJUVBUJPOTXFXJMM
TUVEZJOUIJTDIBQUFS
7.2
253
The Source-Free RC Circuit
'JOBMMZ XFDPOTJEFSGPVSUZQJDBMBQQMJDBUJPOTPG3$BOE3-DJSDVJUT
delay and relay circuits, a photoflBTIVOJU BOEBOBVUPNPCJMFJHOJUJPO
DJSDVJU
iC
C
+
v
iR
R
‒
7.2
The Source-Free RC Circuit
"TPVSDFGSFF3$DJSDVJUPDDVSTXIFOJUTEDTPVSDFJTTVEEFOMZ
EJTDPOOFDUFE Figure 7.1
"TPVSDFGSFF3$DJSDVJU
5IFFOFSHZBMSFBEZTUPSFEJOUIFDBQBDJUPSJTSFMFBTFEUPUIFSFTJTUPST
$POTJEFSBTFSJFTDPNCJOBUJPOPGBSFTJTUPSBOEBOJOJUJBMMZDIBS HFE
A circuit response is the manner in
DBQBDJUPS BTTIP XOJO'JH 5IFSFTJTUPSBOEDBQBDJUPSNBZCFUIF
which the circuit reacts to an
FRVJWBMFOUSFTJTUBODFBOEFRVJ WBMFOUDBQBDJUBODFPGDPNCJOBUJPOTPGSF excitation.
TJTUPSTBOEDBQBDJUPST 0VSPCKFDUJWFJTUPEFUFSNJOFUIFDJSDVJUSFTQPOTF XIJDI GPSQFEBHPHJDSFBTPOT XFBTTVNFUPCFUIFWPMUBHFv U BDSPTTUIF
DBQBDJUPS4JODFUIFDBQBDJUPSJTJOJUJBMMZDIBSHFE XFDBOBTTVNFUIBUBU
UJNFU = UIFJOJUJBMWPMUBHFJT
v = 7
XJUIUIFDPSSFTQPOEJOHWBMVFPGUIFFOFSHZTUPSFEBT
$7
X = @@
"QQMZJOH,$-BUUIFUPQOPEFPGUIFDJSDVJUJO'JHZJFMET
J$ + J3 = By definition, J$ = $Ev∕EUBOEJ3 = v∕35IVT
v = $@@@
Ev+ @@
EU 3
B
PS
Ev+ @@@
v = @@@
C
EU 3$
5IJTJTBfirTUPSEFSEJGGFSFOUJBMFRVBUJPO since only the first deriWBUJWFPG
vJTJOWPMWFE5PTPMWFJU XFSFBSSBOHFUIFUFSNTBT
Ev= − @@@
EU
@@@
v
3$
*OUFHSBUJOHCPUITJEFT XFHFU
MOv = −@@@
U + MO"
3$
XIFSFMO"JTUIFJOUFHSBUJPODPOTUBOU5IVT
U MO@@
v = − @@@
3$
"
5BLJOHQPXFSTPGFQSPEVDFT
v U = "F−U∕3$
#VUGSPNUIFJOJUJBMDPOEJUJPOT v = " = 7)FODF
v U = 7F−U∕3$
5IJTTIPXTUIBUUIFWPMUBHFSFTQPOTFPGUIF3$DJSDVJUJTBOFYQPOFOUJBMEF
DBZPGUIFJOJUJBMWPMUBHF4JODFUIFSFTQPOTFJTEVFUPUIFJOJUJBMFOFSHZTUPSFE
BOEUIFQIZTJDBMDIBSBDUFSJTUJDTPGUIFDJSDVJUBOEOPUEVFUPTPNFF YUFSOBM
WPMUBHFPSDVSSFOUTPVSDF JUJTDBMMFEUIFOBUVSBMSFTQPOTFPGUIFDJSDVJU
254
Chapter 7
First-Order Circuits
The natural response of a circuit refers to the behavior (in terms of voltages
and currents) of the circuit itself, with no external sources of excitation.
The natural response depends on the
nature of the circuit alone, with no
external sources. In fact, the circuit has
a response only because of the energy
initially stored in the capacitor.
v
5IFOBUVSBMSFTQPOTFJTJMMVTUSBUFEHSBQIJDBMMZJO'JH/PUFUIBUBU
U = XFIBWFUIFDPSSFDUJOJUJBMDPOEJUJPOBTJO&R "TUJODSFBTFT UIFWPMUBHFEFDSFBTFTUPXBSE[FSP5IFSBQJEJUZXJUIXIJDIUIFW PMUBHF
EFDSFBTFTJTF YQSFTTFEJOUFSNTPGUIF UJNFDPOTUBOU EFOPUFECZ τ UIF
MPXFSDBTF(SFFLMFUUFSUBV
The time constant of a circuit is the time required for the response to
decay to a factor of 1∕e or 36.8 percent of its initial value.1
V0
V0 e‒t ⁄𝜏
0.368V0
5IJTJNQMJFTUIBUBUU= τ &R CFDPNFT
𝜏
0
t
Figure 7.2
7F−τ∕3$ = 7F− = 7
PS
5IFWPMUBHFSFTQPOTFPGUIF3$DJSDVJU
τ = 3$
*OUFSNTPGUIFUJNFDPOTUBOU &R DBOCFXSJUUFOBT
Values of v (t)∕V0 = e−t∕τ.
U
W U ∕7
τ
τ
τ
τ
τ
v
V0
1.0
0.75
Tangent at t = 0
0.50
v U = 7F−U∕τ
TABLE 7.1
8JUIBDBMDVMBUPSJUJTFBTZUPTIP XUIBUUIFW BMVFPG v U ∕7JTBT TIPXOJO5BCMF*UJTFWJEFOUGSPN5BCMFUIBUUIFWPMUBHFv U JTMFTT
UIBOQFSDFOUPG7BGUFSτ (fivFUJNFDPOTUBOUT 5IVT JUJTDVTUPNBSZUP
BTTVNFUIBUUIFDBQBDJUPSJTGVMMZEJTDIBS HFE PSDIBSged) after fivFUJNF
DPOTUBOUT*OPUIFSXPSET JUUBLFTτ for the circuit to reach its final state
PSTUFBEZTUBUFXIFOOPDIBOHFTUBLFQMBDFXJUIUJNF/PUJDFUIBUGPSFWFSZ
UJNFJOUFSWBMPG τ UIFW oltage is reduced by QFSDFOUPGJUTQSF WJPVT
WBMVF v U + τ = v U ∕F = v U SFHBSEMFTTPGUIFWBMVFPGU
0CTFSWFGSPN&R UIBUUIFTNBMMFSUIFUJNFDPOTUBOU UIFNPSF
SBQJEMZUIFW PMUBHFEFDSFBTFT UIBUJT UIFG BTUFSUIFSFTQPOTF 5IJTJT
JMMVTUSBUFEJO'JH "DJSDVJUXJUIBTNBMMUJNFDPOTUBOUHJ WFTBG BTU
response in that it reaches the steady state (or final state) quickly due to
RVJDLEJTTJQBUJPOPGFOFS HZTUPSFE XIFSFBTBDJSDVJUXJUIBMBS HFUJNF
DPOTUBOUHJWFTBTMP XSFTQPOTFCFDBVTFJUUBL FTMPOHFSUPSFBDITUFBEZ
TUBUF"UBOZSBUF XIFUIFSUIFUJNFDPOTUBOUJTTNBMMPSMBSHF UIFDJSDVJU
reaches steady state in fivFUJNFDPOTUBOUT
8JUIUIFWPMUBHFv U) in Eq. (7.9), we can find the current J3 U
7
vU
J3 U = @@@ = @@@F−U∕τ
3
3
5IFUJNFDPOTUBOUNBZCFWJFXFEGSPNBOPUIFSQFSTQFDUJWF&WBMVBUJOHUIFEFSJWBUJWFPG
v U JO&R BUU = XFPCUBJO
0.37
|
EU ( 7) U=
0.25
0
𝜏
Figure 7.3
2𝜏
3𝜏
4𝜏
5𝜏
(SBQIJDBMEFUFSNJOBUJPOPGUIFUJNF
DPOTUBOUτGSPNUIFSFTQPOTFDVSWF
t (s)
|
F−U∕τ = − @@
@@
E @@@
v = −@@
τ
U=
τ
5IVT UIFUJNFDPOTUBOUJTUIFJOJUJBMSBUFPGEFDBZ PSUIFUJNFUBLFOGPSv∕7UPEFDBZGSPN
VOJUZUP[FSP BTTVNJOHBDPOTUBOUSBUFPGEFDBZ5IJTJOJUJBMTMPQFJOUFSQSFUBUJPOPGUIFUJNF
DPOTUBOUJTPGUFOVTFEJOUIFMBCPSBUPSZUP find τHSBQIJDBMMZGSPNUIFSFTQPOTF DVSWFEJT
played on an oscilloscope. To find τGSPNUIFSFTQPOTFDVSWF ESBXUIFUBOHFOUUPUIFDVSWF
BUU = BTTIPXOJO'JH5IFUBOHFOUJOUFSDFQUTXJUIUIFUJNFBYJTBUU = τ
7.2
The Source-Free RC Circuit
255
v = e‒t ⁄𝜏
V0
1
𝜏=2
𝜏=1
𝜏 = 0.5
0
1
2
3
4
5
t
Figure 7.4
1MPUPGv∕7 = F−U∕τGPSWBSJPVTWBMVFTPGUIFUJNFDPOTUBOU
5IFQPXFSEJTTJQBUFEJOUIFSFTJTUPSJT
7
Q U = vJ3 = @@@F−U∕τ
3
5IFFOFSHZBCTPSCFECZUIFSFTJTUPSVQUPUJNFUJT
t
t 7
X3 U =∫ Q λ Eλ = ∫ @@@F−λ∕τEλ
3
|
U
τ7
$7 − F−U∕τ = − @@@@F−λ∕τ =
@@
3
τ = 3$
/PUJDFUIBUBT U → ∞ w3 ∞ → @$7 XIJDIJTUIFTBNFBT X$ UIFFOFSHZJOJUJBMMZTUPSFEJOUIFDBQBDJUPS5IFFOFSHZUIBUXBTJOJUJBMMZ
TUPSFEJOUIFDBQBDJUPSJTFWFOUVBMMZEJTTJQBUFEJOUIFSFTJTUPS
*OTVNNBSZ
The Key to Working with a Source-Free RC Circuit
Is Finding:
The time constant is the same regardless of what the output is defined
to be.
5IFJOJUJBMWPMUBHFv = 7BDSPTTUIFDBQBDJUPS
5IFUJNFDPOTUBOUτ
8JUIUIFTFUX PJUFNT XFPCUBJOUIFSFTQPOTFBTUIFDBQBDJUPSW PMUBHF
v$ U = v U = v F −U∕τ0ODFUIFDBQBDJUPSWoltage is first obtained, other
WBSJBCMFT DBQBDJUPSDVSSFOUJ$ SFTJTUPSWPMUBHFv3 BOESFTJTUPSDVSSFOUJ3 can be determined. In finding the time constant τ = 3$ 3JTPGUFOUIF
5IFWFOJOFRVJWBMFOUSFTJTUBODFBUUIFUFSNJOBMTPGUIFDBQBDJUPSUIBUJT XFUBLFPVUUIFDBQBDJUPS$ and find 3 = 35IBUJUTUFSNJOBMT
*O'JH MFUv$ = 7'JOEv$ vY BOEJYGPSU>
Solution:
We first need to make the circuit in Fig. 7.5 conform with the standard
3$circuit in Fig. 7.1. We find the equivalent resistance or the Thevenin
When a circuit contains a single
capacitor and several resistors and
dependent sources, the Thevenin
equivalent can be found at the
terminals of the capacitor to form a
simple RC circuit. Also, one can use
Thevenin’s theorem when several
capacitors can be combined to form
a single equivalent capacitor.
Example 7.1
256
Chapter 7
8Ω
5Ω
ix
+
vC
‒
0.1 F
12 Ω
+
vx
‒
First-Order Circuits
resistance at the capacitor terminals. Our objective is always to first ob
UBJODBQBDJUPSWPMUBHFv$'SPNUIJT XFDBOEFUFSNJOFvYBOEJY
5IF ΩBOE ΩSFTJTUPSTJOTFSJFTDBOCFDPNCJOFEUPHJWFB
ΩSFTJTUPS5IJTΩSFTJTUPSJOQBSBMMFMXJUIUIFΩSFTJTUPSDBOCF
DPNCJOFETPUIBUUIFFRVJWBMFOUSFTJTUBODFJT
×
3FR = @@@@@@
= Ω
+ Figure 7.5
'PS&YBNQMF
)FODF UIFFRVJWBMFOUDJSDVJUJTBTTIPXOJO'JH XIJDIJTBOBMPHPVT UP'JH5IFUJNFDPOTUBOUJT
+
v
Req
5IVT
0.1 F
v = v F−U∕τ = F−U∕
7 ‒
v= F−U = F−U7
vY = @@@@@@
+ &RVJWBMFOUDJSDVJUGPSUIFDJSDVJUJO
Fig. 7.5.
'JOBMMZ
Practice Problem 7.1
io
6Ω
v$ = v = F−U
7
'SPN'JH XFDBOVTFWPMUBHFEJWJTJPOUPHFUvYTP
Figure 7.6
12 Ω
τ = 3FR$ = = T
3FGFSUPUIFDJSDVJUJO'JH-FU v$ = 7%FUFSNJOFv$ vY BOE
JPGPSU ≥ 8Ω
+
vx
‒
1
3F
v
JY = @@@Y= F−U"
+
vC
‒
Answer:F−U7 F−U7 −F−U"
Figure 7.7
'PS1SBDUJDF1SPC
Example 7.2
3Ω
20 V +
‒
Figure 7.8
'PS&YBNQMF
t=0
9Ω
5IFTXJUDIJOUIFDJSDVJUJO'JHIBTCFFODMPTFEGPSBMPOHUJNF BOE
JUJTPQFOFEBUU = 'JOEv U GPSU ≥ $BMDVMBUFUIFJOJUJBMFOFSHZTUPSFE
JOUIFDBQBDJUPS
1Ω
+
v
‒
20 mF
Solution:
'PSU< UIFTXJUDIJTDMPTFEUIFDBQBDJUPSJTBOPQFODJSDVJUUPED BT
SFQSFTFOUFEJO'JH B 6TJOHWPMUBHFEJWJTJPO
= 7 v$ U = @@@@@
U<
+
4JODFUIFWPMUBHFBDSPTTBDBQBDJUPSDBOOPUDIBOHFJOTUBOUBOFPVTMZ UIF
WPMUBHFBDSPTTUIFDBQBDJUPSBUU = −JTUIFTBNFBUU = PS
v$ = 7 = 7
7.3
257
The Source-Free RL Circuit
'PSU> UIFTXJUDI JTPQFOFE BOEXFIBWFUIF 3$DJSDVJUTIPXO
JO'JH C </PUJDFUIBUUIF 3$ DJSDVJUJO'JH C JTTPVSDFGSFF
UIFJOEFQFOEFOUTPVSDFJO'JHJTOFFEFEUPQSPWJEF 7PSUIFJOJUJBM
FOFSHZJOUIFDBQBDJUPS>5IFΩBOEΩSFTJTUPSTJOTFSJFTHJWF
3Ω
1Ω
+
20 V +
‒
9Ω
vC (0)
‒
3FR = + = Ω
5IFUJNFDPOTUBOUJT
(a)
1Ω
τ = 3FR$ = × × − = T
5IVT UIFWPMUBHFBDSPTTUIFDBQBDJUPSGPSU≥JT
+
Vo = 15 V
‒
9Ω
v U = v$ F−U∕τ = F−U∕7
PS
20 mF
(b)
Figure 7.9
v U = F−U7
'PS&YBNQMF B U< C U>
5IFJOJUJBMFOFSHZTUPSFEJOUIFDBQBDJUPSJT
$v = @@
××−× = +
X$ = @@
$
*GUIFTXJUDIJO'JHPQFOTBUU = 0, find v U GPSU ≥ BOEX$ Practice Problem 7.2
Answer:F−U7 +
6Ω
24 V +
‒
t=0
+
v
‒
1
6F
12 Ω
Figure 7.10
'PS1SBDUJDF1SPC
The Source-Free RL Circuit
J = *
XJUIUIFDPSSFTQPOEJOHFOFSHZTUPSFEJOUIFJOEVDUPSBT
-* X = @@
"QQMZJOH,7-BSPVOEUIFMPPQJO'JH
v- + v3 = #VUv- = -EJ∕EUBOEv3 = J35IVT
EJ + 3J = -@@
EU
i
L
vL
+
$POTJEFSUIFTFSJFTDPOOFDUJPOPGBSFTJTUPSBOEBOJOEVDUPS BTTIPXOJO
'JH0VSHPBMJTUPEFUFSNJOFUIFDJSDVJUSFTQPOTF XIJDIXFXJMM
BTTVNFUPCFUIFDVSSFOU J U UISPVHIUIFJOEVDUPS 8FTFMFDUUIFJOEVD UPSDVSSFOUBTUIFSFTQPOTFJOPSEFSUPUBLFBEWBOUBHFPGUIFJEFBUIBUUIF
JOEVDUPSDVSSFOUDBOOPUDIBOHFJOTUBOUBOFPVTMZ"UU = XFBTTVNFUIBU
UIFJOEVDUPSIBTBOJOJUJBMDVSSFOU* PS
‒
7.3
Figure 7.11
"TPVSDFGSFF3-DJSDVJU
R
+
vR
‒
4Ω
258
Chapter 7
First-Order Circuits
PS
EJ + @@
3J = @@
EU -
3FBSSBOHJOHUFSNTBOEJOUFHSBUJOHHJWFT
PS
|
iU
JU
U
∫ __
Ei =−∫ @@
3Et
i
*0
0 -
|
U
3U MOJ = − @@
- I
⇒
3U+ MOJ U − MO* = − @@
-
JU
3U
MO@@@
= − @@
*
-
5BLJOHUIFQPXFSTPGF XFIBWF
i(t)
J U = *F−3U∕-
5IJTTIP XTUIBUUIFOBUVSBMSFTQPOTFPGUIF 3- DJSDVJUJTBOF YQP
OFOUJBMEFDBZPGUIFJOJUJBMDVSSFOU 5IFDVSSFOUSFTQPOTFJTTIPXOJO
Fig. *UJTFWJEFOUGSPN&R UIBUUIFUJNFDPOTUBOUGPSUIF 3-DJSDVJUJT
I0
Tangent at t = 0
0.368I0
I0
e‒t ⁄𝜏
-
τ = @@
3 XJUI τBH BJOIB WJOHUIFVOJUPGTFDPOET
XSJUUFOBT
0
𝜏
t
Figure 7.12
5IFDVSSFOUSFTQPOTFPGUIF3-DJSDVJU
The smaller the time constant τ of a
circuit, the faster the rate of decay of
the response. The larger the time
constant, the slower the rate of decay
of the response. At any rate, the
response decays to less than 1 percent
of its initial value (i.e., reaches steady
state) after 5τ
5IVT &R NBZCF −U∕τ
J U = *F 8ith the current in Eq. (7.20), we can find the vPMUBHFBDSPTTUIF
SFTJTUPSBT
v3 U = J3 = *3F−U∕τ
5IFQPXFSEJTTJQBUFEJOUIFSFTJTUPSJT
Q = v3J = * 3F−U∕τ
5IFFOFSHZBCTPSCFECZUIFSFTJTUPSJT
]
U
U
U
X3 U =∫ Q λ Eλ=∫ * 3F−2λ∕τ
Eλ=− @@τ *3F−λ∕τ
0 ,
0
0
-
τ = @@
3
PS
Figure 7.12 shows an initial slope
interpretation may be given to τ
-* − F−U∕τ X3 U = @@
/PUFUIBUBTU → ∞ w3 ∞ → @-* XIJDIJTUIFTBNFBTX- UIFJOJUJBM
FOFSHZTUPSFEJOUIFJOEVDUPSBTJO&R "HBJO UIFFOFSHZJOJUJBMMZ
TUPSFEJOUIFJOEVDUPSJTFWentually dissipated in the resistor
259
The Source-Free RL Circuit
7.3
*OTVNNBSZ
The Key to Working with a Source-Free RL Circuit Is
to Find:
5IFJOJUJBMDVSSFOUJ = *UISPVHIUIFJOEVDUPS
5IFUJNFDPOTUBOUτPGUIFDJSDVJU
8JUIUIFUX PJUFNT XFPCUBJOUIFSFTQPOTFBTUIFJOEVDUPSDVSSFOU 0ODFXFEFUFSNJOFUIFJOEVDUPSDVSSFOUJ- PUIFS
J- U = J U = J F−U∕τ
WBSJBCMFT JOEVDUPSWPMUBHFv- SFTJTUPSWPMUBHFv3 BOESFTJTUPSDVSSFOU
J3 DBOCFPCUBJOFE/PUFUIBUJOHFOFSBM 3JO&R
JTUIF5IFWFOJO
SFTJTUBODFBUUIFUFSNJOBMTPGUIFJOEVDUPS
When a circuit has a single inductor
and several resistors and dependent
sources, the Thevenin equivalent can
be found at the terminals of the
inductor to form a simple RL circuit.
Also, one can use Thevenin’s theorem
when several inductors can be
combined to form a single equivalent
inductor.
Example 7.3
"TTVNJOHUIBU J = " DBMDVMBUF J U BOE JY U JOUIFDJSDVJUPG Fig. 7.13.
Solution:
5IFSFBSFUXPXBZTXFDBOTPMWFUIJTQSPCMFN0OFXBZJTUPPCUBJOUIF
equivalent resistance at the inductor terminals and then use Eq. 5IFPUIFSXBZJTUPTUBSUGSPNTDSBUDICZVTJOH,JSDIIPGGTWPMUBHFMBX
Whichever approach is taken, it is always better to first obtain the induc
UPSDVSSFOU
■ METHOD 1 5IFFRVJWBMFOUSFTJTUBODFJTUIFTBNFBTUIF
5IFWFOJO
4Ω
i
ix
0.5 H
2Ω
+
‒
Figure 7.13
'PS&YBNQMF
SFTJTUBODFBUUIFJOEVDUPSUFSNJOBMT#FDBVTFPGUIFEFQFOEFOUTPVSDF XF
JOTFSUBWPMUBHFTPVSDFXJUIvP = 7BUUIFJOEVDUPSUFSNJOBMTBC BTJO
'JH B 8FDPVMEBMTPJOTFSUB"DVSSFOUTPVSDFBUUIFUFSNJOBMT "QQMZJOH,7-UPUIFUXPMPPQTSFTVMUTJO
J−J + = J−J = − @@
J ⇒ J = @@
⇒
J−J−J = 4VCTUJUVUJOH&R JOUP&R HJWFT
J = −" io
vo = 1 V +
‒
JP = −J = "
4Ω
a
2Ω
i1
i2
4Ω
+
‒
3i1
0.5 H
i1
2Ω
i2
b
(a)
Figure 7.14
4PMWJOHUIFDJSDVJUJO'JH
(b)
+
‒
3i
3i
260
Chapter 7
First-Order Circuits
)FODF
5IFUJNFDPOTUBOUJT
v 3FR = 35I= @@P= @@
Ω
JP @
T
-= @ = @@
τ = @@@
3FR @ 5IVT UIFDVSSFOUUISPVHIUIFJOEVDUPSJT
J U = J F−U∕τ = F− ∕ U" U>
■ METHOD 2 8FNBZEJSFDUMZBQQMZ,7-UPUIFDJSDVJUBTJO
Fig. 7.14(b). For loop 1,
EJ
@@
@@@+ J −J
EU
=
PS
EJ
@@@
+ J−J = EU
'PSMPPQ
J−J−J = ⇒
J J = @@
4VCTUJUVUJOH&R JOUP&R HJWFT
EJ
J = @@@
+ @@
EU
3FBSSBOHJOHUFSNT
EJ
EU
@@@= − @@
J
4JODFJ = J XFNBZSFQMBDFJXJUIJBOEJOUFHSBUF
|
MOJ
PS
JU
J
= −
|
U
2 U __
3 JU
U
MO@@@@
= − @@
J
5BLJOHUIFQPXFSTPGF, we finally obtain
J U = J F− ∕ U= F− ∕ U" U>
XIJDIJTUIFTBNFBTCZ.FUIPE
5IFWPMUBHFBDSPTTUIFJOEVDUPSJT
F− ∕ U7
v = -@@
EJ = −@@ F− ∕ U= − @@@
( )
EU
7.3
261
The Source-Free RL Circuit
4JODFUIFJOEVDUPSBOEUIFΩSFTJTUPSBSFJOQBSBMMFM
v = −F− ∕ U" JY U = @@
U>
Practice Problem 7.3
'JOEJBOEvYJOUIFDJSDVJUPG'JH-FUJ = "
Answer:F−U
" −F−U
7 U>
1Ω
+ vx ‒
i
2H
2Ω
+
‒
6Ω
2vx
Figure 7.15
'PS1SBDUJDF1SPC
Example 7.4
5IFTXJUDIJOUIFDJSDVJUPG'JHIBTCFFODMPTFEGPSBMPOHUJNF"U
U = UIFTXJUDIJTPQFOFE$BMDVMBUFJ U GPSU>
Solution:
8IFOU< UIFTXJUDIJTDMPTFE BOEUIFJOEVDUPSBDUTBTBTIPSUDJSDVJU
UPED5IFΩSFTJTUPSJTTIPSUDJSDVJUFEUIFSFTVMUJOHDJSDVJUJTTIPXO
JO'JH B 5PHFU JJO'JH B XFDPNCJOFUIF ΩBOEΩ
SFTJTUPSTJOQBSBMMFMUPHFU
2Ω
t=0
i(t)
+ 40 V
‒
@@@@@@
×
= Ω
+ )FODF
4Ω
12 Ω
16 Ω
2H
Figure 7.16
'PS&YBNQMF
= "
J = @@@@@
+
i1
8FPCUBJOJ U GSPNJJO'JH B VTJOHDVSSFOUEJWJTJPO CZXSJUJOH
J U = @@@@@@
J = " + 2Ω
4Ω
i(t)
U<
40 V +
‒
12 Ω
4JODFUIFDVSSFOUUISPVHIBOJOEVDUPSDBOOPUDIBOHFJOTUBOUBOFPVTMZ
(a)
J = J − = "
8IFOU> UIFTXJUDIJTPQFOBOEUIFWPMUBHFTPVSDFJTEJTDPOOFDU
FE8FOPXIBWFUIFTPVSDFGSFF 3-DJSDVJUJO'JH C $PNCJOJOH
UIFSFTJTUPST XFIBWF
5IFUJNFDPOTUBOUJT
5IVT
4Ω
i(t)
12 Ω
3FR = + ║ = Ω
-= @@
T
= @@
τ = @@@
3F R J U = J F −U∕τ = F−U
"
16 Ω
2H
(b)
Figure 7.17
4PMWJOHUIFDJSDVJUPG'JH B GPS
U< C GPSU>
262
Chapter 7
Practice Problem 7.4
For the circuit in Fig. 7.18, find J U GPSU>
t=0
Answer:F−U
" U>
8Ω
12 Ω
24 Ω
6A
First-Order Circuits
5Ω
i(t)
2H
Figure 7.18
'PS1SBDUJDF1SPC
Example 7.5
2Ω
*OUIFDJSDVJUTIPwn in Fig. 7.19, find JP vP BOEJGPSBMMUJNF BTTVNJOH
UIBUUIFTXJUDIXBTPQFOGPSBMPOHUJNF
3Ω
+ v
o
10 V
+
‒
‒
t=0
io
i
6Ω
2H
Figure 7.19
'PS&YBNQMF
2Ω
Solution:
It is better to first find the inductor current JBOEUIFOPCUBJOPUIFSRVBOUJ
UJFTGSPNJU
'PSU< UIFTXJUDIJTPQFO4JODFUIFJOEVDUPSBDUTMJLFBTIPSUDJS
DVJUUPED UIF ΩSFTJTUPSJTTIPSUDJSDVJUFE TPUIBUXFIB WFUIFDJSDVJU
TIPXOJO'JH B )FODF JP = BOE
3Ω
+ v
o
10 V +
‒
‒
io
i
6Ω
= " J U = @@@@@
+
U<
vP U = J U = 7 U<
5IVT J = 'PSU> UIFTXJUDIJTDMPTFE TPUIBUUIFW PMUBHFTPVSDFJTTIPSU DJSDVJUFE8FOPXIBWFBTPVSDFGSFF3-DJSDVJUBTTIPwn in Fig. C "UUIFJOEVDUPSUFSNJOBMT
(a)
35I
= ║ = Ω
3Ω
+ v
o
‒
io
6Ω
i
+
vL
‒
TPUIBUUIFUJNFDPOTUBOUJT
- = T
τ = @@@
35I
2H
)FODF
(b)
J U = J F−U∕τ = F−U" Figure 7.20
5IFDJSDVJUJO'JHGPS B U< (b) U>
U>
#FDBVTFUIFJOEVDUPSJTJOQBSBMMFMXJUIUIFBOEΩSFTJTUPST
EJ = − −F−U = F−U7 vP U = −v- = −- @@
EU
BOE
v F−U" JP U = @@-= −@@
U>
U>
7.4
263
Singularity Functions
5IVT GPSBMMUJNF
io U =
{
"
− @@
F−U
"
U<
U> {
7 −U
F 7
vo U =
{
U<
U>
2
i(t)
U<
"
−U
F " U≥ i U =
8FOPUJDFUIBUUIFJOEVDUPSDVSSFOUJTDPOUJOVPVTBU U = XIJMFUIF
DVSSFOUUISPVHIUIF ΩSFTJTUPSESPQTGSPNUP −∕BUU = BOEUIF
WPMUBHFBDSPTTUIFΩSFTJTUPSESPQTGSPNUPBUU = 8FBMTPOPUJDF
that the time constant is the same regardless of what the output is defined
UPCF'JHVSFQMPUTJBOEJP
‒2
3
t
io(t)
Figure 7.21
"QMPUPGJBOEJP
Practice Problem 7.5
%FUFSNJOFJ JP BOEvPGPSBMMUJOUIFDJSDVJUTIPXOJO'JH"TTVNF
UIBUUIFTXJUDIXBTDMPTFEGPSBMPOHUJNF*UTIPVMECFOPUFEUIBUPQFOJOH
3Ω
a switch in series with an ideal current source creates an infinite voltage
BUUIFDVSSFOU TPVSDFUFSNJOBMT$MFBSMZUIJT JTJNQPTTJCMF 'PSUIF QVS
t=0
QPTFTPGQSPCMFNTPMWJOH XFDBOQMBDFBTIVOUSFTJTUPSJOQBSBMMFMXJUI
i
1H
UIFTPVSDF XIJDIOPXNBLFTJUBWPMUBHFTPVSDFJOTFSJFTXJUIBSFTJTUPS i
o
*ONPSFQSBDUJDBMDJSDVJUT EFWJDFTUIBUBDUMJLFDVSSFOUTPVSDFTBSF GPSUIF
+
NPTUQBSU FMFDUSPOJDDJSDVJUT5IFTFDJSDVJUTXJMMBMMPXUIFTPVSDFUPBDU 24 A
vo
2Ω
4Ω
‒
MJLFBOJEFBMDVSSFOUTPVSDFPWFSJUTPQFSBUJOHSBOHFCVUWPMUBHFMJNJUJU XIFOUIFMPBESFTJTUPSCFDPNFTUPPMBSHF BTJOBOPQFODJSDVJU Figure 7.22
Answer:
i=
{
{
" U<
"
io=
−U
−F−U"
U≥ F "
{
vo=
7.4
'PS1SBDUJDF1SPC
7
F−U7
U<
U>
U<
U>
Singularity Functions
#FGPSFHPJOHPOXJUIUIFTFDPOEIBMGPGUIJTDIBQUFS XFOFFEUPEJHSFTT
BOEDPOTJEFSTPNFNBUIFNBUJDBMDPODFQUTUIBUXJMMBJEPVSVOEFSTUBOE JOHPGUSBOTJFOUBOBMZTJT"CBTJDVOEFSTUBOEJOHPGTJOHVMBSJUZGVODUJPOT
XJMMIFMQVTNBLe sense of the response of first-order circuits to a sudden
BQQMJDBUJPOPGBOJOEFQFOEFOUEDWPMUBHFPSDVSSFOUTPVSDF
4JOHVMBSJUZGVODUJPOT BMTPDBMMFETXJUDIJOHGVODUJPOT BSFWFSZVTF
GVMJODJSDVJUBOBMZTJT5IFZTFSWFBTHPPEBQQSPYJNBUJPOTUPUIFTXJUDI
JOHTJHOBMTUIBUBSJTFJODJSDVJUTXJUITXJUDIJOHPQFSBUJPOT
5IFZBSF
IFMQGVMJOUIFOFBU DPNQBDUEFTDSJQUJPOPGTPNFDJSDVJUQIFOPNFOB FT QFDJBMMZUIFTUFQSFTQPOTFPG3$PS3-DJSDVJUTUPCFEJTDVTTFEJOUIFOFYU
sections. By definition,
Singularity functions are functions that either are discontinuous or have
discontinuous derivatives.
264
Chapter 7
First-Order Circuits
u(t)
5IFUISFFNPTUXJEFMZVTFETJOHVMBSJUZGVODUJPOTJODJSDVJUBOBMZTJT
BSFUIFVOJUTUFQ UIFVOJUJNQVMTF BOEUIFVOJUSBNQGVODUJPOT
1
The unit step function u (t) is 0 for negative values of t and 1 for positive
values of t.
*ONBUIFNBUJDBMUFSNT
0
t
Figure 7.23
5IFVOJUTUFQGVODUJPO
{ U<
U> u U =
The unit step function is undefined at U = XIFSFJUDIBOHFTBCSVQU MZGSPNUP*UJTEJNFOTJPOMFTT MJL FPUIFSNBUIFNBUJDBMGVODUJPOT
TVDIBTTJOFBOEDPTJOF'JHVSFEFQJDUTUIFVOJUTUFQGVODUJPO*GUIF
BCSVQUDIBOHFPDDVSTBUU = U XIFSFU> JOTUFBEPGU = UIFVOJUTUFQ
GVODUJPOCFDPNFT
u(t ‒ t0)
1
0
t0
u(t + t0)
1
0
{
U<U
U>U XIJDIJTUIFTBNFBTTBZJOHUIBUV U JTEFMBZFECZUTFDPOET BTTIPXOJO
'JH B 5PHFU&R GSPN&R XFTJNQMZSFQMBDFFWFSZU
CZU−U*GUIFDIBOHFJTBUU = −U UIFVOJUTUFQGVODUJPOCFDPNFT
(a)
‒t0
{
V U−U =
t
t
(b)
Figure 7.24
B 5IFVOJUTUFQGVODUJPOEFMBZFECZU C UIFVOJUTUFQBEWBODFECZU
V U+U =
v U =
U<−U
U>−U NFBOJOHUIBUV U JTBEWBODFECZUTFDPOET BTTIPXOJO'JH C 8FVTFUIFTUFQGVODUJPOUPSFQSFTFOUBOBCSVQUDIBOHFJOWPMUBHFPS
DVSSFOU MJLFUIFDIBOHFTUIBUPDDVSJOUIFDJSDVJUTPGDPOUSPMTZTUFNTBOE
EJHJUBMDPNQVUFST'PSFYBNQMF UIFWPMUBHF
{
7
U<U
U>U NBZCFFYQSFTTFEJOUFSNTPGUIFVOJUTUFQGVODUJPOBT
v U = 7V U−U Alternatively, we may derive
Eqs. (7.25) and (7.26) from Eq. (7.24)
by writing u [f (t)] = 1, f (t) > 0, where
f (t) may be t − t0 or t + t0.
*GXFMFU U = UIFO v U JTTJNQMZUIFTUFQW PMUBHF 7V U " WPMUBHF
TPVSDFPG7V U JTTIPXOJO'JH B JUTFRVJWBMFOUDJSDVJUJTTIPXO
JO'JH C *UJTF WJEFOUJO'JH C UIBUUFSNJOBMT BCBSFTIPSU
DJSDVJUFE v = GPS U< BOEUIBU v = 7BQQFBSTBUUIFUFSNJOBMTGPS
t=0
a
V0 u(t) +
‒
=
a
V0
+
‒
b
b
(a)
Figure 7.25
(b)
B 7PMUBHFTPVSDFPG7V U C JUTFRVJWBMFOUDJSDVJU
7.4
265
Singularity Functions
U>4JNJMBSMZ BDVSSFOUTPVSDFPG*V U JTTIPXOJO'JH B XIJMF
JUTFRVJWBMFOUDJSDVJUJTJO'JH C /PUJDFUIBUGPS U< UIFSFJTBO
PQFODJSDVJU J = BOEUIBUJ = * floXTGPSU>
t=0
a
=
I0u(t)
i
a
I0
b
b
(b)
(a)
Figure 7.26
B $VSSFOUTPVSDFPG*V U C JUTFRVJWBMFOUDJSDVJU
5IFEFSJWBUJWFPGUIFVOJUTUFQGVODUJPOV U JTUIFVOJUJNQVMTFGVOD
UJPOδ U XIJDIXFXSJUFBT
E
EU {
δ U =@@
V U = Undefined, U<
U= U>
5IFVOJUJNQVMTFGVODUJPO‡BMTPLOPXOBTUIFEFMUBGVODUJPO‡JTTIPXO
JO'JH
δ(t)
(∞)
0
t
Figure 7.27
5IFVOJUJNQVMTFGVODUJPO
The unit impulse function δ(t) is zero everywhere except at t = 0, where
it is undefined.
*NQVMTJWFDVSSFOUTBOEW PMUBHFTPDDVSJOFMFDUSJDDJSDVJUTBTBSFTVMU
PGTXJUDIJOHPQFSBUJPOTPSJNQVMTJ WFTPVSDFT "MUIPVHIUIFVOJUJN QVMTFGVODUJPOJTOPUQI ZTJDBMMZSFBMJ[BCMF KVTUMJLFJEFBMTPVSDFT JEFBM
SFTJTUPST FUD JUJTBWFSZVTFGVMNBUIFNBUJDBMUPPM
5IFVOJUJNQVMTFNBZCFSFHBSEFEBTBOBQQMJFEPSSFTVMUJOHTIPDL
*UNBZCFWJTVBMJ[FEBTBWFSZTIPSUEVSBUJPOQVMTFPGVOJUBSFB5IJTNBZ
CFFYQSFTTFENBUIFNBUJDBMMZBT
∫ δ t Et = −
XIFSFU = −EFOPUFTUIFUJNFKVTUCFGPSFU = BOEU = + JTUIFUJNFKVTU
BGUFSU = 'PSUIJTSFBTPO JUJTDVTUPNBSZUPXSJUF EFOPUJOHVOJUBSFB CFTJEFUIFBSSPXUIBUJTVTFEUPTZNCPMJ[FUIFVOJUJNQVMTFGVODUJPO BT
JO'JH5IFVOJUBSFBJTLOPXOBTUIFTUSFOHUIPGUIFJNQVMTFGVOD
UJPO8IFOBOJNQVMTFGVODUJPOIBTBTUSFOHUIPUIFSUIBOVOJUZ UIFBSFB
PGUIFJNQVMTFJTFRVBMUPJUTTUSFOHUI' PSFYBNQMF BOJNQVMTFGVOD UJPOδ U IBTBOBSFBPG'JHVSFTIP XTUIFJNQVMTFGVODUJPOT
δ U + δ U BOE−δ U− 5PJMMVTUSBUFIPXUIFJNQVMTFGVODUJPOBGGFDUTPUIFSGVODUJPOT MFUVT
FWBMVBUFUIFJOUFHSBM
C
∫ f U δ U − U0 Et
B
10δ(t)
5δ(t + 2)
–2
–1
0
1
2
3
‒4δ(t ‒ 3)
Figure 7.28
5ISFFJNQVMTFGVODUJPOT
t
266
Chapter 7
First-Order Circuits
XIFSFB<U<C4JODFδ U−U = FYDFQUBUU = U UIFJOUFHSBOEJT[FSP
FYDFQUBUU5IVT
C
C
a
a
∫ G t δ U−U EU=∫ G U δ U−U EU
b
=G U ∫ δ U−U EU= G U a
PS
b
∫a G U δ U−U EU= G U 5IJTTIPXTUIBUXIFOBGVODUJPOJTJOUFHSBUFEXJUIUIFJNQVMTFGVODUJPO XFPCUBJOUIFWBMVFPGUIFGVODUJPOBUUIFQPJOUXIFSFUIFJNQVMTFPDDVST
5IJTJTBIJHIMZVTFGVMQSPQFSUZPGUIFJNQVMTFGVODUJPOLOP XOBTUIF
TBNQMJOHPSTJGUJOHQSPQFSUZ5IFTQFDJBMDBTFPG&R JTGPS U = 5IFO&R CFDPNFT
r(t)
0+
1
∫ − G U δ U EU= G 0
*OUFHSBUJOHUIFVOJUTUFQGVODUJPO V U SFTVMUTJOUIF VOJUSBNQGVOD
UJPOS U XFXSJUF
U
0
1
t
5IFVOJUSBNQGVODUJPO
{ U U≤
U≥ −∞
PS
Figure 7.29
S U =∫ V λ Eλ= UV(U)
r U =
r (t ‒ t0)
1
The unit ramp function is zero for negative values of t and has a unit
slope for positive values of t.
'JHVSFTIPXTUIFVOJUSBNQGVODUJPO*OHFOFSBM BSBNQJTBGVOD UJPOUIBUDIBOHFTBUBDPOTUBOUSBUF
5IFVOJUSBNQGVODUJPONBZCFEFMBZFEPSBEW BODFEBTTIP XOJO
'JH'PSUIFEFMBZFEVOJUSBNQGVODUJPO
t0 + 1 t
0 t0
(a)
r(t + t0)
{
S U−U =
1
‒t0 + 1 0
t
(b)
Figure 7.30
U≤U
U≥U BOEGPSUIFBEWBODFEVOJUSBNQGVODUJPO
‒t0
U−U
5IFVOJUSBNQGVODUJPO B EFMBZFECZU C BEWBODFECZU
{
U≤−U
S U+U = U≥−U U+U
8FTIPVMELFFQJONJOEUIBUUIFUISFFTJOHVMBSJUZGVODUJPOTJNQVMTF TUFQ BOESBNQ BSFSFMBUFECZEJGGFSFOUJBUJPOBT
EV U
δ U = @@@@@
EU
ES U
V U = @@@@
EU
7.4
Singularity Functions
V U =∫ δ λ Eλ S U =∫ u λ Eλ
267
PSCZJOUFHSBUJPOBT
U
U
−∞
−∞
"MUIPVHIUIFSFBSFNBO ZNPSFTJOHVMBSJUZGVODUJPOT XFBSFPOMZJOUFS FTUFEJOUIFTFUISFF UIFJNQVMTFGVODUJPO UIFVOJUTUFQGVODUJPO BOEUIF
SBNQGVODUJPO BUUIJTQPJOU
Example 7.6
&YQSFTTUIFWPMUBHFQVMTFJO'JHJOUFSNTPGUIFVOJUTUFQ$BMDVMBUF
JUTEFSJWBUJWFBOETLFUDIJU
Solution:
5IFUZQFPGQVMTFJO'JHJTDBMMFEUIF HBUFGVODUJPO *UNBZCF SFHBSEFEBTBTUFQGVODUJPOUIBUTXJUDIFTPOBUPOFWBMVFPG
UBOE
TXJUDIFTPGGBUBOPUIFSWBMVFPGU5IFHBUFGVODUJPOTIPXOJO'JH
TXJUDIFTPOBUU = TBOETXJUDIFTPGGBUU = T*UDPOTJTUTPGUIFTVN
of two unit step functions as shown in Fig. 7.32(a). From the figure, it
JTFWJEFOUUIBU
Gate functions are used along
with switches to pass or block
another signal.
v (t)
10
v U = V U− −V U− = <V U− −V U− >
5BLJOHUIFEFSJWBUJWFPGUIJTHJWFT
@@@
Ev= <δ U− −δ U− >
EU
0
1
2
3
4
Figure 7.31
'PS&YBNQMF
XIJDIJTTIPXOJO'JH C 8FDBOPCUBJO'JH C EJSFDUMZGSPN
'JHCZTJNQMZPCTFSWJOHUIBUUIFSFJTBTVEEFOJODSFBTFCZ7BU
U = TMFBEJOHUP δ U− "U U = T UIFSFJTBTVEEFOEFDSFBTFCZ
7MFBEJOHUP−7δ U− 10u(t ‒ 2)
‒10u(t ‒ 5)
+
10
0
1
2
10
0
t
1
2
3
4
5
‒10
(a)
dv
dt
10
0
1
2
3
4
5
t
‒10
(b)
Figure 7.32
B %FDPNQPTJUJPOPGUIFQVMTFJO'JH C EFSJWBUJWFPGUIFQVMTFJO'JH
t
5
t
268
Chapter 7
Practice Problem 7.6
First-Order Circuits
&YQSFTTUIFDVSSFOUQVMTFJO'JHJOUFSNTPGUIFVOJUTUFQ'JOEJUT
JOUFHSBMBOETLFUDIJU
Answer: <V U − V U − + V U − > " < S U − S U − +
S U− >BNQTFD4FF'JH
i(t)
∫ i dt
10
20
0
2
4
t
‒10
0
Figure 7.33
4
t
Figure 7.34
'PS1SBDUJDF1SPC
Example 7.7
2
*OUFHSBMPGJ U JO'JH
&YQSFTTUIFTBXUPPUIGVODUJPOTIPXOJO'JHJOUFSNTPGTJOHVMBSJUZ
GVODUJPOT
Solution:
There are three ways of solving this problem. The first method is by mere
PCTFSWBUJPOPGUIFHJWFOGVODUJPO XIJMFUIFPUIFSNFUIPETJOWPMWFTPNF
HSBQIJDBMNBOJQVMBUJPOTPGUIFGVODUJPO
v (t)
10
■ METHOD 1
0
#ZMPPLJOHBUUIFTLFUDIPGv U JO'JH JUJTOPU
IBSEUPOPUJDFUIBUUIFHJWFOGVODUJPOv U JTBDPNCJOBUJPOPGTJOHVMBSJUZ
GVODUJPOT4PXFMFU
t
2
v U = v U + v U + ⋯
Figure 7.35
'PS&YBNQMF
5IFGVODUJPOv U JTUIFSBNQGVODUJPOPGTMPQF TIPXOJO'JH B UIBUJT
v U = S U v1 (t)
v1 + v2
10
10
0
2
t
+
v2(t)
0
2
t
=
0
2
‒10
(a)
Figure 7.36
1BSUJBMEFDPNQPTJUJPOPGv U JO'JH
(b)
(c)
t
7.4
269
Singularity Functions
4JODFv U) goes to infinity, we need another function at U = 2s in order UP
HFUv U 8FMFUUIJTGVODUJPOCFv XIJDIJTBSBNQGVODUJPOPGTMPQF− BTTIPXOJO'JH C UIBUJT
v U = −S U− "EEJOHvBOE vHJWFTVTUIFTJHOBMJO'JH D 0CWJPVTMZ UIJTJT
OPUUIFTBNFBTv U JO'JH#VUUIFEJGGFSFODF JTTJNQMZBDPOTUBOU
VOJUTGPSU>T#ZBEEJOHBUIJSETJHOBMv XIFSF
v = −V U− XFHFU v U BTTIPXOJO'JH4VCTUJUVUJOH&RT UISPVHI
JOUP&R HJWFT
v U = S U −S U− −V U−
v1 + v2
+
10
0
2
=
v3 (t)
t
0
2
t
v(t)
10
2
0
‒10
(a)
(b)
Figure 7.37
$PNQMFUFEFDPNQPTJUJPOPGv U JO'JH
■ METHOD 2 "DMPTFPCTFSWBUJPOPG'JHSFWFBMTUIBU v U JT
BNVMUJQMJDBUJPOPGUXPGVODUJPOTBSBNQGVODUJPOBOEBHBUFGVODUJPO
5IVT
v U = U<V U −V U− >
= UV U − UV U−
= S U − U− + V U− = S U − U− V U− −V U−
= S U −S U− − V U−
UIFTBNFBTCFGPSF
■ METHOD 3
5IJTNFUIPEJTTJNJMBSUP.FUIPE8FPCTFSWFGSPN
'JHUIBU v U JTBNVMUJQMJDBUJPOPGBSBNQGVODUJPOBOEBVOJUTUFQ
GVODUJPO BTTIPXOJO'JH5IVT
v U = S U V −U + *GXFSFQMBDFV −U CZ−V U UIFOXFDBOSFQMBDFV −U + CZ−V U− )FODF
v U = S U <−V U− >
which can be simplified as in Method 2 to get the same result.
(c)
t
270
Chapter 7
First-Order Circuits
5r(t)
10
u(‒t + 2)
×
0
2
1
0
t
2
t
Figure 7.38
%FDPNQPTJUJPOPGv U JO'JH
Practice Problem 7.7
3FGFSUP'JH&YQSFTTJ U JOUFSNTPGTJOHVMBSJUZGVODUJPOT
Answer:V U −S U + S U− −S U− "
i(t) (A)
2
0
1
2
3
t (s)
‒2
Figure 7.39
'PS1SBDUJDF1SPC
Example 7.8
(JWFOUIFTJHOBM
{
H U = −
U− 4
U<
<U<
U>
FYQSFTTH U JOUFSNTPGTUFQBOESBNQGVODUJPOT
Solution:
5IFTJHOBMH U) may be regarded as the sum of three functions specified
XJUIJOUIFUISFFJOUFSWBMTU< <U< BOEU>
'PS U < H U NBZCFSFHBSEFEBTNVMUJQMJFECZ V −U XIFSF
V −U = GPSU<BOEGPS U>8JUIJOUIFUJNFJOUFSWBM <U< UIFGVODUJPONBZCFDPOTJEFSFEBT − NVMUJQMJFE CZ B HBUFE GVODUJPO
<V U −V U− >'PSU> UIFGVODUJPONBZCFSFHBSEFEBTU−NVM
UJQMJFECZUIFVOJUTUFQGVODUJPOV U− 5IVT
H U = V −U −<V U −V U− > + U− V U−
= V −U −V U + U− + V U−
= V −U −V U + U− V U−
= V −U −V U + S U−
0OFNBZBWPJEUIFUSPVCMFPGVTJOH V −U CZSFQMBDJOHJUXJUI −V U 5IFO
H U = <−V U >−V U + S U− = −V U + S U−
"MUFSOBUJWFMZ XFNBZQMPUH U BOEBQQMZ.FUIPEGSPN&YBNQMF
7.5
{
*G
h U = −
U−
Step Response of an RC Circuit
U<
<U< <U< U>
271
Practice Problem 7.8
FYQSFTTI U JOUFSNTPGUIFTJOHVMBSJUZGVODUJPOT
Answer:−V U + V U− + S U− −V U− −S U− &WBMVBUFUIFGPMMPXJOHJOUFHSBMTJOWPMWJOHUIFJNQVMTFGVODUJPO
Example 7.9
10
∫ 0 U+U− δ U− EU
∞
∫ <δ U− F −U
DPTU+ δ U+ F−U
TJOU>EU
−∞
Solution:
For the first integral, we apply the sifting property in Eq. (7.32).
10
∫ U+U− δ U− EU= U+U− ∣U= = + − = 0
4JNJMBSMZ GPSUIFTFDPOEJOUFHSBM
∞
∫ <δ U− F −U
DPTU+ δ U+ F−U
TJOU>EU
−∞
= F−UDPTU∣U= + F−UTJOU∣U= −
= F−DPT
+ FTJO − = − = −
Practice Problem 7.9
&WBMVBUFUIFGPMMPXJOHJOUFHSBMT
∞
10
∫−∞ U+U+ δ U+ EU ∫ 0 δ U− π DPTUEU
Answer: −
7.5
Step Response of an RC Circuit
8IFOUIFEDTPVSDFPGBO 3$DJSDVJUJTTVEEFOMZBQQMJFE UIFW PMUBHFPS
DVSSFOUTPVSDF DBOCFNPEFMFEBTBTUFQGVODUJPO BOEUIFSFTQPOTFJT
LOPXOBTBTUFQSFTQPOTF
The step response of a circuit is its behavior when the excitation is the
step function, which may be a voltage or a current source.
272
Chapter 7
t=0
(a)
5IFTUFQSFTQPOTFJTUIFSFTQPOTFPGUIFDJSDVJUEVFUPBTVEEFOBQQMJDB UJPOPGBEDWPMUBHFPSDVSSFOUTPVSDF
$POTJEFSUIF3$DJSDVJUJO'JH B XIJDIDBOCFSFQMBDFECZUIF
DJSDVJUJO'JH C XIFSF 7TJTBDPOTUBOUEDW PMUBHFTPVSDF"HBJO XFTFMFDUUIFDBQBDJUPSWPMUBHFBTUIFDJSDVJUSFTQPOTFUPCFEFUFSNJOFE
8FBTTVNFBOJOJUJBMW PMUBHF 7POUIFDBQBDJUPS BMUIPVHIUIJTJTOPU
OFDFTTBSZGPSUIFTUFQSFTQPOTF4JODFUIFW PMUBHFPGBDBQBDJUPSDBOOPU
DIBOHFJOTUBOUBOFPVTMZ
R
R
Vs
+
‒
C
Vs u(t) +
‒
C
+
v
‒
+
v
‒
(b)
Figure 7.40
First-Order Circuits
"O3$DJSDVJUXJUIWPMUBHFTUFQJOQVU
v − = v + = 7
XIFSFv − JTUIFWPMUBHFBDSPTTUIFDBQBDJUPSKVTUCFGPSFTXJUDIJOHBOE
v + JTJUTWPMUBHFJNNFEJBUFMZBGUFSTXJUDIJOH"QQMZJOH,$- XFIBWF
v −7TV U
$@@@
Ev+ @@@@@@@@@
= 3
EU
PS
7T
v = @@@
@@@
V U Ev+ @@@
EU
3$
3$
XIFSFWJTUIFWPMUBHFBDSPTTUIFDBQBDJUPS'PSU> &R CFDPNFT
7T
v = @@@
@@@
Ev+ @@@
EU 3$ 3$
3FBSSBOHJOHUFSNTHJWFT
v −7T
Ev= − @@@@@@
@@@
3$
EU
PS
EU
@@@@@@
Ev = −@@@
v −7T
3$
*OUFHSBUJOHCPUITJEFTBOEJOUSPEVDJOHUIFJOJUJBMDPOEJUJPOT
|
vU
|
U
U MO v −7T = − @@@
3$ 7
U + MO v U −7T −MO 7−7T = − @@@
3$
PS
v−7T
U MO@@@@@@
= − @@@
7−7T
3$
5BLJOHUIFFYQPOFOUJBMPGCPUITJEFT
v −7T
@@@@@@
τ = 3$
= F−U∕τ 7−7T
v−7T = 7−7T F−U∕τ
PS
v U = 7T + 7−7T F−U∕τ
U>
5IVT
{
v U =
7
−Uτ 7T + 7−7T F
U<
U>
7.5
273
Step Response of an RC Circuit
v(t)
5IJTJTLOPXOBTUIFDPNQMFUFSFTQPOTF PSUPUBMSFTQPOTF PGUIF3$DJS
DVJUUPBTVEEFOBQQMJDBUJPOPGBEDWPMUBHFTPVSDF BTTVNJOHUIFDBQBDJ
UPSJTJOJUJBMMZDIBSHFE5IFSFBTPOGPSUIFUFSNiDPNQMFUFuXJMMCFDPNF
FWJEFOUBMJUUMFMBUFS "TTVNJOHUIBU 7T >7 BQMPUPG v U JTTIP XOJO
'JH
*GXFBTTVNFUIBUUIFDBQBDJUPSJTVODIBSHFEJOJUJBMMZ XFTFU7 = JO&R TPUIBU
U<
v U =
−Uτ 7T −F U>
Vs
{
0
t
Figure 7.41
XIJDIDBOCFXSJUUFOBMUFSOBUJWFMZBT
V0
v U = 7T −F−U∕τ
VU
3FTQPOTFPGBO3$DJSDVJUXJUIJOJUJBMMZ
DIBSHFEDBQBDJUPS
5IJTJTUIFDPNQMFUFTUFQSFTQPOTFPGUIF 3$DJSDVJUXIFOUIFDBQBDJUPS
JTJOJUJBMMZVODIBSHFE5IFDVSSFOUUISPVHIUIFDBQBDJUPSJTPCUBJOFEGSPN
&R VTJOHJ U = $Ev ∕ dt8FHFU
$7 F−U∕τ
J U = $@@@
Ev= @@
T EU τ
τ = 3$ U>
PS
7
J U = @@TF−U∕τ V U 3
'JHVSFTIPXTUIFQMPUTPGDBQBDJUPSW PMUBHFv U BOEDBQBDJUPSDVS SFOUJ U 3BUIFSUIBOHPJOHUISPVHIUIFEFSJWBUJPOTBCPWF UIFSFJTBTZTUFNBU
JDBQQSPBDI‡PSSBUIFS, a shortcut method—for finding the step response
PGBO3$PS3-DJSDVJU-FUVTSFFYBNJOF&R XIJDIJTNPSFHFOFS
BMUIBO&R *UJTFWJEFOUUIBUv U IBTUXPDPNQPOFOUT$MBTTJDBMMZ
UIFSFBSFUXPXBZTPGEFDPNQPTJOHUIJTJOUPUX PDPNQPOFOUTThe first
JTUPCSFBLJUJOUPBiOBUVSBMSFTQPOTFBOEBGPSDFESFTQPOTFBOEUIFTFD
POEJTUPCSFBLJUJOUPBiUSBOTJFOUSFTQPOTFBOEBTUFBEZTUBUFSFTQPOTF
4UBSUJOHXJUIUIFOBUVSBMSFTQPOTFBOEGPSDFESFTQPOTF XFXSJUFUIFUPUBM
PSDPNQMFUFSFTQPOTFBT
$PNQMFUFSFTQPOTF = OBUVSBMSFTQPOTF+ GPSDFESFTQPOTF
TUPSFEFOFSHZ
JOEFQFOEFOUTPVSDF
v(t)
Vs
0
t
(a)
i(t)
Vs
R
PS
v = vO + vG
XIFSF
vO = 7PF−U∕τ
0
t
(b)
BOE
vG= 7T −F−U∕τ
8FBSFGBNJMJBSXJUIUIFOBUVSBMSFTQPOTF vOPGUIFDJSDVJU BTEJTDVTTFE
JO4FDUJPOvGJTLOPXOBTUIFGPSDFESFTQPOTFCFDBVTFJUJTQSPEVDFE
CZUIFDJSDVJUXIFOBOFYUFSOBMiGPSDF BWPMUBHFTPVSDFJOUIJTDBTF JT
BQQMJFE*USFQSFTFOUTXIBUUIFDJSDVJUJTGPSDFEUPEPCZUIFJOQVUFYDJUB
UJPO5IFOBUVSBMSFTQPOTFF WFOUVBMMZEJFTPVUBMPOHXJUIUIFUSBOTJFOU
DPNQPOFOUPGUIFGPSDFESFTQPOTF MFBWJOHPOMZUIFTUFBEZTUBUFDPNQP
OFOUPGUIFGPSDFESFTQPOTF
Figure 7.42
4UFQSFTQPOTFPGBO3$DJSDVJUXJUI
JOJUJBMMZVODIBSHFEDBQBDJUPS B WPMUBHF
SFTQPOTF C DVSSFOUSFTQPOTF
274
Chapter 7
First-Order Circuits
"OPUIFSXBZPGMPPLJOHBUUIFDPNQMFUFSFTQPOTFJTUPCSFBLJOUP
UXPDPNQPOFOUT‡POFUFNQPSBSZBOEUIFPUIFSQFSNBOFOU UIBUJT
$PNQMFUFSFTQPOTF=USBOTJFOUSFTQPOTF+TUFBEZTUBUFSFTQPOTF
UFNQPSBSZQBSU
QFSNBOFOUQBSU
PS
v = vU + vTT
vU = 7P−7T F−U∕τ
B
vTT = 7T
C
XIFSF
BOE
5IFUSBOTJFOUSFTQPOTFvUJTUFNQPSBSZJUJTUIFQPSUJPOPGUIFDPNQMFUF
response that decays to zero as time approaches infinity5IVT
The transient response is the circuit’s temporary response that will die
out with time.
5IFTUFBEZTUBUFSFTQPOTFvTTJTUIFQPSUJPOPGUIFDPNQMFUFSFTQPOTF
UIBUSFNBJOTBGUFSUIFUSBOTJFOUSFQPOTFIBTEJFEPVU5IVT
The steady-state response is the behavior of the circuit a long time after
an external excitation is applied.
This is the same as saying that the
complete response is the sum of the
transient response and the steady-state
response.
The first decomposition of the complete response is in terms of the
TPVSDFPGUIFSFTQPOTFT XIJMFUIFTFDPOEEFDPNQPTJUJPOJTJOUFSNTPG
UIFQFSNBOFODZPGUIFSFTQPOTFT6OEFSDFSUBJODPOEJUJPOT UIFOBUVSBM
SFTQPOTFBOEUSBOTJFOUSFTQPOTFBSFUIFTBNF5IFTBNFDBOCFTBJEBCPVU
UIFGPSDFESFTQPOTFBOETUFBEZTUBUFSFTQPOTF
8IJDIFWFSXBZXFMPPLBUJU UIFDPNQMFUFSFTQPOTFJO&R NBZCFXSJUUFOBT
v U = v ∞ + <v −v ∞ >F−U∕τ
XIFSFv JTUIFJOJUJBMWPMUBHFBUU = + BOEv ∞) is the final or steadyTUBUFWBMVFThus, to find the step response of an3$DJSDVJUSFRVJSFTUISFF
UIJOHT
Once we know x (0), x (∞), and τ,
almost all the circuit problems in this
chapter can be solved using the
formula
Y U = Y ∞ + <Y −Y ∞ >F−U∕τ
5IFJOJUJBMDBQBDJUPSWPMUBHFv The final capacitor voltage v ∞ 5IFUJNFDPOTUBOUτ
8FPCUBJOJUFNGSPNUIFHJWFODJSDVJUGPSU<BOEJUFNTBOEGSPN
UIFDJSDVJUGPS U >0ODFUIFTFJUFNTBSFEFUFSNJOFE XFPCUBJOUIF
7.5
Step Response of an RC Circuit
275
SFTQPOTFVTJOH&R 5IJTUFDIOJRVFFRVBMMZBQQMJFTUP3-DJSDVJUT BTXFTIBMMTFFJOUIFOFYUTFDUJPO
/PUFUIBUJGUIFTXJUDIDIBOHFTQPTJUJPOBUUJNFU = UJOTUFBEPGBU
U = UIFSFJTBUJNFEFMBZJOUIFSFTQPOTFTPUIBU&R CFDPNFT
v U = v ∞ + <v U −v ∞ >F− U−U ∕τ
XIFSFv U JTUIFJOJUJBMWBMVFBUU = U ,FFQJONJOEUIBU&R PS
BQQMJFTPOMZUPTUFQSFTQPOTFT UIBUJT XIFOUIFJOQVUF YDJUBUJPO
JTDPOTUBOU
5IFTXJUDIJO'JHIBTCFFOJOQPTJUJPO"GPSBMPOHUJNF"UU = UIFTXJUDINPWFTUP#%FUFSNJOFv U GPSU>BOEDBMDVMBUFJUTWBMVFBU
U = BOET
3 kΩ
A
B
4 kΩ
t=0
24 V +
‒
5 kΩ
+
v
‒
0.5 mF
+ 30 V
‒
Figure 7.43
'PS&YBNQMF
Solution:
'PSU< UIFTXJUDIJTBUQPTJUJPO "5IFDBQBDJUPS BDUTMJLFBOPQFO
DJSDVJUUPED CVU WJTUIFTBNFBTUIFWPMUBHFBDSPTTUIFL ΩSFTJTUPS
)FODF UIFWPMUBHFBDSPTTUIFDBQBDJUPSKVTUCFGPSF U = JTPCUBJOFECZ
WPMUBHFEJWJTJPOBT
= 7
v − = @@@@@
+
6TJOHUIFGBDUUIBUUIFDBQBDJUPSWPMUBHFDBOOPUDIBOHFJOTUBOUBOFPVTMZ
v = v − = v + = 7
'PSU> UIFTXJUDIJTJOQPTJUJPO#5IF5IFWFOJOSFTJTUBODFDPOOFDUFE
UPUIFDBQBDJUPSJT35I= LΩ BOEUIFUJNFDPOTUBOUJT
τ = 35I$ = ×××− = T
4JODFUIFDBQBDJUPSBDUTMJLFBOPQFODJSDVJUUPEDBUTUFBEZTUBUF v ∞ = 75IVT
v U = v ∞ + <v −v ∞ >F−U∕τ
= + − F−U∕
= −F−U
7
"UU = "UU = v = −F−
= 7
v = −F− = 7
Example 7.10
276
Chapter 7
Practice Problem 7.10
t=0
2Ω
15 V
+
v
‒
+
‒
6Ω
First-Order Circuits
'JOEv U GPSU>JOUIFDJSDVJUPG'JH"TTVNFUIFTXJUDIIBTCFFO
PQFOGPSBMPOHUJNFBOEJTDMPTFEBUU = $BMDVMBUFv U BUU = Answer: + F−U
7GPSBMMU> 7
‒
+ 7.5 V
1
3F
Figure 7.44
'PS1SBDUJDF1SPC
Example 7.11
*O'JH UIFTXJUDIIBTCFFODMPTFEGPSBMPOHUJNFBOEJTPQFOFEBU
U = 'JOEJBOEvGPSBMMUJNF
30u(t) V +
‒
t=0
i
10 Ω
20 Ω
+
v
‒
1
4F
+ 10 V
‒
Figure 7.45
'PS&YBNQMF
Solution:
5IFSFTJTUPSDVSSFOU JDBOCFEJTDPOUJOVPVTBU U = XIJMFUIFDBQBDJUPS
WPMUBHFv cannot. Hence, it is always better to find vBOEUIFOPCUBJO J
GSPNv
By definition of the unit step function,
V U =
i
10 Ω
+
v
‒
20 Ω
+ 10 V
‒
(a)
10 Ω
30 V +
‒
+
v
‒
1
4F
(b)
Figure 7.46
U<
U>
'PSU< UIFTXJUDIJTDMPTFEBOE V U = TPUIBUUIFV U WPMUBHF
TPVSDFJTSFQMBDFECZBTIPSUDJSDVJUBOETIPVMECFSFHBSEFEBTDPOUSJCVU
JOHOPUIJOHUP v4JODFUIFTXJUDIIBTCFFODMPTFEGPSBMPOHUJNF UIF
DBQBDJUPSWPMUBHFIBTSFBDIFETUFBEZTUBUFBOEUIFDBQBDJUPSBDUTMJLFBO
open circuit. Hence, the circuit becomes that shown in Fig. B GPS
U<'SPNUIJTDJSDVJUXFPCUBJO
v = −"
J = − @@@
4JODFUIFDBQBDJUPSWPMUBHFDBOOPUDIBOHFJOTUBOUBOFPVTMZ
v = 7 v = v − = 7
i
20 Ω
{ 4PMVUJPOPG&YBNQMF B GPSU< C GPSU>
'PSU> UIFTXJUDI JTPQFOFEBOEUIF7WPMUBHF TPVSDFJTEJT
DPOOFDUFEGSPNUIFDJSDVJU5IF V U WPMUBHFTPVSDFJTOPXPQFSBUJWF TPUIFDJSDVJUCFDPNFTUIBUTIPXOJO'JH C "GUFSBMPOHUJNF UIF
DJSDVJUSFBDIFTTUFBEZTUBUFBOEUIFDBQBDJUPSBDUTMJLFBOPQFODJSDVJU
BHBJO8FPCUBJOv ∞ CZVTJOHWPMUBHFEJWJTJPO XSJUJOH
v ∞ = @@@@@@@
= 7
+ Step Response of an RC Circuit
7.5
277
5IF5IFWFOJOSFTJTUBODFBUUIFDBQBDJUPSUFSNJOBMTJT
×
Ω
35I= ∥ = @@@@@@@
= @@@
BOEUIFUJNFDPOTUBOUJT
⋅@@
T
τ = 35I$ = @@@
= @@
5IVT
v U = v ∞ + [ v −v ∞ ]F−U∕τ
= + − F − ∕ U= −F−U
7
5PPCUBJOJ XFOPUJDFGSPN'JH C UIBUJJTUIFTVNPGUIFDVSSFOUT
UISPVHIUIFΩSFTJTUPSBOEUIFDBQBDJUPSUIBUJT
v + $@@@
J = @@@
Ev
EU
= + F−U "
= −F−U
+ − − F −U
/PUJDFGSPN'JH C UIBU v + J = 30 is satisfied, as expected.
)FODF
U<
7
v=
−U
−F 7 U ≥ J=
{
{
−"
+F−U
" U<
U>
/PUJDFUIBUUIFDBQBDJUPSWPMUBHFJTDPOUJOVPVTXIJMFUIFSFTJTUPSDVSSFOU
JTOPU
5IFTXJUDIJO'JHJTDMPTFEBU U = 'JOEJ U BOEv U GPSBMMUJNF
/PUFUIBUV −U = GPSU<BOEGPSU>"MTP V −U = −V U 5Ω
+
v
‒
20u(‒t) V +
‒
t=0
i
0.2 F
10 Ω
Figure 7.47
'PS1SBDUJDF1SPC
Answer: J U =
v=
{
{
− +F−1U
" 7
−1U
+F 7
U<
U>
U<
U > ,
3A
Practice Problem 7.11
278
Chapter 7
R
i
t=0
Vs +
‒
+
v(t)
‒
L
(a)
7.6
First-Order Circuits
Step Response of an RL Circuit
$POTJEFSUIF3- DJSDVJUJO'JH B XIJDINBZCFSFQMBDFECZUIF
DJSDVJUJO'JH C "Hain, our goal is to find the inductor current JBT
UIFDJSDVJUSFTQPOTF3BUIFSUIBOBQQMZ,JSDIIPGGTMBXT XFXJMMVTFUIF
TJNQMFUFDIOJRVFJO&RT UISPVHI -FUUIFSFTQPOTFCFUIF
TVNPGUIFUSBOTJFOUSFTQPOTFBOEUIFTUFBEZTUBUFSFTQPOTF
J = JU + JTT
R
i
Vs u(t) +
‒
+
v(t)
‒
L
(b)
Figure 7.48
"O3-DJSDVJUXJUIBTUFQJOQVUWPMUBHF
8FLOPXUIBUUIFUSBOTJFOUSFTQPOTFJTBM XBZTBEFDBZJOHF YQPOFOUJBM UIBUJT
-
τ = @@
3
JU = "F−U∕τ
XIFSF"JTBDPOTUBOUUPCFEFUFSNJOFE
5IFTUFBEZTUBUFSFTQPOTFJTUIFWBMVFPGUIF DVSSFOUBMPOHUJNF
BGUFS UIF TXJUDIJO 'JH B JT DMPTFE8FLOPX UIBU UIFUSBOTJFOU
response essentially dies out after fivFUJNFDPOTUBOUT "UUIBUUJNF UIFJOEVDUPSCFDPNFTBTIPSUDJSDVJU BOEUIFW PMUBHFBDSPTTJUJT[FSP
5IFFOUJSFTPVSDFW PMUBHF7TBQQFBSTBDSPTT 35IVT UIFTUFBEZTUBUF SFTQPOTFJT
7
JTT = @@T
3
4VCTUJUVUJOH&RT BOE JOUP&R HJWFT
7
J = "F−U∕τ + @@T
3
8FOPXEFUFSNJOFUIFDPOTUBOU "GSPNUIFJOJUJBMW BMVFPGJ-FU *CF
UIFJOJUJBMDVSSFOUUISPVHIUIFJOEVDUPS XIJDINBZDPNFGSPNBTPVSDF
PUIFS UIBO 7T 4JODFUIF DVSSFOU UISPVHI UIFJOEVDUPS DBOOPU DIBOHF
JOTUBOUBOFPVTMZ
J + = J − = *
5IVT BUU = &R CFDPNFT
7
* = " + @@T
3
'SPNUIJT XFPCUBJO"BT
i(t)
7
T
" = *−@@
3
4VCTUJUVUJOHGPS"JO&R XFHFU
I0
Vs
R
0
Figure 7.49
t
5PUBMSFTQPOTFPGUIF3-DJSDVJUXJUI
JOJUJBMJOEVDUPSDVSSFOU*
(
7
7
T F−U∕τ
J U = @@T+ *−@@
3
3
5IJTJTUIFDPNQMFUFSFTQPOTFPGUIF 3- DJSDVJU*UJTJMMVTUSBUFEJO
Fig. 7.49. 5IFSFTQPOTFJO&R NBZCFXSJUUFOBT
J U = J ∞ + [ J −J ∞ ]F
−U∕τ
Step Response of an RL Circuit
7.6
XIFSF J BOE J ∞) are the initial and final vBMVFTPG J SFTQFDUJ WFMZ
Thus, to find the step response of an3-DJSDVJUSFRVJSFTUISFFUIJOHT
5IFJOJUJBMJOEVDUPSDVSSFOUJ BUU = The final inductor current J ∞ 5IFUJNFDPOTUBOUτ
8FPCUBJOJUFNGSPNUIFHJWFODJSDVJUGPSU<BOEJUFNTBOEGSPN
UIFDJSDVJUGPS U >0ODFUIFTFJUFNTBSFEFUFSNJOFE XFPCUBJOUIF
SFTQPOTFVTJOH&R ,FFQJONJOEUIBUUIJTUFDIOJRVFBQQMJFTPOMZ
GPSTUFQSFTQPOTFT
"HBJO JGUIFTXJUDIJOHUBL FTQMBDFBUUJNF U = UJOTUFBEPG U = &R CFDPNFT
J U = J ∞ + <J U −J ∞ >F − U−U ∕τ
*G* = UIFO
{
J U = @@
7
T −F−U∕τ
3
PS
U<
U>
B
7
J U = @@T −F−U∕τ V U 3
C
5IJTJTUIFTUFQSFTQPOTFPGUIF3-DJSDVJUXJUIOPJOJUJBMJOEVDUPSDVSSFOU
5IFW PMUBHFBDSPTTUIFJOEVDUPSJTPCUBJOFEGSPN&R VTJOH
v = -EJ∕EU8FHFU
v U = -@@
EJ = 7T@@@
-F−U∕τ
τ3
EU
- τ = @@
3
U>
PS
v U = 7TF−U∕τ
VU
'JHVSFTIPXTUIFTUFQSFTQPOTFTJO&RT BOE i(t)
v(t)
Vs
R
Vs
0
t
(a)
Figure 7.50
0
t
(b)
4UFQSFTQPOTFTPGBO3-DJSDVJUXJUIOPJOJUJBMJOEVDUPS
DVSSFOU B DVSSFOUSFTQPOTF C WPMUBHFSFTQPOTF
279
280
Chapter 7
Example 7.12
'JOEJ U JOUIFDJSDVJUPG'JHGPS U>"TTVNFUIBUUIFTXJUDIIBT
CFFODMPTFEGPSBMPOHUJNF
t=0
2Ω
Solution:
8IFOU< UIFΩSFTJTUPSJTTIPSUDJSDVJUFE BOEUIFJOEVDUPSBDUTMJLF
BTIPSUDJSDVJU5IFDVSSFOUUISPVHIUIFJOEVDUPSBUU = − JF KVTUCFGPSF
U = JT
3Ω
i
10 V
+
‒
First-Order Circuits
1
3H
= A
J − = @@@
Figure 7.51
#FDBVTFUIFJOEVDUPSDVSSFOUDBOOPUDIBOHFJOTUBOUBOFPVTMZ
'PS&YBNQMF
J = J + = J − = "
8IFOU> UIFTXJUDIJTPQFO5IFBOE ΩSFTJTUPSTBSFJOTFSJFT TPUIBU
= "
J ∞ = @@@@@
+
5IF5IFWFOJOSFTJTUBODFBDSPTTUIFJOEVDUPSUFSNJOBMTJT
35I= + = Ω
'PSUIFUJNFDPOTUBOU
@
- = @@ = @@@
T
τ = @@@
35I 5IVT
J U = J ∞ + <J −J ∞ >F−U∕τ
= + − F−U
= + F−U
" U>
$IFDL*O'JH GPSU> 0, KVL must be satisfied; that is,
= J + -@@
EJ
EU
[
]
− F−U
= J + -@@
EJ = < + F−U
> + @@
EU
This confirms the result.
Practice Problem 7.12
i
1.5 H
5IFTXJUDIJO'JHIBTCFFODMPTFEGPSBMPOHUJNF*UPQFOTBU
U = 'JOEJ U GPSU> Answer: + F−U
"GPSBMMU>
5Ω
t=0
Figure 7.52
'PS1SBDUJDF1SPC
10 Ω
6A
Step Response of an RL Circuit
7.6
"UU = TXJUDIJO'JHJTDMPTFE BOETXJUDIJTDMPTFETMBUFS 'JOEJ U GPSU>$BMDVMBUFJGPSU = TBOEU = T
4Ω
S1 t = 0
S2
40 V +
‒
6Ω
P
i
t=4
2Ω
5H
10 V +
‒
Figure 7.53
'PS&YBNQMF
Solution:
8FOFFEUPDPOTJEFSUIFUISFFUJNFJOUFSWBMT U≤ ≤ t ≤ 4 BOEU ≥ TFQBSBUFMZ'PSU< TXJUDIFT4BOE4BSFPQFOTPUIBUJ = 4JODFUIF
JOEVDUPSDVSSFOUDBOOPUDIBOHFJOTUBOUMZ
J − = J = J + = 'PS ≤U≤ 4JTDMPTFETPUIBUUIFBOE ΩSFTJTUPSTBSFJO
TFSJFT 3FNFNCFS BUUIJT UJNF 4JT TUJMMPQFO )FODF BTTVNJOHGPS
OPXUIBU4JTDMPTFEGPSFWFS
5IVT
= " J ∞ = @@@@@
35I
= + = Ω
+
= @@
- = @@@
T
τ = @@@
35I
J U = J ∞ + <J −J ∞ >F −U∕τ
= + − F −U
= −F −U
" ≤ U≤ 'PSU ≥ 4JTDMPTFEUIF7WPMUBHFTPVSDFJTDPOOFDUFE BOEUIF
DJSDVJUDIBOHFT5IJTTVEEFODIBOHFEPFTOPUBGGFDUUIFJOEVDUPSDVSSFOU
CFDBVTFUIFDVSSFOUDBOOPUDIBOHFBCSVQUMZ5IVT UIFJOJUJBMDVSSFOUJT
J = J − = −F− ≃"
To find J ∞ MFUvCFUIFWPMUBHFBUOPEF1JO'JH6TJOH,$−v
v @@@@@@
−v
+ @@@@@@
= @@
⇒
7
v = @@@@
v = @@@
= "
J ∞ = @@
5IF5IFWFOJOSFTJTUBODFBUUIFJOEVDUPSUFSNJOBMTJT
BOE
×
Ω
+ = @@@
35I= ∥ + = @@@@@
= @@@
T
- = @@
τ = @@@
35I
@@
281
Example 7.13
282
Chapter 7
First-Order Circuits
)FODF
J U = J ∞ + <J −J ∞ >F− U− ∕τ U≥
8FOFFE U− JOUIFFYQPOFOUJBMCFDBVTFPGUIFUJNFEFMBZ5IVT
J U = + − F− U− ∕τ τ = @@@
= + F− U− 1VUUJOHBMMUIJTUPHFUIFS
{
J U = −F−U
+ F− U− "UU = U≥
U≤
≤U≤
U≥
J = −F− = "
"UU = J = + F−= "
Practice Problem 7.13
4XJUDI4JO'JHJTDMPTFEBUU = BOETXJUDI4JTDMPTFEBUU = T
$BMDVMBUFJ U GPSBMMU'JOEJ BOEJ t=2
S1
10 Ω
t=0
12 A
15 Ω
Answer:
S2
20 Ω
i(t)
5H
{
J U = −F−U
− F− U− U<
<U<
U>
J = " J = "
Figure 7.54
'PS1SBDUJDF1SPC
7.7
† First-Order Op Amp Circuits
"OPQ BNQDJSDVJU DPOUBJOJOHB TUPSBHFFMFNFOU XJMMFxhibit first-order
CFIBWJPS%JGGFSFOUJBUPSTBOEJOUFHSBUPSTUSFBUFEJO4FDUJPOBSFFYBN
ples of first-order op amp circuits. "HBJO GPSQSBDUJDBMSFBTPOT JOEVDUPST
BSFIBSEMZFWFSVTFEJOPQBNQDJSDVJUTUIFSFGPSF UIFPQBNQDJSDVJUTXF
DPOTJEFSIFSFBSFPGUIF3$UZQF
"TVTVBM XFBOBMZ[FPQBNQDJSDVJUTVTJOHOPEBMBOBMZTJT4PNF UJNFT UIF5IFWFOJOFRVJWBMFOUDJSDVJUJTVTFEUPSFEVDFUIFPQBNQDJS
DVJU UP POF UIBUXF DBO FBTJMZ IBOEMF 5IF GPMMPXJOH UISFF FYBNQMFT
JMMVTUSBUFUIFDPODFQUT The first one deals with a source-free op amp
DJSDVJU XIJMFUIFPUIFSUXPJOWPMWFTUFQSFTQPOTFT5IFUISFFFYBNQMFT
IBWFCFFODBSFGVMMZTFMFDUFEUPDP WFSBMMQPTTJCMF 3$ UZQFTPGPQBNQ
DJSDVJUT EFQFOEJOHPOUIFMPDBUJPOPGUIFDBQBDJUPSXJUISFTQFDUUPUIF PQBNQUIBUJT UIFDBQBDJUPSDBOCFMPDBUFEJOUIFJOQVU UIFPVUQVU PS UIFGFFECBDLMPPQ
7.7
283
First-Order Op Amp Circuits
Example 7.14
'or the op amp circuit in Fig. 7.55(a), find vPGPS U > HJ WFOUIBU v = 7-FU3G = LΩ 3 = LΩ BOE$ = •'
Rf
1
C
+ v ‒
2
3
80 kΩ
80 kΩ
1
‒
+
+ 3V ‒
+
R1
vo
C
2
3
1A
‒
+
20 kΩ
vo
–
(0+)
(a)
(b)
Solution:
5IJTQSPCMFNDBOCFTPMWFEJOUXPXBZT
■ METHOD 1 $POTJEFSUIFDJSDVJUJO'JH B -FUVTEFSJWFUIF
BQQSPQSJBUFEJGGFSFOUJBMFRVBUJPOVTJOHOPEBMBOBMZTJT*GvJTUIFWPMUBHF
BUOPEF BUUIBUOPEF ,$-HJWFT
− v
@@@@@@
= $@@@
Ev
3
EU
#FDBVTF OPEFT BOE NVTUCF BUUIF TBNF QPUFOUJBM UIF QPUFOUJBMBU
OPEFJT[FSP5IVT v − = vPSv = vBOE&R CFDPNFT
v = @@@
Ev+ @@@@
EU
$3
5IJTJTTJNJMBSUP&R C TPUIBUUIFTPMVUJPOJTPCUBJOFEUIFTBNFXBZ
BTJO4FDUJPO JF
v U = 7F−U∕τ
τ = 3$
XIFSF7JTUIFJOJUJBMWPMUBHFBDSPTTUIFDBQBDJUPS#VUv = = 7BOE
τ = × × × ¢ = )FODF
v U = F−U
"QQMZJOH,$-BUOPEFHJWFT
− vP $@@@
Ev= @@@@
3G
EU
PS
Ev
vP = −3G$@@@
EU
Now we can find vBT
vP = − × × × ¢ −F−U
= F−U
7 U
+
vo
‒
(c)
'PS&YBNQMF
20 kΩ
‒
Figure 7.55
‒
+
+v ‒
+
284
Chapter 7
First-Order Circuits
■ METHOD 2 -FUVTBQQMZUIFTIPSUDVUNFUIPEGSPN&R 8F
need to find vP + vP ∞ BOE τ4JODF v + = v ¢ = 7 XFBQQMZ
,$-BUOPEFJOUIFDJSDVJUPG'JH C UPPCUBJO
− vP +
@@@@@@
+ @@@@@@@@@
= PSvP + = 74JODFUIFDJSDVJUJTTPVSDFGSFF v ∞ = 7 To find τ XFOFFEUIFFRVJWBMFOUSFTJTUBODF 3FRBDSPTTUIFDBQBDJUPSUFSNJOBMT*G
XFSFNPWFUIFDBQBDJUPSBOESFQMBDFJUCZB"DVSSFOUTPVSDF XFIBWF
UIFDJSDVJUTIPXOJO'JH D "QQMZJOH,7-UPUIFJOQVUMPPQZJFMET
− v = 5IFO
⇒
v = L7
v = LΩ
3FR = @@
BOEτ = 3FR$ = 5IVT
vP U = vP ∞ + <vP − vP ∞ >F−U∕τ
= + − F−U
= F−U
7 U
BTCFGPSF
Practice Problem 7.14
C
For the op amp circuit in Fig. 7.56, find vPGPSUJGv = 7"TTVNF
UIBU3G = LΩ 3 = LΩ BOE$ = •'
Answer:−F−U
7 U
+ v ‒
Rf
‒
+
R1
+
vo
‒
Figure 7.56
'PS1SBDUJDF1SPC
Example 7.15
%FUFSNJOFv U BOEvP U JOUIFDJSDVJUPG'JH
Solution:
5IJTQSPCMFNDBOCFTPMWFEJOUXPXBZT KVTUMJLFUIFQSFWJPVTFYBNQMF
)PXFWFS XF XJMM BQQMZ POMZ UIF TFDPOE NFUIPE 4JODF XIBU XF BSF
MPPLJOHGPSJTUIFTUFQSFTQPOTF XFDBOBQQMZ&R BOEXSJUF
v U = v ∞ + <v − v ∞ >F−U∕τ U>
7.7
285
First-Order Op Amp Circuits
+ v ‒
where we need only find the time constant τ UIFJOJUJBMWBMVF v BOE
the final value v ∞ /PUJDFUIBUUIJTBQQMJFTTUSJDUMZUPUIFDBQBDJUPSWPMU
BHFEVFBTUFQJOQVU4JODFOPDVSSFOUFOUFSTUIFJOQVUUFSNJOBMTPGUIF
PQBNQ UIFFMFNFOUTPOUIFGFFECBDLMPPQPGUIFPQBNQDPOTUJUVUFBO 3$DJSDVJU XJUI
τ =3$= × × ¢ = 4JODFUIFSFJTOPTUPSBHFFMFNFOUJOUIFJOQVUMPPQ vSFNBJOTDPOTUBOU
GPSBMMU"UTUFBEZTUBUF UIFDBQBDJUPSBDUTMJLFBOPQFODJSDVJUTPUIBUUIF
op amp circuit is a noninverting amplifier. Thus,
#VU
TPUIBU
(
t=0
10 kΩ
'PSU< UIFTXJUDIJTPQFOBOEUIFSFJTOPWPMUBHFBDSPTTUIFDBQBDJUPS
)FODF v = 'PS U> XFPCUBJOUIFWPMUBHFBUOPEFCZWPMUBHF
EJWJTJPOBT
v = @@@@@@@
= 7
+ 1 μF
v = × = 7
vP ∞ = + @@@
v − vP = v
3V +
‒
50 kΩ
v1
20 kΩ
+
‒
20 kΩ
+
vo
‒
Figure 7.57
'PS&YBNQMF
v ∞ = − = −7
4VCTUJUVUJOHτ v BOEv ∞ JOUP&R HJWFT
v U = − + <− − >F−U= F −U
− 7 U>
U>
'SPN&RT BOE XFPCUBJO
vP U = v U −v U = − F−U
7 'JOEv U BOEvP U JOUIFPQBNQDJSDVJUPG'JH
Practice Problem 7.15
Answer: /PUF UIFW PMUBHFBDSPTTUIFDBQBDJUPSBOEUIFPVUQVUW PMUBHF
NVTUCFCPUIFRVBMUP[FSP GPS U < TJODFUIFJOQVUX BT[FSPGPSBMM
U< −F−U V U N7 F −U
− V U N7
100 kΩ
1 μF
10 kΩ
4 mV
t=0
+
‒
+ v ‒
‒
+
+
vo
‒
Figure 7.58
'PS1SBDUJDF1SPC
286
Chapter 7
Example 7.16
'JOEUIFTUFQSFTQPOTF vP U GPSU> 0 in the op amp circuit of Fig. -FUvJ = V U 7 3 = LΩ 3G = LΩ 3 = 3 = LΩ $ = •'
Rf
R1
vi
‒
+
R2
+
‒
First-Order Circuits
R3
+
vo
‒
C
Figure 7.59
'PS&YBNQMF
Solution:
/PUJDFUIBUUIFDBQBDJUPSJO&YBNQMFJTMPDBUFEJOUIFJOQVUMPPQ XIJMFUIFDBQBDJUPSJO&YBNQMFJTMPDBUFEJOUIFGFFECBDL MPPQ*O
UIJTFYBNQMF UIFDBQBDJUPSJTMPDBUFEJOUIFPVUQVUPGUIFPQBNQ"HBJO XFDBOTPMWFUIJTQSPCMFNEJSFDUMZVTJOHOPEBMBOBMZTJT)PXFWFS VTJOH
UIF5IFWFOJOFRVJWBMFOUDJSDVJUNBZTJNQMJGZUIFQSPCMFN
We temporarily remove the capacitor and find the Thevenin equiva
MFOUBUJUTUFSNJOBMT5PPCUBJO 75I, consider the circuit in Fig. B Since the circuit is an inverting amplifier,
3G
7BC = − @@@@vJ
3
#ZWPMUBHFEJWJTJPO
3G
3 ___
3
75I
= @@@@@@@@@@
7BC = −@@@@@@@
vJ
3 + 3
3 + 3 3
Rf
R1
vi
a
‒
+
+
+
‒
Vab
‒
R2
R2
+
R3
VTh
R3
Ro
RTh
‒
b
(a)
(b)
Figure 7.60
0CUBJOJOH75I
BOE3
5I
BDSPTTUIFDBQBDJUPSJO'JH
5PPCUBJO 35IDPOTJEFSUIFDJSDVJUJO'JH C XIFSF 3PJTUIF
PVUQVUSFTJTUBODFPGUIFPQBNQ4JODFXFBSFBTTVNJOHBOJEFBMPQBNQ 3P = BOE
33
35I= 3∥ 3 = @@@@@@@
3 + 3
4VCTUJUVUJOHUIFHJWFOOVNFSJDBMWBMVFT
3G
3 @@@
@@@
V U = −V U
75I= − @@@@@@@
vJ = − @@@
3 + 3 3
33
= LΩ
35I= @@@@@@@
3 + 3
5IF5IFWFOJOFRVJWBMFOUDJSDVJUJTTIPXOJO'JH XIJDIJTTJNJMBS
UP'JH)FODF UIFTPMVUJPOJTTJNJMBSUPUIBUJO&R UIBUJT
5 kΩ
‒2.5u(t) +
‒
Figure 7.61
2 μF
5IFWFOJOFRVJWBMFOUDJSDVJUPGUIFDJSDVJU
JO'JH
vP U = − −F−U∕τ
VU
XIFSFτ = 35I$ = × × × − = 5IVT UIFTUFQSFTQPOTF
GPSU>JT
vP U = F−U
− V U 7
7.8
287
Transient Analysis with PSpice
0CUBJOUIFTUFQSFTQPOTFvP U GPSUIFDJSDVJUJO'JH-FUvJ = V U 7 3 = LΩ 3G = LΩ 3 = 3 = LΩ $ = μ'
Practice Problem 7.16
Rf
−U
Answer: −F
V U 7
R1
7.8
vi
Transient Analysis with PSpice
"TXFEJTDVTTFEJO4FDUJPO UIFUSBOTJFOUSFTQPOTFJTUIFUFNQPSBSZ
SFTQPOTFPGUIFDJSDVJUUIBUTPPOEJTBQQFBST14QJDFDBOCFVTFEUPPCUBJO
UIFUSBOTJFOUSFTQPOTFPGBDJSDVJUXJUITUPSBHFFMFNFOUT4FDUJPO%
JO"QQFOEJY%QSPWJEFTBSFWJFXPGUSBOTJFOUBOBMZTJTVTJOH 14QJDFGPS
8JOEPXT*UJTSFDPNNFOEFEUIBUZPVSFBE4FDUJPO%CFGPSFDPOUJOV
JOHXJUIUIJTTFDUJPO
*GOFDFTTBSZ ED14QJDF analysis is first carried out to determine the
JOJUJBMDPOEJUJPOT 5IFOUIFJOJUJBMDPOEJUJPOTBSFVTFEJOUIFUSBOTJFOU
14QJDFBOBMZTJTUPPCUBJOUIFUSBOTJFOUSFTQPOTFT*UJTSFDPNNFOEFECVU
OPUOFDFTTBSZUIBUEVSJOHUIJTEDBOBMZTJT BMMDBQBDJUPSTTIPVMECFPQFO
DJSDVJUFEXIJMFBMMJOEVDUPSTTIPVMECFTIPSUDJSDVJUFE
+
‒
R2
'PS1SBDUJDF1SPC
PSpice uses “transient” to mean
“function of time.” Therefore, the
transient response in PSpice may not
actually die out as expected.
Example 7.17
4Ω
J = J ∞ = " 3
5I
= J U = J ∞ + <J −J ∞ >F−U∕τ
= −F−U
U>
5PVTF 14QJDF we first draw the schematic as shown in Fig. 7.64.
8FSFDBMMGSPN"QQFOEJY%UIBUUIFQBSUOBNFGPSBDMPTFETXJUDIJT
4X@UDMPTF8FEPOPUOFFEUPTQFDJGZUIFJOJUJBMDPOEJUJPOPGUIFJOEVD UPSCFDBVTF 14QJDFXJMMEFUFSNJOFUIBUGSPNUIFDJSDVJU#ZTFMFDUJOH
"OBMZTJT4FUVQ5SBOTJFOU XFTFU1SJOU4UFQUPNTBOE'JOBM4UFQUP
τ = T"GUFSTBWJOHUIFDJSDVJU XFTJNVMBUFCZTFMFDUJOH "OBMZTJT
4JNVMBUF*OUIF14QJDF"%XJOEPX XFTFMFDU5SBDF"EEBOEEJTQMBZ
m* - BTUIFDVSSFOUUISPVHIUIFJOEVDUPS'JHVSFTIPXTUIFQMPUPG
J U XIJDIBHSFFTXJUIUIBUPCUBJOFECZIBOEDBMDVMBUJPO
+
vo
–
C
Figure 7.62
6TF14QJDF to find the response J U GPSU> 0 in the circuit of Fig. 7.63.
Solution:
4PMWJOHUIJTQSPCMFNCZIBOEHJWFT
τ = ∕ = T TPUIBU
R3
‒
+
i(t)
t=0
2Ω
6A
3H
Figure 7.63
'PS&YBNQMF
2.0 A
1.5 A
1.0 A
tClose = 0
1
2
R2
U1
4
IDC
6A
R1
2
0.5 A
L1
3H
0A
0s
1.0 s
2.0 s
3.0 s
‒I(L1)
Time
0
Figure 7.64
5IFTDIFNBUJDPGUIFDJSDVJUJO'JH
Figure 7.65
'PS&YBNQMFUIFSFTQPOTFPGUIF
DJSDVJUJO'JH
288
Chapter 7
First-Order Circuits
/PUFUIBUUIFOFHBUJWFTJHOPO* - JTOFFEFECFDBVTFUIFDVSSFOU
FOUFSTUISPVHIUIFVQQFSUFSNJOBMPGUIFJOEVDUPS XIJDIIBQQFOTUPCF
UIFOFHBUJWFUFSNJOBMBGUFSPOFDPVOUFSDMPDL w
ise SPUBUJPO"XBZUP
BWPJEUIFOFHBUJWFTJHOJTUPFOTVSFUIBUDVSSFOUFOUFSTQJOPGUIFJOEVD
tor. To obtain this desired direction of positive current flow, the initially
IPSJ[POUBMJOEVDUPSTZNCPMTIPVMECFSPUBUFEDPVOUFSDMPDLXJTF° BOE
QMBDFEJOUIFEFTJSFEMPDBUJPO
Practice Problem 7.17
3Ω
12 V
+
‒
Answer:v U = −F−U 7 U> 5IFSFTQPOTFJTTJNJMBSJOTIBQFUP
UIBUJO'JH
t=0
6Ω
'PSUIFDJSDVJUJO'JH VTF1TQJDF to find v U GPSU>
0.5 F
+
v(t)
‒
Figure 7.66
'PS1SBDUJDF1SPC
Example 7.18
*OUIFDJSDVJUPG'JH B EFUFSNJOFUIFSFTQPOTFv U 12 Ω
t=0
t=0
+ v(t) ‒
0.1 F
30 V +
‒
6Ω
6Ω
3Ω
4A
(a)
+ v (t) ‒
12 Ω
0.1 F
30 V +
‒
6Ω
6Ω
(b)
+ v (t) ‒
10 Ω
0.1 F
10 V +
‒
(c)
Figure 7.67
'PS&YBNQMF0SJHJOBMDJSDVJU B DJSDVJUGPSU> C BOE
SFEVDFEDJSDVJUGPSU> D 7.8
Transient Analysis with PSpice
Solution:
Define.5IFQSPCMFNJTDMFBSMZTUBUFEBOEUIFDJSDVJUJTDMFBSMZ
MBCFMFE
1SFTFOU(JWFOUIFDJSDVJUTIPXOJO'JH B EFUFSNJOFUIF
SFTQPOTFv U "MUFSOBUJWF8FDBOTPMWFUIJTDJSDVJUVTJOHDJSDVJUBOBMZTJT
UFDIOJRVFT OPEBMBOBMZTJT NFTIBOBMZTJT PS14QJDF-FUVTTPMWF
UIFQSPCMFNVTJOHDJSDVJUBOBMZTJTUFDIOJRVFT UIJTUJNF5IFWFOJO
FRVJWBMFOUDJSDVJUT BOEUIFODIFDLUIFBOTXFSVTJOHUXPNFUIPET
PG14QJDF
"UUFNQU 'PSUJNF< UIFTXJUDIPOUIFMFGUJTPQFOBOEUIF
TXJUDIPOUIFSJHIUJTDMPTFE"TTVNFUIBUUIFTXJUDIPOUIFSJHIU
IBTCFFODMPTFEMPOHFOPVHIGPSUIFDJSDVJUUPSFBDITUFBEZTUBUF
UIFOUIFDBQBDJUPSBDUTMJLFBOPQFODJSDVJUBOEUIFDVSSFOUGSPNUIF
4-A source floXTUISPVHIUIFQBSBMMFMDPNCJOBUJPOPGUIFΩBOE
Ω SFTJTUPST ∥ = ∕ = QSPEVDJOHBWPMUBHFFRVBMUP
× = 7 = −v "UU = UIFTXJUDIPOUIFMFGUDMPTFTBOEUIFTXJUDIPO
the right opens, producing the circuit shoXOJO'JH C 5IFFBTJFTUXay to complete the solution is to find the
5IFWFOJOFRVJWBMFOUDJSDVJUBTTFFOCZUIFDBQBDJUPS5IFPQFO
DJSDVJUWPMUBHF XJUIUIFDBQBDJUPSSFNPWFE JTFRVBMUPUIFWPMU
BHFESPQBDSPTTUIFΩSFTJTUPSPOUIFMFGU PS7 UIFWPMUBHF
ESPQTVOJGPSNMZBDSPTTUIFΩSFTJTUPS 7 BOEBDSPTTUIFΩ
SFTJTUPS 7 5IJTJT75I
5IFSFTJTUBODFMPPLJOHJOXIFSFUIF
DBQBDJUPSXBTJTFRVBMUP∥ + =∕ + = Ω XIJDI
JT3FR5IJTQSPEVDFTUIF5IFWFOJOFRVJWBMFOUDJSDVJUTIPXOJO
Fig. 7.67(c). Matching up the boundary conditions (v = −7
BOEv ∞ = 7 BOEτ = 3$ = XFHFU
v U = −F¢U7
&WBMVBUF5IFSFBSFUXPXBZTPGTPMWJOHUIFQSPCMFNVTJOH
14QJDF
■ METHOD 1 One way is to first do the dc 14QJDFBOBMZTJTUP
EFUFSNJOFUIFJOJUJBMDBQBDJUPSWPMUBHF5IFTDIFNBUJDPGUIFSFWFMBOU
DJSDVJUJTJO'JH B 5XPQTFVEPDPNQPOFOU7*&810*/5TBSF JOTFSUFEUPNFBTVSFUIFWPMUBHFTBUOPEFTBOE8IFOUIFDJSDVJU
JTTJNVMBUFE XFPCUBJOUIFEJTQMBZFEWBMVFTJO'JH B BT 7 =
7BOE7 = 75IVT UIFJOJUJBMDBQBDJUPSWPMUBHFJT v = 7−7
= −75IF 14QJDFUSBOTJFOUBOBMZTJTVTFTUIJTWBMVFBMPOHXJUIUIF TDIFNBUJDJO'JH C 0ODFUIFDJSDVJUJO'JH C JTESBXO XFJOTFSUUIFDBQBDJUPSJOJUJBMWPMUBHFBT*$ = −8FTFMFDU"OBMZTJT
4FUVQ5SBOTJFOUBOETFU1SJOU4UFQUPTBOE'JOBM4UFQUPτ = T
"GUFSTBWJOHUIFDJSDVJU XFTFMFDU "OBMZTJT4JNVMBUFUPTJNVMBUFUIF DJSDVJU*OUIF 14QJDF"%XJOEPX XFTFMFDU 5SBDF"EEBOEEJTQMBZ
7 3 m7 3 PS7 $ − 7 $ BTUIFDBQBDJUPSWPMUBHFv U 5IFQMPUPG v U JTTIPXOJO'JH5IJTBHSFFTXJUIUIFSFTVMUPC UBJOFECZIBOEDBMDVMBUJPO v U = −F−U
7
289
290
Chapter 7
0.0000
1
C1
8.0000
First-Order Circuits
10 V
2
0.1
R2
6
R3
6
R4
3
5V
4A
I1
0V
0
‒5 V
(a)
30 V
+
‒
V1
R1
C1
12
0.1
‒10 V
6
R2
R3
6
0s
1.0 s
2.0 s
V(R2 : 2) ‒ V(R3 : 2)
Time
3.0 s
4.0 s
Figure 7.69
3FTQPOTFv U GPSUIFDJSDVJUJO'JH
0
(b)
Figure 7.68
B 4DIFNBUJDGPSEDBOBMZTJTUPHFUv(0), (b) schematic
GPSUSBOTJFOUBOBMZTJTVTFEJOHFUUJOHUIFSFTQPOTFv U ■ METHOD 2 8FDBO TJNVMBUFUIF DJSDVJUJO 'JHEJSFDUMZ TJODF 14QJDFDBOIBOEMFUIFPQFOBOEDMPTFETXJUDIFTBOEEFUFSNJOF
UIFJOJUJBMDPOEJUJPOTBVUPNBUJDBMMZ6TJOHUIJTBQQSPBDI UIFTDIFNBU JDJTESBXOBTTIPXOJO'JH"GUFSESBXJOHUIFDJSDVJU XFTFMFDU
"OBMZTJT4FUVQ5SBOTJFOUBOETFU 1SJOU4UFQUPTBOE 'JOBM4UFQUP
τ = T8FTBWFUIFDJSDVJU UIFOTFMFDU"OBMZTJT4JNVMBUFUPTJNVMBUF
UIFDJSDVJU*OUIF14QJDF"%XJOEPX XFTFMFDU5SBDF"EEBOEEJTQMBZ
7 3 −7 3 BTUIFDBQBDJUPSWPMUBHF v U 5IFQMPUPG v U JTUIF
TBNFBTUIBUTIPXOJO'JH
R1
12
30 V
+
‒
V1
Figure 7.70
tClose = 0
1
2
U1
R2
tOpen = 0
1
2
U2
C1
0.1
6
R3
6
R4
3
I1
4A
0
'PS&YBNQMF
4BUJTGBDUPSZ $MFBSMZ XFIBWFGPVOEUIFWBMVFPGUIFPVUQVU
SFTQPOTFv U BTSFRVJSFECZUIFQSPCMFNTUBUFNFOU$IFDLJOHEPFT
WBMJEBUFUIBUTPMVUJPO8FDBOQSFTFOUBMMUIJTBTBDPNQMFUFTPMVUJPO
UPUIFQSPCMFN
7.9
291
Applications
5IFTXJUDIJO'JHXBTPQFOGPS BMPOHUJNFCVUDMPTFEBU U = *G
J = 10A, find J U GPSU>CZIBOEBOEBMTPCZ14QJDF
Practice Problem 7.18
5Ω
−U
Answer:J U = + F "5IFQMPUPGJ U PCUBJOFECZ14QJDFBOBMZTJT
JTTIPXOJO'JH
12 A
30 Ω
10 A
Figure 7.71
9A
'PS1SBDUJDF1SPC
8A
7A
6A
0s
0.5 s
1.0 s
I (L1)
Time
Figure 7.72
'PS1SBDUJDF1SPC
†Applications
7.9
5IFWBSJPVTEFWJDFTJOXIJDI3$BOE3-DJSDVJUTfind applications include
filtering in dc poXFSTVQQMJFT TNPPUIJOHDJSDVJUTJOEJHJUBMDPNNVOJDB UJPOT EJGGFSFOUJBUPST JOUFHSBUPST EFMBZDJSDVJUT BOESFMBZDJSDVJUT4PNF
PGUIFTFBQQMJDBUJPOTUBLFBEWBOUBHFPGUIFTIPSUPSMPOHUJNFDPOTUBOUT
PGUIF3$PS3-DJSDVJUT8FXJMMDPOTJEFSGPVSTJNQMFBQQMJDBUJPOTIFSF
The first twPBSF3$DJSDVJUT UIFMBTUUXPBSF3-DJSDVJUT
7.9.1
Delay Circuits
"O3$ DJSDVJUDBOCFVTFEUPQSP WJEFWBSJPVTUJNFEFMBZT'JHVSF
TIPXTTVDIBDJSDVJU*UCBTJDBMMZDPOTJTUTPGBO 3$ DJSDVJUXJUIUIFDB QBDJUPSDPOOFDUFEJO QBSBMMFMXJUIBOFPO MBNQ5IFWPMUBHFTPVSDF DBO
QSPWJEFFOPVHIWoltage to fire the lamp. 8IFOUIFTXJUDIJTDMPTFE UIF
DBQBDJUPSWPMUBHFJODSFBTFTHSBEVBMMZUPXBSE7BUBSBUFEFUFSNJOFE
CZUIFDJSDVJUTUJNFDPOTUBOU 3 + 3 $5IFMBNQXJMMBDUBTBOPQFO
R1
+
110 V
‒
Figure 7.73
"O3$EFMBZDJSDVJU
S
R2
C
0.1 μF
70 V
Neon
lamp
i(t)
t=0
6Ω
2H
292
Chapter 7
First-Order Circuits
DJSDVJUBOEOPUFNJUMJHIUVOUJMUIFWPMUBHFBDSPTTJUFYDFFETBQBSUJDVMBS
MFWFM TBZ 78IFOUIFWPMUBHFMFWel is reached, the lamp fires (goes
PO BOEUIFDBQBDJUPSEJTDIBS HFTUISPVHIJU%VFUPUIFMP XSFTJTUBODF
PGUIFMBNQXIFOPO UIFDBQBDJUPSWPMUBHFESPQTGBTUBOEUIFMBNQUVSOT
PGG5IFMBNQBDUTBHBJOBTBOPQFODJSDVJUBOEUIFDBQBDJUPSSFDIBSHFT
#ZBEKVTUJOH 3 XFDBO JOUSPEVDFFJUIFSTIPSUPSMPOHUJNFEFMBZTJOUP
UIFDJSDVJUBOENBL e the lamp fire, recharge, and fire repeatedly eWFSZ
UJNFDPOTUBOUτ = 3 + 3 $ CFDBVTFJUUBLFTBUJNFQFSJPEτUPHFUUIF
DBQBDJUPSWoltage high enough to fire or loXFOPVHIUPUVSOPGG
5IFXBSOJOHCMJOLFSTDPNNPOMZGPVOEPOSPBEDPOTUSVDUJPOTJUFT
BSFPOFFYBNQMFPGUIFVTFGVMOFTTPGTVDIBO3$EFMBZDJSDVJU
Example 7.19
$POTJEFSUIFDJSDVJUJO'JH BOEBTTVNFUIBU 3 = .Ω <
3<.Ω B $BMDVMBUFUIFFYUSFNFMJNJUTPGUIFUJNFDPOTUBOUPG
UIFDJSDVJU C )PXMPOHEPFTJUUBLFGPSUIFMBNQUPHMPw for the first
UJNFBGUFSUIFTXJUDIJTDMPTFE -FU3BTTVNFJUTMBSHFTUWBMVF
Solution:
B 5IFTNBMMFTUWBMVFGPS3JTΩ BOEUIFDPSSFTQPOEJOHUJNFDPOTUBOU
GPSUIFDJSDVJUJT
τ = 3 + 3 $ = × + × × − = T
5IFMBSHFTUWBMVFGPS3JT.Ω BOEUIFDPSSFTQPOEJOHUJNFDPOTUBOU
GPSUIFDJSDVJUJT
τ = 3 + 3 $ = + × × × − = T
5IVT CZQSPQFSDJSDVJUEFTJHO UIFUJNFDPOTUBOUDBOCFBEKVTUFEUPJOUSP
EVDFBQSPQFSUJNFEFMBZJOUIFDJSDVJU
C "TTVNJOHUIBUUIFDBQBDJUPSJTJOJUJBMMZVODIBSHFE v$ = XIJMF
v$ ∞ = #VU
v$ U = v$ ∞ + <v$ −v$ ∞ >F−U∕τ
= <−F−U∕τ
>
XIFSFτ = T BTDBMDVMBUFEJOQBSU B 5IFMBNQHMPXTXIFOv$ = 7
*Gv$ U = 7BUU = U UIFO
= <−F−U
∕τ>
⇒
F−U∕τ = @@@
⇒
@@@
= −F−U
∕τ
PS
FU∕τ = @@@
5BLJOHUIFOBUVSBMMPHBSJUINPGCPUITJEFTHJWFT
= MO = T
U = τ MO@@@
A more general formula for finding UJT
−v ∞
U = τ MO@@@@@@@@@@@
v U −v ∞
The lamp will fire repeatedly every UTFDPOETJGBOEPOMZJGv U <v ∞ 7.9
5IF3$ DJSDVJUJO'JHJTEFTJHOFEUPPQFSBUFBOBMBSNXIJDIBDUJ WBUFTXIFOUIFDVSSFOUUISPVHIJUFYDFFET μ"*G≤3≤LΩ, find
UIFSBOHFPGUIFUJNFEFMBZUIBUUIFWBSJBCMFSFTJTUPSDBODSFBUF
UDIBSHF = 3$
8JUIUIFTXJUDIJOQPTJUJPO UIFDBQBDJUPSW PMUBHFJTEJTDIBS HFE5IF
MPXSFTJTUBODF3PGUIFQIPUPMBNQQFSNJUTBIJHIEJTDIBSHFDVSSFOUXJUI
QFBL* = 7T∕3 in a short duration, as depicted in Fig. C %JTDIBSH
JOHUBLes place in approximately fivFUJNFTUIFUJNFDPOTUBOU
UEJTDIBSHF = 3$
i
v
I1
Vs
0
t
0
(a)
R
S
80 μF
4 kΩ
Figure 7.74
An electronic flBTIVOJUQSP WJEFTBDPNNPOF YBNQMFPGBO 3$ DJSDVJU
5IJTBQQMJDBUJPOFYQMPJUTUIFBCJMJUZPGUIFDBQBDJUPSUPPQQPTFBOZBCSVQU
DIBOHFJOWPMUBHF'JHVSFTIPXTBsimplified DJSDVJU*UDPOTJTUTFT
TFOUJBMMZPGBIJHIWPMUBHFEDTVQQMZ BDVSSFOUMJNJUJOHMBSHFSFTJTUPS3 BOE B DBQBDJUPS $ JOQBSBMMFM XJUI UIF flBTIMBNQ PG MPX SFTJTUBODF 3
8IFOUIFTXJUDIJTJOQPTJUJPO UIFDBQBDJUPSDIBSHFTTMPXMZEVFUPUIF
MBSHFUJNFDPOTUBOU τ = 3$ "TTIPwn in Fig. B UIFDBQBDJUPS
WPMUBHFSJTFTHSBEVBMMZGSPN[FSPUP7T XIJMFJUTDVSSFOUEFDSFBTFTHSBEV
BMMZGSPN * = 7T∕3UP[FSP 5IFDIBSging time is approximately fivF
UJNFTUIFUJNFDPOTUBOU
10 kΩ
Alarm
Photoflash Unit
Practice Problem 7.19
+
9V
‒
Answer:#FUXFFOBOENT
7.9.2
293
Applications
‒I2
(b)
Figure 7.76
B $BQBDJUPSWPMUBHFTIPXJOHTMPXDIBSHFBOEGBTUEJTDIBSHF C DBQBDJUPS
DVSSFOUTIPXJOHMPXDIBSHJOHDVSSFOU* = 7T∕3BOEIJHIEJTDIBSHFDVSSFOU
* = 7T∕3
5IVT UIFTJNQMF3$DJSDVJUPG'JHQSPWJEFTBTIPSUEVSBUJPO IJHI
current pulse. Such a circuit also finds applications in electric spot weld
JOHBOEUIFSBEBSUSBOTNJUUFSUVCF
'PS1SBDUJDF1SPC
R1
1
i
High
+ vs
voltage
‒
dc supply
Figure 7.75
2
R2
C
+
v
‒
Circuit for a flash unit providing slow
DIBSHFJOQPTJUJPOBOEGBTUEJTDIBSHFJO
QPTJUJPO
294
Example 7.20
Chapter 7
First-Order Circuits
An electronic flBTIHVOIBTBDVSSFOUMJNJUJOHLΩSFTJTUPSBOEμ'
FMFDUSPMZUJDDBQBDJUPSDIBS HFEUP 7*GUIFMBNQSFTJTUBODFJT Ω find: (a) the peak charHJOHDVSSFOU C UIFUJNFSFRVJSFEGPSUIFDBQBDJ UPSUPGVMMZDIBSHF D UIFQFBLEJTDIBSHJOHDVSSFOU E UIFUPUBMFOFSHZ
TUPSFEJOUIFDBQBDJUPS BOEUIFBWFSBHFQPXFSEJTTJQBUFECZUIFMBNQ
Solution:
B 5IFQFBLDIBSHJOHDVSSFOUJT
7
* = @@@T= @@@@@@@
= N"
3 × C 'SPN&R UDIBSHF = 3$ = × × × × − = T = NJOVUF
D 5IFQFBLEJTDIBSHJOHDVSSFOUJT
7 * = @@@T= @@@@
= "
3
E 5IFFOFSHZTUPSFEJT
× × − × = +
$7 = @@
8 = @@
T 5IFFOFSHZTUPSFEJOUIFDBQBDJUPSJTEJTTJQBUFEBDSPTTUIFMBNQEVSJOH
UIFEJTDIBSHJOHQFSJPE'SPN&R UEJTDIBSHF = 3$ = × × × − = T
5IVT UIFBWFSBHFQPXFSEJTTJQBUFEJT
8 = XBUUT
Q = @@@@@@
= @@@@
UEJTDIBSHF Practice Problem 7.20
The flash unit of a camera has a 2-mF capacitor charged to 40 V.
