s B N 1 3 9 7 30 1 3 1
BN.lO
0 0
illililt
rililt
Itill
8 {t 2 5 2
A List of Tabt€s
1|b! N9:
PagcN9:
Tirle
8
9
Thc I ernalionalSystemofllrils (Sl)
Derivcd Unils in SI
StandardicdPrefi\es!o SignifyPorversof10
Interprelationof RctcrcnceDircclionsln Fig.1.5
PhlsiologicAlReactlonsto CurctrtLcleh in Hunans
EquarioDslor the DelrosterGrid
Su nary of Resistance
fermsfor DescribtugCircuits
PspiceSensitivltyAnalysisResults
TeNinal Equalionsfor Ideal Inducto6 and Capacilors
lrductors md Capaciton
EquationsforScrics-and Parallel-Connected
'/'tbr
Valueof e
rEqual lo InlegralMultiplesof r
NatNal ResponsePalamelc$ otlhe ParallelRLC Cifcxit
The Responseofa SecondOrdcr Circuills Olerdamped.Underdlmped.or Ciitically Danpcd
In Determiningthe NaruralRcsponseoI a SecondOrdef Circuit.Wc FirsLDeternine Whethcr
i( is OveF.Under.or CriticallyDanped, andThenWe Solvethc AppropriateEquations
of a Sccond-OrderCircnit.WeAPPIythe Appropflate
ln Delernining the StepRespoDsc
Eq uadon$Dependingon the Damping
Values
Impcdincc and Reactance
Values
Adnrittanccand Susceplance
Inpedarce and RclatcdVnLues
Annual EnerSyRequircmcnlsol ElectricHouseholdApplianccs
ThreePowerQuanriticsand lhen Units
An AbbreviatedLisr of LaplaccTranslornPans
An AbbreviatedList of OperationalTranslbrms
Four UsclulTransformPairs
Sumnary oflhc lDomaln EquivalentCircuirs
NumericalValucsot ,"(r)
Input and OtrlputVoltagcMagniludesfor SeveralFrequcDcies
NormalizedGo that o. = I r!d/s) ButterworthPolynomialsup 1othe Fighth Ordef
1.1
1.2
t.l
1.4
2.1,
3.1
4.1
4.2
6.2
7.7
8.1
8.2
8.3
8.4
9.1
9.2
9.3
10.1
10_2
l2.l
12.2
t2.3
13.1
13.2
14.1
15.1
t7.l
17.2
l8.l
18.2
l3
17
l9
r36
217
2r1
23i
2E9
321
321
321
145
350
376
398
4L[
475
.180
,193
5ll)
536
629
708
'713
ot Elencntary Funclions
FourierTransforms
OpelalionalTransfonns
ParametcrConvenionTable
TerminatedTivo PoIt Equalions
7i6
742
GreekAtphabet
I
B
f
p
Beta
K
1
k
Kappa
:
^
Lambda
T
Delta
M
M11
N
Nu
E
€
Epsilon
z
t
Zcra
H
q
Eta
o
0
{
X]
II
Pi
r
Tau
Phi
X
,
;
Signa
o
x
v
q
chi
.f(r) (, > 0-)
T,"e
r(s)
6o
(inpulse)
1
(step)
I
1
F
1
(€xponential)
Gine)
GositrE
1
(danp€dranp)
(danpedsine)
D
(danpedcosine)
!(t)
r(s)
Kf(t\
(F(r)
l(r+t(r)-n(r)+ .
FlG)+&(r)-.3G)+...
df(t')
rFG) - /(0)
dt f(t)
r'.G) - J/(o )
d'f(t)
r'IG) - s' t/(0 ) - t
I f(') dr
f(t-a)u(t-a),a>0
e-- f(t)
r(")
€ -r(r)
F(r + a)
f(at),a>o
tf(t)
dF(r)
{ f(t)
f(t\
t
L*",^
df(o )
",df\o-)
- t - "_.
- dllo-)
d,
dT
d*1 f (0 )
f(,
f(r')
Half 'wave rectif ied sine
Full-wave rectified sine
n,=5.i,[#,''(+))".,r
f\'t=:+:
ftirng'n.r w'v€
no=!^3,.,.!*""*
'io,,o, ii
)
Half-woY€ recrifi €d sine
,r,,
2,4 - 4.4 S cosr@or
"
. -2t lan,
t)
FUI-wlv€ rcctiffed sin€
Fourier
Transforms
of Etem€ntary
Functions
JG)
r(11
6(t) (impulse)
I
2nA6(a)
sgn0) Gis.um)
n\(a) + tljo
e '',0)bosirive tine exponcDtial)
."/,( /) (nesative'lnnccxpooential)
c '' (positive-and negative-rimccxponential)
eii4r(cornplexcxpoienliat)
2,r(o
oo)
,[a(o + oo) + 6(o
j,f6(o + oo) 6(0
o0)]
oo)l
0perationat
Transforms
flt)
F(.t
Kf(t
KI:(u)
f1() f'(t) +.ldr.)
dt tQ)/.tt"
rla)
r,(u)
u@)"
lGtL)
l i f e ) ,, , 0
Fz@) + F1(.a)
ej-tto
r _.
oa) +
tt \a
/
r (^)r(/ ^)dr
l '(.r)
f.(t)
t'l(t)
x (o)H(.,)
;l-F@F,('-'4du
\'! t
dtt)
ELECTRIC
CIRCUITS
EIGHTH
EDITION
W.Nilsson
James
Prcfessor Emeritus
Iowa StateUnirelsitJ
Susan
A. Riedet
Marquette Unive$ity
Upper SaddleRiver,New Je$ey 07458
Dataot File
Libtott of CongresCataloging-in-Publicatiafl
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Preface
The eighth edition of Electlrc Cilclrr is a carefully planned revision to the
most widely used introductory circuits text of the past 25 years.As this
book has evolvedover the yearsto meet the changinglearningstylesof
students,importantly,the underlying teachirg approachesand philoso
phiesremainunchanged.Thegoalsare:
. To build an understandingof conceptsand ideasexplicitlyin termsof
previousleardng.