B )PXNVDIDIBSHFJTPOUIFDBQBDJUPS
b 8IBUJTUIFFOFSHZTUPSFEJOUIFDBQBDJUPS
D If the flash fires in 0.8 ms, what is the average current through
the flashtube?
E How much power is delivered to the flashtube?
"GUFSBQJDUVSFIBTCFFOUBLFO UIFDBQBDJUPSOFFETUPCF
SFDIBSHFECZBQPXFSVOJUUIBUTVQQMJFTBNBYJNVNPGN"
)PXNVDIUJNFEPFTJUUBLFUPDIBSHFUIFDBQBDJUPS
Answer: B N$ C + D " E L8 T
7.9.3
Relay Circuits
"NBHOFUJDBMMZDPOUSPMMFETXJUDIJTDBMMFEB SFMBZ"SFMBZJTFTTFOUJBMMZ
BOFMFDUSPNBHOFUJDEF WJDFVTFEUPPQFOPSDMPTFBTXJUDIUIBUDPOUSPMT
BOPUIFSDJSDVJU'JHVSF B TIP XTBUZQJDBMSFMBZDJSDVJU 5IFDPJM
7.9
Applications
295
DJSDVJUJTBO 3-DJSDVJUMJLFUIBUJO'JH C XIFSF 3BOE-BSFUIF
SFTJTUBODFBOEJOEVDUBODFPGUIFDPJM 8IFOTXJUDI4 in Fig. B JT
DMPTFE UIFDPJMDJSDVJUJTFOFSHJ[FE5IFDPJMDVSSFOUHSBEVBMMZJODSFBTFT
and produces a magnetic field. Eventually the magnetic field is suffi
DJFOUMZTUSPOHUPQVMMUIFNP WBCMFDPOUBDUJOUIFPUIFSDJSDVJUBOEDMPTF
TXJUDI4"UUIJTQPJOU UIFSFMBZJTTBJEUPCFQVMMFEJO5IFUJNFJOUFSWBM
UECFUXFFOUIFDMPTVSFPGTXJUDIFT4BOE4JTDBMMFEUIFSFMBZEFMBZUJNF
3FMBZTXFSFVTFEJOUIFFBSMJFTUEJHJUBMDJSDVJUTBOEBSFTUJMMVTFEGPS
TXJUDIJOHIJHIQPXFSDJSDVJUT
S2
S1
Magnetic field
S1
Vs
R
Coil
Vs
L
(a)
(b)
Figure 7.77
"SFMBZDJSDVJU
5IFDPJMPGBDFSUBJOSFMBZJTPQFSBUFECZB7CBUUFSZ*GUIFDPJMIBTB
SFTJTUBODFPGΩBOEBOJOEVDUBODFPGN)BOEUIFDVSSFOUOFFEFE
UPQVMMJOJTN" DBMDVMBUFUIFSFMBZEFMBZUJNF
Solution:
5IFDVSSFOUUISPVHIUIFDPJMJTHJWFOCZ
J U = J ∞ + <J −J ∞ >F−U∕τ
XIFSF
J = = N"
J ∞ = @@@@
−
× - = @@@@@@@@@
τ = @@
= NT
3
5IVT
J U = <−F−U∕τ>N"
*GJ UE = N" UIFO
= <− F−UE∕τ>
⇒
@@
= −F−U
E∕τ
PS
F−UE∕τ = @@
⇒
FUE∕τ = @@
Example 7.21
296
Chapter 7
First-Order Circuits
#ZUBLJOHUIFOBUVSBMMPHBSJUINPGCPUITJEFT XFHFU
UE = τ MO@@
= MO@@
NT = NT
Alternatively, we may find UEVTJOH
J −J ∞
UE = τ MO@@@@@@@@@@
J UE −J ∞
Practice Problem 7.21
"SFMBZIBTBSFTJTUBODFPG ΩBOEBOJOEVDUBODFPGN)5IF
SFMBZDPOUBDUTDMPTFXIFOUIFDVSSFOUUISPVHIUIFDPJMSFBDIFT μ"
8IBUUJNFFMBQTFTCFUXFFOUIFBQQMJDBUJPOPG7UPUIFDPJMBOEDPO
UBDUDMPTVSF
Answer:μT
7.9.4 Automobile Ignition Circuit
R
i
Vs
Figure 7.78
+
v
‒
L
Spark
plug
Air gap
$JSDVJUGPSBOBVUPNPCJMFJHOJUJPOTZTUFN
5IFBCJMJUZPGJOEVDUPSTUPPQQPTFSBQJEDIBOHFJODVSSFOUNBL FTUIFN
VTFGVMGPSBSDPSTQBSLHFOFSBUJPO "OBVUPNPCJMFJHOJUJPOTZTUFNUBL FT
BEWBOUBHFPGUIJTGFBUVSF
5IFH BTPMJOFFOHJOFPGBOBVUPNPCJMFSFRVJSFTUIBUUIFGVFMBJS
NJYUVSFJOFBDID ZMJOEFSCFJHOJUFEBUQSPQFSUJNFT 5IJTJTBDIJF WFE
CZNFBOTPGBTQBSLQMVH 'JH XIJDIFTTFOUJBMMZDPOTJTUTPGB
QBJSPGFMFDUSPEFTTFQBSBUFECZBOBJSH BQ#ZDSFBUJOHBMBS HFWPMUBHF
UIPVTBOETPGW PMUT CFUXFFOUIFFMFDUSPEFT BTQBSLJTGPSNFEBDSPTT UIFBJSH BQ UIFSFCZJHOJUJOHUIFGVFM#VUIP XDBOTVDIBMBS HFW PMU
BHFCFPCUBJOFEGSPNUIFDBSCBUUFSZ XIJDITVQQMJFTPOMZ7 5IJTJT
BDIJFWFECZNFBOTPGBOJOEVDUPS UIFTQBSLDPJM -4JODFUIFW PMUBHF
BDSPTTUIFJOEVDUPSJT v = -EJ∕EU XFDBONBLFEJ∕EUMBSHFCZDSFBUJOH
BMBSHFDIBOHFJODVSSFOUJOBWFSZTIPSUUJNF8IFOUIFJHOJUJPOTXJUDI
JO'JHJTDMPTFE UIFDVSSFOUUISPVHIUIFJOEVDUPSJODSFBTFTHSBEV BMMZBOESFBDIFTUIFfinal vBMVFPGJ = 7T∕3 XIFSF7T = 7"HBJO UIF
UJNFUBLFOGPSUIFJOEVDUPSUPDIBS ge is fivFUJNFTUIF UJNFDPOTUBOU PG
UIFDJSDVJU τ = -∕3
UDIBSHF = @@
-
3
4JODFBUTUFBEZTUBUF JJTDPOTUBOU EJ∕EU= BOEUIFJOEVDUPSW PMUBHF
v = 8IFOUIFTXJUDITVEEFOMZPQFOT BMBS HFW PMUBHFJTEF WFMPQFE
BDSPTTUIFJOEVDUPS EVFUPUIFSBQJEMZDPMMBQTJOHfield) causing BTQBSL
PSBSDJOUIFBJSH BQ5IFTQBSLDPOUJOVFTVOUJMUIFFOFS HZTUPSFEJOUIF
JOEVDUPSJTEJTTJQBUFEJOUIFTQBSLEJTDIBS HF*OMBCPSBUPSJFT XIFOPOF
JTXPSLJOHXJUIJOEVDUJ WFDJSDVJUT UIJTTBNFFG GFDUDBVTFTBW FSZOBTUZ
TIPDL BOEPOFNVTUFYFSDJTFDBVUJPO
7.10
Summary
"TPMFOPJEXJUISFTJTUBODFΩBOEJOEVDUBODFN)JTVTFEJOBOBVUP NPCJMFJHOJUJPODJSDVJUTJNJMBSUPUIBUJO'JH*GUIFCBUUFSZTVQQMJFT
7, determine: the final current through the solenoid when the switch
JTDMPTFE UIFFOFSHZTUPSFEJOUIFDPJM BOEUIFWPMUBHFBDSPTTUIFBJSHBQ BTTVNJOHUIBUUIFTXJUDIUBLFTμTUPPQFO
297
Example 7.22
Solution:
The final current through the coil is
7 * = @@T= @@@
= "
3
5IFFOFSHZTUPSFEJOUIFDPJMJT
×
-* = @@
× − × = N+
8 = @@
5IFWPMUBHFBDSPTTUIFHBQJT
∆*= × − × @@@@@@@@
= L7
7 = -@@@
∆U
× −
5IFTQBSLDPJMPGBOBVUPNPCJMFJHOJUJPOTZTUFNIBTBN)JOEVD UBODFBOEBΩSFTJTUBODF8JUIBTVQQMZWPMUBHFPG7 DBMDVMBUFUIF
time OFFEFEGPSUIFDPJMUPGVMMZDIBSHF UIFFOFSHZTUPSFEJOUIFDPJM BOE
UIFWPMUBHFEFWFMPQFEBUUIFTQBSLHBQJGUIFTXJUDIPQFOTJOμT
Answer:NT N+ BOEL7
7.10
Summary
5IFBOBMZTJTJOUIJTDIBQUFSJTBQQMJDBCMFUPBO ZDJSDVJUUIBUDBOCF
SFEVDFEUPBOFRVJ WBMFOUDJSDVJUDPNQSJTJOHBSFTJTUPSBOEBTJOHMF
FOFSHZTUPSBHFFMFNFOU JOEVDUPSPSDBQBDJUPS 4VDIBDJSDVJUJT
first-order because its behavior is described by a first-order difGFSFO
UJBMFRVBUJPO8IFOBOBMZ[JOH3$BOE3-DJSDVJUT POFNVTUBMXBZT
LFFQJONJOEUIBUUIFDBQBDJUPSJTBOPQFODJSDVJUUPTUFBEZTUBUFED
DPOEJUJPOTXIJMFUIFJOEVDUPSJTBTIPSUDJSDVJUUPTUFBEZTUBUFED
DPOEJUJPOT
5IFOBUVSBMSFTQPOTFJTPCUBJOFEXIFOOPJOEFQFOEFOUTPVSDFJT
QSFTFOU*UIBTUIFHFOFSBMGPSN
Y U = Y F −U∕τ
XIFSF YSFQSFTFOUT DVSSFOUUISPVHI PSW PMUBHF BDSPTT B SFTJTUPS B DBQBDJUPS PS BO JOEVDUPS BOE Y JT UIF JOJUJBM WBMVF PG Y#F DBVTFNPTUQSBDUJDBMSFTJTUPST DBQBDJUPST BOEJOEVDUPSTBMXBZTIBWF
MPTTFT UIFOBUVSBMSFTQPOTFJTBUSBOTJFOUSFTQPOTF JFJUEJFTPVU
XJUIUJNF
5IFUJNFDPOTUBOU τJTUIFUJNFSFRVJSFEGPSBSFTQPOTFUPEFDBZUP
∕FPGJUTJOJUJBMWBMVF'PS3$DJSDVJUT τ =3$BOEGPS3-DJSDVJUT τ =-∕3
Practice Problem 7.22
298
Chapter 7
First-Order Circuits
5IFTJOHVMBSJUZGVODUJPOTJODMVEFUIFVOJUTUFQ UIFVOJUSBNQGVOD UJPO BOEUIFVOJUJNQVMTFGVODUJPOT5IFVOJUTUFQGVODUJPOV U JT
VU =
{ U<
U>
5IFVOJUJNQVMTFGVODUJPOJT
{
δ U = Undefined,
5IFVOJUSBNQGVODUJPOJT
SU =
{ U
U<
U=
U>
U≤
U≥
5IFTUFBEZTUBUFSFTQPOTFJTUIFCFIB WJPSPGUIFDJSDVJUBGUFSBOJO EFQFOEFOUTPVSDFIBTCFFOBQQMJFEGPSBMPOHUJNF 5IFUSBOTJFOU
SFTQPOTFJTUIFDPNQPOFOUPGUIFDPNQMFUFSFTQPOTFUIBUEJFTPVU
XJUIUJNF
5IFUPUBMPSDPNQMFUFSFTQPOTFDPOTJTUTPGUIFTUFBEZTUBUFSFTQPOTF
BOEUIFUSBOTJFOUSFTQPOTF
5IFTUFQSFTQPOTFJTUIFSFTQPOTFPGUIFDJSDVJUUPBTVEEFOBQQMJDB
UJPOPGBEDDVSSFOUPSW oltage. Finding the step response of a firstPSEFSDJSDVJUSFRVJSFTUIFJOJUJBMW BMVF Y + ), the final vBMVF Y ∞ BOEUIFUJNFDPOTUBOU τ8JUIUIFTFUISFFJUFNT XFPCUBJOUIFTUFQ
SFTQPOTFBT
Y U = Y ∞ + <Y + −Y ∞ >F−U∕τ
"NPSFHFOFSBMGPSNPGUIJTFRVBUJPOJT
Y U = Y ∞ + <Y U+ −Y ∞ >F− U−U ∕τ
0SXFNBZXSJUFJUBT
*OTUBOUBOFPVTWBMVF = 'JOBM + <*OJUJBM−'JOBM>F− U−U ∕τ
14QJDFJTWFSZVTFGVMGPSPCUBJOJOHUIFUSBOTJFOUSFTQPOTFPGBDJSDVJU
'PVSQSBDUJDBMBQQMJDBUJPOTPG 3$BOE3-DJSDVJUTBSFBEFMBZDJSDVJU a photoflBTIVOJU BSFMBZDJSDVJU BOEBOBVUPNPCJMFJHOJUJPODJSDVJU
Review Questions
"O3$DJSDVJUIBT3 = ΩBOE$ = '5IFUJNF
DPOTUBOUJT
B T
E T
C T
T
B T
E T
D T
5IFUJNFDPOTUBOUGPSBO3-DJSDVJUXJUI3 = ΩBOE
- = )JT
B T
E T
C T
T
DBQBDJUPSWPMUBHFUPSFBDIQFSDFOUPGJUTTUFBEZ
TUBUFWBMVFJT
D T
"DBQBDJUPSJOBO3$DJSDVJUXJUI3 = ΩBOE
$ = 4 F is being charged. The time required for the
C T
D T
OPOFPGUIFBCPWF
"O3-DJSDVJUIBT3 = ΩBOE- = )5IFUJNF
OFFEFEGPSUIFJOEVDUPSDVSSFOUUPSFBDIQFSDFOU
PGJUTTUFBEZTUBUFWBMVFJT
B T
E T
C T
D T
OPOFPGUIFBCPWF
299
Problems
*OUIFDJSDVJUPG'JH UIFDBQBDJUPSWPMUBHFKVTU
CFGPSFU = JT
B 7
E 7
C 7
7
D 7
10 V
t=0
'PS3FWJFX2VFTUJPOTBOE
+
v(t)
–
2Ω
7F
'PS3FWJFX2VFTUJPOTBOE
*OUIFDJSDVJUJO'JH v ∞ JT
C 7
7
D 7
C "
"
D "
C "
"
D "
*GvTDIBOHFTGSPN7UP7BUU = XFNBZ
FYQSFTTvTBT
B δ U 7
C V U 7
D V −U + V U 7
E + V U 7
V U −7
5IFQVMTFJO'JH B DBOCFFYQSFTTFEJOUFSNT
PGTJOHVMBSJUZGVODUJPOTBT
'PSUIFDJSDVJUJO'JH UIFJOEVDUPSDVSSFOUKVTU
CFGPSFU = JT
B "
E "
*OUIFDJSDVJUPG'JH J ∞ JT
B "
E "
t=0
B 7
E 7
3Ω
Figure 7.80
Figure 7.79
5H
2Ω
10 A
3Ω
+
‒
i(t)
B V U + V U− 7
C V U −V U− 7
D V U −V U− 7
E V U + V U− 7
"OTXFSTE C D C E B D F D E C
Problems
Section 7.2
The Source-Free RC Circuit
'JOEUIFUJNFDPOTUBOUGPSUIF3$DJSDVJUJO'JH
*OUIFDJSDVJUTIPXOJO'JH
120 Ω
v U = F−U
7 U>
J U = F−UN" U>
50 V +
‒
B 'JOEUIFWBMVFTPG3BOE$
C $BMDVMBUFUIFUJNFDPOTUBOUτ
D %FUFSNJOFUIFUJNFSFRVJSFEGPSUIFWPMUBHFUP
EFDBZIBMGJUTJOJUJBMWBMVFBUU = %FUFSNJOFUIFUJNFDPOTUBOUGPSUIFDJSDVJUJO'JH
'PS1SPC
6 kΩ
+
v
‒
C
50 mF
'PS1SPC
i
Figure 7.81
80 Ω
Figure 7.82
R
12 Ω
50 pF
Figure 7.83
'PS1SPC
25 kΩ
40 kΩ
35 kΩ
300
Chapter 7
First-Order Circuits
5IFTXJUDIJO'JHIBTCFFOJOQPTJUJPO"GPSB
MPOHUJNF"TTVNFUIFTXJUDINPWFTJOTUBOUBOFPVTMZ
GSPN"UP#BUU = 'JOEvGPSU>
v = F−U
7
40 V +
‒
10 μF
2 kΩ
J = F−U
" BOE
+
v
‒
i
Figure 7.84
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETPVSDFGSFF3$DJSDVJUT
R1
i
R2
5IFTXJUDIJO'JHPQFOTBUU = 'JOEvPGPS
U>
R3
t=0
2 kΩ
C
Figure 7.85
15 V
'PS1SPC
Figure 7.88
'PS1SPC
t=0
v +
‒
+
v
‒
C
R
U>
B 'JOE3BOE$
C %FUFSNJOFUIFUJNFDPOTUBOU
D $BMDVMBUFUIFJOJUJBMFOFSHZJOUIFDBQBDJUPS
E 0CUBJOUIFUJNFJUUBLFTUPEJTTJQBUFQFSDFOUPG
UIFJOJUJBMFOFSHZ
5 kΩ A
B
'PSUIFDJSDVJUJO'JH JG
5IFTXJUDIJO'JHIBTCFFODMPTFEGPSBMPOH
UJNF BOEJUPQFOTBUU = 'JOEv U GPSU ≥ +
vo
‒
+
‒
4 kΩ
3 mF
Figure 7.89
'PS1SPC
For the circuit in Fig. 7.90, find vP U GPSU>%FUFS
NJOFUIFUJNFOFDFTTBSZGPSUIFDBQBDJUPSWPMUBHFUP
EFDBZUPPOFUIJSEPGJUTWBMVFBUU = t=0
10 kΩ
t=0
60 V +
‒
+
v(t)
‒
2 kΩ
9 kΩ
40 μF
36 V +
‒
3 kΩ
+
vo
‒
20 μF
Figure 7.86
'PS1SPC
Figure 7.90
'PS1SPC
"TTVNJOHUIBUUIFTXJUDIJO'JHIBTCFFOJO
QPTJUJPO"GPSBMPOHUJNFBOEJTNPWFEUPQPTJUJPO#
BUU = 5IFOBUU = TFDPOE UIFTXJUDINPWFTGSPN
#UP$'JOEv $ U GPSU ≥ Section 7.3
The Source-Free RL Circuit
For the circuit in Fig. 7.91, find JPGPSU>
10 kΩ A
15 V
Figure 7.87
'PS1SPC
+
‒
4Ω
B
500 kΩ
t=0
4H
io
C
2 mF
1 kΩ
24 V +
‒
Figure 7.91
'PS1SPC
4Ω
8Ω
301
Problems
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETPVSDFGSFF3-DJSDVJUT
t=0
%FUFSNJOFUIFUJNFDPOTUBOUGPSFBDIPGUIFDJSDVJUT
JO'JH
R1
v +
‒
R1
L
R2
L1
R2
i(t)
L2
R3
R3
R1
L
R2
(a)
Figure 7.92
(b)
Figure 7.96
'PS1SPC
'PS1SPC
*OUIFDJSDVJUPG'JH
$POTJEFSUIFDJSDVJUPG'JH'JOEvP U JG
J = "BOEv U = v U = F−
U7 U>
J U = F−
UN" U>
1Ω
B 'JOE3 - BOEτ
C $BMDVMBUFUIFFOFSHZEJTTJQBUFEJOUIFSFTJTUBODF
GPS<U<NT
i
+
‒
v(t)
+
v
‒
R
3Ω
+
i (t)
vo(t)
2H
‒
L
Figure 7.97
'PS1SPC
Figure 7.93
'PS1SPC
$BMDVMBUFUIFUJNFDPOTUBOUPGUIFDJSDVJUJO'JH
20 kΩ
'PSUIFDJSDVJUJO'JH EFUFSNJOFvP U XIFO
J = "BOEv U = 2Ω
10 kΩ
0.4 H
40 kΩ
5 mH
30 kΩ
+
i(t)
v(t)
+
‒
3Ω
vo(t)
‒
Figure 7.94
'PS1SPC
'JOEUIFUJNFDPOTUBOUGPSFBDIPGUIFDJSDVJUT
JO'JH
10 Ω
Figure 7.98
'PS1SPC
In the circuit of Fig. 7.99, find J U GPSU>JG
J = "
40 Ω
48 Ω
2Ω
5H
i
160 Ω
40 Ω
20 mH
10 Ω
(a)
Figure 7.95
'PS1SPC
6H
(b)
Figure 7.99
'PS1SPC
0.5i
40 Ω
302
Chapter 7
First-Order Circuits
'PSUIFDJSDVJUJO'JH
Section 7.4
−U
v = F
7
&YQSFTTUIFGPMMPXJOHTJHOBMTJOUFSNTPGTJOHVMBSJUZ
GVODUJPOT
BOE
J = F−U" +
v
‒
{
{
{
Figure 7.100
'PS1SPC
In the circuit of Fig. 7.101, find the value of 3GPS
XIJDIUIFTUFBEZTUBUFFOFSHZTUPSFEJOUIFJOEVDUPS
XJMMCF+
40 Ω
U<
U>
U<
<U<
<U<
U>
<U<
<U<
<U<
0UIFSXJTF
U−
D Y U =
−U
E Z U = −
i
L
{ −
−
C J U =
B v U =
U>
B 'JOE-BOE3
C %FUFSNJOFUIFUJNFDPOTUBOU
D $BMDVMBUFUIFJOJUJBMFOFSHZJOUIFJOEVDUPS
E 8IBUGSBDUJPOPGUIFJOJUJBMFOFSHZJTEJTTJQBUFE
in 10 ms?
R
Singularity Functions
U<
<U<
U>
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOETJOHVMBSJUZGVODUJPOT
&YQSFTTUIFTJHOBMTJO'JHJOUFSNTPG
TJOHVMBSJUZGVODUJPOT
R
60 V +
‒
80 Ω
2H
v1(t)
1
Figure 7.101
v2(t)
'PS1SPC
1
‒1
'JOEJ U BOEv U GPSU>JOUIFDJSDVJUPG'JH
JGJ = "
‒1
2H
0
20 Ω
1Ω
+
v(t)
‒
4
2
t
(b)
(a)
i(t)
5Ω
2
t
0
v3(t)
4
2
v4(t)
Figure 7.102
'PS1SPC
0
2
4
(c)
$POTJEFSUIFDJSDVJUJO'JH(JWFOUIBU
vP = 10 V, find vPBOEvYGPSU>
3Ω
+
vx
‒
Figure 7.103
'PS1SPC
1Ω
6
t
0
1
2
t
‒1
‒2
1
3H
2Ω
+
vo
‒
(d)
Figure 7.104
'PS1SPC
&YQSFTTv U JO'JHJOUFSNTPGTUFQGVODUJPOT
303
Problems
v(t)
'JOEUIFTPMVUJPOUPUIFGPMMPXJOHEJGGFSFOUJBM
FRVBUJPOT
B @@@
Ev+ v = v = −7
EU
C @@
EJ − J = J =
EU
15
10
5
‒1
‒5
0
1
2
3
t
4PMWFGPSvJOUIFGPMMPXJOHEJGGFSFOUJBMFRVBUJPOT TVCKFDUUPUIFTUBUFEJOJUJBMDPOEJUJPO
‒10
B Ev∕EU + v = V U C Ev∕EU−v = V U Figure 7.105
'PS1SPC
@@@
Ev+ v = EU
B 8IBUJTUIFUJNFDPOTUBOUPGUIFDJSDVJU
J U = [S U −S U− −V U− −S U− + S U− + V U U− >"
C 8IBUJTv ∞), the final vBMVFPGv
D *Gv = 2, find v U GPSU ≥ 4LFUDIUIFGPMMPXJOHGVODUJPOT
B Y U = F−UV U−
− U−
C Z U = F
V U
D [ U = DPTUδ U−
&WBMVBUFUIFGPMMPXJOHJOUFHSBMTJOWPMWJOHUIF
JNQVMTFGVODUJPOT
∫−∞
B U δ U− EU
∞
∫−∞
C U DPTπUδ U− EU
"DJSDVJUJTEFTDSJCFECZ
@@
EJ + J = V U
EU
'JOEJ U GPSU>HJWFOUIBUJ = Step Response of an RC Circuit
Section 7.5
$BMDVMBUFUIFDBQBDJUPSWPMUBHFGPSU<BOEU>GPS
FBDIPGUIFDJSDVJUTJO'JH
&WBMVBUFUIFGPMMPXJOHJOUFHSBMT
∞
∫−∞
v = −
"DJSDVJUJTEFTDSJCFECZ
4LFUDIUIFXBWFGPSNSFQSFTFOUFECZ
∞
v =
4Ω
B F−
U δ U− EU
∞
∫−∞
C <δ U + F−Uδ U + DPTπ Uδ U >EU
20 V
+
‒
&WBMVBUFUIFGPMMPXJOHJOUFHSBMT
+
v
‒
1Ω
2F
t=0
t
∫1
B V λ Eλ
(a)
∫0
C S U− EU
2F
∫1
D U− δ U− EU
5IFWPMUBHFBDSPTTBN)JOEVDUPSJT
δ U− N7'JOEUIFJOEVDUPSDVSSFOU BTTVNJOH
UIBUUIFJOEVDUPSJTJOJUJBMMZVODIBSHFE
&WBMVBUFUIFGPMMPXJOHEFSJWBUJWFT
B @@
E<V U− V U + >
EU
C @@
E<S U− V U− >
EU
D @@
E<TJOUV U− >
EU
+ v ‒
12 V +
‒
t=0
4Ω
2A
3Ω
(b)
Figure 7.106
'PS1SPC
'JOEUIFDBQBDJUPSWPMUBHFGPSU<BOEU>GPS
FBDIPGUIFDJSDVJUTJO'JH
304
Chapter 7
3Ω
First-Order Circuits
2Ω
t=0
12 V +
‒
4V +
‒
+
v
‒
3F
5IFTXJUDIJO'JHIBTCFFOJOQPTJUJPOBGPSB
MPOHUJNF"UU = JUNPWFTUPQPTJUJPOC$BMDVMBUF
J U GPSBMMU>
a
t=0
b
(a)
30 V +
‒
t=0
4Ω
2Ω
6A
60 V +
‒
3Ω
5F
'JOEvPJOUIFDJSDVJUPG'JHXIFO
vT = V U) V. Assume that vP = 7
10 kΩ
20 kΩ
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUIFTUFQSFTQPOTFPGBO3$
DJSDVJU
vs +
‒
C
Figure 7.112
+
vo
‒
'PS1SPC
'PSUIFDJSDVJUJO'JH JT U = V U 'JOEv U Figure 7.108
'PS1SPC
B *GUIFTXJUDIJO'JHIBTCFFOPQFOGPSB
MPOHUJNFBOEJTDMPTFEBUU = 0, find vP U C 4VQQPTFUIBUUIFTXJUDIIBTCFFODMPTFEGPSB
MPOHUJNFBOEJTPQFOFEBUU = 'JOEvP U 2Ω
30 V +
‒
4Ω
3F
+
vo
‒
Figure 7.110
'PS1SPC
3F
v
6Ω
+
‒
0.25 F
'PS1SPC
%FUFSNJOFv U GPSU> JOUIFDJSDVJUPG'JHJG
v = 12 Ω
30 Ω
i
80 V +
‒
is
Figure 7.113
'PS1SPC
$POTJEFSUIFDJSDVJUJO'JH'JOEJ U GPSU<
BOEU>
t=0
2Ω
t=0
Figure 7.109
40 Ω
+
vo
‒
3 µF
40 kΩ
t=0
R2
2F
'PS1SPC
Figure 7.107
v +
‒
i
Figure 7.111
+
v
‒
(b)
R1
6Ω
0.5i
2[u(t ‒ 1)] A
50 Ω
+ 24[u(t) ‒ u(t ‒ 1)] V
‒
8Ω
+ v(t) ‒
Figure 7.114
'PS1SPC
50 mF
305
Problems
'JOEv U BOEJ U JOUIFDJSDVJUPG'JH
R1
20 Ω
v +
‒
i
10 Ω
u(‒t) A
0.1 F
+
v
‒
i
t=0
L
R2
Figure 7.118
'PS1SPC
Figure 7.115
'PS1SPC
*GUIFXBWFGPSNJO'JH B JTBQQMJFEUPUIF
circuit of Fig. 7.116(b), find v U "TTVNFv = %FUFSNJOFUIFJOEVDUPSDVSSFOUJ U GPSCPUIU<BOE
U>GPSFBDIPGUIFDJSDVJUTJO'JH
2Ω
3Ω
is (A)
i
2
25 V +
‒
0
1
(a)
t=0
4H
(a)
t (s)
t=0
6Ω
is
4Ω
i
0.5 F
+
v
‒
'PS1SPC
'PS1SPCBOE3FWJFX2VFTUJPO
In the circuit of Fig. 7.117, find JYGPSU>-FU
3 = 3 = LΩ 3 = LΩ BOE$ = N'
0CUBJOUIFJOEVDUPSDVSSFOUGPSCPUIU<BOEU>
JOFBDIPGUIFDJSDVJUTJO'JH
R2
i
ix
R1
C
R3
4Ω
2A
12 Ω
4Ω
t=0
3.5 H
(a)
Figure 7.117
'PS1SPC
Section 7.6
3H
Figure 7.119
Figure 7.116
30 mA
2Ω
(b)
(b)
t=0
4Ω
6A
Step Response of an RL Circuit
3BUIFSUIBOBQQMZJOHUIFTIPSUDVUUFDIOJRVFVTFEJO
4FDUJPO VTF,7-UPPCUBJO&R 6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUIFTUFQSFTQPOTFPGBO3-
DJSDVJU
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
10 V
+
‒
24 V +
‒
i
2H
t=0
2Ω
6Ω
(b)
Figure 7.120
'PS1SPC
3Ω
306
Chapter 7
First-Order Circuits
'JOEv U GPSU<BOEU>JOUIFDJSDVJUPG
Fig. 7.121.
io
'JOEv U GPSU>JOUIFDJSDVJUPG'JHJGUIF
JOJUJBMDVSSFOUJOUIFJOEVDUPSJT[FSP
3Ω
8Ω
5Ω
4u(t) A
t=0
24 V +
‒
+
‒
4io
+
v
‒
2Ω
20 V +
‒
20 Ω
8H
Figure 7.125
'PS1SPC
*OUIFDJSDVJUJO'JH JTDIBOHFTGSPN"UP"
BUU = UIBUJT JT = V −U + V U 'JOEvBOEJ
Figure 7.121
i
'PS1SPC
5Ω
Figure 7.126
6Ω
20 Ω
0.5 H
+
v
‒
+ 20 V
‒
3Ω
6Ω
u(t ‒ 1) V +
‒
'PS1SPC
'JOEJ U BOEJ U GPSU>JOUIFDJSDVJUPG
Fig. 7.123.
i1
6Ω
'PSUIFDJSDVJUJO'JH DBMDVMBUFJ U JGJ = i
Figure 7.122
5A
0.5 H
t=0
2H
+ u(t) V
‒
Figure 7.127
'PS1SPC
i2
5Ω
20 Ω
2.5 H
4H
0CUBJOv U BOEJ U JOUIFDJSDVJUPG'JH
5Ω
i
20 Ω
0.5 H
10u(‒t) V +
‒
Figure 7.123
'PS1SPC
3FXPSL1SPCJGJ = "BOE
v U = V U 7
%FUFSNJOFUIFTUFQSFTQPOTFvP U UPJT = V U "JO
UIFDJSDVJUPG'JH
40 Ω
+ vo(t) ‒
20 Ω
'PS1SPC
%FUFSNJOFUIFWBMVFPGJ- U BOEUIFUPUBMFOFSHZ
EJTTJQBUFECZUIFDJSDVJUGSPNU = TFDUPU = ∞TFD
5IFWBMVFPG iJO(U) is equal to [6 – 6V(U)] A
20 Ω
10 Ω
10 H
iL(t)
Figure 7.124
'PS1SPC
+
v
‒
Figure 7.128
15 H
is
+
v
‒
'PS1SPC
t=0
12 Ω
4Ω
is
For the network shown in Fig. 7.122, find v U GPS
U>
2A
+
v
‒
0.5 H
Figure 7.129
'PS1SPC
iin(t)
307
Problems
t=0
*GUIFJOQVUQVMTFJO'JH B JTBQQMJFEUPUIF
DJSDVJUJO'JH C EFUFSNJOFUIFSFTQPOTFJ U vs(V )
+
‒
4V +
‒
5Ω
10 kΩ
10 kΩ
i
10
vs +
‒
0
1
20 Ω
2H
Figure 7.133
t(s)
'PS1SPC
For the op amp circuit in Fig. 7.134, find vP U GPS
U>
(b)
(a)
Figure 7.130
'PS1SPC
25 mF
10 kΩ
t=0
20 kΩ
100 kΩ
First-order Op Amp Circuits
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
students better understand first-order op amp
DJSDVJUT
+
‒
Section 7.7
+
vo
‒
25 μF
4V +
‒
+
vo
‒
R2
Figure 7.134
C
R1
%FUFSNJOFvPGPSU>XIFOvT = N7JOUIFPQ
BNQDJSDVJUPG'JH
+
‒
vs
'PS1SPC
t=0
+
+
‒
vo
+
‒
vo
‒
vs +
‒
Figure 7.131
5 μF
20 kΩ
'PS1SPC
*Gv = 10 V, find vP U GPSU>JOUIFPQBNQ
DJSDVJUJO'JH-FU3 = LΩBOE$ = •'
R
R
R +
v
‒
R
Figure 7.135
'PS1SPC
'PSUIFPQBNQDJSDVJUJO'JH TVQQPTFvT =
V U 7'JOEv U GPSU>
‒
+
100 kΩ
vo
100 kΩ
C
Figure 7.132
vs +
‒
'PS1SPC
Figure 7.136
0CUBJOvPGPSU>JOUIFDJSDVJUPG'JH
'PS1SPC
‒
+
20 kΩ
20 kΩ 100 µF
+
v(t)
‒
308
Chapter 7
First-Order Circuits
'JOEJPJOUIFPQBNQDJSDVJUJO'JH"TTVNF
UIBUv = −7 3 = LΩ BOE$ = •'
‒
+
Transient Analysis with PSpice
3FQFBU1SPCVTJOH14QJDFPS.VMUJ4JN
C
5IFTXJUDIJO'JHPQFOTBUU = 6TF14QJDF
PS.VMUJ4JNUPEFUFSNJOFv U GPSU>
io
+ v ‒
3u(t) +
‒
Section 7.8
t=0
R
+ v ‒
5Ω
100 mF
Figure 7.137
6Ω
4Ω
5A
+ 30 V
‒
20 Ω
'PS1SPC
'PSUIFPQBNQDJSDVJUPG'JH MFU3 = LΩ 3G = LΩ $ = μ' BOEv = 7'JOEW
C
+ v ‒
‒
+
'PS1SPC
5IFTXJUDIJO'JHNPWFTGSPNQPTJUJPOBUPC
BUU = 6TF14QJDFPS.VMUJ4JN to find J U GPSU>
Rf
R1
Figure 7.141
a
+
vo
4u(t) +
‒
6Ω
4Ω
t=0
‒
108 V +
‒
b
i(t)
3Ω
6Ω
2H
Figure 7.138
'PS1SPC
%FUFSNJOFvP U GPSU>JOUIFDJSDVJUPG'JH
-FUJT = V U μ"BOEBTTVNFUIBUUIFDBQBDJUPSJT
JOJUJBMMZVODIBSHFE
'PS1SPC
*OUIFDJSDVJUPG'JH EFUFSNJOFJP U 10 kΩ
2 μF
‒
+
50 kΩ
is
Figure 7.142
10 Ω
io
+
vo
‒
Figure 7.143
'PS1SPC
In the circuit of Fig. 7.140, find vPBOEJP HJWFOUIBU
vT = <− e−U>V U 7
10 Ω
vs +
‒
+
‒
10 µF
100 kΩ
'PS1SPC
In the circuit of Fig. 7.144, find the value of JPGPSBMM
WBMVFTPG<U
io
10 Ω
vo
io
100 kΩ
'PS1SPC
10 H
35u(t) V +
‒
Figure 7.139
Figure 7.140
10 Ω
2[1 ‒ u(t)] A
10 Ω
1A
25[1 ‒ u(t)] V +
‒
Figure 7.144
'PS1SPC
10 Ω
5Ω
50 mF
309
Comprehensive Problems
3FQFBU1SPCVTJOH14QJDFPS.VMUJ4JN
Section 7.9
Applications
*OEFTJHOJOHBTJHOBMTXJUDIJOHDJSDVJU JUXBTGPVOE
UIBUB•'DBQBDJUPSXBTOFFEFEGPSBUJNFDPO
TUBOUPGNT8IBUWBMVFSFTJTUPSJTOFDFTTBSZGPS
UIFDJSDVJU
"O3$DJSDVJUDPOTJTUTPGBTFSJFTDPOOFDUJPOPGB
7TPVSDF BTXJUDI B.ΩSFTJTUPS BOEB•'
DBQBDJUPS5IFDJSDVJUJTVTFEJOFTUJNBUJOHUIFTQFFE
PGBIPSTFSVOOJOHBLNSBDFUSBDL5IFTXJUDIDMPTFT
XIFOUIFIPSTFCFHJOTBOEPQFOTXIFOUIFIPSTF
crosses the finish line. Assuming that the capacitor
DIBSHFTUP7 DBMDVMBUFUIFTQFFEPGUIFIPSTF
"DBQBDJUPSXJUIBWBMVFPGm'IBTBMFBLBHF
SFTJTUBODFPG.Ɨ)PXMPOHEPFTJUUBLFUIFWPMUBHF
BDSPTTUIFDBQBDJUPSUPEFDBZUPPGUIFJOJUJBMWPMU
BHFUPXIJDIUIFDBQBDJUPSJTDIBSHFE "TTVNFUIBU
UIFDBQBDJUPSJTDIBSHFEBOEUIFOTFUBTJEFCZJUTFMG
"TJNQMFSFMBYBUJPOPTDJMMBUPSDJSDVJUJTTIPXOJO
Fig. 7.145. The neon lamp fires when its voltage
SFBDIFT7BOEUVSOTPGGXIFOJUTWPMUBHFESPQTUP
7*UTSFTJTUBODFJTΩ when on and infinitely
IJHIXIFOPGG
'JHVSFTIPXTBDJSDVJUGPSTFUUJOHUIFMFOHUIPG
UJNFWPMUBHFJTBQQMJFEUPUIFFMFDUSPEFTPGBXFMEJOH
NBDIJOF5IFUJNFJTUBLFOBTIPXMPOHJUUBLFTUIF
DBQBDJUPSUPDIBSHFGSPNUP78IBUJTUIFUJNF
SBOHFDPWFSFECZUIFWBSJBCMFSFTJTUPS
100 kΩ to 1 MΩ
2 μF
12 V
Welding
control
unit
Electrode
Figure 7.146
'PS1SPC
"7EDHFOFSBUPSFOFSHJ[FTBNPUPSXIPTFDPJM
IBTBOJOEVDUBODFPG)BOEBSFTJTUBODFPGΩ
A field discharge resistor of 400 ΩJTDPOOFDUFEJO
QBSBMMFMXJUIUIFNPUPSUPBWPJEEBNBHFUPUIFNP
UPS BTTIPXOJO'JH5IFTZTUFNJTBUTUFBEZ
TUBUF'JOEUIFDVSSFOUUISPVHIUIFEJTDIBSHFSFTJTUPS
NTBGUFSUIFCSFBLFSJTUSJQQFE
B 'PSIPXMPOHJTUIFMBNQPOFBDIUJNFUIFDBQBDJ
UPSEJTDIBSHFT
C 8IBUJTUIFUJNFJOUFSWal between light flBTIFT
Circuit breaker
4 MΩ
+
120 V
‒
6 μF
Neon lamp
Figure 7.145
120 V +
‒
Motor
400 Ω
Figure 7.147
'PS1SPC
'PS1SPC
Comprehensive Problems
5IFDJSDVJUJO'JH B DBOCFEFTJHOFEBT
BOBQQSPYJNBUFEJGGFSFOUJBUPSPSBOJOUFHSBUPS EFQFOEJOHPOXIFUIFSUIFPVUQVUJTUBLFOBDSPTT
the resistor or the capacitor, and also on the time
DPOTUBOUτ =3$PGUIFDJSDVJUBOEUIFXJEUI5PG
UIFJOQVUQVMTFJO'JH C 5IFDJSDVJUJTB
EJGGFSFOUJBUPSJGτ ≪ 5 TBZτ <5 PSBOJOUFHSB
UPSJGτ ≫5 TBZτ >5
B 8IBUJTUIFNJOJNVNQVMTFXJEUIUIBUXJMMBM
MPXBEJGGFSFOUJBUPSPVUQVUUPBQQFBSBDSPTTUIF
DBQBDJUPS
C *GUIFPVUQVUJTUPCFBOJOUFHSBUFEGPSNPGUIFJO
QVU XIBUJTUIFNBYJNVNWBMVFUIFQVMTFXJEUI
DBOBTTVNF
310
Chapter 7
First-Order Circuits
5IFDJSDVJUJO'JHJTVTFECZBCJPMPHZTUVEFOU
UPTUVEZiGSPHLJDLu4IFOPUJDFEUIBUUIFGSPHLJDLFE
BMJUUMFXIFOUIFTXJUDIXBTDMPTFECVULJDLFEWJP
MFOUMZGPSTXIFOUIFTXJUDIXBTPQFOFE.PEFMUIF
GSPHBTBSFTJTUPSBOEDBMDVMBUFJUTSFTJTUBODF"TTVNF
UIBUJUUBLFTN"GPSUIFGSPHUPLJDLWJPMFOUMZ
vi
300 kΩ
vi +
‒
Vm
200 pF
0
(a)
T
t
(b)
50 Ω
Figure 7.148
'PS1SPC
"O3-DJSDVJUNBZCFVTFEBTBEJGGFSFOUJBUPSJGUIF
PVUQVUJTUBLFOBDSPTTUIFJOEVDUPSBOEτ ≪5 TBZ
τ <5 XIFSF5JTUIFXJEUIPGUIFJOQVUQVMTF*G
3 is fixed at 200 kΩ EFUFSNJOFUIFNBYJNVNWBMVF
PG-SFRVJSFEUPEJGGFSFOUJBUFBQVMTFXJUI5 = •T
"OBUUFOVBUPSQSPCFFNQMPZFEXJUIPTDJMMPTDPQFT
XBTEFTJHOFEUPSFEVDFUIFNBHOJUVEFPGUIFJOQVU
WPMUBHFvJ by a factor of 10. As shown in Fig. 7.149,
UIFPTDJMMPTDPQFIBTJOUFSOBMSFTJTUBODF3TBOE
DBQBDJUBODF$T XIJMFUIFQSPCFIBTBOJOUFSOBMSF
TJTUBODF3Q*G3Q is fixed at 6 MΩ, find 3TBOE$TGPS
UIFDJSDVJUUPIBWFBUJNFDPOTUBOUPG•T
Probe
+
vi
‒
Figure 7.149
'PS1SPC
Scope
+
12 V
‒
Switch
Frog
2H
Figure 7.150
'PS1SPC
5PNPWFBTQPUPGBDBUIPEFSBZUVCFBDSPTTUIF
TDSFFOSFRVJSFTBMJOFBSJODSFBTFJOUIFWPMUBHFBDSPTT
the deflection plates, as shown in Fig. 7.151. Given
UIBUUIFDBQBDJUBODFPGUIFQMBUFTJTO' TLFUDIUIF
current flowing through the plates.
v(V)
10
+
Rp
Rs
Cs
vo
‒
Rise time = 2 ms
Drop time = 5 μs
(not to scale)
Figure 7.151
'PS1SPC
t
Second-Order
Circuits
&WFSZPOFXIPDBOFBSOBNBTUFSTEFHSFFJOFOHJOFFSJOHNVTUFBSOB
NBTUFSTEFHSFFJOFOHJOFFSJOHJOPSEFSUPNBYJNJ[FUIFTVDDFTTPGUIFJS
DBSFFS*GZPVXBOUUPEPSFTFBSDI TUBUFPGUIFBSUFOHJOFFSJOH UFBDIJOB
VOJWFSTJUZ PSTUBSUZPVSPXOCVTJOFTT ZPVSFBMMZOFFEUPFBSOB EPDUPSBM
EFHSFF
c h a p t e r
8
‡$IBSMFT,"MFYBOEFS
Enhancing Your Career
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&OHJOFFSTTIPVMEDPOTUBOUMZFEVDBUFUIFNTFMW FT GPSNBMMZBOE JOGPSNBMMZ UBLJOHBEWBOUBHFPGBMMNFBOTPGFEVDBUJPO1FSIBQTUIFSFJT
OPCFUUFSXBZUPFOIBODFZPVSDBSFFSUIBOUPKPJOBQSPGFTTJPOBMTPDJFUZ
TVDIBT*&&&BOECFBOBDUJWFNFNCFS
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TUBOEJOHZPVSHPBMT BEBQUJOHUPDIBOHFT BOUJDJQBUJOHPQQPSUVOJUJFT BOEQMBOOJOH
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311
312
Chapter 8
Second-Order Circuits
Learning Objectives
R
vs
L
+
‒
C
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
%FWFMPQBCFUUFSVOEFSTUBOEJOHPGUIFTPMVUJPOPGHFOFSBM
TFDPOEPSEFSEJGGFSFOUJBMFRVBUJPOT
-FBSOIPw to determine initial and final vBMVFT
6OEFSTUBOEUIFSFTQPOTFJOTPVSDFGSFFTFSJFT3-$DJSDVJUT
6OEFSTUBOEUIFSFTQPOTFJOTPVSDFGSFFQBSBMMFM3-$DJSDVJUT
6OEFSTUBOEUIFTUFQSFTQPOTFPGTFSJFT3-$DJSDVJUT
6OEFSTUBOEUIFTUFQSFTQPOTFPGQBSBMMFM3-$DJSDVJUT
6OEFSTUBOEHFOFSBMTFDPOEPSEFSDJSDVJUT
6OEFSTUBOEHFOFSBMTFDPOEPSEFSDJSDVJUTXJUIPQBNQT
(a)
8.1
is
R
C
L
(b)
R1
vs
+
‒
R2
L1
L2
(c)
R
is
C1
C2
(d)
Figure 8.1
5ZQJDBMFYBNQMFTPGTFDPOEPSEFSDJSDVJUT
B TFSJFT3-$DJSDVJU C QBSBMMFM3-$
DJSDVJU D 3-DJSDVJU E 3$DJSDVJU
Introduction
*OUIFQSFWJPVTDIBQUFSXFDPOTJEFSFEDJSDVJUTXJUIBTJOHMFTUPSBHFFMF
ment (a capacitor or an inductor). Such circuits are first-order because
UIFEJGferential equations describing them are first-order*OUIJTDIBQ UFSXFXJMMDPOTJEFSDJSDVJUTDPOUBJOJOHUXPTUPSBHFFMFNFOUT5IFTFBSF
LOPXOBTTFDPOEPSEFSDJSDVJUTCFDBVTFUIFJSSFTQPOTFTBSFEFTDSJCFECZ
EJGGFSFOUJBMFRVBUJPOTUIBUDPOUBJOTFDPOEEFSJWBUJWFT
5ZQJDBMF YBNQMFTPGTFDPOEPSEFSDJSDVJUTBSF 3-$DJSDVJUT JO
XIJDIUIFUISFFLJOETPGQBTTJWFFMFNFOUTBSFQSFTFOU&YBNQMFTPGTVDI
DJSDVJUTBSFTIPXOJO'JH B BOE C 0UIFSFYBNQMFTBSF3-BOE3$
DJSDVJUT BTTIPXOJO'JH D BOE E *UJTBQQBSFOUGSPN'JHUIBU
BTFDPOEPSEFSDJSDVJUNBZIBWFUXPTUPSBHFFMFNFOUTPGEJGGFSFOUUZQFPS
UIFTBNFUZQF QSPWJEFEFMFNFOUTPGUIFTBNFUZQFDBOOPUCFSFQSFTFOUFE
CZBOFRVJ WBMFOUTJOHMFFMFNFOU "OPQBNQDJSDVJUXJUIUX PTUPSBHF
FMFNFOUTNBZBMTPCFBTFDPOEPSEFSDJSDVJUAs with first-order circuits,
BTFDPOEPSEFSDJSDVJUNBZDPOUBJOTF WFSBMSFTJTUPSTBOEEFQFOEFOUBOE
JOEFQFOEFOUTPVSDFT
A second-order circuit is characterized by a second-order differential
equation. It consists of resistors and the equivalent of two energy storage
elements.
0VSBOBMZTJTPGTFDPOEPSEFSDJSDVJUTXJMMCFTJNJMBSUPUIBUVTFEGPS first-order8e will first consider circuits that are eYDJUFECZUIFJOJUJBM
DPOEJUJPOTPGUIFTUPSBHFFMFNFOUT"MUIPVHIUIFTFDJSDVJUTNBZDPOUBJO
EFQFOEFOUTPVSDFT UIFZBSFGSFFPGJOEFQFOEFOUTPVSDFT5IFTFTPVSDF
GSFFDJSDVJUTXJMMHJWFOBUVSBMSFTQPOTFTBTFYQFDUFE-BUFSXFXJMMDPO
TJEFSDJSDVJUTUIBUBSFF YDJUFECZJOEFQFOEFOUTPVSDFT 5IFTFDJSDVJUT XJMMHJWFCPUIUIFUSBOTJFOUSFTQPOTFBOEUIFTUFBEZTUBUFSFTQPOTF8F
DPOTJEFSPOMZEDJOEFQFOEFOUTPVSDFTJOUIJTDIBQUFS5IFDBTFPGTJOV
TPJEBMBOEFYQPOFOUJBMTPVSDFTJTEFGFSSFEUPMBUFSDIBQUFST
8FCFHJOCZMFBSOJOHIP XUP PCUBJOUIFJOJUJBMDPOEJUJPOTGPS UIF
DJSDVJUW BSJBCMFTBOEUIFJSEFSJ WBUJWFT BTUIJTJTDSVDJBMUPBOBMZ[JOH
TFDPOEPSEFSDJSDVJUT 5IFOXFDPOTJEFSTFSJFTBOEQBSBMMFM 3-$DJS
DVJUTTVDIBTTIPXOJO'JHGPSUIFUXPDBTFTPGFYDJUBUJPOCZJOJUJBM
8.2
313
Finding Initial and Final Values
DPOEJUJPOTPGUIFFOFS HZTUPSBHFFMFNFOUTBOECZTUFQJOQVUT-BUFS XFF YBNJOFPUIFSUZQFTPGTFDPOEPSEFSDJSDVJUT JODMVEJOHPQBNQ
DJSDVJUT 8FXJMMDPOTJEFS 14QJDFBOBMZTJTPGTFDPOEPSEFSDJSDVJUT 'JOBMMZ XFXJMMDPOTJEFSUIFBVUPNPCJMFJHOJUJPOTZTUFNBOETNPPUI JOHDJSDVJUTBTUZQJDBMBQQMJDBUJPOTPGUIFDJSDVJUTUSFBUFEJOUIJTDIBQUFS
Other applications such as resonant circuits and filters will be coWFSFE
JO$IBQUFS
8.2
Finding Initial and Final Values
1FSIBQTUIFNBKPSQSPCMFNTUVEFOUTG BDFJOIBOEMJOHTFDPOEPSEFS
circuits is finding the initial and final conditions on circuit vBSJBCMFT
Students are usually comfortable getting the initial and final vBMVFTPG
vBOEJCVUPGUFOIBWFEJGficulty finding the initial vBMVFTPGUIFJSEFSJWB
UJWFTdv∕dtBOEEJ∕EU'PSUIJTSFBTPO UIJTTFDUJPOJTF YQMJDJUMZEFWPUFE
UPUIFTVCUMFUJFTPGHFUUJOHv J Ev ∕EU EJ ∕EU J ∞ BOEv ∞ 6OMFTTPUIFSXJTFTUBUFEJOUIJTDIBQUFS vEFOPUFTDBQBDJUPSW PMUBHF XIJMFJJTUIFJOEVDUPSDVSSFOU
5IFSFBSFUXPLFZQPJOUTUPLFFQJONJOEJOEFUFSNJOJOHUIFJOJUJBM
DPOEJUJPOT
'JSTU‡BTBM XBZTJODJSDVJUBOBMZTJT‡XFNVTUDBSFGVMMZIBOEMFUIF QPMBSJUZPGWPMUBHFv U BDSPTTUIFDBQBDJUPSBOEUIFEJSFDUJPOPGUIFDVSSFOU
J U UISPVHIUIFJOEVDUPS,FFQJONJOEUIBUvBOEJBSFdefined TUSJDUMZBD
DPSEJOHUPUIFQBTTJWFTJHODPOWFOUJPO TFF'JHTBOE 0OFTIPVME
DBSFGVMMZPCTFSWFIPw these are defined and apply them accordingly
4FDPOE LFFQJONJOEUIBUUIFDBQBDJUPSWPMUBHFJTBMXBZTDPOUJOV
PVTTPUIBU
v + = v − B
BOEUIFJOEVDUPSDVSSFOUJTBMXBZTDPOUJOVPVTTPUIBU
J + = J − C
XIFSFU = −EFOPUFTUIFUJNFKVTUCFGPSFBTXJUDIJOHFWFOUBOEU = +JT
UIFUJNFKVTUBGUFSUIFTXJUDIJOHFWFOU BTTVNJOHUIBUUIFTXJUDIJOHFWFOU
UBLFTQMBDFBUU = Thus, in finding initial conditions, we first focus on those vBSJBCMFT
UIBUDBOOPUDIBOHFBCSVQUMZ DBQBDJUPSWPMUBHFBOEJOEVDUPSDVSSFOU CZ
BQQMZJOH&R 5IFGPMMPXJOHFYBNQMFTJMMVTUSBUFUIFTFJEFBT
Example 8.1
5IFTXJUDIJO'JHIBTCFFODMPTFEGPSBMPOHUJNF*UJTPQFOBUU = 'JOE B J + v + C EJ + ∕EU Ev + ∕EU D J ∞ v ∞ Solution:
4Ω
i
B *GUIFTXJUDIJTDMPTFEBMPOHUJNFCFGPSFU = JUNFBOTUIBUUIFDJS
DVJUIBTSFBDIFEEDTUFBEZTUBUFBUU = "UEDTUFBEZTUBUF UIFJOEVDUPS 12 V +‒
BDUTMJLFBTIPSUDJSDVJU XIJMFUIFDBQBDJUPSBDUTMJLFBOPQFODJSDVJU TP
XFIBWFUIFDJSDVJUJO'JH B BUU = −5IVT
= " Figure 8.2
J − = @@@@@
v − = J − = 7
'PS&YBNQMF
+
0.25 H
2Ω
t=0
0.1 F
+
v
‒
314
Chapter 8
4Ω
12 V
Second-Order Circuits
i
+
‒
4Ω
2Ω
i
4Ω
0.25 H
+ vL ‒
+
v
‒
12 V
+
‒
+
v
‒
0.1 F
+
12 V
+
‒
v
‒
(b)
(a)
i
(c)
Figure 8.3
&RVJWBMFOUDJSDVJUPGUIBUJO'JHGPS B U = − C U = + D U →∞
"TUIFJOEVDUPSDVSSFOUBOEUIFDBQBDJUPSWPMUBHFDBOOPUDIBOHFBCSVQUMZ
J + = J − = " v + = v − = 7
C "U U = + UIFTXJUDIJTPQFOUIFFRVJWBMFOUDJSDVJUJTBTTIPXOJO Fig. 8.3(b). The same current flows through both the inductor and ca
QBDJUPS)FODF
J$ + = J + = "
4JODF$Ev∕EU = J$ Ev∕EU = J$∕$ BOE
J$ +
Ev +
= 7T
@@@@@@
= @@@
= @@@@@
$
EU
4JNJMBSMZ TJODF-EJ∕EU = v- EJ∕EU = v-∕-8FOPXPCUBJOv-CZBQQMZ
JOH,7-UPUIFMPPQJO'JH C 5IFSFTVMUJT
− + J + + v- + + v + = PS
v- + = −− = 5IVT
v- +
EJ +
=
@@@@@
"T
= @@@@@@
= @@@@
EU
D 'PSU> UIFDJSDVJUVOEFSHPFTUSBOTJFODF#VUBT U → ∞ UIFDJSDVJU
SFBDIFTTUFBEZTUBUFBHBJO5IFJOEVDUPSBDUTMJLFBTIPSUDJSDVJUBOEUIF
DBQBDJUPSMJLFBOPQFODJSDVJU TPUIBUUIFDJSDVJUJO'JH C CFDPNFT
UIBUTIPXOJO'JH D GSPNXIJDIXFIBWF
J ∞ = " Practice Problem 8.1
v ∞ = 7
5IFTXJUDIJO'JHXBTPQFOGPSBMPOHUJNFCVUDMPTFEBU
U = %FUFSNJOF B J + v + C EJ + ∕EU Ev + ∕EU D J ∞ v ∞ t=0
10 Ω
2Ω
v
+
‒
1
20 F
0.4 H
i
+ 42 V
‒
Figure 8.4
'PS1SBDUJDF1SPC
Answer: B " 7 C "T 7T D " 7
8.2
315
Finding Initial and Final Values
Example 8.2
*OUIFDJSDVJUPG'JH DBMDVMBUF B J- + v$ + v3 + +
+
+
(b) EJ- ∕EU Ev$ ∕EU Ev3 ∕EU D J- ∞ v$ ∞ v3 ∞ 4Ω
2Ω
3u(t) A
1
2F
+
vR
‒
+
vC
‒
iL
0.6 H
+ 20 V
‒
Figure 8.5
'PS&YBNQMF
Solution:
B 'PS U< V U = "U U = − TJODFUIFDJSDVJUIBTSFBDIFETUFBEZ
TUBUF UIFJOEVDUPSDBOCFSFQMBDFECZBTIPSUDJSDVJU XIJMFUIFDBQBDJUPS
is replaced by an open circuit as shown in Fig. 8.6(a). From this figure
XFPCUBJO
J- − = v3 − = v$ − = − 7
−
"MUIPVHIUIFEFSJWBUJWFTPGUIFTFRVBOUJUJFTBUU = BSFOPUSFRVJSFE JU
JTFWJEFOUUIBUUIFZBSFBMM[FSP TJODFUIFDJSDVJUIBTSFBDIFETUFBEZTUBUF
BOEOPUIJOHDIBOHFT
4Ω
+
vR
a
iL
+
vC
‒
2Ω
+ vo ‒
+ 20 V
‒
4Ω
3A
2Ω
+
vR
‒
1
2F
(b)
Figure 8.6
5IFDJSDVJUJO'JHGPS B U = − C U = +
'PS U > V U = TPUIBUUIFDJSDVJUJTOP XFRVJWBMFOUUPUIBU
JO'JH C 4JODFUIFJOEVDUPSDVSSFOUBOEDBQBDJUPSW PMUBHFDBOOPU
DIBOHFBCSVQUMZ
J- + = J- − = v$ + = v$ − = −7
"MUIPVHIUIFWPMUBHFBDSPTTUIFΩSFTJTUPSJTOPUSFRVJSFE XFXJMMVTF
JUUPBQQMZ,7-BOE,$-MFUJUCFDBMMFEvP"QQMZJOH,$-BUOPEFBJO
'JH C HJWFT
vP +
v3 +
= @@@@@@
+ @@@@@
"QQMZJOH,7-UPUIFNJEEMFNFTIJO'JH C ZJFMET
−v3 + + vP + + v$ + + = iC
+
vC
‒
+ 20 V
‒
‒
(a)
b
iL
+
vL
‒
0.6 H
316
Chapter 8
Second-Order Circuits
4JODFv$ + = −7GSPN&R &R JNQMJFTUIBU
v3 + = vP + 'SPN&RT BOE XFPCUBJO
v3 + = vP + = 7
C 4JODF-EJ-∕EU = vEJ- +
v- +
@@@@@@
= @@@@@@
EU
#VUBQQMZJOH,7-UPUIFSJHIUNFTIJO'JH C HJWFT
v- + = v$ + + = )FODF
EJ +
@@@@@@
- = EU
4JNJMBSMZ TJODF $Ev$∕EU = J$ UIFO Ev$∕EU = J$∕$8FBQQMZ,$-BU
OPEFCJO'JH C UPHFUJ$
v +
@@@@@
P = J$ + + J- + 4JODFvP + = BOEJ- + = J$ + = ∕ = "5IFO
+
Ev
J$ +
$ = 7T
@@@@@@@
= @@@
= @@@@@
$
EU
5PHFUEv3 + ∕EU XFBQQMZ,$-UPOPEFBBOEPCUBJO
vP
v3
= @@
+ @@
5BLJOHUIFEFSJWBUJWFPGFBDIUFSNBOETFUUJOHU = +HJWFT
EvP +
Ev3 +
@@@@@@@
= + @@@@@@
EU
EU
8FBMTPBQQMZ,7-UPUIFNJEEMFNFTIJO'JH C BOEPCUBJO
−v3 + v$ + + vP = "HBJO UBLJOHUIFEFSJWBUJWFPGFBDIUFSNBOETFUUJOHU = +ZJFMET
Ev$ +
EvP +
Ev3 +
− @@@@@@@
+ @@@@@@@
+ @@@@@@
= EU
EU
EU
4VCTUJUVUJOHGPSEv$ + ∕EU = HJWFT
EvP +
Ev3 +
@@@@@@@
= + @@@@@@
EU
EU
'SPN&RT BOE XFHFU
Ev3 +
7T
@@@@@@@
= @@
EU
8.3
317
The Source-Free Series RLC Circuit
We can find EJ3 + ∕EUBMUIPVHIJUJTOPUSFRVJSFE4JODFv3 = J3
EJ3 +
Ev +
= @@
"T
@@@@@@@
@@@@@@
@@
3 = @@
= @@
EU
EU
D "T U → ∞ UIFDJSDVJUSFBDIFTTUFBEZTUBUF8FIBWFUIFFRVJWBMFOU
DJSDVJUJO'JH B FYDFQUUIBUUIF"DVSSFOUTPVSDFJTOPXPQFSBUJWF
#ZDVSSFOUEJWJTJPOQSJODJQMF
"
J- ∞ = @@@@@
= "
+
"
v3 ∞ = @@@@@
× = 7 v$ ∞ = −7
+
'Pr the circuit in Fig. 8.7, find: (a) J- + vC + v3 + C EJ- + ∕EU Ev$ + ∕EU Ev3 + ∕EU D J- ∞ v$ ∞ v3 ∞ + vR ‒
iR
iC
1
5F
4u(t) A
Practice Problem 8.2
iL
5Ω
+
vC
‒
+
vL
‒
2H
6A
Figure 8.7
'PS1SBDUJDF1SPC
Answer: B −" C 7T D −" 7 7
8.3
The Source-Free Series RLC Circuit
"OVOEFSTUBOEJOHPGUIFOBUVSBMSFTQPOTFPGUIFTFSJFT 3-$DJSDVJUJTB
necessary background for future studies in filter design and communica
UJPOTOFUXPSLT
$POTJEFSUIFTFSJFT 3-$DJSDVJUTIP XOJO'JH 5IFDJSDVJUJT
CFJOHFYDJUFECZUIFFOFSHZJOJUJBMMZTUPSFEJOUIFDBQBDJUPSBOEJOEVDUPS
5IFFOFSHZJTSFQSFTFOUFECZUIFJOJUJBMDBQBDJUPSW PMUBHF7BOEJOJUJBM
JOEVDUPSDVSSFOU*5IVT BUU = JEU = 7
v = @@
∫
$ −∞
J = *
L
I0
i
+
V0
‒
B
C
"QQMZJOH,7-BSPVOEUIFMPPQJO'JH
U J τ Eτ = 3J + -@@
EJ
+ @@
∫
EU $ −∞
R
Figure 8.8
"TPVSDFGSFFTFSJFT3-$DJSDVJU
C
318
Chapter 8
Second-Order Circuits
5PFMJNJOBUFUIFJOUFHSBM XFEJGGFSFOUJBUFXJUISFTQFDUUPUBOESFBSSBOHF
UFSNT8FHFU
EJ+ @@
EJ + @@@
J = 3 @@
@@@
-$
EU
EU
5IJTJTBTFDPOEPSEFSEJGGFSFOUJBMFRVBUJPOBOEJTUIFSFBTPOGPSDBMMJOH
UIF3-$DJSDVJUTJOUIJTDIBQUFSTFDPOEPSEFSDJSDVJUT0VSHPBMJTUPTPMW
F
&R 5PTPMW FTVDIBTFDPOEPSEFSEJG GFSFOUJBMFRVBUJPOSFRVJSFT
UIBUXFIBWFUXPJOJUJBMDPOEJUJPOT TVDIBTUIFJOJUJBMW BMVFPGJBOEJUT
first deriWBUJWFPSJOJUJBMW BMVFTPGTPNF JBOEv5IFJOJUJBMWBMVFPGJJT
HJWFOJO&R C 8FHFUUIFJOJUJBMW BMVFPGUIFEFSJ WBUJWFPG JGSPN
&RT B BOE UIBUJT
EJ @@@@@
3J + -
+ 7 = EU
PS
EJ 3* + 7 @@@@@
= −@@
EU
8JUIUIFUXPJOJUJBMDPOEJUJPOTJO&RT C BOE XFDBOOP X
TPMWF&R 0VSF xperience in the preceding chapter on first-order
DJSDVJUTTVHHFTUTUIBUUIFTPMVUJPOJTPGFYQPOFOUJBMGPSN4PXFMFU
J = "FTU
XIFSF"BOETBSFDPOTUBOUTUPCFEFUFSNJOFE4VCTUJUVUJOH&R JOUP
&R BOEDBSSZJOHPVUUIFOFDFTTBSZEJGGFSFOUJBUJPOT XFPCUBJO
"3 TFTU + @@@
"FTU = "TFTU + @@@
-$
PS
3T + @@@
= "FTU(T + @@
-$ )
4JODF J = "FTU is the assumed solution we are trying to find, only the
FYQSFTTJPOJOQBSFOUIFTFTDBOCF[FSP
See Appendix C.1 for the formula to
find the roots of a quadratic equation.
3T + @@@
= T + @@
-$
5IJTRVBESBUJDFRVBUJPOJTLOP XOBTUIF DIBSBDUFSJTUJDFRVBUJPOPGUIF
EJGGFSFOUJBM&R TJODFUIFSPPUTPGUIFFRVBUJPOEJDUBUFUIFDIBSBDUFS
PGJ5IFUXPSPPUTPG&R BSF
3+
T = −@@@
-
___________
3) − @@@
√(@@@
-$
___________
(@@@
3) − @@@
-$
"NPSFDPNQBDUXBZPGFYQSFTTJOHUIFSPPUTJT
√
C
@@@@@@@
@@@@@@@
T = −α + Ŀα − ω
T = −α − Ŀα − ω
3−
T = −@@@
-
B
8.3
The Source-Free Series RLC Circuit
319
XIFSF
3
α = @@@
-
@@@
ω = @@@@
Ŀ
-$ 5IFSPPUTTBOETBSFDBMMFEOBUVSBMGSFRVFODJFT NFBTVSFEJOOFQFST
QFSTFDPOE /QT CFDBVTFUIFZBSFBTTPDJBUFEXJUIUIFOBUVSBMSFTQPOTF
PGUIFDJSDVJU ωJTLOP XOBTUIF SFTPOBOUGSFRVFODZPSTUSJDUMZBTUIF
VOEBNQFEOBUVS BMGSFRVFODZ F YQSFTTFEJOSBEJBOTQFSTFDPOE SBET BOEαJTUIFOFQFSGSFRVFODZFYQSFTTFEJOOFQFSTQFSTFDPOE*OUFSNTPG
αBOEω &R DBOCFXSJUUFOBT
T + αT + ω = B
5IFWBSJBCMFT TBOE ωBSFJNQPSUBOURVBOUJUJFTXFXJMMCFEJTDVTTJOH
UISPVHIPVUUIFSFTUPGUIFUFYU
5IFUXPWBMVFTPGTJO&R JOEJDBUFUIBUUIFSFBSFUXPQPTTJCMF
TPMVUJPOTGPS J FBDIPGXIJDIJTPGUIFGPSNPGUIFBTTVNFETPMVUJPOJO
&R UIBUJT
J = "FTU The neper (Np) is a dimensionless unit
named after John Napier (1550–1617),
a Scottish mathematician.
J = "FTU
The ratio αω0 is known as the damping
ratio ζ.
4JODF&R JTBMJOFBSFRVBUJPO BO ZMJOFBSDPNCJOBUJPOPGUIFUX P
EJTUJODUTPMVUJPOTJBOEJJTBMTPBTPMVUJPOPG&R "DPNQMFUFPS
UPUBMTPMVUJPOPG&R XPVMEUIFSFGPSFSFRVJSFBMJOFBSDPNCJOBUJPOPG
JBOEJ5IVT UIFOBUVSBMSFTQPOTFPGUIFTFSJFT3-$DJSDVJUJT
J U = "FTU + "FTU
XIFSFUIFDPOTUBOUT"BOE"BSFEFUFSNJOFEGSPNUIFJOJUJBMWBMVFTJ BOEEJ ∕EUJO&RT C BOE 'SPN&R XFDBOJOGFSUIBUUIFSFBSFUISFFUZQFTPGTPMVUJPOT
*Gα > ω XFIBWFUIFPWFSEBNQFEDBTF
*Gα = ω XFIBWFUIFDSJUJDBMMZEBNQFEDBTF
*Gα < ω XFIBWFUIFVOEFSEBNQFEDBTF
8FXJMMDPOTJEFSFBDIPGUIFTFDBTFTTFQBSBUFMZ
Overdamped Case (α > ω)
'SPN&RT BOE α>ωJNQMJFT$>-∕38IFOUIJTIBQ
QFOT CPUISPPUTTBOETBSFOFHBUJWFBOESFBM5IFSFTQPOTFJT
J U = "FTU + "FTU
XIJDIEFDBZTBOEBQQSPBDIFT[FSPBTUJODSFBTFT'JHVSF B JMMVTUSBUFT
BUZQJDBMPWFSEBNQFESFTQPOTF
Critically Damped Case (α = ω)
8IFOα = ω $ = -∕3BOE
3
T = T = − α = −@@@
-
The response is overdamped when
the roots of the circuit’s characteristic
equation are unequal and real, critically
damped when the roots are equal and
real, and underdamped when the
roots are complex.
320
Chapter 8
i(t)
'PSUIJTDBTF &R ZJFMET
Second-Order Circuits
J U = "F−αU + "F−αU = "F−αU
0
t
XIFSF" = " + "5IJTDBOOPUCFUIFTPMVUJPO CFDBVTFUIFUXPJOJUJBM
conditions cannot be satisfied with the single constant "8IBU UIFO
DPVMECFXSPOH 0VSBTTVNQUJPOPGBOF YQPOFOUJBMTPMVUJPOJTJODPS rect for the special case of critical damping. Let us go back to Eq. 8IFOα = ω = 3∕- &R CFDPNFT
EJ+ α@@
EJ + αJ = @@@
EU
EU
(a)
PS
i(t)
E
@@
EJ + αJ)+ α(
@@
EJ + αJ)= @@
EU
EU( EU
EJ + αJ
G = @@
EU
*GXFMFU
0
1
α
UIFO&R CFDPNFT
t
EG
@@
+ αG = EU
(b)
i(t)
which is a first-order difGFSFOUJBMFRVBUJPOXJUITPMVUJPOG = "F−αU XIFSF
"JTBDPOTUBOU&RVBUJPO UIFOCFDPNFT
e ‒t
EJ + αJ = " F−αU
@@
EU
0
t
2π
ωd
(c)
Figure 8.9
B 0WFSEBNQFESFTQPOTF C DSJUJDBMMZ
EBNQFESFTQPOTF D VOEFSEBNQFESF
TQPOTF
PS
5IJTDBOCFXSJUUFOBT
FαU@@
EJ + FαUαJ = "
EU
E FαUJ = " @@
EU
*OUFHSBUJOHCPUITJEFTZJFMET
FαUJ = "U + "
PS
J = "U + " F−αU
XIFSF"JTBOPUIFSDPOTUBOU)FODF UIFOBUVSBMSFTQPOTFPGUIFDSJUJDBMMZ
EBNQFEDJSDVJUJTBTVNPGUXPUFSNTBOFHBUJWFFYQPOFOUJBMBOEBOFHB
UJWFFYQPOFOUJBMNVMUJQMJFECZBMJOFBSUFSN PS
J U = " + "U F−αU
"UZQJDBMDSJUJDBMMZEBNQFESFTQPOTFJTTIP XOJO'JH C *OG BDU Fig. C JTBTL FUDIPG J U = UF−αU XIJDISFBDIFTBNBYJNVNW BMVFPG
F−∕αBUU = ∕α POFUJNFDPOTUBOU BOEUIFOEFDBZTBMMUIFXBZUP[FSP
8.3
The Source-Free Series RLC Circuit
321
Underdamped Case (α < ω)
'PSα<ω $<-∕35IFSPPUTNBZCFXSJUUFOBT
@@@@@@@@@@
T = −α + Ŀ− ω − α = −α + KωE
B
T = −α − Ŀ
C
@@@@@@@@@@
− ω − α = −α − KωE
_______
@@@
XIFSFK = Ŀ −BOE
ωE = √ ω− α XIJDIJTDBMMFEUIFEBNQFEGSFRVFODZ
#PUIωBOEωEBSFOBUVSBMGSFRVFODJFTCFDBVTFUIFZIFMQEFUFSNJOFUIF
OBUVSBMSFTQPOTFXIJMF ωJTPGUFODBMMFEUIF VOEBNQFEOBUVS BMGS F
RVFODZ ωEJTDBMMFEUIFEBNQFEOBUVSBMGSFRVFODZ5IFOBUVSBMSFTQPOTF
JT
J U = "F− α−KωE U + "F− α + KωE U
= F−αU "F−KωEU + "F−KωEU 6TJOH&VMFSTJEFOUJUJFT
XFHFU
FKθ = DPTθ + KTJOθ F−Kθ = DPTθ − KTJOθ
J U = F−αU<" DPTωEU + KTJOωEU + " DPTωEU − KTJOωEU >
= F−αU< " + " DPTωEU + K " − " TJOωEU>
3FQMBDJOHDPOTUBOUT " + " BOE K "−" XJUIDPOTUBOUT #BOE # XFXSJUF
J U = F−αU #DPTωEU + #TJOωEU
8JUIUIFQSFTFODFPGTJOFBOEDPTJOFGVODUJPOT JUJTDMFBSUIBUUIF
OBUVSBMSFTQPOTFGPSUIJTDBTFJTF YQPOFOUJBMMZEBNQFEBOEPTDJMMBUPSZ
JOOBUVSF 5IFSFTQPOTFIBTBUJNFDPOTUBOUPG ∕αBOEBQFSJPEPG 5 = π∕ωE'JHVSF D EFQJDUTBUZQJDBMVOEFSEBNQFESFTQPOTF1BSU
B BOE C PG'JHBTTVNFGPSFBDIDBTFUIBUJ = 0ODFUIFJOEVDUPS DVSSFOUJ U JTGPVOEGPSUIF3-$TFSJFTDJSDVJUBT
TIPXOBCPWF PUIFSDJSDVJURVBOUJUJFTTVDIBTJOEJ WJEVBMFMFNFOUWPMUBHFT
DBOFBTJMZCFGPVOE' PSFYBNQMF UIFSFTJTUPSW PMUBHFJTv3 = 3J BOEUIF
JOEVDUPSWPMUBHFJTv- = -EJ∕EU5IFJOEVDUPSDVSSFOUJ U JTTFMFDUFEBTUIF
LFZWariable to be determined first in order to takFBEWantage of Eq. C 8FDPODMVEFUIJTTFDUJPOCZOPUJOHUIFGPMMP XJOHJOUFSFTUJOH QFDV
MJBSQSPQFSUJFTPGBO3-$OFUXPSL
5IFCFIBWJPSPGTVDIBOFUXPSLJTDBQUVSFECZUIFJEFBPGEBNQJOH XIJDIJTUIFHSBEVBMMPTTPGUIFJOJUJBMTUPSFEFOFS HZ BTF WJEFODFE
CZUIFDPOUJOVPVTEFDSFBTF JOUIFBNQMJUVEFPGUIFSFTQPOTF 5IF
EBNQJOHFG GFDUJTEVFUPUIFQSFTFODFPGSFTJTUBODF 35IF OFQFS
GSFRVFODZαEFUFSNJOFTUIFSBUFBUXIJDIUIFSFTQPOTFJTEBNQFE
@@@
Ŀ -$BTUIF
*G3 = UIFOα = BOEXFIBWFBO-$DJSDVJUXJUI∕
VOEBNQFEOBUVSBMGSFRVFODZ4JODFα < ωJOUIJTDBTF UIFSFTQPOTF
JTOPUPOMZVOEBNQFEC VUBMTPPTDJMMBUPSZ5IFDJSDVJUJTTBJEUPCF
MPTTMFTT CFDBVTFUIFEJTTJQBUJOHPSEBNQJOHFMFNFOU 3 JTBCTFOU
#ZBEKVTUJOHUIFWBMVFPG3 UIFSFTQPOTFNBZCFNBEFVOEBNQFE PWFSEBNQFE DSJUJDBMMZEBNQFE PSVOEFSEBNQFE
0TDJMMBUPSZSFTQPOTFJTQPTTJCMFEVFUPUIFQSFTFODFPGUIFUX
P
UZQFTPGTUPSBHFFMFNFOUT)B WJOHCPUI -BOE $BMMPws the floX
R = 0 produces a perfectly sinusoidal
response. This response cannot be
practically accomplished with L and C
because of the inherent losses in them.
See Figs 6.8 and 6.26. An electronic
device called an oscillator can
produce a perfectly sinusoidal
response.
Examples 8.5 and 8.7 demonstrate the
effect of varying R.
The response of a second-order circuit
with two storage elements of the same
type, as in Fig. 8.1(c) and (d), cannot
be oscillatory.
322
What this means in most practical
circuits is that we seek an overdamped
circuit that is as close as possible to
a critically damped circuit.
Example 8.3
Chapter 8
Second-Order Circuits
PGFOFSHZCBDLBOEGPSUICFUXFFOUIFUX P5IFEBNQFEPTDJMMBUJPO
FYIJCJUFECZUIFVOEFSEBNQFESFTQPOTFJTLOP XOBT SJOHJOH*U
TUFNTGSPNUIFBCJMJUZPGUIFTUPSBHFFMFNFOUT -BOE $UPUSBOTGFS
FOFSHZCBDLBOEGPSUICFUXFFOUIFN
0CTFSWFGSPN'JHUIBUUIFX BWFGPSNTPGUIFSFTQPOTFTEJG GFS
*OHFOFSBM JUJTEJG ficult to tell from the wBWFGPSNTUIFEJG GFSFODF
CFUXFFOUIFPWFSEBNQFEBOEDSJUJDBMMZEBNQFESFTQPOTFT5IFDSJUJ
DBMMZEBNQFEDBTFJTUIFCPSEFSMJOFCFUXFFOUIFVOEFSEBNQFEBOE
PWFSEBNQFEDBTFTBOEJUEFDBZTUIFGBTUFTU8JUIUIFTBNFJOJUJBMDPO
EJUJPOT UIFPWFSEBNQFEDBTFIBTUIFMPOHFTUTFUUMJOHUJNF CFDBVTF
it UBL FTUIFMPOHFTUUJNFUPEJTTJQBUFUIFJOJUJBMTUPSFE F OFSHZ*G
we desire the response that approaches the final vBMVFNPTUSBQJEMZ
XJUIPVUPTDJMMBUJPOPSSJOHJOH UIFDSJUJDBMMZEBNQFEDJSDVJUJTUIF
SJHIUDIPJDF
*O'JH 3 = Ω - = ) BOE $ = ∕'$BMDVMBUFUIFDIBSBD UFSJTUJDSPPUTPGUIFDJSDVJU*TUIFOBUVSBMSFTQPOTFP WFSEBNQFE VOEFS
EBNQFE PSDSJUJDBMMZEBNQFE
Solution:
We first calculate
= 3= @@@@
α = @@@
- 5IFSPPUTBSF
= @@@@@@
___
@@@@@
ω = _____
= Ŀ -$ × @@
Ŀ
@@@@@@@
@@@@@@
α − ω
= − ± √ − T = −α ± √
PS
T = − T = −
4JODFα > ω XFDPODMVEFUIBUUIFSFTQPOTFJTPWFSEBNQFE5IJTJTBMTP
FWJEFOUGSPNUIFGBDUUIBUUIFSPPUTBSFSFBMBOEOFHBUJWF
Practice Problem 8.3
*G3 = Ω - = ) BOE$ = N'JO'JH find α ω T BOET8IBU
UZQFPGOBUVSBMSFTQPOTFXJMMUIFDJSDVJUIBWF
Answer: − ± K VOEFSEBNQFE
Example 8.4
'JOEJ U JOUIFDJSDVJUPG'JH "TTVNFUIBUUIFDJSDVJUIBTSFBDIFE
TUFBEZTUBUFBUU = −
Solution:
'PSU< UIFTXJUDIJTDMPTFE5IFDBQBDJUPSBDUTMJLFBOPQFODJSDVJU
XIJMFUIFJOEVDUPSBDUTMJLFBTIVOUFEDJSDVJU5IFFRVJWBMFOUDJSDVJUJT
TIPXOJO'JH B 5IVT BUU = = " J = @@@@@
+
v = J = 7
8.3
t=0
4Ω
10 V
+
‒
i
i
0.02 F
323
The Source-Free Series RLC Circuit
i
4Ω
+
v
‒
6Ω
+
‒
10 V
3Ω
+
v
‒
0.5 H
Figure 8.10
6Ω
0.02 F
(a)
'PS&YBNQMF
5IFDJSDVJUJO'JH B GPSU< C GPSU> XIFSFJ JTUIFJOJUJBMDVSSFOUUISPVHIUIFJOEVDUPSBOEv JTUIFJOJUJBM
WPMUBHFBDSPTTUIFDBQBDJUPS
'PS U > UIFTXJUDIJTPQFOFEBOEUIFW PMUBHFTPVSDFJTEJTDPO OFDUFE5IFFRVJWBMFOUDJSDVJUJTTIPXOJO'JH C XIJDIJTBTPVSDF
GSFFTFSJFT3-$DJSDVJU/PUJDFUIBUUIFΩBOEΩSFTJTUPST XIJDIBSF
JOTFSJFTJO'JHXIFOUIFTXJUDIJTPQFOFE IBWFCFFODPNCJOFEUP
HJWF3 = ΩJO'JH C 5IFSPPUTBSFDBMDVMBUFEBTGPMMPXT
()
= @@@@@@@
@@@
@@@@@@
ω = @@@@
=
Ŀ
-$ @@ × @@@
Ŀ
@@@@@@@
@@@@@@@@
Ŀ
α − ω
= − ± − T = −α ± √
PS
T = − ± K
)FODF UIFSFTQPOTFJTVOEFSEBNQFE α < ω UIBUJT
J U = F−U "DPTU + "TJOU 8FOPXPCUBJO"BOE"VTJOHUIFJOJUJBMDPOEJUJPOT"UU = J = = "
'SPN&R |
EJ = −@@
<3J + v > = −< − > = −"T
@@
EU U= /PUFUIBUv = 7 = −7JTVTFE CFDBVTFUIFQPMBSJUZPG
v in Fig. C JTPQQPTJUFUIBUJO'JH5BLJOHUIFEFSJWBUJWFPGJ U JO&R EJ = −F−U " DPTU + " TJOU
@@
EU
−U
+ F −"TJOU + "DPTU
*NQPTJOHUIFDPOEJUJPOJO&R BUU = HJWFT
− = − " + + − + "
#VU" = GSPN&R 5IFO
− = − + "
⇒
" = 4VCTUJUVUJOHUIFW BMVFTPG "BOE "JO&R ZJFMETUIFDPN QMFUFTPMVUJPOBT
J U = F−U DPTU + TJOU "
0.5 H
(b)
Figure 8.11
3= @@@@@
α = @@@
= - @@ 9Ω
+
v
‒
324
Chapter 8
Practice Problem 8.4
10 Ω
a
5IFDJSDVJUJO'JHIBTSFBDIFETUFBEZTUBUFBU U = −*GUIFNBLF
CFGPSFCSFBLTXJUDINPWFTUPQPTJUJPO CBU U = DBMDVMBUF J U GPS
U> 1
9F
b
t=0
100 V
Second-Order Circuits
Answer:F−U DPTU − TJOU "
i(t)
+
‒
5Ω
1H
Figure 8.12
8.4
'PS1SBDUJDF1SPC
v
+
R
v
‒
Figure 8.13
+
L
I0 v
C
‒
"TPVSDFGSFFQBSBMMFM3-$DJSDVJU
+
V0
‒
The Source-Free Parallel RLC Circuit
1BSBMMFM3-$ circuits find manZQSBDUJDBMBQQMJDBUJPOT OPUBCMZJO DPN
NVOJDBUJPOTOFUXorks and filter designs.
$POTJEFSUIFQBSBMMFM3-$DJSDVJUTIPXOJO'JH"TTVNFJOJUJBM
JOEVDUPSDVSSFOU*BOEJOJUJBMDBQBDJUPSWPMUBHF7
J = * = @@
v U EU
∫
- −∞
B
v = 7
C
#FDBVTFUIFUISFFFMFNFOUTBSFJOQBSBMMFM UIF ZIBWFUIFTBNFWPMUBHFv
BDSPTTUIFN"DDPSEJOHUPQBTTJWFTJHODPOWFOUJPO UIFDVSSFOUJTFOUFS JOHFBDIFMFNFOUUIBUJT UIFDVSSFOUUISPVHIFBDIFMFNFOUJTMFBWJOHUIF
UPQOPEF5IVT BQQMZJOH,$-BUUIFUPQOPEFHJWFT
v + Ev= U v τ Eτ + $
@@
@@
@@@
∫
3
- −∞
EU
5BLJOHUIFEFSJWBUJWFXJUISFTQFDUUPUBOEEJWJEJOHCZ$SFTVMUTJO
Ev
v = @@@
@@@
+ @@@
Ev
+ @@@
EU 3$ EU -$
8e obtain the characteristic equation by replacing the first deriWBUJWF
CZTBOEUIFTFDPOEEFSJWBUJWFCZT#ZGPMMPXJOHUIFTBNFSFBTPOJOH
VTFEJOFTUBCMJTIJOH&RT UISPVHI UIFDIBSBDUFSJTUJDFRVBUJPO
JTPCUBJOFEBT
T + @@@
= T + @@@
3$
-$
5IFSPPUTPGUIFDIBSBDUFSJTUJDFRVBUJPOBSF
@@@@@@@@@@@@
± @@@@
−@@@
T = −@@@@
(3$
) -$ 3$
ð
PS
@@@@@@@
α − ω
T = −α ± √
XIFSF
α = @@@@
3$
@@@ ω = @@@@
Ŀ
-$
8.4
The Source-Free Parallel RLC Circuit
5IFOBNFTPGUIFTF UFSNTSFNBJOUIFTBNFBTJOUIFQSFDFEJOH TFDUJPO BTUIFZQMBZUIFTBNFSPMFJOUIFTPMVUJPO"HBJO UIFSFBSFUISFFQPTTJCMF
TPMVUJPOT EFQFOEJOHPOXIFUIFSα > ω α = ω PSα < ω-FUVTDPO
TJEFSUIFTFDBTFTTFQBSBUFMZ
Overdamped Case (α > ω)
'SPN&R α > ωXIFO-> 3$5IFSPPUTPGUIFDIBSBDUFSJTUJD
FRVBUJPOBSFSFBMBOEOFHBUJWF5IFSFTQPOTFJT
v U = "FTU + "FTU
Critically Damped Case (α = ω)
'PSα = ω - = 3$5IFSPPUTBSFSFBMBOEFRVBMTPUIBUUIFSFTQPOTFJT
v U = " + "U F−αU
Underdamped Case (α < ω)
8IFOα < ω -< 3$*OUIJTDBTFUIFSPPUTBSFDPNQMFYBOENBZCF
FYQSFTTFEBT
XIFSF
T = −α ± KωE
_______
ωE = √ω− α 5IFSFTQPOTFJT
v U = F−αU "DPTωEU + "TJOωEU
5IFDPOTUBOUT "BOE "JOFBDIDBTFDBOCFEFUFSNJOFEGSPNUIF
JOJUJBMDPOEJUJPOT 8F OFFE v BOE Ev ∕EUThe first UFSNJT LOPXO
GSPN&R C 8e find the second term by combining Eqs. (8.27) and
BT
7
Ev @@@
@@@@@
+ * + $
= 3
EU
PS
7 + 3*
Ev @@@@@
= −@@@@@@@@@
3$
EU
5IFWPMUBHFXBWFGPSNTBSFTJNJMBSUPUIPTFTIP XOJO'JHBOEXJMM
EFQFOEPOXIFUIFSUIFDJSDVJUJTPWFSEBNQFE VOEFSEBNQFE PSDSJUJDBMMZ
EBNQFE
)BWJOH GPVOE UIFDBQBDJUPS WPMUBHF v U GPS UIF QBSBMMFM 3-$ DJS
DVJUBTTIPXOBCPWF XFDBOSFBEJMZPCUBJOPUIFSDJSDVJURVBOUJUJFTTVDI
BTJOEJ WJEVBMFMFNFOUDVSSFOUT' PSF YBNQMF UIFSFTJTUPSDVSSFOUJT J3 = v∕3BOEUIFDBQBDJUPSDVSSFOUJTJ$ = $Ev∕EU8FIBWFTFMFDUFEUIF
DBQBDJUPSWPMUBHFv U BTUIFL FZWariable to be determined first in order
UPUBLFBEWantage of Eq. (8.1a). Notice that we first found the inductor
DVSSFOUJ U GPSUIF3-$ series circuit, whereas we first found the capaciUPSWPMUBHFv U GPSUIFQBSBMMFM3-$DJSDVJU
325
326
Example 8.5
Chapter 8
Second-Order Circuits
In the parallel circuit of Fig. 8.13, find v U GPSU> BTTVNJOHv = 5 7 J = - = ) BOE$ = N'$POTJEFSUIFTFDBTFT3 = 1.923 Ω 3 = Ω BOE3 = Ω
Solution:
■ CASE 1 *G3 = Ω
= @@@@@@@@@@@@@@@@@@@
α = @@@@
=
3$ × × × −
= @@@@@@@@@@@@@
@@@
@@@@@@@@@@@@
ω = @@@@
= Ŀ
-$ Ŀ × × −
4JODFα > ωJOUIJTDBTF UIFSFTQPOTFJTPWFSEBNQFE5IFSPPUTPGUIF
DIBSBDUFSJTUJDFRVBUJPOBSF
@@@@@@@
T = −α ± √
α − ω
= − − BOEUIFDPSSFTQPOEJOHSFTQPOTFJT
v U = "F−U + "F−U
8FOPXBQQMZUIFJOJUJBMDPOEJUJPOTUPHFU"BOE"
v = = " + "
Ev v + 3J +
@@@@@
= −@@@@@@@@@@
= −@@@@@@@@@@@@@@@
= −
3$
EU
× × −
#VUEJGGFSFOUJBUJOH&R "UU = Ev= −" F−U − " F−U
@@@
EU
− = −" − "
'SPN&RT BOE XFPCUBJO " = −BOE " = 4VCTUJUVUJOH"BOE"JO&R ZJFMET
v U = −F−U + F−U
■CASE 2 8IFO3 = Ω
= @@@@@@@@@@@@@@@
α = @@@@
=
3$ × × × −
XIJMFω = SFNBJOTUIFTBNF4JODFα = ω = UIFSFTQPOTFJTDSJUJ
DBMMZEBNQFE)FODF T = T = − BOE
v U = " + "U F−U
5PHFU"BOE" XFBQQMZUIFJOJUJBMDPOEJUJPOT
v = = "
Ev v + 3J +
@@@@@
= −@@@@@@@@@@
= −@@@@@@@@@@@@
= −
3$
EU
× × −
#VUEJGGFSFOUJBUJOH&R Ev
@@@
= −" − "U + " F−U
EU
8.4
"UU = The Source-Free Parallel RLC Circuit
− = −" + "
'SPN&RT BOE " = BOE" = −5IVT
v U = − U F−U7
■CASE 3 8IFO3 = Ω
= @@@@@@@@@@@@@@@@@@
α = @@@@
= 3$ × × × −
XIJMFω = SFNBJOTUIFTBNF"Tα < ωJOUIJTDBTF UIFSFTQPOTFJT
VOEFSEBNQFE5IFSPPUTPGUIFDIBSBDUFSJTUJDFRVBUJPOBSF
@@@@@@@
T = −α ± √
α − ω
= − ± K
)FODF
v U = "DPTU + "TJOU F−U
8FOPXPCUBJO"BOE" BT
v = = "
Ev v + 3J +
@@@@@
= −
= − @@@@@@@@@@ = − @@@@@@@@@@@@@@
3$
EU
× × −
#VUEJGGFSFOUJBUJOH&R Ev= −" DPTU − " TJOU − " TJOU + " DPTU F−U
@@@
EU
"UU = − = −" + "
'SPN&RT BOE " = BOE" = −5IVT
v U = DPTU − TJOU F−U
/PUJDFUIBUCZJODSFBTJOHUIFWBMVFPG 3 UIFEFHSFFPGEBNQJOH
EFDSFBTFTBOEUIFSFTQPOTFTEJGGFS'JHVSFQMPUTUIFUISFFDBTFT
v (t) V
5
4
3
2
1
Overdamped
Critically damped
0
‒1
Underdamped
0
Figure 8.14
0.5
1
'PS&YBNQMFSFTQPOTFTGPSUISFFEFHSFFTPGEBNQJOH
1.5 t (s)
327
328
Practice Problem 8.5
Chapter 8
Second-Order Circuits
*O'JH MFU3 = Ω - = ) $ = N' v = J = N"
'JOEv U GPSU> Answer: −UF−U7
Example 8.6
'JOEv U GPSU> JOUIF3-$DJSDVJUPG'JH
30 Ω
40 V
+
‒
0.4 H
t=0
i
50 Ω
20 μF
+
v
‒
Figure 8.15
'PS&YBNQMF
Solution:
8IFOU< UIFTXJUDIJTPQFOUIFJOEVDUPSBDUTMJLFBTIPSUDJSDVJUXIJMF UIFDBQBDJUPSCFIBWFTMJLFBOPQFODJSDVJU5IFJOJUJBMWPMUBHFBDSPTTUIF DBQBDJUPSJTUIFTBNFBTUIFWPMUBHFBDSPTTUIFΩSFTJTUPSUIBUJT
× = 7
v = @@@@@@@
= @@
+ 5IFJOJUJBMDVSSFOUUISPVHIUIFJOEVDUPSJT
J = −@@@@@@@
= −"
+ 5IFEJSFDUJPOPGJJTBTJOEJDBUFEJO'JHUPDPOGPSNXJUIUIFEJSFD UJPOPG *JO'JH XIJDIJTJOBHSFFNFOUXJUIUIFDPOWFOUJPOU IBU
current flows into the positive terminal of an inductor (see Fig. 6.23). 8F
OFFEUPFYQSFTTUIJTJOUFSNTPGEv∕EU TJODFXFBSFMPPLJOHGPSv
Ev v + 3J − × @@@@@
= − @@@@@@@@@@
= −@@@@@@@@@@@@@
=
3$
EU
× × −
8IFOU> UIFTXJUDIJTDMPTFE 5IFWPMUBHFTPVSDFBMPOHXJUIUIF ΩSFTJTUPSJTTFQBSBUFEGSPNUIFSFTUPGUIFDJSDVJU 5IFQBSBMMFM 3-$
DJSDVJUBDUTJOEFQFOEFOUMZPGUIFWPMUBHFTPVSDF BTJMMVTUSBUFEJO'JH
/FYU XFEFUFSNJOFUIBUUIFSPPUTPGUIFDIBSBDUFSJTUJDFRVBUJPOBSF
= @@@@@@@@@@@@@@@@
α = @@@@
= 3$ × × × −
= @@@@@@@@@@@@@
@@@
ω = @@@@
= Ŀ
-$ × × −
@@@@@@@
T = −α ± √
α − ω
@@@@@@@@@@@@@@@@@@
= − ± √ − = − ± PS
T = − T = −
30 Ω
40 V
329
Step Response of a Series RLC Circuit
8.5
0.4 H
+
‒
20 μF
50 Ω
Figure 8.16
5IFDJSDVJUJO'JHXIFOU> 5IFQBSBMMFM
3-$DJSDVJUPOUIFSJHIUIBOETJEFBDUTJOEFQFO
EFOUMZPGUIFDJSDVJUPOUIFMFGUIBOETJEFPGUIF
KVODUJPO
4JODFα > ω XFIBWFUIFPWFSEBNQFESFTQPOTF
v U = "F−U + "F−U
"UU = XFJNQPTFUIFDPOEJUJPOJO&R v = = " + "
⇒
" = − "
5BLJOHUIFEFSJWBUJWFPGv U JO&R Ev
@@@
= −"F−U − "F−U
EU
*NQPTJOHUIFDPOEJUJPOJO&R PS
Ev @@@@@
= = −" − "
EU
= " + "
4PMWJOH&RT BOE HJWFT
" = − " = 5IVT UIFDPNQMFUFTPMVUJPOJO&R CFDPNFT
v U = −F−U + F−U7
Practice Problem 8.6
3FGFSUPUIFDJSDVJUJO'JH'JOEv U GPSU> t=0
Answer: F−U − F−U 7
20 Ω
1.5 A
8.5
Step Response of a Series RLC Circuit
"TXFMFBSOFEJOUIFQSFDFEJOHDIBQUFS UIFTUFQSFTQPOTFJTPCUBJOFECZ
UIFTVEEFOBQQMJDBUJPOPGBEDTPVSDF$POTJEFSUIFTFSJFT 3-$DJSDVJU
TIPXOJO'JH"QQMZJOH,7-BSPVOEUIFMPPQGPSU> EJ+ 3J + v = 7 @@
-
T
EU
4 mF
Figure 8.17
'PS1SBDUJDF1SPC
t=0
Vs
+
‒
#VU
Ev
@@@
J = $
EU
10 H
Figure 8.18
R
L
i
C
+
v
‒
4UFQWPMUBHFBQQMJFEUPBTFSJFT3-$DJSDVJU
330
Chapter 8
Second-Order Circuits
4VCTUJUVUJOHGPSJJO&R BOESFBSSBOHJOHUFSNT
7T
Ev
v = @@@
3
@@@
@@@
+ @@
Ev
+ @@@
-$
-$
EU
EU
which has the same form as Eq. (8.4). More specifically UIFDPFGficients
BSFUIFTBNF BOEUIBUJTJNQPSUBOUJOEFUFSNJOJOHUIFGSFRVFODZQBSBN
FUFST CVUUIFWBSJBCMFJTEJGGFSFOU -JLFXJTF TFF&R )FODF UIF
DIBSBDUFSJTUJDFRVBUJPOGPSUIFTFSJFT 3-$DJSDVJUJTOPUBG GFDUFECZUIF
QSFTFODFPGUIFEDTPVSDF
5IF TPMVUJPO UP &R IBT UXP DPNQPOFOUT UIF USBOTJFOUSF TQPOTFvU U BOEUIFTUFBEZTUBUFSFTQPOTFvTT U UIBUJT
v U = vU U + vTT U 5IFUSBOTJFOUSFTQPOTF vU U JTUIFDPNQPOFOUPGUIFUPUBMSFTQPOTFUIBU
EJFTPVUXJUIUJNF5IFGPSNPGUIFUSBOTJFOUSFTQPOTFJTUIFTBNFBTUIF
GPSNPGUIFTPMVUJPOPCUBJOFEJO4FDUJPOGPSUIFTPVSDFGSFFDJSDVJU HJWFOCZ&RT BOE 5IFSFGPSF UIFUSBOTJFOUSFTQPOTF
vU U GPSUIFPWFSEBNQFE VOEFSEBNQFE BOEDSJUJDBMMZEBNQFEDBTFTBSF
vU U = "FTU + "FTU
vU U = " + "U F−αU
0WFSEBNQFE B
$SJUJDBMMZEBNQFE C
vU U = "DPTωEU + "TJOωEU F−αU
6OEFSEBNQFE D
5IF TUFBEZTUBUFSFTQPOTF JT the final WBMVF PG v U *O UIF DJSDVJU JO
Fig. 8.18, the final vBMVFPGUIFDBQBDJUPSWPMUBHFJTUIFTBNFBTUIFTPVSDF
WPMUBHF7T)FODF
vTT U = v ∞ = 7T
5IVT UIFDPNQMFUFTPMVUJPOTGPSUIFP WFSEBNQFE VOEFSEBNQFE BOE
DSJUJDBMMZEBNQFEDBTFTBSF
v U = 7T + "FTU + "FTU
v U = 7T + " + "U F−αU
B
0WFSEBNQFE
$SJUJDBMMZEBNQFE
−αU
v U = 7T + "DPTωEU + "TJOωEU F 6OEFSEBNQFE
C
D
5IFWBMVFTPGUIFDPOTUBOUT"BOE"BSFPCUBJOFEGSPNUIFJOJUJBMDPOEJ
UJPOTv BOEEv ∕EU,FFQJONJOEUIBU vBOEJBSF SFTQFDUJWFMZ UIF
WPMUBHFBDSPTTUIFDBQBDJUPSBOEUIFDVSSFOUUISPVHIUIFJOEVDUPS5IFSF
GPSF &R POMZBQQMJFTfor finding v#VUPODFUIFDBQBDJUPSWPMUBHF
v$ = vJTLOPXO XFDBOEFUFSNJOFJ = $Ev∕EU XIJDIJTUIFTBNFDVSSFOU
UISPVHIUIFDBQBDJUPS JOEVDUPS BOESFTJTUPS )FODF UIFW PMUBHFBDSPTT
UIFSFTJTUPSJTv3 = J3 XIJMFUIFJOEVDUPSWPMUBHFJTv- = -EJ∕EU
"MUFSOBUJWFMZ UIFDPNQMFUF SFTQPOTFGPS BOZW BSJBCMF Y U DBOCF
GPVOEEJSFDUMZ CFDBVTFJUIBTUIFHFOFSBMGPSN
Y U = YTT U + YU U XIFSFUIFYTT = Y ∞) is the final vBMVFBOEYU U JTUIFUSBOTJFOUSFTQPOTF
The final vBMVFJTGPVOEBTJO4FDUJPO5IFUSBOTJFOUSFTQPOTFIBTUIF
TBNFGPSNBTJO&R BOEUIFBTTPDJBUFEDPOTUBOUTBSFEFUFSNJOFE
GSPN&R CBTFEPOUIFWBMVFTPGY BOEEY ∕EU
8.5
331
Step Response of a Series RLC Circuit
Example 8.7
'or the circuit in Fig. 8.19, find v U BOE J U GPS U> $POTJEFSUIFTF
DBTFT3 = Ω 3 = Ω BOE3 = Ω
Solution:
■CASE 1 8IFO3 = Ω'PSU< UIFTXJUDIJTDMPTFEGPSBMPOH
UJNF5IFDBQBDJUPSCFIBWFTMJLFBOPQFODJSDVJUXIJMFUIFJOEVDUPS
BDUTMJLFBTIPSUDJSDVJU5IFJOJUJBMDVSSFOUUISPVHIUIFJOEVDUPSJT
BOEUIFJOJUJBMWPMUBHFBDSPTTUIFDBQBDJUPSJTUIFTBNFBTUIFWPMUBHF
BDSPTTUIFΩSFTJTUPSUIBUJT
v = J = 7
'PS U > UIFTXJUDIJTPQFOFE TPUIBUXFIB WFUIF ΩSFTJTUPS
EJTDPOOFDUFE8IBUSFNBJOTJTUIFTFSJFT 3-$DJSDVJUXJUIUIFW PMUBHF
TPVSDF5IFDIBSBDUFSJTUJDSPPUTBSFEFUFSNJOFEBTGPMMPXT
= @@@@@@@@@
@@@
@@@@@@@
= ω = @@@@
Ŀ
Ŀ
-$ × @@@@@@@
T = −α ±
= − −
Ŀα − ω
4JODF α > ω XFIBWFUIFPWFSEBNQFEOBUVSBMSFTQPOTF5IFUPUBM
SFTQPOTFJTUIFSFGPSF
v U = vTT + "F−U + "F−U
XIFSFvTT is the steady-state response. It is the final value of the capacitor
WPMUBHF*O'JH vG = 75IVT
v U = + "F−U + "F−U 8FOPw need to find "BOE"VTJOHUIFJOJUJBMDPOEJUJPOT
v = = + " + "
PS
− = " + "
5IFDVSSFOUUISPVHIUIFJOEVDUPSDBOOPUDIBOHFBCSVQUMZBOEJTUIFTBNF
DVSSFOUUISPVHIUIFDBQBDJUPSBUU = +CFDBVTFUIFJOEVDUPSBOEDBQBDJUPS
BSFOPXJOTFSJFT)FODF
Ev @@@@@
J = $
= EU
Ev = = @@@@
= @@
⇒ @@@@@
$ EU
#FGPSFXFVTFUIJTDPOEJUJPO XFOFFEUPUBLFUIFEFSJWBUJWFPG
Eq. (8.7.1).
vJO
Ev= −" F−U− " F−U
@@@
EU
Ev @@@@@
= = −" − "
EU
"UU = 24 V
+
‒
Figure 8.19
i
'PS&YBNQMF
= "
J = @@@@@
+
= 3= @@@@@
α = @@@
- × 1H
R
0.25 F
t=0
+
v
‒
1Ω
332
Chapter 8
Second-Order Circuits
'SPN&RT BOE " = −∕BOE" = ∕4VCTUJUVUJOH"
BOE"JO&R XFHFU
−F−U + F−U 7
v U = + @@
4JODFUIFJOEVDUPSBOEDBQBDJUPSBSFJOTFSJFTGPSU> UIFJOEVDUPS
DVSSFOUJTUIFTBNFBTUIFDBQBDJUPSDVSSFOU)FODF
Ev
@@@
J U = $
EU
.VMUJQMZJOH&R CZ $ = BOETVCTUJUVUJOHUIFWBMVFTPG "
BOE"HJWFT
F−U − F−U "
J U = @@
/PUFUIBUJ = " BTFYQFDUFE
■ CASE 2 8IFO 3 = Ω"HBJO UIFJOJUJBMDVSSFOUUISPVHIUIF
JOEVDUPSJT
= "
J = @@@@@
+
BOEUIFJOJUJBMDBQBDJUPSWPMUBHFJT
v = J = 7
'PSUIFDIBSBDUFSJTUJDSPPUT
3= @@@@@
= α = @@@
- × XIJMFω = SFNBJOTUIFTBNF*OUIJTDBTF T = T = −α = − BOE
XFIBWFUIFDSJUJDBMMZEBNQFEOBUVSBMSFTQPOTF5IFUPUBMSFTQPOTFJT
UIFSFGPSF
v U = vTT + " + "U F−U
BOE BTCFGPSFvTT = 7
v U = + " + "U F−U
To find "BOE" XFVTFUIFJOJUJBMDPOEJUJPOT8FXSJUF
v = = + "
⇒
" = −
4JODFJ = $Ev ∕EU = PS
Ev _____
= = @@@
$
EU
'SPN&R @@@
Ev
= −" − U" + " F−U
EU
Ev _____
= = −" + "
EU
"UU = 8.5
Step Response of a Series RLC Circuit
'SPN&RT BOE " = −BOE " = −5IVT Eq. (8.7.7) becomes
v U = − + U F−U7
5IFJOEVDUPSDVSSFOUJTUIFTBNFBTUIFDBQBDJUPSDVSSFOUUIBUJT
Ev
@@@
J U = $
EU
.VMUJQMZJOH&R CZ $ = BOETVCTUJUVUJOHUIFWBMVFTPG "
BOE"HJWFT
J U = + U F−U"
/PUFUIBUJ = " BTFYQFDUFE
■CASE 3 8IFO3 = Ω5IFJOJUJBMJOEVDUPSDVSSFOUJT
= "
J = @@@@@
+
BOEUIFJOJUJBMWPMUBHFBDSPTTUIFDBQBDJUPSJTUIFTBNFBTUIFWPMUBHF
BDSPTTUIFΩSFTJTUPS
v = J = 7
3= @@@@@
= α = @@@
- × 4JODFα = < ω = XFIBWFUIFVOEFSEBNQFESFTQPOTF
@@@@@@@
T = −α ±
= − ± K
Ŀα − ω
5IFUPUBMSFTQPOTFJTUIFSFGPSF
v U = + "DPTU + "TJOU F−U
8FOPXEFUFSNJOF"BOE"8FXSJUF
v = = + "
⇒
" = −
4JODFJ = $Ev ∕EU = Ev @@@@@
= @@@
= $
EU
#VU
@@@
Ev
= F−U −"TJOU + "DPTU
EU
− F−U "DPTU + "TJOU
"UU = Ev @@@@@
= = − + " − " + EU
4VCTUJUVUJOH" = −HJWFT" = BOE&R CFDPNFT
v U = + TJOU− DPTU F−U7
5IFJOEVDUPSDVSSFOUJT
Ev
@@@
J U = $
EU
333
334
Chapter 8
Second-Order Circuits
.VMUJQMZJOH&R CZ $ = BOETVCTUJUVUJOHUIFWBMVFTPG "
BOE"HJWFT
J U = TJOU + DPTU F−U"
/PUFUIBUJ = " BTFYQFDUFE
Figure 8.20 plots the responses for the three cases. From this figure,
XFPCTFSWFUIBUUIFDSJUJDBMMZEBNQFESFTQPOTFBQQSPBDIFTUIFTUFQJOQVU
PG7UIFGBTUFTU
v (t) V
40
Underdamped
35
30
Critically damped
25
20
15
Overdamped
10
5
0
0
1
2
3
4
5
6
7
8
t (s)
Figure 8.20
'PS&YBNQMFSFTQPOTFGPSUISFFEFHSFFTPGEBNQJOH
Practice Problem 8.7
)BWJOHCFFOJOQPTJUJPO BGPSBMPOHUJNF UIFTXJUDIJO'JHJT
NPWFEUPQPTJUJPOCBUU = 'JOEv U BOEv3 U GPSU >
1Ω
18 V
+
‒
a
2Ω
1
40 F
b
t=0
+
v
‒
2.5 H
10 Ω
‒ vR +
15 V +
‒
Figure 8.21
'PS1SBDUJDF1SPC
Answer: − TJOU + DPTU F−U7 F−UTJOU7
i
Is
t=0
Figure 8.22
R
L
C
1BSBMMFM3-$DJSDVJUXJUIBOBQQMJFE
DVSSFOU
+
v
‒
8.6
Step Response of a Parallel RLC Circuit
$POTJEFSUIFQBSBMMFM 3-$DJSDVJUTIP XOJO'JH 8FXant to find
JEVFUPBTVEEFOBQQMJDBUJPOPGBEDDVSSFOU"QQMZJOH,$-BUUIFUPQ
OPEFGPSU> v + J + $
Ev= * @@
@@@
T
3
EU
Step Response of a Parallel RLC Circuit
8.6
335
#VU
EJ @@
v = -
EU
4VCTUJUVUJOHGPSvJO&R BOEEJWJEJOHCZ-$ XFHFU
*T
E J+ @@@
J = @@@
@@
EJ + @@@
@@@
3$
-$
-$
EU
EU
XIJDIIBTUIFTBNFDIBSBDUFSJTUJDFRVBUJPOBT&R 5IFDPNQMFUFTPMVUJPOUP&R DPOTJTUTPGUIFUSBOTJFOU
S FTQPOTF
JU U BOEUIFTUFBEZTUBUFSFTQPOTFJTTUIBUJT
J U = JU U + JTT U 5IFUSBOTJFOUSFTQPOTFJTUIFTBNFBTXIBUXFIBEJO4FDUJPO 5IF
steady-state response is the final vBMVFPGJ*OUIFDJSDVJUJO'JH UIF
final vBMVFPGUIFDVSSFOUUISPVHIUIFJOEVDUPSJTUIFTBNFBTUIFTPVSDF
DVSSFOU*T5IVT
J U = *T + "FTU + "FTU
J U = *T + " + "U F−αU
0WFSEBNQFE
$SJUJDBMMZEBNQFE
−αU
J U = *T + "DPTωEU + "TJOωEU F 6OEFSEBNQFE
5IFDPOTUBOUT"BOE"JOFBDIDBTFDBOCFEFUFSNJOFEGSPNUIF JOJUJBM
DPOEJUJPOTGPSJBOEEJ∕EU"HBJO XFTIPVMELFFQJONJOEUIBU&R only applies for finding the inductor current J#VUPODFUIFJOEVDUPSDVS
SFOUJ- = JJTLOPwn, we can find v = -EJ∕EU XIJDIJTUIFTBNFW PMUBHF
BDSPTTJOEVDUPS DBQBDJUPS BOESFTJTUPS )FODF UIFDVSSFOUUISPVHIUIF SFTJTUPSJTJ3 = v∕3 XIJMFUIFDBQBDJUPSDVSSFOUJT J$ = $Ev∕EU"MUFSOB
UJWFMZ UIFDPNQMFUFSFTQPOTFGPSBOZWBSJBCMFY U NBZCFGPVOEEJSFDUMZ VTJOH
Y U = YTT U + YU U XIFSFYTTBOEYU are its final value and transient response, respectively.
Example 8.8
In the circuit of Fig. 8.23, find J U BOEJ3 U GPSU> 20 Ω
t=0
i
4A
20 H
iR
20 Ω
8 mF
+
v
‒
+
‒
30u(‒t) V
Figure 8.23
'PS&YBNQMF
Solution:
'PSU< UIFTXJUDIJTPQFO BOEUIFDJSDVJUJTQBSUJUJPOFEJOUPUXPJOEF pendent subcircuits. The 4-A current flows through the inductor, so that
J = "
336
Chapter 8
Second-Order Circuits
4JODFV − U = XIFOU< BOEXIFOU> UIFWPMUBHFTPVSDFJT
PQFSBUJWFGPSU< 5IFDBQBDJUPSBDUTMJLFBOPQFODJSDVJUBOEUIFWPMUBHF
BDSPTTJUJTUIFTBNFBTUIFWPMUBHFBDSPTTUIFΩSFTJTUPSDPOOFDUFEJO
QBSBMMFMXJUIJU#ZWPMUBHFEJWJTJPO UIFJOJUJBMDBQBDJUPSWPMUBHFJT
v = @@@@@@@
= 7
+ 'PSU > UIFTXJUDIJTDMPTFE BOEXFIB WFBQBSBMMFM 3-$DJSDVJU
XJUIBDVSSFOUTPVSDF 5IFWPMUBHFTPVSDFJT[FSPXIJDINFBOTJUBDUT
MJLFBTIPSUDJSDVJU 5IFUX P ΩSFTJTUPSTBSFOP XJOQBSBMMFM 5IFZ
BSFDPNCJOFEUPHJ WF3 = ‖ = Ω5IFDIBSBDUFSJTUJDSPPUTBSF
EFUFSNJOFEBTGPMMPXT
= @@@@@@@@@@@@@@@
α = @@@@
=
3$ × × × −
= @@@@@@@@@@@@@
@@@@@@@@@@@@
@@@
ω = @@@@
= Ŀ
-$ Ŀ × × −
@@@@@@@
@@@@@@@@@@@@@
= − ± Ŀ
− T = −α ±
Ŀα − ω
= − ± PS
T = − T = −
4JODFα > ω XFIBWFUIFPWFSEBNQFEDBTF)FODF
J U = *T + "F−U + "F−U
XIFSF*T = 4 is the final value of J U 8FOPXVTFUIFJOJUJBMDPOEJUJPOTUP
EFUFSNJOF"BOE""UU = J = = + " + "
⇒
" = −"
5BLJOHUIFEFSJWBUJWFPGJ U JO&R EJ = −" F−U − " F−U
@@
EU
TPUIBUBUU = EJ @@@@@
= −" − "
EU
#VU
EJ @@@@@
-
= v = EU
EJ = = @@@
= @@@
⇒ @@@@@
- EU
4VCTUJUVUJOHUIJTJOUP&R BOEJODPSQPSBUJOH&R XFHFU
= − "
⇒
" = 5IVT " = −BOE " = *OTFSUJOH "BOE "JO&R HJWFTUIFDPNQMFUFTPMVUJPOBT
J U = + F−U − F−U "
'SPNJ U XFPCUBJOv U = -EJ∕EUBOE
vU
-
J3 U = @@@
= @@@
@@
EJ = F−U − F−U"
EU
8.7
337
General Second-Order Circuits
Practice Problem 8.8
'JOEJ U BOEv U GPSU >JOUIFDJSDVJUPG'JH
Answer: − DPT U " TJO U 7
3.5u(t) A
8.7
General Second-Order Circuits
8e first determine the initial conditions Y BOE EY ∕EUBOEUIF
final vBMVFY ∞ BTEJTDVTTFEJO4FDUJPO
8FUVSOPGf the independent sources and find the form of the transient
SFTQPOTFYU U CZBQQMZJOH,$-BOE,7-0ODFBTFDPOEPSEFSEJGGFS
FOUJBMFRVBUJPOJTPCUBJOFE XFEFUFSNJOFJUTDIBSBDUFSJTUJD SPPUT%F
QFOEJOHPOXIFUIFSUIFSFTQPOTFJTPWFSEBNQFE DSJUJDBMMZEBNQFE PSVOEFSEBNQFE XFPCUBJOYU U XJUIUXPVOLOPXODPOTUBOUTBTXF
EJEJOUIFQSFWJPVTTFDUJPOT
8FPCUBJOUIFTUFBEZTUBUFSFTQPOTFBT
YTT U = Y ∞ 20 H
0.2 F
Figure 8.24
/PXUIBUXFIBWFNBTUFSFETFSJFTBOEQBSBMMFM3-$DJSDVJUT XFBSFQSF
QBSFEUPBQQMZUIFJEFBTUPBO ZTFDPOEPSEFSDJSDVJUIBWJOHPOFPSNPSF
JOEFQFOEFOUTPVSDFTXJUIDPOTUBOUWBMVFT"MUIPVHIUIFTFSJFTBOEQBSBM
MFM3-$DJSDVJUTBSFUIFTFDPOEPSEFSDJSDVJUTPGHSFBUFTUJOUFSFTU PUIFS
TFDPOEPSEFSDJSDVJUTJODMVEJOHPQBNQTBSFBMTPVTFGVM(JWFOBTFDPOE
PSEFSDJSDVJU XFEFUFSNJOFJUTTUFQSFTQPOTFY U XIJDINBZCFWPMUBHF
PSDVSSFOU CZUBLJOHUIFGPMMPXJOHGPVSTUFQT
i
+
v
‒
'PS1SBDUJDF1SPC
A circuit may look complicated at first.
But once the sources are turned off in
an attempt to find the form of the
transient response, it may be reducible
to a first-order circuit, when the storage
elements can be combined, or to a
parallel/series RLC circuit. If it is
reducible to a first-order circuit, the
solution becomes simply what we had
in Chapter 7. If it is reducible to a
parallel or series RLC circuit, we apply
the techniques of previous sections in
this chapter.
XIFSFY ∞) is the final vBMVFPGY PCUBJOFEJOTUFQ
5IFUPUBMSFTQPOTFJTOPXGPVOEBTUIFTVNPGUIFUSBOTJFOUSFTQPOTF
BOETUFBEZTUBUFSFTQPOTF
Y U = YU U + YTT U e finally determine the constants associated with the transient re
8
TQPOTFCZJNQPTJOHUIFJOJUJBMDPOEJUJPOT Y BOE EY ∕EU EFUFS
NJOFEJOTUFQ
Problems in this chapter can also be
solved by using Laplace transforms,
which are covered in Chapters 15
and 16.
8FDBO BQQMZ UIJTHFOFSBM QSPDFEVSF to find the TUFQ SFTQPOTFPG
BOZTFDPOEPSEFSDJSDVJU JODMVEJOHUIPTFXJUIPQBNQT 5IFGPMMPXJOH
FYBNQMFTJMMVTUSBUFUIFGPVSTUFQT
Example 8.9
'JOEUIFDPNQMFUFSFTQPOTFvBOEUIFOJGPSU > 0 in the circuit of Fig. Solution:
We first find the initial and final values. At U = − UIFDJSDVJUJTBUTUFBEZ
TUBUF5IFTXJUDIJTPQFOUIFFRVJWBMFOUDJSDVJUJTTIPXOJO'JH B It is evident from the figure that
−
v
= 7 −
J
4Ω
12 V
=
i
v + = v − = 7 J + = J − = 2Ω
+
‒
t=0
+
"UU = UIFTXJUDIJTDMPTFEUIFFRVJWBMFOUDJSDVJUJTJO'JH C #Z
UIFDPOUJOVJUZPGDBQBDJUPSWPMUBHFBOEJOEVDUPSDVSSFOU XFLOPXUIBU
1H
Figure 8.25
'PS&YBNQMF
1
2F
+
v
‒
338
Chapter 8
5PHFUEv + ∕EU XFVTF$Ev∕EU = J$PSEv∕EU = J$∕$"QQMZJOH,$-
BUOPEFBJO'JH C
i
4Ω
+
12 V
+
‒
v +
J + = J$ + + @@@@@
= J$ + + @@@
⇒
J$ + = −"
v
‒
)FODF
(a)
4Ω
12 V
1H
+
‒
i
a
2Ω
iC
+
v
‒
0.5 F
Figure 8.26
&RVJWBMFOUDJSDVJUPGUIFDJSDVJUJO'JH
GPS B U< C U> i
1H
=
J ∞ = @@@@@
" +
v ∞ = J ∞ = 7
/FYU XFPCUBJOUIFGPSNPGUIFUSBOTJFOUSFTQPOTFGPSU >#ZUVSO
JOHPGGUIF7WPMUBHFTPVSDF XFIBWFUIFDJSDVJUJO'JH"QQMZ
JOH,$-BUOPEFBJO'JHHJWFT
v+ @@
Ev
@@@
J = @@
EU
"QQMZJOH,7-UPUIFMFGUNFTISFTVMUTJO
v
a
2Ω
Ev +
− = −7T
@@@@@@
= @@@
EU
The final values are obtained when the inductor is replaced by a short
DJSDVJUBOEUIFDBQBDJUPSCZBOPQFODJSDVJUJO'JH C HJWJOH
(b)
4Ω
Second-Order Circuits
+
v
‒
1
2F
4JODF XF BSF JOUFSFTUFE JO v GPS UIF NPNFOU XF TVCTUJUVUF J GSPN
Eq. (8.9.4) into Eq. (8.9.5). We obtain
Ev
Ev+ v = Ev+ @@
@@@
@@@
@@@
+ @@
v + EU EU EU
Figure 8.27
0CUBJOJOHUIFGPSNPGUIFUSBOTJFOU
SFTQPOTFGPS&YBNQMF
EJ + v = @@
J + EU
PS
Ev+ Ev+ v = @@@
@@@
EU
EU
'SPNUIJT XFPCUBJOUIFDIBSBDUFSJTUJDFRVBUJPOBT
T + T + = XJUISPPUTT = −BOET = −5IVT UIFOBUVSBMSFTQPOTFJT
vO U = "F−U + #F−U
XIFSF"BOE#BSFVOLOPXODPOTUBOUTUPCFEFUFSNJOFEMBUFS5IFTUFBEZ
TUBUFSFTQPOTFJT
vTT U = v ∞ = v U = vU + vTT = + "F−U + #F−U
5IFDPNQMFUFSFTQPOTFJT
8FOPXEFUFSNJOF "BOE #VTJOHUIFJOJUJBMWBMVFT'SPN&R v = 4VCTUJUVUJOHUIJTJOUP&R BUU = HJWFT
= + " + #
⇒
" + # = 8.7
339
General Second-Order Circuits
5BLJOHUIFEFSJWBUJWFPGvJO&R @@@
Ev
= −"F−U − #F−U
EU
4VCTUJUVUJOH&R JOUP&R BUU = HJWFT
− = −" − #
⇒
" + # = 'SPN&RT BOE XFPCUBJO
" = # = −
TPUIBU&R CFDPNFT
v U = + F−U − F−U7 U >
'SPN v XFDBOPCUBJOPUIFSRVBOUJUJFTPGJOUFSFTUCZSFGFSSJOHUP
Fig. 8.26(b). To obtain J GPSFYBNQMF
v + @@
Ev
@@@
J = @@
= + F−U − F−U − F−U + F−U
EU
−U
= − F
−U
+ F " U >
/PUJDFUIBUJ = JOBHSFFNFOUXJUI&R %FUFSNJOFvBOE JGPS U >JOUIFDJSDVJUPG'JH 4FFDPNNFOUT
BCPVUDVSSFOUTPVSDFTJO1SBDUJDF1SPC
Practice Problem 8.9
10 Ω
Answer: − F−U 7 − F−U "
4Ω
5A
i
+
v
‒
1
20 F
2H
t=0
Figure 8.28
'PS1SBDUJDF1SPC
Example 8.10
'JOEvP U GPSU >JOUIFDJSDVJUPG'JH
Solution:
5IJTJTBOFYBNQMFPGBTFDPOEPSEFSDJSDVJUXJUIUXPJOEVDUPST8F
first obtain the mesh currents JBOEJ XIJDIIBQQFOUPCFUIFDVSSFOUT
UISPVHIUIFJOEVDUPST 8FOFFEUPPCUBJOUIF initial and final values of
UIFTFDVSSFOUT
'PSU < V U = TPUIBUJ − = = J − 'PSU > V U = TP
UIBUUIFFRVJWBMFOUDJSDVJUJTBTTIPXOJO'JH B %VFUPUIFDPOUJOVJUZ
PGJOEVDUPSDVSSFOU
J + = J − = J + = J − = v- + = vP + = < J + − J + > = +
"QQMZJOH,7-UPUIFMFGUMPPQJO'JH B BUU = = J + + v- + + vP +
3Ω
7u(t) V
+
‒
i1
Figure 8.29
'PS&YBNQMF
1
2H
1Ω
+
vo
‒
i2
1
5H
340
Chapter 8
L1 = 21 H
3Ω
i1
3Ω
+ vL1 ‒
+
‒
7V
Second-Order Circuits
1Ω
i2
+
+
vo
‒
vL2
‒
i2
i1
L 2 = 51 H
7V
+
‒
1Ω
(a)
(b)
Figure 8.30
&RVJWBMFOUDJSDVJUPGUIBUJO'JHGPS B U> C U→ ∞
PS
v- + = 7
4JODF-EJ∕EU = v-
EJ +
v-
= "T
@@@@@@
= @@
= @@@
- @@
EU
EJ +
v-
@@@@@@
= = @@@
-
EU
4JNJMBSMZ TJODF-EJ∕EU = v-
"T U → ∞ UIFDJSDVJUSFBDIFTTUFBEZTUBUF BOEUIFJOEVDUPSTDBOCF
replaced by short circuits, as shown in Fig. 8.30(b). From this figure,
"
J ∞ = J ∞ = @@
/FYU XFPCUBJOUIFGPSNPGUIFUSBOTJFOUSFTQPOTFTCZSFNPWJOHUIF
WPMUBHFTPVSDF BTTIPXOJO'JH"QQMZJOH,7-UPUIFUXPNFTIFT
ZJFMET
1
2H
3Ω
i1
1Ω
i2
1
5H
EJ
@@@
J− J + @@
= EU
EJ
@@@
− J = J + @@
EU
EJ
@@@
J = J + @@
EU
BOE
Figure 8.31
0CUBJOJOHUIFGPSNPGUIFUSBOTJFOU
SFTQPOTFGPS&YBNQMF
'SPN&R 4VCTUJUVUJOH&R JOUP&R HJWFT
EJ
EJ
EJ
@@@
@@@
+ @@
+ @@@
@@@@
− J = J + @@
EU EU EU
EJ
EJ
@@@@
@@@
+ + J = EU
EU
'SPNUIJTXFPCUBJOUIFDIBSBDUFSJTUJDFRVBUJPOBT
T + T + = XIJDIIBTSPPUT T = −BOE T = −)FODF UIFGPSNPGUIFUSBOTJFOU
SFTQPOTFJT
JO = "F−U + #F−U
8.7
341
General Second-Order Circuits
XIFSF"BOE#BSFDPOTUBOUT5IFTUFBEZTUBUFSFTQPOTFJT
"
JTT = J ∞ = @@
'SPN&RT BOE XFPCUBJOUIFDPNQMFUFSFTQPOTFBT
+ "F−U + #F−U
J U = @@
We finally obtain "BOE#GSPNUIFJOJUJBMWBMVFT'SPN&RT BOE
+ " + #
= @@
5BLJOHUIFEFSJWBUJWFPG&R TFUUJOHU = JOUIFEFSJWBUJWF BOE
FOGPSDJOH&R XFPCUBJO
= −" − #
'SPN&RT BOE " = −∕BOE# = −5IVT
− @@
F−U − F−U
J U = @@
8FOP XPCUBJO JGSPN J "QQMZJOH,7-UPUIFMFGUMPPQJO
Fig. 8.30(a) giWFT
EJ
@@@
= J − J + @@
⇒
EU
EJ
@@@
J = − + J + @@
EU
4VCTUJUVUJOHGPSJJO&R HJWFT
J U = − + @@@
− @@@
F−U − F−U + F−U + F−U
− @@@
= @@
F−U + F−U
'SPN'JH
vP U = <J U − J U >
4VCTUJUVUJOH&RT BOE JOUP&R ZJFMET
vP U = F−U − F−U /PUFUIBUvP = BTFYQFDUFEGSPN&R 'PS U > PCUBJO vP U JOUIFDJSDVJUPG'JH
and v
)JOU First find v
Practice Problem 8.10
1Ω
Answer: F−U − F−U 7 U >
v1
1Ω
v2
+ vo ‒
35u(t) V
+
‒
Figure 8.32
1
2F
'PS1SBDUJDF1SPC
1
3F
342
The use of op amps in second-order
circuits avoids the use of inductors,
which are undesirable in some
applications.
Example 8.11
Chapter 8
Second-Order Circuits
8.8
Second-Order Op Amp Circuits
"OPQBNQDJSDVJUXJUIUX PTUPSBHFFMFNFOUTUIBUDBOOPUCFDPNCJOFE
JOUPBTJOHMFFRVJ WBMFOUFMFNFOUJTTFDPOEPSEFS #FDBVTFJOEVDUPSTBSF
CVMLZBOEIFB WZ UIFZBSFSBSFMZVTFEJOQSBDUJDBMPQBNQDJSDVJUT' PS
UIJTSFBTPO XFXJMMPOMZDPOTJEFS3$TFDPOEPSEFSPQBNQDJSDVJUTIFSF
Such circuits find a wide range of applications in devices such as filters
BOEPTDJMMBUPST
5IFBOBMZTJTPGBTFDPOEPSEFSPQBNQDJSDVJUGPMMPXTUIFTBNFGPVS
TUFQTHJWFOBOEEFNPOTUSBUFEJOUIFQSFWJPVTTFDUJPO
*OUIFPQBNQDJSDVJUPG'JH find vP U GPSU >XIFOvT=V U) N7
-FU3 = 3 = LΩ $ = μ' BOE$ = μ'
C2
+ v2 ‒
v1
R1
R2
2
1
vs
+
‒
C1
+
vo
‒
+
‒
vo
Figure 8.33
'PS&YBNQMF
Solution:
"MUIPVHIXFDPVMEGPMMPXUIFTBNFGPVSTUFQTHJWFOJOUIFQSFWJPVTTFD
UJPOUPTPMWFUIJTQSPCMFN XFXJMMTPMWFJUBMJUUMFEJGGFSFOUMZ%VFUP
the voltage follower configuration, the voltage across $JTvP"QQMZJOH
,$-BUOPEF
vT − v
v −v
Ev @@@@@@
@@@@@@
@@@
= $
+ P
3
3
EU
v − vP
Ev
@@@@@@
= $@@@
P
3
EU
v = v − vP
"UOPEF ,$-HJWFT
#VU
8FOPXUSZUPFMJNJOBUFvBOEvJO&RT UP 4VCTUJUVU
JOH&RT BOE JOUP&R ZJFMET
vT − v
EvP
EvP
Ev
@@@@@@
@@@
@@@
@@@
= $
− $
+ $
3
EU
EU
EU
EvP
@@@
v = vP + 3$
EU
'SPN&R 8.8
Second-Order Op Amp Circuits
4VCTUJUVUJOH&R JOUP&R XFPCUBJO
v@@@
v
Ev
EvP
EvP
EvP
EvP
3$ @@@
T
@@@
@@@@
@@@
@@@
= @@@P+ @@@@@
P
+ $
+ 3$$
− $
+ $
3 3
3 EU
EU
EU
EU
EU
PS
vT
Ev
EvP
vP
@@@
@@@@
+ @@@@@
+ @@@@@
P
+ @@@@@@@@
= @@@@@@@@
(
)
3
$
3
$
3
3
$
$
3
3
EU
$$ EU
8JUIUIFHJWFOWBMVFTPG3 3 $ BOE$ &R CFDPNFT
EvP
EvP
@@@
@@@@
+ + vP = vT
EU
EU
5PPCUBJOUIFGPSNPGUIFUSBOTJFOUSFTQPOTF TFUvT = JO&R XIJDI
JTUIFTBNFBTUVSOJOHPGGUIFTPVSDF5IFDIBSBDUFSJTUJDFRVBUJPOJT
T + T + = XIJDIIBTDPNQMFYSPPUTT = − ± K)FODF UIFGPSNPGUIFUSBOTJFOU
SFTQPOTFJT
vPU = F−U "DPTU + #TJOU XIFSF"BOE#BSFVOLOPXODPOTUBOUTUPCFEFUFSNJOFE
"TU → ∞ UIFDJSDVJUSFBDIFTUIFTUFBEZTUBUFDPOEJUJPO BOEUIF capacitors can be replaced by open circuits. Since no current floXTUISPVHI
$BOE$VOEFSTUFBEZTUBUFDPOEJUJPOTBOEOPDVSSFOUDBOFOUFSUIF
JOQVU
terminals of the ideal op amp, current does not floXUISPVHI3BOE3
5IVT
vP ∞ = v ∞ = vT
5IFTUFBEZTUBUFSFTQPOTFJTUIFO
vPTT = vP ∞ = vT = N7 U >
5IFDPNQMFUFSFTQPOTFJT
vP U = vPU + vPTT = + F−U "DPTU + #TJOU N7
5PEFUFSNJOF "BOE # XFOFFEUIFJOJUJBMDPOEJUJPOT'PS U < vT = TPUIBU
vP − = v − = 'PS U > UIF TPVSDF JT PQFSBUJWF )PXFWFS EVFUP DBQBDJUPS WPMUBHF
DPOUJOVJUZ
vP + = v + = 'SPN&R v + = v + + vP + = BOE IFODF GSPN&R EvP +
v − vP
@@@@@@
= = @@@@@@
3$
EU
8FOPXJNQPTF&R POUIFDPNQMFUFSFTQPOTFJO&R BUU = GPS
= + "
⇒
" = −
343
344
Chapter 8
Second-Order Circuits
5BLJOHUIFEFSJWBUJWFPG&R Ev
@@@
P= F−U −"DPTU − #TJOU − "TJOU + #DPTU
EU
4FUUJOHU = BOEJODPSQPSBUJOH&R XFPCUBJO
= −" + #
'SPN&RT BOE " = −BOE# = −5IVT UIFTUFQ
SFTQPOTFCFDPNFT
vP U = − F−U DPTU + TJOU N7 Practice Problem 8.11
R1
vs
+
‒
*OUIFPQBNQDJSDVJUTIPXOJO'JH vT = V U) V, find vP U GPS
U > 0. Assume that 3 = 3 = LΩ $ = μ' BOE$ = μ'
R2
+
‒
+
C1
C2
U >
Answer: − F−U + F−U 7 U >
vo
‒
8.9
Figure 8.34
'PS1SBDUJDF1SPC
PSpice Analysis of RLC Circuits
3-$DJSDVJUTDBOCFBOBMZ[FEXJUIHSFBUFBTFVTJOH 14QJDF KVTUMJLFUIF
3$PS3-DJSDVJUTPG$IBQUFS 5IFGPMMPXJOHUXPFYBNQMFTXJMMJMMVT
USBUFUIJT5IFSFBEFSNBZSFWJFX4FDUJPO%JO"QQFOEJY%PO14QJDF
GPSUSBOTJFOUBOBMZTJT
Example 8.12
5IFJOQVUWPMUBHFJO'JH B JTBQQMJFEUPUIFDJSDVJUJO'JH C 6TF14QJDFUPQMPUv U GPS<U<T
vs
Solution:
12
0
t (s)
2
(a)
60 Ω
vs +
‒
3H
1
27 F
60 Ω
(b)
Figure 8.35
'PS&YBNQMF
+
v
‒
Define. "TUSVFXJUINPTUUFYUCPPLQSPCMFNT UIFQSPCMFNJT
clearly defined.
1SFTFOU 5IFJOQVUJTFRVBMUPBTJOHMFTRVBSFXBWFPGBNQMJUVEF
7XJUIBQFSJPEPGT8FBSFBTLFEUPQMPUUIFPVUQVU VTJOH
14QJDF
"MUFSOBUJWF 4JODFXFBSFSFRVJSFEUPVTF14QJDF UIBUJTUIFPOMZ
BMUFSOBUJWFGPSBTPMVUJPO)PXFWFS XFDBODIFDLJUVTJOHUIFUFDI
OJRVFJMMVTUSBUFEJO4FDUJPO BTUFQSFTQPOTFGPSBTFSJFT3-$
DJSDVJU "UUFNQU 5IFHJWFODJSDVJUJTESBXOVTJOH4DIFNBUJDTBTJO
Fig. 8.36. The pulse is specified using 718-WPMUBHFTPVSDF Cut VPULSE could be used instead. Using the pieceXJTFMJOFBS
GVODUJPO XFTFUUIFBUUSJCVUFTPG718-BT5 = 7 = 5 = 7 = BOETPGPSUI BTTIPXOJO'JH5XP
WPMUBHFNBSLFSTBSFJOTFSUFEUPQMPUUIFJOQVUBOEPVUQVUWPMUBHFT
0ODFUIFDJSDVJUJTESBXOBOEUIFBUUSJCVUFTBSFTFU XFTFMFDU
"OBMZTJT4FUVQ5SBOTJFOUUPPQFOVQUIF5SBOTJFOU"OBMZTJT
EJBMPHCPY"TBQBSBMMFM3-$DJSDVJU UIFSPPUTPGUIFDIBSBDUFSJT
UJDFRVBUJPOBSF −BOE −5IVT XFNBZTFU'JOBM5JNFBTT
GPVSUJNFTUIFNBHOJUVEFPGUIFMPXFSSPPU 8IFOUIFTDIFNBUJD
8.9
345
PSpice Analysis of RLC Circuits
JTTBWFE XFTFMFDU"OBMZTJT4JNVMBUFBOEPCUBJOUIFQMPUTGPSUIF
JOQVUBOEPVUQVUWPMUBHFTVOEFSUIF14QJDF"%XJOEPXBTTIPXO
JO'JH
12 V
10 V
V
T1=0
T2=0.0001
T3=2
T4=2.0001
V1=0
V2=12
V3=12
V4=0
R1
L1
60
3H
8V
V
6V
4V
+
‒
V1
R2
60
0.03703
C1
2V
0V
0s
V(L1:2)
2.0s
1.0s
V(R1:1)
Time
Figure 8.36
Figure 8.37
4DIFNBUJDGPSUIFDJSDVJUJO'JH C 'PS&YBNQMFJOQVUBOEPVUQVU
/PXXFDIFDLVTJOHUIFUFDIOJRVFGSPN4FDUJPO8FDBO
TUBSUCZSFBMJ[JOHUIF5IFWFOJOFRVJWBMFOUGPSUIFSFTJTUPSTPVSDF
DPNCJOBUJPOJT75I = ∕ UIFPQFODJSDVJUWPMUBHFEJWJEFT
FRVBMMZBDSPTTCPUISFTJTUPST = 75IFFRVJWBMFOUSFTJTUBODFJT
30 Ω ‖ 5IVT XFDBOOPXTPMWFGPSUIFSFTQPOTFVTJOH
3 = Ω - = ) BOE$ = ∕ '
8e first need to solvFGPSαBOEω
= @@@@
ω = @@@@@
@@@
4JODFJTHSFBUFSUIBO XFIBWFUIFPWFSEBNQFEDBTF
α = 3∕ - = ∕ = @@@@@@
BOE
T = − ± √
− = − − XIFSF
Ŀ
v =
v ∞ = 7 J =
Ev U
@@@@
J U = $
EU
v U = "F−U + "F−U + v = = " + " + J = = $ −" − "
XIJDIZJFMET" = −"4VCTUJUVUJOHUIJTJOUPUIFBCPWF XFHFU
= " − " + PS" = BOE" = −
v U = −F−U+F−U+ V U 7GPSBMM< U< T
"UU = T v = −F− + F− + = − + + = 7"UU = T v = −F− + + = 7
/PUFUIBUGSPN< U< T 75I = XIJDIJNQMJFTUIBU
v ∞ = 5IFSFGPSF v U = "F− U− + "F− U− V U − 7
"UU = T " + " = − "F− U− − "F− U− J U = @@@@@@@@@@@@@@@@@@@@@@
3.0s
4.0s
346
Chapter 8
Second-Order Circuits
BOE
F− − F−
J = @@@@@@@@@@@@@@@@@
= N"
5IFSFGPSF −" − " = $PNCJOJOHUIFUXPFRVBUJPOT XFHFU −" − − " =
XIJDIMFBETUP" = BOE" = −
v U = F− U− −F− U− V U− 7
"UU = T v = − = 7"UU = T v =
0.7897 V
&WBMVBUF "DIFDLCFUXFFOUIFWBMVFTDBMDVMBUFEBCPWFBOEUIF
QMPUTIPXOJO'JHVSFTIPXTHPPEBHSFFNFOUXJUIJOUIF
PCWJPVTMFWFMPGBDDVSBDZ
4BUJTGBDUPSZ :FT XFIBWFBHSFFNFOUBOEUIFSFTVMUTDBOCF
QSFTFOUFEBTBTPMVUJPOUPUIFQSPCMFN
Practice Problem 8.12
5Ω
i
vs +
‒
Figure 8.38
1 mF
2H
'JOEJ U VTJOH14QJDFGPS< U< TJGUIFQVMTFWPMUBHFJO'JH B JT
BQQMJFEUPUIFDJSDVJUJO'JH
Answer:4FF'JH
3.0 A
2.0 A
'PS1SBDUJDF1SPC
1.0 A
0A
0s
1.0 s
3.0 s
2.0 s
4.0 s
I(L1)
Time
Figure 8.39
1MPUPGJ U GPS1SBDUJDF1SPC
Example 8.13
'PSUIFDJSDVJUJO'JH VTF14QJDFUPPCUBJOJ U GPS< U< T
a
t=0
i(t)
b
4A
5Ω
6Ω
1
42 F
7H
Figure 8.40
'PS&YBNQMF
Solution:
8IFOUIFTXJUDIJTJOQPTJUJPO B UIF ΩSFTJTUPSJTSFEVOEBOU5IF TDIFNBUJDGPSUIJTDBTFJTTIPXOJO'JH B 5PFOTVSFUIBUDVSSFOU 8.9
0.0000
347
PSpice Analysis of RLC Circuits
4.000E+00
I
4A
R1
IDC
5
23.81m
7H
C1
L1
R2
6
IC = 0
C1
23.81m
IC = 4A
7H
L1
0
0
(b)
(a)
Figure 8.41
'PS&YBNQMF B GPSEDBOBMZTJT C GPSUSBOTJFOUBOBMZTJT
J U FOUFSTQJO UIFJOEVDUPSJTSPUBUFEUISFFUJNFTCFGPSFJUJTQMBDFEJOUIF
DJSDVJU5IFTBNFBQQMJFTGPSUIFDBQBDJUPS8FJOTFSUQTFVEP DPNQPOFOUT
7*&810*/5BOE*130#&UPEFUFSNJOFUIFJOJUJBMDBQBDJUPSWPMUBHFBOE
JOJUJBMJOEVDUPSDVSSFOU8FDBSSZPVUBED 14QJDFBOBMZTJTCZTFMFDU JOH"OBMZTJT4JNVMBUF"TTIPXOJO'JH B XFPCUBJOUIFJOJUJBM
DBQBDJUPSWPMUBHFBT7BOEUIFJOJUJBMJOEVDUPSDVSSFOUJ BT"GSPN
UIFEDBOBMZTJT5IFTFJOJUJBMWBMVFTXJMMCFVTFEJOUIFUSBOTJFOUBOBMZTJT
8IFOUIFTXJUDIJTNPWFEUPQPTJUJPOC UIFDJSDVJUCFDPNFTBTPVSDF
GSFFQBSBMMFM 3-$DJSDVJUXJUIUIFTDIFNBUJDJO'JH C 8FTFUUIF
JOJUJBMDPOEJUJPO*$ = GPSUIFDBQBDJUPSBOE*$ = "GPSUIFJOEVDUPS
"DVSSFOUNBSLFSJTJOTFSUFEBUQJOPGUIFJOEVDUPS8FTFMFDU"OBMZTJT
4FUVQ5SBOTJFOUUPPQFOVQUIF 5SBOTJFOU"OBMZTJT EJBMPHCPYBOE
TFU 'JOBM5JNFUPT "GUFSTBWJOHUIFTDIFNBUJD XFTFMFDU "OBMZTJT
5SBOTJFOU'JHVSFTIP XTUIFQMPUPG J U 5IFQMPUBHSFFTXJUI
J U = F−U − F−U" XIJDIJTUIFTPMVUJPOCZIBOEDBMDVMBUJPO
3FGFSUPUIFDJSDVJUJO'JH TFF1SBDUJDF 1SPC 6TF 14QJDFUP
PCUBJOv U GPS< U< Answer:4FF'JH
11 V
10 V
9V
8V
0s
0.5 s
V(C1:1)
1.0 s
Time
Figure 8.43
1MPUPGv U GPS1SBDUJDF1SPC
1.5 s
2.0 s
4.00 A
3.96 A
3.92 A
3.88 A
0s
2.0 s
1.0 s
3.0 s
I(L1)
Time
Figure 8.42
1MPUPGJ U GPS&YBNQMF
Practice Problem 8.13
348
Chapter 8
8.10
Second-Order Circuits
Duality
5IFDPODFQUPGEVBMJUZJTBUJNFTB WJOH FG GPSUFGGFDUJWFNFBTVSFPG TPMWJOH DJSDVJUQSPCMFNT $POTJEFSUIFTJNJMBSJUZ CFUXFFO&RT BOE 5IFUXPFRVBUJPOTBSFUIFTBNF F YDFQUUIBUXFNVTUJOUFS DIBOHFUIFGPMMPXJOHRVBOUJUJFT WPMUBHFBOEDVSSFOU SFTJTUBODF
BOE DPOEVDUBODF DBQBDJUBODF BOE JOEVDUBODF5IVT JU TPNFUJNFT
PDDVSTJODJSDVJUBOBMZTJTUIBUUXPEJGGFSFOUDJSDVJUTIBWFUIFTBNFFRVB
UJPOT BOE TPMVUJPOT FYDFQU UIBU UIF SPMFT PG DFSUBJO DPNQMFNFOUBSZ
FMFNFOUTBSFJOUFSDIBOHFE 5IJTJOUFSDIBOHFBCJMJUZJTLOP XOBTUIF QSJODJQMFPGEVBMJUZ
The duality principle asserts a parallelism between pairs of characterizing
equations and theorems of electric circuits.
TABLE 8.1
Dual pairs.
3FTJTUBODF3
*OEVDUBODF7PMUBHFv
7PMUBHFTPVSDF
/PEF
4FSJFTQBUI
0QFODJSDVJU
,75IFWFOJO
$POEVDUBODF(
$BQBDJUBODF$
$VSSFOUJ
$VSSFOUTPVSDF
.FTI
1BSBMMFMQBUI
4IPSUDJSDVJU
,$/PSUPO
Even when the principle of linearity
applies, a circuit element or variable
may not have a dual. For example,
mutual inductance (to be covered in
Chapter 13) has no dual.
%VBMQBJSTBSFTIP XOJO 5BCMF/PUFUIBUQP XFSEPFTOPUBQQFBSJO
5BCMF CFDBVTFQPXFSIBTOPEVBM5IFSFBTPOGPSUIJTJTUIFQSJODJQMF
PGMJOFBSJUZTJODFQPXFSJTOPUMJOFBS EVBMJUZEPFTOPUBQQMZ"MTPOPUJDF
GSPN5BCMFUIBUUIFQSJODJQMFPGEVBMJUZFYUFOETUPDJSDVJUFMFNFOUT configurations, and theorems.
5XPDJSDVJUTUIBUBSFEFTDSJCFECZFRVBUJPOTPGUIFTBNFGPSN C VU
JOXIJDIUIFWBSJBCMFTBSFJOUFSDIBOHFE BSFTBJEUPCFEVBMUPFBDIPUIFS
Two circuits are said to be duals of one another if they are described
by the same characterizing equations with dual quantities interchanged.
5IFVTFGVMOFTTPGUIFEVBMJUZQSJODJQMFJTTFMGF WJEFOU0ODFXF
LOPXUIFTPMVUJPOUPPOFDJSDVJU XFBVUPNBUJDBMMZIB WFUIFTPMVUJPO
GPSUIFEVBMDJSDVJU*UJTPC WJPVTUIBUUIFDJSDVJUTJO'JHTBOE
BSFEVBM$POTFRVFOUMZ UIFSFTVMUJO&R JTUIFEVBMPGUIBUJO
Eq. 8FNVTUL FFQJONJOEUIBUUIFNFUIPEEFTDSJCFEIFSFGPS
finding a dual is limited to planar circuits. Finding a dual for a nonplanar
DJSDVJUJTCF ZPOEUIFTDPQFPGUIJTUF YUCPPLCFDBVTFOPOQMBOBSDJSDVJUT
DBOOPUCFEFTDSJCFECZBTZTUFNPGNFTIFRVBUJPOT
5o find the dual of a giWFODJSDVJU XFEPOPUOFFEUPXSJUFEP XO
UIFNFTIPSOPEFFRVBUJPOT 8FDBOVTFBHSBQIJDBMUFDIOJRVF(J WFOB
QMBOBSDJSDVJU XFDPOTUSVDUUIFEVBMDJSDVJUCZUBLJOHUIFGPMMPXJOHUISFF
TUFQT
1MBDFBOPEF BUUIF DFOUFSPGFBDI NFTIPGUIF HJWFODJSDVJU 1MBDF
UIFSFGFSFODFOPEF UIFHSPVOE PGUIFEVBMDJSDVJUPVUTJEFUIFHJWFO
DJSDVJU
%SBX MJOFTCFUXFFOUIF OPEFTTVDIUIBU FBDI MJOFDSPTTFT BOFMF NFOU3FQMBDFUIBUFMFNFOUCZJUTEVBM TFF5BCMF 5PEFUFSNJOFUIFQPMBSJUZPGWPMUBHFTPVSDFTBOEEJSFDUJPOPGDVSSFOU
TPVSDFT GPMMPXUIJTSVMF "WPMUBHFTPVSDFUIBUQSPEVDFTBQPTJUJ WF
DMPDLXJTF NFTIDVSSFOUIBTBTJUTEVBMBDVSSFOUTPVSDFXIPTFSFGFS
FODFEJSFDUJPOJTGSPNUIFHSPVOEUPUIFOPOSFGFSFODFOPEF
*ODBTFPGEPVCU POFNBZWFSJGZUIFEVBMDJSDVJUCZXSJUJOHUIFOPEBMPS
NFTIFRVBUJPOT5IFNFTI PSOPEBM FRVBUJPOTPGUIFPSJHJOBMD JSDVJUBSF
TJNJMBSUPUIFOPEBM PSNFTI FRVBUJPOTPGUIFEVBMDJSDVJU 5IFEVBMJUZ
QSJODJQMFJTJMMVTUSBUFEXJUIUIFGPMMPXJOHUXPFYBNQMFT
8.10
349
Duality
Example 8.14
$POTUSVDUUIFEVBMPGUIFDJSDVJUJO'JH
Solution:
As shown in Fig. 8.45(a), we first locate nodes 1 and 2 in the two meshes
BOEBMTPUIFHSPVOEOPEFGPSUIFEVBMDJSDVJU8FESBXBMJOFCFUXFFO
POFOPEFBOEBOPUIFSDSPTTJOHBOFMFNFOU8FSFQMBDFUIFMJOFKPJOJOH
UIFOPEFTCZUIFEVBMTPGUIFFMFNFOUTXIJDIJUDSPTTFT'PSFYBNQMF B
MJOFCFUXFFOOPEFTBOEDSPTTFTB)JOEVDUPS BOEXFQMBDFB'
DBQBDJUPS BOJOEVDUPSTEVBM POUIFMJOF"MJOFCFUXFFOOPEFTBOE
DSPTTJOHUIF7WPMUBHFTPVSDFXJMMDPOUBJOB"DVSSFOUTPVSDF#Z
ESBXJOHMJOFTDSPTTJOHBMMUIFFMFNFOUT XFDPOTUSVDUUIFEVBMDJSDVJU
on UIFHJWFODJSDVJUBTJO'JH B 5IFEVBMDJSDVJUJTSFESBXOJO
Fig. 8.45(b) for clarity.
6V
+
‒
6A
2H
'PS&YBNQMF
2F
1
2H
1
10 mF
Figure 8.44
2Ω
6V +
‒
t=0
t=0
t=0
0.5 Ω
2Ω
2
2F
2
10 mF
10 mH
6A
t=0
0.5 Ω
0
10 mH
0
(b)
(a)
Figure 8.45
B $POTUSVDUJPOPGUIFEVBMDJSDVJUPG'JH C EVBMDJSDVJUSFESBXO
Practice Problem 8.14
%SBXUIFEVBMDJSDVJUPGUIFPOFJO'JH
Answer:4FF'JH
3H
3F
50 mA
10 Ω
Figure 8.46
'PS1SBDUJDF1SPC
4H
50 mV +
‒
0.1 Ω
4F
Figure 8.47
%VBMPGUIFDJSDVJUJO'JH
0CUBJOUIFEVBMPGUIFDJSDVJUJO'JH
Solution:
5IFEVBMDJSDVJUJTDPOTUSVDUFEPOUIFPSJHJOBMDJSDVJUBTJO'JH B We first locate nodes 1 to 3 and the reference node 0. Joining nodes
1 BOE XFDSPTTUIF'DBQBDJUPS XIJDIJTSFQMBDFECZB)
JOEVDUPS
Example 8.15
350
Chapter 8
Second-Order Circuits
5H
10 V +
‒
i1
2F
20 Ω
i2
i3
3A
Figure 8.48
'PS&YBNQMF
+PJOJOHOPEFTBOE XFDSPTTUIF ΩSFTJTUPS XIJDIJTSFQMBDFECZ
B @@@
Ω SFTJTUPS8F LFFQ EPJOHUIJT VOUJMBMM UIFFMFNFOUT BSF DSPTTFE
5IFSFTVMUJTJO'JH B 5IFEVBMDJSDVJUJTSFESBXOJO'JH C 5F
5H
2H
1
10 V
1
+
‒
2F
2H
2
3
20 Ω
3
3A
10 A
1
20 Ω
‒
+
5F
‒
+ 3V
0
1
20 Ω
2
3V
0
10 A
(a)
(b)
Figure 8.49
'PS&YBNQMF B DPOTUSVDUJPOPGUIFEVBMDJSDVJUPG'JH C EVBMDJSDVJUSFESBXO
5PW FSJGZUIFQPMBSJUZPGUIFW PMUBHFTPVSDFBOEUIFEJSFDUJPOPG
UIFDVSSFOUTPVSDF XFNBZBQQMZNFTIDVSSFOUT J J BOE J BMMJOUIF
DMPDLXJTFEJSFDUJPO JOUIFPSJHJOBMDJSDVJUJO'JH5IF7WPMUBHF
TPVSDFQSPEVDFTQPTJUJWFNFTIDVSSFOU J TPUIBUJUTEVBMJTB"DVS
SFOUTPVSDFEJSFDUFEGSPNUP "MTP J = −"JO'JHIBTBTJUT
EVBMv = −7JO'JH C Practice Problem 8.15
'PSUIFDJSDVJUJO'JH PCUBJOUIFEVBMDJSDVJU
Answer:4FF'JH
1
3
5Ω
0.2 F
2A
Figure 8.50
4H
3Ω
'PS1SBDUJDF1SPC
Ω
4F
0.2 H
+
‒
20 V
2V +
‒
Figure 8.51
1
5Ω
%VBMPGUIFDJSDVJUJO'JH
20 A
8.11
8.11
Applications
351
Applications
1SBDUJDBMBQQMJDBUJPOTPG 3-$DJSDVJUTBSFGPVOEJODPOUSPMBOEDPNNV
OJDBUJPOTDJSDVJUTTVDIBTSJOHJOHDJSDVJUT QFBLJOHDJSDVJUT SFTPOBOUDJS cuits, smoothing circuits, and filters. Most of these circuits cannot be
DPWFSFEVOUJMXFUSFBUBDTPVSDFT'PSOPX XFXJMMMJNJUPVSTFMWFTUPUXP
TJNQMFBQQMJDBUJPOTBVUPNPCJMFJHOJUJPOBOETNPPUIJOHDJSDVJUT
8.11.1
Automobile Ignition System
*O4FDUJPO XFDPOTJEFSFEUIFBVUPNPCJMFJHOJUJPOTZTUFNBTB
DIBSHJOHTZTUFN5IBUXBTPOMZBQBSUPGUIFTZTUFN)FSF XFDPOTJEFS
BOPUIFSQBSU‡UIFWPMUBHFHFOFSBUJOHTZTUFN5IFTZTUFNJTNPEFMFECZ
UIF DJSDVJU TIPXO JO 'JH 5IF 7 TPVSDF JT EVF UP UIF CBUUFSZ
BOEBMUFSOBUPS5IFΩSFTJTUPSSFQSFTFOUT UIFSFTJTUBODFPGUIFXJSJOH
5IFJHOJUJPODPJMJTNPEFMFECZUIFN)JOEVDUPS 5IFμ'DBQBDJUPS
LOPXOBTUIFDPOEFOTFSUPBVUPNFDIBOJDT JTJOQBSBMMFMXJUIUIFTXJUDI
LOPXOBTUIF CSFBLJOHQPJOUTPS FMFDUSPOJDJHOJUJPO *OUIFGPMMP XJOH
FYBNQMF XFEFUFSNJOFIPXUIF3-$DJSDVJUJO'JHJTVTFEJOHFO FSBUJOHIJHIWPMUBHF
t=0
1 μF
4Ω
+ v ‒
C
i
+
vL
‒
12 V
8 mH
Spark plug
Ignition coil
Figure 8.52
"VUPNPCJMFJHOJUJPODJSDVJU
"TTVNJOHUIBUUIFTXJUDIJO'JHJTDMPTFEQSJPSUP U = −, find the
JOEVDUPSWPMUBHFv-GPSU >
Solution:
*GUIFTXJUDIJTDMPTFEQSJPSUPU = −BOEUIFDJSDVJUJTJOTUFBEZTUBUF UIFO
J − = @@@
= " v$ − = "UU = + UIFTXJUDIJTPQFOFE5IFDPOUJOVJUZDPOEJUJPOTSFRVJSFUIBU
J + = " +
v$ + = +
8FPCUBJO EJ EUGSPN v- "QQMZJOH,7-UPUIFNFTIBU U = +
ZJFMET
− + J + + v- + + v$ + = − + × + v- + + = ⇒
v- + = Example 8.16
352
Chapter 8
Second-Order Circuits
)FODF
v- +
EJ +
@@@@@
= @@@@@@
= EU
"TU → ∞ UIFTZTUFNSFBDIFTTUFBEZTUBUF TPUIBUUIFDBQBDJUPSBDUTMJLF
BOPQFODJSDVJU5IFO
J ∞ = *GXFBQQMZ,7-UPUIFNFTIGPSU > XFPCUBJO
EJ + @@
UJEU
@@
= 3J + -
∫ + v$ EU $ 5BLJOHUIFEFSJWBUJWFPGFBDIUFSNZJFMET
EJ+ @@
J = 3
@@
@@@
EJ + @@@
EU - EU -$
8FPCUBJOUIFGPSNPGUIFUSBOTJFOUSFTQPOTFCZGPMMPXJOHUIFQSPDFEVSF in Section 8.3. Substituting 3 = Ω - = N) BOE$ = μ' XFHFU
3= α = @@@
-
= × @@@
ω = @@@@
Ŀ
-$
4JODF α < ω UIFSFTQPOTFJTVOEFSEBNQFE5IFEBNQFEOBUVSBMGSF RVFODZJT
_______
ωE = √ω − α≃ω
= × 5IFGPSNPGUIFUSBOTJFOUSFTQPOTFJT
JU U = F−α "DPTωEU + #TJOωE U XIFSF"BOE#BSFDPOTUBOUT5IFTUFBEZTUBUFSFTQPOTFJT
JTT U = J ∞ = TPUIBUUIFDPNQMFUFSFTQPOTFJT
J U = JU U + JTT U = F−U "DPT U + #TJO U 8FOPXEFUFSNJOF"BOE#
J = = " + ⇒
"=
5BLJOHUIFEFSJWBUJWFPG&R @@
EJ = −F−U "DPT U + #TJO U
EU
+ F−U − "TJO U + #DPT U
4FUUJOHU = BOEJODPSQPSBUJOH&R = −" + #
⇒
# = 5IVT
J U = F−U DPT U + TJO U 5IFWPMUBHFBDSPTTUIFJOEVDUPSJTUIFO
EJ = −F−UTJO U
@@
v- U = -
EU
8.11
Applications
353
5IJTIBTBNBYJNVNWBMVFXIFOTJOFJTVOJUZ UIBUJT BU U = π∕
PS U = μT"U UJNF = U UIFJOEVDUPSWPMUBHFSFBDIFTJUTQFBL XIJDIJT
v- U = −F−U = −7
"MUIPVHIUIJTJTGBSMFTTUIBOUIFWPMUBHFSBOHFPGUP 7
required to fire the spark plug in a typical automobile, a device known
BTB USBOTGPSNFS UPCFEJTDVTTFEJO$IBQUFS JTVTFEUPTUFQVQUIF
JOEVDUPSWPMUBHFUPUIFSFRVJSFEMFWFM
Practice Problem 8.16
In Fig. 8.52, find the capacitor voltage v$GPSU >
Answer: − F−UDPT U + F−UTJO U7
8.11.2
Smoothing Circuits
*OBUZQJDBMEJHJUBMDPNNVOJDBUJPOTZTUFN UIFTJHOBMUPCFUSBOTNJUUFE
is first sampled. Sampling refers to the procedure of selecting samples
PGBTJHOBMGPSQSPDFTTJOH BTPQQPTFEUPQSPDFTTJOHUIFFOUJSFTJHOBM
vs (t)
v0(t)
&BDITBNQMFJTDPO WFSUFEJOUPBCJOBSZOVNCFSSFQSFTFOUFECZBTFSJFT p(t)
Smoothing
D/A
PGQVMTFT 5IFQVMTFTBSFUSBOTNJUUFECZBUSBOTNJTTJPOMJOFTVDIBTB
circuit
DPBYJBMDBCMF UXJTUFEQBJS, or optical fiber"UUIFSFDFJWJOHFOE UIFTJH
Figure 8.53
OBMJTBQQMJFEUPBEJHJUBMUPBOBMPH %" DPO WFSUFSXIPTFPVUQVUJTB
"TFSJFTPGQVMTFTJTBQQMJFEUPUIFEJHJUBM
iTUBJSDBTFuGVODUJPO UIBUJT DPOTUBOUBUFBDIUJNFJOUFSW BM*OPSEFSUP
UPBOBMPH %" DPOWFSUFS XIPTFPVUQVU
SFDPWFSUIFUSBOTNJUUFEBOBMPHTJHOBM UIFPVUQVUJTTNPPUIFECZMFUUJOH JTBQQMJFEUPUIFTNPPUIJOHDJSDVJU
JUQBTTUISPVHIBiTNPPUIJOHuDJSDVJU BTJMMVTUSBUFEJO'JH"O3-$
DJSDVJUNBZCFVTFEBTUIFTNPPUIJOHDJSDVJU
5IFPVUQVUPGB%"DPOWFSUFSJTTIPXOJO'JH B *GUIF3-$DJSDVJU
JO'JH C JTVTFEBTUIFTNPPUIJOHDJSDVJU EFUFSNJOFUIFPVUQVU
WPMUBHFvP U vs
10
1
4
vs +
‒
0
‒2
1Ω
1H
3
+
v0
‒
2
1F
t (s)
0
(a)
0
(b)
Figure 8.54
'PS&YBNQMF B PVUQVUPGB%"DPOWFSUFS C BO3-$
TNPPUIJOHDJSDVJU
Solution:
5IJTQSPCMFNJTCFTUTPMWFEVTJOH 14QJDF5IFTDIFNBUJDJTTIPXOJO Fig. 8.55(a). The pulse in Fig. 8.54(a) is specified using the piecewise
Example 8.17
354
Chapter 8
V
T1=0
T2=0.001
T3=1
T4=1.001
T5=2
T6=2.001
T7=3
T8=3.001
V1=0
V2=4
V3=4
V4=10
V5=10
V6= ‒2
V7= ‒2
V8=0
+
‒
Second-Order Circuits
R1
L1
1
1H
V1
V
10 V
5V
1
C1
0V
‒5 V
0s
0
2.0 s
V(V1:+)
4.0 s
V(C1:1)
Time
6.0 s
(b)
(a)
Figure 8.55
'PS&YBNQMF B TDIFNBUJD C JOQVUBOEPVUQVUWPMUBHFT
MJOFBSGVODUJPO5IFBUUSJCVUFTPG7BSFTFUBT5
= 7 = 5 = 7 = 5 = 7 = BOETPPO5PCFBCMFUPQMPUCPUI
JOQVUBOEPVUQVUWPMUBHFT XFJOTFSUUXPWPMUBHFNBSLFSTBTTIPXO8F
TFMFDU"OBMZTJT4FUVQ5SBOTJFOUUPPQFOVQUIF5SBOTJFOU"OBMZTJTEJBMPH
CPYBOETFU 'JOBM5JNFBTT0ODFUIFTDIFNBUJDJTTBWFE XFTFMFDU "OBMZTJT4JNVMBUFUPSVOBOEPCUBJOUIFQMPUTTIPXOJO'JH C Practice Problem 8.17
3FXPSL&YBNQMFJGUIFPVUQVUPGUIF%"DPOWFSUFSJTBTTIPXOJO
'JH
Answer:4FF'JH
8.0 V
vs
8
7
4.0 V
0V
0
‒1
‒3
1
2 3
4
t (s)
2.0 s
V(V1:+)
4.0 s
V(C1:1)
Time
6.0 s
Figure 8.57
Figure 8.56
'PS1SBDUJDF1SPC
8.12
‒4.0 V
0s
3FTVMUPG1SBDUJDF1SPC
Summary
5IFEFUFSNJOBUJPOPGUIFJOJUJBMW BMVFTY BOE EY ∕EU and final
WBMVFY ∞ JTDSVDJBMUPBOBMZ[JOHTFDPOEPSEFSDJSDVJUT
5IF 3-$DJSDVJUJTTFDPOEPSEFSCFDBVTFJUJTEFTDSJCFECZB
TFDPOEPSEFSEJG GFSFOUJBMFRVBUJPO*UTDIBSBDUFSJTUJDFRVBUJPOJT 355
Review Questions
T + αT + ω = XIFSF αJTUIFOFQFSGSFRVFOD ZBOE ωJTUIF
VOEBNQFEOBUVSBMGSFRVFODZ'PSB TFSJFTDJSDVJU α = 3∕@@@ GPSB
QBSBMMFMDJSDVJUα = ∕3$ BOEGPSCPUIDBTFTω = ∕Ŀ-$
*GUIFSFBSFOPJOEFQFOEFOUTPVSDFTJOUIFDJSDVJUBGUFSTXJUDIJOH PS
TVEEFODIBOHF XFSFHBSEUIFDJSDVJUBTTPVSDFGSFF 5IFDPNQMFUF
TPMVUJPOJTUIFOBUVSBMSFTQPOTF
5IFOBUVSBMSFTQPOTFPGBO 3-$DJSDVJUJTP WFSEBNQFE VOEFS EBNQFE PSDSJUJDBMMZEBNQFE EFQFOEJOHPOUIFSPPUTPGUIFDIBSBD
UFSJTUJDFRVBUJPO5IFSFTQPOTFJTDSJUJDBMMZEBNQFEXIFOUIFSPPUT
BSFFRVBM T = TPS α = ω PWFSEBNQFEXIFOUIFSPPUTBSFSFBM
BOEVOFRVBM T ≠ TPSα > ω PSVOEFSEBNQFEXIFOUIFSPPUTBSF
DPNQMFYDPOKVHBUF T = T PSα < ω *GJOEFQFOEFOU TPVSDFTBSF QSFTFOUJOUIFDJSDVJU BGUFSTXJUDIJOH UIFDPNQMFUFSFTQPOTFJTUIFTVNPGUIFUSBOTJFOUSFTQPOTFBOEUIF
TUFBEZTUBUFSFTQPOTF
14QJDFJTVTFEUPBOBMZ[F 3-$DJSDVJUTJOUIFTBNFX BZBTGPS 3$
PS3-DJSDVJUT
5XPDJSDVJUTBSFEVBMJGUIFNFTIFRVBUJPOTUIBUEFTDSJCFPOFDJSDVJU
IBWFUIFTBNFGPSNBTUIFOPEBMFRVBUJPOTUIBUEFTDSJCFUIFPUIFS5IF
BOBMZTJTPGPOFDJSDVJUHJWFTUIFBOBMZTJTPGJUTEVBMDJSDVJU
5IFBVUPNPCJMFJHOJUJPODJSDVJUBOEUIFTNPPUIJOHDJSDVJUBSFUZQJDBM
BQQMJDBUJPOTPGUIFNBUFSJBMDPWFSFEJOUIJTDIBQUFS
Review Questions
'PSUIFDJSDVJUJO'JH UIFDBQBDJUPSWPMUBHFBU
U = − KVTUCFGPSFUIFTXJUDIJTDMPTFE JT
B 7
C 7
D 7
*GUIFSPPUTPGUIFDIBSBDUFSJTUJDFRVBUJPOPGBO3-$
DJSDVJUBSF −BOE − UIFSFTQPOTFJT
B "DPTU + #TJOU F−U
C " + #U F−U
D "F−U + #UF−U
E "F−U + #F−U
E 7
t=0
2Ω
4Ω
12 V +
‒
1H
2F
Figure 8.58
'PS3FWJFX2VFTUJPOTBOE
'PSUIFDJSDVJUJO'JH UIFJOJUJBMJOEVDUPS
DVSSFOU BUU = JT
B "
C "
D "
E "
8IFOBTUFQJOQVUJTBQQMJFEUPBTFDPOEPSEFS
circuit, the final values of the circuit variables are
GPVOECZ
B 3FQMBDJOHDBQBDJUPSTXJUIDMPTFEDJSDVJUTBOE
JOEVDUPSTXJUIPQFODJSDVJUT
C 3FQMBDJOHDBQBDJUPSTXJUIPQFODJSDVJUTBOE
JOEVDUPSTXJUIDMPTFEDJSDVJUT
D %PJOHOFJUIFSPGUIFBCPWF
XIFSF"BOE#BSFDPOTUBOUT
*OBTFSJFT3-$DJSDVJU TFUUJOH3 = XJMMQSPEVDF
B BOPWFSEBNQFESFTQPOTF
C BDSJUJDBMMZEBNQFESFTQPOTF
D BOVOEFSEBNQFESFTQPOTF
E BOVOEBNQFESFTQPOTF
F OPOFPGUIFBCPWF
"QBSBMMFM3-$DJSDVJUIBT- = )BOE$ = '5IF
WBMVFPG3UIBUXJMMQSPEVDFBVOJUZOFQFSfrequency JT
B Ω C Ω
D Ω
E Ω
3FGFSUPUIFTFSJFT3-$DJSDVJUJO'JH 8IBU
LJOEPGSFTQPOTFXJMMJUQSPEVDF
B PWFSEBNQFE
C VOEFSEBNQFE
D DSJUJDBMMZEBNQFE
E OPOFPGUIFBCPWF
356
Chapter 8
1Ω
Second-Order Circuits
1H
R
L
vs +
‒
1F
C
Figure 8.59
L
R
is
C
(a)
(b)
'PS3FWJFX2VFTUJPO
C1
R
$POTJEFSUIFQBSBMMFM3-$DJSDVJUJO'JH 8IBU
UZQFPGSFTQPOTFXJMMJUQSPEVDF
R1
C1
B PWFSEBNQFE
C VOEFSEBNQFE
D DSJUJDBMMZEBNQFE
E OPOFPGUIFBCPWF
R2
is
vs +
‒
C2
(c)
C2
(d)
R1
R1
C
L
R2
vs +
‒
is
1Ω
L
1H
1F
R2
C
(e)
Figure 8.60
L
(f)
Figure 8.61
'PS3FWJFX2VFTUJPO
'PS3FWJFX2VFTUJPO
*OBOFMFDUSJDDJSDVJU UIFEVBMPGSFTJTUBODFJT
.BUDIUIFDJSDVJUTJO'JHXJUIUIFGPMMPXJOH
JUFNT
J first-order circuit
JJ TFDPOEPSEFSTFSJFTDJSDVJU
JJJ TFDPOEPSEFSQBSBMMFMDJSDVJU
JW OPOFPGUIFBCPWF
B DPOEVDUBODF
D DBQBDJUBODF
F TIPSUDJSDVJU
C JOEVDUBODF
E PQFODJSDVJU
"OTXFSTB D C E E D C
J D JJ C F JJJ B JW E G B
C Problems
Section 8.2
Finding Initial and Final Values
For the circuit in Fig. 8.62, find:
B J + BOEv +
C EJ + ∕EUBOEEv + ∕EU
D J ∞ BOEv ∞ 6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
students better understand finding initial and final
WBMVFT
t=0
iR
i
2H
Figure 8.62
'PS1SPC
R2
4Ω
6Ω
12 V +
‒
R1
0.4 F
+
v
‒
v +
‒
Figure 8.63
'PS1SPC
R3
iC
C
t=0
iL
L
357
Problems
3FGFSUPUIFDJSDVJUTIPXOJO'JH $BMDVMBUF
Rs
B J- + v$ + BOEv3 +
C EJ- + ∕EU Ev$ + ∕EU BOEEv3 + ∕EU
D J- ∞ v$ ∞ BOEv3 ∞ +
vR
‒
10 Ω
2u(t) A
C
+
vL
‒
L
Figure 8.67
IL
1
4F
+
‒
+ vR ‒
Vs u(t) +
‒
40 Ω
+
vC
‒
R
1
8H
10 V
'PS1SPC
Section 8.3
Source-Free Series RLC Circuit
"TFSJFT3-$DJSDVJUIBT3 = LΩ - = N) BOE$ = μ'8IBUUZQFPGEBNQJOHJTFYIJCJUFECZ
UIFDJSDVJU
Figure 8.64
'PS1SPC
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFSVOEFS
TUBOETPVSDFGSFF3-$DJSDVJUT
5IFDVSSFOUJOBO3-$DJSDVJUJTEFTDSJCFECZ
In the circuit of Fig. 8.65, find:
B v + BOEJ +
C Ev + ∕EUBOEEJ + ∕EU
D v ∞ BOEJ ∞ 3Ω
+
‒
0.1 F
+
v
‒
5Ω
4u(t) A
Ei + i = Ei+ 1
@@@
@@
EU
EU
(JWFOUIBUi = Ei ∕EU = N"T PCUBJOi U 5IFOBUVSBMSFTQPOTFPGBO3-$DJSDVJUJTEFTDSJCFE
CZUIFEJGGFSFOUJBMFRVBUJPO
Figure 8.65
'PS1SPC
3FGFSUPUIFDJSDVJUJO'JH%FUFSNJOF
B J + BOEv +
C EJ∕ + EUBOEEv + ∕EU
D J ∞ BOEv ∞ i
4Ω
Ev
Ev
@@@
@@@
+ + v = EU
EU
GPSXIJDIUIFJOJUJBMDPOEJUJPOTBSFv = 7BOE
Ev ∕EU = 4PMWFGPSv U *G3 = Ω - = ) XIBUWBMVFPG$XJMMNBLF
BO3-$TFSJFTDJSDVJU
B PWFSEBNQFE
C DSJUJDBMMZEBNQFE
D VOEFSEBNQFE
1H
4u(t) A
*GJ = "BOEEJ ∕EU = 0, find J U GPSU >
5IFEJGGFSFOUJBMFRVBUJPOUIBUEFTDSJCFTUIFDVSSFOUJO
BO3-$OFUXPSLJT
0.25 H
i
4u(–t) V
EJ+ EJ + J = @@@
@@
EU
EU
1
4F
6Ω
+
v
‒
'PSUIFDJSDVJUJO'JH DBMDVMBUFUIFWBMVFPG3
OFFEFEUPIBWFBDSJUJDBMMZEBNQFESFTQPOTF
60 Ω
Figure 8.66
'PS1SPC
In the circuit of Fig. 8.67, find:
B v3 + BOEv- +
C Ev3 + ∕EUBOEEv- + ∕EU
D v3 ∞ BOEv- ∞ R
0.01 F
4H
Figure 8.68
'PS1SPC
5IFTXJUDIJO'JHNPWFTGSPNQPTJUJPO"UP
QPTJUJPO#BUU = QMFBTFOPUFUIBUUIFTXJUDINVTU
358
Chapter 8
Second-Order Circuits
DPOOFDUUPQPJOU#CFGPSFJUCSFBLTUIFDPOOFDUJPOBU
" BNBLFCFGPSFCSFBLTXJUDI -FUv = 0, find
v U GPSU >
30 Ω
A =
t 0
B
80 V +
‒
t=0
100 V +
‒
1Ω
Figure 8.72
10 Ω
'PS1SPC
0CUBJOv U GPSU >JOUIFDJSDVJUPG'JH
Figure 8.69
'PS1SPC
v
10 Ω
J- U = F
−U
− F
'JOEJ U GPSU >JOUIFDJSDVJUPG'JH
t=0
4H
Figure 8.73
'PS1SPC
5IFTXJUDIJOUIFDJSDVJUPG'JHIBTCFFODMPTFE
GPSBMPOHUJNFCVUJTPQFOFEBUU = %FUFSNJOFJ U GPSU >
60 Ω
1
2H
i(t)
2Ω
i(t)
1 mF
30 V +
‒
1F
N"
XIFSFv$BOEJ-BSFUIFDBQBDJUPSWPMUBHFBOE
JOEVDUPSDVSSFOU SFTQFDUJWFMZ%FUFSNJOFUIFWBMVFT
PG3 - BOE$
10 Ω
‒
120 V +
‒
v$ U = − F−U + F−U7
+
t=0
5IFSFTQPOTFTPGBTFSJFT3-$DJSDVJUBSF
−U
1F
0.25 H
4H
+
v(t)
‒
0.25 F
5Ω
40 Ω
30 V
+‒
2.5 H
1
4F
Figure 8.70
'PS1SPC
t=0
Figure 8.74
'PS1SPC
*OUIFDJSDVJUPG'JH UIFTXJUDIJOTUBOUBOFPVTMZ
NPWFTGSPNQPTJUJPO"UP#BUU = 'JOEv U GPSBMM
U≥
t=0
A
$BMDVMBUFv U GPSU >JOUIFDJSDVJUPG'JH
15 Ω
0.25 H
5A
4Ω
6Ω
12 Ω
B
10 Ω
0.04 F
+
v (t)
‒
Figure 8.71
'PS1SPC
t=0
24 V +
‒
60 Ω
+
v
‒
Figure 8.75
'PS1SPC
'JOEUIFWPMUBHFBDSPTTUIFDBQBDJUPSBTBGVODUJPO
PGUJNFGPSU> GPSUIFDJSDVJUJO'JH"TTVNF
TUFBEZTUBUFDPOEJUJPOTFYJTUBUU = −
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
3H
1
27 F
25 Ω
359
Problems
Source-Free Parallel RLC Circuit
Section 8.4
"TTVNJOH3 = LΩ EFTJHOBQBSBMMFM3-$DJSDVJUUIBU
IBTUIFDIBSBDUFSJTUJDFRVBUJPO
*GUIFJOJUJBMDPOEJUJPOTBSFv = = Ev ∕EU, find
vU
"TFSJFT3-$DJSDVJUJTEFTDSJCFECZ
EJ+ 3
EJ + @@
J =
@@@
@@
-
EU $
EU
T + T + = 'PSUIFOFUXPSLJO'JH XIBUWBMVFPG$JT
OFFEFEUPNBLFUIFSFTQPOTFVOEFSEBNQFEXJUI
VOJUZOFQFSGSFRVFODZ α = 10 Ω
20 mH
C
'JOEUIFSFTQPOTFXIFO- = ) 3 = Ω
BOE$ = '-FUJ = EJ ∕EU = 4PMWFUIFGPMMPXJOHEJGGFSFOUJBMFRVBUJPOTTVCKFDUUP
the specified initial conditions
10 mF
B Ev∕EU + v = v = Ev ∕EU = C EJ∕EU + EJ∕EU + J = J = − EJ ∕EU = D Ev∕EU + Ev∕EU + v = v = Ev ∕EU = E EJ∕EU + EJ∕EU + J = J = EJ ∕EU = − Figure 8.76
'PS1SPC
5IFTXJUDIJO'JHNPWFTGSPNQPTJUJPO"UP
QPTJUJPO#BUU = QMFBTFOPUFUIBUUIFTXJUDINVTU
DPOOFDUUPQPJOU#CFGPSFJUCSFBLTUIFDPOOFDUJPOBU
" BNBLFCFGPSFCSFBLTXJUDI %FUFSNJOFJ U GPS
U > 0.
A
20 Ω
t =0
i(t)
10 mF
v$ = − F−U − F−U7 J- U = F−U + F−UN" B
4A
5IFTUFQSFTQPOTFTPGBTFSJFT3-$DJSDVJUBSF
10 Ω
0.25 H
U >
U >
B 'JOE$ C %FUFSNJOFXIBUUZQFPGEBNQJOHJT
FYIJCJUFECZUIFDJSDVJU
$POTJEFSUIFDJSDVJUJO'JH'JOEv- + BOE
v$ + Figure 8.77
'PS1SPC
40 Ω
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETPVSDFGSFF3-$DJSDVJUT
R1
io(t)
L
2u(t)
0.5 H
+
vL
‒
10 Ω
1F
+
vC
‒
+
‒
50 V
Figure 8.79
'PS1SPC
v +
‒
t=0
R2
C
+
vo(t)
‒
For the circuit in Fig. 8.80, find v U GPSU >
Figure 8.78
'PS1SPC
Section 8.5
3u(–t) A
Step Response of a Series RLC Circuit
EJ+
EJ + J = @@@
@@
EU
EU
+ v ‒
4Ω
+‒
(JWFOUIBUJ = BOEEJ ∕EU = TPMWFGPSJ U "CSBODIWPMUBHFJOBO3-$DJSDVJUJTEFTDSJCFECZ
Ev
Ev
@@@
@@@
+ + v = EU
EU
0.04 F
1H
5IFTUFQSFTQPOTFPGBO3-$DJSDVJUJTHJWFOCZ
75u(t) V
Figure 8.80
'PS1SPC
2Ω
360
Chapter 8
Second-Order Circuits
'PSUIFOFUXPSLJO'JH TPMWFGPSJ U GPSU >
'JOEv U GPSU >JOUIFDJSDVJUPG'JH
4.5 A
6Ω
1H
t=0
+
v
‒
10 Ω
i(t)
5Ω
4F
6Ω
6Ω
6u(t) A
t=0
30 V +
‒
Figure 8.81
1
8F
1
2H
10 V +
‒
'PS1SPC
Figure 8.85
$BMDVMBUFJ U GPSU >JOUIFDJSDVJUPG'JH
'PS1SPC
3FGFSUPUIFDJSDVJUJO'JH$BMDVMBUFJ U GPSU > 5(1 ‒ u(t)) A
+ v ‒
35u(‒t) V +
‒
i(t)
i
1
16 F
3
4H
1
4H
1
3F
5Ω
10 Ω
5Ω
Figure 8.82
10 Ω
'PS1SPC
Figure 8.86
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUIFTUFQSFTQPOTFPGTFSJFT
3-$DJSDVJUT
R
+ V
2
‒
%FUFSNJOFv U GPSU >JOUIFDJSDVJUPG'JH
3Ω
C
20u(t) V +
‒
+
+
v
‒
250 mH
4Ω
20u(t) A
t=0
V1 +
‒
'PS1SPC
v(t)
‒
5Ω
500 mF
Figure 8.87
'PS1SPC
5IFTXJUDIJOUIFDJSDVJUPG'JHJTNPWFEGSPN
QPTJUJPOBUPCBUU = "TTVNFUIBUUIFWPMUBHF
BDSPTTUIFDBQBDJUPSJTFRVBMUP[FSPBUU=BOEUIBU
UIFTXJUDIJTBNBLFCFGPSFCSFBLTXJUDI%FUFSNJOF
J U GPSBMMU>
L
Figure 8.83
'PS1SPC
0.02 F 14 Ω
0CUBJOv U BOEJ U GPSU >JOUIFDJSDVJUPG
Fig. 8.84.
2Ω
24 V +
‒
5H
i(t)
i(t)
4Ω
2Ω
‒
v(t)
'PS1SPC
a
t=0
10u(t) A
6Ω
+
6A
200 mF
Figure 8.84
b
2H
Figure 8.88
'PS1SPC
2Ω
+ 18 V
‒
361
Problems
For the network in Fig. 8.89, find J U GPSU >
3Ω
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUIFTUFQSFTQPOTFPGB
QBSBMMFM3-$DJSDVJU
1H
i(t)
i(t)
1Ω
40u(t) A
L
20 V +
‒
40 mF
v +
‒
C
R
Figure 8.89
'PS1SPC
Figure 8.93
'PS1SPC
Given the network in Fig. 8.90, find v U GPSU >
'JOEUIFPVUQVUWPMUBHFvP U JOUIFDJSDVJUPG
Fig. 8.94.
2A
1H
6Ω
4A
1Ω
t=0
1
25 F
t=0
+
v
‒
10 Ω
6A
5Ω
+
vo
‒
10 mF
1H
Figure 8.90
'PS1SPC
Figure 8.94
5IFTXJUDIJO'JHJTPQFOFEBUU = BGUFSUIF
DJSDVJUIBTSFBDIFETUFBEZTUBUF$IPPTF3BOE$
TVDIUIBUα = /QTBOEωE = SBET
10 Ω
t=0
'PS1SPC
Given the circuit in Fig. 8.95, find J U BOEv U GPS
U > 0.
i(t)
R
+ 60 V
‒
0.5 H
1H
C
2Ω
Figure 8.91
'PS1SPC
"TFSJFT3-$DJSDVJUIBTUIFGPMMPXJOHQBSBNFUFST
3 = LΩ - = ) BOE$ = O'8IBUUZQFPG
EBNQJOHEPFTUIJTDJSDVJUFYIJCJU
Section 8.6
Step Response of a Parallel
RLC Circuit
t=0
6V +
‒
Figure 8.95
'PS1SPC
%FUFSNJOFJ U GPSU >JOUIFDJSDVJUPG'JH
In the circuit of Fig. 8.92, find v U BOE
J U GPSU >
[1 + 5u(t)] A
2Ω
+
v
‒
5Ω
i
0.5 F
1H
3[1 – u(t)] A
50 mF
+ 45 V
‒
5H
i(t)
Figure 8.92
'PS1SPC
+
v(t)
‒
1
4F
1Ω
Figure 8.96
'PS1SPC
362
Chapter 8
Second-Order Circuits
For the circuit in Fig. 8.101, find v U GPSU >
"TTVNFUIBUJ + = "
For the circuit in Fig. 8.97, find J U GPSU >
10 Ω
2Ω
i(t)
45 V +
‒
40 Ω
10 mF
9u(t) A
4H
10 H
+
v(t)
‒
i(t)
+
‒
2i(t)
2H
Figure 8.101
Figure 8.97
'PS1SPC
'PS1SPC
'JOEv U GPSU >JOUIFDJSDVJUPG'JH
In the circuit of Fig. 8.102, find J U GPSU >
4Ω
t=0
io
R
+
v
‒
L
i
C
t=0
60 V +
‒
Figure 8.98
6Ω
1
25 F
1
4H
'PS1SPC
Figure 8.102
'PS1SPC
5IFTUFQSFTQPOTFPGBQBSBMMFM3-$DJSDVJUJT
−U
v = + F
DPTU − TJOU 7 U≥
XIFOUIFJOEVDUPSJTN)'JOE3BOE$
Section 8.7
General Second-Order Circuits
(JWFOUIFDJSDVJUTIPXOJO'JH EFUFSNJOFUIF
DIBSBDUFSJTUJDFRVBUJPOPGUIFDJSDVJUBOEUIFWBMVFT
GPSJ U BOEW U GPSBMMU >
"GUFSCFJOHPQFOGPSBEBZ UIFTXJUDIJOUIFDJSDVJU
PG'JHJTDMPTFEBUU = 'JOEUIFEJGGFSFOUJBM
FRVBUJPOEFTDSJCJOHJ U U >
8Ω
t=0
80 Ω
2[1 – u(t)] A
i
120 V +
‒
10 mF
1
18 F
2H
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEHFOFSBMTFDPOEPSEFS
DJSDVJUT
A t=0
R1
*OUIFDJSDVJUPG'JH UIFTXJUDIIBTCFFOJO
QPTJUJPOGPSBMPOHUJNFCVUNPWFEUPQPTJUJPOBU
U = 'JOE
B v + Ev + ∕EU
C v U GPSU≥
2
R3
R2
+
v
‒
C
L
1
8Ω
t=0
i
B
'PS1SPC
12 Ω
Figure 8.103
'PS1SPC
Figure 8.100
+
v(t)
‒
0.25 H
Figure 8.99
I
i(t)
0.25 H
Figure 8.104
'PS1SPC
0.5 Ω
v
+
‒
1F
+
10 V ‒
363
Problems
5IFTXJUDIJO'JHIBTCFFOJOQPTJUJPOGPS
U < 0. At U = JUJTNPWFEGSPNQPTJUJPOUPUIF
UPQPGUIFDBQBDJUPSBUU = 1MFBTFOPUFUIBUUIF
TXJUDIJTBNBLFCFGPSFCSFBLTXJUDI JUTUBZTJO
DPOUBDUXJUIQPTJUJPOVOUJMJUNBLFTDPOUBDUXJUI
UIFUPQPGUIFDBQBDJUPSBOEUIFOCSFBLTUIFDPOUBDUBU
position 1. Given that the initial voltage across the
DBQBDJUPSJTFRVBMUP[FSP EFUFSNJOFv U 4Ω
1 t=0
60 V
v
+
‒
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOETFDPOEPSEFSPQBNQ
DJSDVJUT
C1
R1
4H
1
16 F
16 Ω
R2
vs +
‒
+
vo
C2
‒
Figure 8.109
Figure 8.105
'PS1SPC
0CUBJOJBOEJGPSU >JOUIFDJSDVJUPG'JH
4Ω
6Ω
i1
20u(t) V +
–
i2
2H
2H
'PS1SPC
%FUFSNJOFUIFEJGGFSFOUJBMFRVBUJPOGPSUIFPQBNQ
DJSDVJUJO'JH*Gv + = 7BOE
v + = 0 V, find vPGPSU >-FU3 = LΩBOE
$ = μ'
Figure 8.106
R
'PS1SPC
For the circuit in Prob. 8.5, find JBOEvGPSU >
'JOEUIFSFTQPOTFv3 U GPSU >JOUIFDJSDVJUPG
Fig. 8.107. Let 3 = Ω - = ) BOE$ = N'
C
+
10u(t) A
R
+
v(t)
‒
v1
C
‒
‒
+
C
+
R
v2
‒
+
‒
+
vo
‒
L
Figure 8.110
Figure 8.107
'PS1SPC
'PS1SPC
Section 8.8
+
‒
Second-Order Op Amp Circuits
For the op amp circuit in Fig. 8.108, find the differFOUJBMFRVBUJPOGPSJ U 0CUBJOUIFEJGGFSFOUJBMFRVBUJPOTGPSvP U JOUIFPQ
BNQDJSDVJUPG'JH
C
vs +
‒
Figure 8.108
'PS1SPC
10 μF
‒
+
100 kΩ
i
L
vs +
‒
Figure 8.111
'PS1SPC
100 kΩ
+
‒
R
20 μF
+
vo
‒
364
Chapter 8
Second-Order Circuits
*OUIFPQBNQDJSDVJUPG'JH EFUFSNJOFvP U GPSU >-FUvJO = V U 7 3 = 3 = LΩ $ = $
= μ'
0CUBJOv U GPS< U< TJOUIFDJSDVJUPG'JH
VTJOH14QJDFPS.VMUJ4JN
0.4 F
1H
+
v (t)
‒
20 Ω
6Ω
C1
R2
C2
R1
+
‒
vin
6Ω
13u(t) A
+ 39u(t) V
‒
Figure 8.116
vo
'PS1SPC
Figure 8.112
'PS1SPC
5IFTXJUDIJO'JHIBTCFFOJOQPTJUJPOGPSB
MPOHUJNF"UU = JUJTTXJUDIFEUPQPTJUJPO6TF
14QJDFPS.VMUJ4JN to find J U GPS< U <T
Section 8.9 PSpice Analysis of RLC Circuit
'PSUIFTUFQGVODUJPOvT = V U VTF14QJDFPS
.VMUJ4JN to find the response v U GPS< U< TJO
UIFDJSDVJUPG'JH
2Ω
+
1F
‒
i
2 kΩ
100 μF
'PS1SPC
%FTJHOBQSPCMFN UPCFTPMWFEVTJOH14QJDFPS
.VMUJ4JN UPIFMQPUIFSTUVEFOUTCFUUFSVOEFSTUBOE
TPVSDFGSFF3-$DJSDVJUT
'PS1SPC
(JWFOUIFTPVSDFGSFFDJSDVJUJO'JH VTF14QJDF
PS.VMUJ4JNUPHFUJ U GPS< U< T
5BLFv = 7BOEJ = "
i
1Ω
10 H
2.5 F
Section 8.10
Duality
%SBXUIFEVBMPGUIFDJSDVJUTIPXOJO'JH
4Ω
2Ω
+
v
‒
9V +
‒
Figure 8.114
1Ω
6Ω
3A
Figure 8.118
'PS1SPC
'PS1SPC
'PSUIFDJSDVJUJO'JH VTF14QJDFPS.VMUJ4JN
UPPCUBJOv U GPS< U< T"TTVNFUIBUUIF
DBQBDJUPSWPMUBHFBOEJOEVDUPSDVSSFOUBUU = BSF
CPUI[FSP
6Ω
'PS1SPC
100 mH
1 kΩ
2
t=0
Figure 8.117
v(t)
Figure 8.113
Figure 8.115
10 V +
‒
1
1H
vs +
‒
24 V +
‒
4 kΩ
2H
0CUBJOUIFEVBMPGUIFDJSDVJUJO'JH
12 V
3Ω
0.4 F
+
v
‒
+
‒
4Ω
Figure 8.119
'PS1SPC
0.5 F
10 Ω
+
‒
2H
24 V
365
Comprehensive Problems
'JOEUIFEVBMPGUIFDJSDVJUJO'JH
20 Ω
10 Ω
1F
Applications
"OBVUPNPCJMFBJSCBHJHOJUFSJTNPEFMFECZUIF
DJSDVJUJO'JH%FUFSNJOFUIFUJNFJUUBLFTUIF
voltage across the igniter to reach its first peak after
TXJUDIJOHGSPN"UP#-FU3 = Ω $ = ∕' BOE
- = N)
30 Ω
120 V
‒+
60 V
+‒
4H
Section 8.11
2A
A
B
t=0
12 V +
‒
Figure 8.120
'PS1SPC
%SBXUIFEVBMPGUIFDJSDVJUJO'JH
1Ω
R
'PS1SPC
3Ω
2Ω
0.25 H
Airbag igniter
L
Figure 8.122
5A
1F
C
+
‒
12 V
"MPBEJTNPEFMFEBTBN)JOEVDUPSJOQBSBM
MFMXJUIBΩSFTJTUPS"DBQBDJUPSJTOFFEFEUP
CFDPOOFDUFEUPUIFMPBETPUIBUUIFOFUXPSLJT
DSJUJDBMMZEBNQFEBU)[$BMDVMBUFUIFTJ[FPG
the capacitor.
Figure 8.121
'PS1SPC
Comprehensive Problems
"NFDIBOJDBMTZTUFNJTNPEFMFECZBTFSJFT3-$
DJSDVJU*UJTEFTJSFEUPQSPEVDFBOPWFSEBNQFESFTQPOTF
XJUIUJNFDPOTUBOUTBOENT*GBTFSJFT
LΩ resistor is used, find the values of -BOE$
"OPTDJMMPHSBNDBOCFBEFRVBUFMZNPEFMFECZB
TFDPOEPSEFSTZTUFNJOUIFGPSNPGBQBSBMMFM3-$
DJSDVJU*UJTEFTJSFEUPHJWFBOVOEFSEBNQFEWPMUBHF
BDSPTTBΩSFTJTUPS*GUIFEBNQFEGSFRVFODZJT
L)[BOEUIFUJNFDPOTUBOUPGUIFFOWFMPQFJTT find the necessary values of -BOE$
5IFDJSDVJUJO'JHJTUIFFMFDUSJDBMBOBMPHPG
CPEZGVODUJPOTVTFEJONFEJDBMTDIPPMTUPTUVEZ
DPOWVMTJPOT5IFBOBMPHJTBTGPMMPXT
$ = Volume of fluid in a drug
$ = Volume of blood stream in a specified region
3 = 3FTJTUBODFJOUIFQBTTBHFPGUIFESVHGSPN
UIFJOQVUUPUIFCMPPETUSFBN
3 = 3FTJTUBODFPGUIFFYDSFUJPONFDIBOJTN TVDI
BTLJEOFZ FUD
v = *OJUJBMDPODFOUSBUJPOPGUIFESVHEPTBHF
v U = 1FSDFOUBHFPGUIFESVHJOUIFCMPPETUSFBN
'JOEv U GPSU >HJWFOUIBU$ = μ' $ =
5 μ' 3 = .Ω 3 = .Ω BOEv = V U 7
R1
t=0
+
vo
‒
C2
R2
C1
+
v (t)
‒
Figure 8.123
'PS1SPC
'JHVSFTIPXTBUZQJDBMUVOOFMEJPEFPTDJM
MBUPSDJSDVJU5IFEJPEFJTNPEFMFEBTBOPOMJOFBS
resistor with J% = G v% JF UIFEJPEFDVSSFOUJTB
OPOMJOFBSGVODUJPOPGUIFWPMUBHFBDSPTTUIFEJPEF
%FSJWFUIFEJGGFSFOUJBMFRVBUJPOGPSUIFDJSDVJUJO
UFSNTPGvBOEJ%
R
vs
+
‒
Figure 8.124
'PS1SPC
L
i
+
v
‒
C
ID
+
vD
‒
P A R T
T W O
AC Circuits
9
Sinusoids and Phasors
10
Sinusoidal Steady-State Analysis
11
AC Power Analysis
12
Three-Phase Circuits
13
Magnetically Coupled Circuits
14
Frequency Response
NASA
OUTLINE
c h a p t e r
9
Sinusoids and
Phasors
)FXIPLOPXTOPU BOELOPXTOPUUIBUIFLOPXTOPU JTa fool—shun IJN
)FXIPLOPXTOPU BOELOPXTUIBUIFLOPXTOPU JTBD IJME‡UFBDIIJN
)FXIPLOPXT BOELOPXTOPUUIBUIFLOPXT JT BTMFFQ‡XBLF IJN VQ
)FXIPLOPXT BOELOPXTUIBUIFLOPXT JTXJTF‡GPMMPXIJN
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Enhancing Your Skills and Your Career
ABET EC 2000 criteria (3.d), “an ability to function on
multi-disciplinary teams.”
5IFiBCJMJUZUPGVODUJPOPONVMUJEJTDJQMJOBSZUFBNTuJTJOIFSFOUMZDSJUJDBM
GPSUIFXPSLJOHFOHJOFFS&OHJOFFSTSBSFMZ JGFWFS XPSLCZUIFNTFMWFT
&OHJOFFSTXJMMBMXBZTCFQBSUPGTPNFUFBN0OFPGUIFUIJOHT*MJL FUP
SFNJOETUVEFOUTJTUIBUZPVEPOPUIB WFUPMJLFFWFSZPOFPOBUFBNZPV
KVTUIBWFUPCFBTVDDFTTGVMQBSUPGUIBUUFBN
.PTUGSFRVFOUMZ UIFTFUFBNTJODMVEFJOEJ WJEVBMTGSPNBW BSJFUZPG
FOHJOFFSJOHEJTDJQMJOFT BTXFMMBTJOEJWJEVBMTGSPNOPOFOHJOFFSJOHEJT
DJQMJOFTTVDIBTNBSLeting and finance.
4UVEFOUTDBOFBTJMZEF WFMPQBOEFOIBODFUIJTTLJMMCZX PSLJOHJO
TUVEZHSPVQTJOFWFSZDPVSTFUIFZUBLF$MFBSMZ XPSLJOHJOTUVEZHSPVQT
JOOPOFOHJOFFSJOHDPVSTFT BTXFMMBTFOHJOFFSJOHDPVSTFTPVUTJEFZPVS
EJTDJQMJOF XJMMBMTPHJWFZPVFYQFSJFODFXJUINVMUJEJTDJQMJOBSZUFBNT
1IPUPCZ$IBSMFT"MFYBOEFS
367
368
Chapter 9
Sinusoids and Phasors
Historical
Nikola Tesla m BOEGeorge Westinghouse m (FPSHF8FTUJOHIPVTF1IPUP
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IFMQFEFTUBCMJTIBMUFSOBUJOHDVSSFOUBTUIFQSJNBSZNPEFPGFMFDUSJDJUZ
USBOTNJTTJPOBOEEJTUSJCVUJPO
5PEBZJUJTPCWJPVTUIBUBDHFOFSBUJPOJTXFMMFTUBCMJTIFEBTUIFGPSN
PGFMFDUSJDQPXFSUIBUNBLFTXJEFTQSFBEEJTUSJCVUJPOPGFMFDUSJDQPXFSFGfi
DJFOUBOEFDPOPNJDBM)PXFWFS BUUIFFOEPGUIFUIDFOUVSZXIJDIXBT
UIFCFUUFS‡BDPSED‡X BTIPUMZEFCBUFEBOEIBEF YUSFNFMZPVU TQPLFO
TVQQPSUFSTPOCPUITJEFT5IFEDTJEFXBTMFECZ5IPNBT&EJTPO XIPIBE
FBSOFEBMPUPGSFTQFDUGPSIJTNBOZDPOUSJCVUJPOT1PXFSHFOFSBUJPOVTJOH
BDSFBMMZCF HBOUPC VJMEBGUFSUIFTVDDFTTGVMDPOUSJC VUJPOTPG 5FTMB5IF
SFBMDPNNFSDJBMTVDDFTTJOBDDBNFGSPN(FPS HF8FTUJOHIPVTFBOEUIF PVUTUBOEJOHUFBN JODMVEJOH 5FTMB IFBTTFNCMFE*OBEEJUJPO UX PPUIFS
CJHOBNFTXFSF$'4DPUUBOE#(-BNNF
The most significant contribVUJPOUPUIFFBSMZTVDDFTTPGBDX BTUIF
QBUFOUJOHPGUIFQPMZQIBTFBDNPUPSCZ 5FTMBJO 5IFJOEVDUJPO
NPUPSBOEQPMZQIBTFHFOFSBUJPOBOEEJTUSJC VUJPOTZTUFNTEPPNFEUIF
VTFPGEDBTUIFQSJNFFOFSHZTPVSDF
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
#FUUFSVOEFSTUBOETJOVTPJET
6OEFSTUBOEQIBTPST
6OEFSTUBOEUIFQIBTPSSFMBUJPOTIJQTGPSDJSDVJUFMFNFOUT
,OPXBOEVOEFSTUBOEUIFDPODFQUTPGJNQFEBODFBOE
BENJUUBODF
6OEFSTUBOE,JSDIIPGGTMBXTJOUIFGSFRVFODZEPNBJO
$PNQSFIFOEUIFDPODFQUPGQIBTFTIJGU
6OEFSTUBOEUIFDPODFQUPG"$CSJEHFT
9.1
Introduction
5IVTGBSPVS BOBMZTJTIBTCFFO MJNJUFEGPSUIF NPTUQBSU UPED DJSDVJUT
UIPTFDJSDVJUTF YDJUFECZDPOTUBOUPSUJNFJO WBSJBOUTPVSDFT 8FIB WF
SFTUSJDUFEUIFGPSDJOHGVODUJPOUPEDTPVSDFTGPSUIFTBLFPGTJNQMJDJUZ GPS
QFEBHPHJDSFBTPOT BOEBMTPGPSIJTUPSJDSFBTPOT)JTUPSJDBMMZ EDTPVSDFT
XFSFUIFNBJONFBOTPGQSPWJEJOHFMFDUSJDQPXFSVQVOUJMUIFMBUFT
"UUIFFOEPGUIBUDFOUVSZ UIFCBUUMFPGEJSFDUDVSSFOUWFSTVTBMUFSOBUJOH
DVSSFOUCFHBO#PUIIBEUIFJSBEWPDBUFTBNPOHUIFFMFDUSJDBMFOHJOFFST
PGUIFUJNF#FDBVTFBDJTNPSFFGficient and economical UPUSBOTNJUPWFS
MPOHEJTUBODFT BDTZTUFNTFOEFEVQUIFXJOOFS 5IVT JUJTJOL FFQJOH
XJUIUIFIJTUPSJDBMTFRVFODFPGFWents that we considered dc sources first.
8FOPXCFHJOUIFBOBMZTJTPGDJSDVJUTJOXIJDIUIFTPVSDFW PMUBHFPS
DVSSFOUJTUJNFW BSZJOH*OUIJTDIBQUFS XFBSFQBSUJDVMBSMZJOUFSFTUFEJO
TJOVTPJEBMMZUJNFWBSZJOHFYDJUBUJPO PSTJNQMZ FYDJUBUJPOCZBTJOVTPJE
9.2
Sinusoids
A sinusoid is a signal that has the form of the sine or cosine function.
"TJOVTPJEBMDVSSFOUJTVTVBMMZSFGFSSFEUPBT BMUFSOBUJOHDVSSFOU BD 4VDIBDVSSFOUSFWFSTFTBUSFHVMBSUJNFJOUFSWBMTBOEIBTBMUFSOBUFMZQPTJ
UJWFBOEOFHBUJWFWBMVFT$JSDVJUTESJWFOCZTJOVTPJEBMDVSSFOUPSWPMUBHF
TPVSDFTBSFDBMMFEBDDJSDVJUT
8FBSFJOUFSFTUFEJOTJOVTPJETGPSBOVNCFSPGSFBTPOT'JSTU OBUVSF
JUTFMGJTDIBSBDUFSJTUJDBMMZTJOVTPJEBM 8FF YQFSJFODFTJOVTPJEBMW BSJB
UJPOJOUIFNPUJPOPGBQFOEVMVN UIFWJCSBUJPOPGBTUSJOH UIFSJQQMFT
POUIFPDFBOTVSGBDF BOEUIFOBUVSBMSFTQPOTFPGVOEFSEBNQFETFDPOE PSEFSTZTUFNT UPNFOUJPOCVUBGFX4FDPOE BTJOVTPJEBMTJHOBMJTFBTZUP
HFOFSBUFBOEUSBOTNJU*UJTUIFGPSNPGW PMUBHFHFOFSBUFEUISPVHIPVUUIF
XPSMEBOETVQQMJFEUPIPNFT GBDUPSJFT MBCPSBUPSJFT BOETPPO*UJTUIF
EPNJOBOUGPSNPGTJHOBMJOUIFDPNNVOJDBUJPOTBOEFMFDUSJDQPXFSJO
EVTUSJFT5IJSE UISPVHI'PVSJFSBOBMZTJT BOZQSBDUJDBMQFSJPEJDTJHOBM
DBOCFSFQSFTFOUFECZBTVNPGTJOVTPJET4JOVTPJET UIFSFGPSF QMBZ
BOJNQPSUBOUSPMFJOUIFBOBMZTJTPGQFSJPEJDTJHOBMT-BTUMZ BTJOV TPJEJTFBTZUPIBOEMFNBUIFNBUJDBMMZ 5IFEFSJWBUJWFBOE JOUFHSBMPG
BTJOVTPJEBSFUIFNTFMWF TTJOVTPJET'PSUIFTFBOEPUIFSSFBTPOT UIF TJOVTPJEJTBOFYUSFNFMZJNQPSUBOUGVODUJPOJODJSDVJUBOBMZTJT
"TJOVTPJEBMGPSDJOHGVODUJPOQSPEVDFTCPUIBUSBOTJFOUSFTQPOTF
BOEBTUFBEZTUBUFSFTQPOTF NVDIMJLFUIFTUFQGVODUJPO XIJDIXFTUVE
JFEJO$IBQUFSTBOE5IFUSBOTJFOUSFTQPOTFEJFTPVUXJUIUJNFTPUIBU
POMZUIFTUFBEZTUBUFSFTQPOTFSFNBJOT8IFOUIFUSBOTJFOUSFTQPOTFIBT
CFDPNFOFHMJHJCMZTNBMMDPNQBSFEXJUIUIFTUFBEZTUBUFSFTQPOTF XF
TBZUIBUUIFDJSDVJUJTPQFSBUJOHBUTJOVTPJEBMTUFBEZTUBUF*UJTUIJT TJOV
TPJEBMTUFBEZTUBUFSFTQPOTFUIBUJTPGNBJOJOUFSFTUUPVTJOUIJTDIBQUFS
8FCFHJOXJUIBCBTJDEJTDVTTJPOPGTJOVTPJETBOEQIBTPST8FUIFO
JOUSPEVDFUIFDPODFQUTPGJNQFEBODFBOEBENJUUBODF 5IFCBTJDDJSDVJU
MBXT ,JSDIIPGGTBOE0INT JOUSPEVDFEGPSEDDJSDVJUT XJMMCFBQQMJFE
UPBDDJSDVJUT'JOBMMZ XFDPOTJEFSBQQMJDBUJPOTPGBDDJSDVJUTJOQIBTF
TIJGUFSTBOECSJEHFT
9.2
Sinusoids
$POTJEFSUIFTJOVTPJEBMWPMUBHF
v U = 7NTJOωU
XIFSF
7N = UIFBNQMJUVEFPGUIFTJOVTPJE
ω = UIFBOHVMBSGSFRVFODZJOSBEJBOTT
ωU = UIFBSHVNFOUPGUIFTJOVTPJE
5IFTJOVTPJEJTTIPXOJO'JH B BTBGVODUJPOPGJUTBSHVNFOUBOEJO
'JH C BTBGVODUJPOPGUJNF*UJTF WJEFOUUIBUUIFTJOVTPJESFQFBUT
JUTFMGFWFSZ5TFDPOETUIVT 5JTDBMMFEUIFQFSJPEPGUIFTJOVTPJE'SPN
UIFUXPQMPUTJO'JH XFPCTFSWFUIBUω5 = π
π
5 = @@@
ω
369
370
Chapter 9
Sinusoids and Phasors
Historical
Heinrich Rudorf Hertz m B(FSNBOFYQFSJNFOUBMQIZTJ
DJTU EFNPOTUSBUFEUIBUFMFDUSPNBHOFUJDXBWFTPCFZUIFTBNFGVOEBNFO
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UIFPSZBOEQSFEJDUJPOUIBUTVDIXBWFTFYJTUFE
)FSU[X BTCPSOJOUPBQSPTQFSPVTG BNJMZJO)BNC VSH (FSNBO Z
)FBUUFOEFEUIF6OJ WFSTJUZPG#FSMJOBOEEJEIJTEPDUPSBUFVOEFSUIF
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,BSMTSVIF XIFSFIFCF HBOIJTRVFTUGPSFMFDUSPNBHOFUJDX BWFT)FSU[
TVDDFTTGVMMZHFOFSBUFEBOEEFUFDUFEFMFDUSPNBHOFUJDX BWFTIFXBTUIF
first to shoXUIBUMJHIUJTFMFDUSPNBHOFUJDFOFS HZ*O )FSU[OPUFE
for the first time the photoelectric efGFDUPGFMFDUSPOTJOBNPMFDVMBSTUSVD
UVSF"MUIPVHI)FSU[POMZMJWFEUPUIFBHFPG IJTEJTDPWFSZPGFMFDUSP
NBHOFUJDXBWFTQBWFEUIFXBZGPSUIFQSBDUJDBMVTFPGTVDIXBWFTJOSB
EJP UFMFWJTJPO BOEPUIFSDPNNVOJDBUJPOTZTUFNT5IFVOJUPGGSFRVFODZ UIFIFSU[ CFBSTIJTOBNF
¥)VMUPO"SDIJWFT(FUUZ*NBHFT
v(t)
v(t)
Vm
Vm
0
‒Vm
π
4π
3π
2π
ωt
0
‒Vm
T
2
T
(a)
3T
2
2T
t
(b)
Figure 9.1
"TLFUDIPG7NTJOωU B BTBGVODUJPOPGωU C BTBGVODUJPOPGU
5IFGBDUUIBUv U SFQFBUTJUTFMGFWFSZ5TFDPOETJTTIPXOCZSFQMBDJOHU
CZU + 5JO&R 8FHFU
π v U + 5 = 7NTJOω U + 5 = 7NTJOω(U + @@@
ω)
= 7NTJO ωU + π = 7NTJOωU = v U
)FODF
v U+5 =v U
UIBUJT vIBTUIFTBNF WBMVFBUU + 5BTJUEPFTBU UBOEv U JTTBJEUPCF
QFSJPEJD*OHFOFSBM
A periodic function is one that satisfies f (t ) = f (t + nT ), for all t and for
all integers n.
"TNFOUJPOFE UIF QFSJPE5PG UIFQFSJPEJDGVODUJPO JTUIF UJNFPGPOF
DPNQMFUFDZDMFPSUIFOVNCFSPGTFDPOETQFSD ZDMF5IFSFDJQSPDBMPG
UIJTRVBOUJUZJTUIFOVNCFSPGD ZDMFTQFSTFDPOE LOP XOBTUIF DZDMJD
GSFRVFODZGPGUIFTJOVTPJE5IVT
G = @@
5
9.2
Sinusoids
371
'SPN&RT BOE JUJTDMFBSUIBU
ω = π G
The unit of f is named after the German
physicist Heinrich R. Hertz (1857–1894).
8IJMFωJTJOSBEJBOTQFSTFDPOE SBET GJTJOIFSU[ )[ -FUVTOPXDPOTJEFSBNPSFHFOFSBMFYQSFTTJPOGPSUIFTJOVTPJE
v U = 7NTJO ωU + ϕ XIFSF ωU + ϕ JTUIFBS HVNFOUBOEϕJTUIF QIBTF#PUIBSHVNFOUBOE
QIBTFDBOCFJOSBEJBOTPSEFHSFFT
-FUVTFYBNJOFUIFUXPTJOVTPJET
W U = 7NTJOωU
BOE
v U = 7NTJO ωU + ϕ TIPXOJO'JH5IFTUBSUJOHQPJOUPGv in Fig. 9.2 occurs first in time.