. To emphasizethe relationshipbetween conceptualunderstanding
and problem-solvingapproaches.
. To provide studentswith a strong foundation of engineeringpractices
WHYTHISEDITION?
When planr rg for the eighth editjon ievision of Electic Citcuits, carcfnl
thought was given to how \ve should best update this classictext to improve
upon the success
o{ precedingeditionsandmakethe eightl edition ascom,
pelliry as the first.Througha thoroughreviewprocessthat includedboth
inshuctors and students who currendy use tlectli. C;rc&itsand those who
use other texts, our revision plan was formed. wllat emerged from this
exercisewas a clear pictue of what matters most to instructors and stu
derts. With this feedback in mind, we made the following changes:
. Problem solving is fundamental to the study of circuit analysis.The
authorsput their pdmary eftbrt into updatingand addingnew endof-chapter problems. The result is a ftesh text with approximately
80o/"new or revised problems comparedto the previous edition.
Having a wealth ofnew problemsto assignand work is a key to suc
cessin any circuits course.
. The eighth edition iepresentsa major redesignto the text. Careful
attentionwaspaid toward how to presentthe material tex1,figures,
and aitwork-in a clean,clear mannerthat would facilitatelearning
and encouragereading.The seventhedition was the fi$t introductory circuits text to recognize the changing needs of today's students
with a modern,four-colordesign.The eightheditionrefinesthis color
treatmentwith a morc pedagogjcallycoherentpresentation.
' Navigation was imprcved by the addition of page numbers to the
chapterobjectives,less rcliance on icons whete nameswere more
effective, ard the updated organization of end-of-chapter problems
by sectionlevels Additionally,the layout was enhancedto limit the
instarceswhereExamplesspill over onto multiple pages.
. All artwork, pholos, and imageshave been modernized and enhanced
to present a crisper illustration of the key elementsand application of
circuit analysis.
. Recognizing ttrat moie classpreparation and studying is happening
online wilh t}re use of additional resources, ttre development of
online resources for tle eighth edition represents a sigdfica[t
improvement from the seventh edition. From online, automatically
graded homework, to study aids and an e-book, all of this and more is
now available on an easy{o-navigate website for students and
College textbooks excel at presenting complicaled mateiial in a clear,
straigltforward mannei, Authon and publishers spend countless hous
developingthe bestpossibleleamingaid for studentsand teachhg aid for
instructors. Prcntice Hall is committed to working with authors to create
textbooksandsupporthg resourcesthat enablebetterteachingandbetter
student leamhg. The eigh0l edition of tlectric Cir.cr.itr is one such example,It setthe standardfoi circuitseducation25 yean ago and il continues
that trend today.
HALLMARK
FEATURES
Chapter
Problems
Userc of Electic C cuits have consistently rated the Chapter Problems as
one ofthe book'smost attmctivefeatures.Inthe eighth edition,theie are
ovei 1000problems with approximately 80% that are new or revised from
the previous edition. Problems are organized at the end of each chapter by
PracticalPerspectives
The eighth edition continues the use of Practical Perspective inftoduced
with the chapter openers.They offer examplesof real-world circuits, taken
from real-world devices.Most chapters begin wittr a bdef descdption of a
practical application of the material that follows. Once the chapter mateda1 is presented,the chapter concludes witl a quantitative analysis of the
application along with a Practical Perspective problem. This enables you
to unde$tand how to apply the chapter contents to the solution of a realworld problem.
Assessment
Problems
Each chapter begins with a set of chapter objectives At key points in the
chapter,you are askedto stop and assessyoul masteryof a particular
problems.Ifyou are able to
objectiveby solvingore or more assessment
problems
givel
solve the assessment
for a
objective, you have mastered
that objective.
Examples
Every chapter includes many examples that illustrate the concepts presented in the text in the folm of a numeric example. There are over
130 examples in this text. The examples are intended to illustrate the
applicationoI a particularconcept,and alsoto encouragegood problemsolving skilts.
Equations
andConcepts
Fundamental
Throughout the text, you will seefundamental equatioN and concepts set
apart from the main text.This is done to help you focus on some of the key
piinciples in electric circuits and to help you navigate tbrough the important topics.
Tools
Integrationof Computer
Computer tools can assiststudents in the learning processby pioviding a
visual representation oI a circuit's behavior, validating a calculated solu_
tion, rcducing the computational burden of morc complex circuits, and
iterating toward a desired solution using parameter variation. This computational suppofi is often invaluable in the design process The eighth edition ircludes the support of PSpice, a popular computer tool Chapter
problems suited for explomtion with Pspice are so marked.