5IFSFGPSF XFTBZUIBUvMFBETvCZϕPSUIBUvMBHTvCZϕ*Gϕ≠ XFBMTPTBZUIBU vBOEvBSFPVUPGQIBTF*Gϕ = UIFOvBOEvBSF
TBJEUPCF JOQIBTFUIFZSFBDIUIFJSNJOJNBBOENBYJNBBUF YBDUMZUIF
TBNFUJNF8FDBODPNQBSFvBOEvJOUIJTNBOOFSCFDBVTFUIFZPQFS
BUFBUUIFTBNFGSFRVFODZUIFZEPOPUOFFEUPIBWFUIFTBNFBNQMJUVEF
v 1 = Vm sin ωt
vm
π
ϕ
‒Vm
2π
ωt
v 2 = Vm sin(ωt + ϕ)
Figure 9.2
5XPTJOVTPJETXJUIEJGGFSFOUQIBTFT
"TJOVTPJEDBOCFF YQSFTTFEJOFJUIFSTJOFPSDPTJOFGPSN 8IFO
DPNQBSJOHUXPTJOVTPJET JUJTFYQFEJFOUUPFYQSFTTCPUIBTFJUIFSTJOFPS
DPTJOFXJUIQPTJUJWFBNQMJUVEFT5IJTJTBDIJFWFECZVTJOHUIFGPMMPXJOH
USJHPOPNFUSJDJEFOUJUJFT
TJO " ± # = TJO"DPT# ± DPT"TJO#
DPT " ± # = DPT"DPT# ∓ TJO"TJO#
8JUIUIFTFJEFOUJUJFT JUJTFBTZUPTIPXUIBU
TJO ωU ± ž = −TJOωU
DPT ωU ± ž = −DPTωU
TJO ωU ± ž = ± DPTωU
DPT ωU ± ž = ∓ TJOωU
6TJOHUIFTFSFMBUJPOTIJQT XFDBOUSBOTGPSNBTJOVTPJEGSPNTJOFGPSNUP
DPTJOFGPSNPSWJDFWFSTB
372
Chapter 9
+ cos ωt
‒90°
+ sin ωt
(a)
180°
+ cos ωt
Sinusoids and Phasors
"HSBQIJDBMBQQSPBDINBZCFVTFEUPSFMBUF PSDPNQBSF TJOVTPJET
BTBOBMUFSOBUJ WFUPVTJOHUIFUSJHPOPNFUSJDJEFOUJUJFTJO&RT BOE
$POTJEFS UIF TFU PG BYFT TIPXO JO 'JH B The IPSJ[POUBM
BYJTSFQSFTFOUTUIFNBHOJUVEFPGDPTJOF XIJMFUIFWFSUJDBMBYJT QPJOUJOH
EPXO EFOPUFTUIFNBHOJUVEFPGTJOF"OHMFTBSFNFBTVSFEQPTJUJ WFMZ
counterclockwise from the horizontal, as VTVBMJOQPMBSDPPSEJOBUFT
This HSBQIJDBMUFDIOJRVFDBOCFVTFEUPSFMBUFUX PTJOVTPJET' PSF Y
BNQMF XFTFFJO'JH B UIBU T VCUSBDUJOHžGSPNUIFBS HVNFOUPG
DPTωUHJ WFTTJO ωU P S DPT ωU −ž = TJOωU4JNJMBSMZ BEEJOH ž
UPUIFBSHVNFOUPGTJOωUHJWFT−TJOωU PS TJO ωU + ž = −TJOωU BT
TIPwn in Fig. C 5IFHSBQIJDBMUFDIOJRVFDBOBMTPCFVTFEUPBEEUX PTJOVTPJETPG
UIFTBNFGSFRVFODZXIFOPOFJTJOTJOFGPSNBOEUIFPUIFSJTJODPTJOF
GPSN5PBEE"DPTωUBOE#TJOωU XFOPUFUIBU"JTUIFNBHOJUVEFPG
DPTωUXIJMF #JT UIFNBHOJUVEFPG TJOωU BT TIPXOJO'JH B 5IF
NBHOJUVEFBOEBSHVNFOUPGUIFSFTVMUBOUTJOVTPJEJODPTJOFGPSNJTSFBE
JMZPCUBJOFEGSPNUIFUSJBOHMF5IVT
"DPTωU + #TJOωU = $DPT ωU¢θ XIFSF
+ sin ωt
_______
(b)
Figure 9.3
"HSBQIJDBMNFBOTPGSFMBUJOHDPTJOF
and sine: (a) cos(ωU−ž = TJOωU (b) sin(ωU + ž = −TJOωU
$ = √" + # θ = UBO¢@@
#
"
'PSFYBNQMF XFNBZBEEDPT ωUBOE−TJOωUBTTIPXOJO'JH C BOEPCUBJO
DPTωU¢TJOωU = DPT ωU + ž $PNQBSFEXJUIUIFUSJHPOPNFUSJDJEFOUJUJFTJO&RT BOE
UIFHSBQIJDBMBQQSPBDIFMJNJOBUFTNFNPSJ[BUJPO)P XFWFS XF
NVTUOPUDPOGVTFUIFTJOFBOEDPTJOFBY FTXJUIUIFBY FTGPSDPNQMF Y
OVNCFSTUPCFEJTDVTTFEJOUIFOF YUTFDUJPO4PNFUIJOHFMTFUPOPUFJO
'JHTBOEJTUIBUBMUIPVHIUIFOBUVSBMUFOEFOD ZJTUPIB WFUIF
WFSUJDBMBYJTQPJOUVQ UIFQPTJUJWFEJSFDUJPOPGUIFTJOFGVODUJPOJTEPXO
JOUIFQSFTFOUDBTF
‒4
A
‒θ
cos ωt
5
53.1°
C
B
0
+3
cos ωt
sin ωt
sin ωt
(a)
Figure 9.4
(b)
B "EEJOH"DPTωUBOE#TJOωU C BEEJOHDPTωUBOE−TJOωU
9.2
Sinusoids
373
Example 9.1
'JOEUIFBNQMJUVEF QIBTF QFSJPE BOEGSFRVFODZPGUIFTJOVTPJE
v U = DPT U + ž 7
Solution:
5IFBNQMJUVEFJT7N = 7
5IFQIBTFJTϕ = ž
5IFBOHVMBSGSFRVFODZJTω = SBET
π = @@@
π
5IFQFSJPE5 = @@@
ω = T
= )[
5IFGSFRVFODZJTG = @@
5
(JWFOUIFTJOVTPJEDPT πU +ž DBMDVMBUFJUTBNQMJUVEF QIBTF BOHVMBSGSFRVFODZ QFSJPE BOEGSFRVFODZ
Practice Problem 9.1
Answer: ž SBET NT )[
$BMDVMBUFUIFQIBTFBOHMFCFUXFFO v = −DPT ωU + ž BOE v =
TJO ωU−ž 4UBUFXIJDITJOVTPJEJTMFBEJOH
Solution:
-FUVTDBMDVMBUFUIFQIBTFJOUISFFXBZTThe first twPNFUIPETVTFUSJHP
OPNFUSJDJEFOUJUJFT XIJMFUIFUIJSENFUIPEVTFTUIFHSBQIJDBMBQQSPBDI
■ METHOD 1 *OPSEFSUPDPNQBSF vBOE v XFNVTUF YQSFTT
them in UIFTBNFGPSN*GXFF YQSFTTUIFNJODPTJOFGPSNXJUIQPTJ UJWFBNQMJUVEFT
v = −DPT ωU + ž = DPT ωU + ž−ž
v = DPT ωU−ž PS
v = DPT ωU + ž BOE
v = TJO ωU−ž = DPT ωU−ž−ž
v = DPT ωU−ž *UDBOCFEFEVDFEGSPN&RT BOE UIBUUIFQIBTFEJGGFSFODF
CFUXFFOvBOEvJTž8FDBOXSJUFvBT
v = DPT ωU−ž + ž PS v = DPT ωU + ž $PNQBSJOH&RT BOE TIPXTDMFBSMZUIBUvMFBETv by 30°.
■ METHOD 2 "MUFSOBUJWFMZ XFNBZFYQSFTTvJOTJOFGPSN
v = −DPT ωU + ž = TJO ωU + ž−ž
= TJO ωU−ž = TJO ωU−ž−ž
Example 9.2
374
Chapter 9
cos ωt
50°
v1
10°
Sinusoids and Phasors
#VU v = TJO ωU −ž $PNQBSJOHUIFUX PTIP XTUIBU vMBHT v
by 30°. 5IJTJTUIFTBNFBTTBZJOHUIBUvMFBETvCZž
■ METHOD 3 8FNBZSFHBSEvBTTJNQMZ−DPTωUXJUIBQIBTF
TIJGUPG+ž)FODF vJTBTTIPXOJO'JH4JNJMBSMZ vJTTJOωU
XJUIBQIBTFTIJGUPG −ž BTTIP XOJO'JH*UJTFBTZUPTFFGSPN
Fig. 9.5 that vMFBETvCZž UIBUJT ž−ž−ž
v2
sin ωt
Figure 9.5
'PS&YBNQMF
Practice Problem 9.2
'JOEUIFQIBTFBOHMFCFUXFFO
J = −TJO U + ž BOE
J = DPT U−ž
%PFTJMFBEPSMBHJ
Answer:ž JMFBETJ
9.3
Phasors
4JOVTPJETBSFFBTJMZFYQSFTTFEJOUFSNTPGQIBTPST XIJDIBSFNPSFDPO
WFOJFOUUPXPSLXJUIUIBOTJOFBOEDPTJOFGVODUJPOT
A phasor is a complex number that represents the amplitude and phase
of a sinusoid.
Charles Proteus Steinmetz (1865–1923)
was a German-Austrian mathematician
and electrical engineer.
Appendix B presents a short tutorial on
complex numbers.
1IBTPSTQSPWJEFBTJNQMFNFBOTPGBOBMZ[JOHMJOFBSDJSDVJUTF YDJUFECZ
TJOVTPJEBMTPVSDFTTPMVUJPOTPGTVDIDJSDVJUTXPVMECFJOUSBDUBCMFPUIFS
XJTF5IFOPUJPOPGTPMWJOHBDDJSDVJUTVTJOHQIBTPSTXas first introduced
CZ$IBSMFT4UFJONFU[JO#FGPSFXFDPNQMFUFMZdefine QIBTPSTBOE
BQQMZUIFNUPDJSDVJUBOBMZTJT XFOFFEUPCFUIPSPVHIMZG BNJMJBSXJUI
DPNQMFYOVNCFST
"DPNQMFYOVNCFS[DBOCFXSJUUFOJOSFDUBOHVMBSGPSNBT
[ = Y + KZ
B
@@@
XIFSFK = Ŀ−YJTUIFSFBMQBSUPG[ZJTUIFJNBHJOBSZQBSUPG[*O
UIJTDPOUFYU UIFWBSJBCMFTYBOEZEPOPUSFQSFTFOUBMPDBUJPOBTJOUX P
EJNFOTJPOBMWFDUPSBOBMZTJTCVUSBUIFSUIFSFBMBOEJNBHJOBSZQBSUTPG [
JOUIFDPNQMFYQMBOF/FWFSUIFMFTT XFOPUFUIBUUIFSF BSFTPNFSFTFN
CMBODFTCFUXFFONBOJQVMBUJOHDPNQMFYOVNCFSTBOENBOJQVMBUJOHUXP
EJNFOTJPOBMWFDUPST
5IFDPNQMFYOVNCFS [DBOBMTPCFXSJUUFOJOQPMBSPSF YQPOFOUJBM
GPSNBT
[ = S⧸
ϕ = SF Kϕ
C
9.3
375
Phasors
Historical
Charles Proteus Steinmetz
© Bettmann/Corbis
m B(FSNBO"VTUSJBO
NBUIFNBUJDJBOBOEFOHJOFFS JOUSPEVDFE UIFQIBTPSNFUIPE DPWFSFEJO
UIJTDIBQUFS JOBDDJSDVJUBOBMZTJT)FJTBMTPOPUFEGPSIJTX PSLPOUIF
UIFPSZPGIZTUFSFTJT
4UFJONFU[XBTCPSOJO#SFTMBV (FSNBO Z BOEMPTUIJTNPUIFSBUUIF BHFPGPOF"TBZPVUI IFXBTGPSDFEUPMFBWF(FSNBOZCFDBVTFPGIJT
QPMJUJDBMBDUJ WJUJFTKVTUBTIFX BTBCPVUUPDPNQMFUFIJTEPDUPSBMEJT TFSUBUJPOJONBUIFNBUJDTBUUIF6OJ WFSTJUZPG#SFTMBV)FNJHSBUFEUP
4XJU[FSMBOEBOEMBUFSUPUIF6OJUFE 4UBUFT XIFSFIFX BTFNQMPZFECZ
(FOFSBM&MFDUSJDJO5IBUTBNFZFBS IFQVCMJTIFEBQBQFSJOXIJDI
DPNQMFY OVNCFST XFSF VTFE UP BOBMZ[F BD DJSDVJUT GPS UIF first UJNF
5IJTMFEUPPOFPGIJTNBO ZUFYUCPPLT 5IFPSZBOE$BMDVMBUJPOPGBD 1IFOPNFOB QVCMJTIFECZ.D(SB X)JMMJO*O IFCFDBNF UIFQSFTJEFOUPGUIF "NFSJDBO*OTUJUVUFPG&MFDUSJDBM&OHJOFFST XIJDI MBUFSCFDBNFUIF*&&&
XIFSFSJTUIFNBHOJUVEFPG [ BOEϕJTUIFQIBTFPG [8FOPUJDFUIBU [
DBOCFSFQSFTFOUFEJOUISFFXBZT
[ = Y + KZ
[ = S⧸
ϕ
[ = SFKϕ
3FDUBOHVMBSGPSN
1PMBSGPSN
&YQPOFOUJBMGPSN
5IFSFMBUJPOTIJQCFUXFFOUIFSFDUBOHVMBSGPSNBOEUIFQPMBSGPSNJT
TIPXOJO'JH XIFSFUIFYBYJTSFQSFTFOUTUIFSFBMQBSUBOEUIFZBYJT
SFQSFTFOUTUIFJNBHJOBSZQBSUPGBDPNQMF YOVNCFS(JWFOYBOE Z XF
DBOHFUSBOEϕBT
______
Z
S = √Y + Z ϕ = UBO¢@@
Y
B
0OUIFPUIFSIBOE JGXFLOPXSBOEϕ XFDBOPCUBJOYBOEZBT
Y = SDPTϕ Z = STJOϕ
C
[ = Y + KZ = S⧸
ϕ [ = Y + KZ = S ⧸
ϕ
[ = Y + KZ = S ⧸
ϕ
UIFGPMMPXJOHPQFSBUJPOTBSFJNQPSUBOU
"EEJUJPO
[ + [ = Y + Y + K Z + Z 2j
j
r
y
ϕ
x
Real axis
‒j
"EEJUJPOBOETVCUSBDUJPOPGDPNQMFYOVNCFSTBSFCFUUFSQFSGPSNFE
JOSFDUBOHVMBSGPSNNVMUJQMJDBUJPOBOEEJ WJTJPOBSFCFUUFSEPOFJOQPMBS
GPSN(JWFOUIFDPNQMFYOVNCFST
z
0
5IVT [NBZCFXSJUUFOBT
[ = Y + KZ = S⧸
ϕ = S DPTϕ + KTJOϕ
Imaginary axis
B
‒ 2j
Figure 9.6
3FQSFTFOUBUJPOPGBDPNQMFYOVNCFS
[ = Y + KZ = S⧸
ϕ
376
Chapter 9
Sinusoids and Phasors
4VCUSBDUJPO
[−[ = Y−Y + K Z−Z C
[[ = SS ⧸
ϕ ϕ
D
[
S
@@
= @@ ϕ ¢ϕ E
−ϕ
@@
= @@
F
.VMUJQMJDBUJPO
%JWJTJPO
S ⧸ [
3FDJQSPDBM
[
S⧸
_
_
4RVBSF3PPU
[= √S⧸
ϕ∕
G
[ = Y−KZ = S⧸
−ϕ = SF−Kϕ
H
√
$PNQMFY$POKVHBUF
/PUFUIBUGSPN&R F
@@
= −K
I
K
5IFTFBSFUIFCBTJDQSPQFSUJFTPGDPNQMFYOVNCFSTXFOFFE0UIFSQSPQ
FSUJFTPGDPNQMFYOVNCFSTDBOCFGPVOEJO"QQFOEJY#
5IFJEFBPGQIBTPSSFQSFTFOUBUJPOJTCBTFEPO&VMFS TJEFOUJUZ*O
HFOFSBM
F±Kϕ = DPTϕ ± KTJOϕ
XIJDITIPXTUIBUXFNBZSF HBSEDPTϕBOETJO ϕBTUIFSFBMBOEJNBHJ OBSZQBSUTPGF KϕXFNBZXSJUF
DPTϕ = 3F FKϕ TJOϕ = *N FKϕ B
C
XIFSF3F BOE*N TUBOEGPSUIF SFBMQBSUPGBOEUIFJNBHJOBSZQBSUPG
(JWFOBTJOVTPJE v U = 7NDPT ωU + ϕ XFVTF&R B UPF YQSFTT
v U BT
v U = 7NDPT ωU + ϕ = 3F 7NF K ωU+ϕ v U = 3F 7NF KϕF KωU PS
5IVT
v U = 3F 7F KωU
XIFSF
7 = 7NFKϕ = 7N⧸
ϕ
9.3
377
Phasors
7JTUIVTUIFQIBTPSSFQSFTFOUBUJPOPGUIFTJOVTPJEv U BTXFTBJEF BSMJFS
*OPUIFS XPSET BQIBTPS JTB DPNQMFYSFQSFTFOUBUJPO PGUIFNBHOJUVEF
BOEQIBTFPGBTJOVTPJE&JUIFS&R B PS&R C DBOCFVTFE
UPEFWFMPQUIFQIBTPS CVUUIFTUBOEBSEDPOWFOUJPOJTUPVTF&R B 0OFXBZPGMPPLJOHBU&RT BOE JTUPDPOTJEFSUIFQMPUPG
UIFTJOPS7FKωU = 7NFK ωU+ϕ POUIFDPNQMF YQMBOF "TUJNFJODSFBTFT UIF TJOPSSPUBUFTPOBDJSDMFPGSBEJVT7NBUBOBOHVMBSWFMPDJUZωJOUIFDPVOUFS
DMPDLXJTFEJSFDUJPO BTTIPXOJO'JH B 8FNBZSFHBSEv U BTUIFQSP
KFDUJPOPGUIFTJOPS7FKωUPOUIFSFBMBYJT BTTIPwn in Fig. C 5IFWBMVF
PGUIFTJOPSBUUJNFU = JTUIFQIBTPS7PGUIFTJOVTPJEv U 5IFTJOPSNBZ
CFSFHBSEFEBTBSPUBUJOHQIBTPS5IVT XIFOFWFSBTJOVTPJEJTFYQSFTTFEBT
BQIBTPS UIFUFSNFKωUJTJNQMJDJUMZQSFTFOU*UJTUIFSFGPSFJNQPSUBOU XIFO
EFBMJOHXJUIQIBTPST UPLFFQJONJOEUIFGSFRVFODZωPGUIFQIBTPSPUIFS
XJTFXFDBONBLFTFSJPVTNJTUBLFT
Vm
If we use sine for the phasor instead of
cosine, then v (t ) = V m sin(ωt + ϕ) =
Im( V m e j (ωt+ϕ)) and the corresponding
phasor is the same as that in Eq. (9.24).
v(t) = Re(Ve jωt )
Re
Rotation at ω rad s
A phasor may be regarded as a
mathematical equivalent of a sinusoid
with the time dependence dropped.
Vm
ϕ
t0
t
Im
at t = t0
‒Vm
(a)
(b)
Figure 9.7
3FQSFTFOUBUJPOPG7FKωU B TJOPSSPUBUJOHDPVOUFSDMPDLXJTF C JUTQSPKFDUJPO
POUIFSFBMBYJT BTBGVODUJPOPGUJNF
&RVBUJPO TUBUFTUIBUUPPCUBJOUIFTJOVTPJEDPSSFTQPOEJOHUP
BHJWFOQIBTPS 7 NVMUJQMZUIFQIBTPSCZUIFUJNFG BDUPS FKωUBOEUBL F
UIFSFBMQBSU"TBDPNQMFYRVBOUJUZ BQIBTPSNBZCFFYQSFTTFEJOSFDU
BOHVMBS GPSN QPMBS GPSN PS FYQPOFOUJBM GPSN #FDBVTF BQIBTPS IBT
NBHOJUVEFBOEQIBTF iEJSFDUJPOu JUCFIB WFTBTBWFDUPSBOEJTQSJOUFE
JOCPMEGBDF'PSFYBNQMF QIBTPST7 = 7N⧸
ϕBOE* = *N⧸
−θBSFHSBQIJ
DBMMZSFQSFTFOUFEJO'JH4VDIBHSBQIJDBMSFQSFTFOUBUJPOPGQIBTPST
JTLOPXOBTBQIBTPSEJBHSBN
&RVBUJPOT UISPVHI SF WFBMUIBUUPHFU UIFQIBTPSDPS responding to a sinusoid, we first eYQSFTTUIFTJOVTPJEJOUIFDPTJOFGPSN
TPUIBUUIFTJOVTPJEDBOCFXSJUUFOBTUIFSFBMQBSUPGBDPNQMF YOVNCFS
5IFOXFUBL FPVUUIFUJNFG BDUPS FKωU BOEXIBUF WFSJTMFGUJTUIFQIB TPSDPSSFTQPOEJOHUPUIFTJOVTPJE#ZTVQQSFTTJOHUIFUJNFG BDUPS XF
USBOTGPSNUIFTJOVTPJEGSPNUIFUJNFEPNBJOUPUIFQIBTPSEPNBJO5IJT
USBOTGPSNBUJPOJTTVNNBSJ[FEBTGPMMPXT
v U = 7NDPT ωU + ϕ 5JNFEPNBJO
SFQSFTFOUBUJPO ⇔
7 = 7N⧸
ϕ
1IBTPSEPNBJO
SFQSFTFOUBUJPO
We use lightface italic letters such as
z to represent complex numbers but
boldface letters such as V to represent
phasors, because phasors are vectorlike quantities.
378
Chapter 9
Sinusoids and Phasors
Imaginary axis
V
ω
Vm
Leading direction
ϕ
Real axis
‒θ
Lagging direction
Im
I
ω
Figure 9.8
"QIBTPSEJBHSBNTIPXJOH7 = 7N⧸
ϕBOE* = *N⧸
−θ
(JWFOBTJOVTPJEv U = 7NDPT ωU + ϕ XFPCUBJOUIFDPSSFTQPOEJOH
QIBTPSBT7 = 7N⧸
ϕ&RVBUJPO JTBMTPEFNPOTUSBUFEJO5able XIFSFUIFTJOFGVODUJPOJTDPOTJEFSFEJOBEEJUJPOUPUIFDPTJOFGVODUJPO
'SPN&R XFTFFUIBUUPHFUUIFQIBTPSSFQSFTFOUBUJPOPGBTJOV TPJE XFF YQSFTTJUJODPTJOFGPSNBOEUBL FUIF NBHOJUVEFBOEQIBTF
(JWFOBQIBTPS XFPCUBJOUIFUJNFEPNBJOSFQSFTFOUBUJPOBTUIFDPTJOF
GVODUJPOXJUIUIFTBNFNBHOJUVEFBTUIFQIBTPSBOEUIFBS HVNFOUBTωU
QMVTUIFQIBTFPGUIFQIBTPS5IFJEFBPGFYQSFTTJOHJOGPSNBUJPOJOBMUFS
OBUFEPNBJOTJTGVOEBNFOUBMUPBMMBSFBTPGFOHJOFFSJOH
TABLE 9.1
Sinusoid-phasor transformation.
5JNFEPNBJOSFQSFTFOUBUJPO
1IBTPSEPNBJOSFQSFTFOUBUJPO
7NDPT ωU + ϕ
7N⧸
ϕ
7NTJO ωU + ϕ
7N⧸
ϕ −ž
*NDPT ωU + θ
*N⧸
θ
*NTJO ωU + θ
*N⧸
θ −ž
/PUFUIBUJO&R UIFGSFRVFOD Z PSUJNF G BDUPS FKωUJTTVQ QSFTTFE BOEUIFGSFRVFODZJTOPUFYQMJDJUMZTIPXOJOUIFQIBTPSEPNBJO
SFQSFTFOUBUJPOCFDBVTFωJTDPOTUBOU)PXFWFS UIFSFTQPOTFEFQFOETPO
ω 'PS UIJT SFBTPO UIF QIBTPS EPNBJO JT BMTP LOPXO BT UIF GSFRVFODZ
EPNBJO
'SPN&RT BOE v U = 3F 7FKωU = 7NDPT ωU + ϕ TPUIBU
@@@
Ev= −ω7 TJO ωU + ϕ = ω7 DPT ωU + ϕ + ž
EU
N
N
= 3F ω7NF F F
KωU Kϕ Kž
= 3F Kω7FKωU Phasors
379
5IJTTIPXTUIBUUIFEFSJ WBUJWFv U JTUSBOTGPSNFEUPUIFQIBTPSEPNBJO
BTKω7
Differentiating a sinusoid is equivalent
to multiplying its corresponding phasor
by jω.
9.3
Ev
@@@
EU
⇔
5JNFEPNBJO
Kω7
1IBTPSEPNBJO
4JNJMBSMZ UIFJOUF HSBMPG v U JTUSBOTGPSNFEUPUIFQIBTPSEPNBJOBT
7∕Kω
∫ v EU
5JNFEPNBJO 7
@@@
⇔
Kω
Integrating a sinusoid is equivalent to
dividing its corresponding phasor
by jω.
1IBTPSEPNBJO
&RVBUJPO BMMPXTUIFSFQMBDFNFOUPGBEFSJ WBUJWFXJUISFTQFDU
UPUJNFXJUINVMUJQMJDBUJPOPGKω in the phasor domain, whereas Eq. BMMPXTUIFSFQMBDFNFOUPGBOJOUF HSBMXJUISFTQFDUUPUJNFXJUIEJ WJTJPO
CZKωJOUIFQIBTPSEPNBJO&RVBUJPOT BOE BSFVTFGVMJO
finding the steady-state solution, which does not require knoXJOHUIF
JOJUJBMWBMVFTPGUIFWBSJBCMFJOWPMWFE5IJTJTPOFPGUIF JNQPSUBOUBQ
QMJDBUJPOTPGQIBTPST
#FTJEFTUJNFEJGGFSFOUJBUJPOBOEJOUFHSBUJPO BOPUIFSJNQPSUBOUVTF
PGQIBTPSTJTGPVOEJOTVNNJOHTJOVTPJETPGUIFTBNFGSFRVFODZ. 5IJT
JTCFTUJMMVTUSBUFEXJUIBOFYBNQMF BOE&YBNQMFQSPWides one.
5IFEJGGFSFODFTCFUXFFOv U BOE7TIPVMECFFNQIBTJ[FE
Adding sinusoids of the same
frequency is equivalent to adding their
corresponding phasors.
v U JTUIF JOTUBOUBOFPVTPSUJNFEPNBJO SFQSFTFOUBUJPO XIJMF 7JT
UIFGSFRVFODZPSQIBTPSEPNBJOSFQSFTFOUBUJPO
v U JTUJNFEFQFOEFOU XIJMF 7JTOPU 5IJTG BDUJTPGUFOGPS HPUUFO
CZTUVEFOUT
v U JTBM XBZTSFBMXJUIOPDPNQMF YUFSN XIJMF 7JTHFOFSBMMZ
DPNQMFY
'JOBMMZ XFTIPVMECFBSJONJOEUIBUQIBTPSBOBMZTJTBQQMJFTPOMZXIFO
GSFRVFODZJTDPOTUBOUJUBQQMJFTJONBOJQVMBUJOHUX PPSNPSFTJOVTPJEBM
TJHOBMTPOMZJGUIFZBSFPGUIFTBNFGSFRVFODZ
&WBMVBUFUIFTFDPNQMFYOVNCFST
B ⧸
ž + ⧸
−ž ∕
−ž + −K
⧸
C @@@@@@@@@@@@@@@@@
+ K −K
Solution:
B 6TJOHQPMBSUPSFDUBOHVMBSUSBOTGPSNBUJPO
ž = DPTž + KTJOž = + K
⧸
⧸
−ž = <DPT −ž + KTJO −ž > = −K
"EEJOHUIFNVQHJWFT
⧸
ž + ⧸
−ž = + K = ⧸
ž
Example 9.3
380
Chapter 9
Sinusoids and Phasors
5BLJOHUIFTRVBSFSPPUPGUIJT
⧸
ž + ⧸
−ž ∕ = ⧸
ž
C 6TJOHQPMBSSFDUBOHVMBSUSBOTGPSNBUJPO BEEJUJPO NVMUJQMJDBUJPO BOE
EJWJTJPO
−ž + − K
+ K −K
−K + −K
⧸
@@@@@@@@@@@@@@@@@
= @@@@@@@@@@@@@@@@@
+ K + K
−ž
⧸
−K @@@@@@@@@@@@@
= @@@@@@@@@
=
− + K ⧸
ž
−ž
= ⧸
Practice Problem 9.3
&WBMVBUFUIFGPMMPXJOHDPNQMFYOVNCFST
B < + K − + K −⧸
ž>
ž
+ K + ⧸
C ______________
ž + K
+ ⧸
− + K
Answer: B −−K C + K
Example 9.4
5SBOTGPSNUIFTFTJOVTPJETUPQIBTPST
B J = DPT U−ž "
C v = −TJO U + ž 7
Solution:
B J = DPT U−ž IBTUIFQIBTPS
* = ⧸
−ž "
C 4JODF−TJO" = DPT " + ž
v = −TJO U + ž = DPT U + ž + ž
= DPT U + ž 7
5IFQIBTPSGPSNPGvJT
7 = ⧸
ž 7
Practice Problem 9.4
&YQSFTTUIFTFTJOVTPJETBTQIBTPST
B v = −TJO U − ž 7
C J = −DPT U + ž "
Answer: B 7 = ⧸
ž 7 C * = ⧸
−ž"
9.3
Phasors
'JOEUIFTJOVTPJETSFQSFTFOUFECZUIFTFQIBTPST
381
Example 9.5
B * = − + K"
C 7 = KF−Kž7
Solution:
B * = − + K = ⧸
ž5SBOTGPSNJOHUIJTUPUIFUJNFEPNBJOHJWFT
J U = DPT ωU + ž "
C #FDBVTFK = ⧸
ž
7 = K⧸
−ž = ⧸
ž ⧸
−ž
= ⧸
ž−ž= ⧸
ž 7
$POWFSUJOHUIJTUPUIFUJNFEPNBJOHJWFT
v U = DPT ωU + ž 7
'JOEUIFTJOVTPJETDPSSFTQPOEJOHUPUIFTFQIBTPST
Practice Problem 9.5
B 7 = −⧸
ž 7
C * = K −K "
Answer: B v U = DPT ωU−ž 7PSDPT ωU + ž 7 C J U = DPT ωU + ž "
(JWFO J U = DPT ωU + ž " BOE J U = TJO ωU − ž " find
their sum.
Solution:
)FSFJTBOJNQPSUBOUVTFPGQIBTPST‡GPSTVNNJOHTJOVTPJETPGUIFTBNF
GSFRVFODZ$VSSFOUJ U JTJOUIFTUBOEBSEGPSN*UTQIBTPSJT
* = ⧸
ž
8FOFFEUPFYQSFTTJ U JODPTJOFGPSN5IFSVMFGPSDPOWFSUJOHTJOFUP
DPTJOFJTUPTVCUSBDUž)FODF
J = DPT ωU−ž−ž = DPT ωU−ž
BOEJUTQIBTPSJT
* = ⧸
−ž
*GXFMFUJ = J + J UIFO
* = * + * = ⧸
ž + ⧸
−ž
= + K−−K = −K
= ⧸
−ž"
Example 9.6
382
Chapter 9
Sinusoids and Phasors
5SBOTGPSNJOHUIJTUPUIFUJNFEPNBJO XFHFU
J U = DPT ωU−ž "
Of course, we can find J + JVTJOH&R CVUUIBUJTUIFIBSEXBZ
Practice Problem 9.6
*G v = −TJO ωU −ž 7BOE v = DPT ωU + ž 7 find v =
v + v
Answer:v U = DPT ωU + ž 7
Example 9.7
6TJOHUIFQIBTPSBQQSPBDI EFUFSNJOFUIFDVSSFOU
EFTDSJCFECZUIFJOUFHSPEJGGFSFOUJBMFRVBUJPO
J U JOBDJSDVJU
J + ∫ JEU−@@
EJ = DPT U + ž
EU
Solution:
8F USBOTGPSNFBDI UFSNJO UIFFRVBUJPO GSPNUJNF EPNBJOUPQIBTPS
EPNBJO,FFQJOH&RT BOE JONJOE XFPCUBJOUIFQIBTPS
GPSNPGUIFHJWFOFRVBUJPOBT
#VUω = TP
*−Kω* = ž
* + @@@
⧸
Kω
* −K−K = ⧸
ž
ž @@@@@@@@@@@@
ž
⧸
⧸
* = @@@@@@@
ž"
= ⧸
=
−K ⧸
−ž
$POWFSUJOHUIJTUPUIFUJNFEPNBJO
J U = DPT U + ž "
,FFQJONJOEUIBUUIJTJTPOMZUIFTUFBEZTUBUFTPMVUJPO BOEJUEPFTOPU
SFRVJSFLOPXJOHUIFJOJUJBMWBMVFT
Practice Problem 9.7
'JOE UIFW PMUBHF v U JO BDJSDVJU EFTDSJCFE CZ UIFJOUF HSPEJGGFSFOUJBM
FRVBUJPO
@@@
Ev+ v + ∫ v EU = DPT U−ž
EU
VTJOHUIFQIBTPSBQQSPBDI
Answer:v U = DPT U−ž 7
9.4
9.4
383
Phasor Relationships for Circuit Elements
I
i
Phasor Relationships
for Circuit Elements
+
+
/PXUIBUXFLOPXIPXUPSFQSFTFOUBWPMUBHFPSDVSSFOUJOUIFQIBTPSPS
R
R
v
V
GSFRVFODZEPNBJO POFNBZMF HJUJNBUFMZBTLIP XXFBQQMZUIJTUPDJS ‒
‒
DVJUTJOWPMWJOHUIFQBTTJWFFMFNFOUT3 - BOE$8IBUXFOFFEUPEPJT
UPUSBOTGPSNUIFWPMUBHFDVSSFOUSFMBUJPOTIJQGSPNUIFUJNFEPNBJOUPUIF
v = iR
V = IR
GSFRVFODZEPNBJOGPSFBDIFMFNFOU "HBJO XFXJMMBTTVNFUIFQBTTJ WF
(a)
(b)
TJHODPOWFOUJPO
Figure 9.9
8FCF HJOXJUIUIFSFTJTUPS *GUIFDVSSFOUUISPVHIBSFTJTUPS 3JT 7PMUBHFDVSSFOUSFMBUJPOTGPSBSFTJTUPSJO
J = *NDPT ωU + ϕ UIFWPMUBHFBDSPTTJUJTHJWFOCZ0INTMBXBT
UIF B UJNFEPNBJO C GSFRVFODZEPNBJO
v = J3 = 3*NDPT ωU + ϕ Im
5IFQIBTPSGPSNPGUIJTWPMUBHFJT
V
7 = 3*N⧸
ϕ
#VUUIFQIBTPSSFQSFTFOUBUJPOPGUIFDVSSFOUJT* = *N⧸
ϕ)FODF
7 = 3*
I
TIPXJOHUIBUUIFW PMUBHFDVSSFOUSFMBUJPOGPSUIFSFTJTUPSJOUIFQIBTPS
EPNBJO DPOUJOVFT UP CF 0INT MBX BT JO UIF UJNF EPNBJO 'JHVSF JMMVTUSBUFTUIFWPMUBHFDVSSFOUSFMBUJPOTPGBSFTJTUPS8FTIPVMEOPUFGSPN
&R UIBUWPMUBHFBOEDVSSFOUBSFJOQIBTF BTJMMVTUSBUFEJOUIFQIB
TPSEJBHSBNJO'JH
'PSUIFJOEVDUPS - BTTVNFUIF DVSSFOU UISPVHIJU JT J =
*NDPT ωU + ϕ 5IFWPMUBHFBDSPTTUIFJOEVDUPSJT
v = -@@
EJ = −ω-*NTJO ωU + ϕ EU
ϕ
0
Re
Figure 9.10
1IBTPSEJBHSBNGPSUIFSFTJTUPS
I
i
+
+
v
3FDBMMGSPN&R UIBU −TJO " = DPT " + ž 8FDBOXSJUFUIF
WPMUBHFBT
‒
‒
v = L di
dt
(a)
V = jωLI
v = ω-*NDPT ωU + ϕ + ž XIJDIUSBOTGPSNTUPUIFQIBTPS
7 = ω-*NFK ϕ+ž = ω-*NFKϕFKž = ω-*N⧸
ϕ + ž
(b)
7PMUBHFDVSSFOUSFMBUJPOTGPSBOJOEVDUPSJO
UIF B UJNFEPNBJO C GSFRVFODZEPNBJO
7 = Kω-*
Im
TIPXJOHUIBUUIFW PMUBHFIBTBNBHOJUVEFPG ω-*NBOEBQIBTFPG ϕ + ž5IFWoltage and current are 90° out of phase. Specifically UIF
DVSSFOUMBHTUIFW PMUBHFCZž'JHVSFTIP XTUIFW PMUBHFDVSSFOU
SFMBUJPOTGPSUIFJOEVDUPS'JHVSFTIPXTUIFQIBTPSEJBHSBN
'PSUIFDBQBDJUPS $ BTTVNF UIF WPMUBHF BDSPTT JU JT v =
7NDPT ωU + ϕ 5IFDVSSFOUUISPVHIUIFDBQBDJUPSJT
J = $@@@
Ev
EU
#ZGPMMPXJOHUIFTBNFTUFQTBTXFUPPLGPSUIFJOEVDUPSPSCZBQQMZJOH
&R PO&R XFPCUBJO
L
V
Figure 9.11
#VU*N⧸
ϕ = * BOEGSPN&R FKž = K5IVT
L
* = Kω$7
⇒
* 7 = @@@@
Kω$
ω
V
I
ϕ
0
Re
Figure 9.12
1IBTPSEJBHSBNGPSUIFJOEVDUPS
*MBHT7
Although it is equally correct to say
that the inductor voltage leads the
current by 90°, convention gives the
current phase relative to the voltage.
384
Chapter 9
Sinusoids and Phasors
i
I
+
Im
+
C
v
ω
I
C
V
‒
‒
dv
i = C dt
(a)
I = jωC V
V
ϕ
0
Re
Figure 9.14
(b)
Figure 9.13
7PMUBHFDVSSFOUSFMBUJPOTGPS
a capacitor in the: (a) time
EPNBJO C GSFRVFODZ
EPNBJO
1IBTPSEJBHSBNGPSUIFDBQBDJUPS*
MFBET7
TIPXJOHUIBUUIFDVSSFOUBOEWPMUBHFBSFžPVUPGQIBTF5PCFTQFDJfiD UIFDVSSFOUMFBETUIFWPMUBHFCZž'JHVSFTIPXTUIFWPMUBHFDVSSFOU
SFMBUJPOTGPSUIFDBQBDJUPS'JHHJWFTUIFQIBTPSEJBHSBN5BCMF
TVNNBSJ[FTUIFUJNFEPNBJO BOEQIBTPSEPNBJOSFQSFTFOUBUJPOT PGUIF
DJSDVJUFMFNFOUT
TABLE 9.2
Summary of voltage-current relationships.
&MFNFOU
5JNFEPNBJO
'SFRVFODZEPNBJO
3
v = 3J
v = -@@
EJ
EU
Ev
J = $@@@
EU
7 = 3*
$
Example 9.8
7 = Kω-*
* 7 = @@@@
Kω$
5IFWPMUBHFv = DPT U + ž JTBQQMJFEUPB)JOEVDUPS'JOEUIF
TUFBEZTUBUFDVSSFOUUISPVHIUIFJOEVDUPS
Solution:
'PSUIFJOEVDUPS 7 = Kω-* XIFSF ω = SBETBOE 7 = ⧸
ž7
)FODF
ž
ž
⧸
⧸
7= @@@@@@@@
= ⧸
* = @@@@
−ž"
= @@@@@@@
Kω- K×
⧸
ž
$POWFSUJOHUIJTUPUIFUJNFEPNBJO
J U = DPT U−ž "
Practice Problem 9.8
*GWPMUBHFv = TJO U − ž 7JTBQQMJFEUPBμ'DBQBDJUPS DBMDV
MBUFUIFDVSSFOUUISPVHIUIFDBQBDJUPS
Answer:TJO U + ž N"
9.5
9.5
385
Impedance and Admittance
Impedance and Admittance
*OUIFQSFDFEJOHTFDUJPO XFPCUBJOFEUIFW PMUBHFDVSSFOUSFMBUJPOTGPS
UIFUISFFQBTTJWFFMFNFOUTBT
7 = 3* 7 = Kω-* * 7 = @@@@
Kω$
5IFTFFRVBUJPOTNBZCFXSJUUFOJOUFSNTPGUIFSBUJPPGUIFQIBTPSW PMU
BHFUPUIFQIBTPSDVSSFOUBT
@@
7 = 3 *
@@
7 = Kω- @@
7 = @@@@
*
*
Kω$
'SPNUIFTFUISFFF YQSFTTJPOT XFPCUBJO0IN TMBXJOQIBTPSGPSNGPS
BOZUZQFPGFMFNFOUBT
7 ; = @@
*
PS
7 = ;*
XIFSF;JTBGSFRVFOD ZEFQFOEFOURVBOUJUZLOPXOBT JNQFEBODF NFB
TVSFEJOPINT
The impedance Z of a circuit is the ratio of the phasor voltage V to the
phasor current I, measured in ohms (Ω).
TABLE 9.3
5IFJNQFEBODFSFQSFTFOUTUIFPQQPTJUJPOUIBUUIFDJSDVJUF
YIJCJUTUP
the floXPGTJOVTPJEBMDVSSFOU "MUIPVHIUIFJNQFEBODFJTUIFSBUJPPG
UXPQIBTPST JUJTOPUBQIBTPS CFDBVTFJUEPFTOPUDPSSFTQPOEUPBTJOV TPJEBMMZWBSZJOHRVBOUJUZ
5IFJNQFEBODFTPGSFTJTUPST JOEVDUPST BOEDBQBDJUPSTDBOCFSFBE
JMZPCUBJOFEGSPN&R 5BCMFTVNNBSJ[FTUIFJSJNQFEBODFT
'SPNUIFUBCMFXFOPUJDFUIBU;- = Kω-BOE;$ = −K∕ω$$POTJEFSUXP
FYUSFNFDBTFTPGBOHVMBSGSFRVFOD Z8IFOω = JF GPSEDTPVSDFT ;- = BOE;$ → ∞, confirming what we already knoX‡UIBUUIFJOEVD
UPSBDUTMJLFBTIPSUDJSDVJU XIJMFUIFDBQBDJUPSBDUTMJLFBOPQFODJSDVJU
8IFOω → ∞ JF GPSIJHIGSFRVFODJFT ;- → ∞BOE;$ = JOEJDBU
JOHUIBUUIFJOEVDUPSJTBOPQFODJSDVJUUPIJHIGSFRVFODJFT XIJMFUIF DBQBDJUPSJTBTIPSUDJSDVJU'JHVSFJMMVTUSBUFTUIJT
"TBDPNQMFYRVBOUJUZ UIFJNQFEFODFNBZCFF YQSFTTFEJOSFDUBO
HVMBSGPSNBT
; = 3 ± K9
&MFNFOU *NQFEBODF "ENJUUBODF
3
;=3
-
; = Kω-
$
1 ; = ____
Kω $
1
: = __
3
1 : = ____
Kω: = Kω$
Short circuit at dc
L
Open circuit at
high frequencies
XIFSF3 = 3F;JTUIFSFTJTUBODFBOE9 = *N;JTUIFSFBDUBODF5IF
SFBDUBODF 9 JTKVTUBNBHOJUVEF BQPTJUJ WFWBMVF CVUXIFOVTFEBT BWFDUPS B KJTBTTPDJBUFEXJUIJOEVDUBODFBOEB −KJTBTTPDJBUFEXJUI
DBQBDJUBODF5IVT JNQFEBODF ; = 3 + K9JTTBJEUPCF JOEVDUJWFP S
MBHHJOHTJODFDVSSFOUMBHTW PMUBHF XIJMFJNQFEBODF ; = 3 − K9J T
DBQBDJUJWF PSMFBEJOH CFDBVTF DVSSFOUMFBETW PMUBHF 5IF JNQFEBODF SFTJTUBODF BOE SFBDUBODF BSF BMMNFBTVSFE JO PINT 5IF JNQFEBODF
NBZBMTPCFFYQSFTTFEJOQPMBSGPSNBT
; = ∣;∣⧸
θ
Impedances and admittances
of passive elements.
(a)
Open circuit at dc
C
Short circuit at
high frequencies
(b)
Figure 9.15
&RVJWBMFOUDJSDVJUTBUEDBOEIJHI
GSFRVFODJFT B JOEVDUPS C DBQBDJUPS
386
Chapter 9
Sinusoids and Phasors
$PNQBSJOH&RT BOE XFJOGFSUIBU
; = 3 ± K9 = ∣;∣⧸
θ
XIFSF
_______
∣;∣ = √3 + 9 ±9
θ = UBO¢@@@
3
9 = ∣;∣TJOθ
BOE
3 = ∣;∣DPTθ *UJTTPNFUJNFTDPO WFOJFOUUPX PSLXJUIUIFSFDJQSPDBMPGJNQFE BODF LOPXOBTBENJUUBODF
The admittance Y is the reciprocal of impedance, measured in
siemens (S).
5IFBENJUUBODF :PGBOFMFNFOU PSBDJSDVJU JTUIFSBUJPPGUIFQIBTPS
DVSSFOUUISPVHIJUUPUIFQIBTPSWPMUBHFBDSPTTJU PS
*
= @@
: = @@
; 7
5IFBENJUUBODFTPGSFTJTUPST JOEVDUPST BOEDBQBDJUPSTDBOCFPCUBJOFE
GSPN&R 5IFZBSFBMTPTVNNBSJ[FEJO5BCMF
"TBDPNQMFYRVBOUJUZ XFNBZXSJUF:BT
: = ( + K#
XIFSF( = 3F:JTDBMMFEUIFDPOEVDUBODFBOE# = *N:JTDBMMFEUIFTVT
DFQUBODF"ENJUUBODF DPOEVDUBODF BOETVTDFQUBODFare all FYQSFTTFEJO
the unit of siemens (or mhos). From Eqs. (9.41) and (9.47),
( + K# = @@@@@@
3 + K9
3−K9 @@@@@@@
3−K9
( + K# = @@@@@@
· @@@@@@
=
3 + K9 3−K9 3 + 9 #ZSBUJPOBMJ[BUJPO
&RVBUJOHUIFSFBMBOEJNBHJOBSZQBSUTHJWFT
3 ( = @@@@@@@
3 + 9
9 # = −@@@@@@@
3 + 9
TIPXJOHUIBU (≠∕3BTJUJTJOSFTJTUJ WFDJSDVJUT0GDPVSTF JG 9 = UIFO( = ∕3
9.6
387
Kirchhoff’s Laws in the Frequency Domain
Example 9.9
'JOEv U BOEJ U JOUIFDJSDVJUTIPXOJO'JH
i
Solution:
'SPNUIFWPMUBHFTPVSDFDPTU ω = 5Ω
v s = 10 cos 4t +
‒
7T = ⧸
ž7
0.1 F
5IFJNQFEBODFJT
+
v
‒
Figure 9.16
= + @@@@@@@
; = + @@@@
= −K Ω
Kω$
K×
'PS&YBNQMF
)FODFUIFDVSSFOU
ž
⧸
+ K
7
= @@@@@@@@@@
* = @@@T= @@@@@@@
; −K
+ = + K = ⧸
ž"
5IFWPMUBHFBDSPTTUIFDBQBDJUPSJT
ž
⧸
* = @@@@@@@@@@@@
7 = *;$ = @@@@
Kω$
K×
ž
⧸
= −ž7
= @@@@@@@@@@@@
⧸
ž
⧸
$POWFSUJOH*BOE7JO&RT BOE UPUIFUJNFEPNBJO XFHFU
J U = DPT U + ž "
v U = DPT U−ž 7
/PUJDFUIBUJ U MFBETv U CZžBTFYQFDUFE
Practice Problem 9.9
3FGFSUP'JH%FUFSNJOFv U BOEJ U Answer:TJO U + ž 7 TJO U + ž "
i
v s = 20 sin(10t + 30°) V +
‒
9.6
Kirchhoff’s Laws in the
Frequency Domain
Figure 9.17
'PS1SBDUJDF1SPC
8FDBOOPUEPDJSDVJUBOBMZTJTJOUIFGSFRVFOD ZEPNBJOXJUIPVU,JSDI IPGGTDVSSFOUBOEW PMUBHFMBXT5IFSFGPSF XFOFFEUPF YQSFTTUIFNJO
UIFGSFRVFODZEPNBJO
'PS,7- MFUv v … vOCFUIFWPMUBHFTBSPVOEBDMPTFEMPPQ
5IFO
v + v + ⋯ + vO = *OUIFTJOVTPJEBMTUFBEZTUBUF FBDIW PMUBHFNBZCFXSJUUFOJODPTJOF
GPSN TPUIBU&R CFDPNFT
7NDPT ωU + θ + 7NDPT ωU + θ
+ ⋯ + 7NODPT ωU + θO = 4Ω
0.2 H
+
v
‒
388
Chapter 9
Sinusoids and Phasors
5IJTDBOCFXSJUUFOBT
3F 7NFKθFKωU + 3F 7NFKθFKωU + ⋯ + 3F 7NOFKθOFKωU = PS
3F< 7NFKθ + 7NFKθ + ⋯ + 7NOFKθO FKωU> = *GXFMFU7L = 7NLFKθL UIFO
3F< 7 + 7 + ⋯ + 7O FKωU> = 7 + 7 + ⋯ + 7O = #FDBVTFFKωU≠
JOEJDBUJOHUIBU,JSDIIPGGTWPMUBHFMBXIPMETGPSQIBTPST
#ZGPMMPXJOHBTJNJMBSQSPDFEVSF XFDBOTIPXUIBU,JSDIIPGGTDVS
SFOUMBXIPMETGPSQIBTPST*GXFMFUJ J … JOCFUIFDVSSFOUMFBWJOHPS
FOUFSJOHBDMPTFETVSGBDFJOBOFUXPSLBUUJNFU UIFO
J + J + ⋯ + JO = *G* * … *OBSFUIFQIBTPSGPSNTPGUIFTJOVTPJETJ J … JO UIFO
* + * + ⋯ + *O = XIJDIJT,JSDIIPGGTDVSSFOUMBXJOUIFGSFRVFODZEPNBJO
0ODFXFIBWFTIPXOUIBUCPUI,7-BOE,$-IPMEJOUIFGSFRVFODZ
EPNBJO JUJTFBTZUPEPNBO ZUIJOHT TVDIBTJNQFEBODFDPNCJOBUJPO OPEBMBOENFTIBOBMZTFT TVQFSQPTJUJPO BOETPVSDFUSBOTGPSNBUJPO
9.7
Impedance Combinations
$POTJEFSUIF /TFSJFTDPOOFDUFEJNQFEBODFTTIP XOJO'JH 5IF
TBNFDVSSFOU* floXTUISPVHIUIFJNQFEBODFT"QQMZJOH,7-BSPVOEUIF
MPPQHJWFT
7 = 7 + 7 + ⋯ + 7/ = * ; + ; + ⋯ + ;/ I
Z2
Z1
+
+
V
‒
V1
‒
+
V2
ZN
‒
+
VN
‒
Zeq
Figure 9.18
/JNQFEBODFTJOTFSJFT
5IFFRVJWBMFOUJNQFEBODFBUUIFJOQVUUFSNJOBMTJT
7 = ; + ; + ⋯ + ;
;FR = @@
/
*
PS
;FR = ; + ; + ⋯ + ;/
9.7
389
Impedance Combinations
I
TIPXJOHUIBUUIFUPUBMPSFRVJ WBMFOUJNQFEBODFPGTFSJFTDPOOFDUFE
JNQFEBODFTJTUIFTVNPGUIFJOEJ WJEVBMJNQFEBODFT 5IJTJTTJNJMBSUP
UIFTFSJFTDPOOFDUJPOPGSFTJTUBODFT
*G/ = BTTIP XOJO'JH UIFDVSSFOUUISPVHIUIFJNQFE BODFTJT
+
V
‒
7 * = @@@@@@@
; + ;
Figure 9.19
+
V1
‒
+
V2
‒
Z2
7PMUBHFEJWJTJPO
#FDBVTF7 = ;*BOE7 = ;* UIFO
;
7 7 = @@@@@@@
; + ;
Z1
;
7 = @@@@@@@
7 ; + ;
XIJDIJTUIFWPMUBHFEJWJTJPOSFMBUJPOTIJQ
*OUIFTBNFNBOOFS XFDBOPCUBJOUIFFRVJ WBMFOUJNQFEBODFPS
BENJUUBODFPGUIF /QBSBMMFMDPOOFDUFEJNQFEBODFTTIP XOJO'JH
5IFWPMUBHFBDSPTTFBDIJNQFEBODFJTUIFTBNF "QQMZJOH,$-BUUIF
UPQOPEF
+ ⋯ + @@@
* = * + * + ⋯ + */ = 7 @@@
+ @@@
; ;
;/
I
I
+
I1
I2
IN
V
Z1
Z2
ZN
‒
Zeq
Figure 9.20
/JNQFEBODFTJOQBSBMMFM
5IFFRVJWBMFOUJNQFEBODFJT
* = @@@
+ ⋯ + @@@
+ @@@
@@@
= @@
;FR
7
;
;
;/
BOEUIFFRVJWBMFOUBENJUUBODFJT
:FR = : + : + ⋯ + :/
5IJTJOEJDBUFTUIBUUIFFRVJWBMFOUBENJUUBODFPGBQBSBMMFMDPOOFDUJPOPG
BENJUUBODFTJTUIFTVNPGUIFJOEJWJEVBMBENJUUBODFT
8IFO / = BTTIP XOJO'JH UIFFRVJ WBMFOUJNQFEBODF
CFDPNFT
;;
= @@@@@@@
;FR = @@@
= @@@@@@@@@@@
= @@@@@@@
:FR : + : ∕; + ∕; ; + ;
I
+
I1
I2
V
Z1
Z2
‒
Figure 9.21
$VSSFOUEJWJTJPO
390
Chapter 9
Sinusoids and Phasors
"MTP TJODF
7 = *;FR = *; = *;
UIFDVSSFOUTJOUIFJNQFEBODFTBSF
;
* = @@@@@@@
* ; + ;
;
* = @@@@@@@
* ; + ;
XIJDIJTUIFDVSSFOUEJWJTJPOQSJODJQMF
5IFEFMUBUPXZFBOEXZFUPEFMUBUSBOTGPSNBUJPOT UIBUXFBQQMJFE
UPSFTJTUJ WFDJSDVJUTBSFBMTPW BMJEGPSJNQFEBODFT 8JUISFGFSFODFUP
Fig. 9.22, the conWFSTJPOGPSNVMBTBSFBTGPMMPXT
Zc
a
b
Z2
Z1
n
Za
Zb
Z3
c
Figure 9.22
4VQFSJNQPTFE:BOE∆OFUXPSLT
:∆$POWFSTJPO
;; + ;; + ;;
;B = @@@@@@@@@@@@@@@@@
;
;; + ;; + ;;
;C = @@@@@@@@@@@@@@@@@
;
;; + ;; + ;;
;D = @@@@@@@@@@@@@@@@@
;
∆:$POWFSTJPO
;C;D
; = @@@@@@@@@@@
;B + ;C +
;D
;D;B
; = @@@@@@@@@@@
;B + ;C +
;D
;B;C
; = @@@@@@@@@@@
;B + ;C +
;D 9.7
391
Impedance Combinations
A delta or wye circuit is said to be balanced if it has equal impedances
in all three branches.
8IFOB∆:DJSDVJUJTCBMBODFE &RT BOE CFDPNF
;∆ = ;:
PS
;
;: = @@
∆
XIFSF;: = ; = ; = ;BOE;∆ = ;B = ;C = ;D
"TZPVTFF JOUIJT TFDUJPO UIFQSJODJQMFT PGWPMUBHFEJWJTJPO DVS
SFOUEJWJTJPO DJSDVJUSFEVDUJPO JNQFEBODFFRVJ WBMFODF BOE :∆USBOT
GPSNBUJPOBMMBQQMZUPBDDJSDVJUT$IBQUFSXJMMTIP XUIBUPUIFSD JS
DVJUUFDIOJRVFT‡TVDIBTTVQFSQPTJUJPO OPEBMBOBMZTJT NFTIBOBMZTJT TPVSDFUSBOTGPSNBUJPO UIF 5IFWFOJOUIFPSFN BOEUIF/PSUPOUIFPSFN‡
BSFBMMBQQMJFE UPBD DJSDVJUTJOB NBOOFSTJNJMBS UPUIFJS BQQMJDBUJPOJO
EDDJSDVJUT
Example 9.10
'JOEUIFJOQVUJNQFEBODF PGUIFDJSDVJUJO'JH "TTVNF UIBUUIF
DJSDVJUPQFSBUFTBUω = SBET
2 mF
Solution:
-FU
Z in
; = *NQFEBODFPGUIFN'DBQBDJUPS
; = *NQFEBODFPGUIFΩ SFTJTUPSJOTFSJFTXJUIUIFN'
DBQBDJUPS
; = *NQFEBODFPGUIF)JOEVDUPSJOTFSJFTXJUIUIFΩ
SFTJTUPS
5IFO
= @@@@@@@@@@@@@
; = @@@@
= −KΩ
Kω$ K× × −
= + @@@@@@@@@@@@@
; = + @@@@
= −K Ω
Kω$
K × × −
; = + Kω- = + K × = + K Ω
5IFJOQVUJNQFEBODFJT
−K + K
;JO = ; + ; ‖ ; = −K + @@@@@@@@@@@@@@
+ K
+ K −K
= −K + −KΩ
= −K + @@@@@@@@@@@@@@@@
+ 5IVT
;JO = −KΩ
0.2 H
3Ω
10 mF
Figure 9.23
'PS&YBNQMF
8Ω
392
Chapter 9
Practice Problem 9.10
1 mF
100 Ω
Sinusoids and Phasors
%FUFSNJOFUIFJOQVUJNQFEBODFPGUIFDJSDVJUJO'JHBU
SBET
ω =
8H
Answer: −K Ω
Z in
200 Ω
1 mF
Figure 9.24
'PS1SBDUJDF1SPC
Example 9.11
%FUFSNJOFvP U JOUIFDJSDVJUPG'JH
60 Ω
20 cos(4t ‒ 15°) +
‒
10 mF
+
vo
‒
5H
Solution:
5P EP UIF BOBMZTJT JO UIF GSFRVFODZ EPNBJO XFNVTU first USBOTGPSN
UIFUJNFEPNBJODJSDVJUJO'JHUPUIFQIBTPSEPNBJOFRVJ WBMFOUJO
Fig. 9.26. 5IFUSBOTGPSNBUJPOQSPEVDFT
Figure 9.25
'PS&YBNQMF
⇒
7T = ⧸
−ž7 N'
⇒
@@@@
= @@@@@@@@@@@@
‒j25 Ω
j20 Ω
+
Vo
‒
= −KΩ
Figure 9.26
5IFGSFRVFODZEPNBJOFRVJWBMFOUPGUIF
DJSDVJUJO'JH
-FU
ω=
K × × −
Kω$
60 Ω
20 ‒15° +
‒
vT = DPT U−ž )
⇒
Kω- = K × = KΩ
; = *NQFEBODFPGUIFΩ SFTJTUPS
; = *NQFEBODFPGUIFQBSBMMFMDPNCJOBUJPOPGUIF
N'DBQBDJUPSBOEUIF)JOEVDUPS
5IFO; = ΩBOE
−K × K
; = −K ‖ K = @@@@@@@@@@
= KΩ
−K + K
#ZUIFWPMUBHFEJWJTJPOQSJODJQMF
K
;
7 = @@@@@@@@@
−ž
7P = @@@@@@@
⧸
; + ; T + K
= ⧸
ž ⧸
−ž = ⧸
ž 7
8FDPOWFSUUIJTUPUIFUJNFEPNBJOBOEPCUBJO
vP U = DPT U + ž 7
Practice Problem 9.11
$BMDVMBUFvPJOUIFDJSDVJUPG'JH
0.5 H
50 cos(10t + 30°) +
‒
Figure 9.27
'PS1SBDUJDF1SPC
10 Ω
Answer:vP U = DPT U−ž 7
1
20 F
+
vo
‒
9.7
Impedance Combinations
Example 9.12
'JOEDVSSFOU*JOUIFDJSDVJUPG'JH
‒ j4 Ω
2Ω
I
12 Ω
j4 Ω
8Ω
b
a
50 0° +
‒
c
j6 Ω
‒ j3 Ω
8Ω
Figure 9.28
'PS&YBNQMF
Solution:
5IFEFMUBOFUX PSLDPOOFDUFEUPOPEFT B C BOE DDBOCFDPO WFSUFEUP
the :OFUXPSLPG'JH8FPCUBJOUIF:JNQFEBODFTBTGPMMPXTVTJOH
&R K −K
+ K
;BO = @@@@@@@@@@@@@
= @@@@@@@@
= + K Ω
K + −K + K ;CO = @@@@@
= KΩ −K
;DO = @@@@@@@@
= −K Ω
5IFUPUBMJNQFEBODFBUUIFTPVSDFUFSNJOBMTJT
; = + ;BO + ;CO−K ‖ ;DO + K + = + + K + K ‖ + K
K + K
= + K + @@@@@@@@@@@@@
+ K
žΩ
= + K = ⧸
5IFEFTJSFEDVSSFOUJT
ž
⧸
7 = ____________
* = @@
= ⧸
−ž "
; ⧸
ž
Zan
I
Zcn
n
Zbn
12 Ω
50 0° +
‒
Figure 9.29
a
393
b
c
j6 Ω
‒j3 Ω
5IFDJSDVJUJO'JHBGUFSEFMUBUPXZFUSBOTGPSNBUJPO
8Ω
394
Chapter 9
Practice Problem 9.12
I
Sinusoids and Phasors
'JOE*JOUIFDJSDVJUPG'JH
Answer:⧸
ž "
‒j3 Ω
j4 Ω
j5 Ω
8Ω
45 30° V +
‒
9.8
5Ω
10 Ω
‒j2 Ω
Figure 9.30
'PS1SBDUJDF1SPC
*O$IBQUFSTBOE XFTB XDFSUBJOVTFTPG 3$ 3- BOE 3-$DJSDVJUT
JOEDBQQMJDBUJPOT5IFTFDJSDVJUTBMTPIBWFBDBQQMJDBUJPOTBNPOHUIFN
are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac
CSJEHFDJSDVJUT BOEUSBOTGPSNFST 5IJTMJTUPGBQQMJDBUJPOTJTJOF YIBVT
UJWF8FXJMMDPOTJEFSTPNFPGUIFNMBUFS*UXJMMTVGfice here to observF
UXPTJNQMFPOFT3$QIBTFTIJGUJOHDJSDVJUT BOEBDCSJEHFDJSDVJUT
9.8.1
I
C
+
Vi
R
+
Vo
‒
C
+
Vo
‒
‒
(a)
I
R
+
Vi
‒
(b)
Figure 9.31
4FSJFT3$TIJGUDJSDVJUT B MFBEJOH
PVUQVU C MBHHJOHPVUQVU
vo
Applications
Phase-Shifters
"QIBTFTIJGUJOHDJSDVJUJTPGUFOFNQMP ZFEUPDPSSFDUBOVOEFTJSBCMF
QIBTFTIJGUBMSFBEZQSFTFOUJOBDJSDVJUPSUPQSPEVDFTQFDJBMEFTJSFE
FGGFDUT"O3$DJSDVJUJTTVJUBCMFGPSUIJTQVSQPTFCFDBVTFJUTDBQBDJUPS
DBVTFTUIFDJSDVJUDVSSFOUUPMFBEUIFBQQMJFEW PMUBHF 5XPDPNNPOMZ
VTFE3$DJSDVJUTBSFTIPXOJO'JH 3-DJSDVJUTPSBOZSFBDUJWFDJS
DVJUTDPVMEBMTPTFSWFUIFTBNFQVSQPTF
*O'JH B UIFDJSDVJUDVSSFOU *MFBETUIFBQQMJFEW PMUBHF7JCZ
TPNFQIBTFBOHMFθ XIFSFθ ž EFQFOEJOHPOUIFWBMVFTPG3BOE
$*G9$ = −∕ω$ UIFOUIFUPUBMJNQFEBODFJT; = 3 + K9$ BOEUIFQIBTF
TIJGUJTHJWFOCZ
9
θ = UBO¢@@@
$
3
5IJTTIPXTUIBUUIFBNPVOUPGQIBTFTIJGUEFQFOETPOUIFW BMVFTPG 3 $ BOEUIFPQFSBUJOHGSFRVFOD Z4JODFUIFPVUQVUW PMUBHF7PBDSPTTUIF
SFTJTUPSJTJOQIBTFXJUIUIFDVSSFOU 7PMFBET QPTJUJWFQIBTFTIJGU 7JBT
TIPXOJO'JH B *O'JH C UIFPVUQVUJTUBLFOBDSPTTUIFDBQBDJUPS5IFDVSSFOU
*MFBETUIFJOQVUW PMUBHF7JCZθ CVUUIFPVUQVUW PMUBHFvP U BDSPTTUIF
DBQBDJUPSMBHT OFHBUJWFQIBTFTIJGU UIFJOQVUW PMUBHFvJ U BTJMMVTUSBUFE
JO'JH C vi
vi
vo
t
θ
Phase shift
θ
Phase shift
(a)
Figure 9.32
t
1IBTFTIJGUJO3$DJSDVJUT B MFBEJOHPVUQVU C MBHHJOHPVUQVU
(b)
9.8
395
Applications
8FTIPVMELFFQJONJOEUIBUUIFTJNQMF 3$DJSDVJUTJO'JH
BMTPBDUBTWPMUBHFEJWJEFST5IFSFGPSF BTUIFQIBTFTIJGUθBQQSPBDIFT
ž UIFPVUQVUWP MUBHF 7PBQQSPBDIFT[FSP' PSUIJTSFBTPO UIFTF TJNQMF3$DJSDVJUTBSFVTFEPOMZXIFOTNBMMBNPVOUTPGQIBTFTIJGU BSFSFRVJSFE*GJUJTEFTJSFEUPIB WFQIBTFTIJGUTHSFBUFSUIBOž TJNQMF 3$ OFUXPSLT BSF DBTDBEFE UIFSFCZ QSPWJEJOH B UPUBM QIBTF
TIJGUFRVBMUPUIFTVNPGUIFJOEJ WJEVBMQIBTFTIJGUT*OQSBDUJDF UIF QIBTFTIJGUT EVFUP UIFTUBHFTBSF OPUFRVBM CFDBVTFUIF TVDDFFEJOH
TUBHFTMPBEEPXOUIFFBSMJFSTUBHFTVOMFTTPQBNQTBSFVTFEUPTFQB SBUFUIFTUBHFT
Example 9.13
%FTJHOBO3$DJSDVJUUPQSPWJEFBQIBTFPGžMFBEJOH
Solution:
*GXFTFMFDUDJSDVJUDPNQPOFOUTPGFRVBMPINJDW BMVF TBZ 3 = ∣9$∣ =
20 Ω BUBQBSUJDVMBSGSFRVFOD Z BDDPSEJOHUP&R UIFQIBTFTIJGU
JTFYBDUMZž#ZDBTDBEJOHUX PTJNJMBS3$DJSDVJUTJO'JH B XF
PCUBJOUIFDJSDVJUJO'JH QSPWJEJOHBQPTJUJWFPSMFBEJOHQIBTFTIJGU
PGž BTXFTIBMMTPPOTIPX6TJOHUIFTFSJFTQBSBMMFMDPNCJOBUJPOUFDI
OJRVF ;JO'JHJTPCUBJOFEBT
−K
; = ‖ −K = @@@@@@@@@@@
= −KΩ
−K
‒ j20 Ω
V1
‒ j20 Ω
+
+
Vi
20 Ω
20 Ω
‒
Vo
‒
Z
Figure 9.33
"O3$QIBTFTIJGUDJSDVJUXJUIžMFBEJOH
QIBTFTIJGUGPS&YBNQMF
6TJOHWPMUBHFEJWJTJPO
@@
−K
Ŀ
; ž7J
7 = @@@@@@@
7J = @@@@@@@@
7 = @@@⧸
;−K
−K J
BOE
@@
Ŀ
7P = @@@@@@@@
ž7
7 = @@@⧸
−K 4VCTUJUVUJOH&R JOUP&R ZJFMET
@@
@@
Ŀ
Ŀ
ž7
7P = (@@@
⧸
ž)(@@@
⧸
ž7J)= @@
⧸
J
5IVT UIFPVUQVUMFBET UIFJOQVUCZžC VUJUTNBHOJUVEFJTPOMZBCPVU
QFSDFOUPGUIFJOQVU
Practice Problem 9.13
%FTJHOBO 3$DJSDVJUUPQSP WJEFBžMBHHJOHQIBTFTIJGUPGUIFPVU QVUWPMUBHFSFMBUJWFUPUIFJOQVUW PMUBHF*GBOBDW PMUBHFPG7SNTJT
BQQMJFE XIBUJTUIFPVUQVUWPMUBHF
Answer:'JHVSFTIPXTBUZQJDBMEFTJHO7SNT
10 Ω
10 Ω
+
Vi
‒ j10 Ω
‒
Figure 9.34
'PS1SBDUJDF1SPC
‒j10 Ω
+
Vo
‒
396
Chapter 9
Example 9.14
150 Ω
'PSUIF3-DJSDVJUTIPXOJO'JH B DBMDVMBUFUIFBNPVOUPGQIBTF
TIJGUQSPEVDFEBUL)[
100 Ω
10 mH
Solution:
"UL)[ XFUSBOTGPSNUIFBOEN)JOEVDUBODFTUPUIFDPSSFTQPOE
JOHJNQFEBODFT
5 mH
(a)
150 Ω
100 Ω
V1
+
+
j125.7 Ω
Vi
Sinusoids and Phasors
j62.83 Ω
N)
⇒
N)
⇒
Vo
‒
‒
Z
9- = ω- = π × × × × ¢
= π = Ω
9- = ω- = π × × × × ¢
= π = Ω
$POTJEFSUIFDJSDVJUJO'JH C 5IFJNQFEBODF;JTUIFQBSBMMFMDPN
CJOBUJPOPGKΩ BOE + KΩ)FODF
; = K‖ + K
(b)
K + K
= @@@@@@@@@@@@@@@@@
žΩ
= ⧸
+ K
Figure 9.35
'PS&YBNQMF
6TJOHWPMUBHFEJWJTJPO
ž
⧸
; 7 = @@@@@@@
7J = ____________
7J
+ K
; + = ⧸
ž7J
BOE
K
7P = @@@@@@@@@@@@
ž7
7 = ⧸
+ K
$PNCJOJOH&RT BOE 7P = ⧸
ž ⧸
ž 7J = ⧸
ž7J
TIPXJOHUIBUUIFPVUQVUJTBCPVUQFSDFOUPGUIFJOQVUJONBHOJUVEFCVU
MFBEJOHUIFJOQVUCZž*GUIFDJSDVJUJTUFSNJOBUFECZBMPBE UIFMPBE
XJMMBGGFDUUIFQIBTFTIJGU
Practice Problem 9.14
1 mH
2 mH
+
Vi
+
10 Ω
‒
50 Ω
Vo
3FGFSUPUIF 3-DJSDVJUJO'JH*G V is applied to the input, find
UIFNBHOJUVEFBOEUIFQIBTFTIJGUQSPEVDFEBUL)[4QFDJGZXIFUIFS
UIFQIBTFTIJGUJTMFBEJOHPSMBHHJOH
Answer:7 ž MBHHJOH
‒
Figure 9.36
'PS1SBDUJDF1SPC
9.8.2
AC Bridges
"OBDCSJEHFDJSDVJUJTVTFEJONFBTVSJOHUIFJOEVDUBODF-PGBOJOEVDUPS
PSUIFDBQBDJUBODF$PGBDBQBDJUPS*UJTTJNJMBSJOGPSNUPUIF8IFBUTUPOF
CSJEHFGPSNFBTVSJOHBOVOLOPXOSFTJTUBODF EJTDVTTFEJO4FDUJPO BOEGPMMP XTUIFTBNFQSJODJQMF 5PNFBTVSF -BOE $ IP XFWFS BOBD
9.8
TPVSDFJTOFFEFEBTXFMMBTBOBDNFUFSJOTUFBEPGUIFHBMWBOPNFUFS5IF
BDNFUFSNBZCFBTFOTJUJWFBDBNNFUFSPSWPMUNFUFS
$POTJEFSUIFHFOFSBMBDCSJEHFDJSDVJUEJTQMBZFEJO'JH 5IF
CSJEHFJTCBMBODFE when no current floXTUISPVHIUIFNFUFS5IJTNFBOT
UIBU7 = 7"QQMZJOHUIFWPMUBHFEJWJTJPOQSJODJQMF
;Y
;
7 = @@@@@@@
7 = 7 = @@@@@@@
7
; + ; T
; + ;Y T
397
Applications
Z1
AC
meter
Vs ≈
Z2
5IVT
;Y
;
@@@@@@@
= @@@@@@@
⇒
; + ; ; + ;Y
;; = ;;Y
PS
;
;Y = @@@ ;
;
5IJTJTUIFCBMBODFEFRVBUJPOGPSUIFBDCSJEHFBOEJTTJNJMBSUP&R GPSUIFSFTJTUBODFCSJEHFFYDFQUUIBUUIF3TBSFSFQMBDFECZ;T
Specific ac bridges for measuring -BOE $BSFTIP XOJO'JH XIFSF-YBOE$YBSFUIFVOLOPXOJOEVDUBODFBOEDBQBDJUBODFUPCFNFB
TVSFEXIJMF -TBOE $TBSFBTUBOEBSEJOEVDUBODFBOEDBQBDJUBODF UIF
WBMVFTPGXIJDIBSFLOP XOUPHSFBUQSFDJTJPO *OFBDIDBTF UX PSFTJT
UPST 3BOE3 BSFWBSJFEVOUJMUIFBDNFUFSSFBET[FSP5IFOUIFCSJEHF
JTCBMBODFE'SPN&R XFPCUBJO
3
-Y = @@@-T
3
3
$Y = @@@$T
3
BOE
/PUJDFUIBUUIFCBMBODJOHPGUIFBDCSJEHFTJO'JHEPFTOPUEFQFOE
POUIFGSFRVFODZGPGUIFBDTPVSDF TJODFGEPFTOPUBQQFBSJOUIFSFMBUJPO
TIJQTJO&RT BOE R1
R2
R1
AC
meter
Ls
R2
AC
meter
Lx
Cs
Cx
≈
≈
(a)
(b)
Figure 9.38
Specific ac bridges: (a) for measuring - C GPSNFBTVSJOH$
Z3
Figure 9.37
"HFOFSBMBDCSJEHF
+
V1
‒
+
V2
‒
Zx
398
Example 9.15
Chapter 9
Sinusoids and Phasors
5IFBDCSJEHFDJSDVJUPG'JHCBMBODFTXIFO;JTBLΩSFTJTUPS ;
JTBLΩSFTJTUPS ;JTBQBSBMMFMDPNCJOBUJPOPGB.ΩSFTJTUPSBOE
BQ'DBQBDJUPS BOE G = L)['JOE B UIFTFSJFT DPNQPOFOUTUIBU
NBLFVQ;Y BOE C UIFQBSBMMFMDPNQPOFOUTUIBUNBLFVQ;Y
Solution:
Define.5IFQSPCMFNJTDMFBSMZTUBUFE
1SFTFOU8FBSFUPEFUFSNJOFUIFVOLOPXODPNQPOFOUTTVCKFDUUP
UIFGBDUUIBUUIFZCBMBODFUIFHJWFORVBOUJUJFT(JWFOUIBUBQBSBMMFM
BOETFSJFTFRVJWBMFOUFxists for this circuit, we need to find both.
"MUFSOBUJWF"MUIPVHIUIFSFBSFBMUFSOBUJWFUFDIOJRVFTUIBUDBO
be used to find the unknoXOWBMVFT BTUSBJHIUGPSXBSEFRVBMJUZ
XPSLTCFTU0ODFXFIBWFBOTXFST XFDBODIFDLUIFNCZVTJOH
IBOEUFDIOJRVFTTVDIBTOPEBMBOBMZTJTPSKVTUVTJOH14QJDF
"UUFNQU'SPN&R ;
;Y = @@@;
;
XIFSF;Y = 3Y + K9Y
; = Ω ; = Ω
BOE
3
@@@@@
Kω$
3
; = 3 ‖ @@@@@
= @@@@@@@@@@@
= @@@@@@@@@@
Kω$ 3 + ∕Kω$ + Kω3$
4JODF3 = .ΩBOE$ = Q'
× × @@@@@@@@@@
; = @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
=
+ Kπ × × × × × × − + K
PS
; = −K.Ω
B "TTVNJOHUIBU;YJTNBEFVQPGTFSJFTDPNQPOFOUT XFTVCTUJUVUF
&RT BOE JO&R BOEPCUBJO
−K × 3Y + K9Y = @@@@@
= −K .Ω
&RVBUJOHUIFSFBMBOEJNBHJOBSZQBSUTZJFMET3Y = .ΩBOEB
DBQBDJUJWFSFBDUBODF
= × 9Y = @@@
ω$
PS
= @@@@@@@@@@@@@@@@@@@@@@@
$ = @@@@
= Q'
ω9Y π × × × × 9.8
Applications
C ;YSFNBJOTUIFTBNFBTJO&R CVU3YBOE9YBSFJO
QBSBMMFM"TTVNJOHBO3$QBSBMMFMDPNCJOBUJPO
;Y = −K .Ω
3Y
= 3Y ‖ @@@@@
= @@@@@@@@@@
Kω$Y + Kω3Y$Y
#ZFRVBUJOHUIFSFBMBOEJNBHJOBSZQBSUT XFPCUBJO
3FBM ;Y + *NBH ;Y
+ 3Y = @@@@@@@@@@@@@@@@@@@
= .Ω
= @@@@@@@@@@@@@
3FBM ;Y
*NBH ;Y
$Y = − @@@@@@@@@@@@@@@@@@@@@
ω<3FBM ;Y + *NBH ;Y >
−
= − @@@@@@@@@@@@@@@@@@@@@@
= μ'
π + 8FIBWFBTTVNFEBQBSBMMFM3$DPNCJOBUJPOXIJDIXPSLTJO
this case.
&WBMVBUF-FUVTOPXVTF14QJDFUPTFFJGXFJOEFFEIBWFUIF
DPSSFDUFRVBMJUJFT3VOOJOH14QJDFXJUIUIFFRVJWBMFOUDJSDVJUT BOPQFODJSDVJUCFUXFFOUIFiCSJEHFuQPSUJPOPGUIFDJSDVJU BOEB
WPMUJOQVUWPMUBHFZJFMETUIFGPMMPXJOHWPMUBHFTBUUIFFOETPG
UIFiCSJEHFuSFMBUJWFUPBSFGFSFODFBUUIFCPUUPNPGUIFDJSDVJU
'3&2
7. /@ 71 /@
& & &
& & &
#FDBVTFUIFWPMUBHFTBSFFTTFOUJBMMZUIFTBNF UIFOOPNFBTVSBCMF
current can floXUISPVHIUIFiCSJEHFuQPSUJPOPGUIFDJSDVJUGPS
BOZFMFNFOUUIBUDPOOFDUTUIFUXPQPJOUTUPHFUIFSBOEXFIBWFB
CBMBODFECSJEHF XIJDIJTUPCFFYQFDUFE5IJTJOEJDBUFTXFIBWF
QSPQFSMZEFUFSNJOFEUIFVOLOPXOT
5IFSFJTBWFSZJNQPSUBOUQSPCMFNXJUIXIBUXFIBWFEPOF
%PZPVLOPXXIBUUIBUJT 8FIBWFXIBUDBOCFDBMMFEBOJEFBM iUIFPSFUJDBMuBOTXFS CVUPOFUIBUSFBMMZJTOPUWFSZHPPEJOUIF
SFBMXPSME5IFEJGGFSFODFCFUXFFOUIFNBHOJUVEFTPGUIFVQQFS
JNQFEBODFTBOEUIFMPXFSJNQFEBODFTJTNVDIUPPMBSHFBOEXPVME
OFWFSCFBDDFQUFEJOBSFBMCSJEHFDJSDVJU'PSHSFBUFTUBDDVSBDZ UIFPWFSBMMNBHOJUVEFPGUIFJNQFEBODFTNVTUBUMFBTUCFXJUIJO
UIFTBNFSFMBUJWFPSEFS5PJODSFBTFUIFBDDVSBDy of the solution
PGUIJTQSPCMFN *XPVMESFDPNNFOEJODSFBTJOHUIFNBHOJUVEFPG
UIFUPQJNQFEBODFTUPCFJOUIFSBOHFPGLΩ to 1.5 MΩ0OF
BEEJUJPOBMSFBMXPSMEDPNNFOU5IFTJ[FPGUIFTFJNQFEBODFTBMTP
DSFBUFTTFSJPVTQSPCMFNTJONBLJOHBDUVBMNFBTVSFNFOUT TPUIF
BQQSPQSJBUFJOTUSVNFOUTNVTUCFVTFEJOPSEFSUPNJOJNJ[FUIFJS
MPBEJOH XIJDIXPVMEDIBOHFUIFBDUVBMWPMUBHFSFBEJOHT POUIF
DJSDVJU
4BUJTGBDUPSZ #FDBVTFXFTPMWFEGPSUIFVOLOPXOUFSNTBOEUIFO
UFTUFEUPTFFJGUIFZXPSLFE XFWBMJEBUFEUIFSFTVMUT5IFZDBOOPX
CFQSFTFOUFEBTBTPMVUJPOUPUIFQSPCMFN
399
400
Practice Problem 9.15
Chapter 9
Sinusoids and Phasors
*OUIFBDCSJEHFDJSDVJUPG'JH TVQQPTFUIBUCBMBODFJTBDIJF WFE
XIFO ; JT BL ΩSFTJTUPS ; JT B ΩSFTJTUPS JOTFSJFT XJUIB
μ)JOEVDUPS ;JTBLΩSFTJTUPS BOEG = .)[%FUFSNJOFUIF
TFSJFTDPNQPOFOUTUIBUNBLFVQ;Y
Answer:"ΩSFTJTUPSJOTFSJFTXJUIBμ)JOEVDUPS
9.9
Summary
"TJOVTPJEJTBTJHOBMJOUIFGPSNPGUIFTJOFPSDPTJOFGVODUJPO*U
IBTUIFHFOFSBMGPSN
v U = 7NDPT ωU + ϕ
XIFSF7NJTUIFBNQMJUVEF ω = πGJTUIFBOHVMBSGSFRVFODZ ωU + ϕ JTUIFBSHVNFOU BOEϕJTUIFQIBTF
"QIBTPSJTBDPNQMFYRVBOUJUZUIBUSFQSFTFOUTCPUIUIFNBHOJUVEF BOEUIFQIBTFPGBTJOVTPJE(JWFOUIFTJOVTPJEv U = 7N DPT ωU + ϕ JUTQIBTPS7JT
7 = 7N⧸
ϕ
*OBDDJSDVJUT W PMUBHFBOEDVSSFOUQIBTPSTBM
XBZTIB WFB
fixFE S FMBUJPOUPPOFBOPUIFSBUBO ZNPNFOUPGUJNF*G v U =
7N DPT ωU + ϕW SFQSFTFOUTUIFW PMUBHFUISPVHIBOFMFNFOUBOE J U = *N DPT ωU + ϕJ SFQSFTFOUTUIFDVSSFOUUISPVHIUIFFMFNFOU UIFOϕJ = ϕWJGUIFFMFNFOUJTBSFTJTUPS ϕJMFBET ϕWCZžJGUIF
FMFNFOUJTBDBQBDJUPS BOE ϕJMBHT ϕWCZžJGUIFFMFNFOUJTBO
JOEVDUPS
5IFJNQFEBODF;PGBDJSDVJUJTUIFSBUJPPGUIFQIBTPSWPMUBHFBDSPTT
JUUPUIFQIBTPSDVSSFOUUISPVHIJU
7 = 3 ω + K9 ω
; = @@
*
5IFBENJUUBODF:JTUIFSFDJQSPDBMPGJNQFEBODF
= ( ω + K# ω
: = @@
;
*NQFEBODFTBSFDPNCJOFEJOTFSJFTPSJOQBSBMMFMUIFTBNFX BZBT
SFTJTUBODFTJOTFSJFTPSQBSBMMFMUIBUJT JNQFEBODFT JOTFSJFTBEE
XIJMFBENJUUBODFTJOQBSBMMFMBEE
'PSBSFTJTUPS; = 3 GPSBOJOEVDUPS; = K9 = Kω- BOEGPSBDBQBDJ
UPS; = −K9 = ∕Kω$
#BTJDDJSDVJUMB XT 0INTBOE,JSDIIPG GT BQQMZUPBDDJSDVJUTJO
UIFTBNFNBOOFSBTUIFZEPGPSEDDJSDVJUTUIBUJT
7 = ;*
Σ *L = ,$-
Σ7L = ,7-
401
Problems
5IFUFDIOJRVFTPGWPMUBHFDVSSFOUEJWJTJPO TFSJFTQBSBMMFMDPNCJOB
UJPOPGJNQFEBODFBENJUUBODF DJSDVJUSFEVDUJPO BOE :∆USBOTGPS
NBUJPOBMMBQQMZUPBDDJSDVJUBOBMZTJT
"$DJSDVJUTBSFBQQMJFEJOQIBTFTIJGUFSTBOECSJEHFT
Review Questions
8IJDIPGUIFGPMMPXJOHJTOPUBSJHIUXBZUPFYQSFTT
UIFTJOVTPJE"DPTωU
B "DPTπ GU
D "DPTω U−5 v(t)
C vMFBETv
E vMBHTv
C 'BMTF
5IFJNBHJOBSZQBSUPGJNQFEBODFJTDBMMFE
B SFTJTUBODF
D TVTDFQUBODF
F SFBDUBODF
1
4H
C 'BMTF
+
v o(t)
‒
'PS3FWJFX2VFTUJPO
"TFSJFT3$DJSDVJUIBT∣73∣ = 7BOE∣7$∣ = 7
5IFNBHOJUVEFPGUIFTVQQMZWPMUBHFJT
B −7
C 7
D 7
E 7
"TFSJFT3$-DJSDVJUIBT3 = Ω 9$ = Ω BOE
9- = Ω5IFJNQFEBODFPGUIFDJSDVJUJT
B + KΩ
D − KΩ
F − + KΩ
C BENJUUBODF
E DPOEVDUBODF
5IFJNQFEBODFPGBDBQBDJUPSJODSFBTFTXJUI
JODSFBTJOHGSFRVFODZ
B 5SVF
+
‒
Figure 9.39
5IFWPMUBHFBDSPTTBOJOEVDUPSMFBETUIFDVSSFOU
UISPVHIJUCZž
B 5SVF
1Ω
C L)[
F vBOEvBSFJOQIBTF
C SBET
D SBET
F OPOFPGUIFBCPWF
C IBSNPOJD
E SFBDUJWF
*Gv = TJO ωU + ž BOEv = TJO ωU + ž XIJDIPGUIFTFTUBUFNFOUTBSFUSVF
B vMFBETv
D vMBHTv
B SBET
E ∞ SBET
8IJDIPGUIFTFGSFRVFODJFTIBTUIFTIPSUFSQFSJPE
B LSBET
C "DPT πU∕5
E "TJO ωU−ž
"UXIBUGSFRVFODZXJMMUIFPVUQVUWPMUBHFvP U JO
'JHCFFRVBMUPUIFJOQVUWPMUBHFv U A function that repeats itself after fixFEJOUFSWBMTJT
TBJEUPCF
B BQIBTPS
D QFSJPEJD
C + KΩ
E −−KΩ
"OTXFSTE D C C E B FC E D C
Problems
Section 9.2 Sinusoids
(JWFOUIFTJOVTPJEBMWPMUBHFv U =
DPT U + ž 7, find: (a) the amplitude
7N, (b) the period 5 D UIFGSFRVFODZG BOE
E v U BUU = NT
"DVSSFOUTPVSDFJOBMJOFBSDJSDVJUIBT
JT = DPT π U + ž "
B 8IBUJTUIFBNQMJUVEFPGUIFDVSSFOU
C 8IBUJTUIFBOHVMBSGSFRVFODZ
D 'JOEUIFGSFRVFODZPGUIFDVSSFOU
E $BMDVMBUFJTBUU = NT
&YQSFTTUIFGPMMPXJOHGVODUJPOTJODPTJOFGPSN
B TJO ωU + ž D −TJO ωU + ž
C −TJO U
402
Chapter 9
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOETJOVTPJET
(JWFOv = TJO ωU + ž 7BOEv =
DPT ωU−ž 7 EFUFSNJOFUIFQIBTFBOHMF
CFUXFFOUIFUXPTJOVTPJETBOEXIJDIPOFMBHT
the other
Sinusoids and Phasors
'PSUIFGPMMPXJOHQBJSTPGTJOVTPJET EFUFSNJOF
XIJDIPOFMFBETBOECZIPXNVDI
B v U = DPT U−ž BOE
J U = TJO U + ž
C v U = DPT U + ž BOE
v U = −DPTU
D Y U = DPTU + TJOUBOE
Z U = DPT U−ž
Section 9.3
Phasors
*GG ϕ = DPTϕ + KTJOϕ TIPXUIBUG ϕ = FKϕ
$BMDVMBUFUIFTFDPNQMFYOVNCFSTBOEFYQSFTTZPVS
SFTVMUTJOSFDUBOHVMBSGPSN
⧸
ž
B @@@@@@@@@
+ K
−K
−ž
⧸
C @@@@@@@@@@@@@
+ @@@@@@@@@
− + K
−K + K
−ž + K
D + ⧸
&WBMVBUFUIFGPMMPXJOHDPNQMFYOVNCFSTBOEMFBWF
ZPVSSFTVMUTJOQPMBSGPSN
ž
⧸
B ⧸
ž −K + @@@@@@
(
+K)
ž −ž
+ K − + K
⧸
⧸
C @@@@@@@@@@@@@@@@@
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOEQIBTPST
'JOEUIFQIBTPSTDPSSFTQPOEJOHUPUIFGPMMPXJOH
TJHOBMT
B v U = DPT U−ž 7
C J U = −TJO U + ž N"
D v U = TJO U−ž 7
E J U = −DPT U + ž N"
žBOE: = ⧸
−ž&WBMVBUFUIF
-FU9 = ⧸
GPMMPXJOHRVBOUJUJFTBOEFYQSFTTZPVSSFTVMUTJO
polar form:
B 9+: 9
C 9−: D 9 + : ∕9
&WBMVBUFUIFGPMMPXJOHDPNQMFYOVNCFST
+ K @@@@@@@@
−K
B @@@@@@
+
−K − + K
ž ⧸
−ž
⧸
C @@@@@@@@@@@@@@@@
⧸
−ž −⧸
ž
ǀ
+ K
−K D
ǀ
−K
−K 4JNQMJGZUIFGPMMPXJOHFYQSFTTJPOT
−K − + K
B @@@@@@@@@@@@@@@@@@@@@@
− + K −K + −K
ž + ⧸
−ž −K
⧸
C @@@@@@@@@@@@@@@@@@@@@@@@@@@@
+ K ⧸
ž
+ K @@@@@@@@@@@@@@@@
D @@@@@@@@
Ŀ + K −K ( + K )
&WBMVBUFUIFTFEFUFSNJOBOUT
ǀ
ǀ
+ K
− ⧸
−ž
⧸
ž C
ǀ
−K
K
D
ǀ
−K
− + K B
ǀ
−⧸
−ž
⧸
ž −K
K ǀ
−K
+ K 5SBOTGPSNUIFGPMMPXJOHTJOVTPJETUPQIBTPST
B −DPT U + ž D DPT U + TJO U
C TJO U + ž
5XPWPMUBHFTvBOEvBQQFBSJOTFSJFTTPUIBUUIFJS
TVNJTv = v + v*Gv = DPT U−π∕ 7BOE
v = DPT U + ž 7, find v
0CUBJOUIFTJOVTPJETDPSSFTQPOEJOHUPFBDIPGUIF
GPMMPXJOHQIBTPST
B 7 = ⧸
ž 7 ω = C 7 = + K7 ω = D * = F−Kπ∕" ω = E * = −−K" ω = Using phasors, find:
B DPT U + ž −DPT U−ž
C TJOU + DPT U−ž
D TJOU + DPT U + ž
−TJO U−ž
"MJOFBSOFUXPSLIBTBDVSSFOUJOQVU
DPT U + ž "BOEBWPMUBHFPVUQVU
DPT U + ž 7%FUFSNJOFUIFBTTPDJBUFE
JNQFEBODF
403
Problems
4JNQMJGZUIFGPMMPXJOH
B G U = DPT U + ž −TJO U−ž
C H U = TJOU + DPT U + ž
U
D I U = ∫ DPTU + TJOU EU
"OBMUFSOBUJOHWPMUBHFJTHJWFOCZv U =
DPT U + ž 7. Use phasors to find
U
v U + @@@
Ev−∫ v U EU
−∞
EU
"WPMUBHFv U = DPT U + ž 7JTBQQMJFEUPB
QBSBMMFMDPNCJOBUJPOPGBLΩSFTJTUPSBOEBμ'
DBQBDJUPS'JOEUIFTUFBEZTUBUFDVSSFOUTUISPVHIUIF
SFTJTUPSBOEUIFDBQBDJUPS
"TFSJFT3-$DJSDVJUIBT3 = Ω - = N) BOE
$ = N'*GUIFJOQVUWPMUBHFJTv U = DPTU, find
the current floXJOHUISPVHIUIFDJSDVJU
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEQIBTPSSFMBUJPOTIJQTGPS
DJSDVJUFMFNFOUT
"TTVNFUIBUUIFW BMVFPGUIFJOUF HSBMJT[FSPBU U = −∞
IL
"QQMZQIBTPSBOBMZTJTUPFWBMVBUFUIFGPMMPXJOH
'JOEv U JOUIFGPMMPXJOHJOUFHSPEJGGFSFOUJBM
FRVBUJPOTVTJOHUIFQIBTPSBQQSPBDI
B v U + ∫ v EU = DPTU
Load
(R + jωL)
v +
‒
B v = <TJO U + ž + DPT U−ž >7
C J = <DPT U + ž −TJO U + ž >"
Figure 9.40
'PS1SPC
∫
Ev+ v U + v EU = TJO U + ž
C @@@
EU
6TJOHQIBTPST EFUFSNJOFJ U JOUIFGPMMPXJOH
FRVBUJPOT
B @@
EJ + J U = DPT U−ž
EU
"TFSJFT3-DJSDVJUJTDPOOFDUFEUPB7BD
TPVSDF*GUIFWPMUBHFBDSPTTUIFSFTJTUPSJT7 find the vPMUBHFBDSPTTUIFJOEVDUPS
8IBUWBMVFPGωXJMMDBVTFUIFGPSDFESFTQPOTF vP JO'JHUPCF[FSP
EJ + J U = DPT U + ž "
C ∫ JEU + @@
EU
2Ω
5IFMPPQFRVBUJPOGPSBTFSJFT3-$DJSDVJUHJWFT
@@
EJ + J +
EU
U
∫−∞ JEU = DPTU"
"TTVNJOHUIBUUIFW BMVFPGUIFJOUF HSBMBU U = −∞JT
zero, find J U VTJOHUIFQIBTPSNFUIPE
"QBSBMMFM3-$DJSDVJUIBTUIFOPEFFRVBUJPO
5 mF
50 cos ωt V +
‒
+
vo
20 mH
‒
Figure 9.41
'PS1SPC
∫
@@@
Ev+ v + v EU = DPT U−ž 7
EU
%FUFSNJOF v U VTJOH UIF QIBTPS NFUIPE :PV NBZ
BTTVNFUIBUUIFWBMVFPGUIFJOUFHSBMBUU = −∞JT[FSP
Section 9.4
Section 9.5
Impedance and Admittance
'JOEUIFTUFBEZTUBUFDVSSFOUJJOUIFDJSDVJUPG
Fig. 9.42, when vT U = DPTU7
Phasor Relationships for Circuit
Elements
Determine the current that floXTUISPVHIBOΩ
SFTJTUPSDPOOFDUFEUPBWPMUBHFTPVSDF
vT= DPT U+ž 7
(JWFOUIBUvD =DPT ž 7 XIBUJTUIF
JOTUBOUBOFPVTWPMUBHFBDSPTTBμ'DBQBDJUPSXIFO
UIFDVSSFOUUISPVHIJUJTJ = TJO U + ž "
i
vs +
‒
Figure 9.42
'PS1SPC
10 Ω
5 mF
20 mH
404
Chapter 9
Sinusoids and Phasors
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEJNQFEBODF
i
R1
In the circuit of Fig. 9.47, find JP XIFO
B ω = SBET
D ω = SBET
L
C ω = SBET
io (t)
vs +
‒
C
R2
R3
10 Ω
10 Ω
5 sin (ωt) A
50 mF
2H
Figure 9.47
Figure 9.43
'PS1SPC
'PS1SPC
'JOEv U JOUIF3-$DJSDVJUPG'JH
%FUFSNJOFUIFBENJUUBODF:GPSUIFDJSDVJUJO'JH
10 Ω
1Ω
j30 Ω
Y
10 Ω
‒ j10 Ω
1Ω
115 cos t V +
‒
1H
j10 Ω
Figure 9.44
+
v(t)
‒
1F
Figure 9.48
'PS1SPC
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEBENJUUBODF
30 Ω
i
is(t)
+
v
‒
C
R
$BMDVMBUFvP U JOUIFDJSDVJUPG'JH
50 Ω
50 μF
60 sin 200t V +
‒
0.1 H
(a)
i
R1
v s (t) +
‒
C
Figure 9.49
R2
+
v
‒
'PS1SPC
'JOEDVSSFOU*PJOUIFDJSDVJUTIPXOJO'JH
L
Io
(b)
5 0° A
Figure 9.45
20 Ω
'PS1SPC
'PSUIFDJSDVJUTIPwn in Fig. 9.46, find ;FRBOEVTF
that to find current *-FUω = SBET
I
115 0° V +
‒
Figure 9.46
'PS1SPC
4Ω
j20 Ω
‒ j40 Ω
40 Ω
j40 Ω
Figure 9.50
'PS1SPC
$BMDVMBUFJ U JOUIFDJSDVJUPG'JH
‒j14 Ω
16 Ω
+
v o(t)
‒
i
j25 Ω
160 cos 200t V
Figure 9.51
'PSQSPC
+
‒
5Ω
4Ω
5 mF
10 mH
3Ω
405
Problems
%FUFSNJOFvYJOUIFDJSDVJUPG'JH-FUJT U =
5 cos(100U + ž "
'JOEDVSSFOU*PJOUIFOFUXPSLPG'JH
j2 Ω
j2 Ω
+
‒
0.1 H
Ix
2Ω
j2 Ω
2Ω
20 ° A
Figure 9.52
is (t)
'PS1SPC
*GvT = TJO U + ž 7JOUIFDJSDVJUPG
'JH9.53, find JP
*GUIFWPMUBHFvPBDSPTTUIFΩSFTJTUPSJOUIFDJSDVJU
PG'JHJTDPTU7 PCUBJOJT
2Ω
0.1 F
2Ω
200 mH
100 mF
Figure 9.53
+
vo
‒
1Ω
is
2Ω
Figure 9.58
'PS1SPC
'PS1SPC
*OUIFDJSDVJUPG'JH EFUFSNJOFUIFWBMVF
of JT U is (t)
2Ω
*G7P = ⧸
žV in the circuit of Fig. 9.59, find *T
10 Ω
2 mH
115 cos 2,000t V +
‒
50 μF
20 Ω
Vs +
‒
j5 Ω
10 Ω
+
Vo
‒
‒j10 Ω
Figure 9.59
Figure 9.54
'PS1SPC
'PS1SPC
(JWFOUIBUvT U = TJO U−ž JO'JH EFUFSNJOFJY U 10 Ω
'JOE*PJOUIFDJSDVJUPG'JH
4Ω
30 Ω
Io
ix
vs (t) +
‒
0.2 H
0.5 mF
Figure 9.55
'PS1SPC
2Ω
‒ j2 Ω
160 ‒30° V +
‒
j6 Ω
8Ω
10 Ω
Figure 9.60
'JOEvT U JOUIFDJSDVJUPG'JHJGUIFDVSSFOUJY
UISPVHIUIFΩSFTJTUPSJTTJOU"
2Ω
'PS1SPC
0.5 H
io
+ v
‒ s
Figure 9.56
+
vx
‒
Figure 9.57
'PS1SPC
vs +
‒
20 Ω
1 mF
ix
'PS1SPC
In the circuit of Fig. 9.61, find 7TJG*P = ⧸
ž "
‒ j2 Ω
1Ω
j2 Ω
‒ j1 Ω
2Ω
Figure 9.61
'PS1SPC
j4 Ω
Vs
+‒
j2 Ω
‒ j1 Ω
Io
1Ω
406
Chapter 9
Sinusoids and Phasors
'JOE;JOUIFOFUXPSLPG'JH HJWFOUIBU
7P = ⧸
ž7
12 Ω
20 ‒90° V +
‒
10 Ω
Z
+
Vo
‒
j8 Ω
‒j4 Ω
Figure 9.62
Zin
1 mF
100 mH
Figure 9.66
'PS1SPC
Section 9.7
'PSUIFOFUXork in Fig. 9.66, find ;JO-FUω =
SBET
'PS1SPC
Impedance Combinations
0CUBJO;JOGPSUIFDJSDVJUJO'JH
"Uω = 377 rad/s, find the input impedance of the
DJSDVJUTIPXOJO'JH
25 Ω
50 μF
12 Ω
‒ j50 Ω
Zin
60 mH
j15 Ω
30 Ω
20 Ω
40 Ω
j10 Ω
Figure 9.67
Figure 9.63
'PS1SPC
'PS1SPC
"Uω = SBET PCUBJOUIFJOQVUBENJUUBODFJOUIF
DJSDVJUPG'JH
1Ω
2Ω
Yin
2H
'JOE;FRJOUIFDJSDVJUPG'JH
Zeq
1‒jΩ
1F
1 + j3 Ω
1 + j2 Ω
j5 Ω
Figure 9.64
'PS1SPC
Figure 9.68
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEJNQFEBODFDPNCJOBUJPOT
'PS1SPC
'or the circuit in Fig. 9.69, find the input impedance
;JOBULSBET
R1
50 Ω
L
2 mH
+ v ‒
C
Figure 9.65
'PS1SPC
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
R2
1 μF
Z in
Figure 9.69
'PS1SPC
+
‒
2v
407
Problems
"Uω = rad/s, find the input admittance of each
PGUIFDJSDVJUTJO'JH
'or the circuit in Fig. 9.70, find the vBMVFPG;5
8 Ω ‒ j12 Ω
‒ j16 Ω
20 Ω
ZT
j15 Ω
60 Ω
Yin
10 Ω
10 Ω
60 Ω
10 Ω
‒ j16 Ω
12.5 μF
20 mH
(a)
20 μF
40 Ω
Figure 9.70
'PS1SPC
Yin
60 Ω
30 Ω
'JOE;5BOE7PJOUIFDJSDVJUJO'JH-FUUIF
WBMVFPGUIFJOEVDUBODFFRVBMKΩ
10 mH
(b)
ZT
Figure 9.74
+
‒ j20 Ω
10 Ω
5 30° A
20 Ω
Vo
'PS1SPC
%FUFSNJOF:FRGPSUIFDJSDVJUJO'JH
‒
Yeq
Figure 9.71
3Ω
‒j2 Ω
j1 Ω
‒ j4 Ω
'PS1SPC
%FUFSNJOF;5BOE*GPSUIFDJSDVJUJO'JH
5Ω
Figure 9.75
'PS1SPC
I
4Ω
‒ j6 Ω
3Ω
j4 Ω
'JOEUIFFRVJWBMFOUBENJUUBODF:FRPGUIFDJSDVJUJO
'JH
2Ω
120 10° V +
‒
2S
1S
‒ j3 S
‒ j2 S
j1 S
j5 S
4S
ZT
Figure 9.72
Figure 9.76
'PS1SPC
'PS1SPC
'PSUIFDJSDVJUJO'JH DBMDVMBUF;5BOE7BC
'JOEUIFFRVJWBMFOUJNQFEBODFPGUIFDJSDVJUJO
Fig. 9.77.
10 Ω
j10 Ω
20 Ω
115 90° V +
‒
+
‒ j5 Ω
ZT
Figure 9.73
'PS1SPC
a
b
Vab
j 15 Ω
‒
40 Ω
2Ω
Zeq
Figure 9.77
'PS1SPC
‒ j 10 Ω
5Ω
8Ω
‒ j5 Ω
408
Chapter 9
Sinusoids and Phasors
0CUBJOUIFFRVJWBMFOUJNQFEBODFPGUIFDJSDVJUJO
'JH
j4 Ω
+
2Ω
‒j2 Ω
j2 Ω
Figure 9.81
'PS1SPC
$BMDVMBUFUIFWBMVFPG;BCJOUIFOFUXPSLPG
Fig. 9.79.
‒ j9 Ω
j6 Ω
‒j9 Ω
j6 Ω
j6 Ω
‒j9 Ω
20 Ω
20 Ω
"DPJMXJUIJNQFEBODF + KΩJTDPOOFDUFEJO
TFSJFTXJUIBDBQBDJUJWFSFBDUBODF95IFTFSJFT
DPNCJOBUJPOJTDPOOFDUFEJOQBSBMMFMXJUIBSFTJT
UPS3(JWFOUIBUUIFFRVJWBMFOUJNQFEBODFPGUIF
SFTVMUJOHDJSDVJUJT⧸
žΩ, find the vBMVFPG3BOE9
B $BMDVMBUFUIFQIBTFTIJGUPGUIFDJSDVJUJO'JH
C 4UBUFXIFUIFSUIFQIBTFTIJGUJTMFBEJOHPS
MBHHJOH PVUQVUXJUISFTQFDUUPJOQVU D %FUFSNJOFUIFNBHOJUVEFPGUIFPVUQVUXIFOUIF
JOQVUJT7
20 Ω
10 Ω
40 Ω
30 Ω
+
b
Vi
Figure 9.79
j10 Ω
j30 Ω
j60 Ω
‒
'PS1SPC
%FUFSNJOFUIFFRVJWBMFOUJNQFEBODFPGUIFDJSDVJUJO
'JH
‒ j4 Ω
2Ω
‒ j6 Ω
4Ω
j6 Ω
j8 Ω
j8 Ω
'PS1SPC
$POTJEFSUIFQIBTFTIJGUJOHDJSDVJUJO'JH-FU
7J = 7PQFSBUJOHBU)['JOE
B 7PXIFO3JTNBYJNVN
C 7PXIFO3JTNJOJNVN
D UIFWBMVFPG3UIBUXJMMQSPEVDFBQIBTFTIJGUPGž
0 < R < 100 Ω
j12 Ω
50 Ω
+
Figure 9.80
vi
'PS1SPC
‒
Applications
200 mH
+
vo
‒
Figure 9.83
%FTJHOBO3-DJSDVJUUPQSPWJEFBžMFBEJOHQIBTF
TIJGU
%FTJHOBDJSDVJUUIBUXJMMUSBOTGPSNBTJOVTPJEBM
WPMUBHFJOQVUUPBDPTJOVTPJEBMWPMUBHFPVUQVU
'PSUIFGPMMPXJOHQBJSTPGTJHOBMT EFUFSNJOFJGv
MFBETPSMBHTvBOECZIPXNVDI
B v = DPT U−ž v = TJOU
C v = DPT U + ž v = TJOU
D v = −DPTU +
Vo
‒
Figure 9.82
b
Section 9.8
+
Vo
‒
10 Ω
‒
'PS1SPC
a
Vi
Zeq
Figure 9.78
a
B $BMDVMBUFUIFQIBTFTIJGUBU.)[
C 'JOEUIFGSFRVFODZXIFSFUIFQIBTFTIJGUJTž
100 nF
‒j Ω
1Ω
3FGFSUPUIF3$DJSDVJUJO'JH
v = TJOU
'PS1SPC
5IFBDCSJEHFJO'JHJTCBMBODFEXIFO3 =
Ω 3 = Ω 3 = LΩ BOE$ = μ'
'JOE3YBOE$Y"TTVNF3BOE$BSFJOTFSJFT
"DBQBDJUBODFCSJEHFCBMBODFTXIFO3 = Ω 3 = LΩ BOE$T = μ'8IBUJT$Y UIF
DBQBDJUBODFPGUIFDBQBDJUPSVOEFSUFTU
"OJOEVDUJWFCSJEHFCBMBODFTXIFO3 = LΩ 3 = Ω BOE-T = N)8IBUJTUIFWBMVFPG
-Y UIFJOEVDUBODFPGUIFJOEVDUPSVOEFSUFTU
409
Comprehensive Problems
5IFBDCSJEHFTIPXOJO'JHJTLOPXOBTB
.BYXFMMCSJEHFBOEJTVTFEGPSBDDVSBUFNFBTVSFNFOU
PGJOEVDUBODFBOESFTJTUBODFPGBDPJMJOUFSNTPGB
TUBOEBSEDBQBDJUBODF$T4IPXUIBUXIFOUIFCSJEHF
JTCBMBODFE
-Y = 33$T
BOE
5IFBDCSJEHFDJSDVJUPG'JHJTDBMMFEB8JFO
CSJEHF*UJTVTFEGPSNFBTVSJOHUIFGSFRVFODZPGB
TPVSDF4IPXUIBUXIFOUIFCSJEHFJTCBMBODFE
________
G = @@@@@@@@@@@@
π√33$$
3
3Y = @@@3
3
'JOE -YBOE 3YGPS 3 = L Ω 3 = L Ω 3 = LΩ BOE$T = μ'
R1
R1
R3
Cs
R3
AC
meter
AC
meter
R2
Lx
R2
R4
C2
Rx
C4
Figure 9.85
Figure 9.84
8JFOCSJEHF'PS1SPC
.BYXFMMCSJEHF'PS1SPC
Comprehensive Problems
‒j20 Ω
5IFDJSDVJUTIPXOJO'JHJTVTFEJOBUFMFWJTJPO
SFDFJWFS8IBUJTUIFUPUBMJNQFEBODFPGUIJTDJSDVJU
j30 Ω
120 Ω
‒ j20 Ω
250 Hz ≈
240 Ω
j95 Ω
‒ j84 Ω
Figure 9.88
'PS1SPC
Figure 9.86
'PS1SPC
5IFOFUXPSLJO'JHJTQBSUPGUIFTDIFNBUJD
EFTDSJCJOHBOJOEVTUSJBMFMFDUSPOJDTFOTJOHEFWJDF
8IBUJTUIFUPUBMJNQFEBODFPGUIFDJSDVJUBUL)[
50 Ω
10 mH
2 μF
80 Ω
"OJOEVTUSJBMMPBEJTNPEFMFEBTBTFSJFTDPNCJOBUJPO
PGBOJOEVDUPSBOEBSFTJTUBODFBTTIPwn in Fig. 9.89.