Emphasis
Design
The eighth edition continues to support the emphasison the design of circuits in many ways First, several of the Practical Perspective discussions
focus on the design aspects of the circuits. The accompanying Chapter
Problems continue the discussion of the design issuesin these pnctical
examples.Second,desigl-odentedChapterPrcblemshave been labeled
explicitly, enabling students and instructols to identify those problems
with a design focus.Third, the identification of problems suited to explo_
ration wit}I Pspice suggestsdesign opporturdties using this soflware
Accuracy
All text and problems in the eig.hthedition have undergone our strict hallmark triple accuracychecking process,to ensure the most error-ftee book
possible.
ANDINSTRUCTORS
FORSTUDENTS
RESOURCES
ha[[.con/ni
lsson
www.pren
The eighth edition of Elsctlic Circuits comeswith Prentice Hall's poweful
new suite of student and instructol online resources
ForStudentsi
Online homework and Dractice with immediate feedback and inte
grarede-bookuslDgPH UradeAssl\l
. Online Study Guide that highlights the key conceptsof electric circuits
. Additional book and coursespecificresouces
,
ForInstructors:
. Assignable, automatically graded online homework with PH
cradeAssist
. Digital version of all figur€s from the book
. Interactive Learning classroomPowerPoint slides
. Sample Chapter Tests
. Additional book and coursespecificresources
ADDITIONAL
OFFLINE
RESOURCES
ForStudents:
. Student Study Pack-This new resoutce teachesstudents techniques
for solvingproblemspresentedin the text. Organizedby concepts,
this is avaluableproblem-solvingresourcefor all levelsof students.
. Introduction to Pspice Manual-Updated for the eighth edition, the
manual comeswith the latest available releaseof the software on CD.
ForInstructors:
. Instructor Solutions Manual-Fully worked out solutions to end-ofchapter problems.
' Instructor Problem Bank A tremendous new resoruce with many
additional problems and conespo ding solutions to problems not
found in the text.This is a great tool for qeating homework and exams.
0rdering0ptionsl
Ek!:tric Citcuits wilh PH Grade Assist Online Homework Access:
ISBN 0-13-514291-1
Elecftic Citcuits wft SfitdentStudt Pa&:ISBN 0,13 51,4290-3
Eled c Circuits \NithInrrcduction to Pspice Ma utl..ISBN 0-13,514292-X
PREREQUISITES
In writing tlte fint 12 chapters of the text, we have assumed tlat the
reader has taken a course in elementary differential and integral calculus.
We have also assumedthat the rcader has had an introductory physics
coune, at either the high school or university level, that iritroduces the
concepts ol energy,power, electric charge,electric current, electdc poten,
tial, and electromagnetic fields. In writing the final six chapters,we have
assumedthe studenthas had,or is enrolledin, an introductorycoune in
differential equations.
c0uRsE
0m0Ns
The lext hasbeen designedfor usein a one-semestei,
two-semester,
or a
thiee-quartersequence.
. Single-semestetcourserAlter coveringChapters1-4 and (hapten 6 10
(omitting Sections?.7 and 8.5) the instructor can choosefrom Chaprer5
(operational amplifiels), Chapter 11 (tbree-phasecircuirs).Chapters13
and 14 (hplac€ methods), and Chapter 18 (Tko-Port Circuits) to
develop the desiredemphasis
. Ttro-semestersequencerAssuming three lectuies per week, tle fiIst
nine chapters can be covered during the first semester, leaving
ChaptersI0- l8 for thesecond.emesler.
' Academicquartu schedrlerThe book can be subdividedinto three
parts:Chapten 1-6, Chapters7-12,and Chapters13-18.
The introduction to operational amplifier circuits can be omitted vithout
interfercnce by the rcader going to the subsequentchapten. For examplq if
Chapter 5 is omitted, the instructor can simply skip Section7.7,Sectron8-5,
Chapter 15, and those problerns and assessingobjective problems in the
chaptercfolowing Chapter 5 that pertain to operational amplifiers.
There are seveml appendixesat the end of the book to help readers
make effective useof t]left matlematical background.Appendix A review$
Cramer's method of solving simultaneous lirear equations and simple
matrix algebra;complex numbers are reviewed ir Appendix B;Appendix C
contains additional mateiial on magnetically coupled coils and ideal transfomels; Appendi{ D contabs a biiel discussior of the decibeliAppendixE
is dedicated to Bode diagEmsiAppendix F is devoted to an abbreviated
table of trigonometdc identities that are useful in circuit analysis;and an
abbreviated table of useful integrals is given in Appendix G Appendix H
prcvides answersto selectedsuggestedprcblems
ACKNOWLEDGMENTS
We continue to expressour appreciation for the contributions of Norman
Wittels of Woicester Polytechnic Institute. His contributions to the
Practical Perspectives geatly enhanced both this and the previous two
editions. Jacob Chacko, a transmission and distribution engineer at the
Ames Municipal Electric System, also contributed to t}le Practical
Perspecrives.
Special tlanks also to Robert Yahn (USAF), Stephen
O'Conner(USAF), andwilliam Oliver (BostonUniversity)for their con
tinued intercst in and suggestionsfor this book.