$BMDVMBUFUIFWBMVFPGBDBQBDJUPS$BDSPTTUIFTFSJFT
DPNCJOBUJPOTPUIBUUIFOFUJNQFEBODFJTSFTJTUJWFBUB
GSFRVFODZPGL)[
100 Ω
10 Ω
C
5 mH
Figure 9.87
'PS1SPC
Figure 9.89
'PS1SPC
"TFSJFTBVEJPDJSDVJUJTTIPXOJO'JH
B 8IBUJTUIFJNQFEBODFPGUIFDJSDVJU
C *GUIFGSFRVFODZXFSFIBMWFE XIBUXPVMECFUIF
JNQFEBODFPGUIFDJSDVJU
"OJOEVTUSJBMDPJMJTNPEFMFEBTBTFSJFTDPNCJOBUJPO
PGBOJOEVDUBODF-BOESFTJTUBODF3 BTTIPXO
in Fig. 9.90. Since an ac vPMUNFUFSNFBTVSFT
only the magnitude of a sinusoid, the folloXJOH
410
Chapter 9
Sinusoids and Phasors
NFBTVSFNFOUTBSFUBLFOBU)[XIFOUIFDJSDVJU
PQFSBUFTJOUIFTUFBEZTUBUF
∣7T∣ = 7 ∣7∣ = 7 ∣7P∣ = 7
6TFUIFTFNFBTVSFNFOUTUPEFUFSNJOFUIFWBMVFTPG-
BOE3
80 Ω
Coil
+ V ‒
+
1
R
Vs +
‒
Vo
L
‒
"USBOTNJTTJPOMJOFIBTBTFSJFTJNQFEBODFPG
; = ⧸
ž ΩBOEBTIVOUBENJUUBODFPG: =
⧸
ž μ4'JOE B UIFDIBSBDUFSJTUJDJNQFEBODF
@@@@@
@@@
;P = Ŀ;∕:
C UIFQSPQBHBUJPODPOTUBOUγ = Ŀ;:
"QPXFSUSBOTNJTTJPOTZTUFNJTNPEFMFEBTTIPXO
JO'JH(JWFOUIFTPVSDFWPMUBHFBOEDJSDVJU
FMFNFOUT
7T = ⧸
ž 7 TPVSDFJNQFEBODF
ZT = + K Ω MJOFJNQFEBODF
;U = + K Ω BOEMPBEJNQFEBODF
Z- = + K Ω find the load current *-
Zs
Figure 9.90
Zℓ
'PS1SPC
IL
'JHVSFTIPXTBTFSJFTDPNCJOBUJPOPGBO
JOEVDUBODFBOEBSFTJTUBODF*GJUJTEFTJSFEUPDPOOFDU
BDBQBDJUPSJOQBSBMMFMXJUIUIFTFSJFTDPNCJOBUJPO
TVDIUIBUUIFOFUJNQFEBODFJTSFTJTUJWFBUL)[ XIBUJTUIFSFRVJSFEWBMVFPG$
vs +
‒
Zℓ
Source
Figure 9.92
10 Ω
C
2 mH
Figure 9.91
'PS1SPC
ZL
'PS1SPC
Transmission line Load
c h a p t e r
Sinusoidal SteadyState Analysis
10
5ISFFNFOBSFNZGSJFOET‡IFUIBUMPWFTNF IFUIBUIBUFTNF IFUIBUJT
JOEJGGFSFOUUPNF8IPMPWFTNF UFBDIFTNFUFOEFSOFTTXIPIBUFTNF UFBDIFTNFDBVUJPOXIPJTJOEJGGFSFOUUPNF UFBDIFTNFTFMGSFMJBODF
‡+&%JOHFS
Enhancing Your Career
Career in Software Engineering
4PGUXBSFFOHJOFFSJOHJTUIBUBTQFDUPGFOHJOFFSJOHUIBUEFBMTXJUIUIF
practical application of scientific knoXMFEHFJOUIFEFTJHO DPOTUSVDUJPO BOEW BMJEBUJPOPGDPNQVUFSQSPHSBNTBOEUIFBTTPDJBUFEEPDVNFOUB UJPOSFRVJSFEUPEF WFMPQ PQFSBUF BOENBJOUBJOUIFN*UJTBCSBODIPG
FMFDUSJDBMFOHJOFFSJOHUIBUJTCFDPNJOHJODSFBTJOHMZJNQPSUBOUBTNPSF
BOENPSFEJTDJQMJOFTSFRVJSFPOFGPSNPGTPGUXBSFQBDLBHFPSBOPUIFSUP
QFSGPSNSPVUJOFUBTLTBOEBTQSPHSBNNBCMFNJDSPFMFDUSPOJDTZTUFNTBSF
VTFEJONPSFBOENPSFBQQMJDBUJPOT
5IFSPMFPGBTPGUX BSFFOHJOFFSTIPVMEOPUCFDPOGVTFEXJUI
UIBUPGBDPNQVUFSTDJFOUJTUUIFTPGUX BSFFOHJOFFSJTBQSBDUJUJPOFS not BUIFPSFUJDJBO "TPGUX BSFFOHJOFFSTIPVMEIB WFHPPEDPNQVUFS -
QSPHSBNNJOHTLJMMTBOECFGBNJMJBSXJUIQSPHSBNNJOHMBOHVBHFT JOQBSUJDVMBS$++ XIJDIJTCFDPNJOHJODSFBTJOHMZQPQVMBS#FDBVTFIBSEXBSF
BOETPGUXBSFBSFDMPTFMZJOUFSMJOL FE JUJTFTTFOUJBMUIBUBTPGUX BSFFOHJOFFSIBWFBUIPSPVHIVOEFSTUBOEJOHPGIBSEXBSFEFTJHO.PTUJNQPSUBOU UIFTPGUXBSFFOHJOFFSTIPVMEIB WFTPNFTQFDJBMJ[FELOP XMFEHFPGUIF
BSFBJOXIJDIUIFTPGUXBSFEFWFMPQNFOUTLJMMJTUPCFBQQMJFE
"MM JO BMM UIF field PG TPGUXBSF FOHJOFFSJOH PGGFST B HSFBU DBSFFS
UPUIPTFXIPFOKP ZQSPHSBNNJOHBOEEF WFMPQJOHTPGUX BSFQBDLBHFT
5IF IJHIFS SFXBSET XJMM HP UP UIPTF IBWJOH UIF CFTU QSFQBSBUJPO XJUI
UIFNPTUJOUFSFTUJOHBOEDIBMMFOHJOHPQQPSUVOJUJFTHPJOHUPUIPTFXJUI
HSBEVBUFFEVDBUJPO
"UISFF-EJNFOTJPOBMQSJOUJOHPGUIF
PVUQVUPGBO"VUP$"%NPEFMPGB
NASA flywheel.
¥"OTPGU$PSQPSBUJPO
411
412
Chapter 10
Sinusoidal Steady-State Analysis
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
"OBMZ[FFMFDUSJDBMDJSDVJUTJOUIFGSFRVFODZEPNBJOVTJOH
OPEBMBOBMZTJT
"OBMZ[FFMFDUSJDBMDJSDVJUTJOUIFGSFRVFODZEPNBJOVTJOH
NFTIBOBMZTJT
"QQMZUIFTVQFSQPTJUJPOQSJODJQMFUPGSFRVFODZEPNBJO
FMFDUSJDBMDJSDVJUT
"QQMZTPVSDFUSBOTGPSNBUJPOJOGSFRVFODZEPNBJODJSDVJUT
6OEFSTUBOEIPX5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUTDBO
CFVTFEJOUIFGSFRVFODZEPNBJO
"OBMZ[FFMFDUSJDBMDJSDVJUTXJUIPQBNQT
10.1
Introduction
*O$IBQUFS XFMFBSOFEUIBUUIFGPSDFEPSTUFBEZ -TUBUFSFTQPOTFPGDJSDVJUTUPTJOVTPJEBMJOQVUTDBOCFPCUBJOFECZVTJOHQIBTPST8FBMTPLOPX
UIBU 0INT BOE ,JSDIIPGGT MBXT BSF BQQMJDBCMF UP BD DJSDVJUT *O UIJT
DIBQUFS XFX BOUUPTFFIP XOPEBMBOBMZTJT NFTIBOBMZTJT 5IFWFOJOT
UIFPSFN /PSUPOTUIFPSFN TVQFSQPTJUJPO BOETPVSDFUSBOTGPSNBUJPOT
BSFBQQMJFEJOBOBMZ[JOHBDDJSDVJUT4JODFUIFTFUFDIOJRVFTXFSFBMSFBEZ
JOUSPEVDFEGPSEDDJSDVJUT PVSNBKPSFGGPSUIFSFXJMMCFUPJMMVTUSBUFXJUI
FYBNQMFT
"OBMZ[JOHBDDJSDVJUTVTVBMMZSFRVJSFTUISFFTUFQT
Steps to Analyze AC Circuits:
5SBOTGPSNUIFDJSDVJUUPUIFQIBTPSPSGSFRVFODZEPNBJO
4PMWFUIFQSPCMFNVTJOHDJSDVJUUFDIOJRVFT OPEBMBOBMZTJT NFTIBOBMZTJT TVQFSQPTJUJPO FUD 5SBOTGPSNUIFSFTVMUJOHQIBTPSUPUIFUJNFEPNBJO
Frequency domain analysis of an ac
circuit via phasors is much easier
than analysis of the circuit in the
time domain.
Step 1 is not necessary if the problem is specified in the frequencZ
EPNBJO*OTUFQ UIFBOBMZTJTJTQFSGPSNFEJOUIFTBNFNBOOFSBTED
DJSDVJUBOBMZTJTFYDFQUUIBUDPNQMFYOVNCFSTBSFJOWPMWFE)BWJOHSFBE
$IBQUFS XFBSFBEFQUBUIBOEMJOHTUFQ
5PXBSEUIFFOEPGUIFDIBQUFS XFMFBSOIP XUPBQQMZ 14QJDFJO
TPMWJOHBDDJSDVJUQSPCMFNT 8e finally apply ac circuit analysis to twP
QSBDUJDBMBDDJSDVJUTPTDJMMBUPSTBOEBDUSBOTJTUPSDJSDVJUT
10.2
Nodal Analysis
5IFCBTJTPGOPEBMBOBMZTJTJT,JSDIIPG GTDVSSFOUMB X4JODF,$-JT
WBMJEGPSQIBTPST BTEFNPOTUSBUFEJO4FDUJPO XFDBOBOBMZ[FBDDJSDVJUTCZOPEBMBOBMZTJT5IFGPMMPXJOHFYBNQMFTJMMVTUSBUFUIJT
10.2
Nodal Analysis
Example 10.1
'JOEiYJOUIFDJSDVJUPG'JHVTJOHOPEBMBOBMZTJT
10 Ω
1H
ix
20 cos 4t V +
‒
2ix
0.1 F
0.5 H
Figure 10.1
'PS&YBNQMF
Solution:
We first convert the circuit to the frequency domain:
0° ⧸__
Ω = SBET
KΩ- = K
KΩ- = K
'
⇒ @@@@
= −K
KΩ$
5IVT UIFGSFRVFODZEPNBJOFRVJWBMFOUDJSDVJUJTBTTIPXOJO'JH
DPTU
)
)
10 Ω
V1
⇒
⇒
⇒
j4 Ω
V2
Ix
20 0° V +
‒
‒ j2.5 Ω
2Ix
j2 Ω
Figure 10.2
'SFRVFODZEPNBJOFRVJWBMFOUPGUIFDJSDVJUJO'JH
"QQMZJOH,$-BUOPEF
− 7 @@@@@@
7− 7
7
@@@@@@@
+ @@@@@@@
=
K
− K
PS
"UOPEF
+ K 7 + K7= 7−7 @@@
7
*Y + @@@@@@@
= K
K
#VU*Y = 7∕−K4VCTUJUVUJOHUIJTHJWFT
7
7−7 @@@
7
@@@@@
+ @@@@@@@
= −K
K
K
#ZTJNQMJGZJOH XFHFU
7 + 7 = &RVBUJPOT BOE DBOCFQVUJONBUSJYGPSNBT
+ K K 7
=
7 8FPCUBJOUIFEFUFSNJOBOUTBT
+ K K
∆=
= − K
[
ǀ
][ ] [ ]
ǀ
413
414
Chapter 10
Sinusoidal Steady-State Analysis
∆ =
ǀ
ǀ
ǀ
ǀ
K
+ K = ∆ =
= −
∆
7 = @@@= @@@@@@@
= ∕18.43° V
∆ − K
∆
−
7 = @@@= @@@@@@@
= ∕198.3° V
∆ − K
5IFDVSSFOU*YJTHJWFOCZ
7
∕18.43°
*Y = @@@@@
= @@@@@@@@@@@@
= ∕108.4°"
−K
∕−90°
5SBOTGPSNJOHUIJTUPUIFUJNFEPNBJO
iY = DPT U + ž "
Practice Problem 10.1
Using nodal analysis, find vBOEvJOUIFDJSDVJUPG'JH
0.2 F
v1
2Ω
25 cos (2t) A
4Ω
v2
+
vx
‒
2H
+
‒
3vx
Figure 10.3
'PS1SBDUJDF1SPC
Answer: v t = DPT U + ž 7
v t = DPT U + ž 7
Example 10.2
$PNQVUF7BOE7JOUIFDJSDVJUPG'JH
10 45° V
+‒
V1
1
4Ω 2
‒j3 Ω
3 0° A
V2
j6 Ω
12 Ω
Figure 10.4
'PS&YBNQMF
Solution:
/PEFTBOEGPSNBTVQFSOPEFBTTIPXOJO'JH"QQMZJOH,$-
BUUIFTVQFSOPEFHJWFT
7 @@@
7
7
= @@@@
+ + @@@
−K K PS
= K7 + −K 7
10.3
415
Supernode
V2
V1
‒j3 Ω
3A
Mesh Analysis
j6 Ω
12 Ω
Figure 10.5
"TVQFSOPEFJOUIFDJSDVJUPG'JH
#VUBWPMUBHFTPVSDFJTDPOOFDUFECFUXFFOOPEFTBOE TPUIBU
7 = 7 + ∕45°
4VCTUJUVUJOH&R JO&R SFTVMUTJO
−∕135° = + K 7
⇒
7 = ∕−° 7
'SPN&R 7 = 7 + ∕45°= ∕−° 7
Practice Problem 10.2
$BMDVMBUF7BOE7JOUIFDJSDVJUTIPXOJO'JH
4Ω
V1
75 0° V +
‒
100 60° V
+‒
j4 Ω
V2
‒ j1 Ω
2Ω
Figure 10.6
'PS1SBDUJDF1SPC
Answer:7 = ∕69.66°7 7 = ∕165.72°7
10.3
Mesh Analysis
,JSDIIPGGTWPMUBHFMBX ,7- GPSNTUIFCBTJTPGNFTIBOBMZTJT 5IF
WBMJEJUZPG,7-GPSBDDJSDVJUTXBTTIPXOJO4FDUJPOBOEJTJMMVTUSBUFE
JOUIFGPMMP XJOHFYBNQMFT,FFQJONJOEUIBUUIFW FSZOBUVSFPGVTJOH
NFTIBOBMZTJTJTUIBUJUJTUPCFBQQMJFEUPQMBOBSDJSDVJUT
%FUFSNJOFDVSSFOU*PJOUIFDJSDVJUPG'JHVTJOHNFTIBOBMZTJT
Solution:
"QQMZJOH,7-UPNFTI XFPCUBJO
+ K−K *− −K *−K* = Example 10.3
416
Chapter 10
Sinusoidal Steady-State Analysis
4Ω
I3
5 0° A
j 10 Ω
I2
I1
8Ω
Io
‒ j2 Ω
+ 20 90° V
‒
‒ j2 Ω
Figure 10.7
'PS&YBNQMF
'PSNFTI
−K−K *− −K *− −K * + ∕90° = 'PSNFTI * = 4VCTUJUVUJOHUIJTJO&RT BOE XFHFU
+ K * + K* = K
K* + −K * = −K−K
&RVBUJPOT BOE DBOCFQVUJONBUSJYGPSNBT
[
][ ] [ ]
K
*
K
+ K
=
K − K * −K GSPNXIJDIXFPCUBJOUIFEFUFSNJOBOUT
∆=
∆ =
ǀ
ǀ
ǀ
K
+ K
= + K −K + = K − K
ǀ
+ K K
= −K = ∕−°
K −K
°
∆
∕−
* = @@@= @@@@@@@@@@@@@@
= ∕−°"
∆
5IFEFTJSFEDVSSFOUJT
*P = −* = ∕144.78°"
Practice Problem 10.3
10 0° A
‒j2 Ω
Answer:∕65.45°"
6Ω
Io
8Ω
Figure 10.8
j4 Ω
'PS1SBDUJDF1SPC
'JOE*PJO'JHVTJOHNFTIBOBMZTJT
+ 50 30° V
‒
10.3
Mesh Analysis
Example 10.4
4PMWFGPS7PJOUIFDJSDVJUPG'JHVTJOHNFTIBOBMZTJT
4 0° A
‒j4 Ω
10 0° V +
‒
6Ω
j5 Ω
8Ω
+
Vo
‒
‒j2 Ω
3 0° A
Figure 10.9
'PS&YBNQMF
Solution:
"TTIPXOJO'JH NFTIFTBOEGPSNBTVQFSNFTIEVFUPUIF
DVSSFOUTPVSDFCFUXFFOUIFNFTIFT'PSNFTI ,7-HJWFT
− + −K *− −K *−* = PS
'PSNFTI
−K * + K*−* = * = −
'PSUIFTVQFSNFTI
−K *−* + + K *−K* = %VFUPUIFDVSSFOUTPVSDFCFUXFFONFTIFTBOE BUOPEF"
* = * + ■ METHOD 1 *OTUFBEPGTPMWJOHUIFBCPWFGPVSFRVBUJPOT XFSF EVDFUIFNUPUXPCZFMJNJOBUJPO
$PNCJOJOH&RT BOE −K *−* = + K
$PNCJOJOH&RT UP −* + + K * = −−K
I3 A
‒j4 Ω
I3
I1
Figure 10.10
‒ j2 Ω
+
Vo
‒
"OBMZTJTPGUIFDJSDVJUJO'JH
Supermesh
I4
4A
8Ω
10 V +
‒
I4
6Ω
j5 Ω
I2
3A
417
418
Chapter 10
Sinusoidal Steady-State Analysis
'SPN&RT BOE XFPCUBJOUIFNBUSJYFRVBUJPO
[
][ ] [
]
*
− K
−
+ K
=
− + K * −−K 8FPCUBJOUIFGPMMPXJOHEFUFSNJOBOUT
∆=
ǀ
ǀ
ǀ
− K
−
= + K−K + − = −K
− + K
+ K
−−K
ǀ
−
= + K + K−−−K
+ K = −−K
$VSSFOU*JTPCUBJOFEBT
∆ =
−−K
∆
* = @@@= @@@@@@@@@@
= ∕274.5°"
∆
−K
5IFSFRVJSFEWPMUBHF7PJT
7P = −K *−* = −K ∕274.5° + = −−K = ∕222.32°7
■ METHOD 2 8FDBOVTF ."5-"#UPTPMWF&RT UP
(10.4.4). We first cast the equations as
PS
][ ] [ ]
[
*
−K K
−
−
*
=
− −K −K + K *
−
* B
"* = #
#ZJOWFSUJOH" XFDBOPCUBJO*BT
* = "¢#
8FOPXBQQMZ."5-"#BTGPMMPXT
"< K K K K K >
#<>
*JOW " #
*
J
J
J
7P K * * 7P
J
BTPCUBJOFEQSFWJPVTMZ
C
10.4
419
Superposition Theorem
Practice Problem 10.4
$BMDVMBUFDVSSFOU*PJOUIFDJSDVJUPG'JH
Answer:∕5.94°"
Io
10 Ω
‒ j4 Ω
j8 Ω
2.4 0° A
10.4
60 0° V +
‒
Superposition Theorem
4JODFBDDJSDVJUTBSFMJOFBS UIFTVQFSQPTJUJPOUIFPSFNBQQMJFTUPBD DJSDVJUTUIFTBNFXBZJUBQQMJFTUPEDDJSDVJUT5IFUIFPSFNCFDPNFT JNQPSUBOUJGUIFDJSDVJUIBTTPVSDFTPQFSBUJOHBU EJGGFSFOUGSFRVFODJFT
*OUIJTDBTF TJODFUIFJNQFEBODFTEFQFOEPOGSFRVFODZ XFNVTUIBWF
BEJGGFSFOUGSFRVFODZEPNBJODJSDVJUGPSFBDIGSFRVFODZ5IFUPUBMSF TQPOTFNVTUCFPCUBJOFECZBEEJOHUIFJOEJWJEVBMSFTQPOTFTJOUIFUJNF
EPNBJO*UJTJODPSSFDUUPUSZUPBEEUIFSFTQPOTFTJOUIFQIBTPSPSGSFRVFODZEPNBJO8IZ #FDBVTFUIFFYQPOFOUJBMGBDUPS eKωUJTJNQMJDJU
JOTJOVTPJEBMBOBMZTJT BOEUIBUGBDUPSXPVMEDIBOHFGPSFWFSZBOHVMBS
GSFRVFODZ ω*UXPVMEUIFSFGPSFOPUNBLFTFOTFUPBEESFTQPOTFTBU EJGGFSFOUGSFRVFODJFTJOUIFQIBTPSEPNBJO5IVT XIFOBDJSDVJUIBT TPVSDFTPQFSBUJOHBUEJGGFSFOUGSFRVFODJFT POFNVTUBEEUIFSFTQPOTFT
EVFUPUIFJOEJWJEVBMGSFRVFODJFTJOUIFUJNFEPNBJO
‒j6 Ω
5Ω
Figure 10.11
'PS1SBDUJDF1SPC
Example 10.5
Use the superposition theorem to find *PJOUIFDJSDVJUJO'JH
Solution:
-FU
*P = *′P + *″P
XIFSF*′PBOE*″PBSFEVFUPUIFWPMUBHFBOEDVSSFOUTPVSDFT SFTQFDUJWFMZ
To find *′P DPOTJEFSUIFDJSDVJUJO'JH B *GXFMFU;CFUIFQBSBMMFM
DPNCJOBUJPOPG−KBOE + K UIFO
4Ω
−K + K
; = @@@@@@@@@@@@
= −K
−K + + K
j10 Ω
+
‒
K
K
*′P = @@@@@@@@@
= @@@@@@@@@@@
−K + ; −K
(a)
PS
*′P = − + K
+ K *−K* + K* = 5A
8Ω
−K * + K* + K* = I2
‒j2 Ω
I1
(b)
* = Figure 10.12
I"o
‒j2 Ω
'PSNFTI
I3
j10 Ω
'PSNFTI
4Ω
5PHFU*″P DPOTJEFSUIFDJSDVJUJO'JH C 'PSNFTI
j20 V
‒ j2 Ω
8Ω
BOEDVSSFOU*′PJT
I'o
‒ j2 Ω
4PMVUJPOPG&YBNQMF
420
Chapter 10
Sinusoidal Steady-State Analysis
'SPN&RT BOE −K * + K* + K = &YQSFTTJOH*JOUFSNTPG*HJWFT
* = + K *−
4VCTUJUVUJOH&RT BOE JOUP&R XFHFU
+ K < + K *−>−K + K* = PS
−K
* = @@@@@@@@
= −K
$VSSFOU*″PJTPCUBJOFEBT
*″P = −* = − + K
'SPN&RT BOE XFXSJUF
*P = *′P + *″P = − + K = ∕144.78°"
XIJDIBHSFFTXJUIXIBUXFHPUJO&YBNQMF*UTIPVMECFOPUFEUIBU
BQQMZJOHUIFTVQFSQPTJUJPOUIFPSFNJTOPUUIFCFTUXBZUPTPMWFUIJTQSPC
MFN*UTFFNTUIBUXFIBWFNBEFUIFQSPCMFNUXJDFBTIBSEBTUIFPSJHJ OBMPOFCZVTJOHTVQFSQPTJUJPO)PXFWFS JO&YBNQMF TVQFSQPTJ UJPOJTDMFBSMZUIFFBTJFTUBQQSPBDI
Practice Problem 10.5
'JOEDVSSFOU *PJOUIFDJSDVJUPG'JHVTJOHUIFTVQFSQPTJUJPO
UIFPSFN
Answer:∕65.45°"
Example 10.6
'JOEvPPGUIFDJSDVJUPG'JHVTJOHUIFTVQFSQPTJUJPOUIFPSFN
2H
1Ω
4Ω
+ vo ‒
10 cos 2t V +
‒
2 sin 5t A
0.1 F
+ 5V
‒
Figure 10.13
'PS&YBNQMF
Solution:
#FDBVTFUIFDJSDVJUPQFSBUFTBUUISFFEJGGFSFOUGSFRVFODJFT ω = GPSUIF
EDWPMUBHFTPVSDF POFXBZUPPCUBJOBTPMVUJPOJTUPVTFTVQFSQPTJUJPO XIJDICSFBLTUIFQSPCMFNJOUPTJOHMF-GSFRVFODZQSPCMFNT4PXFMFU
vP = v + v + v
10.4
421
Superposition Theorem
XIFSFvJTEVFUPUIF-7EDWPMUBHFTPVSDF vJTEVFUPUIFDPTU7
WPMUBHFTPVSDF BOEvJTEVFUPUIFTJOU"DVSSFOUTPVSDF
5Pfind v XFTFUUP[FSPBMMTPVSDFTFYDFQUUIF-7EDTPVSDF8F
recall that at steady state, a capacitor is an open circuit to dc XIJMFBO
inductor is a short circuit to dc. There is an alternative way PGMPPLJOH
BUUIJT#FDBVTFω = Kω- = ∕Kω$ = ∞&JUIFSXBZ UIFFRVJWBMFOU
DJSDVJUJTBTTIPXOJO'JH B #ZWPMUBHFEJWJTJPO
= 7
−v = @@@@@
+
To find v XFTFUUP[FSPCPUIUIF-7TPVSDFBOEUIFTJOUDVSSFOU
TPVSDFBOEUSBOTGPSNUIFDJSDVJUUPUIFGSFRVFODZEPNBJO
DPTU
⇒
∕0° )
⇒
Kω- = KΩ
'
⇒
@@@@
= −KΩ
ω = SBET
Kω$
5IFFRVJWBMFOUDJSDVJUJTOPXBTTIPXOJO'JH C -FU
−K×
; = −K∥ = @@@@@@@ = −K
−K
1Ω
j4 Ω
4Ω
+ v1 ‒
+
‒
5V
10 0° V
+
‒
1Ω
I1
4Ω
+ V3 ‒
+ V2 ‒
‒ j5 Ω
(a)
1Ω
j10 Ω
(b)
Figure 10.14
2 ‒90° A
‒j2 Ω
4Ω
(c)
4PMVUJPOPG&YBNQMF B TFUUJOHBMMTPVSDFTUP[FSPFYDFQUUIF-7EDTPVSDF C TFUUJOHBMMTPVSDFTUP[FSPFYDFQUUIFBD
WPMUBHFTPVSDF D TFUUJOHBMMTPVSDFTUP[FSPFYDFQUUIFBDDVSSFOUTPVSDF
#ZWPMUBHFEJWJTJPO
7 = @@@@@@@@@
∕0° = @@@@@@@@@@@@@
= ∕−30.79°
+ K + ;
+ K
*OUIFUJNFEPNBJO
v = DPT U−ž 5PPCUBJOv XFTFUUIFWPMUBHFTPVSDFTUP[FSPBOEUSBOTGPSNXIBU
JTMFGUUPUIFGSFRVFODZEPNBJO
TJOU
⇒
∕−90° )
⇒
Kω- = KΩ
'
⇒
@@@@
= −KΩ
Kω$
ω = SBET
422
Chapter 10
Sinusoidal Steady-State Analysis
5IFFRVJWBMFOUDJSDVJUJTJO'JH D -FU
−K×
; = −K∥ = @@@@@@@ = −KΩ
−K
#ZDVSSFOUEJWJTJPO
K
∕−90° "
* = @@@@@@@@@@@
K + + ; K
7 = *× = @@@@@@@@@
−K = ∕−80°7
+ K
*OUIFUJNFEPNBJO
v = DPT U−ž = TJO U + ž 7
4VCTUJUVUJOH&RT UP JOUP&R XFIBWF
vP t = − + DPT U−ž + TJO U + ž 7
Practice Problem 10.6
$BMDVMBUF vPJOUIFDJSDVJUPG'JHVTJOHUIFTVQFSQPTJUJPO
UIFPSFN
8Ω
+
vo
‒
75 sin 5t V +
‒
0.2 F
1H
6 cos 10t A
Figure 10.15
'PS1SBDUJDF1SPC
Answer:TJO U−ž + DPT U−ž 7
10.5
Source Transformation
"T'JHTIPXT TPVSDFUSBOTGPSNBUJPOJOUIFGSFRVFODZEPNBJO JOWPMWFTUSBOTGPSNJOHBWPMUBHFTPVSDFJOTFSJFTXJUIBOJNQFEBODFUP
BDVSSFOUTPVSDFJOQBSBMMFMXJUIBOJNQFEBODF PSWJDFWFSTB"TXFHP
GSPNPOFTPVSDFUZQFUPBOPUIFS XFNVTULFFQUIFGPMMPXJOHSFMBUJPOTIJQ
JONJOE
7T = ;T*T
⇔
7
*T = @@@T
;T 10.5
423
Source Transformation
Zs
a
a
Vs +
‒
Zs
Is
Vs = ZsIs
b
Is =
b
Vs
Zs
Figure 10.16
4PVSDFUSBOTGPSNBUJPO
$BMDVMBUF 7YJOUIFDJSDVJUPG'JHVTJOHUIFNFUIPEPGTPVSDF
USBOTGPSNBUJPO
5Ω
4Ω
‒ j13 Ω
3Ω
2 0 ‒90° V +
‒
Example 10.7
j4 Ω
10 Ω
+
Vx
‒
Figure 10.17
'PS&YBNQMF
Solution:
8FUSBOTGPSNUIFWPMUBHFTPVSDFUPBDVSSFOUTPVSDFBOEPCUBJOUIFDJS DVJUJO'JH B XIFSF
∕−90°
*T = @@@@@@@@
= ∕−90° = −K"
5IFQBSBMMFMDPNCJOBUJPOPG-ΩSFTJTUBODFBOE + K JNQFEBODFHJWFT
+ K
; = @@@@@@@@
= + KΩ
+ K
$POWFSUJOHUIFDVSSFOUTPVSDFUPBWPMUBHFTPVSDFZJFMETUIFDJSDVJUJO
Fig. 10.18(b), where
7T = *T; = −K + K = −K7
4Ω
‒ j13 Ω
3Ω
Is = ‒ j4 A
5Ω
j4 Ω
10 Ω
2.5 Ω
+
Vx
‒
j 1.25 Ω
Vs = 5 ‒ j10 V +
‒
(a)
Figure 10.18
4PMVUJPOPGUIFDJSDVJUJO'JH
#ZWPMUBHFEJWJTJPO
7Y = @@@@@@@@@@@@@@@@@@@@@@@
−K = ∕−28°7
+ + K + −K
4Ω
‒ j13 Ω
10 Ω
(b)
+
Vx
‒
424
Chapter 10
Practice Problem 10.7
Sinusoidal Steady-State Analysis
'JOE *PJOUIFDJSDVJUPG'JHVTJOHUIFDPODFQUPGTPVSDF
USBOTGPSNBUJPO
j1 Ω
2Ω
Io
4Ω
2.4 90° A
j5 Ω
1Ω
‒ j3 Ω
‒ j2 Ω
Figure 10.19
'PS1SBDUJDF1SPC
Answer:∕99.46°"
10.6
ZTh
a
Linear
circuit
a
VTh
+
‒
b
b
Figure 10.20
5IFWFOJOFRVJWBMFOU
5IFWFOJOTBOE/PSUPOTUIFPSFNTBSFBQQMJFEUPBDDJSDVJUTJOUIFTBNF
XBZBTUIFZBSFUPEDDJSDVJUT5IFPOMZBEEJUJPOBMFGGPSUBSJTFTGSPNUIF
OFFEUPNBOJQVMBUFDPNQMFYOVNCFST5IFGSFRVFODZEPNBJOWFSTJPOPG
B5IFWFOJOFRVJWBMFOUDJSDVJUJTEFQJDUFEJO'JH XIFSFBMJOFBS
DJSDVJUJTSFQMBDFECZBWPMUBHFTPVSDFJOTFSJFTXJUIBOJNQFEBODF5IF
/PSUPOFRVJWBMFOUDJSDVJUJTJMMVTUSBUFEJO'JH XIFSFBMJOFBSDJS DVJUJTSFQMBDFECZBDVSSFOUTPVSDFJOQBSBMMFMXJUIBOJNQFEBODF,FFQ
JONJOEUIBUUIFUXPFRVJWBMFOUDJSDVJUTBSFSFMBUFEBT
a
Linear
circuit
Thevenin and Norton
Equivalent Circuits
75I = ;/*/ ;5I = ;/ a
IN
b
Figure 10.21
/PSUPOFRVJWBMFOU
Example 10.8
ZN
b
KVTUBTJOTPVSDFUSBOTGPSNBUJPO75IJTUIFPQFO-DJSDVJUWPMUBHFXIJMF*/
JTUIFTIPSU-DJSDVJUDVSSFOU
*GUIFDJSDVJUIBTTPVSDFTPQFSBUJOHBUEJG GFSFOUGSFRVFODJFT TFF
&YBNQMF GPSFYBNQMF UIF5IFWFOJOPS /PSUPOFRVJWBMFOUDJSDVJU
NVTUCFEFUFSNJOFEBUFBDIGSFRVFOD Z5IJTMFBETUPFOUJSFMZEJG GFSFOU
FRVJWBMFOUDJSDVJUT POFGPSFBDIGSFRVFOD Z OPUPOFFRVJ WBMFOUDJSDVJU
XJUIFRVJWBMFOUTPVSDFTBOEFRVJWBMFOUJNQFEBODFT
0CUBJOUIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTB-CPGUIFDJSDVJUJO'JH
d
‒ j6 Ω
120 75° V +
‒
e
4Ω
a
b
j12 Ω
8Ω
f
Figure 10.22
'PS&YBNQMF
c
10.6
425
Thevenin and Norton Equivalent Circuits
Solution:
We find ;5ICZTFUUJOHUIFWPMUBHFTPVSDFUP[FSP"TTIPXOJO
Fig. B UIF -ΩSFTJTUBODFJTOPXJOQBSBMMFMXJUIUIF −KSFBD UBODF TPUIBUUIFJSDPNCJOBUJPOHJWFT
−K×
; = −K∥ = @@@@@@@ = −KΩ
−K
4JNJMBSMZ UIF -ΩSFTJTUBODFJTJOQBSBMMFMXJUIUIF KSFBDUBODF BOE
UIFJSDPNCJOBUJPOHJWFT
K× ; = ∥ K = @@@@@@@
= + KΩ
+ K
d
f,d
8Ω
I1
f,d
‒ j6 Ω
‒ j6 Ω
4Ω
a
j12 Ω
120 75° V +
‒
a
ZTh
c
f
(b)
Figure 10.23
Solution of the circuit in Fig. 10.22: (a) finding ;5I, (b) finding 75I
5IF5IFWFOJOJNQFEBODFJTUIFTFSJFTDPNCJOBUJPOPG
that is,
;BOE ;
;5I = ; + ; = −KΩ
To find 75I DPOTJEFSUIFDJSDVJUJO'JH C $VSSFOUT*BOE*
BSFPCUBJOFEBT
∕75° * = @@@@@@@@
" −K
∕75°
* = @@@@@@@@
"
+ K
"QQMZJOH,7-BSPVOEMPPQCDEFBCJO'JH C HJWFT
75I−* + −K * = PS
∕75°
∕75° + 90°
75I = * + K* = @@@@@@@@
+ @@@@@@@@@@@@@
+ K
−K
= ∕3.43° + ∕201.87°
= −−K = ∕220.31°7
b
c
j12 Ω
8Ω
(a)
+ VTh ‒
e
b
e
I2
4Ω
426
Chapter 10
Practice Problem 10.8
Sinusoidal Steady-State Analysis
'JOEUIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTB-CPGUIFDJSDVJUJO'JH
j2 Ω
6Ω
a
75 20° V +
‒
b
10 Ω
‒j4 Ω
Figure 10.24
'PS1SBDUJDF1SPC
Answer: ;5I = −KΩ 75I = ∕−51.57°7
Example 10.9
'JOEUIF 5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO'JHBTTFFOGSPN
UFSNJOBMTB-C
j3 Ω
4Ω
a
Io
15 0° A
2Ω
0.5Io
‒ j4 Ω
b
Figure 10.25
'PS&YBNQMF
Solution:
To find 75I XFBQQMZ,$-BUOPEFJO'JH B = *P + *P
⇒
*P = "
"QQMZJOH,7-UPUIFMPPQPOUIFSJHIU -IBOETJEFJO'JH B XF
PCUBJO
−*P −K + *P + K + 75I = PS
75I = −K − + K = −K
5IVT UIF5IFWFOJOWPMUBHFJT
75I = ∕−90°7
0.5Io
1
4 + j3 Ω
4 + j3 Ω
2
+
Io
15 A
2 ‒ j4 Ω
0.5Io
a
VTh
‒
Is
+
2 ‒ j4 Ω
0.5Io
Solution of the problem in Fig. 10.25: (a) finding 75I, (b) finding ;5I
Vs
‒
b
(a)
Figure 10.26
a
Vs
Io
(b)
b
Is = 3 0° A
10.6
427
Thevenin and Norton Equivalent Circuits
5PPCUBJO;5I XFSFNPWFUIFJOEFQFOEFOUTPVSDF%VFUPUIFQSFT FODFPGUIFEFQFOEFOUDVSSFOU TPVSDF XF DPOOFDUB-"DVSSFOUTPVSDF
JTBOBSCJUSBSZWBMVFDIPTFOGPSDPOWFOJFODFIFSF BOVNCFSEJWJTJCMF
by UIFTVNPGDVSSFOUTMFBWJOHUIFOPEF UPUFSNJOBMT B-CBTTIPXOJO
Fig. 10.26(b). At the node, KCL gives
= *P + *P
⇒
*P = "
"QQMZJOH,7-UPUIFPVUFSMPPQJO'JH C HJWFT
7T = *P + K + −K = −K
5IF5IFWFOJOJNQFEBODFJT
−K
7
;5I = @@@T= @@@@@@@
= −KΩ
*T
%FUFSNJOFUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO'JHBTTFFO
GSPNUIFUFSNJOBMTa-C
Practice Problem 10.9
j4 Ω
8Ω
Answer: ;5I = ⧸
−žΩ 75I = ⧸
ž7
+
Vo
‒
a
‒ j2 Ω
8 0° A
4Ω
0.2Vo
b
Figure 10.27
'PS1SBDUJDF1SPC
Example 10.10
0CUBJODVSSFOU*PJO'JHVTJOH/PSUPOTUIFPSFN
a
5Ω
8Ω
‒ j2 Ω
Io
3 0° A
20 Ω
10 Ω
40 90° V +
‒
j15 Ω
j4 Ω
b
Figure 10.28
'PS&YBNQMF
Solution:
Our first PCKFDUJWFJT to find the Norton equivalent BUUFSNJOBMT B-C;/
JTGPVOEJOUIFTBNFXBZBT ;5I8FTFUUIFTPVSDFTUP[FSPBTTIPXO
in Fig. 10.29(a). As evident from the figure, the (8 −K BOE + K JNQFEBODFTBSFTIPSU-DJSDVJUFE TPUIBU
;/ = Ω
5PHFU */ XFTIPSU -DJSDVJUUFSNJOBMT B-CBTJO'JH C BOE BQQMZNFTIBOBMZTJT/PUJDFUIBUNFTIFTBOEGPSNBTVQFSNFTI
CFDBVTFPGUIFDVSSFOUTPVSDFMJOLJOHUIFN'PSNFTI
−K + + K *− −K *− + K * = 428
Chapter 10
a
Sinusoidal Steady-State Analysis
a
I2
I3
a
IN
5
8
‒j2
ZN
I2
5
10
8
j40 +
‒
j4
‒ j2
I1
Io
3
I3
10
20
3 + j8
5
j15
j4
b
b
(a)
b
(c)
(b)
Figure 10.29
Solution of the circuit in Fig. 10.28: (a) finding ;/, (b) finding 7/ D DBMDVMBUJOH*P
'PSUIFTVQFSNFTI
−K * + + K *− + K * = "UOPEFB EVFUPUIFDVSSFOUTPVSDFCFUXFFONFTIFTBOE
* = * + "EEJOH&RT BOE HJWFT
−K + * = ⇒
* = K
'SPN&R * = * + = + K
5IF/PSUPODVSSFOUJT
*/ = * = + K "
'JHVSF D TIPXTUIF/PSUPOFRVJWBMFOUDJSDVJUBMPOHXJUIUIFJN QFEBODFBUUFSNJOBMTB-C#ZDVSSFOUEJWJTJPO
+ K
*P = @@@@@@@@@@@
ž"
*/ = @@@@@@
= ⧸
+ + K
+ K
Practice Problem 10.10
%FUFSNJOFUIF/PSUPOFRVJWBMFOUPGUIFDJSDVJUJO'JHBTTFFOGSPN
UFSNJOBMTB-C. Use the equivalent to find *P
4Ω
8Ω
20 0° V
+
‒
j2 Ω
1Ω
‒j3 Ω
4 ‒90° A
a
Io
10 Ω
‒ j5 Ω
b
Figure 10.30
'PS1SBDUJDF1SPCBOE1SPC
Answer: ;/ = + KΩ */ = ⧸
−ž" *P = ⧸
−ž"
10.7
10.7
429
Op Amp AC Circuits
Op Amp AC Circuits
5IFUISFFTUFQTTUBUFEJO4FDUJPOBMTPBQQMZUPPQBNQDJSDVJUT BT
MPOHBTUIFPQBNQJTPQFSBUJOHJOUIFMJOFBSSFHJPO"TVTVBM XFXJMM
BTTVNFJEFBMPQBNQT 4FF4FDUJPO "TEJTDVTTFEJO$IBQUFS UIF
LFZUPBOBMZ[JOHPQBNQDJSDVJUTJTUPLFFQUXPJNQPSUBOUQSPQFSUJFTPG
BOJEFBMPQBNQJONJOE
/PDVSSFOUFOUFSTFJUIFSPGJUTJOQVUUFSNJOBMT
5IFWPMUBHFBDSPTTJUTJOQVUUFSNJOBMTJT[FSP
5IFGPMMPXJOHFYBNQMFTXJMMJMMVTUSBUFUIFTFJEFBT
%FUFSNJOF vP t GPSUIFPQBNQDJSDVJUJO'JH B JG
DPTU7
Example 10.11
vT =
20 kΩ
20 kΩ
10 kΩ
vs +
‒
‒ j10 kΩ
0.1 μF
10 kΩ
10 kΩ
‒
vo
+
0.2 μF
V1
10 kΩ
3 0° V +
‒
‒j5 kΩ
(a)
(b)
'PS&YBNQMF B UIFPSJHJOBMDJSDVJUJOUIFUJNFEPNBJO C JUTGSFRVFODZEPNBJOFRVJWBMFOU
Solution:
We first transform the circuit to the frequency domain, as shown in
Fig. C XIFSF 7T = ⧸
ž ω = SBET"QQMZJOH,$-BU
node 1, we obtain
⧸
7 −7
7 − @@@@@@@
7 @@@@@@
@@@@@@@@@
+ = @@@@
+ P
−K
PS
= + K 7−7P
"UOPEF ,$-HJWFT
−7P
7 − @@@@@@
@@@@@@
=
−K
XIJDIMFBETUP
7 = −K7P
4VCTUJUVUJOH&R JOUP&R ZJFMET
= −K + K 7P−7P = −K 7P
)FODF
7P = @@@@@@
ž
= ⧸
−K
vP t = DPT U + ž 7
0V
2
1
Figure 10.31
ž−7
Vo
‒
+
Vo
430
Chapter 10
Practice Problem 10.11
Sinusoidal Steady-State Analysis
'JOE vPBOE iPJOUIFPQBNQDJSDVJUPG'JH-FU
DPTU7
10 kΩ
10 nF
20 kΩ
vs +
‒
vT =
‒
io
+
vo
20 nF
Figure 10.32
'PS1SBDUJDF1SPC
Answer: TJO U7 TJO Uμ"
Example 10.12
$PNQVUFUIFDMPTFE -MPPQH BJOBOEQIBTFTIJGUGPSUIFDJSDVJUJO
Fig. "TTVNF UIBUR = 3 = LΩ $ = μ' $ = μ' BOE
ω = SBET
C2
C1
R1
Solution:
5IFGFFECBDLBOEJOQVUJNQFEBODFTBSFDBMDVMBUFEBT
R2
‒
+
vs +
‒
+
vo
‒
Figure 10.33
3
;G = 3 @@@@@
= @@@@@@@@@@
Kω$ + Kω3$
+ Kω3$
= @@@@@@@@@@
;J = 3 + @@@@@
Kω$
Kω$
4JODFUIFDJSDVJUJO'JH JTBOinverting amplifier, UIFDMPTFE-MPPQ
HBJOJTHJWFOCZ
'PS&YBNQMF
;G
−Kω$3
7
( = @@@P= −@@ = @@@@@@@@@@@@@@@@@@@@@@
7T
;J
+ Kω3$ + Kω3$
4VCTUJUVUJOHUIFHJWFOWBMVFTPGR 3 $ $ BOEω XFPCUBJO
−K
( = @@@@@@@@@@@@@
1ž
= ⧸
+ K + K
5IVT UIFDMPTFE-MPPQHBJOJTBOEUIFQIBTFTIJGUJTž
Practice Problem 10.12
+
‒
vs +
‒
R
Figure 10.34
'PS1SBDUJDF1SPC
C
R
vo
0CUBJOUIFDMPTFE-MPPQHBJOBOEQIBTFTIJGUGPSUIFDJSDVJUJO'JH
-FU3 = LΩ $ = μ' BOEω = SBET
Answer: −ž
10.8
10.8
AC Analysis Using PSpice
431
AC Analysis Using PSpice
14QJDFBGGPSETBCJHSFMJFGGSPNUIFUFEJPVTUBTLPGNBOJQVMBUJOHDPNQMFYOVNCFSTJOBDDJSDVJUBOBMZTJT5IFQSPDFEVSFGPSVTJOH 14QJDF
GPSBDBOBMZTJTJTRVJUFTJNJMBSUPUIBUSFRVJSFEGPSEDBOBMZTJT5IF SFBEFSTIPVMESFBE4FDUJPO%JO"QQFOEJY%GPSBSFWJFXPG14QJDF
DPODFQUTGPSBDBOBMZTJT"$DJSDVJUBOBMZTJTJTEPOFJOUIFQIBTPS
PSGSFRVFODZEPNBJO BOEBMMTPVSDFTNVTUIBWFUIFTBNFGSFRVFODZ
"MUIPVHIBDBOBMZTJTXJUI 14QJDFJOWPMWFTVTJOH"$4XFFQ PVS BOBMZTJTJOUIJTDIBQUFSSFRVJSFTBTJOHMFGSFRVFODZf = ω∕π 5IFPVUput file of 14QJDFDPOUBJOTWPMUBHFBOEDVSSFOUQIBTPST*GOFDFTTBSZ UIFJNQFEBODFTDBOCFDBMDVMBUFEVTJOHUIFWPMUBHFTBOEDVSSFOUTJO
the output file.
Example 10.13
0CUBJOvPBOEiPJOUIFDJSDVJUPG'JHVTJOH14QJDF
50 mH
4 kΩ
io
8 sin(1000t + 50°) V +
‒
2 μF
0.5io
2 kΩ
+
vo
‒
Figure 10.35
'PS&YBNQMF
Solution:
We first convert the sine function to cosine.
TJO U + ž = DPT U + ž−ž
= DPT U−ž
5IFGSFRVFODZGJTPCUBJOFEGSPNωBT
ω = @@@@@
G = @@@
= )[
π
π
5IFTDIFNBUJDGPSUIFDJSDVJUJTTIPXOJO'JH/PUJDFUIBUUIF
DVSSFOU-DPOUSPMMFEDVSSFOUTPVSDF'JTDPOOFDUFETVDIUIBUJUTDVSSFOU
flows from node 0 to node 3 in conformity with the original circuit in
'JH4JODFXFPOMZXBOUUIFNBHOJUVEFBOEQIBTFPGvPBOEJP XF
TFUUIFBUUSJCVUFTPG*13*/5BOE713*/5FBDIUP "$ = ZFT ."( =
ZFT 1)"4& = ZFT"TBTJOHMF -GSFRVFODZBOBMZTJT XFTFMFDU "OBMZTJT
4FUVQ"$4XFFQBOEFOUFS 5PUBM1UT = 4UBSU'SFR = BOE
'JOBM'SFR = "GUFSTBWJOHUIFTDIFNBUJD XFTJNVMBUFJUCZ TFMFDUJOH "OBMZTJT4JNVMBUF The output file includes the source freRVFODZJOBEEJUJPOUPUIFBUUSJCVUFT DIFDLFEGPSUIFQTFVEPDPNQPOFOUT
*13*/5BOE713*/5
'3&2
& *. 7@13*/5 *1 7@13*/5
&o
o& '3&2
& 7. & 71 o& 432
Chapter 10
Sinusoidal Steady-State Analysis
R1
2
L1
50mH
4k
IPRINT
ACMAG=8
+
ACPHASE=‒40 ‒ V
AC=ok
MAG=ok
PHASE=ok
3
AC=yes
MAG=yes
PHASE=ok
C1
2u
F1
2k
R2
GAIN=0.5
0
Figure 10.36
5IFTDIFNBUJDPGUIFDJSDVJUJO'JH
From this output file, we obtain
7P = ⧸
−ž7 *P = ⧸
−žN"
XIJDIBSFUIFQIBTPSTGPS
vP = DPT U−ž = TJO U−ž 7
BOE
JP = DPT U−ž N"
Practice Problem 10.13
6TF14QJDFUPPCUBJOvPBOEJPJOUIFDJSDVJUPG'JH
io
2 kΩ
120 cos 3000t V +
‒
3 kΩ
2H
1 μF
+
vo
‒
+
‒
2vo
1 kΩ
Figure 10.37
'PS1SBDUJDF1SPC
Answer: DPT U−ž 7 DPT U−ž N"
Example 10.14
'JOE7BOE7JOUIFDJSDVJUPG'JH
Solution:
Define.*OJUTQSFTFOUGPSN UIFQSPCMFNJTDMFBSMZTUBUFE"HBJO XFNVTUFNQIBTJ[FUIBUUJNFTQFOUIFSFXJMMTBWFMPUTPGUJNFBOE
FYQFOTFMBUFSPO0OFUIJOHUIBUNJHIUIBWFDSFBUFEBQSPCMFN
GPSZPVJTUIBU JGUIFSFGFSFODFXBTNJTTJOHGPSUIJTQSPCMFN ZPV
XPVMEUIFOOFFEUPBTLUIFJOEJWJEVBMBTTJHOJOHUIFQSPCMFNXIFSF
10.8
433
AC Analysis Using PSpice
‒ j2
0.2Vx
3 0° A
1Ω
j2 Ω
V1
2Ω
+
Vx
‒
‒j1 Ω
V2
j2 Ω
2Ω
+ 18 30° V
‒
‒ j1 Ω
Figure 10.38
'PS&YBNQMF
JUJTUPCFMPDBUFE*GZPVDPVMEOPUEPUIBU UIFOZPVXPVMEOFFEUP
BTTVNFXIFSFJUTIPVMECFBOEUIFODMFBSMZTUBUFXIBUZPVEJEBOE
XIZZPVEJEJU
1SFTFOU5IFHJWFODJSDVJUJTBGSFRVFODZEPNBJODJSDVJUBOEUIF
VOLOPXOOPEFWPMUBHFT7BOE7BSFBMTPGSFRVFODZEPNBJO
WBMVFT$MFBSMZ XFOFFEBQSPDFTTUPTPMWFGPSUIFTFVOLOPXOTJO
UIFGSFRVFODZEPNBJO
"MUFSOBUJWF8FIBWFUXPEJSFDUBMUFSOBUJWFTPMVUJPOUFDIOJRVFT
UIBUXFDBOFBTJMZVTF8FDBOEPBTUSBJHIUGPSXBSEOPEBM
BOBMZTJTBQQSPBDIPSVTF14QJDF4JODFUIJTFYBNQMFJTJOB
TFDUJPOEFEJDBUFEUPVTJOH14QJDFUPTPMWFQSPCMFNT XFXJMM
use 14QJDF to find 7BOE78FDBOUIFOVTFOPEBMBOBMZTJT
to check the answer
"UUFNQU5IFDJSDVJUJO'JHJTJOUIFUJNFEPNBJO XIFSFBT
UIFPOFJO'JHJTJOUIFGSFRVFODZEPNBJO4JODFXFBSFOPU
HJWFOBQBSUJDVMBSGSFRVFODZBOE14QJDFSFRVJSFTPOF XFTFMFDUBOZ
GSFRVFODZDPOTJTUFOUXJUIUIFHJWFOJNQFEBODFT'PSFYBNQMF JG
XFTFMFDUω = SBET UIFDPSSFTQPOEJOHGSFRVFODZJTG = ω∕π =
)[8FPCUBJOUIFWBMVFTPGUIFDBQBDJUBODF $ = ∕ω9$ BOEJOEVDUBODFT - = 9-∕ω .BLJOHUIFTFDIBOHFT
results in the schematic in Fig. 10.39. 5PFBTFXJSJOH XFIBWF
FYDIBOHFEUIFQPTJUJPOTPGUIFWPMUBHF-DPOUSPMMFEDVSSFOUTPVSDF
AC=ok
MAG=ok
PHASE=yes
C1
AC=ok
MAG=ok
PHASE=yes
0.5C
1
ACMAG=3A
‒
‒
L1
L2
R3
2
2H
2H
2
GAIN=0.2
I1
R1
1 C2
1C G1 + ‒ G
ACPHASE=0
Figure 10.39
R2
4DIFNBUJDGPSUIFDJSDVJUJOUIF'JH
C3
1C ACMAG=18V
ACPHASE=30
+
‒
V1
434
Chapter 10
Sinusoidal Steady-State Analysis
(BOEUIF + KΩJNQFEBODF/PUJDFUIBUUIFDVSSFOUPG(
floXTGSPNOPEFUPOPEF XIJMFUIFDPOUSPMMJOHWPMUBHFJT
BDSPTTUIFDBQBDJUPS$ BTSFRVJSFEJO'JH5IFBUUSJCVUFT
PGQTFVEPDPNQPOFOUT713*/5BSFTFUBTTIPXO"TBTJOHMFGSFRVFODZBOBMZTJT XFTFMFDU"OBMZTJT4FUVQ"$4XFFQBOEFOUFS
5PUBM1UT = 4UBSU'SFR = BOE'JOBM'SFR = "GUFSTBWJOHUIFTDIFNBUJD XFTFMFDU"OBMZTJT4JNVMBUF
to simulate the circuit. When this is done, the output file includes
'3&2
&o
7. & 71 o& '3&2
&
7. & 71 o& GSPNXIJDIXFPCUBJO
7 = ⧸
−ž7
BOE
7 = ⧸
−ž7
&WBMVBUF0OFPGUIFNPTUJNQPSUBOUMFTTPOTUPCFMFBSOFEJTUIBU
XIFOVTJOHQSPHSBNTTVDIBT14QJDFZPVTUJMMOFFEUPWBMJEBUF
UIFBOTXFS5IFSFBSFNBOZPQQPSUVOJUJFTGPSNBLJOHBNJTUBLF JODMVEJOHDPNJOHBDSPTTBOVOLOPXOiCVHuJO14QJDFUIBUZJFMET
JODPSSFDUSFTVMUT
4P IPXDBOXFWBMJEBUFUIJTTPMVUJPO 0CWJPVTMZ XFDBO
SFXPSLUIFFOUJSFQSPCMFNXJUIOPEBMBOBMZTJT BOEQFSIBQTVTJOH
."5-"# UPTFFJGXFPCUBJOUIFTBNFSFTVMUT5IFSFJTBOPUIFS
XBZXFXJMMVTFIFSF8SJUFUIFOPEBMFRVBUJPOTBOETVCTUJUVUFUIF
BOTXFSTPCUBJOFEJOUIF14QJDF solution, and see if the nodal
equations are satisfied.
5IFOPEBMFRVBUJPOTGPSUIJTDJSDVJUBSFHJWFOCFMPX/PUFXF
IBWFTVCTUJUVUFE7 = 7YJOUPUIFEFQFOEFOUTPVSDF
7− @@@@@@
7−7
7 − @@@@@@@
7 −7
− + @@@@@@
+ + + 7 + @@@@@@@
= −K
+ K
−K
+ K + −K + + K 7
− −K + K 7 = + K 7− + K 7 = ⧸
ž 7−⧸
ž7 = /PX UPDIFDLUIFBOTXFS XFTVCTUJUVUFUIF14QJDFBOTXFST
JOUPUIJT
⧸
ž×⧸
−ž−⧸
ž×⧸
−ž
= ⧸
−1ž−⧸
−ž
= −K− + K
= −K
<"OTXFSDIFDLT>
4BUJTGBDUPSZ "MUIPVHIXFVTFEPOMZUIFFRVBUJPOGSPNOPEF
UPDIFDLUIFBOTXFS UIJTJTNPSFUIBOTBUJTGBDUPSZUPWBMJEBUFUIF
BOTXFSGSPNUIF14QJDFTPMVUJPO8FDBOOPXQSFTFOUPVSXPSLBT
BTPMVUJPOUPUIFQSPCMFN
10.9
Applications
Practice Problem 10.14
0CUBJO7YBOE*YJOUIFDJSDVJUEFQJDUFEJO'JH
48 0° V
+‒
1Ω
j2 Ω
j2 Ω
Vx
‒ j0.25
1Ω
Ix
+
‒
‒j1 Ω
2Ω
16 60° A
4Ix
Figure 10.40
'PS1SBDUJDF1SPC
Answer: ⧸
ž7 ⧸
158ž"
10.9
Applications
5IF DPODFQUT MFBSOFE JO UIJT DIBQUFS XJMM CF BQQMJFE JO MBUFS DIBQUFST
UPDBMDVMBUFFMFDUSJDQPXFSBOEEFUFSNJOFGSFRVFODZSFTQPOTF5IFDPO DFQUTBSFBMTPVTFEJOBOBMZ[JOHDPVQMFEDJSDVJUT UISFF-QIBTFDJSDVJUT BD
transistor circuits, filters, oscillators, and other ac circuits. In this section,
XFBQQMZUIFDPODFQUTUPEFWFMPQUXPQSBDUJDBMBDDJSDVJUTUIFDBQBDJ UBODFNVMUJQMJFSBOEUIFTJOFXBWFPTDJMMBUPST
10.9.1
Capacitance Multiplier
5IFPQBNQDJSDVJUJO'JHJTLOPXOBTB DBQBDJUBODFNVMUJQMJFS GPSSFBTPOTUIBUXJMMCFDPNFPCWJPVT4VDIBDJSDVJUJTVTFEJOJOUFHSBUFE
DJSDVJUUFDIOPMPHZUPQSPEVDFBNVMUJQMFPGBTNBMMQIZTJDBMDBQBDJUBODF
$ when a large capacitance is needed. The circuit in Fig. DBOCF
used to multiply capacitance values by a factor up to 1,000. For FYBNQMF B-Q'DBQBDJUPSDBOCFNBEFUPCFIBWFMJLFB-O'DBQBDJUPS
Vi
Ii
1
+
Zi
R2
‒
A2
+
A1
+
$BQBDJUBODFNVMUJQMJFS
0V
2
‒
Vi
‒
Figure 10.41
R1
C
435
Vo
436
Chapter 10
Sinusoidal Steady-State Analysis
In Fig. 10.41, the first op amp operates as a vPMUBHFGPMMPXFS XIJMF
UIF TFDPOE POF JT BO JOWFSUJOH amplifier 5IF WPMUBHF GPMMPXFS JTPMBUFTUIFDBQBDJUBODFGPSNFECZUIFDJSDVJUGSPNUIFMPBEJOHJNQPTFE
CZUIFJOWFSUJOHamplifier4JODFOPDVSSFOUFOUFSTUIFJOQVUUFSNJOBMT
PGUIFPQBNQ UIFJOQVUDVSSFOU *J floXTUISPVHIUIFGFFECBDLDBQBDJ UPS)FODF BUOPEF
7J−7P
*J = @@@@@@@
= Kω$ 7J−7P ∕Kω$
"QQMZJOH,$-BUOPEFHJWFT
7 − @@@@@@
−7P
@@@@@@
J =
3
3
PS
3
7P = −@@@7
3 J
4VCTUJUVUJOH&R JOUP HJWFT
3
*J = Kω$ + @@@ 7J
3 )
(
PS
(
3 )
*
3
@@
J= Kω + @@@ $
7
;J = @@J= @@@@@
*J Kω$FR
7J
5IFJOQVUJNQFEBODFJT
XIFSF
3
$FR = + @@@ $
3 )
(
5IVT CZBQSPQFSTFMFDUJPOPGUIFWBMVFTPG3BOE3 UIFPQBNQDJSDVJU
JO'JHDBOCFNBEFUPQSPEVDFBOFGGFDUJWFDBQBDJUBODFCFUXFFO
UIFJOQVUUFSNJOBMBOEHSPVOE XIJDIJTBNVMUJQMFPGUIFQIZTJDBMDBQBDJ
UBODF$5IF TJ[FPG UIFFGGFDUJWF DBQBDJUBODFJT QSBDUJDBMMZMJNJUFE CZ
UIFJOWFSUFEPVUQVUWPMUBHFMJNJUBUJPO5IVT UIFMBSHFSUIFDBQBDJUBODF
NVMUJQMJDBUJPO UIFTNBMMFSJTUIFBMMPXBCMFJOQVUWPMUBHFUPQSFWFOUUIF
PQBNQTGSPNSFBDIJOHTBUVSBUJPO
"TJNJMBSPQBNQDJSDVJUDBOCFEFTJHOFEUPTJNVMBUFJOEVDUBODF
4FF1SPC There is also an op amp circuit configuration to create
BSFTJTUBODFNVMUJQMJFS
Example 10.15
$BMDVMBUF$FRJO'JHXIFO3 = LΩ 3 = .Ω BOE$ = O'
Solution:
'SPN&R 3
× $FR = + @@@ $ = + @@@@@@@@
( × )O' = O'
3 )
(
10.9
437
Applications
%FUFSNJOFUIFFRVJWBMFOUDBQBDJUBODFPGUIFPQBNQDJSDVJUJO'JH
JG3 = LΩ 3 = .Ω BOE$ = O'
Practice Problem 10.15
Answer:μ'
10.9.2
Oscillators
8FLOPXUIBUEDJTQSPEVDFECZCBUUFSJFT#VUIPXEPXFQSPEVDFBD 0OFXBZJTVTJOHPTDJMMBUPST XIJDIBSFDJSDVJUTUIBUDPOWFSUEDUPBD
An oscillator is a circuit that produces an ac waveform as output when
powered by a dc input.
5IFPOMZFYUFSOBMTPVSDFBOPTDJMMBUPSOFFETJTUIFEDQPXFSTVQQMZ
*SPOJDBMMZ UIFEDQPXFSTVQQMZJTVTVBMMZPCUBJOFECZDPO WFSUJOHUIFBD
TVQQMJFECZUIFFMFDUSJDVUJMJUZDPNQBO ZUPED)BWJOHHPOFUISPVHIUIF
USPVCMFPGDPOWFSTJPO POFNBZXPOEFSXIZXFOFFEUPVTFUIFPTDJMMBUPS
UPDPOWFSUUIFEDUPBDBH BJO5IFQSPCMFNJTUIBUUIFBDTVQQMJFECZUIF
VUJMJUZDPNQBOZPQFSBUFTBUBQSFTFUGSFRVFOD y of 60 )[JOUIF6OJUFE
4UBUFT )[JOTPNFPUIFSOBUJPOT XIFSFBTNBO ZBQQMJDBUJPOTTVDI
BTFMFDUSPOJDDJSDVJUT DPNNVOJDBUJPOTZTUFNT BOENJDSP XBWFEFWJDFT
SFRVJSFJOUFSOBMMZHFOFSBUFEGSFRVFODJFTUIBUSBOHFGSPNUP()[PS
IJHIFS0TDJMMBUPSTBSFVTFEGPSHFOFSBUJOHUIFTFGSFRVFODJFT
*OPSEFSGPSTJOFX BWFPTDJMMBUPSTUPTVTUBJOPTDJMMBUJPOT UIF ZNVTU
NFFUUIF#BSLIBVTFODSJUFSJB
This corresponds to ω = 2πf =
377 rad/s.
5IFPWFSBMMHBJOPGUIFPTDJMMBUPSNVTUCFVOJUZPSHSFBUFS5IFSFGPSF MPTTFTNVTUCFDPNQFOTBUFEGPSCZBOBNQMJGZJOHEFWJDF
5IFPWFSBMMQIBTFTIJGU GSPNJOQVUUPPVUQVUBOECBDLUPUIFJOQVU NVTUCF[FSP
5ISFFDPNNPOUZQFTPGTJOFXBWFPTDJMMBUPSTBSFQIBTF -TIJGU UXJO 5 BOE8JFO -CSJEHFPTDJMMBUPST)FSFXFDPOTJEFSPOMZUIF8JFO
-CSJEHF
PTDJMMBUPS
5IF8JFOCSJEHFPTDJMMBUPSJTXJEFMZVTFEGPSHFOFSBUJOHTJOVTPJET
JOUIFGSFRVFODZSBOHFCFMPX.)[*UJTBO3$PQBNQDJSDVJUXJUIPOMZ
BGFXDPNQPOFOUT FBTJMZUVOBCMFBOEFBTZUPEFTJHO "TTIPXOJO'JH
UIFPTDJMMBUPSFTTFOUJBMMZDPOTJTUTPGBOPOJOWerting amplifier with
UXPGFFECBDLQBUIT5IFQPTJUJWFGFFECBDLQBUIUPUIFOPOJOWFSUJOHJOQVU
DSFBUFTPTDJMMBUJPOT XIJMFUIFOF HBUJWFGFFECBDLQBUIUPUIFJO WFSUJOH
JOQVUDPOUSPMTUIFHain. If we define the impedances of the 3$TFSJFTBOE
QBSBMMFMDPNCJOBUJPOTBT;TBOE;Q UIFO
K
= 3 −@@@@
;T = 3 + @@@@@
Kω$
ω$
3
;Q = 3‖ @@@@@
= @@@@@@@@@@
Kω$ + Kω3$
Negative feedback
path to control gain
Rf
Rg
‒
+
+
v2 R2
‒
;Q
7
@@@= @@@@@@@
7P ;T + ;Q
+
vo
‒
C2
5IFGFFECBDLSBUJPJT
C1
R1
Figure 10.42
Positive feedback path
to create oscillations
8JFO-CSJEHFPTDJMMBUPS
438
Chapter 10
Sinusoidal Steady-State Analysis
4VCTUJUVUJOH&RT BOE JOUP&R HJWFT
7
3
@@@= @@@@@@@@@@@@@@@@@@@@@@@@@@
7P
K
3 + 3−@@@@
+ Kω3$
ω$ )
(
$
ω3
= @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
ω 3$ + 3$ + 3$ + K ω3$3$−
5PTBUJTGZUIFTFDPOE#BSLIBVTFODSJUFSJPO 7NVTUCFJOQIBTFXJUI7P XIJDIJNQMJFTUIBUUIFSBUJPJO&R NVTUCFQVSFMZSFBM)FODF UIFJNBHJOBSZQBSUNVTUCF[FSP4FUUJOHUIFJNBHJOBSZQBSUFRVBMUP[FSP
HJWFTUIFPTDJMMBUJPOGSFRVFODZωPBT
ωP3$3$− = PS
@@@@@@@@@
ωP = @@@@@@@@@@
Ŀ 33$$
*ONPTUQSBDUJDBMBQQMJDBUJPOT 3 = 3 = 3BOE$ = $ = $ TPUIBU
= πG ωP = @@@
P
3$
PS
GP = @@@@@
π3$
4VCTUJUVUJOH&R BOE3 = 3 = 3 $ = $ = $JOUP&R ZJFMET
7
@@@= @@
7P Thus, in order to satisfy the first Barkhausen DSJUFSJPO UIFPQBNQNVTU
DPNQFOTBUFCZQSPWJEJOHBHBJOPGPSHSFBUFSTPUIBUUIFPWFSBMMHBJOJT
at least 1 or unity. We recall that for a noninverting amplifier,
3G
7
@@@P= + @@@= 3H
7
PS
3G = 3H
%VFUPUIFJOIFSFOUEFMBZDBVTFECZUIFPQBNQ 8JFO-CSJEHFPTDJMMBUPSTBSFMJNJUFEUPPQFSBUJOHJOUIFGSFRVFODZSBOHFPG.)[PSMFTT
Example 10.16
%FTJHOB8JFO-CSJEHFDJSDVJUUPPTDJMMBUFBUL)[
Solution:
6TJOH&R XFPCUBJOUIFUJNFDPOTUBOUPGUIFDJSDVJUBT
= @@@@@@@@@@@@@
3$ = @@@@
= × ¢
πGP π × × *GXFTFMFDU 3 = L Ω UIFOXFDBOTFMFDU $ = Q'UPTBUJTGZ
Eq. 4JODFUIFHBJONVTUCF 3G∕3H = 8FDPVMETFMFDU 3G = LΩXIJMF3H = LΩ
439
Review Questions
*OUIF8JFO-CSJEHFPTDJMMBUPSDJSDVJUJO'JH MFU3 = 3 = 2.5 LΩ $ = $ = O'%FUFSNJOFUIFGSFRVFODZGPPGUIFPTDJMMBUPS
Practice Problem 10.16
Answer:L)[
10.10
Summary
8FBQQMZOPEBMBOE NFTIBOBMZTJTUP BDDJSDVJUT CZBQQMZJOH ,$-
BOE,7-UPUIFQIBTPSGPSNPGUIFDJSDVJUT
*OTPMWJOHGPSUIFTUFBEZ -TUBUFSFTQPOTFPGBDJSDVJUUIBUIBTJOEF QFOEFOUTPVSDFTXJUIEJGGFSFOUGSFRVFODJFT FBDIJOEFQFOEFOUTPVSDF
NVTUCFDPOTJEFSFETFQBSBUFMZ5IFNPTUOBUVSBMBQQSPBDIUPBOBMZ[JOHTVDIDJSDVJUTJTUPBQQMZUIFTVQFSQPTJUJPOUIFPSFN "TFQBSBUF
QIBTPSDJSDVJUGPSFBDIGSFRVFODZNVTUCFTPMWFEJOEFQFOEFOUMZ BOE
UIFDPSSFTQPOEJOHSFTQPOTFTIPVMECFPCUBJOFEJOUIFUJNFEPNBJO
5IFPWFSBMMSFTQPOTFJTUIFTVNPGUIFUJNFEPNBJOSFTQPOTFTPGBMM
UIFJOEJWJEVBMQIBTPSDJSDVJUT
5IFDPODFQUPGTPVSDFUSBOTGPSNBUJPOJTBMTPBQQMJDBCMFJOUIFGSF RVFODZEPNBJO
5IF5IFWFOJOFRVJWBMFOUPGBOBDDJSDVJUDPOTJTUTPGBWPMUBHFTPVSDF
75IJOTFSJFTXJUIUIF5IFWFOJOJNQFEBODF;5I
5IF/PSUPOFRVJWBMFOUPGBOBDDJSDVJUDPOTJTUTPGBDVSSFOUTPVSDF*/
JOQBSBMMFMXJUIUIF/PSUPOJNQFEBODF;/ =;5I 14QJDFJTBTJNQMFBOEQPXFSGVMUPPMGPSTPMWJOHBDDJSDVJUQSPCMFNT
*USFMJFWFTVTPGUIFUFEJPVTUBTLPGXPSLJOHXJUIUIFDPNQMFYOVNCFSTJOWPMWFEJOTUFBEZ-TUBUFBOBMZTJT
5IFDBQBDJUBODFNVMUJQMJFSBOEUIFBDPTDJMMBUPSQSPWJEFUXPUZQJDBM
BQQMJDBUJPOT GPSUIFDPODFQUT QSFTFOUFE JOUIJTDIBQUFS " DBQBDJUBODFNVMUJQMJFSJTBOPQBNQDJSDVJUVTFEJOQSPEVDJOHBNVMUJQMFPG
BQIZTJDBMDBQBDJUBODF"OPTDJMMBUPSJTBEFWJDFUIBUVTFTBEDJOQVU
UPHFOFSBUFBOBDPVUQVU
Review Questions
5IFWPMUBHF7PBDSPTTUIFDBQBDJUPSJO'JHJT
5IFWBMVFPGUIFDVSSFOU*P in the circuit of Fig. JT
B ⧸ ž7
C ⧸ ž7
B ⧸ 0ž"
C ⧸ −ž"
D ⧸ −ž7
E ⧸ −ž7
D ⧸ 0ž"
E −"
1Ω
10 0° V +
‒
Figure 10.43
'PS3FWJFX2VFTUJPO
‒j1 Ω
+
Vo
‒
Io
3 0° A
Figure 10.44
'PS3FWJFX2VFTUJPO
j8 Ω
‒j2 Ω
440
Chapter 10
Sinusoidal Steady-State Analysis
6TJOHOPEBMBOBMZTJT UIFWBMVFPG7PJOUIFDJSDVJUPG
'JHJT
'PSUIFDJSDVJUJO'JH UIF5IFWFOJO
JNQFEBODFBUUFSNJOBMTB-CJT
B −7
C −7
B Ω
C −KΩ
D 7
E 7
D + KΩ
E + KΩ
F −KΩ
Vo
1Ω
j6 Ω
1H
a
‒ j3 Ω
4 90° A
5 cos t V +
‒
1F
b
Figure 10.48
Figure 10.45
'PS3FWJFX2VFTUJPOTBOE
'PS3FWJFX2VFTUJPO
*OUIFDJSDVJUPG'JH DVSSFOUJ U JT
B DPTU"
C TJOU"
E TJOU"
F DPT U−ž "
i(t)
B ⧸ −ž7
C ⧸ ž7
D ⧸ −ž7
E ⧸ 45ž7
3FGFSUPUIFDJSDVJUJO'JH5IF/PSUPO
FRVJWBMFOUJNQFEBODFBUUFSNJOBMTB-CJT
1F
1H
10 cos t V +
‒
D DPTU"
*OUIFDJSDVJUPG'JH UIF5IFWFOJOWPMUBHFBU
UFSNJOBMTB-CJT
B −KΩ
C −KΩ
D KΩ
E KΩ
1Ω
‒ j2 Ω
Figure 10.46
a
'PS3FWJFX2VFTUJPO
6 0° V +
‒
3FGFSUPUIFDJSDVJUJO'JHBOEPCTFSWFUIBUUIF
UXPTPVSDFTEPOPUIBWFUIFTBNFGSFRVFODZ5IF
DVSSFOUJY U DBOCFPCUBJOFECZ
j4 Ω
b
Figure 10.49
'PS3FWJFX2VFTUJPOTBOE
B TPVSDFUSBOTGPSNBUJPO
C UIFTVQFSQPTJUJPOUIFPSFN
D 14QJDF
5IF/PSUPODVSSFOUBUUFSNJOBMTB-CJOUIFDJSDVJUPG
'JHJT
1Ω
1H
ix
sin 2t V +
‒
Figure 10.47
'PS3FWJFX2VFTUJPO
1F
+ sin 10t V
‒
B ⧸ ž"
C ⧸ −ž"
D ⧸ ž"
E ⧸ ž"
14QJDFDBOIBOEMFBDJSDVJUXJUIUXPJOEFQFOEFOU
TPVSDFTPGEJGGFSFOUGSFRVFODJFT
B 5SVF
C 'BMTF
"OTXFSTD B E B C D B B E C
441
Problems
Problems
Section 10.2
%FUFSNJOF7YJO'JH
Nodal Analysis
+ Vx
%FUFSNJOFJJOUIFDJSDVJUPG'JH
1Ω
i
2 cos 10t V +
‒
20 Ω
1F
0.2Vx
+ 60 0° V
‒
1Ω
1H
‒
20 Ω
j10 Ω
Figure 10.55
'PS1SPC
Figure 10.50
'PS1SPC
Use nodal analysis to find 7JOUIFDJSDVJUPG
Fig. 10.56.
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEOPEBMBOBMZTJT
2Ω
4 0° V +
‒
40 Ω
‒j5 Ω
j4 Ω
+
Vo
‒
120 ‒15° V +
‒
j20 Ω
V
‒ j30 Ω
6 30° A
50 Ω
Figure 10.51
'PS1SPC
%FUFSNJOFvPJOUIFDJSDVJUPG'JH
1
12 F
4Ω
16 sin 4t V
+
vo
‒
+
‒
Figure 10.56
'PS1SPC
2H
1Ω
2 cos 4t A
6Ω
Use nodal analysis to find current JPJOUIFDJSDVJUPG
'JH-FUJT = DPT U + ž "
0.1 vo
Figure 10.52
io
'PS1SPC
$PNQVUFvP U JOUIFDJSDVJUPG'JH
ix
24 cos (4t + 45°) V +
‒
is
0.25 F
1H
0.5ix
1Ω
+
vo
‒
Figure 10.53
20 Ω
‒
50 μF
100 mH
Figure 10.57
'PS1SPC
'JOEJPJOUIFDJSDVJUPG'JH
io
25 cos(4 × 103t) V +
‒
'PS1SPC
+
Use nodal analysis to find vP in the circuit of Fig. 10.58.
'PS1SPC
Figure 10.54
vo
40 Ω
2 kΩ
20 Ω
2 μF
0.25 H
50 μF
10 mH
io
+
‒
10io
10 cos 103t V
+
‒
Figure 10.58
'PS1SPC
20 Ω
4io
30 Ω
+
vo
‒
442
Chapter 10
Sinusoidal Steady-State Analysis
Use nodal analysis to find vPJOUIFDJSDVJUPG
Fig. 10.59. Let ω = LSBET
$BMDVMBUFUIFWPMUBHFBUOPEFTBOEJOUIFDJSDVJU
PG'JHVTJOHOPEBMBOBMZTJT
j4 Ω
2 µF
36 sin ωt A
+
+
vx
2 kΩ
0.1 vx 4 kΩ
50 mH
‒
20 30° A
1
vo
‒
2
‒ j2 Ω
10 Ω
‒ j5 Ω
j2 Ω
Figure 10.59
'PS1SPC
Using nodal analysis, find JP U JOUIFDJSDVJUJO
'JH
0.25 F
2H
2Ω
1H
Figure 10.63
'PS1SPC
4PMWFGPSUIFDVSSFOU*JOUIFDJSDVJUPG'JH
VTJOHOPEBMBOBMZTJT
5 0° A
io
8 sin (2t + 30°) V +
‒
0.5 F
j1 Ω
2Ω
cos 2t A
I
20 ‒90° V +
‒
Figure 10.60
‒ j2 Ω
4Ω
2I
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEOPEBMBOBMZTJT
Figure 10.64
'PS1SPC
Use nodal analysis to find 7YJOUIFDJSDVJUTIPXOJO
'JH
2io
j4 Ω
R2
+ V
io
R1
is
C
L
%FUFSNJOF7YJOUIFDJSDVJUPG'JHVTJOHBOZ
NFUIPEPGZPVSDIPJDF
‒j2 Ω
'PS1SPC
5Ω
‒j3 Ω
3 45° A
'PS1SPC
'PS1SPC
Figure 10.62
2 0° A
‒
Figure 10.65
Figure 10.61
+
40 30° V ‒
x
8Ω
+
Vx 3 Ω
‒
#ZOPEBMBOBMZTJT PCUBJODVSSFOU*PJOUIFDJSDVJUPG
'JH
j6 Ω
10 Ω
j4 Ω
5 0° A
+
100 20° V ‒
3Ω
Figure 10.66
'PS1SPC
Io
2Ω
1Ω
‒j2 Ω
443
Problems
6TFOPEBMBOBMZTJTUPPCUBJO7P in the circuit of Fig. 10.67 below.
8Ω
+
Vx
‒
4 45° A
2Ω
j6 Ω
4Ω
j5 Ω
2Vx
‒ j1 Ω
‒ j2 Ω
+
Vo
‒
Figure 10.67
'PS1SPC
'PSUIFDJSDVJUJO'JH EFUFSNJOF7P∕7T
0CUBJO7PJO'JHVTJOHOPEBMBOBMZTJT
R1
j2 Ω
12 0° V
4Ω
+‒
+
Vo
‒
2Ω
R2
Vs +
‒
‒j4 Ω
C
L
+
Vo
‒
0.2Vo
Figure 10.71
'PS1SPC
Figure 10.68
'PS1SPC
6TJOHOPEBMBOBMZTJTPCUBJO7JOUIFDJSDVJUPG
Fig. 10.72.
3FGFSUP'JH*GvT U = 7NTJOωUBOE
vP U = "TJO ωU + ϕ EFSJWFUIFFYQSFTTJPOTGPS
" and ϕ
j𝜔L
Vs +
‒
R
+
‒
vs(t)
R
+
vo(t)
‒
C
L
+
1
j𝜔C
1
j𝜔C
‒
V
Figure 10.72
'PS1SPC
Figure 10.69
'PS1SPC
Section 10.3
For each of the circuits in Fig. 10.70, find 7P∕7JGPS
ω = ω → ∞ BOEω = ∕-$
Mesh Analysis
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOENFTIBOBMZTJT
4PMWFGPSJPJO'JHVTJOHNFTIBOBMZTJT
R
L
R
+
C
Vi
‒
(a)
Figure 10.70
'PS1SPC
+
+
Vo
Vi
‒
‒
C
4Ω
+
L
Vo
‒
2H
io
10 cos 2t V +
‒
(b)
Figure 10.73
'PS1SPC
0.25 F
+
‒
6 sin 2t V
444
Chapter 10
Sinusoidal Steady-State Analysis
Use mesh analysis to find current JPJOUIFDJSDVJUPG
'JH
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOENFTIBOBMZTJT
1 µF
2 kΩ
io
10 cos 103t V +
‒
jXL1
+ 20 sin 103t V
‒
0.4 H
R3
R2
I1
R1
Figure 10.74
I2
jXL3
jXL2
'PS1SPC
+‒
Using mesh analysis, find *BOE*JOUIFDJSDVJUPG
'JH
‒ jXC
Vs
Figure 10.77
'PS1SPC
j10 Ω
60 30° V +
‒
I1
40 Ω
+ 75 0° V
‒
I2
‒ j20 Ω
Figure 10.75
Use mesh analysis to find vPJOUIFDJSDVJUPG
Fig. 10.78. Let vT = DPT U + ž 7
vT = DPTU7
'PS1SPC
*OUIFDJSDVJUPG'JH EFUFSNJOFUIFNFTI
DVSSFOUTJBOEJ-FUv = DPTU7BOE
v = DPT U−ž 7
1Ω
1H
vs1 +
‒
i1
400 mH
300 mH
50 µF
200 mH
+
vo
‒
10 Ω
+ v
‒ s2
1Ω
1H
Figure 10.78
1F
v1 +
‒
20 Ω
+ v2
‒
i2
1Ω
Figure 10.76
'PS1SPC
6TFNFTIBOBMZTJTUPEFUFSNJOFDVSSFOU*PJOUIF
DJSDVJUPG'JHCFMPX
'PS1SPC
80 Ω
+
50 120° V ‒
Figure 10.79
'PS1SPC
‒ j40 Ω
Io
j60 Ω
‒ j40 Ω
20 Ω
+
‒
30 ‒30° V
445
Problems
%FUFSNJOF7PBOE*PJOUIFDJSDVJUPG'JHVTJOH
NFTIBOBMZTJT
6TJOHNFTIBOBMZTJT PCUBJO*PJOUIFDJSDVJUTIPXO
JO'JH
Io
j4 Ω
10 ‒30° A
2Ω
+
Vo
‒
3Vo
‒
+
Io
‒ j2 Ω
2 0° A
j2 Ω
2Ω
1Ω
Figure 10.80
+
‒
‒j4 Ω
10 90° V
1Ω
4 0° A
'PS1SPC
Figure 10.83
'PS1SPC
$PNQVUF*JO1SPCVTJOHNFTIBOBMZTJT
'JOE* * * BOE*YJOUIFDJSDVJUPG'JH
Use mesh analysis to find *PJO'JH GPS
&YBNQMF 10 Ω
20 Ω
$BMDVMBUF*PJO'JH GPS1SBDUJDF1SPC VTJOHNFTIBOBMZTJT
$PNQVUF7PJOUIFDJSDVJUPG'JHVTJOHNFTI
BOBMZTJT
‒ j3 Ω
j4 Ω
4 90° A
2Ω
2Ω
+
Vo
‒
‒ j15 Ω
I3
j16 Ω
Ix
12 64° V
I1
+
‒
I2
‒ j25 Ω
8Ω
Figure 10.84
'PS1SPC
2Ω
+ 12 0° V
‒
Section 10.4
Superposition Theorem
'JOEJPJOUIFDJSDVJUTIPXOJO'JHVTJOH
TVQFSQPTJUJPO
2 0° A
Figure 10.81
4Ω
'PS1SPC
2Ω
io
Use mesh analysis to find currents * * BOE*JO
UIFDJSDVJUPG'JH
25 cos 4t V +
‒
+ 20 V
‒
1H
Figure 10.85
'PS1SPC
I1
120 ‒90° V +
‒
I2
'JOEvPGPSUIFDJSDVJUJO'JH BTTVNJOHUIBU
JT(U) = TJO U + DPT U "
Z
Z = 80 ‒ j35 Ω
‒
120 ‒30° V +
Figure 10.82
'PS1SPC
I3
Z
is(t)
Figure 10.86
'PS1SPC
10 Ω
5H
+
vo
‒
446
Chapter 10
Sinusoidal Steady-State Analysis
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUIFTVQFSQPTJUJPOUIFPSFN
Io
jXL
V1 +
‒
4PMWFGPSvP U JOUIFDJSDVJUPG'JHVTJOHUIF
TVQFSQPTJUJPOQSJODJQMF
R2
6Ω
+ V
‒
2
‒ jXC
R1
18 cos 3t V +
‒
1
12 F
2H
+
vo
‒
+ 15 V
‒
6 sin 2t A
Figure 10.91
Figure 10.87
'PS1SPC
'PS1SPC
Using the superposition principle, find JYJOUIF
DJSDVJUPG'JH
1
8 F
10 cos(2t + 10°) A
3Ω
ix
1Ω
+ 20 cos(2t ‒ 60°) V
‒
4H
%FUFSNJOFJPJOUIFDJSDVJUPG'JH VTJOHUIF
TVQFSQPTJUJPOQSJODJQMF
Figure 10.88
10 sin(t ‒ 30°) V
+
‒
1
6F
24 V
2H
‒+
io
2Ω
4Ω
2 cos 3t
Figure 10.92
'PS1SPC
'PS1SPC
6TFUIFTVQFSQPTJUJPOQSJODJQMFUPPCUBJOvYJOUIF
DJSDVJUPG'JH-FUvT = TJOU7BOE
JT = DPT U + ž "
'JOEJPJOUIFDJSDVJUPG'JHVTJOHTVQFSQPTJUJPO
20 μF
20 Ω
vs +
‒
50 mF
20 Ω
+
vx
‒
is
150 cos 2000t V +
‒
io
40 mH
6 sin 4000t A
80 Ω
100 Ω
60 Ω
+ 72 V
‒
Figure 10.89
'PS1SPC
Figure 10.93
'PS1SPC
Use superposition to find J U JOUIFDJSDVJUPG
Fig. 10.90.
i
+ 3 sin 4t V
‒
20 Ω
25 cos(20t + 15°) +
‒
300 mH
Figure 10.90
'PS1SPC
Source Transformation
Using source transformation, find JJOUIFDJSDVJUPG
'JH
20 Ω
8 cos(10t + 30°) V +
‒
Section 10.5
Figure 10.94
'PS1SPC
i
20 Ω
5 mF
1H
447
Problems
‒ j5 Ω
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTVOEFSTUBOETPVSDFUSBOTGPSNBUJPO
vs(t) +
‒
R2
C
8Ω
12 0° A
L
R1
a
j10 Ω
b
+
vo
‒
(b)
Figure 10.98
'PS1SPC
Figure 10.95
'PSFBDIPGUIFDJSDVJUTJO'JH PCUBJO5IFWFOJO
and Norton equivalent circuits at terminals B-C
'PS1SPC
Use source transformation to find *PJOUIFDJSDVJUPG
1SPC
j4 Ω
6Ω
Use the method of source transformation to find *YJO
UIFDJSDVJUPG'JH
a
‒ j2 Ω
2 0° A
2Ω
‒ j2 Ω
j4 Ω
b
ix
60 0° V +
‒
(a)
4Ω
6Ω
5 90° A
30 Ω
‒j3 Ω
120 45° V +
‒
Figure 10.96
'PS1SPC
j10 Ω
60 Ω
a
‒j5 Ω
b
Use the concept of source transformation to find 7P
JOUIFDJSDVJUPG'JH
‒ j3 Ω
4Ω
80 0° V +
‒
2Ω
j2 Ω
Figure 10.99
'PS1SPC
j4 Ω
‒ j2 Ω
+
Vo
‒
Figure 10.97
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOE5IFWFOJOBOE/PSUPO
FRVJWBMFOUDJSDVJUT
R1
‒ jXC
Vs +
‒
'PS1SPC
3FXPSL1SPCVTJOHTPVSDFUSBOTGPSNBUJPO
Section 10.6
(b)
'JOEUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUTBU
UFSNJOBMTB-CGPSFBDIPGUIFDJSDVJUTJO'JH
25 30° V +
‒
10 Ω
jXL
Figure 10.100
'PS1SPC
Thevenin and Norton
Equivalent Circuits
j20 Ω
a
For the circuit depicted in Fig. 10.101, find the
5IFWFOJOFRVJWBMFOUDJSDVJUBUUFSNJOBMTB-C
a
b
b
Figure 10.101
'PS1SPC
10 Ω
‒ j10 Ω
3 0° A
‒j10 Ω
(a)
R2
30 90° V +
‒
448
Chapter 10
Sinusoidal Steady-State Analysis
$BMDVMBUFUIFPVUQVUJNQFEBODFPGUIFDJSDVJUTIPXO
JO'JH
‒j2 Ω
0CUBJOUIF/PSUPOFRVJWBMFOUPGUIFDJSDVJUEFQJDUFE
JO'JHBUUFSNJOBMTB-C
10 Ω
10 Ω
50 mH
a
+ Vo ‒
+ 160 sin (200t + 60°) V
‒
j40 Ω
0.2Vo
500 µF
b
Figure 10.106
Figure 10.102
'PS1SPC
'PS1SPC
'JOEUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO
Fig. 10.103 as seen from:
B UFSNJOBMTB-C
For the circuit shown in Fig. 10.107, find the Norton
FRVJWBMFOUDJSDVJUBUUFSNJOBMTB-C
C UFSNJOBMTD-E
c
d
j10 Ω
10 Ω
‒ j20 Ω
20 0° V +
‒
a
3 60° A
10 Ω
2 0° V
b
a
b
‒j30 Ω
j80 Ω
Figure 10.107
Figure 10.103
'PS1SPC
'PS1SPC
'JOEUIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTB-CPGUIF
DJSDVJUJO'JH
4Ω
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOE/PSUPOTUIFPSFN
a
Ix
vs(t)
R
+‒
io
‒ j3 Ω
15 0° A
40 Ω
60 Ω
1.5Ix
L
C1
C2
b
Figure 10.104
Figure 10.108
'PS1SPC
'PS1SPC
Using Thevenin’s theorem, find vPJOUIFDJSDVJUPG
'JH
2vo
+‒
10 Ω
"UUFSNJOBMTB-C PCUBJO5IFWFOJOBOE/PSUPO
FRVJWBMFOUDJSDVJUTGPSUIFOFUXPSLEFQJDUFEJO
Fig. 10.109. Take ω = SBET
10 µF 12 cos ωt V
‒+
‒j10 Ω
2 0° A
10 Ω
j5 Ω
j5 Ω
10 Ω
+
vo
‒
2 sin ωt A
+
vo
‒
10 Ω
1
2H
a
2vo
b
Figure 10.105
'PS1SPC
Figure 10.109
'PS1SPC
449
Problems
'JOEUIF5IFWFOJOBOE/PSUPOFRVJWBMFOUDJSDVJUTBU
UFSNJOBMTB-CJOUIFDJSDVJUPG'JH
R2
C
R1
‒j5 Ω
12 Ω
13 Ω
90 30° V +
‒
a
vs
+
vo
‒
b
j6 Ω
10 Ω
‒
+
+
‒
Figure 10.113
'PS1SPC
8Ω
'JOEvPJOUIFPQBNQDJSDVJUPG'JH
Figure 10.110
'PS1SPC
+
‒
'JOEUIF5IFWFOJOFRVJWBMFOUBUUFSNJOBMTB-CJOUIF
DJSDVJUPG'JH
io
+
vo
3
+
‒
1
F
20
4io
Figure 10.111
‒
Figure 10.114
'PS1SPC
b
$PNQVUFJP U JOUIFPQBNQDJSDVJUJO'JHJG
vT = DPT U 7
50 kΩ
'PS1SPC
vs +
‒
Op Amp AC Circuits
'PSUIFJOUFHSBUPSTIPXOJO'JH PCUBJO
7P∕7T'JOEvP U XIFOvT U = 7NTJOωUBOE
ω = ∕3$
R
‒
+
+
‒
io
1 nF
100 kΩ
Figure 10.115
'PS1SPC
If the input impedance is defined as ;JO = 7T∕*T find the input impedance of the op amp circuit in
'JHXIFO3 = LΩ 3 = LΩ $ = O' $ = O' BOEω = SBET
C
vs +
‒
vo
10 kΩ
a
1 H vo
‒
Section 10.7
0.5 µF
+
12 cos(2t + 30°) V ‒
2 kΩ
4Ω
3 sin10t V +
‒
+
C1
+
vo
‒
Figure 10.112
'PS1SPC
Is
R1
Vs +
‒
C2
Zin
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEPQBNQTJO"$DJSDVJUT
R2
Figure 10.116
'PS1SPC
+
+
‒-
Vo
450
Chapter 10
Sinusoidal Steady-State Analysis
&WBMVBUFUIFWPMUBHFHBJO"W = 7P∕7TJOUIFPQBNQ
DJSDVJUPG'JH'JOE"WBUω = ω → ∞ ω = ∕3$ BOEω = ∕3$
%FUFSNJOF7PBOE*PJOUIFPQBNQDJSDVJUPG
Fig. 10.119.
io
20 kΩ
C2
R2
R1
C1
Vs +
‒
‒
+
‒ j4 kΩ
+
+
6 30° V ‒
Vo
‒
+
‒ j2 kΩ
Vo
Figure 10.119
'PS1SPC
'PS1SPC
In the op amp circuit of Fig. 10.118, find the closedMPPQHBJOBOEQIBTFTIJGUPGUIFPVUQVUWPMUBHFXJUI
SFTQFDUUPUIFJOQVUWPMUBHFJG$ = $ = O' 3 = 3 = LΩ 3 = LΩ 3 = LΩ BOE
ω = SBET
$PNQVUFUIFDMPTFE-MPPQHBJO7P∕7TGPSUIFPQBNQ
DJSDVJUPG'JH
R3
R1
R1
vs +
‒
C2
+
‒
vs +
‒
+
‒
‒
Figure 10.117
C1
10 kΩ
R4
R2
+
C2
R2
+
vo
C1
‒
+
vo
R3
‒
Figure 10.120
'PS1SPC
‒
%FUFSNJOFvP t JOUIFPQBNQDJSDVJUJO'JH
CFMPX
Figure 10.118
'PS1SPC
20 kΩ
10 kΩ
+
10 sin(400t) V ‒
0.25 µF
0.5 µF
+
‒
10 kΩ
40 kΩ
20 kΩ
Figure 10.121
'PS1SPC
vo
451
Problems
2Ω
'PSUIFPQBNQDJSDVJUJO'JH PCUBJO7P
6Ω
200 kΩ
‒ j 100 kΩ
200 0° µA
is
‒ j 200 kΩ
4Ω
100 kΩ
‒
+
‒
+
50 kΩ
8Ω
4 µF
10 mH
+
vo
‒
+
Vo Figure 10.125
‒ 'PS1SPC
0CUBJO7PJOUIFDJSDVJUPG'JHVTJOH14QJDF
PS.VMUJ4JN
Figure 10.122
'PS1SPC
‒j2 Ω
0CUBJOvP U GPSUIFPQBNQDJSDVJUJO'JHJG
vT = DPT U−ž 7
0.1 µF
‒
+
vs +
‒
10 kΩ
‒
+
+
vo
Figure 10.123
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEQFSGPSNJOH"$BOBMZTJT
XJUI14QJDFPS.VMUJ4JN
‒ jXC
R2
AC Analysis Using PSpice
+
6TF14QJDFPS.VMUJ4JNUPEFUFSNJOF7PJOUIF
circuit of Fig. 10.124. Assume ω = SBET
24 0° V +
‒
4 0° A
10 Ω
40 Ω
j4 Ω
Vx
‒
jXL
Is
R4
Figure 10.127
25 Ω
30 Ω
+
R3 Vo
‒
0.25Vx
R1
'PS1SPC
‒j2 Ω
+
Vo
‒
'PS1SPC
‒
Section 10.8
2Ω
Figure 10.126
0.2 µF
20 kΩ
+
Vx
‒
1Ω
3 0° A
50 kΩ
2Vx
j4 Ω
'PS1SPC
+
Vo
‒
6TF14QJDFPS.VMUJ4JN to find 7 7 BOE7JOUIF
OFUXPSLPG'JH
8Ω
V1
Figure 10.124
'PS1SPC
j 10 Ω
60 30° V +
‒
4PMWF1SPCVTJOH14QJDFPS.VMUJ4JN
6TF14QJDFPS.VMUJ4JN to find vP U JOUIFDJSDVJUPG
'JH-FUJT = DPT U "
Figure 10.128
'PS1SPC
‒j4 Ω
V2
j 10 Ω
‒ j4 Ω
V3
4 0° A
452
Chapter 10
Sinusoidal Steady-State Analysis
%FUFSNJOF7 7 BOE7JOUIFDJSDVJUPG
Fig. 10.129 using 14QJDFPS.VMUJ4JN
'JHVSFTIPXTB8JFO-CSJEHFOFUXPSL4IPX
UIBUUIFGSFRVFODZBUXIJDIUIFQIBTFTIJGUCFUXFFO
UIFJOQVUBOEPVUQVUTJHOBMTJT[FSPJTG = @@@
3$ π
BOEUIBUUIFOFDFTTBSZHBJOJT"W = 7P∕7J = BU
that frequency.
j10 Ω
V1
‒j4 Ω
2Ω
8Ω
j6 Ω
4 0° A
1Ω
V2
V3
‒ j2 Ω
2 0° A
R
R1
C
Vi +
‒
Figure 10.129
'PS1SPC
+ Vo ‒
C
R2
R
6TF14QJDFPS.VMUJ4JN to find vPBOEJPJOUIF
circuit of Fig. 10.130 below.
4Ω
Figure 10.132
'PS1SPC
20 mF
2H
io
6 cos 4t V +
‒
0.5vo
+
‒
4io
10 Ω
25 mF
+
vo
‒
Figure 10.130
'PS1SPC
$POTJEFSUIFPTDJMMBUPSJO'JH
Section 10.9 Applications
B %FUFSNJOFUIFPTDJMMBUJPOGSFRVFODZ
C 0CUBJOUIFNJOJNVNWBMVFPG3GPSXIJDI
PTDJMMBUJPOUBLFTQMBDF
5IFPQBNQDJSDVJUJO'JHJTDBMMFEBO
JOEVDUBODFTJNVMBUPS4IPXUIBUUIFJOQVU
JNQFEBODFJTHJWFOCZ
7JO
;JO = @@@
= Kω-FR
*JO
XIFSF
333
-FR = @@@@@@
3$
80 kΩ
20 kΩ
R1
R2
‒
+
Figure 10.131
'PS1SPC
R3
C
‒
+
R4
‒
+
0.4 mH 2 nF
Iin
+ V
in
‒
10 kΩ
Figure 10.133
'PS1SPC
R
453
Problems
5IFPTDJMMBUPSDJSDVJUJO'JHVTFTBOJEFBM
op amp.
'JHVSFTIPXTB)BSUMFZPTDJMMBUPS4IPXUIBU
UIFGSFRVFODZPGPTDJMMBUJPOJT
B $BMDVMBUFUIFNJOJNVNWBMVFPG3PUIBUXJMM
DBVTFPTDJMMBUJPOUPPDDVS
C 'JOEUIFGSFRVFODZPGPTDJMMBUJPO
@@@@@@@@@
GP = @@@@@@@@@@@@@
πĿ$ - + - Rf
1 MΩ
100 kΩ
‒
+
Ri
‒
+
Vo
Ro
C
10 µH
10 kΩ
2 nF
L2
L1
Figure 10.136
Figure 10.134
")BSUMFZPTDJMMBUPSGPS1SPC
'PS1SPC
'JHVSFTIPXTB$PMQJUUTPTDJMMBUPS4IPXUIBU
UIFPTDJMMBUJPOGSFRVFODZJT
B 4IPXUIBU
@@@@
GP = @@@@@@@@
πĿ -$5
XIFSF$5 = $$∕ $ + $ "TTVNF3J≫9$
Rf
Ri
‒
+
3FGFSUPUIFPTDJMMBUPSJO'JH
7
@@@
= @@@@@@@@@@@@@@@@@@
7P
+ K ω-∕3−3∕ω-
C %FUFSNJOFUIFPTDJMMBUJPOGSFRVFODZGP
D 0CUBJOUIFSFMBUJPOTIJQCFUXFFO3BOE3JO
PSEFSGPSPTDJMMBUJPOUPPDDVS
Vo
R2
R1
L
C2
‒
+
C1
V2
Figure 10.135
"$PMQJUUTPTDJMMBUPSGPS1SPC
)JOU4FUUIFJNBHJOBSZQBSUPGUIFJNQFEBODFJOUIF
GFFECBDLDJSDVJUFRVBMUP[FSP
%FTJHOB$PMQJUUTPTDJMMBUPSUIBUXJMMPQFSBUFBUL)[
L
Figure 10.137
'PS1SPC
Vo
L
R
R
c h a p t e r
AC Power Analysis
11
'PVSUIJOHTDPNFOPUCBDLUIFTQPLFOXPSEUIFTQFEBSSPXUJNFQBTU
UIFOFHMFDUFEPQQPSUVOJUZ
‡"M)BMJG0NBS*CO
Enhancing Your Career
Career in Power Systems
5IFEJTDPWFSZPGUIFQSJODJQMFPGBOBDHFOFSBUPSCZ.JDIBFM'BSBEBZJO
XBTBNBKPSCSFBLUISPVHIJOFOHJOFFSJOHJUQSPWJEFEBDPOWFOJFOU
XBZPGHFOFSBUJOHUIFFMFDUSJDQP XFSUIBUJTOFFEFEJOF WFSZFMFDUSPOJD FMFDUSJDBM PSFMFDUSPNFDIBOJDBMEFWJDFXFVTFOPX
&MFDUSJDQPXFSJTPCUBJOFECZDPO WFSUJOHFOFSHZGSPNTPVSDFTTVDI
BTGPTTJMGVFMT H BT PJM BOEDPBM OVDMFBSGVFM VSBOJVN I ZESPFOFSHZ
XBUFS GBMMJOHUISPVHIBIFBE HFPUIFSNBM FOFSHZ IPU XBUFS TUFBN XJOEFOFSHZ UJEBMFOFS HZ BOECJPNBTTFOFS HZ XBTUFT 5IFTFWBSJPVT
XBZTPGHFOFSBUJOHFMFDUSJDQP wer are studied in detail in the field of
QPXFSFOHJOFFSJOH XIJDIIBTCFDPNFBOJOEJTQFOTBCMFTVCEJTDJQMJOF
PGFMFDUSJDBMFOHJOFFSJOH"OFMFDUSJDBMFOHJOFFSTIPVMECFG BNJMJBSXJUI
UIFBOBMZTJT HFOFSBUJPO USBOTNJTTJPO EJTUSJC VUJPO BOEDPTUPGFMFDUSJD
QPXFS
5IFFMFDUSJDQP XFSJOEVTUSZJTBW FSZMBSHFFNQMP ZFSPGFMFDUSJDBM
FOHJOFFST 5IFJOEVTUSZJODMVEFTUIPVTBOETPGFMFDUSJDVUJMJUZTZTUFNT
SBOHJOHGSPNMBSHF JOUFSDPOOFDUFETZTUFNTTFSWJOHMBS HFSFHJPOBMBSFBT
UPTNBMMQPXFSDPNQBOJFTTFSWJOHJOEJ WJEVBMDPNNVOJUJFTPSG BDUPSJFT
%VFUPUIFDPNQMF YJUZPGUIFQP XFSJOEVTUSZ UIFSFBSFOVNFSPVTFMFD USJDBMFOHJOFFSJOHKPCTJOEJG GFSFOUBSFBTPGUIFJOEVTUSZQP XFSQMBOU
HFOFSBUJPO USBOTNJTTJPOBOEEJTUSJC VUJPO NBJOUFOBODF SFTFBSDI EBUB
acquisition and floXDPOUSPM BOENBOBHFNFOU4JODFFMFDUSJDQP XFSJT
VTFEF WFSZXIFSF FMFDUSJDVUJMJUZDPNQBOJFTBSFF WFSZXIFSF PG GFSJOH
FYDJUJOHUSBJOJOHBOETUFBEZFNQMP ZNFOUGPSNFOBOEX PNFOJOUIPV TBOETPGDPNNVOJUJFTUISPVHIPVUUIFXPSME
"QPMFUZQFUSBOTGPSNFSXJUIBMPX
WPMUBHF UISFFXJSFEJTUSJCVUJPOTZTUFN
¥%FOOJT8JTF(FUUZ*NBHFT3'
455
456
Chapter 11
AC Power Analysis
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
'VMMZVOEFSTUBOEJOTUBOUBOFPVTBOEBWFSBHFQPXFS
6OEFSTUBOEUIFCBTJDTPGNBYJNVNBWFSBHFQPXFS
6OEFSTUBOEFGGFDUJWFPSSNTWBMVFTBOEIPXUPDBMDVMBUFUIFN
BOEUPVOEFSTUBOEUIFJSJNQPSUBODF
6OEFSTUBOEBQQBSFOUQPXFS DPNQMFYQPXFS QPXFS BOESFBD
UJWFQPXFSBOEQPXFSGBDUPS
6OEFSTUBOEQPXFSGBDUPSDPSSFDUJPOBOEUIFJNQPSUBODFPGJUTVTF
11.1
Introduction
0VSFGGPSUJOBDDJSDVJUBOBMZTJTTPG BSIBTCFFOGPDVTFENBJOMZPODBM DVMBUJOHWPMUBHFBOEDVSSFOU0VSNBKPSDPODFSOJOUIJTDIBQUFSJTQPXFS
BOBMZTJT
1PXFSBOBMZTJTJTPGQBSBNPVOUJNQPSUBODF1P XFSJTUIFNPTU
JNQPSUBOURVBOUJUZJOFMFDUSJDVUJMJUJFT FMFDUSPOJD BOEDPNNVOJDBUJPO
TZTUFNT CFDBVTFTVDITZTUFNTJO WPMWFUSBOTNJTTJPOPGQP XFSGSPN
POF QPJOUUPBOPUIFS "MTP F WFSZJOEVTUSJBMBOEIPVTFIPMEFMFDUSJDBM EFWJDF‡FWFSZGBO NPUPS MBNQ QSFTTJOHJSPO 57 QFSTPOBMDPNQVUFS‡
IBTBQP XFSSBUJOHUIBUJOEJDBUFTIP XNVDIQP XFSUIFFRVJQNFOUSF RVJSFTF YDFFEJOHUIFQP XFSSBUJOHDBOEPQFSNBOFOUEBNBHFUPBO
BQQMJBODF5IFNPTUDPNNPOGPSNPGFMFDUSJDQP XFSJTPS)[BD
QPXFS5IFDIPJDFPGBDP WFSEDBMMP XFEIJHIWPMUBHFQPXFSUSBOTNJT
TJPOGSPNUIFQPXFSHFOFSBUJOHQMBOUUPUIFDPOTVNFS
8FXJMMCF gin by defining and deriWJOH JOTUBOUBOFPVTQPXFS BOE
BWFSBHFQPXFS8FXJMMUIFOJOUSPEVDFPUIFSQPXFSDPODFQUT"TQSBDUJDBM
BQQMJDBUJPOTPGUIFTFDPODFQUT XFXJMMEJTDVTTIPXQPXFSJTNFBTVSFEBOE
SFDPOTJEFSIPXFMFDUSJDVUJMJUZDPNQBOJFTDIBSHFUIFJSDVTUPNFST
11.2
Instantaneous and Average Power
"TNFOUJPOFEJO$IBQUFS UIFJOTUBOUBOFPVTQPXFSQ U BCTPSCFECZBO
FMFNFOUJTUIFQSPEVDUPGUIFJOTUBOUBOFPVTW PMUBHFv U BDSPTTUIFFMF NFOUBOEUIFJOTUBOUBOFPVTDVSSFOUJ U UISPVHIJU"TTVNJOHUIFQBTTJWF
TJHODPOWFOUJPO
Q U =v U J U
We can also think of the instantaneous
power as the power absorbed by the
element at a specific instant of time.
Instantaneous quantities are denoted
by lowercase letters.
The instantaneous power (in watts) is the power at any instant of time.