There were many hard-working people behind the scenesat our publisher who deserve our thanks and gmtitude for their efforts on behalf of
the eigl[h editiorl At Pretrtice Hall, we would like to thank Michael
McDonald, Rose Kernan, Xiaohong Zhu, Lisa McDowell, Jonathan
Boylan, David A. Georye, Tim Galigan, and Scott Disanno for their continued suppo and a ton of really hard work. The authon would also like
to acknowledge the stafl at GEX Publishing Servicesfor their dedication
and hard woik in typesetting this text.
The many revisions of the text were guided by carelul and thorough
reviews tom professors.Our heanJelt thanks to:
. Paul PanayotatoER tgerc University
. Evan Goldsrein, University ofwathington
. Kalpathy B. Sundaram, Uniretsity of Central Flotidtl
. Andrew K. Chan, feta, A&M Unitersity
. A. S^taai-Jazi,Viryinia PolytechnicInstitute and State University
. Clifford H. Grigg, Rose-Hulman Institute of Technology
. Karl Bithdnger, Univetsit! ofwashington
. Carl Wells, Iryr8fttnglonState Unirercity
' Aydir I.Karsilayan,kasA&M UnireNity
. Ramakant Sdvastava,University of Florida
. Michel M. Mala$iz, Univetsity of Mithigan, Ann Arbor
. Christopher Hoople, Rocheskr Institute ofTechnolosy
. SannasiRamanan, RochesterItlstitute ofTechnologj
. Gary A. Hallock, U/.ivenitt) ol Texasat Austin
The authom would also like to thanl Ramakant S vastava of the
Udversityol !loflda.aDdlhe AccuraryRe!iewTeamal CEX Publisbing
Senices, for their help in checking ttre text and all the problems in the
eighthedition.
We are deeply indebted to the many instructors and students who
have offered positive feedback and suggestionsfor improvement. We use
as many of these suggestionsas possible to continue to improve the content, the pedagogy, and the presentation. We are honorcd to have the
opportunity to impacr the educational expeiience ol the many thousands
of future engheers who will tum the pagesof this text.
JaMEsW. NlssoN
A, REDEL
SUSAN
BriefContents
Chapter
1
Chapter
2
Chapter
3
Chapter4
Chapter5
Chapter
6
Chapter
7
Chapter
8
Chapter
9
Chapter
10
Chapter
11
Chapter
12
Chapter
13
Chapter
14
Chapter
15
Chapter
16
Chapter
17
Chapter
18
Appendix
A
AppendixB
Appendix
C
Appendix
D
Appendix
E
Appendix
F
Appendix
G
Appendix
H
List of Examplesxiii
Preface xvii
CircuitVariables 2
CircuitEtements 22
SimpteResistiveCircuits 56
Techniques
of CircuitAnalysis 92
The0perational
Amplifier 154
Inductance,
Capacitance,
andMutualInductance186
Response
of First-orderRLandRCCircuits 228
NaturalandStepResponses
of RtCCircuits 284
Sinusoidal
Steady-State
Analysis 330
SinusoidalSteady-State
PowerCalculations390
Balanced
Three-Phase
Circuits 432
Introductionto the LaplaceTransform 466
TheLaplaceTransform
in CircuitAnalysis 506
Introductionto Frequency
SelectiveCircuits 566
ActiveFitterCircuits 606
FourierSeries 656
TheFourierTransform 698
Two-Port
Circuits 730
TheSotutionof LinearSimultaneous
Equations759
Complex
Numbers781
Moreon Magneticatly
Coupted
CoilsandldealTransformers
787
TheDecibel 797
BodeDiagrams799
An Abbreviated
Tableof Trigonometric
Identities 817
An Abbreviated
Tableof Integrals 819
Answers
to Selected
Problems821
Index 839
Contents
Listof Examptesxiii
Prefacexvii
Chapter1 CircuitVariables 2
1.1
1.2
1.3
1.4
1.5
1.6
Etectri.atEngineering:
An 0verview 3
TheInterrationatSystemof Units 8
CircuitAnalysis!