*UJTUIFSBUFBUXIJDIBOFMFNFOUBCTPSCTFOFSHZ
$POTJEFSUIFHFOFSBMDBTFPGJOTUBOUBOFPVTQP XFSBCTPSCFECZBO
BSCJUSBSZDPNCJOBUJPOPGDJSDVJUFMFNFOUTVOEFSTJOVTPJEBMFYDJUBUJPO BT
11.2
457
Instantaneous and Average Power
TIPXOJO'JH-FUUIFW PMUBHFBOEDVSSFOUBUUIFUFSNJOBMTPGUIF
DJSDVJUCF
v U = VNDPT ωU + θv B
J U = *NDPT ωU + θJ C
XIFSFVNBOE*NBSFUIFBNQMJUVEFT PSQFBLWBMVFT BOEθvBOEθJBSFUIF
QIBTFBOHMFTPGUIFWPMUBHFBOEDVSSFOU SFTQFDUJWFMZ5IFJOTUBOUBOFPVT
QPXFSBCTPSCFECZUIFDJSDVJUJT
Q U = v U J U = VN*NDPT ωU + θv DPT ωU + θJ 8FBQQMZUIFUSJHPOPNFUSJDJEFOUJUZ
DPT"DPT# = @@
<DPT
"− # + DPT " + # >
BOEFYQSFTT&R BT
V N*NDPT ωU + θv + θJ Q U = @@
V N*NDPT θv− θJ + @@
5IJTTIPXTVTUIBUUIFJOTUBOUBOFPVTQP XFSIBTUXPQBSUTThe first part
JTDPOTUBOUPS UJNFJOEFQFOEFOU *UTWBMVFEFQFOETPO UIFQIBTFEJG GFS
FODFCFUXFFOUIFWPMUBHFBOEUIFDVSSFOU5IFTFDPOEQBSUJTBTJOVTPJEBM
GVODUJPOXIPTFGSFRVFODZJTω XIJDIJTUXJDFUIFBOHVMBSGSFRVFODZPG
UIFWPMUBHFPSDVSSFOU
"TLFUDIPGQ U JO&R JTTIPXOJO'JH XIFSF5 = π∕ω
JTUIFQFSJPEPGW PMUBHFPSDVSSFOU 8FPCTFSW FUIBU Q U JTQFSJPEJD Q U = Q U + 5 BOEIBTBQFSJPEPG5 = 5∕ TJODFJUTGSFRVFODZJTUXJDF
UIBUPGWPMUBHFPSDVSSFOU8FBMTPPCTFSWFUIBUQ U JTQPTJUJWFGPSTPNF
QBSUPGFBDID ZDMFBOEOF HBUJWFGPSUIFSFTUPGUIFD ZDMF8IFO Q U JT
QPTJUJWF QPXFSJTBCTPSCFECZUIFDJSDVJU 8IFOQ U JTOFHBUJWF QPXFS
JTBCTPSCFECZUIFTPVSDFUIBUJT QPXFSJTUSBOTGFSSFEGSPNUIFDJSDVJUUP
UIFTPVSDF5IJTJTQPTTJCMFCFDBVTFPGUIFTUPSBHFFMFNFOUT DBQBDJUPST
BOEJOEVDUPST JOUIFDJSDVJU
p(t)
1
V I
2 m m
1
V I cos(θv ‒ θi )
2 m m
0
T
2
T
t
Figure 11.2
5IFJOTUBOUBOFPVTQPXFSQ U FOUFSJOHBDJSDVJU
5IFJOTUBOUBOFPVTQPXFSDIBOHFTXJUIUJNFBOEJTUIFSFGPSFEJGficult
UPNFBTVSF5IFBWFSBHFQPXFSJTNPSFDPOWFOJFOUUPNFBTVSF*OG BDU UIFXBUUNFUFS UIFJOTUSVNFOUGPSNFBTVSJOHQPXFS SFTQPOETUPBWFSBHF
QPXFS
The average power, in watts, is the average of the instantaneous power
over one period.
i(t)
Sinusoidal
source
Figure 11.1
+
v (t)
‒
Passive
linear
network
4JOVTPJEBMTPVSDFBOEQBTTJWFMJOFBSDJSDVJU
458
Chapter 11
AC Power Analysis
5IVT UIFBWFSBHFQPXFSJTHJWFOCZ
1 Q U EU
1 = __
∫
5 5
"MUIPVHI&R TIPXTUIFBWFSBHJOHEPOFPWFS5 XFXPVMEHFUUIF
TBNFSFTVMUJGXFQFSGPSNFEUIFJOUFHSBUJPOPWFSUIFBDUVBMQFSJPEPGQ U XIJDIJT5 = 5∕
4VCTUJUVUJOHQ U JO&R JOUP&R HJWFT
5
1 = @@
∫ @@
V * DPT θv− θJ EU
5 N N
5
@@
+ @@
∫ V * DPT ωU + θv + θJ EU
5 N N
5
= @@
V N*NDPT θv− θJ @@
∫ EU
5 5
+ @@
V N*N@@
∫ DPT ωU + θv + θJ EU
5 The first inteHSBOEJTDPOTUBOU BOEUIFBWFSBHFPGBDPOTUBOUJTUIFTBNF
DPOTUBOU5IFTFDPOEJOUFHSBOEJTBTJOVTPJE 8FLOPXUIBUUIFBWFSBHF
PGBTJOVTPJEPWFSJUTQFSJPEJT[FSPCFDBVTFUIFBSFBVOEFSUIFTJOVTPJE
EVSJOHBQPTJUJ WFIBMGDZDMFJTDBODFMFECZUIFBSFBVOEFSJUEVSJOHUIF
GPMMPXJOHOFHBUJWFIBMGDZDMF5IVT UIFTFDPOEUFSNJO&R W BO
JTIFTBOEUIFBWFSBHFQPXFSCFDPNFT
1 = @@
VN*NDPT θv− θJ 4JODFDPT θW− θi = DPT θJ− θv XIBUJTJNQPSUBOUJTUIFEJGGFSFODFJOUIF
QIBTFTPGUIFWPMUBHFBOEDVSSFOU
/PUFUIBUQ U JTUJNFWBSZJOHXIJMF1EPFTOPUEFQFOEPOUJNF 5P
find the instantaneous poXFS XFNVTUOFDFTTBSJMZIBWFv U BOEJ U JOUIF
time domain. But we can find the aWFSBHFQPXFSXIFOWPMUBHFBOEDVS
SFOUBSFFYQSFTTFEJOUIFUJNFEPNBJO BTJO&R PSXIFOUIFZBSF
FYQSFTTFEJOUIFGSFRVFODZEPNBJO5IFQIBTPSGPSNTPGv U BOEJ U JO
&R BSF7 = VN⧸
θvBOE* = *N⧸
θJ SFTQFDUJWFMZ1JTDBMDVMBUFEVTJOH
&R PSVTJOHQIBTPST7BOE*5PVTFQIBTPST XFOPUJDFUIBU
@@
7*
θv¢θJ
= @@
VN*N⧸
= @@
V N*N<DPT θv− θJ + KTJO θv− θJ >
8FSFDPHOJ[FUIFSFBMQBSUPGUIJTF YQSFTTJPOBTUIFB WFSBHFQP XFS 1
BDDPSEJOHUP&R 5IVT
1 = @@
3F<7* > = @@
VN*NDPT θv− θJ
$POTJEFSUXPTQFDJBMDBTFTPG&R 8IFOθv = θJ UIFWPMUBHF
BOEDVSSFOUBSFJOQIBTF 5IJTJNQMJFTBQVSFMZSFTJTUJ WFDJSDVJUPSSFTJT UJWFMPBE3 BOE
1 = @@
V N*N = @@
*N3 = @@
∣*∣
3
11.2
Instantaneous and Average Power
459
XIFSF∣*∣ = *×* &RVBUJPO TIPXTUIBUBQVSFMZSFTJTUJWFDJSDVJU
BCTPSCTQPXFSBUBMMUJNFT8IFOθv− θJ = ±ž XFIBWFBQVSFMZSFBD
UJWFDJSDVJU BOE
V * DPTž = 1 = @@
N N
TIPXJOHUIBUBQVSFMZSFBDUJ WFDJSDVJUBCTPSCTOPB
TVNNBSZ
WFSBHFQP XFS*O
A resistive load (R ) absorbs power at all times, while a reactive load
(L or C ) absorbs zero average power.
Example 11.1
(JWFOUIBU
v U = DPT U + ž 7 BOE J U = DPT U− ž "
find the instantaneous poXFSBOEUIFB WFSBHFQP XFSBCTPSCFECZUIF
QBTTJWFMJOFBSOFUXPSLPG'JH
Solution:
5IFJOTUBOUBOFPVTQPXFSJTHJWFOCZ
Q = vJ = DPT U + ž DPT U− ž
"QQMZJOHUIFUSJHPOPNFUSJDJEFOUJUZ
DPT"DPT# = @@
<DPT " + # + DPT "− # >
HJWFT
Q = <DPT U + ž + DPTž>
PS
Q U = + DPT U + ž 8
5IFBWFSBHFQPXFSJT
DPT<ž− −ž >
1 = @@
VN*NDPT θv− θJ = @@
= DPTž = 8
XIJDIJTUIFDPOTUBOUQBSUPGQ U BCPWF
$BMDVMBUFUIFJOTUBOUBOFPVTQPXFSBOE BWFSBHFQPXFSBCTPSCFECZUIF
QBTTJWFMJOFBSOFUXPSLPG'JHJG
v U = DPT U + ž 7
BOE
Practice Problem 11.1
J U = TJO U + ž "
Answer: + DPT U− ž L8 L8
$BMDVMBUFUIFBWFSBHFQPXFSBCTPSCFECZBOJNQFEBODF ; = − KΩ
XIFOBWPMUBHF7 = ⧸
žJTBQQMJFEBDSPTTJU
Solution:
5IFDVSSFOUUISPVHIUIFJNQFEBODFJT
ž @@@@@@@@@@@@
ž
⧸
⧸
7 = @@@@@@@@
= ⧸
* = @@
ž"
=
; − K ⧸
−ž
Example 11.2
460
Chapter 11
AC Power Analysis
5IFBWFSBHFQPXFSJT
V * DPT θ − θ = @@
1 = @@
DPT − ž = 8
v
J
N N
Practice Problem 11.2
"DVSSFOU* = ⧸
ž"flows UISPVHIBOJNQFEBODF; = ⧸
¢žΩ
'JOEUIFBWFSBHFQPXFSEFMJWFSFEUPUIFJNQFEBODF
Answer:L8
Example 11.3
I
'PSUIFDJSDVJUTIPwn in Fig. 11.3, find the aWFSBHFQPXFSTVQQMJFECZUIF
TPVSDFBOEUIFBWFSBHFQPXFSBCTPSCFECZUIFSFTJTUPS
4Ω
5 30° V +
‒
‒j2 Ω
Figure 11.3
'PS&YBNQMF
Solution:
5IFDVSSFOU*JTHJWFOCZ
ž @@@@@@@@@@@@@
ž
⧸
⧸
= ⧸
* = @@@@@@
ž"
=
− K ⧸
−ž
5IFBWFSBHFQPXFSTVQQMJFECZUIFWPMUBHFTPVSDFJT
DPT ž− ž = 8
1 = @@
5IFDVSSFOUUISPVHIUIFSFTJTUPSJT
ž"
*3 = * = ⧸
BOEUIFWPMUBHFBDSPTTJUJT
73 = *3 = ⧸
ž7
5IFBWFSBHFQPXFSBCTPSCFECZUIFSFTJTUPSJT
= 8
1 = @@
XIJDIJTUIFTBNFBTUIFBWFSBHFQPXFSTVQQMJFE;FSPBWFSBHFQPXFSJT
BCTPSCFECZUIFDBQBDJUPS
Practice Problem 11.3
3Ω
440 45° V +
‒
Figure 11.4
'PS1SBDUJDF1SPC
j1 Ω
*OUIFDJSDVJUPG'JH DBMDVMBUFUIFBWFSBHFQPXFSBCTPSCFECZUIF
SFTJTUPSBOEJOEVDUPS'JOEUIFBWFSBHFQPXFSTVQQMJFECZUIFWPMUBHF
TPVSDF
Answer:L8 8 L8
11.2
461
Instantaneous and Average Power
Example 11.4
%FUFSNJOFUIFBWFSBHFQPXFSHFOFSBUFECZFBDITPVSDFBOEUIF BWFSBHF
QPXFSBCTPSCFECZFBDIQBTTJWFFMFNFOUJOUIFDJSDVJUPG'JH B 20 Ω
‒ j5 Ω
2
4
1
4 0° A
j10 Ω
3
5 +
‒ 60 30° V
4 0° A
+
V1
‒
(a)
+
V2
I1
‒
j10 Ω
(b)
Figure 11.5
'PS&YBNQMF
Solution:
8FBQQMZNFTIBOBMZTJTBTTIPXOJO'JH C 'PSNFTI
*= "
'PSNFTI
K− K *− K*+ ⧸
ž= *= "
PS
K*= −⧸
ž + K
⇒
*= −⧸
¢ž + = ⧸
ž"
For the voltage source, the current flowing from it is *= ⧸
ž"
BOEUIFWPMUBHFBDSPTTJUJT⧸
ž7 TPUIBUUIFBWFSBHFQPXFSJT
1= @@
DPT ž− ž = 8
'PMMPXJOHUIFQBTTJWFTJHODPOWFOUJPO TFF'JH UIJTBWFSBHFQPXFS
JTBCTPSCFECZUIFTPVSDF JOWJFXPGUIFEJSFDUJPOPG *BOEUIFQPMBSJUZ
PGUIFWPMUBHFTPVSDF5IBUJT UIFDJSDVJUJTEFMJWFSJOHBWFSBHFQPXFSUP
UIFWPMUBHFTPVSDF
'PSUIFDVSSFOUTPVSDF UIFDVSSFOUUISPVHIJUJT *= ⧸
žBOEUIF
WPMUBHFBDSPTTJUJT
‒ j5 Ω
20 Ω
7= *+ K *− * = + K − − K
= + K = ⧸
ž7
5IFBWFSBHFQPXFSTVQQMJFECZUIFDVSSFOUTPVSDFJT
DPT ž− = −8
1= −@@
*UJTOFHBUJWFBDDPSEJOHUPUIFQBTTJWFTJHODPOWFOUJPO NFBOJOHUIBUUIF
DVSSFOUTPVSDFJTTVQQMZJOHQPXFSUPUIFDJSDVJU
žBOEUIFWPMUBHF
'PSUIFSFTJTUPS UIFDVSSFOUUISPVHIJUJT *= ⧸
BDSPTTJUJT*= ⧸
ž TPUIBUUIFQPXFSBCTPSCFECZUIFSFTJTUPSJT
1= @@
= 8
I2
+ 60 30° V
‒
462
Chapter 11
AC Power Analysis
'PSUIFDBQBDJUPS UIFDVSSFOUUISPVHIJUJT *= ⧸
žBOEUIFWPMU
¢ž ⧸
ž = ⧸
ž¢ž5IF
BHFBDSPTTJUJT −K*= ⧸
BWFSBHFQPXFSBCTPSCFECZUIFDBQBDJUPSJT
1 = @@
D
PT −ž = 'PS UIF JOEVDUPS UIF DVSSFOU UISPVHI JU JT * − * = − K =
⧸
¢ž5IFWPMUBHFBDSPTTJUJTK *− * = ⧸
¢ž ž
)FODF UIFBWFSBHFQPXFSBCTPSCFECZUIFJOEVDUPSJT
1 = @@
D
PTž = /PUJDFUIBUUIFJOEVDUPSBOEUIFDBQBDJUPSBCTPSC[FSPBWFSBHFQPXFS
BOEUIBUUIFUPUBMQPXFSTVQQMJFECZUIFDVSSFOUTPVSDFFRVBMTUIFQPXFS
BCTPSCFECZUIFSFTJTUPSBOEUIFWPMUBHFTPVSDF PS
1+ 1 + 1+ 1+ 1= − + + + + = JOEJDBUJOHUIBUQPXFSJTDPOTFSWFE
Practice Problem 11.4
Calculate the average power absorbed by each of the five elements in the
DJSDVJUPG'JH
j4 Ω
8Ω
40 0° V +
‒
‒ j2 Ω
+ 20 90° V
‒
Figure 11.6
'PS1SBDUJDF1SPC
Answer:7 7PMUBHFTPVSDF −8K77PMUBHFTPVSDF−8
SFTJTUPS8PUIFST8
11.3
Maximum Average Power Transfer
*O4FDUJPOXFTPMW FEUIFQSPCMFNPGNBYJNJ[JOHUIFQP XFSEFMJW
FSFECZBQPXFSTVQQMZJOHSFTJTUJWFOFUXPSLUPBMPBE3-3FQSFTFOUJOH
UIFDJSDVJUCZJUT 5IFWFOJOFRVJ WBMFOU XFQSP WFEUIBUUIFNBYJNVN QPXFSXPVMECFEFMJWFSFEUPUIFMPBEJGUIFMPBESFTJTUBODFJTFRVBMUPUIF
5IFWFOJOSFTJTUBODF3- = 35I8FOPXFYUFOEUIBUSFTVMUUPBDDJSDVJUT
$POTJEFSUIFDJSDVJUJO'JH XIFSFBOBDDJSDVJUJTDPOOFDUFEUP
BMPBE;-BOEJTSFQSFTFOUFECZJUT5IFWFOJOFRVJWBMFOU5IFMPBEJTVTV
BMMZSFQSFTFOUFE CZBO JNQFEBODF XIJDI NBZNPEFMBO FMFDUSJDNPUPS BOBOUFOOB B57 BOETPGPSUI*OSFDUBOHVMBSGPSN UIF5IFWFOJOJNQFE
BODF;5IBOEUIFMPBEJNQFEBODF;-BSF
;5I = 35I + K95I
B
;- = 3- + K9-
C
11.3
463
Maximum Average Power Transfer
5IFDVSSFOUUISPVHIUIFMPBEJT
75I
75I
* = @@@@@@@@
= @@@@@@@@@@@@@@@@@@@@@
;5I + ;35I +K95I + 3- + K9-
Linear
circuit
'SPN&R UIFBWFSBHFQPXFSEFMJWFSFEUPUIFMPBEJT
(a)
∣75I∣ 3-∕
∣*∣3 = @@@@@@@@@@@@@@@@@@@@@@
1 = @@
35I + 3- + 95I + 9- I
Z Th
VTh +
‒
0VSPCKFDUJ WFJTUPBEKVTUUIFMPBEQBSBNFUFST 3-BOE 9-TPUIBU 1JT
NBYJNVN5PEPUIJTXFTFU ∂1∕∂3-BOE ∂1∕∂9-FRVBMUP[FSP'SPN
&R XFPCUBJO
ZL
(b)
Figure 11.7
∣75I∣3- 95I + 9∂1= −@@@@@@@@@@@@@@@@@@@@@@@
@@@@
∂9< 35I + 3- + 95I + 9- >
ZL
B
'JOEJOHUIFNBYJNVNBWFSBHFQPXFS
USBOTGFS B DJSDVJUXJUIBMPBE C UIF
5IFWFOJOFRVJWBMFOU
∣75I∣< 35I + 3- + 95I + 9- − 3- 35I + 3- >
@@@@
∂1= @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
∂3< 35I + 3- + 95I + 9- >
C
4FUUJOH∂1∕∂9-UP[FSPHJWFT
9- = −95I
BOETFUUJOH∂1∕∂3-UP[FSPSFTVMUTJO
_______________
3- = √ 35I + 95I + 9- $PNCJOJOH&RT BOE MFBETUPUIFDPODMVTJPOUIBUGPSNBYJ
NVNBWFSBHFQPXFSUSBOTGFS ;-NVTUCFTFMFDUFETPUIBU 9- = −95IBOE
3- = 35I JF
;- = 3- + K9- = 35I − K95I = ; 5I
For maximum average power transfer, the load impedance ZL must be
equal to the complex conjugate of the Thevenin impedance ZTh.
5IJTSFTVMUJTLOPXOBTUIFNBYJNVNBWFSBHFQPXFSUSBOTGFSUIFPSFNGPS
UIFTJOVTPJEBMTUFBEZTUBUF4FUUJOH3- = 35IBOE9- = −95IJO&R HJWFTVTUIFNBYJNVNBWFSBHFQPXFSBT
∣75I∣
1NBY = @@@@@
35I *OBTJUVBUJPOJOXIJDIUIFMPBEJTQVSFMZSFBM UIFDPOEJUJPOGPSNBYJ
NVNQP XFSUSBOTGFSJTPCUBJOFEGSPN&R CZTFUUJOH 9- = UIBUJT
_________
3- = √35I + 9 5I = ∣;5I∣
When ZL = Z*Th, we say that the load is
matched to the source.
464
Chapter 11
AC Power Analysis
5IJTNFBOTUIBUGPSNBYJNVNB WFSBHFQPXFSUSBOTGFSUPBQVSFMZSFTJT UJWFMPBE UIFMPBEJNQFEBODF PSSFTJTUBODF JTFRVBMUPUIFNBHOJUVEFPG
UIF5IFWFOJOJNQFEBODF
Example 11.5
4Ω
j5 Ω
8Ω
10 0° V +
‒
%FUFSNJOFUIFMPBEJNQFEBODF ;-UIBUNBYJNJ[FTUIFB WFSBHFQP XFS
ESBXOGSPNUIFDJSDVJUPG'JH 8IBUJTUIFNBYJNVNB WFSBHF
QPXFS
ZL
‒ j6 Ω
Solution:
'JSTUXFPCUBJOUIF5IFWFOJOFRVJWBMFOUBUUIFMPBEUFSNJOBMT5PHFU;5I consider the circuit shown in Fig. 11.9(a). We find
− K
;5I = K + ║ − K = K + @@@@@@@@@
= + KΩ
+ − K
Figure 11.8
'PS&YBNQMF
j5 Ω
4Ω
8Ω
‒ j6 Ω
j5 Ω
4Ω
Z Th
8Ω
10 V +
‒
(a)
‒ j6 Ω
+
VTh
‒
(b)
Figure 11.9
'JOEJOHUIF5IFWFOJOFRVJWBMFOUPGUIFDJSDVJUJO'JH
To find 75I DPOTJEFSUIFDJSDVJUJO'JH C #ZWPMUBHFEJWJTJPO
− K
75I = @@@@@@@@@
= ⧸
¢ž7
+ − K
5IFMPBEJNQFEBODFESBXTUIFNBYJNVNQPXFSGSPNUIFDJSDVJUXIFO
;- = ; 5I = − KΩ
"DDPSEJOHUP&R UIFNBYJNVNBWFSBHFQPXFSJT
∣75I∣ @@@@@@@@
1NBY = @@@@@
=
= 8
35I
Practice Problem 11.5
‒j4 Ω
8Ω
Figure 11.10
j10 Ω
12 A
'PS1SBDUJDF1SPC
5Ω
ZL
'PSUIFDJSDVJU shown in Fig. 11.10, find the load impedance ;-UIBU
BCTPSCTUIFNBYJNVNBWFSBHFQPXFS$BMDVMBUFUIBUNBYJNVNBWFSBHF
QPXFS
Answer:− KΩ 8
11.4
465
Effective or RMS Value
Example 11.6
In the circuit in Fig. 11.11, find the value of 3-UIBUXJMMBCTPSCUIF NBYJNVNBWFSBHFQPXFS$BMDVMBUFUIBUQPXFS
Solution:
We first find the Thevenin equivalent at the terminals of 3-
K − K
;5I = − K ║K = @@@@@@@@@@@@@
= + KΩ
K + − K
#ZWPMUBHFEJWJTJPO
40 Ω ‒ j30 Ω
150 30° V +
‒
j20 Ω
Figure 11.11
'PS&YBNQMF
K
75I = @@@@@@@@@@@@@
ž = ⧸
ž7
⧸
K + − K
5IFWBMVFPG3-UIBUXJMMBCTPSCUIFNBYJNVNBWFSBHFQPXFSJT
_____________
3- = ∣;5I∣ = √ + = Ω
5IFDVSSFOUUISPVHIUIFMPBEJT
ž
⧸
75I
* = @@@@@@@@
= @@@@@@@@@@@@@
= ⧸
ž"
;5I + 3- + K
5IFNBYJNVNBWFSBHFQPXFSBCTPSCFECZ3-JT
∣*∣3 = @@
= 8
1NBY = @@
*O'JH UIFSFTJTUPS 3-JTBEKVTUFEVOUJMJUBCTPSCTUIFNBYJNVN
BWFSBHFQPXFS$BMDVMBUF3-BOEUIFNBYJNVNBWFSBHFQPXFSBCTPSCFE
CZJU
80 Ω
220 60° V +
‒
j60 Ω
90 Ω
‒j30 Ω
RL
Figure 11.12
'PS1SBDUJDF1SPC
Answer:Ω 8
11.4
Effective or RMS Value
5IFJEFBPGFGGFDUJWFWBMVFBSJTFTGSPNUIFOFFEUPNFBTVSFUIFFGGFDUJWFOFTT
PGBWPMUBHFPSDVSSFOUTPVSDFJOEFMJWFSJOHQPXFSUPBSFTJTUJWFMPBE
The effective value of a periodic current is the dc current that delivers
the same average power to a resistor as the periodic current.
Practice Problem 11.6
RL
466
Chapter 11
i(t)
v(t)
+
‒
R
*O'JH UIFDJSDVJUJO B JTBDXIJMFUIBUPG C JTED0VSPCKFDUJ WFJT
to find *FGGUIBUXJMMUSBOTGFSUIFTBNFQP XFSUPSFTJTUPS 3BTUIFTJOVTPJE J
5IFBWFSBHFQPXFSBCTPSCFECZUIFSFTJTUPSJOUIFBDDJSDVJUJT
5
5
3
1 = @@
∫ J3EU = @@
∫ JEU
5 5 1 = * FGG 3
XIJMFUIFQPXFSBCTPSCFECZUIFSFTJTUPSJOUIFEDDJSDVJUJT
(a)
I eff
+
V eff
‒
AC Power Analysis
R
&RVBUJOHUIFFYQSFTTJPOTJO&RT BOE BOETPMWJOHGPS*FGG XFPCUBJO
_______
*FGG =
(b)
Figure 11.13
'JOEJOHUIFFGGFDUJWFDVSSFOU B BDDJSDVJU C EDDJSDVJU
√@@5∫ J EU
5
5IFFGGFDUJWFWBMVFPGUIFW PMUBHFJTGPVOEJOUIFTBNFX BZBTDVSSFOU
UIBUJT
________
VFGG =
√
5vEU
@@
∫
5 5IJTJOEJDBUFTUIBUUIFFGGFDUJWFWBMVFJTUIF TRVBSF SPPUPGUIFNFBO PS
BWFSBHF PGUIFTRVBSFPGUIFQFSJPEJDTJHOBM5IVT UIFFGGFDUJWFWBMVFJT
PGUFOLOPXOBTUIFSPPUNFBOTRVBSFWBMVF PSSNTWBMVFGPSTIPSUBOE
XFXSJUF
*FGG = *SNT VFGG = VSNT
'PSBOZQFSJPEJDGVODUJPOY U JOHFOFSBM UIFSNTWBMVFJTHJWFOCZ
________
9SNT =
√@@5∫ Y EU 5
The effective value of a periodic signal is its root mean square (rms) value.
Equation 11.27 states that to find the rms vBMVFPGY U), we first find
JUTTRVBSFY and then find the NFBOPGUIBU PS
5
@@
∫ xEU
5 @@@@@@
BOEUIFOUIFTRVBSFSPPU Ŀ PGUIBUNFBO5IFSNTWBMVFPGBDPO
TUBOUJTUIFDPOTUBOUJUTFMG'PSUIFTJOVTPJEJ U = *NDPTωU UIFFGGFDUJWF
PSSNTWBMVFJT
______________
*SNT =
√@@5∫ I DPT ωUEU
5
N
___________________
=
√
*@@
*N
N
+ DPTωU EU= @@@
@@
∫ 5 @@
5
Ŀ
4JNJMBSMZ GPSv U = VNDPTωU
VN
@@
VSNT = @@@
Ŀ
,FFQJONJOEUIBU&RT BOE BSFPOMZW BMJEGPSTJOVTPJEBM
TJHOBMT
11.4
467
Effective or RMS Value
5IFBWFSBHFQPXFSJO&R DBOCFXSJUUFOJOUFSNTPGUIFSNT
WBMVFT
VN @@@
*N
V * DPT θ − θ = @@@
@@ @@DPT θv− θJ 1 = @@
v
J
N N
ð
Ŀ
= VSNT*SNTDPT θv− θJ
4JNJMBSMZ UIFBWFSBHFQPXFSBCTPSCFECZBSFTJTUPS3JO&R DBO
CFXSJUUFOBT
7SNT
1 = *SNT
3 = @@@@
3
8IFOBTJOVTPJEBMWoltage or current is specified, it is often in terms
PGJUTNBYJNVN PSQFBL W BMVFPSJUTSNTW BMVF TJODFJUTB WFSBHFWBMVF
JT[FSP5IFQPXFSJOEVTUSJFTTQFDJGZQIBTPSNBHOJUVEFTJOUFSNTPGUIFJS
SNTWBMVFTSBUIFSUIBOQFBLW BMVFT'PSJOTUBODF UIF 7BWBJMBCMFBU
FWFSZIPVTFIPMEJTUIFSNTWBMVFPGUIFWPMUBHFGSPNUIFQPXFSDPNQBOZ
*UJTDPOWFOJFOUJOQPXFSBOBMZTJTUPF YQSFTTWPMUBHFBOEDVSSFOUJOUIFJS
SNTWBMVFT"MTP BOBMPHW PMUNFUFSTBOEBNNFUFSTBSFEFTJHOFEUPSFBE
EJSFDUMZUIFSNTWBMVFPGWPMUBHFBOEDVSSFOU SFTQFDUJWFMZ
%FUFSNJOFUIFSNTW BMVFPGUIFDVSSFOUX BWFGPSNJO'JH*GUIF
DVSSFOUJTQBTTFEUISPVHIB ΩSFTJTUPS, find the aWFSBHFQPXFSBCTPSCFE
CZUIFSFTJTUPS
Solution:
5IFQFSJPEPGUIFXBWFGPSNJT 5 = 0WFSBQFSJPE XFDBOXSJUFUIF
DVSSFOUXBWFGPSNBT
JU =
U
−
<U<
<U<
5IFSNTWBMVFJT
________
*SNT =
√
5
iEU=
@@
∫
5 |
1 t)EU+ −)EU __
∫
]
4 [ ∫
Example 11.7
i(t)
10
0
2
4
6
8
10
t
‒10
Figure 11.14
'PS&YBNQMF
_____________
+ ) = "
| √@@(@@@@
@@
= @@
U + U =
5IFQPXFSBCTPSCFECZBΩSFTJTUPSJT
1 = *SNT3 = = 8
'JOEUIFSNTWBMVFPGUIFDVSSFOUXBWFGPSNPG'JH*GUIF DVSSFOU
flows through a 9-ΩSFTJTUPS DBMDVMBUFUIFBWFSBHFQPXFSBCTPSCFECZ
UIFSFTJTUPS
Answer:" 8
Practice Problem 11.7
i(t)
16
0
1
2
Figure 11.15
3
'PS1SBDUJDF1SPC
4
5
6
t
468
Chapter 11
Example 11.8
AC Power Analysis
5IFXBWFGPSNTIPXOJO'JHJTBIBMGX BWe rectified sine wBWF
'JOEUIFSNTWBMVFBOEUIFBNPVOUPGBWFSBHFQPXFSEJTTJQBUFEJOBΩ
SFTJTUPS
v (t)
10
Solution:
5IFQFSJPEPGUIFWPMUBHFXBWFGPSNJT5 = π BOE
0
π
2π
3π
vU =
t
Figure 11.16
'PS&YBNQMF
{
TJOU <U<π
π <U<π
5IFSNTWBMVFJTPCUBJOFEBT
[
5
π
]
π
v U EU = @@@
TJOU EU + EU VSNT
= @@
∫
∫
∫π
5 π #VUTJOU = @@ −DPTU )FODF
]
U−@@@@@
π@@@@
V SNT = @@@
TJOU
∫ −DPTU EU = @@@
π π (
) π −@@
= @@@
TJOπ −)= π (
π
VSNT = 7
5IFBWFSBHFQPXFSBCTPSCFEJT
VSNT @@@
1 = @@@@
= = 8
3
Practice Problem 11.8
Find the rms value of the full-wave rectified sine wave in Fig. 11.17.
$BMDVMBUFUIFBWFSBHFQPXFSEJTTJQBUFEJOBΩSFTJTUPS
v (t)
Answer:7 8
100
0
π
Figure 11.17
'PS1SBDUJDF1SPC
2π
3π
t
11.5
Apparent Power and Power Factor
*O4FDUJPOXFTBXUIBUJGUIFWPMUBHFBOEDVSSFOUBUUIFUFSNJOBMTPG
BDJSDVJUBSF
v U = VNDPT ωU + θv BOE
J U = *NDPT ωU + θJ PS JOQIBTPSGPSN 7 = VN⧸
θvBOE* = *N⧸
θJ UIFBWFSBHFQPXFSJT
V * DPT θ −θ 1 = @@
v
J
N N
*O4FDUJPO XFTBXUIBU
1 = VSNT*SNTDPT θv−θJ = 4DPT θv−θJ 8FIBWFBEEFEBOFXUFSNUPUIFFRVBUJPO
4 = VSNT*SNT
11.5
Apparent Power and Power Factor
469
5IFB WFSBHFQP XFS JTBQSPEVDU PGUX PUFSNT 5IF QSPEVDU VSNT *SNTJT
LOPXOBTUIF BQQBSFOUQPXFS4 5IFG BDUPSDPT θv − θJ JTDBMMFEUIF
QPXFSGBDUPS QG The apparent power (in VA) is the product of the rms values of voltage
and current.
5IFBQQBSFOUQPXFSJTTPDBMMFECFDBVTFJUTFFNTBQQBSFOUUIBUUIFQPXFS
TIPVMECFUIFW PMUBHFDVSSFOUQSPEVDU CZBOBMPHZXJUIEDSFTJTUJ WFDJS
DVJUT*UJTNFBTVSFEJOWPMUBNQFSFTPS7"UPEJTUJOHVJTIJUGSPNUIFBWFS
BHFPSSFBMQPXFS XIJDIJTNFBTVSFEJOXBUUT5IFQPXFSGBDUPSJTEJNFO
TJPOMFTT TJODFJUJTUIFSBUJPPGUIFBWFSBHFQPXFSUPUIFBQQBSFOUQPXFS
1 = DPT θ − θ
QG = @@
v
J
4
5IFBOHMFθv− θJJTDBMMFEUIF QPXFSGBDUPSBOHMF CFDBVTFJUJTUIF
BOHMFXIPTFDPTJOFJTUIFQP XFSGBDUPS5IFQPXFSGBDUPSBOHMFJTFRVBM
UPUIFBOHMFPGUIFMPBEJNQFEBODFJG7JTUIFWPMUBHFBDSPTTUIFMPBEBOE
*JTUIFDVSSFOUUISPVHIJU5IJTJTFWJEFOUGSPNUIFGBDUUIBU
θv @@@
VN⧸
V
7 = @@@@@
= N⧸θv¢θJ
; = @@
*N
*
*N⧸
θJ
"MUFSOBUJWFMZ TJODF
7@@= V θ 7SNT = @@@
SNT⧸ v
Ŀ
B
*@@ = * θ *SNT = @@@
SNT⧸ J
Ŀ
C
7SNT @@@@
V
7 = @@@@
; = @@
= SNT⧸θv¢θJ
*SNT
*
*SNT
BOE
UIFJNQFEBODFJT
The power factor is the cosine of the phase difference between voltage
and current. It is also the cosine of the angle of the load impedance.
'SPN&R UIFQPXFSGBDUPSNBZCFTFFOBTUIBUG BDUPSCZXIJDI
UIFBQQBSFOUQPXFSNVTUCFNVMUJQMJFEUPPCUBJOUIFSFBMPSB
WFSBHFQPXFS
5IFWBMVFPGQGSBOHFTCFUXFFO[FSPBOEVOJUZ ' PSBQVSFMZSFTJTUJ WF
MPBE UIFWPMUBHFBOEDVSSFOUBSFJOQIBTF TPUIBUθv−θJ = BOEQG = 5IJTJNQMJFTUIBUUIFBQQBSFOUQP XFSJTFRVBMUPUIFB WFSBHFQPXFS'PS
BQVSFMZSFBDUJWFMPBE θv−θJ = ±žBOEQG = *OUIJTDBTFUIFB WFS
BHFQPXFSJT[FSP*OCFUXFFOUIFTFUX PFYUSFNFDBTFT QGJTTBJEUPCF
MFBEJOHPSMBHHJOH-FBEJOHQPXFSGBDUPSNFBOTUIBUDVSSFOUMFBETW PMU
BHF XIJDIJNQMJFTBDBQBDJUJ WFMPBE-BHHJOHQP XFSGBDUPSNFBOTUIBU
DVSSFOUMBHTWPMUBHF JNQMZJOHBOJOEVDUJWFMPBE1PXFSGBDUPSBGGFDUTUIF
From Eq. (11.36), the power factor
may also be regarded as the ratio of
the real power dissipated in the load
to the apparent power of the load.
470
Chapter 11
AC Power Analysis
FMFDUSJDCJMMTDPOTVNFSTQBZUIFFMFDUSJDVUJMJUZDPNQBOJFT BTXFXJMMTFF
JO4FDUJPO
Example 11.9
"TFSJFTDPOOFDUFEMPBEESB XTBDVSSFOU J U = DPT πU + ž "
XIFOUIFBQQMJFEW PMUBHFJT v U = DPT πU − ž 7 'JOE UIF
BQQBSFOUQPXFSBOEUIFQPXFSGBDUPSPGUIFMPBE%FUFSNJOFUIFFMFNFOU
WBMVFTUIBUGPSNUIFTFSJFTDPOOFDUFEMPBE
Solution:
5IFBQQBSFOUQPXFSJT
@@
4 = VSNT*SNT = @@@@
@@@
@@= 7"
Ŀ
Ŀ
5IFQPXFSGBDUPSJT
QG = DPT θv − θJ = DPT −ž − ž = MFBEJOH
5IFQGJTMFBEJOHCFDBVTFUIFDVSSFOUMFBETUIFWPMUBHF5IFQGNBZBMTP
CFPCUBJOFEGSPNUIFMPBEJNQFEBODF
¢ž
⧸
7 = @@@@@@@@@
= ⧸
; = @@
¢ž = − KΩ
*
⧸
ž
QG = DPT −ž = MFBEJOH
5IFMPBEJNQFEBODF;DBOCFNPEFMFECZBΩSFTJTUPS JOTFSJFT
XJUIBDBQBDJUPSXJUI
9$ = − = −@@@
ω$
PS
= @@@@@@@@@
$ = @@@@
= μ'
ω × π
Practice Problem 11.9
0CUBJOUIFQPXFSGBDUPSBOEUIFBQQBSFOUQPXFSPGBMPBEXIPTF
JNQFEBODF JT ; = + K ΩXIFO UIFBQQMJFEWPMUBHF JT v U =
DPT U + ž 7
Answer:MBHHJOH ⧸
ž7"
Example 11.10
%FUFSNJOFUIFQPXFSGBDUPSPGUIFFOUJSFDJSDVJUPG'JHBTTFFOCZ
UIFTPVSDF$BMDVMBUFUIFBWFSBHFQPXFSEFMJWFSFECZUIFTPVSDF
Solution:
5IFUPUBMJNQFEBODFJT
−K × ; = + ║ −K = + @@@@@@@ ¢žΩ
= − K = ⧸
− K
11.6
471
Complex Power
6Ω
5IFQPXFSGBDUPSJT
QG = DPT − = MFBEJOH
30 0° V rms +
‒
TJODFUIFJNQFEBODFJTDBQBDJUJWF5IFSNTWBMVFPGUIFDVSSFOUJT
⧸
ž
7SNT @@@@@@@@@
=
= ⧸
*SNT = @@@@
ž"
;
⧸
¢ž
‒ j2 Ω
4Ω
Figure 11.18
'PS&YBNQMF
5IFBWFSBHFQPXFSTVQQMJFECZUIFTPVSDFJT
1 = 7SNT*SNT QG = = 8
PS
1 = * SNT
3 = = 8
XIFSF3JTUIFSFTJTUJWFQBSUPG;
$BMDVMBUFUIFQPXFSGBDUPSPGUIFFOUJSFDJSDVJUPG'JHBTTFFOCZ
UIFTPVSDF8IBUJTUIFBWFSBHFQPXFSTVQQMJFECZUIFTPVSDF
Practice Problem 11.10
10 Ω
8Ω
Answer:MBHHJOH L8
165 0° V rms +
‒
11.6
'PS1SBDUJDF1SPC
7* 4 = @@
I
+
V
BTTVNJOHUIFQBTTJ WFTJHODPO WFOUJPO TFF'JH *OUFSNTPGUIF
SNTWBMVFT
‒
4 = 7SNT* SNT
7@@= 7 θ 7SNT = @@@
SNT⧸ v
Ŀ
*@@ = * θ *SNT = @@@
SNT⧸ i
Ŀ
XIFSF
BOE
‒ j6 Ω
Figure 11.19
Complex Power
$POTJEFSBCMFFGGPSUIBTCFFOFYQFOEFEPWFSUIFZFBSTUPFYQSFTTQPXFS
SFMBUJPOTBTTJNQMZBTQPTTJCMF1P XFSFOHJOFFSTIBWFDPJOFEUIFUFSN
DPNQMFYQPXFS XIJDIUIFy use to find the total efGFDUPGQBSBMMFMMPBET
$PNQMFYQPXFSJTJNQPSUBOUJOQP XFSBOBMZTJTCFDBVTFJUDPOUBJOT BMM
UIFJOGPSNBUJPOQFSUBJOJOHUPUIFQPXFSBCTPSCFECZBHJWFOMPBE
$POTJEFSUIFBDMPBEJO'JH(J WFOUIFQIBTPSGPSN 7 =
VN⧸
θWBOE * = *N⧸
θJPGWPMUBHFv U BOEDVSSFOU J U UIF DPNQMFYQPXFS
4BCTPSCFECZUIFBDMPBEJTUIFQSPEVDUPGUIFW PMUBHFBOEUIFDPNQMF Y
DPOKVHBUFPGUIFDVSSFOU PS
j4 Ω
Figure 11.20
Load
Z
5IFWPMUBHFBOEDVSSFOUQIBTPSTBTTPDJBUFE
XJUIBMPBE
472
When working with the rms values of
currents or voltages, we may drop the
subscript rms if no confusion will be
caused by doing so.
Chapter 11
AC Power Analysis
5IVT XFNBZXSJUF&R BT
4 = 7SNT*SNT⧸
θv¢θJ
= 7SNT*SNTDPT θv − θJ + K7SNT*SNTTJO θv − θJ 5IJTFRVBUJPODBOBMTPCFPCUBJOFEGSPN&R 8FOPUJDFGSPN &R UIBUUIFNBHOJUVEFPGUIFDPNQMFYQPXFSJTUIFBQQBSFOUQPXFS
IFODF UIFDPNQMF YQP XFSJTNFBTVSFEJOW PMUBNQFSFT 7" "MTP XF OPUJDFUIBUUIFBOHMFPGUIFDPNQMFYQPXFSJTUIFQPXFSGBDUPSBOHMF
5IFDPNQMFYQPXFSNBZCFFYQSFTTFEJOUFSNTPGUIFMPBEJNQFE
BODF;'SPN&R UIFMPBEJNQFEBODF;NBZCFXSJUUFOBT
7SNT @@@@
7
7 = @@@@
; = @@
= SNT⧸θv¢θJ
*SNT
*
*SNT
7SNT
4 = * SNT ; = @@@@
= 7SNT* SNT ;
5IVT 7SNT = ;*SNT4VCTUJUVUJOHUIJTJOUP&R HJWFT
4JODF; = 3 + K9 &R CFDPNFT
4 = *SNT
3 + K9 = 1 + K2
XIFSF1BOE 2BSF UIFSFBM BOEJNBHJOBSZQBSUT PGUIF DPNQMFYQPXFS
UIBUJT
1 = 3F 4 = *SNT
3
2 = *N 4 = *SNT
9
1JTUIFB WFSBHFPSSFBMQP XFSBOEJUEFQFOETPOUIFMPBE TSFTJTUBODF
3 2EFQFOETPOUIFMPBE TSFBDUBODF 9BOEJTDBMMFEUIF SFBDUJWF PS
RVBESBUVSF QPXFS
$PNQBSJOH&R XJUI&R XFOPUJDFUIBU
1 = 7SNT *SNTDPT θv − θJ 2 = 7SNT *SNTTJO θv − θJ 5IFSFBMQPXFS1JTUIFB WFSBHFQPXFSJOXBUUTEFMJWFSFEUPBMPBEJUJT UIFPOMZVTFGVMQP XFS*UJTUIFBDUVBMQP XFSEJTTJQBUFECZUIFMPBE 5IF
SFBDUJWFQPXFS2JTBNFBTVSFPGUIFFOFSHZFYDIBOHFCFUXFFOUIFTPVSDF
BOEUIFSFBDUJWFQBSUPGUIFMPBE5IFVOJUPG2JTUIFWPMUBNQFSFSFBDUJWF
7"3 UPEJTUJOHVJTIJUGSPNUIFSFBMQP XFS XIPTFVOJUJTUIFX BUU8F
LOPXGSPN$IBQUFSUIBUFOFS HZTUPSBHFFMFNFOUTOFJUIFSEJTTJQBUFOPS TVQQMZQPXFS CVUFYDIBOHFQPXFSCBDLBOEGPSUIXJUIUIFSFTUPGUIFOFU
XPSL*OUIFTBNFX BZ UIFSFBDUJWFQPXFSJTCFJOHUSBOTGFSSFECBDLBOE
GPSUICFUXFFOUIFMPBEBOEUIFTPVSDF*USFQSFTFOUTBMPTTMFTTJOUFSDIBOHF
CFUXFFOUIFMPBEBOEUIFTPVSDF/PUJDFUIBU
2 = GPSSFTJTUJWFMPBET VOJUZQG 2< GPSDBQBDJUJWFMPBET MFBEJOHQG 2GPSJOEVDUJWFMPBET MBHHJOHQG 5IVT
Complex power (in VA) is the product of the rms voltage phasor and the
complex conjugate of the rms current phasor. As a complex quantity, its real
part is real power P and its imaginary part is reactive power Q.
11.6
473
Complex Power
*OUSPEVDJOHUIFDPNQMFYQPXFSFOBCMFTVTUPPCUBJOUIFSFBMBOESFBDUJWF
QPXFSTEJSFDUMZGSPNWPMUBHFBOEDVSSFOUQIBTPST
$PNQMFY1PXFS = 4 = 1 + K2 = 7SNT *SNT
= ∣7SNT∣ ∣*SNT∣⧸
θv¢θJ
_______
"QQBSFOU1PXFS = 4 = ∣4∣ = ∣7SNT∣∣*SNT∣ = √ 1 + 2
3FBM1PXFS = 1 = 3F 4 = 4DPT θv − θJ
3FBDUJWF1PXFS = 2 = *N 4 = 4TJO θv − θJ
1 = DPT θ − θ
1PXFS'BDUPS = @@
v
J
4
5IJTTIP XTIP XUIFDPNQMF YQP XFSDPOUBJOT BMMUIFSFMF WBOUQP XFS
JOGPSNBUJPOJOBHJWFOMPBE
*UJT BTUBOEBSEQSBDUJDFUPSFQSFTFOU 4 1 BOE2JOUIF GPSNPG B
USJBOHMF LOPXOBTUIF QPXFSUSJBOHMF TIPXOJO'JH B 5IJTJT
TJNJMBSUPUIFJNQFEBODFUSJBOHMFTIP XJOHUIFSFMBUJPOTIJQCFUXFFO
; 3 BOE 9 JMMVTUSBUFEJO'JH C 5IFQP XFSUSJBOHMFIBTGPVS
JUFNT‡UIFBQQBSFOUDPNQMFYQPXFS SFBMQPXFS SFBDUJWFQPXFS BOEUIF
QPXFSGBDUPSBOHMF(JWFOUXPPGUIFTFJUFNT UIFPUIFSUXPDBOFBTJMZCF
PCUBJOFEGSPNUIFUSJBOHMF"TTIPXOJO'JH XIFO4MJFT JOUIF
first quadrant, we haWFBOJOEVDUJWFMPBEBOEBMBHHJOHQG8IFO4MJFTJO
UIFGPVSUIRVBESBOU UIFMPBEJTDBQBDJUJWFBOEUIFQGJTMFBEJOH*UJTBMTP
QPTTJCMFGPSUIFDPNQMF YQPXFSUPMJFJOUIFTFDPOEPSUIJSERVBESBOU
5IJTSFRVJSFTUIBUUIFMPBEJNQFEBODFIB WFBOFHBUJWFSFTJTUBODF XIJDI
JTQPTTJCMFXJUIBDUJWFDJSDVJUT
S contains all power information of a
load. The real part of S is the real
power P ; its imaginary part is the
reactive power Q ; its magnitude is the
apparent power S; and the cosine of
its phase angle is the power factor pf.
Im
S
Q
| Z|
X
θ
θ
P
R
(a)
(b)
+Q (lagging pf )
S
θv ‒ θi
θv ‒ θi
P
Re
Figure 11.21
B 1PXFSUSJBOHMF C JNQFEBODFUSJBOHMF
S
‒Q (leading pf )
Figure 11.22
1PXFSUSJBOHMF
5IFW PMUBHFBDSPTTBMPBEJT v U = DPT ωU − ž 7BOEUIFDVS SFOUUISPVHIUIFFMFNFOUJOUIFEJSFDUJPOPGUIFW PMUBHFESPQJT J U =
DPT ωU + ž "'JOE B UIFDPNQMF YBOEBQQBSFOUQP XFST C UIFSFBMBOESFBDUJ WFQPXFST BOE D UIFQP XFSGBDUPSBOEUIFMPBE
JNQFEBODF
Example 11.11
474
Chapter 11
AC Power Analysis
Solution:
B 'PSUIFSNTWBMVFTPGUIFWPMUBHFBOEDVSSFOU XFXSJUF
@@⧸
7SNT = @@@
¢ž Ŀ
@@⧸
*SNT = @@@
ž
Ŀ
5IFDPNQMFYQPXFSJT
@@⧸
@@⧸
4 = 7SNT* SNT = @@@
¢ž @@@
¢ž = ⧸
¢ž7"
)
( Ŀ )( Ŀ 5IFBQQBSFOUQPXFSJT
4 = ∣4∣ = 7"
C 8FDBOFYQSFTTUIFDPNQMFYQPXFSJOSFDUBOHVMBSGPSNBT
4 = ⧸
¢ž = <DPT −ž + KTJO −ž > = − K
4JODF4 = 1 + K2 UIFSFBMQPXFSJT
1 = 8
XIJMFUIFSFBDUJWFQPXFSJT
2 = −7"3
D 5IFQPXFSGBDUPSJT
QG = DPT −ž = MFBEJOH
*UJTMFBEJOH CFDBVTFUIFSFBDUJWFQPXFSJTOFHBUJWF5IFMPBEJNQFEBODFJT
¢ž
⧸
7 = @@@@@@@@@
= ⧸
; = @@
¢žΩ
* ⧸
ž
XIJDIJTBDBQBDJUJWFJNQFEBODF
Practice Problem 11.11
'PSBMPBE 7SNT = ⧸
ž7 *SNT = ⧸
ž"%FUFSNJOF B UIFDPN QMFYBOEBQQBSFOUQPXFST C UIFSFBM BOESFBDUJWF QPXFST BOE D UIF
QPXFSGBDUPSBOEUIFMPBEJNQFEBODF
Answer: B ⧸
ž7" 7" C 8 7"3 D MBHHJOH + K Ω
Example 11.12
"MPBE ;ESB XTL7 "BUBQP XFSG BDUPSPGMBHHJOHGSPNB
7SNTTJOVTPJEBMTPVSDF$BMDVMBUF B UIFBWFSBHFBOESFBDUJWFQPXFST
EFMJWFSFEUPUIFMPBE C UIFQFBLDVSSFOU BOE D UIFMPBEJNQFEBODF
Solution:
B (JWFOUIBUQG = DPT θ = XFPCUBJOUIFQPXFSBOHMFBT
θ = DPT¢ = ž*GUIFBQQBSFOUQPXFSJT 4 = 7" UIFO
UIFBWFSBHFPSSFBMQPXFSJT
1 = 4DPTθ = × = L8
11.7
475
Conservation of AC Power
XIJMFUIFSFBDUJWFQPXFSJT
2 = 4TJOθ = × = L7"
C 4JODFUIFQGJTMBHHJOH UIFDPNQMFYQPXFSJT
4 = 1 + K2 = + KL7"
'SPN4 = 7SNT* SNT XFPCUBJO
+ K
4 = @@@@@@@@@@@@@
* SNT = @@@@
ž"
= + K" = ⧸
7SNT
⧸
ž
5IVT*SNT = ⧸
¢žBOEUIFQFBLDVSSFOUJT
__
__
*N = √*SNT = √ = "
D 5IFMPBEJNQFEBODF
ž
⧸
7SNT @@@@@@@@@@@
= ⧸
; = @@@@
=
žΩ
*SNT ⧸
¢ž
XIJDIJTBOJOEVDUJWFJNQFEBODF
"TJOVTPJEBMTPVSDFTVQQMJFTL7"3SFBDUJWFQPXFSUPMPBE
; =
⧸
¢ž Ω%FUFSNJOF B UIFQPXFSGBDUPS C UIFBQQBSFOUQPXFS
EFMJWFSFEUPUIFMPBE BOE D UIFSNTWPMUBHF
Practice Problem 11.12
Answer: B MFBEJOH C L7" D L7
11.7
Conservation of AC Power
5IFQSJODJQMFPGDPOTFSWBUJPOPGQPXFSBQQMJFTUPBDDJSDVJUTBTXFMMBTUP
EDDJSDVJUT TFF4FDUJPO 5PTFFUIJT DPOTJEFSUIFDJSDVJUJO'JH B XIFSFUXPMPBE
JNQFEBODFT;BOE ;BSFDPOOFDUFEJOQBSBMMFMBDSPTTBOBDTPVSDF 7
,$-HJWFT
* = * + *
5IFDPNQMFYQPXFSTVQQMJFECZUIFTPVSDFJT GSPNOPXPO VOMFTTPUI
erwise specified, all vBMVFTPGWPMUBHFTBOEDVSSFOUTXJMMCFBTTVNFEUP
CFSNTWBMVFT
4 = 7* = 7 * + * = 7* + 7* = 4 + 4
I
V +
‒
I1
I2
Z1
Z2
(a)
Figure 11.23
I
V +
‒
Z1
Z2
+V ‒
+V ‒
1
2
(b)
"OBDWPMUBHFTPVSDFTVQQMJFEMPBETDPOOFDUFEJO B QBSBMMFM C TFSJFT
In fact, we already saw in Examples
11.3 and 11.4 that average power is
conserved in ac circuits.
476
Chapter 11
AC Power Analysis
XIFSF4BOE 4EFOPUFUIFDPNQMF YQPXFSTEFMJWFSFEUPMPBET ;BOE
; SFTQFDUJWFMZ
*GUIFMPBETBSFDPOOFDUFEJOTFSJFTXJUIUIFWPMUBHFTPVSDF BTTIPXO
JO'JH C ,7-ZJFMET
7 = 7 + 7
5IFDPNQMFYQPXFSTVQQMJFECZUIFTPVSDFJT
4 = 7* = 7 + 7 * = 7* + 7* = 4 + 4 XIFSF4BOE 4EFOPUFUIFDPNQMF YQPXFSTEFMJWFSFEUPMPBET ;BOE
; SFTQFDUJWFMZ
8FDPODMVEFGSPN&RT BOE UIBUXIFUIFSUIFMPBET
BSFDPOOFDUFEJOTFSJFTPSJOQBSBMMFM PSJOHFOFSBM UIFUPUBMQP
XFS
TVQQMJFECZUIFTPVSDFFRVBMTUIFUPUBMQPXFSEFMJWFSFEUPUIFMPBE5IVT JOHFOFSBM GPSBTPVSDFDPOOFDUFEUP/MPBET
4 = 4 + 4 + ⋯ + 4/
In fact, all forms of ac power are
conserved: instantaneous, real,
reactive, and complex.
5IJTNFBOTUIBUUIFUPUBMDPNQMF YQPXFSJOBOFUXPSLJTUIFTVNPGUIF
DPNQMFYQPXFSTPGUIFJOEJWJEVBMDPNQPOFOUT 5IJTJTBMTPUSVFPGSFBM
QPXFSBOESFBDUJWFQPXFS CVUOPUUSVFPGBQQBSFOUQPXFS 5IJTFYQSFTTFT
UIFQSJODJQMFPGDPOTFSWBUJPOPGBDQPXFS
The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual
loads.
'SPNUIJTXFJNQMZUIBUUIFSFBM PSSFBDUJWF QPwer floXGSPNTPVSDFTJO
BOFUXPSLFRVBMTUIFSFBM PSSFBDUJWF QPXFSfloXJOUPUIFPUIFSFMFNFOUT
JOUIFOFUXPSL
Example 11.13
'JHVSFTIP XTBMPBECFJOHGFECZBW PMUBHFTPVSDFUISPVHIB
USBOTNJTTJPOMJOF 5IFJNQFEBODFPGUIFMJOFJTSFQSFTFOUFECZUIF
+ K ΩJNQFEBODFBOEBSFUVSOQBUI'JOEUIFSFBMQPXFSBOESFBDUJWF
QPXFSBCTPSCFECZ B UIFTPVSDF C UIFMJOF BOE D UIFMPBE
I
4Ω
15 Ω
220 0° V rms +
‒
Source
j2 Ω
‒ j10 Ω
Line
Load
Figure 11.24
'PS&YBNQMF
Solution:
5IFUPUBMJNQFEBODFJT
; = + K + ¢K = ¢K = ⧸
¢žΩ
11.7
477
Conservation of AC Power
5IFDVSSFOUUISPVHIUIFDJSDVJUJT
⧸
ž
7
= ⧸
* = @@@T= @@@@@@@@@@@@@
ž"SNT
; ⧸
¢ž
B 'PSUIFTPVSDF UIFDPNQMFYQPXFSJT
4T = 7T* = ⧸
ž ⧸
¢ž
= ⧸
¢ž = −K 7"
'SPNUIJT XFPCUBJOUIFSFBMQPXFSBT8BOEUIFSFBDUJWFQPXFS
BT7"3 MFBEJOH C 'PSUIFMJOF UIFWPMUBHFJT
7MJOF = + K * = ⧸
ž ⧸
ž
= ⧸
ž7SNT
5IFDPNQMFYQPXFSBCTPSCFECZUIFMJOFJT
4MJOF = 7MJOF* = ⧸
ž ⧸
¢ž
= ⧸
ž = + K7"
PS
4MJOF = ∣*∣;MJOF = + K = + K7"
5IBUJT UIFSFBMQPXFSJT8BOEUIFSFBDUJWFQPXFSJT7"3
MBHHJOH D 'PSUIFMPBE UIFWPMUBHFJT
7- = −K * = ⧸
¢ž ⧸
ž
= ⧸
¢ž7SNT
5IFDPNQMFYQPXFSBCTPSCFECZUIFMPBEJT
4- = 7-* = ⧸
¢ž ⧸
¢ž
= ⧸
¢ž = −K 7"
5IFSFBMQPXFSJT8BOEUIFSFBDUJWFQPXFSJT7"3 MFBEJOH /PUFUIBU 4T = 4MJOF + 4- BTFYQFDUFE8FIBWFVTFEUIFSNTWBMVFTPG WPMUBHFTBOEDVSSFOUT
*OUIFDJSDVJUJO'JH UIF ΩSFTJTUPSBCTPSCTBOBWFSBHFQPXFS
PG8'JOE 7BOEUIFDPNQMFYQPXFSPGFBDICSBODIPGUIFDJSDVJU
8IBUJTUIFPWFSBMMDPNQMFYQPXFSPGUIFDJSDVJU "TTVNFUIFDVSSFOU
UISPVHIUIFΩSFTJTUPSIBTOPQIBTFTIJGU
Answer: ⧸
ž 7 SNT UIF ΩSFTJTUPS 7"UIF −K ΩJNQFEBODF −K7"UIF + K ΩJNQFEBODF
+ K7"PWFSBMM−K7"
Practice Problem 11.13
20 Ω
V
+
‒
Figure 11.25
30 Ω
j20 Ω
‒j10 Ω
60 Ω
'PS1SBDUJDF1SPC
478
Chapter 11
Example 11.14
*OUIFDJSDVJUPG'JH ; = ⧸
¢žΩBOE; = ⧸
žΩ$BMDV
MBUFUIFUPUBM B BQQBSFOUQPXFS C SFBMQPXFS D SFBDUJWFQPXFS BOE
E QG TVQQMJFECZUIFTPVSDFBOETFFOCZUIFTPVSDF
It
120 10° V rms +
‒
AC Power Analysis
I1
I2
Z1
Z2
Solution:
5IFDVSSFOUUISPVHI;JT
⧸
ž
7= @@@@@@@@
* = @@@
ž"SNT
= ⧸
; ⧸
¢ž
Figure 11.26
'PS&YBNQMF
XIJMFUIFDVSSFOUUISPVHI;JT
⧸
ž
7= @@@@@@@@
* = @@@
¢ž"SNT
= ⧸
;
⧸
ž
5IFDPNQMFYQPXFSTBCTPSCFECZUIFJNQFEBODFTBSF
7SNT
= ⧸
4 = @@@@
= @@@@@@@
¢ž = −K7"
⧸
ž
;
7SNT
@@@@@@@@
= ⧸
4 = @@@@
=
ž = + K7"
⧸
¢ž
;
5IFUPUBMDPNQMFYQPXFSJT
4U = 4 + 4 = + K7"
B 5IFUPUBMBQQBSFOUQPXFSJT
_____________
∣4U∣ = √ + = 7"
C 5IFUPUBMSFBMQPXFSJT
1U = 3F 4U = 8PS1U = 1 + 1
D 5IFUPUBMSFBDUJWFQPXFSJT
2U = *N 4U = 7"3PS2U = 2 + 2
E 5IFQG = 1U∕∣4U∣ = ∕ = MBHHJOH We may cross check the result by finding the complex power 4TTVQQMJFE
CZUIFTPVSDF
*U = * + * = + K + −K
= −K = ⧸
¢ž"SNT
4T = 7* U = ⧸
ž ⧸
ž
= ⧸
ž = + K7"
XIJDIJTUIFTBNFBTCFGPSF
Practice Problem 11.14
5XPMPBETDPOOFDUFEJOQBSBMMFMBSFSFTQFDUJWFMZL8BUBQGPG
MFBEJOHBOEL8BUBQGPGMBHHJOH$BMDVMBUFUIFQGPGUIFDPN CJOFEUXPMPBET'JOEUIFDPNQMFYQPXFSTVQQMJFECZUIFTPVSDF
Answer: MFBEJOH −KL7"
11.8
11.8
479
Power Factor Correction
Power Factor Correction
.PTUEPNFTUJDMPBET TVDIBTX BTIJOHNBDIJOFT BJSDPOEJUJPOFST BOE
SFGSJHFSBUPST BOEJOEVTUSJBMMPBET TVDIBTJOEVDUJPONPUPST BSFJOEVD UJWFBOEPQFSBUFBUBMP XMBHHJOHQP XFSG BDUPS"MUIPVHIUIFJOEVDUJ WF
OBUVSFPGUIFMPBEDBOOPUCFDIBOHFE XFDBOJODSFBTFJUTQPXFSGBDUPS
The process of increasing the power factor without altering the voltage or
current to the original load is known as power factor correction.
4JODFNPTUMPBETBSFJOEVDUJ WF BTTIPXOJO'JH B BMPBE T
QPXFSGBDUPSJTJNQSPWFEPSDPSSFDUFECZEFMJCFSBUFMZJOTUBMMJOHBDBQBDJ
UPSJOQBSBMMFMXJUIUIFMPBE BTTIPXOJO'JH C 5IFFGGFDUPGBEE
JOHUIFDBQBDJUPSDBOCFJMMVTUSBUFEVTJOHFJUIFSUIFQPXFSUSJBOHMFPSUIF
QIBTPSEJBHSBNPGUIFDVSSFOUTJO WPMWFE'JHVSFTIP XTUIFMBUUFS XIFSFJUJTBTTVNFEUIBUUIFDJSDVJUJO'JH B IBTBQPXFSGBDUPSPG
DPTθ XIJMFUIFPOFJO'JH C IBTBQPXFSGBDUPSPGDPTθ*UJT
FWJEFOUGSPN'JHUIBUBEEJOHUIFDBQBDJUPSIBTDBVTFEUIFQIBTF
BOHMFCFUXFFOUIFTVQQMJFEW PMUBHFBOEDVSSFOUUPSFEVDFGSPN θUPθ UIFSFCZJODSFBTJOHUIFQPXFSGBDUPS8FBMTPOPUJDFGSPNUIFNBHOJUVEFT
PGUIFWFDUPSTJO'JHUIBUXJUIUIFTBNFTVQQMJFEW PMUBHF UIFDJS
DVJUJO'JH B ESB XTMBSHFSDVSSFOU*-UIBOUIFDVSSFOU *ESBXOCZ
UIFDJSDVJUJO'JH C 1PXFSDPNQBOJFTDIBSHFNPSFGPSMBSHFSDVS
SFOUT CFDBVTFUIFZSFTVMUJOJODSFBTFEQPXFSMPTTFT CZBTRVBSFEGBDUPS TJODF1 = *-3 Therefore, it is beneficial to both the poXFSDPNQBOZBOE
UIFDPOTVNFSUIBUFWFSZFGGPSUJTNBEFUPNJOJNJ[FDVSSFOUMFWFMPSLFFQ
UIFQPXFSGBDUPSBTDMPTFUPVOJUZBTQPTTJCMF#ZDIPPTJOHBTVJUBCMFTJ[F
GPSUIFDBQBDJUPS UIFDVSSFOUDBOCFNBEFUPCFDPNQMFUFMZJOQIBTFXJUI
UIFWPMUBHF JNQMZJOHVOJUZQPXFSGBDUPS
IL
+
V
Inductive
load
V
An inductive load is modeled as a
series combination of an inductor and
a resistor.
IC
I
+
Alternatively, power factor correction
may be viewed as the addition of a
reactive element (usually a capacitor)
in parallel with the load in order to
make the power factor closer to unity.
IL
Inductive
load
IC
θ1
C
θ2
V
IC
I
‒
‒
(a)
Figure 11.27
(b)
1PXFSGBDUPSDPSSFDUJPO B PSJHJOBMJOEVDUJWFMPBE C JOEVDUJWFMPBEXJUIJNQSPWFEQPXFSGBDUPS
IL
Figure 11.28
1IBTPSEJBHSBNTIPXJOHUIFFGGFDUPG
BEEJOHBDBQBDJUPSJOQBSBMMFMXJUIUIF
JOEVDUJWFMPBE
8FDBOMPPLBUUIFQPXFSGBDUPSDPSSFDUJPOGSPNBOPUIFSQFSTQFDUJWF
$POTJEFSUIFQP XFSUSJBOHMFJO'JH*GUIFPSJHJOBMJOEVDUJ WFMPBE
IBTBQQBSFOUQPXFS4 UIFO
1 = 4DPTθ 2 = 4TJOθ = 1UBOθ
480
Chapter 11
AC Power Analysis
*GXFEFTJSFUPJODSFBTFUIFQP XFSGBDUPSGSPNDPT θUPDPT θXJUIPVU
BMUFSJOHUIFSFBMQPXFS JF 1 = 4DPTθ UIFOUIFOFXSFBDUJWFQPXFSJT
QC
2 = 1UBOθ
S1
Q1
S2
Q2
θ1 θ2
P
5IFSFEVDUJPOJOUIFSFBDUJ WFQP XFSJTDBVTFECZUIFTIVOUDBQBDJUPS
UIBUJT
2$ = 2−2 = 1 UBOθ−UBOθ #VUGSPN&R 2$ = 7 SNT
∕9$ = ω$7 SNT
5IFWBMVFPGUIFSFRVJSFE
TIVOUDBQBDJUBODF$JTEFUFSNJOFEBT
Figure 11.29
1PXFSUSJBOHMFJMMVTUSBUJOHQPXFSGBDUPS
DPSSFDUJPO
2$
1 UBOθ−UBOθ
$ = @@@@@
= @@@@@@@@@@@@@@@
ω7 SNT
ω7 SNT
/PUFUIBUUIFSFBMQP XFS1EJTTJQBUFECZUIFMPBEJTOPUBG GFDUFECZUIF
QPXFSG BDUPSDPSSFDUJPOCFDBVTFUIFB WFSBHFQPXFSEVFUPUIFDBQBDJ UBODFJT[FSP
"MUIPVHIUIFNPTUDPNNPOTJUVBUJPOJOQSBDUJDFJTUIBUPGBOJOEVD
UJWFMPBE JUJTBMTPQPTTJCMFUIBUUIFMPBEJTDBQBDJUJ WFUIBUJT UIFMPBE
JTPQFSBUJOHBUBMFBEJOHQPXFSGBDUPS*O UIJTDBTF BOJOEVDUPSTIPVME
CFDPOOFDUFEBDSPTTUIFMPBEGPSQP XFSGBDUPSDPSSFDUJPO 5IFSFRVJSFE
TIVOUJOEVDUBODF-DBOCFDBMDVMBUFEGSPN
7SNT
7 @@@@
2- = @@@@
= SNT
9ω-
⇒
7SNT
- = @@@@
ω2-
XIFSF2- = 2−2 UIFEJG GFSFODFCFUXFFOUIFOF XBOEPMESFBDUJ WF
QPXFST
Example 11.15
8IFODPOOFDUFEUPB7 SNT )[QP XFSMJOF BMPBEBCTPSCT
L8BUBMBHHJOHQP XFSG BDUPSPG'JOEUIFW BMVFPGDBQBDJUBODF
OFDFTTBSZUPSBJTFUIFQGUP
Solution:
*GUIFQG = UIFO
DPTθ = ⇒
θ = ž
XIFSFθJTUIFQIBTFEJGGFSFODFCFUXFFOWPMUBHFBOEDVSSFOU8FPCUBJO
UIFBQQBSFOUQPXFSGSPNUIFSFBMQPXFSBOEUIFQGBT
1 4 = @@@@@
= @@@@@
= 7"
DPTθ
5IFSFBDUJWFQPXFSJT
2 = 4TJOθ = TJO = 7"3
8IFOUIFQGJTSBJTFEUP
DPTθ = ⇒
θ = ž
11.9
Applications
481
5IFSFBMQPXFS1IBTOPUDIBOHFE#VUUIFBQQBSFOUQPXFSIBTDIBOHFE
JUTOFXWBMVFJT
1 = 7"
4 = @@@@@
= @@@@@
DPTθ 5IFOFXSFBDUJWFQPXFSJT
2 = 4TJOθ = 7"3
5IFEJGGFSFODFCFUXFFOUIFOFXBOEPME SFBDUJWFQPXFSTJTEVFUPUIF
QBSBMMFMBEEJUJPOPGUIFDBQBDJUPSUPUIFMPBE5IFSFBDUJWFQPXFSEVFUP
UIFDBQBDJUPSJT
2$ = 2−2 = − = 7"3
BOE
2$
$ = @@@@@
= μ'
= @@@@@@@@@@@@@
ω7SNT
π × × /PUF$BQBDJUPSTBSFOPSNBMMZQVSDIBTFEGPSWPMUBHFTUIFZFYQFDUUPTFF
*OUIJTDBTF UIFNBYJNVNWPMUBHFUIJTDBQBDJUPSXJMMTFFJTBCPVU7
QFBL8FXPVMETVHHFTUQVSDIBTJOHBDBQBDJUPSXJUIBWPMUBHFSBUJOH
FRVBMUP TBZ 7
'JOEUIFWBMVFPGQBSBMMFMDBQBDJUBODFOFFEFEUPDPSSFDUBMPBEPG
L7"3BUMBHHJOHQGUPVOJUZQG"TTVNFUIBUUIFMPBEJTTVQ QMJFECZB7 SNT )[MJOF
Practice Problem 11.15
Answer:N'
11.9
Applications
*OUIJTTFDUJPO XFDPOTJEFSUXPJNQPSUBOUBQQMJDBUJPOBSFBTIPXQPXFS
JT NFBTVSFE BOE IPX FMFDUSJD VUJMJUZ DPNQBOJFT EFUFSNJOF UIF DPTU PG
FMFDUSJDJUZDPOTVNQUJPO
11.9.1
Power Measurement
5IFB WFSBHFQP XFSBCTPSCFECZBMPBEJTNFBTVSFECZBOJOTUSVNFOU
DBMMFEUIFXBUUNFUFS
Reactive power is measured by an
instrument called the varmeter. The
varmeter is often connected to the
load in the same way as the wattmeter.
The wattmeter is the instrument used for measuring the average power.
'JHVSFTIP XTBX BUUNFUFSUIBU DPOTJTUTFTTFOUJBMMZ PGUX P
DPJMTUIFDVSSFOUDPJM BOEUIFWPMUBHFDPJM"DVSSFOUDPJMXJUIWFSZMPX
JNQFEBODF JEFBMMZ[FSP JTDPOOFDUFEJOTFSJFTXJUIUIFMPBE 'JH BOESFTQPOETUPUIFMPBEDVSSFOU5IFWPMUBHFDPJMXJUIWFSZIJHIJNQFE
BODF (ideally infinite) is connected in parallel with the load as shoXOJO
'JHBOESFTQPOETUPUIFMPBEW PMUBHF5IFDVSSFOUDPJMBDUTMJL FB
TIPSUDJSDVJUCFDBVTFPGJUTMPXJNQFEBODFUIFWPMUBHFDPJMCFIBWFTMJLF
Some wattmeters do not have coils;
the wattmeter considered here is the
electromagnetic type.
482
Chapter 11
AC Power Analysis
i
i
±
Current coil
±
±
+
v
‒
R
Voltage coil
+
v
‒
ZL
Figure 11.31
5IFXBUUNFUFSDPOOFDUFEUPUIFMPBE
i
±
Figure 11.30
"XBUUNFUFS
BOPQFODJSDVJUCFDBVTFPGJUTIJHIJNQFEBODF "TBSFTVMU UIFQSFTFODF
PGUIFX BUUNFUFSEPFTOPUEJTUVSCUIFDJSDVJUPSIB WFBOFG GFDUPOUIF
QPXFSNFBTVSFNFOU
8IFOUIFUXPDPJMTBSFFOFSHJ[FE UIFNFDIBOJDBMJOFSUJBPGUIFNPW
JOHTZTUFNQSPEVDFTBdeflection BOHMFUIBUJTQSPQPSUJPOBMUPUIFBWFSBHF
WBMVFPGUIFQSPEVDUv U J U *GUIFDVSSFOUBOEWPMUBHFPGUIFMPBEBSFv U
= 7NDPT ωU + θv BOE J U = *NDPT ωU + θJ UIFJSDPSSFTQPOEJOHSNT
QIBTPSTBSF
7N
*N
@@ ⧸
@@⧸
7SNT = @@@
BOE
*SNT = @@@
θJ
θv
Ŀ
ð
BOEUIFXBUUNFUFSNFBTVSFTUIFBWFSBHFQPXFSHJWFOCZ
1 = |7SNT| |*SNT| DPT θv −θJ = 7SNT*SNTDPT θv −θJ "T TIPXO JO 'JH FBDI XBUUNFUFSDPJMIBT UXP UFSNJOBMT
XJUIPOFNBSL FE±5o ensure upscale deflection, the ±UFSNJOBMPG UIFDVSSFOUDPJMJTUPXBSEUIFTPVSDF XIJMFUIF±UFSNJOBMPGUIFWPMU
BHFDPJMJTDPOOFDUFEUPUIFTBNFMJOFBTUIFDVSSFOUDPJM3F WFSTJOH
CPUIDPJMDPOOFDUJPOTTUJMMSFTVMUTJOVQTDBMFdeflection. )PXFWFS SF
WFSTJOHPOFDPJMBOEOPUUIFPUIFSSFTVMUTJOEPwnscale deflection and
OPXBUUNFUFSSFBEJOH
Example 11.16
'JOEUIFXBUUNFUFSSFBEJOHPGUIFDJSDVJUJO'JH
12 Ω
j10 Ω
±
±
150 0° V rms +
‒
8Ω
‒ j6 Ω
Figure 11.32
'PS&YBNQMF
Solution:
Define.The problem is clearly defined. Interestingly UIJTJTBQSPC
MFNXIFSFUIFTUVEFOUDPVMEBDUVBMMZW BMJEBUFUIFSFTVMUTCZEPJOH
UIFQSPCMFNJOUIFMBCPSBUPSZXJUIBSFBMXBUUNFUFS
11.9
Applications
1SFTFOUThis problem consists of finding the aWFSBHFQPXFSEFMJW
FSFEUPBMPBECZBOFYUFSOBMTPVSDFXJUIBTFSJFTJNQFEBODF
"MUFSOBUJWF5IJTJTBTUSBJHIUGPSXBSEDJSDVJUQSPCMFNXIFSFBMMXF
need to do is find the magnitude and phase of the current through the
MPBEBOEUIFNBHOJUVEFBOEUIFQIBTFPGUIFWPMUBHFBDSPTTUIFMPBE
5IFTFRVBOUJUJFTDPVMEBMTPCFGPVOECZVTJOH14QJDF XIJDIXFXJMM
VTFBTBDIFDL
"UUFNQU*O'JH UIFX BUUNFUFSSFBETUIFB WFSBHFQPXFSBC
TPSCFECZUIF −K ΩJNQFEBODFCFDBVTFUIFDVSSFOUDPJMJTJO
TFSJFTXJUIUIFJNQFEBODFXIJMFUIFW PMUBHFDPJMJTJOQBSBMMFMXJUI
JU5IFDVSSFOUUISPVHIUIFDJSDVJUJT
ž
⧸
= @@@@@@@
*SNT = @@@@@@@@@@@@@@@@@@
"
+ K
+ K + −K
5IFWPMUBHFBDSPTTUIF −K ΩJNQFEBODFJT
−K
7SNT = *SNT −K = @@@@@@@@@@
7
+ K
5IFDPNQMFYQPXFSJT
−K @@@@@@@
−K
4 = 7SNT* SNT = @@@@@@@@@@
· = @@@@@@@@@@@
+ K
−K
+ = −K7"
5IFXBUUNFUFSSFBET
1 = 3F 4 = 8
&WBMVBUF8FDBODIFDLPVSSFTVMUTCZVTJOH14QJDF
AC=ok
MAG=ok
PHASE=yes
AC=ok
MAG=ok
IPRINT
PHASE=yes
R1
L1
12
10
R2
ACMAG=150V +
ACPHASE=0
‒
8
V1
C2
0.16667
4JNVMBUJPOZJFMET
'3&2
&
*. 7@13*/5
& *1 7@13*/5
& '3&2
&
7. /@
& 71 /@
& BOE
483
484
Chapter 11
AC Power Analysis
5PDIFDLPVSBOTXFS BMMXFOFFEJTUIFNBHOJUVEFPGUIFDVSSFOU
A) floXJOHUISPVHIUIFMPBESFTJTUPS
1 = *- 3 = = 8
"TFYQFDUFE UIFBOTXFSEPFTDIFDL
4BUJTGBDUPSZ 8FIB WFTBUJTG BDUPSJMZTPMW FEUIFQSPCMFNBOEUIF
SFTVMUTDBOOPXCFQSFTFOUFEBTBTPMVUJPOUPUIFQSPCMFN
Practice Problem 11.16
For the circuit in Fig. 11.33, find the wattmeter reading.
4Ω
±
‒ j2 Ω
±
120 30° V rms +
‒
j9 Ω
12 Ω
Figure 11.33
'PS1SBDUJDF1SPC
Answer:L8
11.9.2 Electricity Consumption Cost
In Section 1.7, we considered a simplified model of the wBZUIFDPTUPG FMFDUSJDJUZDPOTVNQUJPOJTEFUFSNJOFE#VUUIFDPODFQUPGQP XFSGBDUPS
XBTOPUJODMVEFEJOUIFDBMDVMBUJPOT/PXXFDPOTJEFSUIFJNQPSUBODFPG
QPXFSGBDUPSJOFMFDUSJDJUZDPOTVNQUJPODPTU
-PBETXJUIMPXQPXFSGBDUPSTBSFDPTUMZUPTFSWFCFDBVTFUIFZSFRVJSF
MBSHFDVSSFOUT BTFYQMBJOFEJO4FDUJPO5IFJEFBMTJUVBUJPOXPVMECF
UPESBXNJOJNVNDVSSFOUGSPNBTVQQMZTPUIBU4 = 1 2 = BOEQG = "MPBEXJUIOPO[FSP2NFBOTUIBUFOFSgy floXTCBDLBOEGPSUICFUXFFO
UIFMPBEBOEUIFTPVSDF HJ WJOHSJTFUPBEEJUJPOBMQP XFSMPTTFT*OWJF X
PGUIJT QPXFSDPNQBOJFTPGUFOFODPVSBHFUIFJSDVTUPNFSTUPIBWFQPXFS
GBDUPSTBTDMPTFUPVOJUZBTQPTTJCMFBOEQFOBMJ[FTPNFDVTUPNFSTXIPEP
OPUJNQSPWFUIFJSMPBEQPXFSGBDUPST
6UJMJUZDPNQBOJFTEJWJEFUIFJSDVTUPNFSTJOUPDBUFHPSJFTBTSFTJEFO
UJBM EPNFTUJD DPNNFSDJBM BOEJOEVTUSJBM PSBTTNBMMQP XFS NFEJVN
QPXFS BOEMBSHFQPXFS5IFZIBWFEJGGFSFOUSBUFTUSVDUVSFTGPSFBDIDBU
FHPSZ5IFBNPVOUPGFOFSHZDPOTVNFEJOVOJUTPGLJMPXBUUIPVST L8I JTNFBTVSFEVTJOHBLJMP XBUUIPVSNFUFSJOTUBMMFEBUUIFDVTUPNFS T
QSFNJTFT
"MUIPVHIVUJMJUZDPNQBOJFTVTFEJGGFSFOUNFUIPETGPSDIBSHJOHDVT
UPNFST UIFUBSJG GPSDIBS HFUPBDPOTVNFSJTPGUFOUX PQBSU The first
part is fixFEBOEDPSSFTQPOETUPUIFDPTUPGHFOFSBUJPO USBOTNJTTJPO BOEEJTUSJCVUJPOPGFMFDUSJDJUZUPNFFUUIFMPBESFRVJSFNFOUTPGUIFDPO TVNFST5IJTQBSUPGUIFUBSJG GJTHFOFSBMMZF YQSFTTFEBTBDFSUBJOQSJDF 11.9
Applications
485
QFSL8PGNBYJNVNEFNBOE0SJUNBZCFCBTFEPOL7"PGNBYJNVN
EFNBOE UPBDDPVOUGPSUIFQPXFSGBDUPS QG PGUIFDPOTVNFS"QGQFO
BMUZDIBSHFNBZCFJNQPTFEPOUIFDPOTVNFSXIFSFCZBDFSUBJOQFSDFOU
BHFPGL8PSL7"NBYJNVNEFNBOEJTDIBSHFEGPSFWFSZGBMMJOQG
CFMPXBQSFTDSJCFEWBMVF TBZPS0OUIFPUIFSIBOE BQGDSFEJU
NBZCFHJWFOGPSFWFSZUIBUUIFQGFYDFFETUIFQSFTDSJCFEWBMVF
5IFTFDPOEQBSUJTQSPQPSUJPOBMUPUIFFOFS HZDPOTVNFEJOL8IJ U
NBZCFJOHSBEFEGPSN GPSFxample, the first 100 kWh at 16 cents/L8I UIF
OFYUL8IBUDFOUTL8IBOETPGPSUI5IVT UIFCJMMJTEFUFSNJOFE
CBTFEPOUIFGPMMPXJOHFRVBUJPO
5PUBM$PTU = 'JYFE$PTU + $PTUPG&OFSHZ
" NBOVGBDUVSJOH JOEVTUSZ DPOTVNFT .8I JOPOF NPOUI *G UIF
NBYJNVNEFNBOEJT L8 DBMDVMBUFUIFFMFDUSJDJUZCJMMCBTFEPOUIF
GPMMPXJOHUXPQBSUSBUF
Example 11.17
%FNBOEDIBSHFQFSNPOUIQFSL8PGCJMMJOHEFNBOE
Energy charge: 8 cents per kWh for the first 50,000 kWh, 5 cents
QFSL8IGPSUIFSFNBJOJOHFOFSHZ
Solution:
5IFEFNBOEDIBSHFJT
× = The energy charge for the first 50,000 kWh is
× = 5IFSFNBJOJOHFOFSHZJT L8I− L8I = L8I BOE
UIFDPSSFTQPOEJOHFOFSHZDIBSHFJT
× = "EEJOHUIFSFTVMUTPG&RT UP HJWFT
5PUBMCJMMGPSUIFNPOUI = + + = *UNBZBQQFBSUIBUUIFDPTUPGFMFDUSJDJUZJTUPPIJHI#VUUIJTJTPGUFOBTNBMM
GSBDUJPOPGUIFPWFSBMMDPTUPGQSPEVDUJPOPGUIFHPPETNBOVGBDUVSFEPSUIF selling price of the finished product.
5IFNPOUIMZSFBEJOHPGBQBQFSNJMMTNFUFSJTBTGPMMPXT
.BYJNVNEFNBOE L8
&OFSHZDPOTVNFE.8I
6TJOHUIFUXPQBSUSBUFJO&YBNQMF DBMDVMBUFUIFNPOUIMZCJMMGPS
UIFQBQFSNJMM
Answer: Practice Problem 11.17
486
Example 11.18
Chapter 11
AC Power Analysis
"L8MPBETVQQMJFEBUL7 SNT PQFSBUFTIPVSTBNPOUIBU
QFSDFOUQPXFSGBDUPS$BMDVMBUFUIFBWFSBHFDPTUQFSNPOUICBTFEPO
this simplified tarifG
&OFSHZDIBSHFDFOUTQFSL8I
1PXFSGBDUPSQFOBMUZQFSDFOUPGFOFSHZDIBSHFGPSFWFSZ
UIBUQGGBMMTCFMPX
1PXFSGBDUPSDSFEJUQFSDFOUPGFOFSHZDIBSHFGPSFWFSZ
UIBUQGFYDFFET
Solution:
5IFFOFSHZDPOTVNFEJT
8 = L8 × I = L8I
5IFPQFSBUJOHQPXFS GBDUPSQG = = JT × CFMPX UIFQSF
TDSJCFEQPXFSGBDUPSPG4JODFUIFSFJTQFSDFOUFOFSHZDIBSHFGPS
FWFSZ UIFSFJTBQPXFSGBDUPSQFOBMUZDIBSHFPGQFSDFOU5IJT BNPVOUTUPBOFOFSHZDIBSHFPG
× ∆8 = × @@@@@@
= L8I
5IFUPUBMFOFSHZJT
8U = 8 + ∆8 = + = L8I
5IFDPTUQFSNPOUIJTHJWFOCZ
$PTU = DFOUT × 8U = × = Practice Problem 11.18
"OL8JOEVDUJPOGVSOBDFBUQPXFSGBDUPS PQFSBUFTIPVST
QFSEBZGPSEBZTJOBNPOUI%FUFSNJOFUIFFMFDUSJDJUZCJMMQFSNPOUI
CBTFEPOUIFUBSJGGJO&YBNQMF
Answer: 11.10
Summary
5IFJOTUBOUBOFPVTQPXFSBCTPSCFECZBOFMFNFOUJTUIFQSPEVDUPG
UIFFMFNFOUTUFSNJOBMWPMUBHFBOEUIFDVSSFOUUISPVHIUIFFMFNFOU
Q = vJ
"WFSBHFPSSFBMQP XFS1 JOX BUUT JTUIFB WFSBHFPGJOTUBOUBOFPVT
QPXFSQ
5
QEU
1 = @@
∫
5 @@
*Gv U = 7N@@DPT ωU + θv BOEJ U = *NDPT ωU + θJ UIFOVSNT = VN∕Ŀ
BOE
*SNT = *N∕Ŀ
7 * DPT θ −θ = 7 * DPT θ −θ
1 = @@
v
J
SNT SNT
v
J
N N
11.10
Summary
*OEVDUPSTBOE DBQBDJUPSTBCTPSCOPB WFSBHFQPXFS XIJMFUIFB WFS
BHFQPXFSBCTPSCFECZBSFTJTUPSJT ∕ *N3 = *SNT
3
.BYJNVNBWFSBHFQPXFSJTUSBOTGFSSFEUPBMPBEXIFOUIFMPBEJN
QFEBODFJTUIFDPNQMF YDPOKVH BUFPGUIF 5IFWFOJOJNQFEBODFBT
TFFOGSPNUIFMPBEUFSNJOBMT ;- = ;5I 5IFFGGFDUJWFWBMVFPGBQFSJPEJDTJHOBMY U JTJUTSPPUNFBOTRVBSF
SNT WBMVF
________
9FGG= 9SNT =
√@@5 ∫ Y EU
5
'PSBTJOVTPJE
UIFFG GFDUJWFPSSNTW BMVFJTJUTBNQMJUVEFEJ WJEFE
@@
CZĿ
5IFQPXFSGBDUPSJTUIFDPTJOFPGUIFQIBTFEJGGFSFODFCFUXFFOWPMU
BHFBOEDVSSFOU
QG = DPT θv −θJ
*UJTBMTPUIFDPTJOFPGUIFBOHMFPGUIFMPBEJNQFEBODFPSUIFSBUJP
PGSFBMQPXFSUPBQQBSFOUQPXFS5IFQGJTMBHHJOHJGUIFDVSSFOUMBHT
WPMUBHF JOEVDUJWFMPBE BOEJTMFBEJOHXIFOUIFDVSSFOUMFBETW PMU
BHF DBQBDJUJWFMPBE "QQBSFOUQPXFS4 JO7" JTUIFQSPEVDUPGUIFSNTWBMVFTPGWPMUBHF
BOEDVSSFOU
4 = 7SNT*SNT
_______
*UJTBMTPHJWFOCZ4 = ∣4∣ = √1 + 2 XIFSF1JTUIFSFBMQPXFSBOE
2JTSFBDUJWFQPXFS
3FBDUJWFQPXFS JO7"3 JT
7 * TJO θ −θ = 7 * TJO θ − θ
2 = @@
v
J
SNT SNT
v
J
N N
$PNQMFYQPXFS4 JO7" JTUIFQSPEVDUPGUIFSNTW PMUBHFQIBTPS
BOEUIFDPNQMFYDPOKVHBUFPGUIFSNTDVSSFOUQIBTPS *UJTBMTPUIF
DPNQMFYTVNPGSFBMQPXFS1BOESFBDUJWFQPXFS2
4 = 7SNT* SNT = 7SNT*SNT⧸
θW − θJ = 1 + K2
"MTP
7SNT
@@@@
4 = *SNT
; =
;
5IFUPUBMDPNQMF YQPXFSJOBOFUX PSLJTUIFTVNPGUIFDPNQMF Y
QPXFSTPGUIFJOEJWJEVBMDPNQPOFOUT5PUBMSFBMQPXFSBOESFBDUJWF
QPXFSBSFBMTP SFTQFDUJWFMZ UIFTVNTPGUIFJOEJWJEVBMSFBMQPXFST
BOEUIFSFBDUJ WFQPXFST CVUUIFUPUBMBQQBSFOUQP XFSJTOPUDBMDV MBUFECZUIFQSPDFTT
1PXFSGBDUPSDPSSFDUJPOJTOFDFTTBSZGPSFDPOPNJDSFBTPOTJUJTUIF
QSPDFTTPGJNQSP WJOHUIFQP XFSG BDUPSPGBMPBECZSFEVDJOHUIF
PWFSBMMSFBDUJWFQPXFS
5IFXBUUNFUFSJTUIFJOTUSVNFOUGPSNFBTVSJOHUIFBWFSBHFQPXFS
&OFSHZDPOTVNFEJTNFBTVSFEXJUIBLJMPXBUUIPVSNFUFS
487
488
Chapter 11
AC Power Analysis
Review Questions
5IFBWFSBHFQPXFSBCTPSCFECZBOJOEVDUPSJT[FSP
B 5SVF
C 'BMTF
5IF5IFWFOJOJNQFEBODFPGBOFUXPSLTFFOGSPNUIF
MPBEUFSNJOBMTJT + KΩ'PSNBYJNVNQPXFS
USBOTGFS UIFMPBEJNQFEBODFNVTUCF
B − + KΩ
D −KΩ
C −−KΩ
E + KΩ
5IFBNQMJUVEFPGUIFWPMUBHFBWBJMBCMFJOUIF)[ 7QPXFSPVUMFUJOZPVSIPNFJT
B 7
D 7
C D F OPOFPGUIFTF
"RVBOUJUZUIBUDPOUBJOTBMMUIFQPXFSJOGPSNBUJPOJO
BHJWFOMPBEJTUIF
B QPXFSGBDUPS
D BWFSBHFQPXFS
F DPNQMFYQPXFS
C BQQBSFOUQPXFS
E SFBDUJWFQPXFS
3FBDUJWFQPXFSJTNFBTVSFEJO
B XBUUT
D 7"3
C 7"
E OPOFPGUIFTF
*OUIFQPXFSUSJBOHMFTIPXOJO'JH B UIF
SFBDUJWFQPXFSJT
B 7"3MFBEJOH
D 7"3MFBEJOH
1000 VAR
60°
500 W
(a)
(b)
Figure 11.34
'PS3FWJFX2VFTUJPOTBOE
C 7
E 7
*GUIFMPBEJNQFEBODFJT−K UIFQPXFSGBDUPSJT
−45°
B ⧸
E 30°
C 7"3MBHHJOH
E 7"3MBHHJOH
'PSUIFQPXFSUSJBOHMFJO'JH C UIFBQQBSFOU
QPXFSJT
B 7"
D 7"3
C 7"3
E 7"3
"TPVSDFJTDPOOFDUFEUPUISFFMPBET; ; BOE;
JOQBSBMMFM8IJDIPGUIFTFJTOPUUSVF
B 1 = 1 + 1 + 1
D 4 = 4 + 4 + 4
C 2 = 2 + 2 + 2
E 4 = 4 + 4 + 4
5IFJOTUSVNFOUGPSNFBTVSJOHBWFSBHFQPXFSJTUIF
B WPMUNFUFS
D XBUUNFUFS
F LJMPXBUUIPVSNFUFS
C BNNFUFS
E WBSNFUFS
"OTXFSTB D D E F D E B D D
Problems1
Section 11.2 Instantaneous and Average Power
*Gv U = DPTU7BOEJ U =
−TJO U−ž " DBMDVMBUFUIFJOTUBOUBOFPVT
QPXFSBOEUIFBWFSBHFQPXFS
Given the circuit in Fig. 11.35, find the average
QPXFSTVQQMJFEPSBCTPSCFECZFBDIFMFNFOU
30 Ω
200 0° V +
‒
Figure 11.35
'PS1SPC
"MPBEDPOTJTUTPGBΩSFTJTUPSJOQBSBMMFMXJUIB
μ'DBQBDJUPS*GUIFMPBEJTDPOOFDUFEUPBWPMUBHF
TPVSDFvT U = DPTU, find the average power
EFMJWFSFEUPUIFMPBE
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEJOTUBOUBOFPVTBOEBWFSBHF
QPXFS
j50 Ω
R1
‒ j10 Ω
Vs +
‒
jXL
Figure 11.36
'PS1SPC
Starting with problem 11.22, unless otherwise specified, assume that all values of currents and voltages are rms.
R2
‒ jXC
489
Problems
TTVNJOHUIBUvT = DPT U−ž 7JOUIFDJSDVJUPG
Fig. 11.37, find the average power delivered to each
PGUIFQBTTJWFFMFNFOUT
1Ω
'PSUIFPQBNQDJSDVJUJO'JH 7T = ⧸
30°7
'JOEUIFBWFSBHFQPXFSBCTPSCFECZUIFLΩ
SFTJTUPS
2Ω
vs +
‒
3H
+
‒
Vs
0.25 F
+
‒
10 kΩ
j6 kΩ
2 kΩ
Figure 11.37
20 kΩ
‒ j12 kΩ
j4 kΩ
'PS1SPC
Figure 11.41
'PSUIFDJSDVJUJO'JH JT = DPT U"'JOE
UIFBWFSBHFQPXFSBCTPSCFECZUIFΩSFTJTUPS
In the op amp circuit in Fig. 11.42, find the total
BWFSBHFQPXFSBCTPSCFECZUIFSFTJTUPST
20i x
ix
is
'PS1SPC
+ ‒
R
50 Ω
20 mH
40 μF
10 Ω
+
‒
+
‒
R
Vo cos ωt V +
‒
R
Figure 11.38
'PS1SPC
Figure 11.42
'PS1SPC
Given the circuit of Fig. 11.39, find the average
QPXFSBCTPSCFECZUIFΩSFTJTUPS
4Ω
20 45° V +
‒
0.1 Vo
'PSUIFOFUXPSLJO'JH BTTVNFUIBUUIFQPSU
JNQFEBODFJT
3 @@@@@@@@@@@
;BC = @@@@@@@@@@@
¢UBO¢ω3$
⧸
Ŀ + ω3$
‒ j5 Ω
Io
+
‒
8Io
j5 Ω
10 Ω
+
Vo
‒
'JOEUIFBWFSBHFQPXFSDPOTVNFECZUIFOFUXPSL
XIFO3 = LΩ $ = O' BOEJ =
TJO U + ž N"
Figure 11.39
i
'PS1SPC
Linear
network
*OUIFDJSDVJUPG'JH EFUFSNJOFUIFBWFSBHF
QPXFSBCTPSCFECZUIFΩSFTJTUPS
Io
6 0° A
Figure 11.40
'PS1SPC
a
v
‒
b
Figure 11.43
‒ j20 Ω
j10 Ω
+
'PS1SPC
0.5Io
40 Ω
Section 11.3
Maximum Average Power Transfer
'PSUIFDJSDVJUTIPXOJO'JH EFUFSNJOFUIF
MPBEJNQFEBODFZLGPSNBYJNVNQPXFSUSBOTGFS
(to ZL $BMDVMBUFUIFNBYJNVNQPXFSBCTPSCFECZ
UIFMPBE
490
Chapter 11
AC Power Analysis
j2 Ω
‒ j3 Ω
4Ω
200 0° V
$BMDVMBUFUIFWBMVFPG;-JOUIFDJSDVJUPG'JH
JOPSEFSGPS;-UPSFDFJWFNBYJNVNBWFSBHFQPXFS
8IBUJTUIFNBYJNVNBWFSBHFQPXFSSFDFJWFECZ;-
+
‒
‒ j10 Ω
5Ω
ZL
30 Ω
ZL
5 90° A
Figure 11.44
'PS1SPC
j20 Ω
40 Ω
5IF5IFWFOJOJNQFEBODFPGBTPVSDFJT;5I =
+ KΩ XIJMFUIFQFBL5IFWFOJOWPMUBHFJT
75I = + K7%FUFSNJOFUIFNBYJNVN
BWBJMBCMFBWFSBHFQPXFSGSPNUIFTPVSDF
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOENBYJNVNBWFSBHFQPXFS
USBOTGFSUPBMPBEZ
Figure 11.48
'PS1SPC
'JOEUIFWBMVFPG;-JOUIFDJSDVJUPG'JHGPS
NBYJNVNQPXFSUSBOTGFS
C
+
‒
R1
40 Ω
vs +
‒
L
R2
40 Ω
Z
j40 Ω
60 0° V
‒ j10 Ω
80 Ω
5 0° A
ZL
Figure 11.45
'PS1SPC
Figure 11.49
*n the circuit of Fig. 11.46, find the value of ;-UIBU
XJMMBCTPSCUIFNBYJNVNQPXFSBOEUIFWBMVFPGUIF
NBYJNVNQPXFS
‒ j1 Ω
1Ω
12 0° V
+
‒
+
Vo
‒
j1 Ω
'PS1SPC
5IFWBSJBCMFSFTJTUPS3JOUIFDJSDVJUPG'JH
JTBEKVTUFEVOUJMJUBCTPSCTUIFNBYJNVNBWFSBHF
QPXFS'JOE3BOEUIFNBYJNVNBWFSBHFQPXFS
BCTPSCFE
‒ j100 Ω
100 Ω
2Vo
ZL
100 Ω
100 0° V +
‒
Figure 11.46
R
j100 Ω
'PS1SPC
For the circuit in Fig. 11.47, find the value of ;-UIBU
XJMMSFDFJWFUIFNBYJNVNQPXFSGSPNUIFDJSDVJU
5IFODBMDVMBUFUIFQPXFSEFMJWFSFEUPUIFMPBE;-
50 sin 4t V +
‒
Figure 11.47
'PS1SPC
'PS1SPC
5IFMPBESFTJTUBODF3-JO'JHJTBEKVTUFEVOUJM
JUBCTPSCTUIFNBYJNVNBWFSBHFQPXFS$BMDVMBUF
UIFWBMVFPG3-BOEUIFNBYJNVNBWFSBHFQPXFS
0.5 vo
2Ω
Figure 11.50
4Ω
+
vo
‒
1
F
20
1H
ZL
4 0° A
Figure 11.51
'PS1SPC
‒ j10 Ω 40 Ω
+
V1
‒
V1/40
j20 Ω
R
491
Problems
"TTVNJOHUIBUUIFMPBEJNQFEBODFJTUPCFQVSFMZ
SFTJTUJWF XIBUMPBETIPVMECFDPOOFDUFEUPUFSNJOBMT
BCPGUIFDJSDVJUTJO'JHTPUIBUUIFNBYJNVN
QPXFSJTUSBOTGFSSFEUPUIFMPBE
f (t)
10
‒ j10 Ω
100 Ω
a
‒1
0
1
2
3
4
5
t
‒10
40 Ω
120 60° V +
‒
'JOEUIFSNTWBMVFPGUIFTJHOBMTIPXOJO'JH
50 Ω
2 90° A
j30 Ω
Figure 11.56
'PS1SPC
b
Figure 11.52
'JOEUIFFGGFDUJWFWBMVFPGUIFWPMUBHFXBWFGPSNJO
'JH
'PS1SPC
Section 11.4
Effective or RMS Value
v(t)
'JOEUIFSNTWBMVFPGUIFPGGTFUTJOFXBWFTIPXOJO
'JH
i(t)
4
20
10
0
π
0
2π
3π
t
2
4
6
8
10
t
Figure 11.57
Figure 11.53
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
students better understand how to find the rms value
PGBXBWFTIBQF
$BMDVMBUFUIFSNTWBMVFPGUIFDVSSFOUXBWFGPSNPG
'JH
'PS1SPC
i(t)
v (t)
Vp
5
0
0
T/3
2T/3
T
4T/3
5
10
15
20
25
t
Figure 11.58
t
'PS1SPC
Figure 11.54
'PS1SPC
%FUFSNJOFUIFSNTWBMVFPGUIFXBWFGPSNJO
'JH
'JOEUIFSNTWBMVFPGUIFWPMUBHFXBWFGPSNPG
Fig. 11.59 as well as the average power absorbed by
BΩSFTJTUPSXIFOUIFWPMUBHFJTBQQMJFEBDSPTTUIF
SFTJTUPS
v(t)
v (t)
5
8
0
‒5
Figure 11.55
'PS1SPC
1
2
3
4
t
0
Figure 11.59
'PS1SPC
2
5
7
10
12
t
492
Chapter 11
AC Power Analysis
$BMDVMBUFUIFFGGFDUJWFWBMVFPGUIFDVSSFOUXBWF
GPSNJO'JHBOEUIFBWFSBHFQPXFSEFMJWFSFE
UPBΩSFTJTUPSXIFOUIFDVSSFOUSVOTUISPVHIUIF
SFTJTUPS
%FUFSNJOFUIFSNTWBMVFGPSUIFXBWFGPSNJO
'JH
i(t)
8A
i(t)
30
0
0
5
10
15
20
25
30 t
‒30
1
2
3
4
5
6
7
8
t
9 10
Figure 11.64
'PS1SPC
'JOEUIFFGGFDUJWFWBMVFPGG U) defined in Fig. 11.65.
Figure 11.60
'PS1SPC
f (t)
$PNQVUFUIFSNTWBMVFPGUIFXBWFGPSNEFQJDUFEJO
'JH
‒1
6
0
1
2
3
4
t
5
Figure 11.65
v (t)
'PS1SPC
2
0
‒1
2
4
6
8
10
t
Figure 11.61
0OFDZDMFPGBQFSJPEJDWPMUBHFXBWFGPSNJTEFQJDUFE
JO'JH'JOEUIFFGGFDUJWFWBMVFPGUIFWPMUBHF
/PUFUIBUUIFDZDMFTUBSUTBUU = BOEFOETBUU = T
v (t)
'PS1SPC
'JOEUIFSNTWBMVFPGUIFTJHOBMTIPXOJO'JH
30
20
v (t)
2
10
0
1
2
3
4
5
t
0
‒4
1
2
3
4
5
6
t
Figure 11.66
'PS1SPC
Figure 11.62
'PS1SPC
0CUBJOUIFSNTWBMVFPGUIFDVSSFOUXBWFGPSNTIPXO
JO'JH
i (t)
$BMDVMBUFUIFSNTWBMVFGPSFBDIPGUIFGPMMPXJOH
GVODUJPOT
B J U = "
C v U = + DPTU7
D J U = −TJOU" E v U = TJOU + DPTU7
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOEIPXUPEFUFSNJOFUIFSNTWBMVFPGUIF
TVNPGNVMUJQMFDVSSFOUT
10t 2
10
Section 11.5
0
Figure 11.63
'PS1SPC
1
2
3
4
5
t
Apparent Power and Power Factor
For the power system in Fig. 11.67, find: (a) the
BWFSBHFQPXFS C UIFSFBDUJWFQPXFS D UIFQPXFS
GBDUPS/PUFUIBU7JTBOSNTWBMVF
493
Problems
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTVOEFSTUBOE
DPNQMFYQPXFS
+
440 V, 60 Hz
‒
124 0° Ω
'JOEUIFDPNQMFYQPXFSEFMJWFSFECZvTUPUIFOFU
XPSLJO'JH-FUvT = DPTU7
20 ‒ j25 Ω
80 + j60 Ω
30 Ω
Figure 11.67
40 μF
20 Ω
ix
'PS1SPC
vs +
‒
"OBDNPUPSXJUIJNQFEBODF;- = + KΩJTTVQ
QMJFECZB7 )[TPVSDF B 'JOEQG 1 BOE
2 C %FUFSNJOFUIFDBQBDJUPSSFRVJSFEUPCFDPO
OFDUFEJOQBSBMMFMXJUIUIFNPUPSTPUIBUUIFQPXFS
GBDUPSJTDPSSFDUFEUPVOJUZ
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOEBQQBSFOUQPXFSBOEQPXFSGBDUPS
0CUBJOUIFQPXFSGBDUPSGPSFBDIPGUIFDJSDVJUTJO
'JH4QFDJGZFBDIQPXFSGBDUPSBTMFBEJOHPS
MBHHJOH
60 mH
+
‒
4ix
Figure 11.69
'PS1SPC
5IFWPMUBHFBDSPTTBMPBEBOEUIFDVSSFOUUISPVHIJU
BSFHJWFOCZ
v U = + DPTU7
J U = −TJOU"
'JOE
B UIFSNTWBMVFTPGUIFWPMUBHFBOEPGUIFDVSSFOU
C UIFBWFSBHFQPXFSEJTTJQBUFEJOUIFMPBE
j5 Ω
4Ω
‒j2 Ω
‒ j2 Ω
'PSUIFGPMMPXJOHWPMUBHFBOEDVSSFOUQIBTPST DBMDVMBUFUIFDPNQMFYQPXFS BQQBSFOUQPXFS SFBM
QPXFS BOESFBDUJWFQPXFS4QFDJGZXIFUIFSUIFQGJT
MFBEJOHPSMBHHJOH
B 7 = ⧸
30° 7SNT * = ⧸
60° "SNT
C 7 = ⧸
−10°7SNT
(a)
‒j1 Ω
4Ω
1Ω
j2 Ω
j1 Ω
(b)
Figure 11.68
For each of the following cases, find the complex
QPXFS UIFBWFSBHFQPXFS BOEUIFSFBDUJWFQPXFS
B v U = TJO U + ž 7
J U = TJO U "
C v U = TJO U+ ž 7
J U = TJO U + ž "
D 7 = ⧸
90°7SNT ; = ⧸
45° Ω
E * = ⧸
60°"SNT ; = ⧸
60°Ω
'PS1SPC
Section 11.6
−25° "SNT
* = ⧸
D 7 = ⧸
0° 7SNT * = ⧸
−15° "SNT
E 7 = ⧸
45° 7SNT * = ⧸
90° "SNT
Complex Power
"7SNT )[TPVSDFJTBQQMJFEUPBMPBE
JNQFEBODF;5IFBQQBSFOUQPXFSFOUFSJOHUIFMPBE
JT7"BUBQPXFSGBDUPSPGMBHHJOH
B $BMDVMBUFUIFDPNQMFYQPXFS
C 'JOEUIFSNTDVSSFOUTVQQMJFEUPUIFMPBE
D %FUFSNJOF;
E "TTVNJOHUIBU; = 3 + Kω-, find the vBMVFTPG
3BOE-
%FUFSNJOFUIFDPNQMFYQPXFSGPSUIFGPMMPXJOH
DBTFT
B 1 = 8 2 = 7"3 DBQBDJUJWF
C 2 = 7"3 QG = MFBEJOH
D 4 = 7" 2 = 7"3 JOEVDUJWF
E VSNT = 7 1 = L8
∣;∣ = Ω JOEVDUJWF
494
Chapter 11
AC Power Analysis
I
'JOEUIFDPNQMFYQPXFSGPSUIFGPMMPXJOHDBTFT
B 1 = L8 QG = MBHHJOH
C 4 = L7" 1 = L8 DBQBDJUJWF
D 7SNT = ⧸
20°7 *SNT = ⧸
−50° "
E 7SNT = ⧸
30°7 ; = + KΩ
0CUBJOUIFPWFSBMMJNQFEBODFGPSUIFGPMMPXJOHDBTFT
B 1 = 8 QG = MFBEJOH
VSNT = 7
C 1 = 8 2 = 7"3 JOEVDUJWF
*SNT = "
D 4 = ⧸
60° 7" 7 = ⧸
45° 7
A
+
120 30° V
B
C
‒
Figure 11.72
'PS1SPC
Section 11.7
Conservation of AC Power
For the network in Fig. 11.73, find the complex
QPXFSBCTPSCFECZFBDIFMFNFOU
'PSUIFFOUJSFDJSDVJUJO'JH DBMDVMBUF
10 Ω
B UIFQPXFSGBDUPS
j10 Ω
C UIFBWFSBHFQPXFSEFMJWFSFECZUIFTPVSDF
‒ j20 Ω
1 0° A
D UIFSFBDUJWFQPXFS
20 Ω
E UIFBQQBSFOUQPXFS
F UIFDPNQMFYQPXFS
Figure 11.73
'PS1SPC
2Ω
‒j5 Ω
50 50° V +
‒
j6 Ω
10 Ω
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEUIFDPOTFSWBUJPOPG"$
QPXFS
‒ jXC
jXL
8Ω
V1 +
‒
Figure 11.70
'PS1SPC
R
+ V
‒ 2
Figure 11.74
*OUIFDJSDVJUPG'JH EFWJDF"SFDFJWFTL8
BUQGMBHHJOH EFWJDF#SFDFJWFTL7"BUQG
MFBEJOH XIJMFEFWJDF$JTJOEVDUJWFBOEDPOTVNFT
1 kW and receives 500 VAR.
'PS1SPC
0CUBJOUIFDPNQMFYQPXFSEFMJWFSFECZUIFTPVSDFJO
UIFDJSDVJUPG'JH
3Ω
B %FUFSNJOFUIFQPXFSGBDUPSPGUIFFOUJSFTZTUFN
C 'JOE*HJWFOUIBU7T = ⧸
45°7SNT
I
5Ω
‒j2 Ω
6Ω
A
+
Vs
2 30° A
j4 Ω
Figure 11.75
B
C
‒
Figure 11.71
'PS1SPC
For the circuit in Fig. 11.76, find the average, SFBDUJWF BOEDPNQMFYQPXFSEFMJWFSFECZUIFEFQFOEFOU
DVSSFOUTPVSDF
'PS1SPC
*OUIFDJSDVJUPG'JH MPBE"SFDFJWFTL7"
BUQGMFBEJOH-PBE#SFDFJWFTL7"BUQG
MBHHJOH#PY$JTBOJOEVDUJWFMPBEUIBUDPOTVNFT
1 kW and receives 500 VAR.