An overview t0
VoltageandCurrent 11
TheldeatEasicCircuitEtement 12
PowerandEnergy 14
Sunnoty 16
Prcblehs 17
Chapter
2 CircuitElements22
2.1
2,2
2,3
2.4
2.5
PrdcticqlPerspective:
E[ectri.slSoIety 23
VoltageandCuffentSources24
Etectrical
Resistan(e
(0hm'sLaw) 28
Constrlction
of a CircuitModet 32
Kirchhoff's
Laws 36
Anatysis
of a CircuitContaining
Dependent
Souraes42
ProcticolPetspective:
ElectficolSolety 46
Sumndry 47
Prcblens 48
Chapter
3 SimpleResistive
Circuits 56
ProcticolPerspedive,A neor Window
DeIrcster57
nesistors
in Series 58
Resistors
in Paraltel 59
TheVoltage-Divider
andCurrent-Divider
LITCUIIS
bZ
VottageDivisionandCurrentDivision 65
Measuring
VoltageandCurrent 68
Measuring
Resistance-The
Wheatstone
Bridge 71
3 . 7 Delta-to-Wye
(Pi-to-Tee)
EquivatentCircuits
73
PructicalPe5pedive:A RearWindow
Defrcster 76
Sumnory 79
Problems 80
Chapter
4 Techniques
of CircuitAnalysis 92
Pnctical Pe6pedive:Circuitswith Reolistic
Resistors 93
4.7 Terminology94
4.2 Introduction
to the Node-Vol.tage
Method 97
4.3 TheNode-Vottage
Methodand0ependent
Sourcest00
4.4 TheNode-Voltage
Metbod:SomeSpecial
Cases 101
4.5 Introduction
to the Mesh-Current
MethodJ05
4.6 TheMesh-Current
MethodandDependent
SourcesJoZ
4.7 TheMesh-Current
Method:SorneSpecial
Cases 109
the Node-Vottage
MethodVersus
the
Mesh-CurrentMethod
112
4.9 Source
Transfoimations116
4.10 Th6v€nin
andNortonEquivalents
119
4.11 Moreon Deriyinga ThAvenin
Equivatent12J
4.12 Ma)dmum
Power-fransfer126
4.13 Superposition
129
Pncticol Pe6pectivetCitcuitswith Realistic
Resistors 133
Sunnary 137
Problens 138
Chapter
5 TheOperational
Amplifier 154
5.1
5.2
5.3
5.4
5,5
5.6
5.7
PrdcEcqlPe6pective:StruinGqges 155
operational
AmptifierTerninals 156
lerminalVottages
andCurrents 156
TheInverting-Amptifier
Circuit 161
TheSumming-Amptifier
Circuit 163
TheNoninverting-Amptifier
Circuit 164
TheDifference-A$plifier
Cir.uit J65
A MoreReatistic
Modelfor the 0perational
Ampliliet 170
ProcticalPerspective:
Sttuin 6ages 173
Sunnary 175
Problens 176
Chapter6 Inductance.
Capacitance,
andMutualInductance186
Pnctical Perspective:
ProximitySwitches 187
TheInductor 188
TheCapacitor195
Series-Paraltel
Cornbinations
of Inductance
andCapacitance200
6.4 MutuatInductance203
6.5 A CloserLookat MutualIndudance 207
PructicalPerspective:
Prcxinity Sdtches 214
Sunmary 217
Prcblens 218
Chapter9 SinusoidaI
Steady-State
Analysis 330
6.1
9.1
9.2
9.3
9.4
9.5
9.6
9.7
Chapter7 Response
of First-0rderfl
andflf, Circuits 228
7.1
7.2
7.3
7,4
7,5
7.6
7.7
PructicolPerspedive:A FloshingLight
Citcuit 229
TheNaturalResDonse
of an fl Circuit 230
TheNaturalResponse
of an 8CCircuit 2j6
TheStepResponse
ot fl. andRCCircuits 240
A Generat
Solutionfor SteDandNatural
Responses248
Sequentiat
Switching 254
Unbounded
Response258
TheIntegratingAmplifiet 260
ProdicdlPe6pective:A FlqshingLight
Circuit 263
Sumnory 265
froDens
zbb
Chapter8 NaturalandStepResponses
of RLCCircuits 284
8.1
8,2
8.3
8.4
8.5
PtocticalPetspedive:Anlgnition Cir.uit 285
Introduction
to the NaturalResDonse
of a
Palattet
nIC Circuit 286
TheFormsof the NaturalResDonse
of a
Paratl.et
RICCircuit 291
TheStepResponse
of a Parallel
RICCircuit 301
TheNaturatandStepResponse
of a Series
ilc Circuit 308
A Circuitwith TwoIntegratingAmplifiers 312
Ptodicol Percpective:
An lgnition Circuit 317
Summam Szu
Probte;321
9.8
9.9
9.10
9.11
9.12
PracticdlPetspective:
A Househotd
Distribution
Circuit 331
TheSinusoidal.
Sowce 332
TheSinusoidal.
ResDonse335
ThePhasor 337
ThePassive
CircuitElements
in the Freouencv
Domain 342
Kirchhoff!Lawsin the Frequency
Donain 346
Series.Parallel.
andDelta-to-Wye
SimDlifi(ations348
Source
Transformations
andTh6venin-Norton
Eouivalent
Circuits 355
lhe Node-Vottage
Method 359
TheMesh-Current
Method 360
TheTransformer361
TheldealTransformer365
PhasorDiagrams-t72
Pnctical Percoective:
A HouseholdDistribution
Circait 375
Sunnary 375
rnoDtems J/b
Chapter10 Sinusoidal
Steady-State
PowerCalculations390
10.1
10.2
10.3
10.4
10.5
10.6
PructicolPerspedive:HeotingAppliancs 391
Instantaneous
Povrer392
Average
andReactive
Power 394
ThermsValueandPowerCatculations399
Comptex
Power 401
PowerCatcllations 403
Maximum
Power
Transfer 410
ProcticdlPerspedive:HeqtingApptisnces 417
Sumnoty 419
Prcblent 420
11 Balanced
Chapter
Three-Phase
Circuits 432
11.1
11.2
11.3
11.4
11.5
ProcticolPe6pective:Trunsnission
ondDlstibution oI Electic Powet 433
Balanced
Three-Phase
Voltages4j.
Three-Phase
VoltageSources435
Anatysis
of the Wye-Wye
Circuit 436
Anatysis
of the Wye-Delta
Circuit 442
PowerCa[.ulations
in Balanced
Three-Phase
Circsits 445
11.6 Measuring
Average
Powerin Three-Phase
Llrcut$
452
Ptsdicol Perspedive:Transnission
on.t
DisttibutionoJElectic Powet 455
Su nary 456
ftootens
45/
Chapter
12 Introduction
to the Laptace
Transform 466
12.1
12.2
12.3
12.4
12.5
12.6
12.7
12.8
12.9
Definition of the LaplaceTransform 46l
TheStep Function 468
TheImputseFunction 470
functionaITransfofids 474
0perationatTransfortf.s 475
Apptyingthe LaplaceTransform 481
InverseTransforms 482
Potesand Zerosof F(s) 494
Initial- and Finat-Value
Theorems 495
Sumnaty 498
Ptoblens 499
Chapter
13 TheLaplace
Transform
in
CircuitAnalysis 506
Pncticol Pe$pective:SurgeSuppressors507
1 3 . 1 CircuitElernents
ifl the s Domain 508
13.2 f,ircuitAnatysis
in the s Domain 5Jl
13.3 Applicatiohs512
13.4 TheTransfer
function 526
TheTransfer
Function
in Partiatfraction
Expansions528
13.6 TheTransfer
Functionandthe Convotution
Integrat 537
13,7 TheTransfer
Fundionandthe Steady-state
SinusoidalResponse
537
13.8 TheImpulsefunctionin Ciiclit Analysis 540
Procticd[Pe6pective:SurgeSuppressors548
Summary549
Probtems 550
Chapter
14 Introduction
to Frequency
SelectiveCircuits 566
PrccticalPe6pective:PushbuttonTelephone
LtrcutEs 5b/
14.1 SomePretiminaries
568
14.2 Low-Pass
Fillets 570
14.3 High-Pass
Filtets 577
14.4 Bandpass
Fitters 582
14.5 Bandreject
Fitters 593
ProcticqlPerspe.tive:PushbuttonTe[ephone
Circuits 598
Sunnary 599
Problens 599
Chapter15 ActiveFilterCircuits 606
Prdcticdl PeBpedive: BqssVolume
Confiol 607
15.1 First-order
Low-Pass
and High-Pass
75.2
15.3
75,4
15.5
iitters 608
ScaLing612
0p AmpSandpass
andBandreje€t
Fitters 615
fligherorder0p AmpFitters 622
Narrowband
Bandpass
andBandreje€t
ntters
DJD
PructicolPerspective.
EdssVolune
Control 642
Sunndry 644
Prcblens 646
Chapter
16 Fourier
Series 656
16.1 FourierSeriesAnalysis:
An overview 658
16.2 ThefourierCoefficients659
16.3 TheEffectof Symmetry
on the Fourier
Coefficients662
16.4 An Alternative
Trigonometric
Formof the
FourierSeries 668
16.5 AnApptiration670
16.6 Average-Power
Calculations
with Periodic
funcDonso/5
16.7 ThermsValueof a PeriodicFunclion 678
16.8 TheExponentiat
Formof the Fourier
sede. 679
16.9 AmptitudeandPhaseSpe.tra 682
Sunnory 685
Prcblens 686
Chapter
17 TheFourier
Transform698
17.1 TheDerivation
of the FourierTransform699
17.2 TheConvergence
of the FourierIntegral 70J
17.3 usingLaplace
Transforms
to FindFourier
franstofifis 703
17.4 FourierTransforrns
in the Limit 706
17.5 SomeMathematical
Properties7r8
17.6 0perational
Transfo(ms710
17.7 CircuitApplications714
17.8 Parseval!
Theorem 717
Sunnary 724
Problens 725
Chapter18 Two-Port
Circuits 730
gquations 731
18.1 TheTerminat
18.2 TheTwo-PortPa'amelers 732
18.3 Anatysis
of the Terminated
Two,Port
Cit.uit 741
18.4 Interconneded
Two-Poft
Ctr.uits 747
Sunnaty 751
Prcblens 752
AppendixA TheSolutionof Linear
SimultaneousEquations
759
A.1 Pretiminary
Steps /59
4.2 Crame/sMethod 760
4.3 TheCharacteristic
Determinant760
4.4 TheNurnerator
Determinant760
4.5 TheEvaluation
of a Determinant761
4,6 Mati.es 764
A.7 MatrirAtgebra 765
A.8 Identity,Adjoint,andInverseMatri(es ZZ0
4.9 Partitioned
Matrices 772
4.10 Apptications776
Appendix
B Complex
Numbers781
8.1 Notation 781
8.2 TheGraphirat
Representation
ot a Complex
Nurnber 782
8.3 Arithmeti.operations /83
8.4 UsefulIdentities 785
8,5 TheIntegerPowerof a Complex
Number 785
8.6 TheRootsof a Complex
Number 286
AppendixC Moreon Magnetically
Coupled
CoilsandIdeal
Transformers787
C,1 Equivatent
Circuitsfor Magnetically
Coupted
Coils 787
C.2 TheNeedfor ldealTransformers
in the
Equivalent
Circuits 792
Appendix
D TheDecibet 797
Appendix
E BodeDiagrams799
E.1
E.2
E,3
E.4
E.5
8.6
E.7
E.8
Reat,First-0rder
PolesandZeros 299
Straight-lineAmptitudePlots 800