B %FUFSNJOF*
C $BMDVMBUFUIFQPXFSGBDUPSPGUIFDPNCJOBUJPO
4Ω
50 25° V +
‒
Figure 11.76
'PS1SPC
1Ω
‒ j1 Ω
+
Vo
‒
2Ω
j2 Ω
2Vo
495
Problems
0CUBJOUIFDPNQMFYQPXFSEFMJWFSFEUPUIFLΩ
SFTJTUPSJO'JHCFMPX
‒ j3 kΩ
500 Ω Io
30 0° mV +
‒
20Io
j1 kΩ
4 kΩ
10 kΩ
Figure 11.77
'PS1SPC
$BMDVMBUFUIFSFBDUJWFQPXFSJOUIFJOEVDUPSBOE
DBQBDJUPSJOUIFDJSDVJUPG'JH
Given the circuit in Fig. 11.80, find *PBOEUIFPWFSBMM
DPNQMFYQPXFSTVQQMJFE
Io
100 Ω j100 Ω
100 90° V
100 0° mA
100 Ω
+
‒
1.2 kW
0.8 kVAR (cap)
2 kVA
0.707 pf leading
4 kW
0.9 pf lagging
‒ j200 Ω
+
20 0° V ‒
Figure 11.80
'PS1SPC
Figure 11.78
'PS1SPC
For the circuit in Fig. 11.81, find 7T
For the circuit in Fig. 11.79, find 7PBOEUIFJOQVU
QPXFSGBDUPS
0.2 Ω j0.04 Ω
0.3 Ω
j0.15 Ω
+
Vs +
‒
+
6 0° A
20 kW
0.8 pf lagging
Vo
16 kW
0.9 pf lagging
‒
10 W
0.9 pf lagging
15 W
0.8 pf leading
‒
Figure 11.81
'PS1SPC
Figure 11.79
'PS1SPC
'JOE*PJOUIFDJSDVJUPG'JH
Io
440 0° V
+
‒
Figure 11.82
'PS1SPC
120 V rms
16 kW
12 kW
15 kVAR
0.866 pf leading 0.85 pf lagging 0.866 pf lagging
496
Chapter 11
AC Power Analysis
%FUFSNJOF*TJOUIFDJSDVJUPG'JH JGUIF
WPMUBHFTPVSDFTVQQMJFTL8BOEL7"3
MFBEJOH $PNQVUFUIFDPNQMFYQPXFSTVQQMJFECZUIFDVSSFOU
TPVSDFJOUIFTFSJFT3-$DJSDVJUJO'JH
R
8Ω
+
‒
Is
Io cos ωt
120 0° V
'PS1SPC
Figure 11.83
Section 11.8
'PS1SPC
*OUIFPQBNQDJSDVJUPG'JH vT = DPTU7
'JOEUIFBWFSBHFQPXFSEFMJWFSFEUPUIFLΩ
SFTJTUPS
vs +
‒
C
Figure 11.87
j12 Ω
100 kΩ
L
Power Factor Correction
3FGFSUPUIFDJSDVJUTIPXOJO'JH
B 8IBUJTUIFQPXFSGBDUPS
C 8IBUJTUIFBWFSBHFQPXFSEJTTJQBUFE
D 8IBUJTUIFWBMVFPGUIFDBQBDJUBODFUIBUXJMMHJWF
BVOJUZQPXFSGBDUPSXIFODPOOFDUFEUPUIFMPBE
+
‒
1 nF
50 kΩ
440 V rms +
60 Hz ‒
C
Z = 20 + j15 Ω
Figure 11.84
'PS1SPC
Figure 11.88
0CUBJOUIFBWFSBHFQPXFSBCTPSCFECZUIFΩ
SFTJTUPSJOUIFPQBNQDJSDVJUJO'JH
‒j200 kΩ 200 kΩ
‒j50 kΩ 50 kΩ
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFSVOEFS
TUBOEQPXFSGBDUPSDPSSFDUJPO
‒
+
5 0° V +
‒
10 Ω
j10 Ω
Figure 11.85
'PS1SPC
'PSUIFPQBNQDJSDVJUJO'JH DBMDVMBUF
B UIFDPNQMFYQPXFSEFMJWFSFECZUIFWPMUBHFTPVSDF
C UIFBWFSBHFQPXFSEJTTJQBUFEJOUIFΩSFTJTUPS
‒ j150 kΩ 200 kΩ
‒ j30 kΩ 40 kΩ
25 0° V +
‒
Figure 11.86
'PS1SPC
'PS1SPC
5ISFFMPBETBSFDPOOFDUFEJOQBSBMMFMUPB
⧸
0°7SNTTPVSDF-PBEBCTPSCTL7"3
BUQG = MBHHJOH MPBEBCTPSCTL8BOE
L7"3MFBEJOH BOEMPBEBCTPSCTL8BU
QG = B 'JOEUIFFRVJWBMFOUJNQFEBODF
C $BMDVMBUFUIFQPXFSGBDUPSPGUIFQBSBMMFM
DPNCJOBUJPO D %FUFSNJOFUIFDVSSFOUTVQQMJFE
CZUIFTPVSDF
5XPMPBETDPOOFDUFEJOQBSBMMFMESBXBUPUBMPG
L8BUQGMBHHJOHGSPNB7SNT )[
MJOF0OFMPBEBCTPSCTL8BUBQGMBHHJOH
%FUFSNJOF B UIFQGPGUIFTFDPOEMPBE C UIF
QBSBMMFMFMFNFOUSFRVJSFEUPDPSSFDUUIFQGUP
MBHHJOHGPSUIFUXPMPBET
"7SNT)[TVQQMZTFSWFTBMPBEUIBUJT
L8 SFTJTUJWF L7"3 DBQBDJUJWF BOE
L7"3 JOEVDUJWF 'JOE
‒
+
10 Ω
j10 Ω
B UIFBQQBSFOUQPXFS
C UIFDVSSFOUESBXOGSPNUIFTVQQMZ
D UIFL7"3SBUJOHBOEDBQBDJUBODFSFRVJSFEUP
JNQSPWFUIFQPXFSGBDUPSUPMBHHJOH
E UIFDVSSFOUESBXOGSPNUIFTVQQMZVOEFSUIFOFX
QPXFSGBDUPSDPOEJUJPOT
497
Problems
"7SNT)[TPVSDFTVQQMJFTUXPMPBET
DPOOFDUFEJOQBSBMMFM BTTIPXOJO'JH
8IBUJTUIFSFBEJOHPGUIFXBUUNFUFSJOUIFOFUXPSL
PG'JH
B 'JOEUIFQPXFSGBDUPSPGUIFQBSBMMFMDPNCJOBUJPO
6Ω
C $BMDVMBUFUIFWBMVFPGUIFDBQBDJUBODFDPOOFDUFE
JOQBSBMMFMUIBUXJMMSBJTFUIFQPXFSGBDUPSUPVOJUZ
4H
±
±
120 cos 2t V +
‒
Load 1
24 kW
pf = 0.8
lagging
Load 2
40 kW
pf = 0.95
lagging
0.1 F
15 Ω
Figure 11.92
'PS1SPC
'JOEUIFXBUUNFUFSSFBEJOHPGUIFDJSDVJUTIPXOJO
'JH
Figure 11.89
'PS1SPC
10 Ω
$POTJEFSUIFQPXFSTZTUFNTIPXOJO'JH
$BMDVMBUF
5Ω
±
±
170 sin 4t V +
‒
B UIFUPUBMDPNQMFYQPXFS
1H
1
12 F
4Ω
Figure 11.93
C UIFQPXFSGBDUPS
D UIFQBSBMMFMDBQBDJUBODFOFDFTTBSZUPFTUBCMJTIB
VOJUZQPXFSGBDUPS
'PS1SPC
%FUFSNJOFUIFXBUUNFUFSSFBEJOHPGUIFDJSDVJUJO
'JH
440 0° V
10 Ω
+
‒
̶ j 30 Ω
10 Ω
20 Ω io
10 Ω
40 Ω
j10 Ω
±
10 mH
±
10 cos 100t +
‒
Figure 11.90
'PS1SPC
2 io
Figure 11.94
'PS1SPC
Section 11.9
Applications
5IFDJSDVJUPG'JHQPSUSBZTBXBUUNFUFS
DPOOFDUFEJOUPBOBDOFUXPSL
0CUBJOUIFXBUUNFUFSSFBEJOHPGUIFDJSDVJUJO
'JH
B 'JOEUIFNBHOJUVEFPGUIFMPBEDVSSFOU
C $BMDVMBUFUIFXBUUNFUFSSFBEJOH
4 Ω ‒ j3 Ω
12 0° V +
‒
Figure 11.91
'PS1SPC
±
WM
±
j2 Ω
8Ω
3 30° A
240 V +
‒
Figure 11.95
'PS1SPC
│Z L│ = 20 Ω
pf = 0.8
500 μF
498
Chapter 11
AC Power Analysis
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFSVOEFS
TUBOE
IPXUPDPSSFDUQPXFSGBDUPSUPWBMVFTPUIFSUIBOVOJUZ
C $BMDVMBUFUIFDIBSge per kWh with a flBUSBUF
UBSJGGJGUIFSFWFOVFUPUIFVUJMJUZDPNQBOZJTUP
SFNBJOUIFTBNFBTGPSUIFUXPQBSUUBSJGG
"7SNT)[TPVSDFTVQQMJFTBQBSBMMFMDPNCJ
OBUJPOPGBL8IFBUFSBOEBL7"JOEVDUJPO
NPUPSXIPTFQPXFSGBDUPSJT%FUFSNJOF
"SFHVMBSIPVTFIPMETZTUFNPGBTJOHMFQIBTFUISFF
XJSFDJSDVJUBMMPXTUIFPQFSBUJPOPGCPUI7BOE
7 )[BQQMJBODFT5IFIPVTFIPMEDJSDVJUJT
NPEFMFEBTTIPXOJO'JH$BMDVMBUF
B UIFTZTUFNBQQBSFOUQPXFS
C UIFTZTUFNSFBDUJWFQPXFS
B UIFDVSSFOUT* * BOE*O
D UIFL7"SBUJOHPGBDBQBDJUPSSFRVJSFEUPBEKVTU
UIFTZTUFNQPXFSGBDUPSUPMBHHJOH
C UIFUPUBMDPNQMFYQPXFSTVQQMJFE
E UIFWBMVFPGUIFDBQBDJUPSSFRVJSFE
D UIFPWFSBMMQPXFSGBDUPSPGUIFDJSDVJU
0TDJMMPTDPQFNFBTVSFNFOUTJOEJDBUFUIBUUIFQFBL
WPMUBHFBDSPTTBMPBEBOEUIFQFBLDVSSFOUUISPVHI
0°7BOE⧸
25°"
JUBSF SFTQFDUJWFMZ ⧸
%FUFSNJOF
B UIFSFBMQPXFS
I1
120
0° V +
‒
C UIFBQQBSFOUQPXFS
In
120 Ω
Lamp
20 Ω
D UIFSFBDUJWFQPXFS
E UIFQPXFSGBDUPS
"DPOTVNFSIBTBOBOOVBMDPOTVNQUJPOPG
.8IXJUIBNBYJNVNEFNBOEPG.7"
5IFNBYJNVNEFNBOEDIBSHFJTQFSL7"QFS
BOOVN BOEUIFFOFSHZDIBSHFQFSL8IJTDFOUT
B %FUFSNJOFUIFBOOVBMDPTUPGFOFSHZ
120
10 Ω
+
0° V ‒
I2
Kitchen range
Refrigerator
15 mH
Figure 11.96
'PS1SPC
Comprehensive Problems
"USBOTNJUUFSEFMJWFSTNBYJNVNQPXFSUPan BOUFOOB
XIFOUIFBOUFOOBJTBEKVTUFEUPSFQSFTFOUBMPBE
PGΩSFTJTUBODFJOTFSJFTXJUIBOJOEVDUBODFPG
4 μH. If the transmitter operates at 4.12 MHz, find
JUTJOUFSOBMJNQFEBODF
of capacitors is required to operate the turbineHFOFSBUPSCVULFFQJUGSPNCFJOHPWFSMPBEFE
5IFOBNFQMBUFPGBOFMFDUSJDNPUPSIBTUIFGPMMPXJOH
JOGPSNBUJPO
-JOFWPMUBHF7SNT
-JOFDVSSFOU"SNT
-JOFGSFRVFODZ)[
1PXFS8
*OB57USBOTNJUUFS BTFSJFTDJSDVJUIBTBOJNQFEBODF
PGLΩBOEBUPUBMDVSSFOUPGN"*GUIFWPMUBHF
BDSPTTUIFSFTJTUPSJT7 XIBUJTUIFQPXFSGBDUPS
PGUIFDJSDVJU
"DFSUBJOFMFDUSPOJDDJSDVJUJTDPOOFDUFEUPB7
BDMJOF5IFSPPUNFBOTRVBSFWBMVFPGUIFDVSSFOU
ESBXOJT" XJUIBQIBTFBOHMFPGž
B 'JOEUIFUSVFQPXFSESBXOCZUIFDJSDVJU
C $BMDVMBUFUIFBQQBSFOUQPXFS
"OJOEVTUSJBMIFBUFSIBTBOBNFQMBUFUIBUSFBET
7)[L7"QGMBHHJOH
%FUFSNJOF
B UIFBQQBSFOUBOEUIFDPNQMFYQPXFS
C UIFJNQFEBODFPGUIFIFBUFS
"L8UVSCJOFHFOFSBUPSPGQPXFSGBDUPS
PQFSBUFTBUUIFSBUFEMPBE"OBEEJUJPOBMMPBEPG
300 kW at 0.8 power factor is added.What kVAR
%FUFSNJOFUIFQPXFSGBDUPS MBHHJOH PGUIFNPUPS
'JOEUIFWBMVFPGUIFDBQBDJUBODF$UIBUNVTUCF
DPOOFDUFEBDSPTTUIFNPUPSUPSBJTFUIFQGUPVOJUZ
"TTIPXOJO'JH B7GFFEFSMJOFTVQQMJFT
BOJOEVTUSJBMQMBOUDPOTJTUJOHPGBNPUPSESBXJOH
90 kW at 0.8 pf (inductive), a capacitor with a SBUJOH
PGL7"3 BOEMJHIUJOHESBXJOHL8
B $BMDVMBUFUIFUPUBMSFBDUJWFQPXFSBOEBQQBSFOU
QPXFSBCTPSCFECZUIFQMBOU
C %FUFSNJOFUIFPWFSBMMQG
D 'JOEUIFNBHOJUVEFPGUIFDVSSFOUJOUIFGFFEFS
MJOF
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
499
Comprehensive Problems
Amplifier
550 V +
‒
90 kW
pf = 0.8
20 kVAR
10 kW
Coupling capacitor
Speaker
Vin
Figure 11.97
(a)
'PS1SPC
10 Ω
"GBDUPSZIBTUIFGPMMPXJOHGPVSNBKPSMPBET
r "NPUPSSBUFEBUIQ QGMBHHJOH
IQ = L8 r "IFBUFSSBUFEBUL8 QG
r 5FO8MJHIUCVMCT
r "TZODISPOPVTNPUPSSBUFEBUL7"3 QG
MFBEJOH
B $BMDVMBUFUIFUPUBMSFBMBOESFBDUJWFQPXFS
C 'JOEUIFPWFSBMMQPXFSGBDUPS
".7"TVCTUBUJPOPQFSBUFTBUGVMMMPBEBUQPXFS
GBDUPS*UJTEFTJSFEUPJNQSPWFUIFQPXFSGBDUPSUP
CZJOTUBMMJOHDBQBDJUPST"TTVNFUIBUOFXTVCTUBUJPO
BOEEJTUSJCVUJPOGBDJMJUJFTDPTUQFSL7"JOTUBMMFE BOEDBQBDJUPSTDPTUQFSL7"JOTUBMMFE
B $BMDVMBUFUIFDPTUPGDBQBDJUPSTOFFEFE
C 'JOEUIFTBWJOHTJOTVCTUBUJPODBQBDJUZSFMFBTFE
D "SFDBQBDJUPSTFDPOPNJDBMGPSSFMFBTJOHUIF
BNPVOUPGTVCTUBUJPODBQBDJUZ
40 nF
4Ω
vs
80 mH
Amplifier
Speaker
(b)
Figure 11.98
'PS1SPC
A power amplifier has an output impedance of
+ KΩ*UQSPEVDFTBOPMPBEPVUQVUWPMUBHFPG
7BU)[
B %FUFSNJOFUIFJNQFEBODFPGUIFMPBEUIBU
BDIJFWFTNBYJNVNQPXFSUSBOTGFS
C $BMDVMBUFUIFMPBEQPXFSVOEFSUIJTNBUDIJOH
DPOEJUJPO
"QPXFSUSBOTNJTTJPOTZTUFNJTNPEFMFEBTTIPXO
JO'JH*G7T = ⧸
0° rms, find the average
QPXFSBCTPSCFECZUIFMPBE
j1 Ω
0.1 Ω
"DPVQMJOHDBQBDJUPSJTVTFEUPCMPDLEDDVSSFOUGSPN
an amplifier as shown in Fig. 11.98(a). The amplifier
BOEUIFDBQBDJUPSBDUBTUIFTPVSDF XIJMFUIFTQFBLFS
JTUIFMPBEBTJO'JH C B "UXIBUGSFRVFODZJTNBYJNVNQPXFSUSBOTGFSSFE
UPUIFTQFBLFS
C *GVT = 7SNT IPXNVDIQPXFSJTEFMJWFSFE
UPUIFTQFBLFSBUUIBUGSFRVFODZ
100 Ω
Vs +
‒
Source
Figure 11.99
'PS1SPC
j20 Ω
j1 Ω
0.1 Ω
Line
Load
c h a p t e r
Three-Phase Circuits
12
)FXIPDBOOPUGPSHJWFPUIFSTCSFBLTUIFCSJEHFPWFSXIJDIIFNVTUQBTT
IJNTFMG
‡()FSCFSU
Enhancing Your Skills and Your Career
ABET EC 2000 criteria (3.e), “an ability to identify, formulate,
and solve engineering problems.”
%FWFMPQJOHBOEFOIBODJOHZPVSiBCJMJUZUPJEFOUJGZ
GPSNVMBUF BOE
TPMWFFOHJOFFSJOHQSPCMFNTuJTBQSJNBSZGPDVTPGUF YUCPPL' PMMPX
JOHPVSTJYTUFQQSPCMFNTPMWJOHQSPDFTTJTUIFCFTUX BZUPQSBDUJDF UIJTTLJMM0VSSFDPNNFOEBUJPOJTUIBUZPVVTFUIJTQSPDFTTXIFOF WFS
QPTTJCMF:PVNBZCFQMFBTFEUPMFBSOUIBUUIJTQSPDFTTX PSLTXFMMGPS OPOFOHJOFFSJOHDPVSTFT
ABET EC 2000 criteria (f), “an understanding of professional
and ethical responsibility.”
i"OVOEFSTUBOEJOHPGQSPGFTTJPOBMBOEFUIJDBMSFTQPOTJCJMJUZuJTSFRVJSFE
PGFWFSZFOHJOFFS5PTPNFF YUFOU UIJTVOEFSTUBOEJOHJTW FSZQFSTPOBM
GPSFBDIPGVT-FUVTJEFOUJGZTPNFQPJOUFSTUPIFMQZPVEF WFMPQUIJT
VOEFSTUBOEJOH0OFPGNZG BWPSJUFFYBNQMFTJTUIBUBOFOHJOFFSIBTUIF
SFTQPOTJCJMJUZUPBOTXFSXIBU*DBMMUIFiVOBTLFERVFTUJPOu'PSJOTUBODF BTTVNFUIBUZPVPXOBDBSUIBUIBTBQSPCMFNXJUIUIFUSBOTNJTTJPO*O
UIFQSPDFTTPGTFMMJOHUIBUDBS UIFQSPTQFDUJWFCVZFSBTLTZPVJGUIFSFJT
BQSPCMFNJOUIFSJHIUGSPOUXIFFMCFBSJOH:PVBOTXFSOP)PXFWFS BT
BOFOHJOFFS ZPVBSFSFRVJSFEUPJOGPSNUIFCVZFSUIBUUIFSFJTBQSPCMFN
XJUIUIFUSBOTNJTTJPOXJUIPVUCFJOHBTLFE
:PVSSFTQPOTJCJMJUZCPUIQSPGFTTJPOBMMZBOEFUIJDBMMZJTUPQFSGPSN
in BNBOOFSUIBUEPFTOPUIBSNUIPTF BSPVOEZPVBOEUPXIPNZPV BSF
SFTQPOTJCMF$MFBSMZ EFWFMPQJOHUIJTDBQBCJMJUZXJMMUBL FUJNFBOENB UVSJUZPOZPVSQBSU*SFDPNNFOEQSBDUJDJOHUIJTCZMPPLJOHGPSQSPGFT TJPOBMBOEFUIJDBMDPNQPOFOUTJOZPVSEBZUPEBZBDUJWJUJFT
1IPUPCZ$IBSMFT"MFYBOEFS
501
502
Chapter 12
Three-Phase Circuits
Learning Objectives
#ZVTJOHUIFJOGPSNBUJPOBOEFYFSDJTFTJOUIJTDIBQUFSZPVXJMMCF
BCMFUP
6OEFSTUBOECBMBODFEUISFFQIBTFWPMUBHFT
"OBMZ[FCBMBODFEXZFXZFDJSDVJUT
6OEFSTUBOEBOEBOBMZ[FCBMBODFEXZFEFMUBDJSDVJUT
"OBMZ[FCBMBODFEEFMUBEFMUBDJSDVJUT
6OEFSTUBOEBOEBOBMZ[FCBMBODFEEFMUBXZFDJSDVJUT
&YQMBJOBOEBOBMZ[FQPXFSJOCBMBODFEUISFFQIBTFDJSDVJUT
"OBMZ[FVOCBMBODFEUISFFQIBTFDJSDVJUT
12.1
Introduction
4PGBSJOUIJTUFYU XFIBWFEFBMUXJUITJOHMFQIBTFDJSDVJUT"TJOHMF
QIBTFBDQPXFSTZTUFNDPOTJTUTPGBHFOFSBUPSDPOOFDUFEUISPVHIBQBJS
PGXJSFT BUSBOTNJTTJPOMJOF UPBMPBE'JHVSF B EFQJDUTBTJOHMF
QIBTFUXPXJSFTZTUFN XIFSF 7QJTUIFSNTNBHOJUVEFPGUIFTPVSDF
WPMUBHFBOEϕJTUIFQIBTF8IBUJTNPSFDPNNPOJOQSBDUJDFJTBTJOHMF
QIBTFUISFFXJSFTZTUFN TIPXOJO'JH C *UDPOUBJOTUXPJEFOUJDBM
TPVSDFT FRVBMNBHOJUVEFBOEUIFTBNFQIBTF UIBUBSFDPOOFDUFEUPUXP
MPBETCZUXPPVUFSXJSFTBOEUIFOFVUSBM'PSFYBNQMF UIFOPSNBMIPVTF
IPMETZTUFNJTBTJOHMFQIBTFUISFFXJSFTZTUFNCFDBVTFUIFUFSNJOBM
WPMUBHFTIBWFUIFTBNFNBHOJUVEFBOEUIFTBNFQIBTF4VDIBTZTUFN
BMMPXTUIFDPOOFDUJPOPGCPUIBOE7BQQMJBODFT
Historical note: Thomas Edison
invented a three-wire system, using
three wires instead of four.
Vp ϕ +
‒
Vp ϕ +
‒
ZL
Vp ϕ +
‒
a
A
n
N
b
B
ZL1
ZL2
(b)
(a)
Figure 12.1
4JOHMFQIBTFTZTUFNT B UXPXJSFUZQF C UISFFXJSFUZQF
Vp 0° +
‒
Vp ‒90° +
‒
Figure 12.2
a
A
n
N
b
B
5XPQIBTFUISFFXJSFTZTUFN
ZL1
ZL2
$JSDVJUTPSTZTUFNTJOXIJDIUIFBDTPVSDFTPQFSBUFBUUIFTBNFGSF RVFODZCVUEJGGFSFOUQIBTFTBSFLOPXOBTQPMZQIBTF'JHVSFTIPXT
BUXPQIBTFUISFFXJSFTZTUFN BOE'JHTIP XTBUISFFQIBTFGPVS
XJSFTZTUFN"TEJTUJODUGSPNBTJOHMFQIBTFTZTUFN BUXPQIBTFTZTUFN
JTQSPEVDFECZBHFOFSBUPSDPOTJTUJOHPGUX PDPJMTQMBDFEQFSQFOEJDVMBS
UPFBDIPUIFSTPUIBUUIFWPMUBHFHFOFSBUFECZPOFMBHTUIFPUIFSCZž
#ZUIFTBNFUPLFO BUISFFQIBTFTZTUFNJTQSPEVDFECZBHFOFSBUPSDPO
TJTUJOHPGUISFFTPVSDFTIBWJOHUIFTBNFBNQMJUVEFBOEGSFRVFODZCVUPVU
PGQIBTFXJUIFBDIPUIFSCZž#FDBVTFUIFUISFFQIBTFTZTUFNJTCZ
GBSUIFNPTUQSF WBMFOUBOENPTUFDPOPNJDBMQPMZQIBTFTZTUFN EJTDVT TJPOJOUIJTDIBQUFSJTNBJOMZPOUISFFQIBTFTZTUFNT
5ISFFQIBTFTZTUFNTBSFJNQPSUBOUGPSBUMFBTUUISFFSFBTPOT'JSTU OFBSMZBMMFMFDUSJDQP XFSJTHFOFSBUFEBOEEJTUSJC VUFEJOUISFFQIBTF 12.2
503
Balanced Three-Phase Voltages
Historical
Nikola Tesla m XBTB$SPBUJBO"NFSJDBOFOHJOFFSXIPTF
inventions—among them the induction motor and the first polyphase
ac power system—greatly influenced the settlement of the ac versus dc
EFCBUFJOGBWPSPGBD)FXBTBMTPSFTQPOTJCMFGPSUIFBEPQUJPOPG)[
BTUIFTUBOEBSEGPSBDQPXFSTZTUFNTJOUIF6OJUFE4UBUFT
#PSOJO"VTUSJB)VOHBSZ OPX$SPBUJB UPBDMFSHZNBO 5FTMBIBE
an incredible memory and a keen affinity for mathematics. He moved
to the United States in 1884 and first worked for Thomas Edison. At
UIBUUJNF UIFDPVOUSZXBTJOUIFiCBUUMFPGUIFDVSSFOUTuXJUI(FPSHF
8FTUJOHIPVTF m QSPNPUJOHBDBOE5IPNBT&EJTPOSJHJEMZ
MFBEJOHUIFEDGPSDFT5FTMBMFGU&EJTPOBOEKPJOFE8FTUJOHIPVTFCF DBVTF PG IJT JOUFSFTU JO BD 5ISPVHI8FTUJOHIPVTF 5FTMB HBJOFE UIF
SFQVUBUJPOBOEBDDFQUBODFPGIJTQPMZQIBTFBDHFOFSBUJPO USBOTNJTTJPO BOEEJTUSJCVUJPOTZTUFN)FIFMEQBUFOUTJOIJTMJGFUJNF)JTPUIFS
JOWFOUJPOTJODMVEFIJHIWPMUBHFBQQBSBUVT UIFUFTMBDPJM BOEBXJSFMFTT
transmission system. The unit of magnetic flux density, the tesla, was
OBNFEJOIPOPSPGIJN
BUUIFPQFSBUJOHGSFRVFOD ZPG)[ PS ω = SBET JOUIF6OJUFE
4UBUFTPS)[ PS ω = SBET JOTPNFPUIFSQBSUTPGUIFX PSME
8IFOPOFQIBTFPSUX PQIBTFJOQVUTBSFSFRVJSFE UIF ZBSFUBLFOGSPN
UIFUISFFQIBTFTZTUFNSBUIFSUIBOHFOFSBUFEJOEFQFOEFOUMZ&WFOXIFO
NPSFUIBOUISFFQIBTFTBSFOFFEFE‡TVDIBTJOUIFBMVNJOVNJOEVTUSZ XIFSFQIBTFTBSFSFRVJSFEGPSNFMUJOHQVSQPTFT‡UIF ZDBOCFQSP WJEFECZNBOJQVMBUJOHUIFUISFFQIBTFTTVQQMJFE4FDPOE UIFJOTUBOUB OFPVTQPXFSJOBUISFFQIBTFTZTUFNDBOCFDPOTUBOU OPUQVMTBUJOH BT
XFXJMMTFFJO4FDUJPO5IJTSFTVMUTJOVOJGPSNQPXFSUSBOTNJTTJPO
BOEMFTTWJCSBUJPOPGUISFFQIBTFNBDIJOFT5IJSE GPSUIFTBNFBNPVOU
PGQPXFS UIF UISFFQIBTFTZTUFNJTNPSF FDPOPNJDBMUIBOUIFTJOHMF
QIBTF5IFBNPVOUPGXJSFSFRVJSFEGPSBUISFFQIBTFTZTUFNJTMFTTUIBO
UIBUSFRVJSFEGPSBOFRVJWBMFOUTJOHMFQIBTFTZTUFN
8FCFHJOXJUIBEJTDVTTJPOPGCBMBODFEUISFFQIBTFW PMUBHFT5IFO
we analyze each of the four possible configurations of balanced threeQIBTFTZTUFNT 8FBMTPEJTDVTTUIFBOBMZTJTPGVOCBMBODFEUISFFQIBTF
TZTUFNT8FMFBSOIPXUPVTF14QJDFGPS8JOEPXTUPBOBMZ[FBCBMBODFE
PSVOCBMBODFEUISFFQIBTFTZTUFN'JOBMMZ XFBQQMZUIFDPODFQUTEFWFM
PQFEJOUIJTDIBQUFSUPUISFFQIBTFQP XFSNFBTVSFNFOUBOESFTJEFOUJBM
FMFDUSJDBMXJSJOH
12.2
Balanced Three-Phase Voltages
5ISFFQIBTFWPMUBHFTBSFPGUFOQSPEVDFEXJUIBUISFFQIBTFBDHFOFSB UPS PSBMUFSOBUPS XIPTFDSPTTTFDUJPOBMWJFXJTTIPXOJO'JH5IF
HFOFSBUPSCBTJDBMMZDPOTJTUTPGBSPUBUJOHNBHOFU DBMMFEUIF SPUPS TVS
SPVOEFECZBTUBUJPOBSZXJOEJOH DBMMFEUIFTUBUPS 5ISFFTFQBSBUFXJOE
JOHTPSDPJMTXJUIUFSNJOBMT BB′ CC′ BOE DD′BSFQIZTJDBMMZQMBDFE
žBQBSUBSPVOEUIFTUBUPS5FSNJOBMT BBOEB′ GPSFYBNQMF TUBOEGPS
POFPGUIFFOETPGDPJMTHPJOHJOUPBOEUIFPUIFSFOEDPNJOHPVUPGUIF
-JCSBSZPG$POHSFTT
<-$64;>
Vp 0°
‒+
Vp ‒120°
‒+
Vp +120°
‒+
a
A
ZL1
b
B
ZL2
c
C
ZL3
n
N
Figure 12.3
5ISFFQIBTFGPVSXJSFTZTUFN
504
Chapter 12
Three-Phase Circuits
a
Threephase b
output
c
b′
c
N
a′
Stator
Rotor
a
S
b
c′
n
Figure 12.4
"UISFFQIBTFHFOFSBUPS
0
120°
ωt
240°
Figure 12.5
5IFHFOFSBUFEWPMUBHFTBSFžBQBSU
GSPNFBDIPUIFS
page. As the rotor rotates, its magnetic field “cuts” the flux from the three
DPJMTBOEJOEVDFTWPMUBHFTJOUIFDPJMT#FDBVTFUIFDPJMTBSFQMBDFEž
BQBSU UIFJOEVDFEWPMUBHFTJOUIFDPJMTBSFFRVBMJONBHOJUVEFCVUPVUPG
QIBTFCZž 'JH 4JODFFBDIDPJMDBOCFSFHBSEFEBTBTJOHMF
QIBTFHFOFSBUPSCZJUTFMG UIFUISFFQIBTFHFOFSBUPSDBOTVQQMZQPXFSUP
CPUITJOHMFQIBTFBOEUISFFQIBTFMPBET
"UZQJDBMUISFFQIBTFTZTUFNDPOTJTUTPGUISFFWPMUBHFTPVSDFTDPO
OFDUFEUPMPBETCZUISFFPSGPVSXJSFT PSUSBOTNJTTJPOMJOFT 5ISFF
QIBTFDVSSFOUTPVSDFTBSFWFSZTDBSDF "UISFFQIBTFTZTUFNJTFRVJWBMFOU
UPUISFFTJOHMFQIBTFDJSDVJUT 5IFW PMUBHFTPVSDFTDBOCFFJUIFSXZF
DPOOFDUFEBTTIPXOJO'JH B PSEFMUBDPOOFDUFEBTJOFig. C a
+ V
an
‒
+
‒
+
‒
Vcn
n
Vbn
b
a
Vca
+
‒
Van(t) Vbn(t) Vcn(t)
+ Vab
‒
‒+
b
Vbc
c
(a)
c
(b)
Figure 12.6
5ISFFQIBTFWPMUBHFTPVSDFT B :DPOOFDUFETPVSDF C ∆DPOOFDUFE
TPVSDF
-FUVTDPOTJEFSUIFXZFDPOOFDUFEWPMUBHFTJO'JH B GPSOPX
5IFWPMUBHFT7BO 7CO BOE7DOBSFSFTQFDUJWFMZCFUXFFOMJOFT B C BOE
D BOEUIFOFVUSBMMJOFO5IFTFWPMUBHFTBSFDBMMFEQIBTFWPMUBHFT*GUIF
WPMUBHFTPVSDFTIBWFUIFTBNFBNQMJUVEFBOEGSFRVFODZωBOEBSFPVUPG
QIBTFXJUIFBDIPUIFSCZž UIFWPMUBHFTBSFTBJEUPCFCBMBODFE5IJT
JNQMJFTUIBU
As a common tradition in power
systems, voltage and current in this
chapter are in rms values unless
otherwise stated.
7BO + 7CO + 7DO = ]7BO] = ]7CO] = ]7DO]
5IVT
Balanced phase voltages are equal in magnitude and are out of phase
with each other by 120°.
12.2
505
Balanced Three-Phase Voltages
#FDBVTFUIFUISFFQIBTFWPMUBHFTBSFžPVU PGQIBTFXJUIFBDI
PUIFS UIFSFBSFUX PQPTTJCMFDPNCJOBUJPOT0OFQPTTJCJMJUZJTTIP XOJO
'JH B BOEFYQSFTTFENBUIFNBUJDBMMZBT
7BO = 7Q⧸
0ž
7CO = 7Q⧸
−ž
7DO = 7Q⧸
−ž = 7Q⧸
+ž
Vcn
ω
120°
120°
Van
‒120°
Vbn
XIFSF 7QJTUIFFGGFDUJWFPSSNTWBMVFPGUIFQIBTFWPMUBHFT5IJTJT
LOPXOBTUIFBCDTFRVFODFPSQPTJUJWFTFRVFODF*OUIJTQIBTFTFRVFODF 7BOMFBET7CO XIJDIJOUVSOMFBET 7DO5IJTTFRVFODFJTQSPEVDFEXIFO
UIFSPUPSJO'JHSPUBUFTDPVOUFSDMPDLXJTF5IFPUIFSQPTTJCJMJUZJT
TIPXOJO'JH C BOEJTHJWFOCZ
7BO = 7Q⧸
ž
7DO = 7Q⧸
−ž
7CO = 7Q⧸
−ž = 7Q⧸
+ž
Vbn
ω
120°
120°
‒120°
5IJTJTDBMMFEUIFBDCTFRVFODFPSOFHBUJWFTFRVFODF'PSUIJTQIBTFTF
RVFODF 7BOMFBET7DO XIJDIJOUVSOMFBET7CO5IFBDCTFRVFODFJTQSP
EVDFEXIFOUIFSPUPSJO'JHSPUBUFTJOUIFDMPDLXJTFEJSFDUJPO*UJT
easy to show that the voltages in Eqs. (12.3) or (12.4) satisfy Eqs. BOE 'PSFYBNQMF GSPN&R (a)
Vcn
Van
(b)
Figure 12.7
1IBTFTFRVFODFT B BCDPSQPTJUJWF
TFRVFODF C BDCPSOFHBUJWFTFRVFODF
7BO + 7CO + 7DO = 7Q⧸
ž + 7Q⧸
−ž + 7Q⧸
+ž
= 7Q −−K− + K =
The phase sequence is the time order in which the voltages pass through
their respective maximum values.
5IFQIBTFTFRVFODFJTEFUFSNJOFECZUIFPSEFSJOXIJDIUIFQIBTPSTQBTT
through a fixFEQPJOUJOUIFQIBTFEJBHSBN
*O'JH B BTUIFQIBTPSTSPUBUFJOUIFDPVOUFSDMPDLXJTF EJSFD
UJPOXJUIGSFRVFOD Z ω UIF ZQBTTUISPVHIUIFIPSJ[POUBMBYJTJOBTF RVFODFBCDBCDB5IVT UIFTFRVFODFJT BCDPSCDBPSDBC4JNJMBSMZ GPSUIFQIBTPSTJO'JH C BTUIF ZSPUBUFJOUIFDPVOUFSDMPDLXJTF
EJSFDUJPO UIFZQBTTUIFIPSJ[POUBMBYJTJOBTFRVFODF BDCBDCB 5IJT
EFTDSJCFTUIF BDCTFRVFODF 5IFQIBTFTFRVFODFJTJNQPSUBOUJOUISFF
QIBTFQPXFSEJTUSJCVUJPO*UEFUFSNJOFTUIFEJSFDUJPOPGUIFSPUBUJPOPGB
NPUPSDPOOFDUFEUPUIFQPXFSTPVSDF GPSFYBNQMF
-JLFUIFHFOFSBUPSDPOOFDUJPOT BUISFFQIBTFMPBEDBOCFFJUIFS
XZFDPOOFDUFEPSEFMUBDPOOFDUFE EFQFOEJOHPOUIFFOEBQQMJDBUJPO
'JHVSF B TIP XTBXZFDPOOFDUFEMPBE BOE'JH C TIP XTB
EFMUBDPOOFDUFEMPBE5IFOFVUSBMMJOFJO'JH B NBZPSNBZOPUCF
UIFSF EFQFOEJOHPOXIFUIFSUIFTZTUFNJTGPVS PSUISFFXJSF "OE PG
DPVSTF BOFVUSBMDPOOFDUJPOJTUPQPMPHJDBMMZJNQPTTJCMFGPSBEFMUBDPO OFDUJPO "XZFPSEFMUBDPOOFDUFEMPBEJTTBJEUPCFVOCBMBODFEJGUIF
QIBTFJNQFEBODFTBSFOPUFRVBMJONBHOJUVEFPSQIBTF
The phase sequence may also be
regarded as the order in which the
phase voltages reach their peak (or
maximum) values with respect to time.
Reminder: As time increases, each
phasor (or sinor) rotates at an angular
velocity ω.
506
Chapter 12
a
b
Three-Phase Circuits
A balanced load is one in which the phase impedances are equal in
magnitude and in phase.
Z2
Z1
n
'PSBCBMBODFEXZFDPOOFDUFEMPBE
;1 = ;2 = ;3 = ;:
Z3
c
XIFSF;:JTUIFMPBEJNQFEBODFQFSQIBTF'PSB
CBMBODFEEFMUBDPOOFDUFE
MPBE
(a)
a
;B = ;C = ;D = ;∆
Zb
Zc
b
Za
c
(b)
Figure 12.8
5Xo possible three-phase load configuraUJPOT B B:DPOOFDUFEMPBE C B
∆DPOOFDUFEMPBE
Reminder: A Y-connected load consists
of three impedances connected to a
neutral node, while a ∆-connected
load consists of three impedances
connected around a loop. The load is
balanced when the three impedances
are equal in either case.
Example 12.1
XIFSF;∆JTUIFMPBEJNQFEBODFQFSQIBTFJOUIJTDBTF8FSFDBMMGSPN
&R UIBU
; ;: = @@
∆
TPXFLOPXUIBUBXZFDPOOFDUFEMPBEDBOCFUSBOTGPSNFEJOUPBEFMUB
DPOOFDUFEMPBE PSWJDFWFSTB VTJOH&R #FDBVTFCPUIUIFUISFFQIBTFTPVSDFBOEUIFUISFFQIBTFMPBEDBO
CFFJUIFSXZFPSEFMUBDPOOFDUFE XFIBWFGPVSQPTTJCMFDPOOFDUJPOT
;∆ = ;:
PS
r ::DPOOFDUJPO JF :
DPOOFDUFETPVSDFXJUIB:DPOOFDUFEMPBE r :∆DPOOFDUJPO
r ∆∆DPOOFDUJPO
r ∆:DPOOFDUJPO
*OTVCTFRVFOUTFDUJPOT XFXJMMDPOTJEFSFBDI PGUIFTFQPTTJCMFDPO figurations.
*UJTBQQSPQSJBUFUPNFOUJPOIFSFUIBUBCBMBODFEEFMUBDPOOFDUFEMPBEJT
NPSFDPNNPOUIBOBCBMBODFEXZFDPOOFDUFEMPBE5IJTJTEVFUPUIFFBTF
XJUIXIJDIMPBETNBZCFBEEFEPSSFNP WFEGSPNFBDIQIBTFPGBEFMUB
DPOOFDUFEMPBE5IJTJTWFSZEJGficult with a wye-connected load because the
OFVUSBMNBZOPUCFBDDFTTJCMF0OUIFPUIFSIBOE EFMUBDPOOFDUFETPVSDFT BSFOPUDPNNPOJOQSBDUJDFCFDBVTFPGUIFDJSDVMBUJOHDVSSFOUUIBUXJMMSFTVMU
JOUIFEFMUBNFTIJGUIFUISFFQIBTFWPMUBHFTBSFTMJHIUMZVOCBMBODFE
%FUFSNJOFUIFQIBTFTFRVFODFPGUIFTFUPGWPMUBHFT
vBO = DPT ωU + ž
vCO = DPT ωU−ž vDO = DPT ωU−ž
Solution:
5IFWPMUBHFTDBOCFFYQSFTTFEJOQIBTPSGPSNBT
7BO = ⧸
10ž7 7CO = ⧸
−ž7 7DO = ⧸
−ž7
8FOPUJDFUIBU7BOMFBET7DOCZžBOE7DOJOUVSOMFBET7COCZž
)FODF XFIBWFBOBDCTFRVFODF
Practice Problem 12.1
(JWFOUIBU7CO = ⧸
ž V, find 7BOBOE7DO BTTVNJOHBQPTJUJWF BCD TFRVFODF
Answer:⧸
150ž7 ⧸
−ž7
12.3
12.3
507
Balanced Wye-Wye Connection
Balanced Wye-Wye Connection
8FCFHJOXJUIUIF::TZTUFN CFDBVTFBOZCBMBODFEUISFFQIBTFTZT UFNDBOCFSFEVDFEUPBOFRVJWBMFOU::TZTUFN5IFSFGPSF BOBMZTJTPG
UIJTTZTUFNTIPVMECFSFHBSEFEBTUIFLFZUPTPMWJOHBMMCBMBODFEUISFF
QIBTFTZTUFNT
A balanced Y-Y system is a three-phase system with a balanced
Y-connected source and a balanced Y-connected load.
$POTJEFSUIFCBMBODFEGPVSXJSF::TZTUFNPG'JH XIFSFB
:DPOOFDUFEMPBEJTDPOOFDUFEUPB :DPOOFDUFETPVSDF8FBTTVNFB CBMBODFEMPBETPUIBUMPBEJNQFEBODFTBSFFRVBM"MUIPVHIUIFJNQFE
BODF;:JTUIFUPUBMMPBEJNQFEBODFQFSQIBTF JUNBZBMTPCFSF HBSEFE
BTUIFTVNPGUIFTPVSDFJNQFEBODF ;T MJOFJNQFEBODF ;ê BOEMPBE
JNQFEBODF ;-GPSFBDIQIBTF TJODFUIFTFJNQFEBODFTBSFJOTFSJFT "TJMMVTUSBUFEJO'JH ;TEFOPUFTUIFJOUFSOBMJNQFEBODFPGUIF QIBTFXJOEJOHPGUIFHFOFSBUPS ;êJTUIFJNQFEBODFPGUIFMJOFKPJO JOHBQIBTFPGUIFTPVSDFXJUIBQIBTFPGUIFMPBE;-JTUIFJNQFEBODF
PGFBDIQIBTFPGUIFMPBEBOE ;OJTUIFJNQFEBODFPGUIFOFVUSBMMJOF
5IVT JOHFOFSBM
;: = ;T + ;ê + ;-
Zl
a
A
Zs
ZL
+
‒
Van
Zn
n
N
‒ Vbn
+
Vcn +‒
ZL
Zs
Zs
c
b
ZL
C
B
Zl
Zl
Figure 12.9
"CBMBODFE::TZTUFN TIPXJOHUIFTPVSDF MJOF BOEMPBEJNQFEBODFT
;TBOE;êBSFPGUFOWFSZTNBMMDPNQBSFEXJUI;- TPPOFDBOBTTVNFUIBU
;: = ;-JGOPTPVSDFPSMJOFJNQFEBODFJTHJWFO*OBOZFWFOU CZMVNQ JOHUIFJNQFEBODFTUPHFUIFS UIF::TZTUFNJO'JHDBOCFTJNQMJ
fied to that shown in Fig. 12.10.
"TTVNJOHUIFQPTJUJ WFTFRVFODF UIF QIBTFW PMUBHFT PSMJOFUP OFVUSBMWPMUBHFT BSF
7BO = 7Q⧸
ž
7CO = 7Q⧸
−ž 7DO = 7Q⧸
+ž
Ia
a
Van
Vcn
+
‒
n
‒
+
ZY
In
N
‒ V
bn
+
c
Ib
Ic
A
ZY
ZY
C
B
b
Figure 12.10
#BMBODFE::DPOOFDUJPO
508
Chapter 12
Three-Phase Circuits
5IFMJOFUPMJOFWPMUBHFTPSTJNQMZ MJOFWPMUBHFT 7BC 7CD BOE 7DBBSF
SFMBUFEUPUIFQIBTFWPMUBHFT'PSFYBNQMF
7BC = 7BO + 7OC = 7BO−7CO = 7Q⧸
ž−7Q⧸
−ž
@@
@@
Ŀ
+ K@@@
7Q⧸
= 7Q + @@
= Ŀ
3ž
4JNJMBSMZ XFDBOPCUBJO
@@
B
7Q⧸
7CD = 7CO−7DO = Ŀ
−ž
C
7Q⧸
7DB = 7DO−7BO = Ŀ
−ž
D
@@
@@
UJNFTUIFNBHOJUVEFPG
5IVT UIFNBHOJUVEFPGUIFMJOFWPMUBHFT7-JTĿ
UIFQIBTFWPMUBHFT7Q PS
@@
7Q
7- = Ŀ
7Q = ]7BO] = ]7CO] = ]7DO]
7- = ]7BC] = ]7CD] = ]7DB]
XIFSF
BOE
Vcn
Vab = Van + Vnb
Vnb
30°
Van
"MTPUIFMJOFWPMUBHFTMFBEUIFJSDPSSFTQPOEJOHQIBTFWPMUBHFTCZž
'JHVSF B JMMVTUSBUFTUIJT'JHVSF B BMTPTIPXTIPXUPEFUFS
NJOF7BCGSPNUIFQIBTFWPMUBHFT XIJMF'JH C TIPXTUIFTBNF
GPSUIFUISFFMJOFWPMUBHFT/PUJDFUIBU 7BCMFBET 7CDCZž BOE 7CD
MFBET7DBCZž TPUIBUUIFMJOFWPMUBHFTTVNVQUP[FSPBTEPUIFQIBTF
WPMUBHFT
"QQMZJOH,7-UPFBDIQIBTFJO'JH XFPCUBJOUIFMJOFDVS SFOUTBT
−ž
7BO⧸
7CO @@@@@@@@@@
7BO
*C = @@@
=
= *B⧸
−ž
*B = @@@
;:
;:
;:
−ž
7BO⧸
7DO @@@@@@@@@@
*D = @@@
=
= *B⧸
−ž
;:
;:
Vbn
8FDBOSFBEJMZJOGFSUIBUUIFMJOFDVSSFOUTBEEVQUP[FSP
(a)
Vca
Vcn
Vab
*B +*C + *D = *O = − *B + *C + *D = B
7O/ = ;O*O = C
TPUIBU
Van
Vbn
Vbc
(b)
Figure 12.11
PS
1IBTPSEJBHSBNTJMMVTUSBUJOHUIFSFMBUJPO
TIJQCFUXFFOMJOFWPMUBHFTBOEQIBTF
WPMUBHFT
UIBUJT UIFWPMUBHFBDSPTTUIFOFVUSBMXJSFJT[FSP5IFOFVUSBMMJOFDBO
UIVTCFSFNPWFEXJUIPVUBGGFDUJOHUIFTZTUFN*OGBDU JOMPOHEJTUBODF
QPXFSUSBOTNJTTJPO DPOEVDUPSTJONVMUJQMFTPGUISFFBSFVTFEXJUIUIF
FBSUIJUTFMGBDUJOHBTUIFOFVUSBMDPOEVDUPS1PXFSTZTUFNTEFTJHOFEJO
UIJTXBZBSFXFMMHSPVOEFEBUBMMDSJUJDBMQPJOUTUPFOTVSFTBGFUZ
8IJMF UIF MJOF DVSSFOU JT UIF DVSSFOUJO FBDI MJOF UIF QIBTFDVSSFOU JTUIFDVSSFOUJOFBDIQIBTFPGUIFTPVSDFPSMPBE*OUIF ::TZTUFN UIF MJOFDVSSFOUJTUIFTBNFBTUIFQIBTFDVSSFOU 8FXJMMVTFTJOHMFTVCTDSJQUT 12.3
509
Balanced Wye-Wye Connection
GPSMJOFDVSSFOUTCFDBVTFJUJTOBUVSBMBOEDPO WFOUJPOBMUPBTTVNFUIBUMJOF currents floXGSPNUIFTPVSDFUPUIFMPBE
"OBMUFSOBUJWFXBZPGBOBMZ[JOHBCBMBODFE::TZTUFNJTUPEPTP
POBiQFSQIBTFuCBTJT 8FMPPLBUPOFQIBTF TBZQIBTF B BOEBOBMZ[F
UIFTJOHMFQIBTFFRVJWBMFOUDJSDVJUJO'JH5IFTJOHMFQIBTFBOBMZ
TJTZJFMETUIFMJOFDVSSFOU*BBT
Ia
a
A
Van +
‒
ZY
n
N
Figure 12.12
"TJOHMFQIBTFFRVJWBMFOUDJSDVJU
7BO
*B = @@@
;: 'SPN*B XFVTFUIFQIBTFTFRVFODFUPPCUBJOPUIFSMJOFDVSSFOUT5IVT BTMPOHBTUIFTZTUFNJTCBMBODFE XFOFFEPOMZBOBMZ[FPOFQIBTF
We NBZEPUIJTFWFOJGUIFOFVUSBMMJOFJTBCTFOU BTJOUIFUISFFXJSF TZTUFN
$BMDVMBUFUIFMJOFDVSSFOUTJOUIFUISFFXJSF::TZTUFNPG'JH
5 ‒ j2 Ω
a
+
‒
A
110 0° V
10 + j8 Ω
110 ‒240° V +‒
c
‒ 110 ‒120° V
+
5 ‒ j2 Ω
b
B
5 ‒ j2 Ω
C
10 + j8 Ω
10 + j8 Ω
100
Figure 12.13
5ISFFXJSF::TZTUFNGPS&YBNQMF
Solution:
5IFUISFFQIBTFDJSDVJUJO'JHJTCBMBODFEXFNBZSFQMBDFJU
with JUTTJOHMFQIBTFFRVJWBMFOUDJSDVJUTVDIBTJO'JH8FPCUBJO
*BGSPNUIFTJOHMFQIBTFBOBMZTJTBT
7BO
*B = @@@
;:
XIFSF;: = −K + + K = + K = ⧸
21.8ž)FODF
ž
⧸
*B = @@@@@@@@@@@@
−ž"
= ⧸
⧸
21.8ž
*OBTNVDIBTUIFTPVSDFWPMUBHFTJO'JHBSFJOQPTJUJWFTFRVFODF UIFMJOFDVSSFOUTBSFBMTPJOQPTJUJWFTFRVFODF
*C = *B⧸
−ž = ⧸
−ž"
*D = *B⧸
−ž = ⧸
−ž" = ⧸
98.2ž"
Example 12.2
510
Practice Problem 12.2
Chapter 12
Three-Phase Circuits
":DPOOFDUFECBMBODFEUISFFQIBTFHFOFSBUPSXJUIBOJNQFEBODFPG
+ KΩQFSQIBTFJTDPOOFDUFEUPB:DPOOFDUFECBMBODFEMPBEXJUI
BOJNQFEBODFPG + KΩQFSQIBTF5IFMJOFKPJOJOHUIFHFOFSBUPSBOE
UIFMPBEIBTBOJNQFEBODFPG + KΩQFSQIBTF"TTVNJOHBQPTJ UJWFTFRVFODF GPSUIF TPVSDFWPMUBHFT BOEUIBU 7BO = ⧸
30ž7 find:
(a) the line voltages, (b) the line currents.
Answer: (a) 207.8⧸
60ž7 ⧸
−ž7 ⧸
−ž7
(b) 3.75⧸
−ž" ⧸
−ž" ⧸
ž"
12.4
Balanced Wye-Delta Connection
A balanced Y-∆ system consists of a balanced Y-connected source
feeding a balanced ∆-connected load.
This is perhaps the most practical
three-phase system, as the three-phase
sources are usually Y-connected while
the three-phase loads are usually
∆-connected.
5IFCBMBODFE :EFMUBTZTUFNJTTIP XOJO'JH XIFSFUIF
TPVSDFJT:DPOOFDUFEBOEUIFMPBEJT ∆DPOOFDUFE5IFSFJT PGDPVSTF OPOFVUSBMDPOOFDUJPOGSPNTPVSDFUPMPBEGPSUIJTDBTF "TTVNJOHUIF
QPTJUJWFTFRVFODF UIFQIBTFWPMUBHFTBSFBHBJO
7BO = 7Q⧸
ž
7CO = 7Q⧸
−ž 7DO = 7Q⧸
+ž
"TTIPXOJO4FDUJPO UIFMJOFWPMUBHFTBSF
@@
@@
7Q⧸
7Q⧸
7BC = Ŀ
3ž = 7"# 7CD = Ŀ
−ž = 7#$
@@
Ŀ
7Q⧸
7DB = ž = 7$"
TIPXJOHUIBUUIFMJOFWPMUBHFTBSFFRVBMUPUIFWPMUBHFTBDSPTTUIFMPBE
impedances for this system configuration. From these voltages, we can
PCUBJOUIFQIBTFDVSSFOUTBT
7#$
7$"
7"#
*#$ = @@@@
*$" = @@@@
*"# = @@@@
;Ĵ
;Ĵ
;Ĵ
5IFTFDVSSFOUTIBWFUIFTBNFNBHOJUVEFCVUBSFPVUPGQIBTFXJUIFBDI
PUIFSCZž
Ia
a
Van
+
‒
n
Vcn +‒
IAB
‒ Vbn
+
c
b
Ib
Ic
Figure 12.14
#BMBODFE:∆DPOOFDUJPO
A
ZΔ
ZΔ
ZΔ
B
ICA
C
IBC
12.4
511
Balanced Wye-Delta Connection
"OPUIFSXBZUPHFUUIFTFQIBTFDVSSFOUTJTUPBQQMZ ,7-'PSFY
BNQMF BQQMZJOH,7-BSPVOEMPPQB"#COBHJWFT
−7BO + ;Ĵ*"# + 7CO = PS
7BO−7CO @@@
7
7"#
*"# = @@@@@@@@@
= BC= @@@@
;Ĵ
;Ĵ
;Ĵ
which is the same as Eq. (12.21). This is the more general way of finding
UIFQIBTFDVSSFOUT
5IFMJOFDVSSFOUTBSFPCUBJOFEGSPNUIFQIBTFDVSSFOUTCZBQQMZJOH
,$-BUOPEFT" # BOE$5IVT
*B = *"#−*$" *C = *#$− *"# *D = *$"−*#$ 4JODF*$" = *"#⧸
−ž
*B = *"#−*$" = *"# −⧸
−ž
@@
⧸−ž
= *"# + −K = *"#Ŀ
@@
UJNFTUIFNBHOJ
TIPXJOHUIBUUIFNBHOJUVEF*-PGUIFMJOFDVSSFOUJTĿ
UVEF*QPGUIFQIBTFDVSSFOU PS
@@
*Q
*- = Ŀ
*- = ]*B] = ]*C] = ]*D]
XIFSF
Ic
BOE
*Q = ]*"#] = ]*#$] = ]*$"]
ICA
"MTP UIFMJOFDVSSFOUTMBHUIFDPSSFTQPOEJOHQIBTFDVSSFOUTCZž BTTVNJOHUIFQPTJUJWFTFRVFODF'JHVSFJTBQIBTPSEJBHSBNJMMVT
USBUJOHUIFSFMBUJPOTIJQCFUXFFOUIFQIBTFBOEMJOFDVSSFOUT
"OBMUFSOBUJWFXBZPGBOBMZ[JOHUIF :∆DJSDVJUJTUPUSBOTGPSNUIF
∆DPOOFDUFEMPBEUPBOFRVJ WBMFOU:DPOOFDUFEMPBE6TJOHUIF ∆:
USBOTGPSNBUJPOGPSNVMBJO&R ;
;: = @@@Ĵ
30°
IAB
30°
Ia
30°
Ib
IBC
Figure 12.15
1IBTPSEJBHSBNJMMVTUSBUJOHUIFSFMBUJPOTIJQ
CFUXFFOQIBTFBOEMJOFDVSSFOUT
"GUFSUIJTUSBOTGPSNBUJPO XFOPXIBWFB::TZTUFNBTJO'JH
5IFUISFFQIBTF:∆TZTUFNJO'JHDBOCFSFQMBDFECZUIFTJOHMF
QIBTFFRVJWBMFOUDJSDVJUJO'JH5IJTBMMPXTVTUPDBMDVMBUFPOMZ
UIFMJOFDVSSFOUT5IFQIBTFDVSSFOUTBSFPCUBJOFEVTJOH&R BOE
VUJMJ[JOHUIFGBDUUIBUFBDIPGUIFQIBTFDVSSFOUTMFBETUIFDPSSFTQPOEJOH
MJOFDVSSFOUCZž
"CBMBODFEBCDTFRVFODF:DPOOFDUFETPVSDFXJUI 7BO = ⧸
ž7JT
DPOOFDUFEUPB∆DPOOFDUFECBMBODFEMPBE + K ΩQFSQIBTF$BMDV
MBUFUIFQIBTFBOEMJOFDVSSFOUT
Ia
ZΔ
3
Van +
‒
Figure 12.16
"TJOHMFQIBTFFRVJWBMFOUDJSDVJUPGBCBM
BODFE:∆DJSDVJU
Example 12.3
512
Chapter 12
Three-Phase Circuits
Solution:
5IJTDBOCFTPMWFEJOUXPXBZT
■ METHOD 1 5IFMPBEJNQFEBODFJT
;Ĵ = + K = ⧸
26.57žΩ
*GUIFQIBTFWPMUBHF7BO = ⧸
ž UIFOUIFMJOFWPMUBHFJT
@@
@@
⧸30ž = Ŀ
7BC = 7BOĿ
⧸ž+ž = 7"#
PS
7"# = ⧸
ž7
5IFQIBTFDVSSFOUTBSF
ž
⧸
7"# @@@@@@@@@@@@
*"# = @@@@
=
ž"
= ⧸
;Ĵ
⧸
ž
*#$ = *"#⧸
−ž = ⧸
−ž"
*$" = *"#⧸
+ž = ⧸
133.43ž"
5IFMJOFDVSSFOUTBSF
@@
@@
⧸−ž = Ŀ
⧸
*B = *"#Ŀ
ž− ž
= ⧸
−ž"
*C = *B⧸
−ž = ⧸
−ž"
*D = *B⧸
+ž = ⧸
ž"
■ METHOD 2 "MUFSOBUJWFMZ VTJOHTJOHMFQIBTFBOBMZTJT
1ž
⧸
7BO
*B = @@@@@
= @@@@@@@@@@@@@
−ž"
= ⧸
;Ĵ∕ ⧸
ž
BTBCPWF0UIFSMJOFDVSSFOUTBSFPCUBJOFEVTJOHUIFBCDQIBTFTFRVFODF
Practice Problem 12.3
0OFMJOFWPMUBHFPGBCBMBODFE:DPOOFDUFETPVSDFJT
7"# = ⧸
−ž7
*GUIFTPVSDFJTDPOOFDUFEUPB ∆DPOOFDUFEMPBEPG ⧸
4žΩ, find the
QIBTFBOEMJOFDVSSFOUT"TTVNFUIFBCDTFRVFODF
Answer:⧸
−ž" ⧸
−ž" ⧸
6ž" ⧸
−ž" ⧸
ž" ⧸
3ž"
12.5
Balanced Delta-Delta Connection
A balanced ∆-∆ system is one in which both the balanced source and
balanced load are ∆-connected.
5IFTPVSDFBTXFMMBTUIFMPBENBZCFEFMUBDPOOFDUFEBTTIP XO
JO'JH0VSHPBMJTUPPCUBJOUIFQIBTFBOEMJOFDVSSFOUTBTVTVBM
12.5
Balanced Delta-Delta Connection
Ia
a
A
IAB
Vca
‒
+
+
‒
Vab
Ib
c
‒+
Vbc
b
513
ZΔ
ZΔ
Ic
B
ICA
C
IBC
ZΔ
Figure 12.17
"CBMBODFE∆-∆DPOOFDUJPO
"TTVNJOHBQPTJUJWFTFRVFODF UIFQIBTFWPMUBHFTGPSBEFMUBDPOOFDUFE
TPVSDFBSF
7BC = 7Q⧸
ž
7CD = 7Q⧸
−ž 7DB = 7Q⧸
+ž
5IFMJOFWPMUBHFTBSFUIFTBNFBTUIFQIBTFWPMUBHFT'SPN'JH BTTVNJOHUIFSFJTOPMJOFJNQFEBODFT UIFQIBTFWPMUBHFTPGUIFEF MUB
connected source are equal to the voltages across the impedances; that JT
7BC = 7"# 7CD = 7#$ 7DB = 7$"
)FODF UIFQIBTFDVSSFOUTBSF
7#$ @@@
7
7
7"# @@@
*"# = @@@@
= BC
*#$ = @@@@
= CD
;Ĵ
;Ĵ
;Ĵ
;Ĵ
7$" @@@
7
*$" = @@@@
= DB
;Ĵ
;Ĵ
#FDBVTFUIFMPBEJTEFMUBDPOOFDUFEKVTUBTJOUIFQSFWJPVTTFDUJPO TPNF
PGUIFGPSNVMBTEFSJWFEUIFSFBQQMZIFSF5IFMJOFDVSSFOUTBSFPCUBJOFE
GSPNUIFQIBTFDVSSFOUTCZBQQMZJOH,$-BUOPEFT" # BOE$ BTXFEJE
JOUIFQSFWJPVTTFDUJPO
*B = *"#−*$" *C = *#$−*"# *D = *$"− *#$
"MTP BTTIPXOJOUIFMBTUTFDUJPO FBDIMJOFDVSSFOUMBHTUIFDPSSF
@@TQPOE
UJNFT
JOHQIBTFDVSSFOUCZžUIFNBHOJUVEF*-PGUIFMJOFDVSSFOUJTĿ
UIFNBHOJUVEF*QPGUIFQIBTFDVSSFOU
@@
*Q
*- = Ŀ
"OBMUFSOBUJWFXBZPGBOBMZ[JOHUIF∆∆DJSDVJUJTUPDPOWFSUCPUI
UIFTPVSDFBOEUIFMPBEUPUIFJS :FRVJWBMFOUT8FBMSFBEZLOP XUIBU
;: = ;Ĵ∕5PDPOWFSUB∆DPOOFDUFETPVSDFUPB:DPOOFDUFETPVSDF TFFUIFOFYUTFDUJPO
"CBMBODFE ∆DPOOFDUFEMPBEIB WJOHBOJNQFEBODF − K ΩJT
DPOOFDUFEUPB
∆DPOOFDUFE QPTJUJ WFTFRVFODFHFOFSBUPSIB WJOH
7BC = ⧸
ž7$BMDVMBUFUIFQIBTFDVSSFOUTPGUIFMPBEBOEUIFMJOF
DVSSFOUT
Example 12.4
514
Chapter 12
Three-Phase Circuits
Solution:
5IFMPBEJNQFEBODFQFSQIBTFJT
;Ĵ = −K = ⧸
−žΩ
4JODF7"# = 7BC UIFQIBTFDVSSFOUTBSF
ž
⧸
7"# @@@@@@@@@@
*"# = @@@@
=
36.87ž"
= ⧸
;Ĵ
⧸
−ž
*#$ = *"#⧸
−ž = ⧸
−ž"
*$" = *"#⧸
+ž = ⧸
156.87ž"
'PSB EFMUBMPBE UIFMJOF DVSSFOU BMXBZTMBHT
UIFDPSSFTQPOEJOHQIBTF
@@
UJNFTUIBUPGUIFQIBTFDVSSFOU
DVSSFOUCZžBOEIBTBNBHOJUVEF Ŀ
)FODF UIFMJOFDVSSFOUTBSF
@@
Practice Problem 12.4
@@
⧸−ž = ⧸
*B = *"#Ŀ
36.87ž Ŀ
⧸−ž
= ⧸
6.87ž"
*C = *B⧸
−ž = ⧸
−ž"
*D = *B⧸
+ž = ⧸
126.87ž"
"QPTJUJWFTFRVFODF CBMBODFE ∆DPOOFDUFETPVSDFTVQQMJFTBCBMBODFE
∆DPOOFDUFEMPBE*GUIFJNQFEBODFQFSQIBTFPGUIFMPBEJT + KΩ
BOE*B = ⧸
35° A, find *"#BOE7"#
65ž" ⧸
98.69ž7
Answer:⧸
12.6
Balanced Delta-Wye Connection
A balanced ∆-Y system consists of a balanced ∆-connected source
feeding a balanced Y-connected load.
$POTJEFSUIF ∆:DJSDVJUJO'JH "HBJO BTTVNJOHUIF BCD
TFRVFODF UIFQIBTFWPMUBHFTPGBEFMUBDPOOFDUFETPVSDFBSF
7BC = 7Q⧸
ž 7CD = 7Q⧸
−ž
7DB = 7Q⧸
+ž
5IFTFBSFBMTPUIFMJOFWPMUBHFTBTXFMMBTUIFQIBTFWPMUBHFT
8FDBOPCUBJOUIFMJOFDVSSFOUTJONBO ZXBZT0OFXBZJTUPBQQMZ
,7-UPMPPQB"/#CBJO'JH XSJUJOH
−7BC + ;:*B−;:*C = PS
;: *B−*C = 7BC = 7Q⧸
ž
5IVT
ž
7Q⧸
*B−*C = @@@@@
;:
12.6
Ia
a
515
Balanced Delta-Wye Connection
A
ZY
Vca
‒
+
+ Vab
‒
N
Ib
c
‒+
ZY
B
b
Vbc
ZY
C
Ic
Figure 12.18
"CBMBODFE∆:DPOOFDUJPO
#VU *CMBHT *BCZž TJODFXFBTTVNFEUIF
*C = *B⧸
−ž)FODF
BCDTFRVFODFUIBUJT *B−*C = *B −⧸
−ž
@@
@@
Ŀ
+ K@@@
⧸3ž
= *B + @@
= *BĿ
4VCTUJUVUJOH&R JOUP&R HJWFT
@@
⧸−ž
7Q∕Ŀ
*B = @@@@@@@@@@@
;:
'SPNUIJT XFPCUBJOUIFPUIFSMJOFDVSSFOUT *CBOE*DVTJOHUIFQPTJUJWF
QIBTFTFRVFODF JF *C = *B⧸
−ž *D = *B⧸
+ž5IFQIBTFDVSSFOUT
BSFFRVBMUPUIFMJOFDVSSFOUT
"OPUIFSX BZUPPCUBJOUIFMJOFDVSSFOUTJTUPSFQMBDFUIFEFMUB
DPOOFDUFETPVSDFXJUIJUTFRVJ WBMFOUXZFDPOOFDUFE TPVSDF BTTIP XO
JO'JH*O4FDUJPO XFGPVOEUIBUUIFMJOFUPMJOFWPMUBHFTPG
BXZFDPOOFDUFETPVSDFMFBEUIFJSDPSSFTQPOEJOHQIBTFWPMUBHFTCZž
5IFSFGPSF XFPCUBJOFBDIQIBTFWPMUBHFPGUIFFRVJWBMFOUXZFDPOOFDUFE
TPVSDFCZEJWJEJOHUIFDPSSFTQPOEJOHMJOFWPMUBHFPGUIFEFMUBDPOOFDUFE
@@
BOETIJGUJOHJUTQIBTFCZ −ž5IVT UIFFRVJWBMFOUXZF
TPVSDFCZĿ
DPOOFDUFETPVSDFIBTUIFQIBTFWPMUBHFT
7Q
@@⧸
7BO = @@@
−ž
Ŀ
7Q
@@⧸−ž 7CO = @@@
Ŀ
7Q
@@⧸
7DO = @@@
+ž
Ŀ
@@
XIJDIJTUIFTBNFBT&R +‒
n
Vcn
+‒
Vbn
+ V
‒ ab
‒+
b
Vbc
Figure 12.19
5SBOTGPSNJOHB∆DPOOFDUFETPVSDFUPBO
FRVJWBMFOU:DPOOFDUFETPVSDF
Ia
Vp ‒30°
+ V
an
‒
Vca +‒
c
*GUIFEFMUBDPOOFDUFETPVSDFIBTTPVSDFJNQFEBODF ;TQFSQIBTF UIF
FRVJWBMFOUXZFDPOOFDUFETPVSDFXJMMIBWFBTPVSDFJNQFEBODFPG ;T∕
QFSQIBTF BDDPSEJOHUP&R 0ODFUIFTPVSDFJTUSBOTGPSNFEUPXZF UIFDJSDVJUCFDPNFTBXZF
XZFTZTUFN 5IFSFGPSF XFDBOVTFUIFFRVJ WBMFOUTJOHMFQIBTFDJSDVJU
TIPXOJO'JH GSPNXIJDIUIFMJOFDVSSFOUGPSQIBTFBJT
⧸
−ž
7Q∕Ŀ
*B = @@@@@@@@@@@
;:
a
√3
+
‒
Figure 12.20
5IFTJOHMFQIBTFFRVJWBMFOUDJSDVJU
ZY
516
Chapter 12
Three-Phase Circuits
"MUFSOBUJWFMZ XFNBZUSBOTGPSNUIFXZFDPOOFDUFEMPBEUPBO
FRVJWBMFOUEFMUBDPOOFDUFEMPBE 5IJTSFTVMUTJOBEFMUBEFMUBTZTUFN XIJDIDBOCFBOBMZ[FEBTJO4FDUJPO/PUFUIBU
7Q
@@⧸
7"/ = *B;: = @@@
−ž
Ŀ
7#/ = 7"/⧸
−ž 7$/ = 7"/⧸
+ž
"TTUBUFEFBSMJFS UIFEFMUBDPOOFDUFEMPBEJTNPSFEFTJSBCMFUIBO
UIFXZFDPOOFDUFEMPBE*UJTFBTJFSUPBMUFSUIFMPBETJOBOZPOFQIBTFPG
UIFEFMUBDPOOFDUFEMPBET BTUIFJOEJWJEVBMMPBETBSFDPOOFDUFEEJSFDUMZ
BDSPTTUIFMJOFT)P XFWFS UIFEFMUBDPOOFDUFETPVSDFJTIBSEMZVTFEJO
QSBDUJDFCFDBVTFBOZTMJHIUJNCBMBODFJOUIFQIBTFWPMUBHFTXJMMSFTVMUJO
VOXBOUFEDJSDVMBUJOHDVSSFOUT
5BCMFQSFTFOUTBTVNNBSZPGUIFGPSNVMBT GPSQIBTFDVSSFOUT
BOEWPMUBHFTBOEMJOFDVSSFOUTBOEWPMUBHFTGPSUIFGPVSDPOOFDUJPOT4UV
EFOUTBSFBEWJTFE OPUUP NFNPSJ[FUIFGPSNVMBT CVUUPVOEFSTUBOEIP X
TABLE 12.1
Summary of phase and line voltages/currents for
balanced three-phase systems.1
$POOFDUJPO
1IBTFWPMUBHFTDVSSFOUT
-JOFWPMUBHFTDVSSFOUT
@@
7BO = 7Q⧸
ž
7
7BC = Ŀ
Q⧸
ž
+ž
7DO = 7Q⧸
7DB = 7BC⧸
+ž
4BNFBTMJOFDVSSFOUT
::
−ž
7CO = 7Q⧸
7CD = 7BC⧸
−ž
*B = 7BO∕;:
*C = *B⧸
−ž
:∆
ž
7BO = 7Q⧸
−ž
7CO = 7Q⧸
+ž
7DO = 7Q⧸
*"# = 7"#∕;Ĵ
*#$ = 7#$∕;Ĵ
*$" = 7$"∕;Ĵ
∆∆
ž
7BC = 7Q⧸
@@
7Q⧸
7BC = 7"# = Ŀ
3ž
7CD = 7#$ = 7BC⧸
−ž
7DB = 7$" = 7BC⧸
+ž
@@
⧸−ž
*B = *"#Ŀ
*C = *B⧸
−ž
*D = *B⧸
+ž
4BNFBTQIBTFWPMUBHFT
−ž
7CD = 7Q⧸
+ž
7DB7Q⧸
*"#7BC∕;Ĵ
*#$7CD∕;Ĵ
*$"7DB∕;Ĵ
Ĵ:
+ž
*D = *B⧸
ž
7BC7Q⧸
@@
⧸
*B*"#Ŀ
−ž
*C*B⧸
−ž
*D*B⧸
+ž
4BNFBTQIBTFWPMUBHFT
−ž
7CD7Q⧸
+ž
7DB7Q⧸
4BNFBTMJOFDVSSFOUT
7Q⧸
−ž
@@ *B = @@@@@@@@
Ŀ
;
:
*C = *B⧸
−ž
+ž
*D = *B⧸
1PTJUJWFPSBCDTFRVFODFJTBTTVNFE
12.7
Power in a Balanced System
517
UIFZBSFEFSJWFE5IFGPSNVMBTDBOBMXBZTCFPCUBJOFECZEJSFDUMZBQQMZ
JOH,$-BOE,7-UPUIFBQQSPQSJBUFUISFFQIBTFDJSDVJUT
"CBMBODFE:DPOOFDUFEMPBEXJUIBQIBTFJNQFEBODFPG + KΩJT
TVQQMJFECZBCBMBODFE QPTJUJWFTFRVFODF∆DPOOFDUFETPVSDFXJUIBMJOF
WPMUBHFPG7$BMDVMBUFUIFQIBTFDVSSFOUT6TF7BCBTBSFGFSFODF
Example 12.5
Solution:
5IFMPBEJNQFEBODFJT
;: = + K = ⧸
žΩ
BOEUIFTPVSDFWPMUBHFJT
7BC = ⧸
ž7
8IFOUIF∆DPOOFDUFETPVSDFJTUSBOTGPSNFEUPB:DPOOFDUFETPVSDF
7BC
@@
7BO = @@@
−ž = ⧸
−ž7
⧸
Ŀ
5IFMJOFDVSSFOUTBSF
−ž
⧸
7BO @@@@@@@@@@
*B = @@@
=
−ž"
= ⧸
;:
⧸
ž
*C = *B⧸
−ž = ⧸
−ž"
*D = *B⧸
ž = ⧸
ž"
XIJDIBSFUIFTBNFBTUIFQIBTFDVSSFOUT
*OBCBMBODFE ∆:DJSDVJU 7BC = ⧸
žBOE ;: = + K Ω
$BMDVMBUFUIFMJOFDVSSFOUT
Answer:⧸
−ž" ⧸
+ž" ⧸
ž"
12.7
Power in a Balanced System
-FUVTOPXDPOTJEFSUIFQPXFSJOBCBMBODFEUISFFQIBTFTZTUFN8F
CFHJOCZFYBNJOJOHUIFJOTUBOUBOFPVTQPXFSBCTPSCFECZUIFMPBE5IJT
SFRVJSFTUIBUUIFBOBMZTJTCFEPOFJOUIFUJNFEPNBJO'PSB:DPOOFDUFE
MPBE UIFQIBTFWPMUBHFTBSF
@@
@@
7QDPTωU 7QDPT ωU−ž
v"/ = Ŀ
v#/ = Ŀ
@@
7QDPT ωU + ž v$/ = Ŀ
@@
JTOFDFTTBSZCFDBVTF7Q has been defined as the rms
XIFSFUIFGBDUPSĿ
WBMVFPGUIFQIBTFWPMUBHF*G ;: = ;⧸
θ UIFQIBTFDVSSFOUTMBHCFIJOE
UIFJSDPSSFTQPOEJOHQIBTFWPMUBHFTCZθ5IVT
@@
@@
*QDPT ωU−θ JB = Ŀ
JC = Ŀ
*QDPT ωU−θ −ž @@
Ŀ
*QDPT ωU−θ + ž
JD = Practice Problem 12.5
518
Chapter 12
Three-Phase Circuits
XIFSF*QJTUIFSNTWBMVFPGUIFQIBTFDVSSFOU5IFUPUBMJOTUBOUBOFPVT
QPXFSJOUIFMPBEJTUIFTVNPGUIFJOTUBOUBOFPVTQPXFSTJOUIFUISFF
QIBTFTUIBUJT
Q = QB + QC + QD = v"/JB + v#/JC + v$/JD
= 7Q*Q<DPTωUDPT ωU−θ
+ DPT ωU−ž DPT ωU−θ −ž
+ DPT ωU + ž DPT ωU−θ + ž >
"QQMZJOHUIFUSJHPOPNFUSJDJEFOUJUZ
<DPT " + # + DPT "−# >
DPT"DPT# = @@
HJWFT
Q = 7Q*Q<DPTθ + DPT ωU−θ + DPT ωU−θ −ž
+DPT ωU−θ + ž >
= 7Q*Q<DPTθ + DPTα + DPTα DPTž + TJOα TJOž
+ DPTα DPTž−TJOα TJOž>
XIFSFα = ωU−θ
DPTα = 7 * DPTθ
= 7Q*Q DPTθ + DPTα + −@@
Q Q
5IVTUIFUPUBMJOTUBOUBOFPVTQPXFSJOBCBMBODFEUISFFQIBTFTZTUFNJT
DPOTUBOU‡JUEPFTOPUDIBOHFXJUIUJNFBTUIFJOTUBOUBOFPVTQPXFSPG
FBDIQIBTFEPFT5IJTSFTVMUJTUSVFXIFUIFSUIFMPBEJT:PS
∆DPOOFDUFE
5IJTJTPOFJNQPSUBOUSFBTPOGPSVTJOHBUISFFQIBTFTZTUFNUPHFOFSBUF
BOEEJTUSJCVUFQPXFS8FXJMMMPPLJOUPBOPUIFSSFBTPOBMJUUMFMBUFS
4JODFUIFUPUBMJOTUBOUBOFPVTQP XFSJTJOEFQFOEFOUPGUJNF UIF
BWFSBHFQP XFSQFSQIBTF 1Q GPS FJUIFS UIF ∆DPOOFDUFEMPBEPSUIF :DPOOFDUFEMPBEJTQ∕ PS
[
]
1Q = 7Q*QDPTθ
BOEUIFSFBDUJWFQPXFSQFSQIBTFJT
2Q = 7Q*QTJOθ
5IFBQQBSFOUQPXFSQFSQIBTFJT
4Q = 7Q*Q
5IFDPNQMFYQPXFSQFSQIBTFJT
4Q = 1Q + K2Q = 7Q*Q
XIFSF7QBOE*QBSFUIFQIBTFWPMUBHFBOEQIBTFDVSSFOUXJUINBHOJUVEFT
7QBOE*Q SFTQFDUJWFMZ5IFUPUBMBWFSBHFQPXFSJTUIFTVNPGUIFBWFSBHF
QPXFSTJOUIFQIBTFT
@@
7-*-DPTθ
1 = 1B + 1C + 1D = 1Q = 7Q*QDPTθ = Ŀ
@@
'PSB:DPOOFDUFEMPBE @@ *- = *QCVU 7- = Ŀ
7Q XIFSFBTGPSB *QCVU 7- = 7Q5IVT &R BQQMJFTGPS
∆DPOOFDUFEMPBE *- = Ŀ
CPUI :DPOOFDUFE BOE ∆DPOOFDUFE MPBET 4JNJMBSMZ UIF UPUBMSFBDUJWF
QPXFSJT
@@
7-*-TJOθ
2 = 7Q*QTJOθ = 2Q = Ŀ
12.7
519
Power in a Balanced System
BOEUIFUPUBMDPNQMFYQPXFSJT
7Q
4 = 4Q = 7Q* Q = *Q;Q = @@@@
;Q XIFSF;Q = ;Q⧸
θJTUIFMPBEJNQFEBODFQFSQIBTF ;QDPVMECF;:PS;Ĵ "MUFSOBUJWFMZ XFNBZXSJUF&R BT
@@
7-*-⧸
4 = 1 + K2 = Ŀ
θ
3FNFNCFSUIBU7Q *Q 7- BOE*-BSFBMMSNTWBMVFTBOEUIBUθJTUIFBOHMF
PGUIFMPBEJNQFEBODFPSUIFBOHMFCFUXFFOUIFQIBTFWPMUBHFBOEUIF
QIBTFDVSSFOU
"TFDPOENBKPSBEW BOUBHFPGUISFFQIBTFTZTUFNTGPSQP XFSEJT
USJCVUJPOJTUIBUUIFUISFFQIBTFTZTUFNVTFTBMFTTFSBNPVOUPGXJSF UIBOUIFTJOHMFQIBTFTZTUFNGPSUIFTBNFMJOFWPMUBHF7-BOEUIFTBNF
BCTPSCFEQPXFS1-8FXJMMDPNQBSFUIFTFDBTFTBOEBTTVNFJOCPUI
UIBUUIFXJSFTBSFPGUIFTBNFNBUFSJBM FH DPQQFSXJUISFTJTUJ WJUZ
ρ PGUIFTBNFMFOHUI ê BOEUIBUUIFMPBETBSFSFTJTUJ WF JF VOJUZ
QPXFSGBDUPS 'PSUIFUX PXJSFTJOHMFQIBTFTZTUFNJO'JH B *- = 1-∕7- TPUIFQPXFSMPTTJOUIFUXPXJSFTJT
1
1MPTT = *-3 = 3@@@
-
7-
R′
IL
R
PL
R
VL
‒
Ia
+
+
Singlephase
source
Load
Threephase
balanced
source
R′
Ib
R′
Ic
VL 0°
‒
+
VL ‒120°
‒
Transmission lines
Transmission lines
(a)
(b)
Figure 12.21
$PNQBSJOHUIFQPXFSMPTTJO B BTJOHMFQIBTFTZTUFN BOE C BUISFFQIBTFTZTUFN
'PSUIFUISFFXJSFUISFFQIBTFTZTUFNJO'JH
C *′- = ]*B] = ]*C] =
@@
Ŀ
*
=
1
∕
7
GSPN&R
5IFQPXFSMPTTJOUIFUISFFXJSFTJT
] D] 1-
1-
@@@
1′MPTT = *′- 3′ = 3′@@@@
=
3′
7-
7-
&RVBUJPOT BOE TIPXUIBUGPSUIFTBNFUPUBMQPXFSEFMJW FSFE1-BOETBNFMJOFWPMUBHF7
1MPTT @@@
@@@@
= 3
1′MPTT 3′
Threephase
balanced
load
520
Chapter 12
Three-Phase Circuits
#VUGSPN$IBQUFS 3 = ρê∕πSBOE3′ = ρê∕πS′ XIFSFSBOES′BSFUIF
SBEJJPGUIFXJSFT5IVT
1MPTT @@@@
@@@@
= S′ 1′MPTT
S
*GUIFTBNFQPXFSMPTTJTUPMFSBUFEJOCPUITZTUFNT UIFO S = S′5IF
SBUJPPGNBUFSJBMSFRVJSFEJTEFUFSNJOFECZUIFOVNCFSPGXJSFTBOEUIFJS
WPMVNFT TP
.BUFSJBMGPSTJOHMFQIBTF @@@@@@@
πSê
S
@@@@@@@@@@@@@@@@@@@@@
=
= @@@@
.BUFSJBMGPSUISFFQIBTF πS′ ê
S′
= = @@
TJODFS = S′&RVBUJPO TIPXTUIBUUIFTJOHMFQIBTFTZTUFNVTFT
QFSDFOUNPSFNBUFSJBMUIBOUIFUISFFQIBTFTZTUFNPSUIBUUIFUISFF
QIBTFTZTUFNVTFTPOMZQFSDFOUPGUIFNBUFSJBMVTFEJOUIFFRVJWBMFOU
TJOHMFQIBTFTZTUFN*OPUIFSXPSET DPOTJEFSBCMZMFTTNBUFSJBMJTOFFEFE
UPEFMJWFSUIFTBNFQPXFSXJUIBUISFFQIBTFTZTUFNUIBOJTSFRVJSFEGPS
BTJOHMFQIBTFTZTUFN
Example 12.6
3FGFSUPUIFDJSDVJUJO'JH JO&YBNQMF %FUFSNJOFUIFUPUBM
BWFSBHFQPXFS SFBDUJWFQPXFS BOEDPNQMFYQPXFSBUUIFTPVSDFBOEBU
UIFMPBE
Solution:
It is sufficient to consider one phase, as the system is balanced. For
QIBTFB
7Q = ⧸
ž7
BOE
*Q = ⧸
−ž"
5IVT BUUIFTPVSDF UIFDPNQMFYQPXFSBCTPSCFEJT
4T = −7Q* Q = − ⧸
ž ⧸
ž
= −⧸
ž = − + K 7"
5IFSFBMPSBWFSBHFQPXFSBCTPSCFEJT−8BOEUIFSFBDUJWFQPXFS
JT−7"3
"UUIFMPBE UIFDPNQMFYQPXFSBCTPSCFEJT
4- = ]*Q];Q
XIFSF;Q = + K = ⧸
38.66žBOE*Q = *B = ⧸
−ž)FODF
4- = ⧸
ž = ⧸
= + K 7"
5IFSFBMQPXFSBCTPSCFEJT8BOEUIFSFBDUJWFQPXFSBCTPSCFE
JT7"35IFEJGGFSFODFCFUXFFOUIFUXPDPNQMFYQPXFSTJTBC TPSCFECZUIFMJOFJNQFEBODF −K Ω5PTIPXUIBUUIJTJTUIFDBTF we find the complex power absorbed by the line as
4ê = ]*Q];ê = −K = −K7"
XIJDIJTUIFEJGGFSFODFCFUXFFO 4TBOE4-UIBUJT 4T + 4ê + 4- = BT
FYQFDUFE
12.7
Power in a Balanced System
'PSUIF::DJSDVJUJO1SBDUJDF1SPC DBMDVMBUFUIFDPNQMFYQPXFS
BUUIFTPVSDFBOEBUUIFMPBE
521
Practice Problem 12.6
Answer:− + K 7" + K 7"
"UISFFQIBTFNPUPSDBOCFSFHBSEFEBTBCBMBODFE:MPBE"UISFFQIBTF
NPUPSESBXTL8XIFOUIFMJOFWPMUBHFJT7BOEUIFMJOFDVSSFOU
JT"%FUFSNJOFUIFQPXFSGBDUPSPGUIFNPUPS
Solution:
5IFBQQBSFOUQPXFSJT
@@
Example 12.7
@@
7-*- = Ŀ
= 7"
4 = Ŀ
4JODFUIFSFBMQPXFSJT
1 = 4DPTθ = 8
UIFQPXFSGBDUPSJT
1 = @@@@@@@
QG = DPTθ = @@
= 4 $BMDVMBUFUIFMJOFDVSSFOUSFRVJSFEGPSBL8UISFFQIBTFNPUPSIBWJOH
BQPXFSGBDUPSPGMBHHJOHJGJUJTDPOOFDUFEUPBCBMBODFETPVSDFXJUI
BMJOFWPMUBHFPG7
Practice Problem 12.7
Answer:"
5XPCBMBODFE MPBETBSFDPOOFDUFEUP BL7 SNT)[MJOF BT
TIPXOJO'JH B -PBEESB XTL8BUBQP XFSGBDUPSPG
MBHHJOH XIJMFMPBEESBXTL7"3BUBQPXFSGBDUPSPGMBHHJOH
"TTVNJOHUIFBCDTFRVFODF EFUFSNJOF B UIFDPNQMFY SFBM BOESFBD
UJWFQPXFSTBCTPSCFECZUIFDPNCJOFEMPBE C UIFMJOFDVSSFOUT BOE
D UIFL7"3SBUJOHPGUIFUISFFDBQBDJUPST∆DPOOFDUFEJOQBSBMMFMXJUI
UIFMPBEUIBUXJMMSBJTFUIF QPXFSGBDUPSUPMBHHJOHBOE UIFDBQBDJ
UBODFPGFBDIDBQBDJUPS
Solution:
B 'PSMPBE HJWFOUIBU11 = L8BOEDPTθ1 = UIFOTJOθ1 = )FODF
11
L8
41 = @@@@@
= @@@@@@
= L7"
DPTθ1
BOE21 = 41TJOθ1 = = L7"35IVT UIFDPNQMFYQPXFSEVF
UPMPBEJT
41 = 11 + K21 = + KL7"
Example 12.8
522
Chapter 12
Three-Phase Circuits
'PSMPBE JG22 = L7"3BOEDPTθ2 = UIFOTJOθ2 = 8Ffind
22
L7"
42 = @@@@@
= @@@@@@@
= L7"
TJOθ2
BOE12 = 42DPTθ2 = L85IFSFGPSFUIFDPNQMFYQPXFSEVF
UPMPBEJT
Balanced
load 1
42 = 12 + K22 = + KL7"
Balanced
load 2
'SPN&RT BOE UIFUPUBMDPNQMFYQPXFSBCTPSCFECZ
UIFMPBEJT
(a)
4 = 41 + 42 = + KL7" = ⧸
žL7"
C
C
C
XIJDIIBTBQPXFSGBDUPSPGDPTž = MBHHJOH5IFSFBMQPXFS
JTUIFOL8 XIJMFUIFSFBDUJWFQPXFSJTL7"3
@@
7-*- UIFMJOFDVSSFOUJT
C 4JODF4 = Ŀ
4
@@ *- = @@@@@
Ŀ
7 8FBQQMZUIJTUPFBDIMPBE LFFQJOHJONJOEUIBUGPSCPUIMPBET 7- =
240 kV. For load 1,
@@
*- = @@@@@@@@@@
= N"
Ŀ
Combined
load
(b)
Figure 12.22
'PS&YBNQMF B 5IFPSJHJOBM
CBMBODFEMPBET C UIFDPNCJOFEMPBE
XJUIJNQSPWFEQPXFSGBDUPS
4JODFUIFQPXFSGBDUPSJTMBHHJOH UIFMJOFDVSSFOUMBHTUIFMJOFWPMUBHFCZ
θ1 = DPT¢ = ž5IVT
*B = ⧸
−ž
'PSMPBE
@@
*- = @@@@@@@@@@
= N"
Ŀ
BOEUIF MJOFDVSSFOUMBHT UIFMJOF WPMUBHFCZ θ2 = DPT¢ = ž
)FODF
*B = ⧸
−ž
5IFUPUBMMJOFDVSSFOUJT
*B = *B + *B = ⧸
−ž + ⧸
−ž
= −K + −K
= −K = ⧸
−žN"
"MUFSOBUJWFMZ XFDPVME PCUBJOUIFDVSSFOU GSPN UIFUPUBMDPNQMF Y
QPXFSVTJOH&R BT
@@
*- = @@@@@@@@@@
= N"
Ŀ
BOE
−žN"
*B = ⧸
XIJDIJTUIFTBNFBTCFGPSF5IFPUIFSMJOFDVSSFOUT *CBOE*DB DBOCF
PCUBJOFEBDDPSEJOHUPUIFBCDTFRVFODF JF *C = ⧸
−žN"
BOE*D = ⧸
žN" D We can find the SFBDUJWFQPXFSOFFEFEUPCSJOHUIF QPXFSGBDUPS UP
MBHHJOHVTJOH&R 2$ = 1 UBOθPME−UBOθOFX
12.8
523
Unbalanced Three-Phase Systems
XIFSF1 = L8 θPME = ž BOEθOFX = DPT¢ = ž)FODF
2$ = UBOž−UBOž = L7"3
5IJTSFBDUJWFQPXFSJTGPSUIFUISFFDBQBDJUPST'PSFBDIDBQBDJUPS UIF
SBUJOH2′$ = L7"3'SPN&R UIFSFRVJSFEDBQBDJUBODFJT
2′$
$ = @@@@@@
ω7SNT
4JODFUIFDBQBDJUPSTBSF ∆DPOOFDUFEBTTIPXOJO'JH C 7SNTJO
the above formula is the line-to-line or line voltage, which is 240 L7
5IVT
$ = @@@@@@@@@@@@@@@
= Q'
π "TTVNFUIBUUIFUXPCBMBODFEMPBETJO 'JH B BSFTVQQMJFECZBO
7SNT)[MJOF-PBEJT:DPOOFDUFEXJUI+ KΩQFSQIBTF XIJMFMPBEJTBCBMBODFEUISFFQIBTFNPUPSESBXJOHL8BUBQPXFS
GBDUPSPGMBHHJOH"TTVNJOHUIFBCDTFRVFODF DBMDVMBUF B UIFDPN
QMFYQPXFSBCTPSCFECZUIFDPNCJOFEMPBE C UIFL7"3SBUJOHPGFBDI
PGUIFUISFFDBQBDJUPST∆DPOOFDUFEJOQBSBMMFMXJUIUIFMPBEUPSBJTF UIF
QPXFSGBDUPSUPVOJUZ BOE D UIFDVSSFOUESBXOGSPNUIFTVQQMZBUVOJUZ
QPXFSGBDUPSDPOEJUJPO
Practice Problem 12.8
Answer: B + KL7" C L7"3 D "
12.8
Unbalanced Three-Phase Systems
5IJTDIBQUFSXPVMECFJODPNQMFUFXJUIPVUNFOUJPOJOHVOCBMBODFEUISFF
QIBTFTZTUFNT"OVOCBMBODFETZTUFNJTDBVTFECZUXPQPTTJCMFTJUVB UJPOT 5IFTPVSDFWPMUBHFTBSFOPUFRVBMJONBHOJUVEFBOEPSEJGGFS
JOQIBTFCZBOHMFTUIBUBSFVOFRVBM PS MPBEJNQFEBODFTBSFVOFRVBM
5IVT
Ia
A
ZA
VAN
An unbalanced system is due to unbalanced voltage sources or an
unbalanced load.
5PTJNQMJGZBOBMZTJT XFXJMMBTTVNFCBMBODFETPVSDFWPMUBHFT CVUBO
VOCBMBODFEMPBE
6OCBMBODFEUISFFQIBTFTZTUFNTBSFTPMWFECZEJSFDUBQQMJDBUJPOPG
NFTIBOEOPEBMBOBMZTJT'JHVSFTIP XTBOF YBNQMFPGBOVOCBM BODFEUISFFQIBTFTZTUFNUIBUDPOTJTUTPGCBMBODFETPVSDFW PMUBHFT OPU
TIPwn in the figure) and an unbalanced :DPOOFDUFEMPBE TIPXOJOUIF
figure). Since the load is unbalanced, ;" ;# BOE;$BSFOPUFRVBM5IF
MJOFDVSSFOUTBSFEFUFSNJOFECZ0INTMBXBT
7"/
*B = @@@@
;" 7#/
*C = @@@@
;# 7$/
*D = @@@@
;$ In
N
Ib
VBN
Ic
B
VCN
ZB
ZC
C
Figure 12.23
6OCBMBODFEUISFFQIBTF:DPOOFDUFEMPBE
A special technique for handling
unbalanced three-phase systems is the
method of symmetrical components,
which is beyond the scope of this text.
524
Chapter 12
Three-Phase Circuits
5IJTTFUPGVOCBMBODFEMJOFDVSSFOUTQSPEVDFTDVSSFOUJOUIFOFVUSBMMJOF XIJDIJTOPU[FSPBTJOBCBMBODFETZTUFN"QQMZJOH,$-BUOPEF
/HJWFT
UIFOFVUSBMMJOFDVSSFOUBT
*O = − *B + *C + *D *OBUISFFXJSFTZTUFNXIFSFUIFOFVUSBMMJOFJTBCTFOU XFDBOTUJMM
find the line currents *B *C BOE*DVTJOHNFTIBOBMZTJT"UOPEF/ ,$-
NVTUCFsatisfied so UIBU*B + *C + *D = JOUIJTDBTF5IFTBNFDPVMECF
EPOFGPSBOVOCBMBODFE ∆: :∆ PS ∆∆UISFFXJSFTZTUFN "TNFO
UJPOFEFBSMJFS JOMPOHEJTUBODFQP XFSUSBOTNJTTJPO DPOEVDUPSTJONVM UJQMFTPGUISFF NVMUJQMFUISFFXJSFTZTUFNT BSFVTFE XJUIUIFFBSUIJUTFMG
BDUJOHBTUIFOFVUSBMDPOEVDUPS
5PDBMDVMBUFQP XFSJOBOVOCBMBODFEUISFFQIBTFTZTUFNSFRVJSFT
that we find the poXFSJOFBDIQIBTFVTJOH&RT UP 5IF
UPUBMQPXFSJTOPUTJNQMZUISFFUJNFTUIFQPXFSJOPOFQIBTFCVUUIFTVN
PGUIFQPXFSTJOUIFUISFFQIBTFT
Example 12.9
5IFVOCBMBODFE:MPBEPG'JHIBTCBMBODFEWPMUBHFTPG7BOE
UIF BDCTFRVFODF$BMDVMBUFUIFMJOFDVSSFOUTBOEUIFOFVUSBMDVSSFOU
5BLF;" = Ω ;# = + KΩ ;$ = −KΩ
Solution:
6TJOH&R UIFMJOFDVSSFOUTBSF
ž
⧸
*B = @@@@@@@
= ⧸
ž"
ž @@@@@@@@@@@@
ž
⧸
⧸
*C = @@@@@@@@@
=
ž"
= ⧸
+ K
⧸
ž
−ž @@@@@@@@@@
−ž
⧸
⧸
*D = @@@@@@@@@@
=
−ž"
= ⧸
−K
⧸
−ž
6TJOH&R UIFDVSSFOUJOUIFOFVUSBMMJOFJT
*O = − *B + *C + *D = − − + K + −K
= − + K = ⧸
ž"
Practice Problem 12.9
Ia
A
Ib
Ic
‒j5 Ω
B
Figure 12.24
Answer:⧸
−ž" ⧸
−ž" ⧸
ž"
8Ω
10 Ω
j6 Ω
16 Ω
5IFVOCBMBODFE∆MPBEPG'JHJTTVQQMJFECZCBMBODFEMJOFUPMJOF
WPMUBHFTPG7JOUIFQPTJUJWFTFRVFODF'JOEUIFMJOFDVSSFOUT5BLF
7BCBTSFGFSFODF
C
6OCBMBODFE∆MPBEGPS1SBDUJDF1SPC. 12.8
Unbalanced Three-Phase Systems
'or the unbalanced circuit in Fig. 12.25, find: (a) the line currents, (b) the
UPUBMDPNQMFYQPXFSBCTPSCFECZUIFMPBE BOE D UIFUPUBMDPNQMFx QPXFS
BCTPSCFECZUIFTPVSDF
Ia
a
120 0° rms +
‒
j5 W
I1
n
120 120° rms +‒
A
N
‒ 120 ‒120° rms
+
Ib
c
b
‒ j 10 W
10 W
C
B
I2
Ic
Figure 12.25
'PS&YBNQMF
Solution:
(a) We use mesh analysis to find the required currents. For mesh 1,
⧸
−ž−⧸
ž + + K *1−*2 = PS
@@
⧸ž
+ K *1−*2 = Ŀ
'PSNFTI
⧸
ž−⧸
−ž + −K *2−*1 = PS
@@
⧸−ž
−*1 + −K *2 = Ŀ
][ ] [
]
&RVBUJPOT BOE GPSNBNBUSJYFRVBUJPO
[
@@
⧸
ž
Ŀ
+ K *1
−
= @@@@@@@@@@@
@@
−
−K *2
⧸−ž
Ŀ
5IFEFUFSNJOBOUTBSF
|
|
+ K −
∆ = = −K = ⧸
−ž
−
−K
|
@@
|
⧸
ž −
Ŀ
∆1 = = −K
@@
Ŀ
⧸−ž −K
= ⧸
−ž
|
@@
|
Ŀ
⧸ž
+ K ∆2 = = −K
@@
−
Ŀ
⧸−ž
= ⧸
−ž
5IFNFTIDVSSFOUTBSF
−ž
⧸
∆
*1 = @@@1 = @@@@@@@@@@@@ = "
∆
⧸
−ž
−ž
⧸
∆
ž"
*2 = @@@2 = @@@@@@@@@@@@@
= ⧸
∆
⧸
−ž
525
Example 12.10
526
Chapter 12
Three-Phase Circuits
5IFMJOFDVSSFOUTBSF
*B = *1 = " *D = −*2 = ⧸
−ž"
*C = *2−*1 = + K− = ⧸
ž"
C 8FDBOOPXDBMDVMBUFUIFDPNQMFYQPXFSBCTPSCFECZUIFMPBE'PS
QIBTF"
4" = ]*B];" = K = K 7"
'PSQIBTF#
4# = ]*C];# = = 7"
'PSQIBTF$
4$ = ]*D];$ = −K = −K 7"
5IFUPUBMDPNQMFYQPXFSBCTPSCFECZUIFMPBEJT
4- = 4" + 4# + 4$ = −K7"
(c) We check the result above by finding the power absorbed by the
TPVSDF'PSUIFWPMUBHFTPVSDFJOQIBTFB
4B = −7BO* B = − ⧸
ž = −7"
'PSUIFTPVSDFJOQIBTFC
4C = −7CO* C = − ⧸
−ž ⧸
−ž
= −⧸
ž = −K7"
'PSUIFTPVSDFJOQIBTFD
4D = −7CO* D = − ⧸
ž ⧸
ž
= −⧸
ž = − + K7"
5IFUPUBMDPNQMFYQPXFSBCTPSCFECZUIFUISFFQIBTFTPVSDFJT
4T = 4B + 4C + 4D = − + K7"
TIPXJOHUIBU 4T + 4- = 0 and confirming the conservation principle of
BDQPXFS
Practice Problem 12.10
'JOEUIFMJOFDVSSFOUTJOUIFVOCBMBODFEUISFFQIBTFDJSDVJUPG'JH
BOEUIFSFBMQPXFSBCTPSCFECZUIFMPBE
a
440 ‒120° rms V +‒
A
+ 440 0° rms V
‒
‒+
c
440 120° rms V
b
‒j5 Ω
B
10 Ω
j 10 Ω
Figure 12.26
'PS1SBDUJDF1SPC
Answer:⧸
ž" ⧸
−ž" ⧸
−ž" L8
C
PSpice for Three-Phase Circuits
12.9
527
PSpice for Three-Phase Circuits
12.9
14QJDFDBOCFVTFEUPBOBMZ[FUISFFQIBTFCBMBODFEPSVOCBMBODFEDJS DVJUTJOUIFTBNFXBZJUJTVTFEUPBOBMZ[FTJOHMFQIBTFBDDJSDVJUT)PX
FWFS BEFMUBDPOOFDUFETPVSDFQSFTFOUTUXPNBKPSQSPCMFNTUP 14QJDF
'JSTU BEFMUBDPOOFDUFETPVSDFJTBMPPQPGWPMUBHFTPVSDFT‡XIJDI
14QJDFEPFTOPUMJLF5PBWPJEUIJTQSPCMFN XFJOTFSUBSFTJTUPSPGOFH MJHJCMFSFTJTUBODF TBZ μΩQFSQIBTF JOUPFBDIQIBTFPGUIFEFMUB
DPOOFDUFETPVSDF4FDPOE UIFEFMUBDPOOFDUFETPVSDFEPFTOPUQSPWJEFB
DPOWFOJFOUOPEFGPSUIFHSPVOEOPEF XIJDIJTOFDFTTBSZUPSVO 14QJDF
5IJT QSPCMFN DBOCF FMJNJOBUFE CZ JOTFSUJOH CBMBODFE XZFDPOOFDUFE
MBSHFSFTJTUPST TBZ .ΩQFSQIBTF JOUIF EFMUBDPOOFDUFETPVSDFTP
UIBUUIFOFVUSBMOPEFPGUIFXZFDPOOFDUFESFTJTUPSTTFSWFTBTUIFHSPVOE
OPEF&YBNQMFXJMMJMMVTUSBUFUIJT
'PSUIFCBMBODFE :∆DJSDVJUJO'JH VTF 14QJDF to find the line
DVSSFOU*B" UIFQIBTFWPMUBHF7"# BOEUIFQIBTFDVSSFOU*"$"TTVNFUIBU
UIFTPVSDFGSFRVFODZJT)[
100 0° V
‒+
a
1Ω
A
100 Ω
100 ‒120° V
‒+
n
b
1Ω
100 Ω
0.2 H
B
100 Ω
100 120° V
‒+
c
1Ω
0.2 H
0.2 H
C
Figure 12.27
'PS&YBNQMF
Solution:
5IFTDIFNBUJDJTTIPXOJO'JH5IFQTFVEPDPNQPOFOUT*13*/5
BSFJOTFSUFEJOUIFBQQSPQSJBUFMJOFTUPPCUBJO
*B"BOE*"$ XIJMF713*/5
JTJOTFSUFECFUXFFOOPEFT"BOE#UPQSJOUEJGGFSFOUJBMWPMUBHF
7"#
8FTFUUIFBUUSJCVUFTPG*13*/5BOE713*/5FBDIUP
"$ = ZFT ."( = ZFT 1)"4& = ZFT UPQSJOUPOMZUIFNBHOJUVEFBOEQIBTFPG the DVSSFOUTBOEWPMUBHFT"TBTJOHMFGSFRVFODZBOBMZTJT XFTFMFDU
"OBMZTJT4FUVQ"$4XFFQBOEFOUFS 5PUBM1UT = 4UBSU'SFR = BOE'JOBM'SFR = 0ODFUIFDJSDVJUJTTBWFE JUJTTJNVMBUFECZTFMFDU
JOH"OBMZTJT4JNVMBUF. The output file includes the following:
'3&2
& 7 " # & 71 " #
& '3&2
& *. 7@13*/5 *1 7@13*/5
& & '3&2
& *. 7@13*/5 *1 7@13*/5
& & Example 12.11
528
Chapter 12
ACMAG = 100 V
ACPHASE = 0
AC = yes
MAG = yes
PHASE = yes
Three-Phase Circuits
AC = yes
MAG = yes
PHASE = yes
IPRINT
R1
+
‒
A
V1
ACMAG = 100
ACPHASE = ‒120
1
R2
+
‒
V2
ACMAG = 100 V
ACPHASE = 120
100
0.2H
B
L1
R5
100
0.2H
C
L3
1
R3
+
‒
V3
R4
1
R6
100
IPRINT
0.2H
L2
AC = yes
MAG = yes
PHASE = yes
0
Figure 12.28
4DIFNBUJDGPSUIFDJSDVJUJO'JH
'SPNUIJT XFPCUBJO
*B" = ⧸
−ž"
7"# = ⧸
ž7 *"$ = ⧸
−ž"
Practice Problem 12.11
3FGFSUPUIFCBMBODFE::DJSDVJUPG'JH6TF 14QJDF to find the
MJOFDVSSFOU*C#BOEUIFQIBTFWPMUBHF7"/5BLFG = )[
120 60° V
‒+
a
2Ω
1.6 mH
A
10 Ω
120 ‒60° V
n
‒+
b
2Ω
1.6 mH
B
10 Ω
10 mH
10 mH
N
10 Ω
120 180° V
‒+
c
2Ω
10 mH
1.6 mH
C
Figure 12.29
'PS1SBDUJDF1SPC
Answer:⧸
ž7 ⧸
−ž"
Example 12.12
$POTJEFSUIFVOCBMBODFE ∆∆DJSDVJUJO'JH6TF 14QJDFU P
find the generator current *BC UIFMJOFDVSSFOU *C# BOEUIFQIBTF DVSSFOU*#$
12.9
a
A
2Ω
+ 208
‒
+
‒
208 130° V
529
PSpice for Three-Phase Circuits
b
j5 Ω
50 Ω
10° V
2Ω
+ 208 ‒110° V
‒
2Ω
‒ j40 Ω
B
j5 Ω
j30 Ω
j5 Ω
c
C
Figure 12.30
'PS&YBNQMF
Solution:
Define.The problem and solution process are clearly defined.