MoreAccurate
AmplitudePtots 8ra
Straight-Line
Phase
AngtePlots 805
BodeDiagrarns:
Complex
PolesandZeros 807
AmotitudePtots 809
Correcting
Straight-Line
AmplitudePtots 810
Phase
AnglePtots 813
AppendixF An Abbreviated
Table
of Trigonometric
Identities 817
AppendixG An Abbreviated
Tableof Integrals 819
AppendixH Answers
to Selected
Problems821
Index 839
Listof Examptes
2
Chapter
of IdeatSources26
2.1 lestingInterconne.tions
of IdeatIndependent
2.2 Testinglnterconnedions
Sources27
andDependent
for a
vottage,
Current.andPower
2.3 Catcutating
Circuit 31
SimpleResistive
2.4 Construding
a CircuitModelof a
Ftashtight33
a Circuit[{ode[Basedon Terminal
2.5 Constructing
Measurements
35
CurrentLaw 39
2.6 UsingKirchhoff's
2.7 UsingKirchhofftVottageLaw 39
Laws
2.8 Applyingohm'sLawandKirchhoff's
Current 40
to Findah Unknown
2.9 Constructing
a Cir€uitl4odelBasedonTerninal
Measurements
41
2.10 Applyingohm'sLawand(i(hhoff's Laws
Vottage 44
to Findan Unknown
Law
2.11 ApFlyingohm'sLawandKirchhoff's
in an Arnplifiertircuit 45
3
Chapter
61
3.1 ApptyingSeries-ParattetSirnptification
Circuit 63
3.2 Anatyzing
the vottage-Divider
Circuit 64
3.3 Analyzing
a Current-Divider
3.4 UsingVoltageDivisionandCurrentDivision
to Solvea Circuit 67
Ammeter 69
3.5 Usinga d Arsonval
Vottmeter Z0
3.6 Usinga dArsonvat
Transform75
3.7 Applyinga Delta-to-Wye
4.6
4.7
4.8
4,9
4.10
4.11
4,12
4.13
Chapter5
an 0p AmpCircuit J60
5.1 AnatyzinE
6
Chapter
6.1
6.2
6.3
6.4
6.5
6.6
4
Chapter
4.1
4.4
4.5
ldentifying Node.Eranch,Mesh,and Loop
in a Circuit 95
Method 99
Usingthe Node-Vottage
Methodwith
Usingthe Node-Vol.tage
Dependent
Sources 100
Meihod 106
Usingthe Mesh-Cuffent
Methodwith
Usingthe Mesh-C{rrrent
DependentSources 108
Method
tlnderstanding
the Node-Voltage
Method 113
Versus
Mesh-Curent
andMesh-Current
[omparingthe Node-Vottage
Methods115
to Sotve
Transformations
UsingSource
a Cit.uit 117
Transformation
UsingSpecialSource
118
Techniques
Equivalent
of a Circuit
Findingthe Th6venin
Source 122
with a Dependent
Equivatent
Usinga Test
Findingthe Thdvenin
SoUwrce
124
for MadmumPower
the Condition
Catcutating
Transfer 128
to Solvea Circuit 132
UsingSuperposition
Giventhe Current,
Determining
the Vottage,
of an Indlctor 189
at the Terminats
the Current,Giventhe Vottage.
Determining
at the Terminats
of antndu.tor 191
Powet
Vottage.
Determining
the Current,
andEnergy
for an Inductor 193
Power,
and
Detefmining
Current,Vottage.
197
for a capacitor
Energy
Finding?).p, and,lr)Inducedby a Triangutar
CurrentPutsefor a Capac.tlor198
for a Circuit
Equations
FindingMesh-Current
coits
206
with Magneticatty
Coupled
7
Chapter
of an
the NaturalResponse
Determining
RLCircuit 234
of an
the Natual Response
7.2 oetermining
Inductors
235
iI Circuitwith Parallel
of an
the NaturalResponse
7,3 Determining
nfCircuit 238
7.1
xiv
Lkt of Exanptes
7.4
Determining
the NaturalResponse
of an
RCCircuitwith SeriesCapacitors2-39
7.5 Determining
the StepResponse
of an
RLcir.uit 244
7.6 Determining
the StepResponse
of an
RCCitcuit247
7.7 Usingthe General
SolutionMethodto Findan
RCCircuit'sStepResponse250
7.8 Usingthe Generat
SotutionMethodwith Zero
lnitiat Conditions251
7.9 Usingthe General
SolutionMethodto Findan
,(LLrrcurl5
>Iepxesponsez5l
7.10 Determining
the StepResponse
of a Circuit
with Magneticatl.y
Coupted
Coils 253
7.11 Anatyzing
an Rl Circuitthat hasSequentiat
Switching 255
7.72 Analyzing
an RCCirclit that hasSequential
Switching 257
7,13 Finding
the Unbounded
Response
in an
RCCircuit 259
7.14 Anatyzing
an IntegratingAnptifier 261
7.15 Anatyzing
an IntegratingAmptifierthat has
Seqoentiat
Switching 262
Chapter
8
8.1
Findingthe Rootsof the Characteristic
Equation
of a ParallelRLCCircuit 290
8.2 Findingthe overdamped
NaturalResponse
of a
Paraltet
RaCCircuit 293
8.3 Catrulating
Bran(hCurrents
in the Natural
Response
of a Panllel RLCCir.uit 294
8.4 Findingthe Underdamped