1SFTFOU8e are to find the generator current floXJOHGSPNBUPC the line current floXJOHGSPNCUP# and the phase current floXJOH
GSPN#UP$
"MUFSOBUJWF"MUIPVHIUIFSFBSFEJGGFSFOUBQQSPBDIFTUPTPMWJOH
UIJTQSPCMFN UIFVTFPG14QJDFJTNBOEBUFE5IFSFGPSF XFXJMMOPU
VTFBOPUIFSBQQSPBDI
"UUFNQU"TNFOUJPOFEBCPWF XFBWPJEUIFMPPQPGWPMUBHF
TPVSDFTCZJOTFSUJOHBμΩTFSJFTSFTJTUPSJOUIFEFMUBDPOOFDUFE
TPVSDF5PQSPWJEFBHSPVOEOPEF XFJOTFSUCBMBODFEXZF
DPOOFDUFESFTJTUPST .ΩQFSQIBTF JOUIFEFMUBDPOOFDUFE
TPVSDF BTTIPwn in the schematic in Fig. 12.31. 5ISFF*13*/5
QTFVEPDPNQPOFOUTXJUIUIFJSBUUSJCVUFTBSFJOTFSUFEUPCFBCMF
R7
1u
AC = yes
MAG = yes
PHASE = yes
R5
R4
V3
ACPHASE = 130
+
‒
ACMAG = 208 V
1u
2
5
R10
PRINT2
1Meg
ACMAG = 208 V V1
+
PRINT1
‒
IPRINT
ACPHASE = 10
AC
= yes
R8
L2 MAG = yes
R2
1Meg
PHASE = yes
5
2
1Meg
ACMAG = 208 V
ACPHASE = ‒110
Figure 12.31
L1
IPRINT
R9
R6
R1
4DIFNBUJDGPSUIFDJSDVJUJO'JH
+
‒
1u
L4
V2
R3
AC = yes
MAG = yes
PHASE = yes
L3
2
5
50
C1
0.025
30
IPRINT
PRINT3
530
Chapter 12
Three-Phase Circuits
UPHFUUIFSFRVJSFEDVSSFOUT*BC *C# BOE*#$4JODFUIFPQFSBUJOH
GSFRVFODZJTOPUHJWFOBOEUIFJOEVDUBODFTBOEDBQBDJUBODFT
should be specified instead of impedances, we assume ω = SBET
TPUIBUG = ∕π = )[5IVT
9- = @@@
ω BOE
$ = @@@@
ω9$
8FTFMFDU"OBMZTJT4FUVQ"$4XFFQBOEFOUFS5PUBM1UT = 4UBSU'SFR = BOE'JOBM'SFR = 0ODFUIF
TDIFNBUJDJTTBWFE XFTFMFDU"OBMZTJT4JNVMBUFUPTJNVMBUFUIF
circuit. The output file includes:
'3&2
&
*. 7@13*/5 *1 7@13*/5
& & '3&2
&
*. 7@13*/5 *1 7@13*/5
& & '3&2
&
*. 7@13*/5 *1 7@13*/5
& & XIJDIZJFMET
*BC = ⧸
−ž " *C# = ⧸
ž" BOE
*#$ = ⧸
ž"
&WBMVBUF8FDBODIFDLPVSSFTVMUTCZVTJOHNFTIBOBMZTJT
Let the loop B"#CCFMPPQ UIFMPPQC#$DCFMPPQ BOEUIF
MPPQ"$# be loop 3, with the three loop currents all floXJOHJO
UIFDMPDLXJTFEJSFDUJPO8FUIFOFOEVQXJUIUIFGPMMPXJOHMPPQ
FRVBUJPOT
-PPQ
+ K *1− + K *2− *3 = ⧸
ž = + K
-PPQ
− + K *1 + + K *2− K *3 = ⧸
−ž
= −−K
-PPQ
− *1− K *2 + −K *3 = 6TJOH."5-"#UPTPMWFUIJTXFHFU
; = < J
J J
J J
J >
J
;=
JJ
J + JJ
JJ
12.9
PSpice for Three-Phase Circuits
7 < J J >
7
F J
J
* JOW ; 7
*
J
J
J
*C# = −*1 + *2 = − + K + + K
= − + K = ⧸
ž"
"OTXFSDIFDLT
*#$ = *2−*3 = + K − + K
= − + K = ⧸
ž"
"OTXFSDIFDLT
/PXUPTPMWFGPS*BC*GXFBTTVNFBTNBMMJOUFSOBMJNQFEBODFGPS
FBDITPVSDF XFDBOPCUBJOBSFBTPOBCMZHPPEFTUJNBUFGPS*BC
"EEJOHJOJOUFSOBMSFTJTUPSTPGΩ BOEBEEJOHBGPVSUIMPPQ
BSPVOEUIFTPVSDFDJSDVJU XFOPXHFU
-PPQ
+ K *1− + K *2− *3 −*4 = ⧸
ž
= + K
-PPQ
− + K *1 + + K *2− K *3−*4
= ⧸
−ž = −−K
-PPQ
− *1− K *2 + −K *3 = -PPQ
− *1− *2 + *4 = ; < J
J J
J J J J >
;=
+ JJ JJJ
JJ
7 = < + J J >
531
532
Chapter 12
Three-Phase Circuits
7
F J
J
* JOW ; 7
*
J
J
J
J
*BC = −*1 + *4 = − + K + + K
= −−K = ⧸
−ž" "OTXFSDIFDLT
4BUJTGBDUPSZ 8FIBWFBTBUJTGBDUPSZTPMVUJPOBOEBOBEFRVBUF
DIFDLGPSUIFTPMVUJPO8FDBOOPXQSFTFOUUIFSFTVMUTBTBTPMVUJPO
UPUIFQSPCMFN
'PSUIFVOCBMBODFEDJSDVJUJO'JH VTF14QJDF to find the generator
DVSSFOU*DB UIFMJOFDVSSFOU*D$ BOEUIFQIBTFDVSSFOU*"#
a
A
10 Ω
+ 220 ‒30° V
‒
220 90° V
+
‒
Practice Problem 12.12
j10 Ω
b
B
10 Ω
+ 220 ‒150° V
‒
c
10 Ω
‒j 10 Ω
C
Figure 12.32
'PS1SBDUJDF1SPC
Answer:⧸
−ž" ⧸
ž" ⧸
−ž"
12.10
Applications
#PUIXZFBOEEFMUBTPVSDFDPOOFDUJPOTIBWFJNQPSUBOUQSBDUJDBMBQQMJDB
UJPOT5IFXZFTPVSDFDPOOFDUJPOJTVTFEGPSMPOHEJTUBODFUSBOTNJTTJPO
PGFMFDUSJDQPXFS XIFSFSFTJTUJWFMPTTFT *3 TIPVMECFNJOJNBM5IJT
12.10
533
Applications
@@
JTEVFUPUIFGBDUUIBUUIFXZFDPOOFDUJPOHJWFTBMJOFWPMUBHFUIBUJT Ŀ HSFBUFSUIBOUIFEFMUBDPOOFDUJPOIFODF
GPSUIFTBNFQPXFS UIFMJOF
@@
TNBMMFS*OBEEJUJPO EFMUBDPOOFDUFEBSFBMTPVOEFTJSBCMF
DVSSFOUJT Ŀ
EVFUPUIFQPUFOUJBMPGIBWJOHEJTBTUSPVTDJSDVMBUJOHDVSSFOUT4PNFUJNFT VTJOHUSBOTGPSNFST XFDSFBUF UIFFRVJWBMFOU PGEFMUBDPOOFDU TPVSDF
5IJTDPOWFSTJPOGSPNUISFFQIBTFUPTJOHMFQIBTFJTSFRVJSFEJOSFTJEFO
UJBMXJSJOH CFDBVTFIPVTFIPMEMJHIUJOHBOEBQQMJBODFTVTFTJOHMFQIBTF
QPXFS5ISFFQIBTFQPXFSJTVTFEJOJOEVTUSJBMXJSJOHXIFSFBMBSHF
QPXFSJTSFRVJSFE*OTPNFBQQMJDBUJPOT JUJTJNNBUFSJBMXIFUIFSUIFMPBE
JTXZFPSEFMUBDPOOFDUFE'PSFYBNQMF CPUIDPOOFDUJPOTBSFTBUJTGBD UPSZXJUIJOEVDUJPONPUPST*OGBDU TPNFNBOVGBDUVSFSTDPOOFDUBNPUPS
JOEFMUBGPS7BOEJOXZFGPS7TPUIBUPOFMJOFPGNPUPSTDBOCF
SFBEJMZBEBQUFEUPUXPEJGGFSFOUWPMUBHFT
)FSFXFDPOTJEFSUXPQSBDUJDBMBQQMJDBUJPOTPGUIPTFDPODFQUTDP W
FSFE JO UIJT DIBQUFS QPXFS NFBTVSFNFOU JO UISFFQIBTF DJSDVJUT BOE
SFTJEFOUJBMXJSJOH
12.10.1
Three-Phase Power Measurement
4FDUJPOQSFTFOUFEUIFXBUUNFUFSBTUIFJOTUSVNFOUGPSNFBTVSJOH
UIFBWFSBHF PSSFBM QPXFSJOTJOHMFQIBTFDJSDVJUT"TJOHMFXBUUNFUFS DBOBMTPNFBTVSFUIFBWFSBHFQPXFSJOBUISFFQIBTFTZTUFNUIBUJTCBM BODFE TPUIBU 11 = 12 = 13UIFUPUBMQPXFSJTUISFFUJNFTUIFSFBEJOH
PGUIBUPOFXBUUNFUFS)PXFWFS UXPPSUISFFTJOHMFQIBTFXBUUNFUFST BSFOFDFTTBSZUPNFBTVSFQPXFSJGUIFTZTUFNJTVOCBMBODFE5IFUISFF
XBUUNFUFSNFUIPE PGQPXFSNFBTVSFNFOU TIPXOJO'JH XJMM
XPSLSFHBSEMFTTPGXIFUIFSUIFMPBEJTCBMBODFEPSVOCBMBODFE XZFPS
EFMUBDPOOFDUFE5IF UISFFXBUUNFUFSNFUIPEJT XFMMTVJUFEGPS QPXFS
NFBTVSFNFOUJOBUISFFQIBTFTZTUFNXIFSFUIFQPXFSGBDUPSJTDPO TUBOUMZDIBOHJOH5IFUPUBMBWFSBHFQPXFSJTUIFBMHFCSBJDTVNPGUIF UISFFXBUUNFUFSSFBEJOHT
15 = 11 + 12 + 13
±
a
W1
±
±
b
W2
±
O
c
±
W3
Three-phase
load (wye
or delta,
balanced or
unbalanced)
±
Figure 12.33
5ISFFXBUUNFUFSNFUIPEGPSNFBTVSJOH
UISFFQIBTFQPXFS
XIFSF11 12 BOE 13DPSSFTQPOEUPUIFSFBEJOHTPGXBUUNFUFST 81 82 BOE 83 SFTQFDUJWFMZ /PUJDF UIBUUIF DPNNPOPSSFGFSFODF QPJOU PJO
Fig. 12.33 is selected arbitrarily. If the load is wye-connected, point P
DBOCFDPOOFDUFEUPUIFOFVUSBMQPJOU O'PSBEFMUBDPOOFDUFEMPBE point PDBOCFDPOOFDUFEUPBOZQPJOU*GQPJOU PJTDPOOFDUFEUPQPJOU
C GPSFYBNQMF UIFWPMUBHFDPJMJOXBUUNFUFS 82SFBET[FSPBOE 12 = JOEJDBUJOHUIBUXBUUNFUFS82JTOPUOFDFTTBSZ5IVT UXPXBUUNFUFSTBSF
sufficient to measure the total power.
5IFUXPXBUUNFUFSNFUIPEJTUIFNPTUDPNNPOMZVTFENFUIPEGPS
UISFFQIBTFQPXFSNFBTVSFNFOU5IFUXPXBUUNFUFSTNVTUCFQSPQFSMZ
DPOOFDUFEUPBO ZUXPQIBTFT BTTIP XOUZQJDBMMZJO'JH/PUJDF
UIBUUIFDVSSFOUDPJMPGFBDIX BUUNFUFSNFBTVSFTUIFMJOFDVSSFOU XIJMF
UIFSFTQFDUJWFWPMUBHFDPJMJTDPOOFDUFECFUXFFOUIFMJOFBOEUIFUIJSE
MJOFBOENFBTVSFTUIFMJOFW PMUBHF"MTPOPUJDFUIBUUIF ±UFSNJOBMPG
UIFWPMUBHFDPJMJTDPOOFDUFEUPUIFMJOFUPXIJDIUIFDPSSFTQPOEJOHDVS SFOU DPJM JT DPOOFDUFE "MUIPVHI UIF JOEJWJEVBM XBUUNFUFST OP MPOHFS
SFBEUIFQP XFSUBLFOCZBO ZQBSUJDVMBSQIBTF UIFBMHFCSBJDTVNPGUIF
UXPXBUUNFUFSSFBEJOHTFRVBMTUIFUPUBMB WFSBHFQPXFSBCTPSCFECZUIF
MPBE SF HBSEMFTTPGXIFUIFSJUJTXZFPSEFMUBDPOOFDUFE CBMBODFEPS
a
±
W1
±
b
c
±
W2
Three-phase
load (wye
or delta,
balanced or
unbalanced)
±
Figure 12.34
5XPXBUUNFUFSNFUIPEGPSNFBTVSJOH
UISFFQIBTFQPXFS
534
Chapter 12
Three-Phase Circuits
VOCBMBODFE5IFUPUBMSFBMQPXFSJTFRVBMUPUIFBMHFCSBJDTVNPGUIFUXP
XBUUNFUFSSFBEJOHT
15 = 11 + 12
8FXJMMTIPXIFSFUIBUUIFNFUIPEXPSLTGPSBCBMBODFEUISFFQIBTF
TZTUFN
$POTJEFSUIFCBMBODFE XZFDPOOFDUFEMPBEJO'JH0VSPC
KFDUJWFJTUPBQQMZUIFUXPXattmeter method to find the aWFSBHFQPXFS
BCTPSCFECZUIFMPBE "TTVNFUIFTPVSDFJTJOUIF BCDTFRVFODFBOE
UIFMPBEJNQFEBODF;: = ;:⧸ θ%VFUPUIFMPBEJNQFEBODF FBDIWPMU
BHFDPJMMFBETJUTDVSSFOUDPJMCZ θ TPUIBUUIFQP XFSGBDUPSJTDPT θ
8FSFDBMMUIBUFBDIMJOFWPMUBHFMFBETUIFDPSSFTQPOEJOHQIBTFWPMUBHF
CZž 5IVT UIFUPUBMQIBTFEJG GFSFODFCFUXFFOUIFQIBTFDVSSFOU *BBOEMJOFW PMUBHF 7BCJT θ + ž BOEUIFB WFSBHFQP XFSSFBECZ
XBUUNFUFS81JT
11 = 3F<7BC* B> = 7BC*BDPT θ + ž = 7-*-DPT θ + ž W1
a
+
b
Ib
‒
ZY
‒
ZY
ZY
Vcb
c
Ia
±
±
Vab
+
±
W2
±
Ic
Figure 12.35
5XPXBUUNFUFSNFUIPEBQQMJFEUPBCBMBODFEXZFMPBE
Similarly, we can show that the average power read by wattmeter 2 is
12 = 3F<7DC* D> = 7DC*DDPT θ −ž = 7-*-DPT θ −ž 8FOPXVTFUIFUSJHPOPNFUSJDJEFOUJUJFT
DPT " + # = DPT"DPT#−TJO"TJO#
DPT "−# = DPT"DPT# + TJO"TJO#
to find the sum and the difference of the two wattmeter readings in
&RT BOE 11 + 12 = 7-*-<DPT θ + ž + DPT θ −ž >
= 7-*- DPTθ DPTž−TJOθ TJOž
+ DPTθ DPTž + TJOθ
TJOž
@@
7-*-DPTθ
= 7-*-DPTžDPTθ = Ŀ
@@
$PNQBSJOH&R XJUI&R TIPXTUIBU
TJODFDPTž = Ŀ
UIFTVNPGUIFXBUUNFUFSSFBEJOHTHJWFTUIFUPUBMBWFSBHFQPXFS
15 = 11 + 12
12.10
Applications
535
4JNJMBSMZ
11−12 = 7-*-<DPT θ + ž −DPT θ −ž >
= 7M*- DPTθ DPTž−TJOθ TJOž
−DPTθ DPTž−TJOθ TJOž = −7-*-TJOžTJOθ
12−11 = 7-*-TJOθ
TJODFTJOž = $PNQBSJOH&R XJUI&R TIPXTUIBU
UIFEJGGFSFODFPGUIFXBUUNFUFSSFBEJOHTJTQSPQPSUJPOBMUPUIFUPUBMSFBD
UJWFQPXFS PS
@@
25 =Ŀ
12−11 'SPN&RT BOE UIFUPUBMBQQBSFOUQPXFSDBOCF
obtained as
@@@@@@@@
45 = Ŀ15 + 25
%JWJEJOH&R CZ&R HJWFTUIFUBOHFOUPGUIFQPXFSGBD tor angle as
@@ 1 −1
2
1
@@@@@@@
UBOθ = @@@5= Ŀ
2
15
12 + 11
GSPNXIJDIXFDBOPCUBJOUIFQPXFSGBDUPSBTQG
= DPT θ5IVT UIF
UXPXBUUNFUFSNFUIPEOPUPOMZQSPWJEFTUIFUPUBMSFBMBOESFBDUJWFQPX
FST JUDBOBMTPCFVTFEUPDPNQVUFUIFQPXFSGBDUPS'SPN&RT BOE XFDPODMVEFUIBU
*G12 = 11 UIFMPBEJTSFTJTUJWF
*G1211 UIFMPBEJTJOEVDUJWF
*G1211 UIFMPBEJTDBQBDJUJWF
"MUIPVHIUIFTFSFTVMUTBSFEFSJWFEGSPNBCBMBODFEXZFDPOOFDUFEMPBE UIFZBSFFRVBMMZWBMJEGPSBCBMBODFEEFMUBDPOOFDUFEMPBE)PXFWFS UIFUXPXBUUNFUFSNFUIPEDBOOPUCFVTFEGPSQPXFSNFBTVSFNFOUJOB
UISFFQIBTFGPVSXJSFTZTUFNVOMFTTUIFDVSSFOUUISPVHIUIFOFVUSBMMJOF
JT[FSP8FVTFUIFUISFFXBUUNFUFSNFUIPEUPNFBTVSFUIFSFBMQPXFSJO
BUISFFQIBTFGPVSXJSFTZTUFN
5ISFFXBUUNFUFST81 82 BOE83BSFDPOOFDUFE SFTQFDUJWFMZ UPQIBTFT
B C BOE D UP NFBTVSF UIF UPUBM QPXFS BCTPSCFE CZ UIFVOCBMBODFE
XZFDPOOFDUFEMPBEJO&YBNQMF TFF'JH B 1SFEJDUUIF
XBUUNFUFSSFBEJOHT C 'JOEUIFUPUBMQPXFSBCTPSCFE
Solution:
1BSUPGUIFQSPCMFNJTBMSFBEZTPMWFEJO&YBNQMF"TTVNFUIBUUIF
XBUUNFUFSTBSFQSPQFSMZDPOOFDUFEBTJO'JH
Example 12.13
536
Chapter 12
Three-Phase Circuits
Ia
A
W1 +
‒
‒
VCN
VBN
Ib
W2
+
15 Ω
VAN
‒
In
Ic
N
10 Ω
6Ω
‒j8 Ω
j5 Ω
+
C
B
W3
Figure 12.36
'PS&YBNQMF
B 'SPN&YBNQMF
7"/ = ⧸
ž 7#/ = ⧸
ž 7$/ = ⧸
−ž7
*C = ⧸
ž *D = ⧸
−ž"
XIJMF
*B = ⧸
ž 8FDBMDVMBUFUIFXBUUNFUFSSFBEJOHTBTGPMMPXT
11 = 3F 7"/* B = 7"/*BDPT θ7"/−θ*B
= × × DPT ž−ž = 8
12 = 3F 7#/* C = 7#/*CDPT θ7#/−θ*C
= × × DPT ž−ž = 8
13 = 3F 7$/* D = 7$/*DDPT θ7$/−θ*D
= × × DPT −ž + ž = 8
C 5IFUPUBMQPXFSBCTPSCFEJT
15 = 11 + 12 + 13 = + + = 8
8Fcan find UIFQPXFS BCTPSCFECZUIF SFTJTUPSTJO'JH BOEVTF
that to check or confirm this result
15 = ]*B] + ]*C] + ]*D] = + + = + + = 8
XIJDIJTFYBDUMZUIFTBNFUIJOH
Practice Problem 12.13
3FQFBU&YBNQMFGPSUIFOFUXPSLJO'JH TFF1SBDUJDF
Prob. )JOU$POOFDUUIFSFGFSFODFQPJOUP in Fig. 12.33 to point #
Answer: B L8 8 L8 C L8
Example 12.14
5IFUXPXBUUNFUFSNFUIPEQSPEVDFTX BUUNFUFSSFBEJOHT 11 = 8
BOE12 = 8XIFODPOOFDUFEUPBEFMUBDPOOFDUFEMPBE*GUIFMJOF
WPMUBHFJT7 DBMDVMBUF B UIFQFSQIBTFBWFSBHFQPXFS C UIFQFS
QIBTFSFBDUJWFQPXFS D UIFQPXFSGBDUPS BOE E UIFQIBTFJNQFEBODF
12.10
Applications
537
Solution:
8FDBOBQQMZUIFHJWFOSFTVMUTUPUIFEFMUBDPOOFDUFEMPBE
B 5IFUPUBMSFBMPSBWFSBHFQPXFSJT
15 = 1 + 12 = + = 8
5IFQFSQIBTFBWFSBHFQPXFSJTUIFO
1 = 8
1Q = @@
5
C 5IFUPUBMSFBDUJWFQPXFSJT
@@
@@
12−11 = Ŀ
25 = Ŀ
− = 7"3
TPUIBUUIFQFSQIBTFSFBDUJWFQPXFSJT
2 = 7"3
2Q = @@
5
D 5IFQPXFSBOHMFJT
2
θ = UBO¢@@@
5= UBO¢@@@@@
= ž
15
)FODF UIFQPXFSGBDUPSJT
DPTθ = MBHHJOH
*UJTBMBHHJOHQGCFDBVTF25JTQPTJUJWFPS1211
E 5IFQIBTFJNQFEBODFJT;Q = ;Q⧸
θ8FLOPXUIBUθJTUIFTBNFBTUIF
QGBOHMFUIBUJT θ = ž
7Q
;Q = @@@
*Q
8FSFDBMMUIBUGPSBEFMUBDPOOFDUFEMPBE Eq. (12.46),
1Q = 7Q*QDPTθ
⇒
7Q = 7- = 7'SPN
*Q = @@@@@@@@@@@
= "
× )FODF
7Q
;Q = @@@= @@@@@
= Ω
*Q BOE
;Q = ⧸
žΩ
-FUUIFMJOFWPMUBHF 7- = 7BOEUIFXBUUNFUFSSFBEJOHTPGUIF
CBMBODFETZTUFNJO'JHCF 11 = −8BOE 12 = 8
%FUFSNJOF
B UIFUPUBMBWFSBHFQPXFS
C UIFUPUBMSFBDUJWFQPXFS
D UIFQPXFSGBDUPS
E UIFQIBTFJNQFEBODF
*TUIFJNQFEBODFJOEVDUJWFPSDBQBDJUJWF
Answer: B 8 C L7 "3 D E ⧸
žΩ JOEVDUJWF
Practice Problem 12.14
538
Example 12.15
Chapter 12
Three-Phase Circuits
5IFUISFFQIBTFCBMBODFEMPBEJO'JHIBTJNQFEBODFQFSQIBTFPG
;: = + KΩ*GUIFMPBEJTDPOOFDUFEUP7MJOFT QSFEJDUUIFSFBE JOHTPGUIFXBUUNFUFST81BOE82'JOE15BOE25
Solution:
5IFJNQFEBODFQFSQIBTFJT
;: = + K = ⧸
žΩ
TPUIBUUIFQGBOHMFJTž4JODFUIFMJOFWPMUBHF7- = 7 UIFMJOF
DVSSFOUJT
@@
7Q @@@@@@@
∕Ŀ
@@@@
*- =
=
= "
;
] :]
5IFO
11 = 7-*-DPT θ + ž = × × DPT ž + ž
= 8
12 = 7-*-DPT θ −ž = × × DPT ž−ž
= 8
5IVT XBUUNFUFSSFBET8 XIJMFXBUUNFUFSSFBET8
4JODF 12 11 UIFMPBEJTJOEVDUJWF5IJTJTFWJEFOUGSPNUIFMPBE ;:
JUTFMG/FYU
15 = 11 + 12 = L8
BOE
@@
@@
12−11 = Ŀ
7"3 = L7"3
25 = Ŀ
Practice Problem 12.15
*GUIFMPBEJO'JHJTEFMUBDPOOFDUFEXJUIJNQFEBODFQFSQIBTFPG
;Q = KΩBOE7- = 7 QSFEJDUUIFSFBEJOHTPGUIFXBUUNFUFST
81BOE82$BMDVMBUF15BOE25
Answer:8 L8 L8 L7"3
12.10.2 Residential Wiring
*OUIF6OJUFE4UBUFT NPTUIPVTFIPMEMJHIUJOHBOEBQQMJBODFTPQFSBUFPO
7 )[ TJOHMFQIBTFBMUFSOBUJOHDVSSFOU 5IFFMFDUSJDJUZNBZBMTP
CFTVQQMJFEBU PS7 EFQFOEJOHPOUIFMPDBUJPO 5IFMPDBM
QPXFSDPNQBOZTVQQMJFTUIFIPVTFXJUIBUISFFXJSFBDTZTUFN 5ZQJ
DBMMZ BTJO'JH UIFMJOFWPMUBHFPG TBZ 7JTTUFQQFEEPXO
UP7XJUIBUSBOTGPSNFS NPSFEFUBJMTPOUSBOTGPSNFSTJOUIFOFYU
DIBQUFS 5IFUISFFXJSFTDPNJOHGSPNUIFUSBOTGPSNFSBSFUZQJDBMMZDPM
PSFESFE IPU CMBDL IPU BOEXIJUF OFVUSBM "TTIPXOJO'JH UIFUXP7WPMUBHFTBSFPQQPTJUFJOQIBTFBOEIFODFBEEVQUP[FSP
5IBUJT 78 = ⧸
ž 7# = ⧸
ž 73 = ⧸
ž= −7#
7#3 = 7#−73 = 7#− −7# = 7# = ⧸
ž
12.10
Applications
Step-down
transformer
Circuit
#1
120 V
Wall of
house
Circuit
#2
120 V
Fuse
Circuit
#3
240 V
Fuses
Switch
Fuse
Light
pole
Watt-hour meter
Grounded metal
stake
Ground
Figure 12.37
"IPVTFIPMEQPXFSTZTUFN
4PVSDF".BSDVTBOE$.5IPNTPO &MFDUSJDJUZGPS5FDIOJDJBOT OEFEJUJPO ¥ Q1FBSTPO&EVDBUJPO *OD 6QQFS4BEEMF3JWFS /+
To other houses
Black
(hot) B
White
(neutral)
W
Ground
R
Red (hot)
+
120 V
‒
‒
120 V
+
Transformer
120 V
lights
120 V
appliance
240 V
appliance
120 V
lights
120 V
appliance
House
Figure 12.38
4JOHMFQIBTFUISFFXJSFSFTJEFOUJBMXJSJOH
4JODFNPTUBQQMJBODFTBSFEFTJHOFEUPPQFSBUFXJUI 7 UIFMJHIU JOHBOEBQQMJBODFTBSFDPOOFDUFEUPUIF7MJOFT BTJMMVTUSBUFEJO
Fig. GPSB SPPN/PUJDF JO'JH UIBUBMM BQQMJBODFTBSFDPO
Lamps
OFDUFEJOQBSBMMFM)FB WZBQQMJBODFTUIBUDPOTVNFMBS HFDVSSFOUT TVDI
BTBJSDPOEJUJPOFST EJTIXBTIFST PWFOT BOEMBVOESZNBDIJOFT BSFDPO Switch
OFDUFEUPUIF7QPXFSMJOF
Base outlets
#FDBVTFPGUIFEBOHFSTPGFMFDUSJDJUZ IPVTFXJSJOHJTDBSFGVMMZ
SFHVMBUFECZBDPEFESB XOCZMPDBMPSEJOBODFTBOECZUIF/BUJPOBM Neutral
&MFDUSJDBM$PEF /&$ 5PBWPJEUSPVCMF JOTVMBUJPO HSPVOEJOH GVTFT 120 volts
BOEDJSDVJUCSFBLFSTBSFVTFE.PEFSOXJSJOHDPEFTSFRVJSFBUIJSEXJSF
Ungrounded conductor
GPSBTFQBSBUFHSPVOE 5IFHSPVOEXJSFEPFTOPUDBSSZQP XFSMJLFUIF
OFVUSBMXJSFCVUFOBCMFTBQQMJBODFTUPIBWFBTFQBSBUFHSPVOEDPOOFD
UJPO'JHVSFTIPXTUIFDPOOFDUJPOPGUIFSFDFQUBDMFUPB7 Figure 12.39
SNTMJOFBOEUPUIFHSPVOE "TTIP wn in the figure, the neutral line "UZQJDBMXJSJOHEJBHSBNPGBSPPN
4PVSDF".BSDVTBOE$.5IPNTPO JTDPOOFDUFEUPUIFHSPVOE UIFFBSUI BUNBO ZDSJUJDBMMPDBUJPOT "M &MFDUSJDJUZGPS5FDIOJDJBOT OEFEJUJPO UIPVHI UIF HSPVOE MJOF TFFNT SFEVOEBOU HSPVOEJOHJT JNQPSUBOU GPS ¥ Q1FBSTPO&EVDBUJPO *OD NBOZSFBTPOT'JSTU JUJTSFRVJSFECZ/&$4FDPOE HSPVOEJOHQSPWJEFT 6QQFS4BEEMF3JWFS /+
539
540
Chapter 12
Three-Phase Circuits
Fuse or circuit breaker
Hot wire
Receptacle
120 V rms +
‒
To other appliances
Neutral wire
Ground wire
Power system
Service
ground
panel ground
Figure 12.40
$POOFDUJPOPGBSFDFQUBDMFUPUIFIPUMJOFBOEUPUIFHSPVOE
BDPOWFOJFOUQBUIUPHSPVOEGPSMJHIUOJOHUIBUTUSJL FTUIFQP XFSMJOF
5IJSE HSPVOETNJOJNJ[FUIFSJTLPGFMFDUSJDTIPDL8IBUDBVTFTTIPDL
JTUIFQBTTBHFPGDVSSFOUGSPNPOFQBSUPGUIFCPEZUPBOPUIFS 5IF
IVNBOCPEZJTMJL FBCJHSFTJTUPS 3*G 7JTUIFQPUFOUJBMEJG GFSFODF
between the body and UIFHSPVOE UIFDVSSFOUUISPVHIUIFCPEZJT EFUFSNJOFECZ0INTMBXBT
7
* = @@
3
5IFWBMVFPG 3WBSJFTGSPNQFSTPOUPQFSTPOBOEEFQFOETPOXIFUIFS UIFCPEZJTXFUPSESZ)PXHSFBUPSIPXEFBEMZUIFTIPDLJTEFQFOET POUIFBNPVOUPGDVSSFOU UIFQBUIXBZPGUIFDVSSFOUUISPVHIUIFCPEZ BOEUIFMFOHUIPGUJNFUIFCPEZJTF YQPTFEUPUIFDVSSFOU$VSSFOUT MFTTUIBON"NBZOPUCFIBSNGVMUPUIFCPEZ CVUDVSSFOUTHSFBUFS than 10 N"DBODBVTFTF WFSFTIPDL "NPEFSOTBGFUZEF WJDFJTUIF HSPVOEGBVMUDJSDVJUJOUFSSVQUFS ('$* VTFEJOPVUEPPSDJSDVJUTBOEJO
CBUISPPNT XIFSFUIFSJTLPGFMFDUSJDTIPDLJTHSFBUFTU*UJTFTTFOUJBMMZ
BDJSDVJUCSFBL FSUIBUPQFOTXIFOUIFTVNPGUIFDVSSFOUT J3 J8 BOE
J#UISPVHIUIFSFE XIJUF BOEUIFCMBDLMJOFTJTOPUFRVBMUP[FSP PS J3 + J8 + J#≠
5IFCFTUX BZUPB WPJEFMFDUSJDTIPDLJTUPGPMMP XTBGFUZHVJEF MJOFTDPODFSOJOHFMFDUSJDBMTZTUFNTBOEBQQMJBODFT)FSFBSFTPNFPG
UIFN
r /FWFSBTTVNFUIBUBOFMFDUSJDBMDJSDVJUJTEFBE "MXBZTDIFDLUP
be sure.
r 6TFTBGFUZEF WJDFTXIFOOFDFTTBSZ BOEXFBSTVJUBCMFDMPUIJOH
JOTVMBUFETIPFT HMPWFT FUD r /FWFSVTFUX PIBOETXIFOUFTUJOHIJHIW PMUBHFDJSDVJUT TJODFUIF
DVSSFOUUISPVHIPOFIBOEUPUIFPUIFSIBOEIBTBEJSFDUQBUIUISPVHI
ZPVSDIFTUBOEIFBSU
r %POPUUPVDIBOFMFDUSJDBMBQQMJBODFXIFOZPVBSFXFU3FNFNCFS
UIBUXBUFSDPOEVDUTFMFDUSJDJUZ
r #FFYUSFNFMZDBSFGVMXIFOXPSLJOHXJUIFMFDUSPOJDBQQMJBODFTTVDI
BTSBEJPBOE 57CFDBVTFUIFTFBQQMJBODFTIB WFMBS HFDBQBDJUPST
JOUIFN 5IFDBQBDJUPSTUBL FUJNFUPEJTDIBS HFBGUFSUIFQP XFSJT
EJTDPOOFDUFE
r "MXBZTIB WFBOPUIFSQFSTPOQSFTFOUXIFOX PSLJOHPOBXJSJOH
TZTUFN KVTUJODBTFPGBOBDDJEFOU
Review Questions
12.11
541
Summary
5IFQIBTFTFRVFODFJTUIFPSEFSJOXIJDIUIFQIBTFW PMUBHFTPGB
UISFFQIBTFHFOFSBUPSPDDVSXJUISFTQFDUUPUJNF*OBOBCDTFRVFODF
PGCBMBODFETPVSDFW PMUBHFT 7BOMFBET 7COCZž XIJDIJOUVSO
MFBET 7DOCZž*OBO BDCTFRVFODFPGCBMBODFEW PMUBHFT 7BO
MFBET7DOCZž XIJDIJOUVSOMFBET7COCZž
"CBMBODFEXZFPSEFMUBDPOOFDUFEMPBEJTPOFJOXIJDIUIFUISFF
QIBTFJNQFEBODFTBSFFRVBM
5IFFBTJFTUXBZUPBOBMZ[FBCBMBODFEUISFFQIBTFDJSDVJUJTUPUSBOT
GPSNCPUIUIFTPVSDFBOEUIFMPBEUPB::TZTUFNBOEUIFOBOBMZ[F
UIFTJOHMFQIBTFFRVJWBMFOUDJSDVJU 5BCMFQSFTFOUTBTVNNBSZ
PGUIFGPSNVMBTGPSQIBTFDVSSFOUTBOEWPMUBHFTBOEMJOFDVSSFOUTBOE
Woltages for the four possible configurations.
5IFMJOFDVSSFOU *- is the current floXJOHGSPNUIFHFOFSBUPSUPUIF
MPBEJOFBDI USBOTNJTTJPOMJOFJOBUISFFQIBTF TZTUFN 5IFMJOF
WPMUBHF7-JTUIFW PMUBHFCFUXFFOFBDIQBJSPGMJOFT F YDMVEJOHUIF
OFVUSBMMJOFJGJUF YJTUT5IFQIBTFDVSSFOU *Q is the current floXJOH
UISPVHIFBDIQIBTFJOBUISFFQIBTFMPBE5IFQIBTFWPMUBHF7QJTUIF
WPMUBHFPGFBDIQIBTF'PSBXZFDPOOFDUFEMPBE
@@
7Q
7- = Ŀ
BOE
*- = *Q
'PSBEFMUBDPOOFDUFEMPBE
7- = 7Q
@@
*- = Ŀ
*Q
BOE
5IFUPUBMJOTUBOUBOFPVTQP XFSJOBCBMBODFEUISFFQIBTFTZTUFNJT
DPOTUBOUBOEFRVBMUPUIFBWFSBHFQPXFS
5IFUPUBMDPNQMF YQP XFSBCTPSCFECZBCBMBODFEUISFFQIBTF
:DPOOFDUFEPS∆DPOOFDUFEMPBEJT
@@
4 = 1 + K2 = Ŀ
7-*-⧸
θ
XIFSFθJTUIFBOHMFPGUIFMPBEJNQFEBODFT
"OVOCBMBODFEUISFFQIBTFTZTUFNDBOCFBOBMZ[FEVTJOHOPEBMPS
NFTIBOBMZTJT
14QJDFJTVTFEUPBOBMZ[FUISFFQIBTFDJSDVJUTJOUIFTBNFX BZBTJU
JTVTFEGPSBOBMZ[JOHTJOHMFQIBTFDJSDVJUT
5IFUPUBMSFBMQPXFSJTNFBTVSFEJOUISFFQIBTFTZTUFNTVTJOHFJUIFS
UIFUISFFXBUUNFUFSNFUIPEPSUIFUXPXBUUNFUFSNFUIPE
3FTJEFOUJBMXJSJOHVTFTB7 TJOHMFQIBTF UISFFXJSFTZTUFN
Review Questions
8IBUJTUIFQIBTFTFRVFODFPGBUISFFQIBTFNPUPS
GPSXIJDI7"/ = ⧸
−ž7BOE
7#/ = ⧸
ž7
B BCD
C BDC
−ž UIFO
*GJOBOBDCQIBTFTFRVFODF 7BO = ⧸
7DOJT
8IJDIPGUIFTFJTOPUBSFRVJSFEDPOEJUJPOGPSB
CBMBODFETZTUFN
B ]7BO] = ]7CO] = ]7DO]
C *B + *C + *D = D 7BO + 7CO + 7DO = −ž
B ⧸
C ⧸
ž
E 4PVSDFWPMUBHFTBSFžPVUPGQIBTFXJUIFBDI
PUIFS
−ž
D ⧸
E ⧸
1ž
F -PBEJNQFEBODFTGPSUIFUISFFQIBTFTBSFFRVBM
542
Chapter 12
Three-Phase Circuits
*OB:DPOOFDUFEMPBE UIFMJOFDVSSFOUBOEQIBTF
DVSSFOUBSFFRVBM
B 5SVF
C 'BMTF
*OB∆DPOOFDUFEMPBE UIFMJOFDVSSFOUBOEQIBTF
DVSSFOUBSFFRVBM
B 5SVF
C 'BMTF
*OB::TZTUFN BMJOFWPMUBHFPG7QSPEVDFTB
QIBTFWPMUBHFPG
B 7
E 7
C 7
F 7
D 7
C 7
F 7
B 5SVF
C 'BMTF
*OBCBMBODFEUISFFQIBTFDJSDVJU UIFUPUBM
JOTUBOUBOFPVTQPXFSJTFRVBMUPUIFBWFSBHFQPXFS
B 5SVF
C 'BMTF
5IFUPUBMQPXFSTVQQMJFEUPBCBMBODFE∆MPBEJT
GPVOEJOUIFTBNFXBZBTGPSBCBMBODFE:MPBE
B 5SVF
*OB∆-∆TZTUFN BQIBTFWPMUBHFPG7QSPEVDFT
BMJOFWPMUBHFPG
B 7
E 7
8IFOB:DPOOFDUFEMPBEJTTVQQMJFECZWPMUBHFT
JOBCDQIBTFTFRVFODF UIFMJOFWPMUBHFTMBHUIF
DPSSFTQPOEJOHQIBTFWPMUBHFTCZž
D 7
C 'BMTF
"OTXFSTB B D B C F D C B B
Problems1
VP 0° V
Section 12.2 Balanced Three-Phase Voltages
‒+
*G7BC = 7JOBCBMBODFE:DPOOFDUFEUISFF
phase generator, find the phase voltages, assuming
UIFQIBTFTFRVFODFJT
B BCD
(JWFOBCBMBODFE:DPOOFDUFEUISFFQIBTFHFOFSBUPS
XJUIBMJOFUPMJOFWPMUBHFPG7BC⧸
45ž7BOE
7CD⧸
165ž7 EFUFSNJOFUIFQIBTFTFRVFODF
BOEUIFWBMVFPG7DB
"UISFFQIBTFTZTUFNXJUIBCDTFRVFODFBOE
7- = 7GFFETB:DPOOFDUFEMPBEXJUI
;- = ⧸
30žΩ'JOEUIFMJOFDVSSFOUT
'PSB:DPOOFDUFEMPBE UIFUJNFEPNBJOFYQSFTTJPOT
GPSUISFFMJOFUPOFVUSBMWPMUBHFTBUUIFUFSNJOBMTBSF
W"/ = DPT ωU + ž 7
W#/ = DPT ωU–ž 7
W$/ = DPT ωU + ž 7
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOECBMBODFEXZFXZF
DPOOFDUFEDJSDVJUT
VP 120° V
‒+
A
R
j XL
b
B
R
j XL
c
C
R
N
j XL
Figure 12.41
'PS1SPC
0CUBJOUIFMJOFDVSSFOUTJOUIFUISFFQIBTFDJSDVJUPG
'JHPOUIFOFYUQBHF
*OBCBMBODFEUISFFQIBTF::TZTUFN UIF
source is an BDCTFRVFODFPGWPMUBHFTBOE
7DO = ⧸
ž7SNT5IFMJOFJNQFEBODFQFS
QIBTFJT + K Ω, while the per-phase JNQFEBODF
PGUIFMPBEJT + K Ω$BMDVMBUFUIFMJOF
currents and the load WPMUBHFT
"CBMBODFE::GPVSXJSFTZTUFNIBTQIBTFWPMUBHFT
8SJUFUIFUJNFEPNBJOFYQSFTTJPOTGPSUIFMJOFUP
MJOFWPMUBHFTW"# v#$ BOEW$"
Section 12.3 Balanced Wye-Wye Connection
‒+
n
C BDC
8IBUJTUIFQIBTFTFRVFODFPGBCBMBODFEUISFF
QIBTFDJSDVJUGPSXIJDI7BO = ⧸
3ž7BOE
7DO = ⧸
−ž7 'JOE7CO
VP ‒120° V
a
7BO = ⧸
ž 7CO = ⧸
−ž
ž7
7DO = ⧸
5IFMPBEJNQFEBODFQFSQIBTFJT + KΩ, and the
MJOFJNQFEBODFQFSQIBTFJT + KΩ4PMWFGPS
the line currents and neutral current.
3FNFNCFSUIBUVOMFTTTUBUFEPUIFSXJTF BMMHJWFOWPMUBHFTBOEDVSSFOUTBSFSNTWBMVFT
543
Problems
Ia
a
A
440 0° V +
‒
6 ‒ j8 Ω
n
N
6 ‒ j8 Ω
6 ‒ j8 Ω
440 120° V +‒
‒
+ 440 ‒120° V
Ib
Ic
Figure 12.42
'PS1SPC
'PSUIFDJSDVJUJO'JH EFUFSNJOFUIFDVSSFOUJO
UIFOFVUSBMMJOF
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEXZFEFMUBDPOOFDUFE
DJSDVJUT
Ia
2Ω
880 0° V
+
‒
2Ω
‒+
10 + j5 Ω
VP 0° V
ZΔ
n
VP 120° V +‒
20 Ω
+
‒
880 120° V
+
‒
25 ‒ j10 Ω
880 ‒120° V
A
a
‒ V
P
+
‒120° V
Ib
c
b
2Ω
ZΔ
Ic
ZΔ
C
B
Figure 12.43
Figure 12.45
'PS1SPC
'PS1SPC
Section 12.4
*OUIFCBMBODFEUISFFQIBTF:∆TZTUFNJO
Fig. 12.46, find the line current *-BOEUIFBWFSBHF
QPXFSEFMJWFSFEUPUIFMPBE
Balanced Wye-Delta Connection
*OUIF:∆TZTUFNTIPXOJO'JH UIFTPVSDFJT
BQPTJUJWFTFRVFODFXJUI7BO = ⧸
ž7BOEQIBTF
JNQFEBODF;Q = –KΩ$BMDVMBUFUIFMJOFWPMUBHF
7-BOEUIFMJOFDVSSFOU*-
110 0° V rms
‒+
110 ‒120° V rms
Van
‒+
Vbn
n
‒+
Vcn
‒+
Figure 12.44
'PS1SPC
‒+
a
110 120° V rms
Zp
b
‒+
Zp
Zp
2Ω
2Ω
2Ω
9 ‒ j6 Ω
9 ‒ j6 Ω
9 ‒ j6 Ω
Figure 12.46
'PS1SPC
c
0CUBJOUIFMJOFDVSSFOUTJOUIFUISFFQIBTFDJSDVJUPG
'JHPOUIFOFYUQBHF
544
Chapter 12
Three-Phase Circuits
1 + j2 Ω
A
a
ZL
+ 100 0° V
‒
n
ZL
C
100 120° V +‒
+ 100 ‒120° V
‒
b
c
B
ZL = 12 + j 12 Ω
1 + j2 Ω
1 + j2 Ω
Figure 12.47
'PS1SPC
5IFDJSDVJUJO'JHJTFYDJUFECZBCBMBODFE
UISFFQIBTFTPVSDFXJUIBMJOFWPMUBHFPG7
*G;M = + KΩ ;∆ = −KΩ BOE
;: = + KΩ EFUFSNJOFUIFNBHOJUVEFPG
UIFMJOFDVSSFOUPGUIFDPNCJOFEMPBET
a
Zl
Section 12.5
Balanced Delta-Delta Connection
'PSUIF∆∆DJSDVJUPG'JH DBMDVMBUFUIFQIBTF
BOEMJOFDVSSFOUT
ZY
ZΔ
ZΔ
Zl
a
30 Ω
ZY
+ 173 0° V
‒
b
Zl
173 120° V
ZY
j 10 Ω
b
+
‒
c
ZΔ
A
+ 173 ‒120° V
‒
'PS1SPC
j 10 Ω
"CBMBODFEEFMUBDPOOFDUFEMPBEIBTBQIBTFDVSSFOU
*"$ = ⧸
−ž"
B %FUFSNJOFUIFUISFFMJOFDVSSFOUTBTTVNJOHUIBU
UIFDJSDVJUPQFSBUFTJOUIFQPTJUJWFQIBTFTFRVFODF
C $BMDVMBUFUIFMPBEJNQFEBODFJGUIFMJOFWPMUBHF
JT7"# = ⧸
ž7
"QPTJUJWFTFRVFODFXZFDPOOFDUFETPVSDFXIFSF
7BO =⧸
ž7 JTDPOOFDUFEUPBEFMUBDPOOFDUFEMPBE
XIFSF;-= +K Ɨ%FUFSNJOFUIFMJOFDVSSFOUT
c
C
Figure 12.50
'PS1SPC
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOECBMBODFEEFMUBEFMUB
DPOOFDUFEDJSDVJUT
žV in the network of Fig. 12.49,
*G7BO = ⧸
find the load phase currents *"# *#$ BOE*$"
12 Ω
Three-phase,
Y-connected
generator
'PS1SPC
A
IAB
j9 Ω
j9 Ω
12 Ω
b
B
c
Ia
A
a
Figure 12.49
j10 Ω
30 Ω
Figure 12.48
(+) phase
sequence
30 Ω
B
12 Ω
j9 Ω
VL 120° V +‒
+
‒
VL
0° V
ZL
ZL
ICA
Ib
‒+
C
VL
Figure 12.51
'PS1SPC
‒120° V
Ic
B
C
IBC
ZL
545
Problems
5ISFF7HFOFSBUPSTGPSNBEFMUBDPOOFDUFE
TPVSDFUIBUJTDPOOFDUFEUPBCBMBODFEEFMUB
DPOOFDUFEMPBEPG;- K ΩQFSQIBTFBT
TIPXOJO'JH%FUFSNJOFUIFWBMVFPG*#$BOE
*B"8IBUJTUIFQGPGUIFMPBE
Section 12.6
Balanced Delta-Wye Connection
*OUIFDJSDVJUPG'JH JG7BC = ⧸
ž 7CD = ⧸
−ž 7DB = ⧸
130° V, find the line
DVSSFOUT
a
a
Ia
3 + j2 Ω
Ib
3 + j2 Ω
Ic
A
‒+
ZL
+ 440 120° V
‒
b
B
+ V
ab
‒
ZL
C
Vca
+
‒
440 ‒120° V +‒
c
3 + j2 W
b
ZL
440 0° V
10 ‒ j 8 Ω
10 ‒ j 8 W
+ V
‒
bc
Figure 12.52
10 ‒ j 8 Ω
c
'PS1SPC
Figure 12.54
'PS1SPC
'JOEUIFMJOFDVSSFOUT*B" *C# BOE*D$JOUIF
UISFFQIBTFOFUXPSLPG'JH5BLF
;L = +K87) ΩBOE;M= +K Ω
"CBMBODFEEFMUBDPOOFDUFETPVSDFJTDPOOFDUFE
UPBCBMBODFEEFMUBDPOOFDUFEMPBEXIFSF;-=
+K ΩBOE;M= +K Ω(JWFOUIBUUIFMPBE
WPMUBHFTBSF7"#=⧸
ž7 7#$=⧸
ž7 BOE7$"=⧸
−ž7$BMDVMBUFUIFTPVSDFWPMU
BHFT7BC 7CD BOE7DB
"CBMBODFEEFMUBDPOOFDUFETPVSDFIBTQIBTFWPMUBHF
7BC = ⧸
ž7BOEBQPTJUJWFQIBTFTFRVFODF*G
UIJTJTDPOOFDUFEUPBCBMBODFEEFMUBDPOOFDUFEMPBE find the line and phase currents. Take the load imQFEBODFQFSQIBTFBT⧸
žΩBOEMJOFJNQFEBODF
QFSQIBTFBT + KΩ
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOECBMBODFEEFMUBDPOOFDUFE
TPVSDFTEFMJWFSJOHQPXFSUPCBMBODFEXZFDPOOFDUFE
MPBET
I aA
a
A
R
Three-phase,
Δ-connected
generator
(+) phase
sequence
‒ jXC
N
I bB
b
I cC
c
R
B
‒ jXC
Figure 12.55
'PS1SPC
Zl
a
240 ‒120° V +‒
240 0° V
Figure 12.53
'PS1SPC
ZL
+ 240 120° V
‒
‒+
c
A
b
Zl
Zl
ZL
B
C
ZL
‒j XC
R
C
546
Chapter 12
Three-Phase Circuits
%FTJHOBQSPCMFNUPIFMQPUIFSTUVEFOUTCFUUFS
VOEFSTUBOEQPXFSJOBCBMBODFEUISFFQIBTF
TZTUFN
"∆DPOOFDUFETPVSDFTVQQMJFTQPXFSUPB:
DPOOFDUFEMPBEJOBUISFFQIBTFCBMBODFETZTUFN
(JWFOUIBUUIFMJOFJNQFEBODFJT + KΩQFSQIBTF
XIJMFUIFMPBEJNQFEBODFJT + KΩ per phase, find
UIFNBHOJUVEFPGUIFMJOFWPMUBHFBUUIFMPBE"TTVNF
UIFTPVSDFQIBTFWPMUBHF7BC = ⧸
ž7SNT
5IFMJOFUPMJOFWPMUBHFTJOB:MPBEIBWFBNBHOJUVEF
of 880 V and are in the positive sequence at 60 Hz. If
UIFMPBETBSFCBMBODFEXJUI;1 = ;2 = ;3 = ⧸
ž find all line currents and phase voltages.
Section 12.7 Power in a Balanced System
"CBMBODFEUISFFQIBTF:∆TZTUFNIBT7BO = ⧸
ž
7SNTBOE;∆ = + KΩ*GUIFMJOFJNQFEBODFQFS
QIBTFJT + KΩ, find the total complex power
EFMJWFSFEUPUIFMPBE
*O'JH UIFSNTWBMVFPGUIFMJOFWPMUBHFJT
208 V. Find the average power delivered to the load.
"UISFFQIBTFTPVSDFEFMJWFSTL7"UPBXZF
DPOOFDUFEMPBEXJUIBQIBTFWPMUBHFPG7BOEB
QPXFSGBDUPSPGMBHHJOH$BMDVMBUFUIFTPVSDFMJOF
DVSSFOUBOEUIFTPVSDFMJOFWPMUBHF
"CBMBODFEXZFDPOOFDUFEMPBEXJUIBQIBTF
JNQFEBODFPG–KΩJTDPOOFDUFEUPBCBMBODFE
UISFFQIBTFHFOFSBUPSXJUIBMJOFWPMUBHFPG7
%FUFSNJOFUIFMJOFDVSSFOUBOEUIFDPNQMFYQPXFS
BCTPSCFECZUIFMPBE
5ISFFFRVBMJNQFEBODFT + KΩFBDI BSF
EFMUBDPOOFDUFEUPB7SNT UISFFQIBTFDJSDVJU
"OPUIFSUISFFFRVBMJNQFEBODFT + KΩFBDI BSFXZFDPOOFDUFEBDSPTTUIFTBNFDJSDVJUBUUIF
TBNFQPJOUT%FUFSNJOF
B UIFMJOFDVSSFOU
a
A
+ V V
a
‒
b
b
n
‒+
‒
+
c
B
ZL
C UIFUPUBMDPNQMFYQPXFSTVQQMJFEUPUIFUXP
MPBET
ZL
N
D UIFQPXFSGBDUPSPGUIFUXPMPBETDPNCJOFE
Z L = 30 45°
Vc
"7 UISFFQIBTFUSBOTNJTTJPOMJOFIBTBO
JNQFEBODFPG + KΩQFSQIBTF*GJUTVQQMJFTBMPBE
of 1 MVA at 0.75 power factor (lagging), find:
C
Figure 12.56
'PS1SPC
B UIFDPNQMFYQPXFS
C UIFQPXFSMPTTJOUIFMJOF
D UIFWPMUBHFBUUIFTFOEJOHFOE
"CBMBODFEEFMUBDPOOFDUFEMPBEJTTVQQMJFECZB
)[UISFFQIBTFTPVSDFXJUIBMJOFWPMUBHFPG
7&BDIMPBEQIBTFESBXTL8BUBMBHHJOH
QPXFSGBDUPSPG'JOE
5IFUPUBMQPXFSNFBTVSFEJOBUISFFQIBTFTZTUFN
GFFEJOHBCBMBODFEXZFDPOOFDUFEMPBEJTL8
at a power factor of 0.6 leading. If the line voltage
is 440 V, calculate the line current *-BOEUIFMPBE
JNQFEBODF;:
B UIFMPBEJNQFEBODFQFSQIBTF
C UIFMJOFDVSSFOU
D UIFWBMVFPGDBQBDJUBODFOFFEFEUPCFDPOOFDUFE
JOQBSBMMFMXJUIFBDIMPBEQIBTFUPNJOJNJ[FUIF
DVSSFOUGSPNUIFTPVSDF
Given the circuit in Fig. 12.57 below, find the total
DPNQMFYQPXFSBCTPSCFECZUIFMPBE
1Ω
110 0° V +‒
9Ω
110 240° V
‒+
‒
+
j2 Ω
110 120° V
Figure 12.57
'PS1SPC
1Ω
j2 Ω
9Ω
j12 Ω
j12 Ω
j 12 Ω
1Ω
j2 Ω
1Ω
j2 Ω
9Ω
547
Problems
'JOEUIFSFBMQPXFSBCTPSCFECZUIFMPBEJO'JH
4 + j2
a
A
+
100 90° V ‒
100 ‒150° V
‒
+
n
36 + j28 Ω
N
100 ‒30° V
‒
36 + j28 Ω
+
c
b
4 + j2
36 + j 28 Ω
B
C
4 + j2
Figure 12.58
'PS1SPC
For the three-phase circuit in Fig. 12.59, find the
BWFSBHFQPXFSBCTPSCFECZUIFEFMUBDPOOFDUFEMPBE
XJUI;Ĵ = + KΩ
100 0° V rms
‒+
100 ‒120° V rms
‒+
100 120° V rms
‒+
1Ω
j 0.5 Ω
1Ω
j0.5 Ω
1Ω
j0.5 Ω
QPXFSGSPNBUISFFQIBTFTPVSDFXJUIBMJOFWPMUBHF
PG7"TTVNF[FSPMJOFJNQFEBODF
5IFGPMMPXJOHUISFFQBSBMMFMDPOOFDUFEUISFFQIBTF
MPBETBSFGFECZBCBMBODFEUISFFQIBTFTPVSDF
-PBEL7" QGMBHHJOH
-PBEL7" QGMFBEJOH
-PBEL7" VOJUZQG
*GUIFMJOFWPMUBHFJTL7 DBMDVMBUFUIFMJOF
DVSSFOUBOEUIFQPXFSGBDUPSPGUIFTPVSDF"TTVNF
UIBUUIFMJOFJNQFEBODFJT[FSP
"CBMBODFE QPTJUJWFTFRVFODFXZFDPOOFDUFETPVSDF
IBT7BO = ⧸
ž7SNTBOETVQQMJFTBOVOCBMBODFE
EFMUBDPOOFDUFEMPBEWJBBUSBOTNJTTJPOMJOFXJUI
JNQFEBODF + KΩQFSQIBTF
B $BMDVMBUFUIFMJOFDVSSFOUTJG;"# = + KΩ ;#$ = Ω ;$" = –KΩ
C 'JOEUIFDPNQMFYQPXFSTVQQMJFECZUIFTPVSDF
&BDIQIBTFMPBEDPOTJTUTPGBΩSFTJTUPSBOEB
ΩJOEVDUJWFSFBDUBODF8JUIBMJOFWPMUBHFPG
480 V rms, calculate the average power taken by the
MPBEJG
ZΔ
ZΔ
ZΔ
Figure 12.59
'PS1SPC
"CBMBODFEEFMUBDPOOFDUFEMPBEESBXTL8BUBQPXFS
GBDUPSPGMBHHJOH*GUIFUISFFQIBTFTZTUFNIBTBO
effective line voltage of 400 V, find the line current.
"CBMBODFEUISFFQIBTFHFOFSBUPSEFMJWFSTL8UP
BXZFDPOOFDUFEMPBEXJUIJNQFEBODF–KΩQFS
QIBTF'JOEUIFMJOFDVSSFOU*-BOEUIFMJOFWPMUBHF7-
B UIFUISFFQIBTFMPBETBSFEFMUBDPOOFDUFE
C UIFMPBETBSFXZFDPOOFDUFE
"CBMBODFEUISFFQIBTFTPVSDFXJUI7- = 7SNT
JTTVQQMZJOHL7"BUQPXFSGBDUPSMBHHJOHUP
UXPXZFDPOOFDUFEQBSBMMFMMPBET*GPOFMPBEESBXT
L8BUVOJUZQPXFSGBDUPS DBMDVMBUFUIFJNQFEBODF
QFSQIBTFPGUIFTFDPOEMPBE
Section 12.8
Unbalanced Three-Phase Systems
$POTJEFSUIFXZFEFMUBTZTUFNTIPXOJO'JH
-FU;= 100 Ω ;=KΩ BOE;=–KΩ
%FUFSNJOFUIFQIBTFDVSSFOUT *"# *#$ BOE*$" BOE
UIFMJOFDVSSFOUT *B" *C# BOE*D$
3FGFSUP'JH0CUBJOUIFDPNQMFYQPXFS
BCTPSCFECZUIFDPNCJOFEMPBET
"UISFFQIBTFMJOFIBTBOJNQFEBODFPG + KΩQFS
QIBTF5IFMJOFGFFETBCBMBODFEEFMUBDPOOFDUFEMPBE XIJDIBCTPSCTBUPUBMDPNQMFYQPXFSPG + KL7"
*GUIFMJOFWPMUBHFBUUIFMPBEFOEIBTBNBHOJUVEFPG
7 DBMDVMBUFUIFNBHOJUVEFPGUIFMJOFWPMUBHFBU
UIFTPVSDFFOEBOEUIFTPVSDFQPXFSGBDUPS
"CBMBODFEXZFDPOOFDUFEMPBEJTDPOOFDUFEUPUIF
HFOFSBUPSCZBCBMBODFEUSBOTNJTTJPOMJOFXJUIBO
JNQFEBODFPG + KΩQFSQIBTF*GUIFMPBEJT
SBUFEBUL8 QPXFSGBDUPSMBHHJOH 7
line voltage, find the line voltage at the generator.
"UISFFQIBTFMPBEDPOTJTUTPGUISFFΩSFTJT
UPSTUIBUDBOCFXZFPSEFMUBDPOOFDUFE%FUFSNJOF
XIJDIDPOOFDUJPOXJMMBCTPSCUIFNPTUBWFSBHF
a
+
120 90° V ‒
n
120 ‒150° V +‒
c
A
‒ 120 ‒30° V
+
Ib
b
Ic
Figure 12.60
'PS1SPC
Z2
Z1
B
C
Z3
548
Chapter 12
Three-Phase Circuits
"GPVSXJSFXZFXZFDJSDVJUIBT
ž 7BO = ⧸
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTUPCFUUFSVOEFSTUBOEVOCBMBODFEUISFFQIBTF
TZTUFNT
7CO = ⧸
ž
−ž7
7DO = ⧸
*GUIFJNQFEBODFTBSF
ž ;"/ = ⧸
VP
žΩ
;DO = ⧸
A
a
;#/ = ⧸
ž
j XL
0° V +‒
‒+
find the current in the neutral line.
6TJOH'JH EFTJHOBQSPCMFNUIBUXJMMIFMQ
PUIFSTUVEFOUTCFUUFSVOEFSTUBOEVOCBMBODFEUISFF
QIBTFTZTUFNT
VP
120° V +‒
b
B
VP ‒ 120° V
‒j XC
R
C
c
Figure 12.63
'PS1SPC
Ia
+
‒
Vp 0°
Vp 120°
j XL
Vp ‒120°
‒
+
‒
+
%FUFSNJOFUIFMJOFDVSSFOUTGPSUIFUISFFQIBTF
DJSDVJUPG'JH-FU7B = ⧸
ž 7C =
⧸
−ž 7D = ⧸
ž7
R1
Ib
Ia
R2
Va +
‒
Ic
80 + j50 Ω
60 ‒ j40 Ω
20 + j30 Ω
Figure 12.61
‒
+
Vc +‒
'PS1SPC
"CBMBODFEUISFFQIBTF:TPVSDFXJUI71 =
880 V rms drives a Y-connected three-phase load
XJUIQIBTFJNQFEBODF;" = Ω ;# = + KΩ BOE;$ = KΩ$BMDVMBUFUIFMJOFDVSSFOUTBOEUPUBM
DPNQMFYQPXFSEFMJWFSFEUPUIFMPBE"TTVNFUIBU
UIFOFVUSBMTBSFDPOOFDUFE
"UISFFQIBTFTVQQMZ XJUIUIFMJOFUPMJOFWPMUBHF
PG7SNT IBTUIFVOCBMBODFEMPBEBTTIPXO
in Fig. 12.62. Find the line currents and the total
DPNQMFYQPXFSEFMJWFSFEUPUIFMPBE
Ib
Ic
Figure 12.64
'PS1SPC
Section 12.9 PSpice for Three-Phase Circuits
4PMWF1SPCVTJOH14QJDFPS.VMUJ4JN
5IFTPVSDFJO'JHJTCBMBODFEBOEFYIJCJUTB
QPTJUJWFQIBTFTFRVFODF*GG = )[ VTF14QJDFPS
.VMUJ4JN to find 7"/ 7#/ BOE7$/
A
15 + j8.66 Ω
N
15 + j 8.66 Ω
a
15 + j8.66 Ω
A
100 0° V +‒
n
B
Figure 12.62
'PS1SPC
10 Ω
+‒
C
c
Figure 12.65
'PS1SPC
‒+
b
B 40 Ω
N
0.2 mF
10 mF
C
549
Problems
6TF14QJDFPS.VMUJ4JN to find currents *B"BOE*"$
JOUIFVOCBMBODFEUISFFQIBTFTZTUFNTIPXOJO
'JH-FU
6TF14QJDFPS.VMUJ4JNUPEFUFSNJOF*PJOUIF
TJOHMFQIBTF UISFFXJSFDJSDVJUPG'JH-FU
;1 = –KΩ ;2 = + KΩ BOE
;3 = + KΩ
;M = + K ;2 = –KΩ 4Ω
220 0° V +
‒
220 0° V +
‒
;1 = + KΩ
Io
220 0° V
Z1
4Ω
‒+
Z3
‒+
Z1
a
220 ‒120° V
Z2
4Ω
;3 = Ω
A
Z1
Z1
B
b
Z3
Z2
Figure 12.66
220 120° V
'PS1SPC
‒+
(JWFOUIFDJSDVJUJO'JH VTF14QJDFPS
.VMUJ4JNUPEFUFSNJOFDVSSFOUT*B"BOEWPMUBHF7#/
Z1
C
c
Figure 12.69
'PS1SPC
'PSUIFDJSDVJUJO'JH VTF14QJDFPS.VMUJ4JN
to find the line currents and the phase currents.
240 0° V
‒+
a
4Ω
j3 Ω
‒j36 Ω
240 ‒120° V
n
‒+
b
4Ω
j3 Ω
"CBMBODFEUISFFQIBTFDJSDVJUJTTIPXOJO'JH
POUIFOFYUQBHF6TF14QJDFPS.VMUJ4JNto find the
MJOFDVSSFOUT*B" *C# BOE*D$
j 15 Ω
A 10 Ω
‒j36 Ω
B
10 Ω
j 15 Ω
10 Ω
j 15 Ω
Section 12.10
N
‒j36 Ω
240 120° V
‒+
c
4Ω
j3 Ω
C UIFDVSSFOUT*1 *2 *3 BOE*O
'PS1SPC
D UIFSFBEJOHTPGUIFXBUUNFUFST
E UIFUPUBMQPXFSBCTPSCFECZUIFMPBE
6TJOH'JH EFTJHOBQSPCMFNUPIFMQPUIFS
TUVEFOUTCFUUFSVOEFSTUBOEIPXUPVTF14QJDFPS
.VMUJ4JNUPBOBMZ[FUISFFQIBTFDJSDVJUT
a
+
‒
Lline
A
R
VL 0° V
Lline
Rline
B
C
"TTIPXOJO'JH BUISFFQIBTFGPVSXJSF
MJOFXJUIBQIBTFWPMUBHFPG7SNTBOEQPTJ
UJWFQIBTFTFRVFODFTVQQMJFTBCBMBODFENPUPSMPBE
BUL7"BUQGMBHHJOH5IFNPUPSMPBEJT
DPOOFDUFEUPUIFUISFFNBJOMJOFTNBSLFEB C BOE
D*OBEEJUJPO JODBOEFTDFOUMBNQT VOJUZQG BSF
DPOOFDUFEBTGPMMPXTL8GSPNMJOFDUPUIF
OFVUSBM L8GSPNMJOFCUPUIFOFVUSBM BOEL8
GSPNMJOFDUPUIFOFVUSBM
N
+
‒
c
'PS1SPC
Rline
b
+
‒
Figure 12.68
"UISFFQIBTF GPVSXJSFTZTUFNPQFSBUJOHXJUI
B7MJOFWPMUBHFJTTIPXOJO'JH5IF
TPVSDFWPMUBHFTBSFCBMBODFE5IFQPXFSBCTPSCFE
CZUIFSFTJTUJWFXZFDPOOFDUFEMPBEJTNFBTVSFECZ
UIFUISFFXBUUNFUFSNFUIPE$BMDVMBUF
B UIFWPMUBHFUPOFVUSBM
C
Figure 12.67
VL 120° V
Applications
VL ‒120° V
Rline
Lline
C
L
B *GUISFFXBUUNFUFSTBSFBSSBOHFEUPNFBTVSFUIF
QPXFSJOFBDIMJOF DBMDVMBUFUIFSFBEJOHPGFBDI
NFUFS
C 'JOEUIFNBHOJUVEFPGUIFDVSSFOUJOUIFOFVUSBM
MJOF
"OBTUFSJTLJOEJDBUFTBDIBMMFOHJOHQSPCMFN
550
Chapter 12
Three-Phase Circuits
a
0.6 Ω
j 0.5 Ω
A
0.2 Ω
30 Ω
j1 Ω
0.2 Ω
+
‒
240 10° V
j1 Ω
b
‒ j 20 Ω
0.6 Ω
j 0.5 Ω
B
+ 240 ‒110° V
+
-‒
+
‒
240 130° V
30 Ω
‒ j20 Ω
30 Ω
j1 Ω
‒ j 20 Ω
0.2 Ω
0.6 Ω
j 0.5 Ω
c
C
Figure 12.70
'PS1SPC
a
I1
b
W1
c
I2
W2
In
48 Ω
n
I3
Motor load
260 kVA,
0.85 pf, lagging
d
40 Ω
60 Ω
W3
Figure 12.71
'PS1SPC
.FUFSSFBEJOHTGPSBUISFFQIBTFXZFDPOOFDUFE
BMUFSOBUPSTVQQMZJOHQPXFSUPBNPUPSJOEJDBUFUIBU
UIFMJOFWPMUBHFTBSF7 UIFMJOFDVSSFOUTBSF" BOEUIFUPUBMMJOFQPXFSJTL8'JOE
24 kW 15 kW 9 kW
Lighting loads
Figure 12.72
'PS1SPC
5IFUXPXBUUNFUFSNFUIPEHJWFT11 = 8
BOE12 = −8GPSBUISFFQIBTFNPUPSSVOOJOH
POB7MJOF"TTVNFUIBUUIFNPUPSMPBEJT
XZFDPOOFDUFEBOEUIBUJUESBXTBMJOFDVSSFOUPG
"$BMDVMBUFUIFQGPGUIFNPUPSBOEJUTQIBTF
JNQFEBODF
B UIFMPBEJO7"
*O'JH UXPXBUUNFUFSTBSFQSPQFSMZDPOOFDUFE
UPUIFVOCBMBODFEMPBETVQQMJFECZBCBMBODFE
TPVSDFTVDIUIBU7BC = ⧸
ž7XJUIQPTJUJWFQIBTF
TFRVFODF
C UIFMPBEQG
B %FUFSNJOFUIFSFBEJOHPGFBDIXBUUNFUFS
D UIFQIBTFDVSSFOU
C $BMDVMBUFUIFUPUBMBQQBSFOUQPXFSBCTPSCFECZ
UIFMPBE
E UIFQIBTFWPMUBHF
"DFSUBJOTUPSFDPOUBJOTUISFFCBMBODFEUISFFQIBTF
MPBET5IFUISFFMPBETBSF
a
-PBEL7"BUQGMBHHJOH
-PBEL7"BUQGMBHHJOH
-PBEL8BUVOJUZQG
b
5IFMJOFWPMUBHFBUUIFMPBEJT7SNTBU)[ BOEUIFMJOFJNQFEBODFJT + KΩ%FUFSNJOF
UIFMJOFDVSSFOUBOEUIFDPNQMFYQPXFSEFMJWFSFEUP
UIFMPBET
W1
0
A
20 Ω
B
12 Ω
10 Ω
c
Figure 12.73
'PS1SPC
W2
‒ j10 Ω
j5 Ω
C
551
Comprehensive Problems
*GXBUUNFUFST81BOE82BSFQSPQFSMZDPOOFDUFE
SFTQFDUJWFMZCFUXFFOMJOFTBBOECBOEMJOFTC
BOEDUPNFBTVSFUIFQPXFSBCTPSCFECZUIFEFMUB
DPOOFDUFEMPBEJO'JH QSFEJDUUIFJSSFBEJOHT
±
±
±
±
Z = 60 ‒ j30 Ω
Figure 12.75
Z
Z = 10 + j30 Ω
+
‒
Z
±
+
‒
208 ‒60° V
W2
240 ‒120° V
Z
W2
W1
240 ‒60° V +
‒
Z
±
208 0° V +
‒
For the circuit displayed in Fig. 12.74, find the
XBUUNFUFSSFBEJOHT
±
W1
'PS1SPC
"NBOIBTBCPEZSFTJTUBODFPGΩ)PXNVDI
current flows through his ungrounded body:
±
B XIFOIFUPVDIFTUIFUFSNJOBMTPGB7
BVUPCBUUFSZ
Figure 12.74
C when he sticks his finger into a 120-V light sockFU
'PS1SPC
1SFEJDUUIFXBUUNFUFSSFBEJOHTGPSUIFDJSDVJUJO
'JH
4IPXUIBUUIF*3MPTTFTXJMMCFIJHIFSGPSB7
BQQMJBODFUIBOGPSB7BQQMJBODFJGCPUIIBWFUIF
TBNFQPXFSSBUJOH
Comprehensive Problems
"UISFFQIBTFHFOFSBUPSTVQQMJFEL7"BUBQPXFS
GBDUPSPGMBHHJOH*G 8BSFEFMJWFSFEUP
UIFMPBEBOEMJOFMPTTFTBSF8QFSQIBTF XIBU
BSFUIFMPTTFTJOUIFHFOFSBUPS
"UISFFQIBTF7 L8 L7"JOEVDUJWF
MPBEPQFSBUFTBU)[BOEJTXZFDPOOFDUFE*UJT
EFTJSFEUPDPSSFDUUIFQPXFSGBDUPSUPMBHHJOH
8IBUWBMVFPGDBQBDJUPSTIPVMECFQMBDFEJOQBSBMMFM
XJUIFBDIMPBEJNQFEBODF
"CBMBODFEUISFFQIBTFHFOFSBUPSIBTBOBCDQIBTF
TFRVFODFXJUIQIBTFWPMUBHF7BO = ⧸
ž7
5IFHFOFSBUPSGFFETBOJOEVDUJPONPUPSXIJDINBZ
CFSFQSFTFOUFECZBCBMBODFE:DPOOFDUFEMPBE
XJUIBOJNQFEBODFPG + KΩQFSQIBTF'JOEUIF
MJOFDVSSFOUTBOEUIFMPBEWPMUBHFT"TTVNFBMJOF
JNQFEBODFPGΩQFSQIBTF
"CBMBODFEUISFFQIBTFTPVSDFGVSOJTIFTQPXFSUP
UIFGPMMPXJOHUISFFMPBET
-PBEL7"BUQGMBHHJOH
-PBEVOLOPXO
-PBEL8BUQGMFBEJOH
*GUIFMJOFDVSSFOUJT"SNT UIFMJOFWPMUBHFBU
UIFMPBEJT7SNT BOEUIFDPNCJOFEMPBEIBTB
QGMBHHJOH EFUFSNJOFUIFVOLOPXOMPBE
"QSPGFTTJPOBMDFOUFSJTTVQQMJFECZBCBMBODFE
UISFFQIBTFTPVSDF5IFDFOUFSIBTGPVSCBMBODFE
UISFFQIBTFMPBETBTGPMMPXT
-PBEL7"BUQGMFBEJOH
-PBEL8BUVOJUZQG
-PBEL7"BUQGMBHHJOH
-PBEL8BOEL7"3 JOEVDUJWF
*GUIFMJOFJNQFEBODFJT + KΩQFSQIBTF
and the line voltage at the loads is 480 V, find the
NBHOJUVEFPGUIFMJOFWPMUBHFBUUIFTPVSDF
"CBMBODFEUISFFQIBTFTZTUFNIBTBEJTUSJCVUJPO
XJSFXJUIJNQFEBODF + KΩQFSQIBTF5IFTZTUFN
TVQQMJFTUXPUISFFQIBTFMPBETUIBUBSFDPOOFDUFEJO
parallel. The first is a balanced wye-connected load
UIBUBCTPSCTL7"BUBQPXFSGBDUPSPGMBH
HJOH5IFTFDPOEMPBEJTBCBMBODFEEFMUBDPOOFDUFE
MPBEXJUIJNQFEBODFPG + KΩQFSQIBTF*G
the magnitude of the line voltage at the loads is
2400 V rms, calculate the magnitude of the line
WPMUBHFBUUIFTPVSDFBOEUIFUPUBMDPNQMFYQPXFS
TVQQMJFEUPUIFUXPMPBET
"DPNNFSDJBMMZBWBJMBCMFUISFFQIBTFJOEVDUJWF
NPUPSPQFSBUFTBUBGVMMMPBEPGIQ IQ=
746 W) at 95 percent efficiency at a lagging power
552
Chapter 12
Three-Phase Circuits
GBDUPSPG5IFNPUPSJTDPOOFDUFEJOQBSBMMFM
UPBL8CBMBODFEUISFFQIBTFIFBUFSBUVOJUZ
QPXFSGBDUPS*GUIFNBHOJUVEFPGUIFMJOFWPMUBHFJT
480 V rms, calculate the line current.
'JHVSFEJTQMBZTBUISFFQIBTFEFMUBDPOOFDUFE
NPUPSMPBEXIJDIJTDPOOFDUFEUPBMJOFWPMUBHF
of 440 V and draws 4 kVA at a power factor of
72 perDFOUMBHHJOH*OBEEJUJPO BTJOHMFL7"3
DBQBDJUPSJTDPOOFDUFECFUXFFOMJOFTBBOEC XIJMF
B8MJHIUJOHMPBEJTDPOOFDUFECFUXFFOMJOFD
BOEOFVUSBM"TTVNJOHUIFBCDTFRVFODFBOEUBLJOH
7BO = 7Q⧸
0°, find the magnitude and phase angle of
DVSSFOUT*B *C *D BOE*O
'PSUIFTJOHMFQIBTFUISFFXJSFTZTUFNJO'JH find currents *B" *C# BOE*O/
a
440 0° V rms +
‒
n
440 0° V rms +
‒
1Ω
1Ω
A
24 ‒ j 2 Ω
N
15 + j 4 Ω
1Ω
b
B
Figure 12.77
'PS1SPC
Ia
a
b
c
Ib
1.8 kVAR
Ic
In
d
Motor load
4 kVA,
pf = 72%, lagging
800 W lighting load
Figure 12.76
$POTJEFSUIFTJOHMFQIBTFUISFFXJSFTZTUFNTIPXO
JO'JH'JOEUIFDVSSFOUJOUIFOFVUSBMXJSFBOE
UIFDPNQMFYQPXFSTVQQMJFECZFBDITPVSDF5BLF
7TBTB⧸
ž7 )[TPVSDF
1Ω
Vs +
‒
2Ω
20 Ω
15 Ω
'PS1SPC
%FTJHOBUISFFQIBTFIFBUFSXJUITVJUBCMFTZNNFUSJD
MPBETVTJOHXZFDPOOFDUFEQVSFSFTJTUBODF"TTVNF
UIBUUIFIFBUFSJTTVQQMJFECZB7MJOFWPMUBHF
BOEJTUPHJWFL8PGIFBU
Vs +
‒
Figure 12.78
'PS1SPC
1Ω
30 Ω
50 mH
Magnetically
Coupled Circuits
*GZPVXPVMEJODSFBTFZPVSIBQQJOFTTBOEQSPMPOHZPVSMJGF GPSHFUZPVS
OFJHICPSTGBVMUT'PSHFUUIFQFDVMJBSJUJFTPGZPVSGSJFOET BOEPOMZ
SFNFNCFSUIFHPPEQPJOUTXIJDI NBLFZPVGPOEPGUIFN0CMJUFSBUF
FWFSZUIJOHEJTBHSFFBCMFGSPNZFTUFSEBZXSJUFVQPOUPEBZTDMFBOTIFFU
UIPTFUIJOHTMPWFMZBOEMPWBCMF
c h a p t e r
13
‡"OPOZNPVT
Enhancing Your Career
Career in Electromagnetics
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JDT UIBUEFBMTXJUIUIFBOBMZTJTBOEBQQMJDBUJPOPGFMFDUSJDBOENBHOFUJD
fields. In electromagnetics, electric circuit analysis is applied at loX
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5IFQSJODJQMFTPG&.BSFBQQMJFEJOWBSJPVTBMMJFEEJTDJQMJOFT TVDIBT
FMFDUSJDNBDIJOFT FMFDUSPNFDIBOJDBMFOFS HZDPOWFSTJPO SBEBSNFUFPSPMP HZ SFNPUFTFOTJOH TBUFMMJUFDPNNVOJDBUJPOT CJPFMFDUSPNBHOFUJDT FMFDUSP
NBHOFUJDJOUFSGFSFODFBOEDPNQBUJCJMJUZ, plasmas, and fiber optics. EM de
WJDFTJODMVEFFMFDUSJDNPUPSTBOEHFOFSBUPST USBOTGPSNFST FMFDUSPNBHOFUT NBHOFUJDMFWJUBUJPO BOUFOOBT SBEBST NJDSPXBWFPWFOT NJDSPXBWFEJTIFT TVQFSDPOEVDUPST BOEFMFDUSPDBSEJPHSBNT 5IFEFTJHOPGUIFTFEF WJDFTSF
RVJSFTBUIPSPVHILOPXMFEHFPGUIFMBXTBOEQSJODJQMFTPG&.
&.JTSFHBSEFEBTPOFPGUIFNPSFEJG ficult disciplines in electrical
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FOHJOFFSTTQFDJBMJ[FJOUIJTBSFB&MFDUSJDBMFOHJOFFSTXIPTQFDJBMJ[FJO
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WFSBMDPNNVOJDBUJPOT
JOEVTUSJFT
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TBUFMMJUFT¥%7(FUUZ*NBHFT3'
553
554
Chapter 13
Magnetically Coupled Circuits
Historical
James Clerk Maxwell m BHSBEVBUFJONBUIFNBUJDTGSPN
$BNCSJEHF6OJWFSTJUZ JOXSPUFBNPTUSFNBSLBCMFQBQFSJOXIJDI
he mathematically unified the laws of Faraday and Ampere. This relation
TIJQCFUXFFOUIFFMFDUSJDfield and magnetic field served as the basis for
what was later called electromagnetic fields and waves, a major field of
TUVEZJOFMFDUSJDBMFOHJOFFSJOH5IF*OTUJUVUFPG&MFDUSJDBMBOE&MFDUSPO JDT&OHJOFFST *&&& VTFTBHSBQIJDBMSFQSFTFOUBUJPOPGUIJTQSJODJQMFJO
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represents the electromagnetic field. This relationship is commonly
LOPXOBT UIFSJHIUIBOESVMF .BYXFMMXBTBWFSZBDUJWFUIFPSFUJDJBO
BOETDJFOUJTU)FJTCFTULOPXOGPSUIFi.BYXFMMFRVBUJPOTu5IFNBY well, a unit of magnetic flux, was named after him.
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Learning Objectives
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BCMFUP
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IPXUPBOBMZ[FDJSDVJUTDPOUBJOJOHNVUVBMMZDPVQMFEJOEVDUPST
6OEFSTUBOEIPXFOFSHZJTTUPSFEJONVUVBMMZDPVQMFEDJSDVJUT
6OEFSTUBOEIPXMJOFBSUSBOTGPSNFSTXPSLBOEIPXUPBOBMZ[F
DJSDVJUTDPOUBJOJOHUIFN
6OEFSTUBOEIPXJEFBMUSBOTGPSNFSTXPSLBOEIPXUPBOBMZ[F
DJSDVJUTDPOUBJOJOHUIFN
6OEFSTUBOEIPXJEFBMBVUPUSBOTGPSNFSTXPSLBOELOPXIPX
UPBOBMZ[FUIFNXIFOVTFEJOBWBSJFUZPGDJSDVJUT
13.1
Introduction
5IFDJSDVJUTXFIBWFDPOTJEFSFETPGBSNBZCFSFHBSEFEBTDPOEVDUJWFMZ
DPVQMFE CFDBVTFPOFMPPQBG GFDUTUIFOFJHICPSJOHMPPQUISPVHIDVSSFOU
DPOEVDUJPO 8IFOUX PMPPQTXJUIPSXJUIPVUDPOUBDUTCFUXFFOUIFN
BGGFDUFBDIPUIFSUISPVHIUIF magnetic field HFOFSBUFECZ POFPGUIFN UIFZBSFTBJEUPCFNBHOFUJDBMMZDPVQMFE
5IFUSBOTGPSNFSJTBOFMFDUSJDBMEF WJDFEFTJHOFEPOUIFCBTJTPG
UIFDPODFQUPGNBHOFUJDDPVQMJOH*UVTFTNBHOFUJDBMMZDPVQMFEDPJMTUP
USBOTGFSFOFSHZGSPNPOFDJSDVJUUPBOPUIFS5SBOTGPSNFSTBSFLFZDJSDVJU
FMFNFOUT5IFZBSFVTFEJOQPXFSTZTUFNTGPSTUFQQJOHVQPSTUFQQJOH
EPXOBDWPMUBHFTPSDVSSFOUT5IFZBSFVTFEJOFMFDUSPOJDDJSDVJUTTVDIBT
SBEJPBOEUFMFWJTJPOSFDFJWFSTGPSTVDIQVSQPTFTBTJNQFEBODFNBUDIJOH JTPMBUJOHPOFQBSUPGBDJSDVJUGSPNBOPUIFS BOEBHBJOGPSTUFQQJOHVQPS
EPXOBDWPMUBHFTBOEDVSSFOUT
8FXJMMCFHJOXJUIUIFDPODFQUPGNVUVBMJOEVDUBODFBOEJOUSPEVDF
UIFEPUDPOWFOUJPOVTFEGPSEFUFSNJOJOHUIFW PMUBHFQPMBSJUJFTPGJOEVD
UJWFMZDPVQMFEDPNQPOFOUT#BTFEPOUIFOPUJPOPGNVUVBMJOEVDUBODF 13.2
555
Mutual Inductance
XFUIFOJOUSPEVDFUIFDJSDVJUFMFNFOULOPXOBTUIFUSBOTGPSNFS8FXJMM
DPOTJEFSUIFMJOFBSUSBOTGPSNFS UIFJEFBMUSBOTGPSNFS UIFJEFBMBVUPUSBOT
GPSNFS BOEUIFUISFFQIBTFUSBOTGPSNFS'JOBMMZ BNPOHUIFJSJNQPSUBOU
BQQMJDBUJPOT XFMPPLBUUSBOTGPSNFST BTJTPMBUJOHBOENBUDIJOH EFWJDFT
BOEUIFJSVTFJOQPXFSEJTUSJCVUJPO
13.2
Mutual Inductance
8IFO UXP JOEVDUPST PS DPJMT BSF JO B DMPTF QSPYJNJUZ UP FBDI PUIFS the magnetic flux caused by current in one coil links with the other coil,
UIFSFCZJOEVDJOHW PMUBHFJOUIFMBUUFS 5IJTQIFOPNFOPOJTLOP XOBT
NVUVBMJOEVDUBODF
Let us first consider a single inductor BDPJMXJUI / UVSOT8IFO
DVSSFOUJfloXTUISPVHIUIFDPJM BNBHOFUJDflux ϕJTQSPEVDFEBSPVOEJU
'JH "DDPSEJOHUP'BSBEBZTMBX UIFWPMUBHFWJOEVDFEJOUIFDPJM
JTQSPQPSUJPOBMUPUIFOVNCFSPGUVSOT/BOEUIFUJNFSBUFPGDIBOHFPG
the magnetic flux ϕUIBUJT
Eϕ
W=/___
EU
ϕ
+
i(t)
v
‒
Figure 13.1
Magnetic flux produced by a single coil
XJUI/UVSOT
#VUUIFflux ϕJTQSPEVDFECZDVSSFOUJTPUIBUBOZDIBOHFJOϕJTDBVTFE
CZBDIBOHFJOUIFDVSSFOU)FODF &R DBOCFXSJUUFOBT
PS
Eϕ EJ
W=/___
__
EJ EU
W=-@@
EJ
EU
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Eϕ
-=/@@@
EJ
5IJTJOEVDUBODFJTDPNNPOMZDBMMFETFMGJOEVDUBODF CFDBVTF JUSFMBUFT
UIFWPMUBHFJOEVDFEJOBDPJMCZBUJNFWBSZJOHDVSSFOUJOUIFTBNFDPJM
/PXDPOTJEFSUXPDPJMTXJUITFMGJOEVDUBODFT -BOE -UIBUBSFJO
DMPTFQSPYJNJUZXJUIFBDIPUIFS 'JH $PJMIBT /UVSOT XIJMF
DPJMIBT /UVSOT' PSUIFTBL FPGTJNQMJDJUZ BTTVNFUIBUUIFTFDPOE
JOEVDUPSDBSSJFTOPDVSSFOUThe magnetic flux ϕFNBOBUJOHGSPNDPJM
IBTUXPDPNQPOFOUT0OFDPNQPOFOU ϕMJOLTPOMZDPJM BOEBOPUIFS
DPNQPOFOUϕMJOLTCPUIDPJMT)FODF
ϕ=ϕ+ϕ
"MUIPVHIUIFUX PDPJMTBSFQI ZTJDBMMZTFQBSBUFE UIF ZBSFTBJEUPCF
NBHOFUJDBMMZDPVQMFE. Since the entire flux ϕMJOLTDPJM UIFW PMUBHF
JOEVDFEJODPJMJT
Eϕ
W=/@@@@
EU
Only flux ϕMJOLTDPJM TPUIFWPMUBHFJOEVDFEJODPJMJT
Eϕ
W=/@@@@
EU
L1
+
i1(t)
Figure 13.2
ϕ 11
L2
ϕ12
+
v1
v2
‒
‒
N1 turns N2 turns
.VUVBMJOEVDUBODF.PGDPJMXJUI
S FTQFDUUPDPJM
556
Chapter 13
Magnetically Coupled Circuits
"Hain, as the fluxFTBSFDBVTFECZUIFDVSSFOU J floXJOHJODPJM Eq. (13.6) can be written as
Eϕ EJ
EJ
W = /@@@@@@@
= -@@@
EJ EU
EU
XIFSF - = / Eϕ∕EJ JT UIF TFMGJOEVDUBODF PG DPJM 4JNJMBSMZ Eq. (13.7) can be written as
Eϕ @@@
EJ
EJ
W = /@@@@
= .@@@
EJ
EU
EU
XIFSF
Eϕ
. = /@@@@
EJ
.JTLOPXOBTUIF NVUVBMJOEVDUBODFPGDPJMXJUISFTQFDUUPDPJM
4VCTDSJQUJOEJDBUFTUIBUUIFJOEVDUBODF.SFMBUFTUIFWPMUBHFJOEVDFE
JODPJMUPUIFDVSSFOUJODPJM5IVT UIFPQFODJSDVJUNVUVBMWPMUBHF PS
JOEVDFEWPMUBHF BDSPTTDPJMJT
L1
+
ϕ22
+
v1
v2
‒
‒
i2(t)
N1 turns N2 turns
Figure 13.3
4VQQPTFXFOPXMFUDVSSFOUJ floXJODPJM XIJMFDPJMDBSSJFTOP
DVSSFOU 'JH The magnetic flux ϕFNBOBUJOHGSPNDPJMDPNQSJTFT
flux ϕ that links only coil 2 and flux ϕUIBUMJOLTCPUIDPJMT)FODF
L2
ϕ21
EJ
W= .@@@ EU
.VUVBMJOEVDUBODF.PGDPJMXJUI
S FTQFDUUPDPJM
ϕ= ϕ+ ϕ
The entire flux ϕMJOLTDPJM TPUIFWPMUBHFJOEVDFEJODPJMJT
Eϕ
Eϕ EJ
EJ
W= /@@@@
= /@@@@@@@
= -@@@
EU
EJ EU
EU
XIFSF-= /Eϕ∕EJJTUIFTFMGJOEVDUBODF of coil 2. Since only flux
ϕMJOLTDPJM UIFWPMUBHFJOEVDFEJODPJMJT
Eϕ
Eϕ @@@
EJ
EJ
W= /@@@@
= .@@@
= /@@@@
EU
EJ EU
EU
XIFSF
Eϕ
.= /@@@@
EJ
XIJDIJTUIFNVUVBMJOEVDUBODFPGDPJMXJUISFTQFDUUPDPJM5IVT UIF
PQFODJSDVJUNVUVBMWPMUBHFBDSPTTDPJMJT
EJ
W= .@@@ EU
8FXJMMTFFJOUIFOFYUTFDUJPOUIBU.BOE.BSFFRVBMUIBUJT
.=.=.
BOEXFSFGFSUP.BTUIFNVUVBMJOEVDUBODFCFUXFFOUIFUX PDPJMT-JLF
TFMGJOEVDUBODF- NVUVBMJOEVDUBODF.JTNFBTVSFEJOIFOSZT ) ,FFQ
JONJOEUIBUNVUVBMDPVQMJOHPOMZFYJTUTXIFOUIFJOEVDUPSTPSDPJMTBSF
JODMPTFQSPYJNJUZ BOEUIFDJSDVJUTBSFESJ WFOCZUJNF WBSZJOHTPVSDFT
8FSFDBMMUIBUJOEVDUPSTBDUMJLFTIPSUDJSDVJUTUPED
'SPNUIFUXPDBTFTJO'JHTBOE XFDPODMVEFUIBUNVUVBM
JOEVDUBODFSFTVMUTJGBW PMUBHFJTJOEVDFECZBUJNFW BSZJOHDVSSFOUJO
BOPUIFSDJSDVJU*UJTUIFQSPQFSUZPGBOJOEVDUPSUPQSPEVDFBW PMUBHFJO
SFBDUJPOUPBUJNFWBSZJOHDVSSFOUJOBOPUIFSJOEVDUPSOFBSJU5IVT
13.2
557
Mutual Inductance
Mutual inductance is the ability of one inductor to induce a voltage
across a neighboring inductor, measured in henrys (H).
"MUIPVHINVUVBMJOEVDUBODF .JTBM XBZTBQPTJUJ WFRVBOUJUZ UIF
NVUVBMWPMUBHF.EJ∕EUNBZCFOFHBUJWFPSQPTJUJWF KVTUMJLFUIFTFMGJO
EVDFEWPMUBHF-EJ∕EU)PXFWFS VOMJLFUIFTFMGJOEVDFE-EJ∕EU XIPTF
QPMBSJUZJTEFUFSNJOFECZUIFSFGFSFODFEJSFDUJPOPGUIFDVSSFOUBOEUIF
SFGFSFODFQPMBSJUZPGUIFW PMUBHF BDDPSEJOHUPUIFQBTTJ WFTJHODPOWFO
UJPO UIFQPMBSJUZPGNVUVBMW PMUBHF.EJ∕EUJTOPUFBTZUPEFUFSNJOF CFDBVTFGPVSUFSNJOBMTBSFJOWPMWFE5IFDIPJDFPGUIFDPSSFDUQPMBSJUZGPS
.EJ∕EUJTNBEFCZFYBNJOJOHUIFPSJFOUBUJPOPSQBSUJDVMBSXBZJOXIJDI
CPUIDPJMTBSFQIZTJDBMMZXPVOEBOEBQQMZJOH-FO[TMBXJODPOKVODUJPO
XJUIUIFSJHIUIBOESVMF4JODFJUJTJODPOWFOJFOUUPTIPXUIFDPOTUSVDUJPO
EFUBJMTPGDPJMTPOBDJSDVJUTDIFNBUJD XFBQQMZUIFEPUDPOWFOUJPOJODJS
DVJUBOBMZTJT#ZUIJTDPOWFOUJPO BEPUJTQMBDFEJOUIFDJSDVJUBUPOFFOE
PGFBDIPGUIFUXPNBHOFUJDBMMZDPVQMFEDPJMTUPJOEJDBUFUIFEJSFDUJPOPG
the magnetic flux if current enters that dotted terminal of the coil. 5IJTJT
JMMVTUSBUFEJO'JH (JWFOBDJSDVJU UIFEPUTBSFBMSFBEZQMBDFECFTJEF
UIFDPJMTTP UIBUXFOFFE OPUCPUIFSBCPVUIP XUPQMBDFUIFN5IFEPUT
BSFVTFEBMPOHXJUIUIFEPUDPO WFOUJPOUPEFUFSNJOFUIFQPMBSJUZPGUIF
NVUVBMWPMUBHF5IFEPUDPOWFOUJPOJTTUBUFEBTGPMMPXT
If a current enters the dotted terminal of one coil, the reference polarity
of the mutual voltage in the second coil is positive at the dotted terminal
of the second coil.
M
i1
+
v2 = M
di1
dt
‒
(a)
M
i1
+
"MUFSOBUJWFMZ
v 2 = ‒M
If a current leaves the dotted terminal of one coil, the reference polarity
of the mutual voltage in the second coil is negative at the dotted terminal
of the second coil.
5IVT UIFSFGFSFODFQPMBSJUZPGUIFNVUVBMW PMUBHFEFQFOETPOUIFSFG FSFODFEJSFDUJPOPGUIFJOEVDJOHDVSSFOUBOEUIFEPUTPOUIFDPVQMFE
DPJMT"QQMJDBUJPOPGUIFEPUDPO WFOUJPOJTJMMVTUSBUFEJOUIFGPVSQBJST
PGNVUVBMMZDPVQMFEDPJMTJO'JH'P
SUIFDPVQMFEDPJMTJO
'JH B UIFTJHOPGUIFNVUVBMWPMUBHFWJTEFUFSNJOFECZUIFSFGFS
FODFQPMBSJUZGPSWBOEUIFEJSFDUJPOPGJ4JODFJFOUFSTUIFEPUUFEUFSNJ
OBMPGDPJMBOEWJTQPTJUJWFBUUIFEPUUFEUFSNJOBMPGDPJM UIFNVUVBM
WPMUBHFJT +.EJ ∕EU'PSUIFDPJMTJO 'JH C UIF DVSSFOUJFOUFST
‒
(b)
M
di2
dt
v 1 = ‒M
‒
(c)
M
ϕ21
+
v1
‒
ϕ11
i2
+
ϕ12
i1
di1
dt
i2
+
v2
‒
ϕ22
i2
+
v1 = M
di2
dt
‒
(d)
Coil 1
Figure 13.4
*MMVTUSBUJPOPGUIFEPUDPOWFOUJPO
Coil 2
Figure 13.5
&YBNQMFTJMMVTUSBUJOHIPXUPBQQMZUIF
dot conWFOUJPO
558
Chapter 13
UIFEPUUFEUFSNJOBM PGDPJMBOE WJTOF HBUJWF at the EPUUFEUFSNJOBM
PGDPJM)FODF UIFNVUVBMW PMUBHFJT −.EJ ∕EU5IFTBNFSFBTPOJOH
BQQMJFTUPUIFDPJMTJO'JHT D BOE E 'JHVSFTIPXTUIFEPUDPOWFOUJPOGPSDPVQMFEDPJMTJOTFSJFT'PS
UIFDPJMTJO'JH B UIFUPUBMJOEVDUBODFJT
M
i
i
L1
(+)
Magnetically Coupled Circuits
L2
(a)
-=-+ -+ .
4FSJFTBJEJOHDPOOFDUJPO
M
i
'PSUIFDPJMTJO'JH C
i
L1
(‒)
-=-+ -− .
L2
4FSJFTPQQPTJOHDPOOFDUJPO
/PXUIBUXFLOPXIPXUPEFUFSNJOFUIFQPMBSJUZPGUIFNVUVBMW PMUBHF XFBSFQSFQBSFEUPBOBMZ[FDJSDVJUTJO WPMWJOHNVUVBMJOEVDUBODF"TUIF
first eYBNQMF DPOTJEFSUIFDJSDVJUJO'JH B "QQMZJOH,7-UPDPJM
HJWFT
EJ
EJ
W=J3+-@@@+.@@@
B
EU
EU
'PSDPJM ,7-HJWFT
EJ
EJ
W=J3+-@@@+.@@@
C
EU
EU
8FDBOXSJUF&R JOUIFGSFRVFODZEPNBJOBT
(b)
Figure 13.6
%PUDPOWFOUJPOGPSDPJMTJOTFSJFT
UIFTJHOJOEJDBUFTUIFQPMBSJUZPG
UIFNVUVBMWPMUBHF B TFSJFT
BJEJOHDPOOFDUJPO C TFSJFT
PQQPTJOHDPOOFDUJPO
7= 3+ Kω- *+ Kω.*
B
7=Kω.*+ 3+ Kω- *
C
"TBTFDPOEF YBNQMF DPOTJEFSUIFDJSDVJUJO'JH C 8FBOBMZ[F
UIJTJOUIFGSFRVFODZEPNBJO"QQMZJOH,7-UPDPJM XFHFU
7= ;+Kω- * − Kω.*
B
= ¢Kω.*+ ;-+ Kω- *
C
'PSDPJM ,7-ZJFMET
&RVBUJPOT BOE BSFTPMW FEJOUIFVTVBMNBOOFSUPEFUFS NJOFUIFDVSSFOUT
0OFPGUIFNPTUJNQPSUBOUUIJOHTJONBLJOHTVSFPOFTPMW FTQSPC
MFNTBDDVSBUFMZJTUPCFBCMFUPDIFDLFBDITUFQEVSJOHUIFTPMVUJPOQSP DFTTBOEUPNBL FTVSFBTTVNQUJPOTDBOCFW erified. 5PPPGUFO TPMWJOH
NVUVBMMZDPVQMFEDJSDVJUTSFRVJSFTUIFQSPCMFNTPMW FSUPUSBDLUX PPS
NPSFTUFQTNBEFBUPODFSF HBSEJOHUIFTJHOBOEW BMVFTPGUIFNVUVBMMZ
JOEVDFEWPMUBHFT
M
R1
v1
+
‒
i1
L1
L2
(a)
Figure 13.7
i2
jωM
Z1
R2
+ v
2
‒
V
+
‒
I1
jωL 1
jωL 2
I2
ZL
(b)
5JNFEPNBJOBOBMZTJTPGBDJSDVJUDPOUBJOJOHDPVQMFEDPJMT B BOEGSFRVFODZEPNBJOBOBMZTJTPGBDJSDVJUDPOUBJO
JOHDPVQMFEDPJMT C 13.2
559
Mutual Inductance
M
jωL 1
I1
L1
L2
I2
jωL 2
jωMI 2
(a)
jωMI 1
jωL 1
jωMI 2
jωL 2
‒
+
(b)
+
‒
jωMI 1
(c)
Figure 13.8
.PEFMUIBUNBLFTBOBMZTJTPGNVUVBMMZDPVQMFEFBTJFSUPTPMWF
&YQFSJFODFIBTTIP XOUIBUJGXFCSFBLUIFQSPCMFNJOUPTUFQTPG
TPMWJOHGPSUIFWBMVFBOEUIFTJHOJOUPTFQBSBUFTUFQT UIFEFDJTJPOTNBEF
BSFFBTJFSUPUSBDL 8FTVHHFTUUIBUNPEFM 'JHVSF C CFVTFE
XIFOBOBMZ[JOHDJSDVJUTDPOUBJOJOHBNVUVBMMZD PVQMFEDJSDVJUTIPXOJO
Figure 13.8(a):
/PUJDFUIBUXFIBWFOPUJODMVEFEUIFTJHOTJOUIFNPEFM 5IFSFBTPO
for that is that we first determine the vBMVFPGUIFJOEVDFEWPMUBHFTBOEUIFO
XFEFUFSNJOFUIFBQQSPQSJBUFTJHOT$MFBSMZ*JOEVDFTBWPMUBHFXJUIJOUIF
TFDPOEDPJMSFQSFTFOUFECZUIFWBMVFKω*BOE* JOEVDFTBWPMUBHFPGKω*
in the first coil. Once we haWFUIFWBMVFTXFOFxt use both circuits to find
UIFDPSSFDUTJHOTGPSUIFEFQFOEFOUTPVSDFTBTTIPXOJO'JHVSF D 4JODF*FOUFST-BUUIF EPUUFEFOE JUJOEVDFT BWPMUBHFJO-UIBU
USJFTUPGPSDFBDVSSFOUPVUPGUIFEPUUFEFOEPG -XIJDINFBOTUIBUUIF
TPVSDFNVTUIBWFBQMVTPOUPQBOEBNJOVTPOUIFCPUUPNBTTIP XOJO
'JHVSF D *MFBWFTUIFEPUUFEFOEPG-XIJDINFBOTUIBUJUJOEVDFT
BWPMUBHFJO -XIJDIUSJFTUPGPSDFBDVSSFOUJOUPUIFEPUUFEFOEPG -
SFRVJSJOHBEFQFOEFOUTPVSDFUIBUIBTBQMVTPOUIFCPUUPNBOEBNJOVT
POUPQBTTIPXOJO'JHVSF D /P XBMMXFIBWFUPEPJTUPBOBMZ[F
BDJSDVJUXJUIUXPEFQFOEFOUTPVSDFT5IJTQSPDFTTBMMPXTZPVUPDIFDL
FBDIPGZPVSBTTVNQUJPOT
"UUIJTJOUSPEVDUPSZMFWFMXFBSFOPUDPODFSOFEXJUIUIFEFUFSNJOBUJPO
PGUIFNVUVBMJOEVDUBODFTPGUIFDPJMTBOEUIFJSEPUQMBDFNFOUT-JL F3 - BOE$ DBMDVMBUJPOPG.XPVMEJOWPMWFBQQMZJOHUIFUIFPSZPGFMFDU SPNBH
OFUJDTUPUIFBDUVBMQI ZTJDBMQSPQFSUJFTPGUIFDPJMT*OUIJTUF YU XFBTTVNF
UIBUUIFNVUVBMJOEVDUBODFBOEUIFQMBDFNFOUPGUIFEPUTBSFUIFiHJ WFOT
PGUIFDJSDVJUQSPCMFN MJLFUIFDJSDVJUDPNQPOFOUT3 - BOE$
Example 13.1
$BMDVMBUFUIFQIBTPSDVSSFOUT*BOE*JOUIFDJSDVJUPG'JH
j3 Ω
‒j4 Ω
12 0° V
+
‒
j5 Ω
I1
‒j4 Ω
j6 Ω
I2
(a)
Figure 13.9
'PS&YBNQMF
Solution:
'PSMPPQ ,7-HJWFT
¢ ¢K K *¢K*=
12 Ω
12 0° +
‒
j5
I1
j3I2
j6
‒
+
+
‒
(b)
j3I1
I2
12 Ω
560
Chapter 13
Magnetically Coupled Circuits
PS
K*− K*= 'PSMPPQ ,7-HJWFT
¢K* K *=
PS
+ K *
* = @@@@@@@@@
= − K *
K
4VCTUJUVUJOHUIJTJO&R XFHFU
K+ − K *= −K *=
PS
*=@@@@@
=⧸
ž"
− K
'SPN&RT BOE Practice Problem 13.1
*= − K *= ⧸
¢ž ⧸
ž
= 13.01 ⧸
−49.39ž"
%FUFSNJOFUIFWPMUBHFVoJOUIFDJSDVJUPG'JH
j1 Ω
4Ω
120 45° V
+
‒
I1
j8 Ω
j5 Ω
+
10 Ω Vo
‒
I2
Figure 13.10
'PS1SBDUJDF1SPC
Answer:⧸
¢ž7
Example 13.2
$BMDVMBUFUIFNFTIDVSSFOUTJOUIFDJSDVJUPG'JH
4Ω
‒j3 Ω
j8 Ω
j2 Ω
100 0° V
+
‒
I1
j6 Ω
I2
5Ω
Figure 13.11
'PS&YBNQMF
Solution:
5IFLFZUPBOBMZ[JOHBNBHOFUJDBMMZDPVQMFEDJSDVJUis knowing the QPMBS
ity of the mutual voltage. We need to apply the dot rule. In Fig. TVQQPTFDPJMJTUIFPOFXIPTFSFBDUBODFJT Ω BOEDPJMJTUIFPOF XIPTFSFBDUBODFJT Ω. To figure out the polarity of the mutual voltage
JODPJMEVFUPDVSSFOU* XFPCTFSWFUIBU*MFBWFTUIFEPUUFEUFSNJOBMPG
DPJM4JODFXFBSFBQQMZJOH,7-JOUIFDMPDLXJTFEJSFDUJPO JUJNQMJFT UIBUUIFNVUVBMWPMUBHFJTOFHBUJWF UIBUJT ¢K*
13.2
561
Mutual Inductance
"MUFSOBUJWFMZ, it might be best to figure out the mutual vPMUBHFCZ
SFESBXJOHUIFSFMF WBOUQPSUJPOPGUIFDJSDVJU BTTIP XOJO'JH XIFSFJUCFDPNFTDMFBSUIBUUIFNVUVBMWPMUBHFJT7=¢K*
5IVT GPSNFTIJO'JH ,7-HJWFT
¢+ * − K+ K − K*− K* = ‒ +
j6
I1
PS
j2(I1‒I2)
j8
I2
j2I2
‒
+
= + K *− K*
Similarly, to figure out the mutual voltage in coil 2 due to current * DPO Figure 13.12
TJEFSUIFSFMFWBOUQPSUJPOPGUIFDJSDVJU BTTIPXOJO'JH"QQMZJOH .PEFMGPS&YBNQMFTIPXJOHUIF
UIFEPUDPOWFOUJPOHJWFTUIFNVUVBMWPMUBHFBT7=¢K*"MTP DVSSFOU QPMBSJUZPGUIFJOEVDFEWPMUBHFT
*TFFTUIFUXPDPVQMFEDPJMTJOTFSJFTJO 'JH TJODFJUMFBWFTUIF
EPUUFEUFSNJOBMTJOCPUIDPJMT &R BQQMJFT5IFSFGPSF GPSNFTI
JO'JH ,7-HJWFT
= ¢K*− K*+ K+ K+ K×+ *
PS
=¢K*+ +K *
1VUUJOH&RT BOE JONBUSJYGPSN XFHFU
[ ] [
][ ]
¢K
*
+ K
=
¢K + K * 5IFEFUFSNJOBOUTBSF
ǀ
ǀ
ǀ
ǀ
¢K = K
Ĵ= + K
¢K + K
ǀ
¢K
= + K
+ K
Δ=
ǀ
+ K = K
¢K Δ=
5IVT XFPCUBJOUIFNFTIDVSSFOUTBT
°
⧸
+K
Δ
*=@@@
=@@@@@@@@@@@
°"
= ⧸
=@@@@@@@@@@@@
Δ
+K
⧸
°
°
⧸
K
Δ
*=@@@
=@@@@@@@@
°"
= ⧸
=@@@@@@@@@
Δ + K ⧸
°
%FUFSNJOFUIFQIBTPSDVSSFOUT*BOE*JOUIFDJSDVJUPG'JH
5Ω
100 60° V
+
‒
j2 Ω
j3 Ω
I1
j6 Ω
I2
‒j4 Ω
Figure 13.13
'PS1SBDUJDF1SPC
Answer:* =⧸°" * =⧸°"
Practice Problem 13.2
562
Chapter 13
13.3
Magnetically Coupled Circuits
Energy in a Coupled Circuit
*O$IBQUFS XFTBXUIBUUIFFOFSHZTUPSFEJOBOJOEVDUPSJTHJWFOCZ
X= @@
-J
M
i1
i2
+
v1
+
L1
‒
Figure 13.14
L2
v2
‒
5IFDJSDVJUGPSEFSJWJOHFOFSHZTUPSFE
JOBDPVQMFEDJSDVJU
8F OPXX BOUUP EFUFSNJOF UIFFOFSHZ TUPSFE JONBHOFUJDBMMZ DPVQMFE
DPJMT
$POTJEFSUIFDJSDVJUJO'JH 8FBTTVNFUIBUDVSSFOUT JBOE J
BSF[FSPJOJUJBMMZ TPUIBUUIFFOFS gy stored in the coils is zero. If we let J
JODSFBTFGSPN[FSPUP*XIJMFNBJOUBJOJOHJ = UIFQPXFSJODPJMJT
EJ
Q U = vJ = J-@@@
EU
BOEUIFFOFSHZTUPSFEJOUIFDJSDVJUJT
*
-* X = ∫ QEU= -∫ JEJ = @@
*GXFOPXNBJOUBJOJ = *BOEJODSFBTFJGSPN[FSPUP* UIFNVUVBMWPMU
BHFJOEVDFEJODPJMJT .EJ∕EU XIJMFUIFNVUVBMW PMUBHFJOEVDFE JO
DPJMJT[FSP TJODFJEPFTOPUDIBOHF5IFQPXFSJOUIFDPJMTJTOPX
EJ
EJ
EJ
Q U = J.@@@
+ Jv = *.@@@
+ J-@@@
EU
EU
EU
BOEUIFFOFSHZTUPSFEJOUIFDJSDVJUJT
*
*
X = ∫ QEU=.*∫ EJ + -∫ JEJ
= .** + @@
-* 5IFUPUBMFOFS HZTUPSFEJOUIFDPJMTXIFOCPUI JBOE JIB WFSFBDIFE
DPOTUBOUWBMVFTJT
X= X + X = @@
-*+ @@
-* + .**
*GXFSFWerse the order by which the currents reach their final vBMVFT UIBU
is, if we first increase JGSPN[FSPUP*BOEMBUFSJODSFBTFJGSPN[FSPUP
* UIFUPUBMFOFSHZTUPSFEJOUIFDPJMTJT
X= @@
-* + @@
-* + .**
#FDBVTFUIFUPUBMFOFSHZTUPSFETIPVMECFUIFTBNFSFHBSEMFTTPGIPXXF
reach the final conditions, comparing Eqs. (13.28) and (13.29) leads us
UPDPODMVEFUIBU
BOE
. = . = .
B
-*+ .**
X=@@
-* + @@
C
5IJTFRVBUJPOXBTEFSJWFECBTFEPOUIFBTTVNQUJPOUIBUUIFDPJMDVSSFOUT
CPUIFOUFSFEUIFEPUUFEUFSNJOBMT*GPOFDVSSFOUFO