NaturatResponse
of
a Parattet
nlC Circuit 297
8.5 Findingthe CriticattyDamped
Naturat
ResDonse
of a ParallelRICCircuit 300
8.6 Findingthe overdamped
StepResponse
of a
Parattet
f,lCCircuit 304
8.7 Findingthe Underdamped
step Response
of a
Parattet
RICCircuit 305
8.8 Findingthe CriticaltyDamped
StepResponse
of a Parattet
faf Circuit 305
8.9 Comparing
theThree-Step
Response
Forms 306
8.10 FindingStepResponse
of a Paratl.et
ilc Cir.uit
with Initiat Storedlnergy 306
8.11 Finding
the Underdamped
NaturaI
Response
of
a SeriesRaf,Circuit 310
8.12 Finding
the Underdarnped
StepResponse
of a
SeriesRtCCircuit -?11
8,13 Anatyzing
TwoCascaded
lntegrating
Amptifiers 3l.t
8.14 Anatyzing
TwoCascaded
IntegratingAmplifiers
with Feedback
Resistors316
Chapter
9
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
9.12
9.13
9.14
9.15
9.16
Findingthe Characteristi€s
of a Sinusoidal
Cufient 334
findingthe Characteristics
of a Sinusoidal
Vottage 334
Translating
a SineExpression
to a Cosine
Expression-t.t4
Catcutating
the rmsVatueof a Triangutar
Waveforrn3J5
Adding
Cosines
UsingPhasors341
Impedances
in Series 349
Combining
Combining
Impedances
i'n Seriesandin
Paraltet 351
Usinga Delta-to-Wye
Transform
in the
F equency
Domain 353
Perforrnihg
lransformations
Source
ir| the
Frequency
Domain 356
Findinga Th6venin
Equivalent
in the
Frequency
Domain 35l
Usingthe Node-Vottage
Methodin the
Frequenqr
oomain 359
Usingthe Mesh-Current
l4ethodin the
Frequenqf
Donain 360
Anatyzing
a Linearlransformer
in the
Frequenqr
Domain 364
Anatyzing
an IdealTransformer
Circuitin the
trequency
Domain 370
UsingPhasorDiagrams
to Analyzea
Citcuit 372
UsingPhasorDiagrams
to AnalyzeCapacitive
LoadihgEffects 373
Chapter
10
10.1 Cattulating
Average
andReadivePower 396
10.2 MakingPowerCalculations
Involving
HousehotdAppliances
398
10,3 oetermining
Average
PowerDetivered
to a
Resistor
by a Sinusoidat
Voltage 400
Listof Examphs xv
10.4
10.5
10.6
10,7
Power 402
Catcutating
Complex
Average
Power 406
Calcutating
andReactive
Power
in Pardllet
Loads 40l
Calculating
Balancing
PowerDelivered
with Power
Absorbed
in an acCircuit 408
10,8 DeteminingMa$mumPower
Transfer
without
LoadRestrictions41-t
10.9 DeteminingMaximum
Power
Transfer
with
LoadImpedance
Restri.llon 414
10.10FindingMaximum
Power
Transfer
with
I pedance
AogleRestidions 414
10.11FindingMaximunPower
Transfer
in a Circuit
with an IdealTransformer4J5
11
Chapter
11,1 Anatyzing
a Wye-Wye
Cit.uiI 440
11.2 Anatyzing
a Wye-Detta
Circuit 444
11,3 Catcul.ating
Powerin a Three-Phase
Wye-Wye
Circuit 449
11.4 Catculating
Por,rer
in a Three-Phase
Wye-Delta
14.4
14.5
14.6
14.7
Loading
the Seriesna fiigh-Pass
Fitter 580
Designing
a Bandpass
Fitter 587
Designing
a Parattet
RaCBandpass
Fitter 588
Determining
Effed of a Nonideal
Voltage
Source
on a fLC Bahdpass
Fitter 589
14.8 Designing
a SeriesRtCBandreject
Fitter 596
15
Chapter
15.1 0esigning
a Low-Pass
0p AmpFitter 609
15.2 Designing
a High-Pass
0p AmpFitter 611
15,3 Scal.ing
a SeriesRLCCir.uit 613
15.4 Scating
a Prototype
Low-Pass
0p Arnp
FiLlet 614
15.5 Designing
a Broadband
Bandpass
0p Anp
Filtet 618
15.6 Designing
Bandreject
a Sroadband
0p Amp
Filtet 621
15.7 Designing
a Fourth-order
Low-Pass
0p Amp
fitrer
bz5
15.8 CaLcutating
Butterworth
Transfer
Functions628
11.5 Catcllating
Three-Phase
Powerwith
15.9 Designing
a Fourth-order
Low-Pass
an Unspecified
Load 450
ButterworthFilter 631
11.6 Computing
Wattmeter
Readings
in Three-Phase 15.10Determining
the orderof a Eutterworth
Citcuits 454
Filtet 634
15.11AnAlternate
Approach
to Determining
12
Chapter
the orderof a Eutterworth
fitter 634
12.1 UsingSteptunctionsto Represent
a Function 15.12Designing
a High-4Bandpass
Fitter 638
oT ilntre uuralon 4/u
15.13Designing
a High-oEandreject
Fitter 64,
tltcull
45u
13
Chapter
13.1 Deriving
the Transfer
tundionof a Circuit 527
13.2 Anatyzing
the Transfer
Fundion
of a Circuit 529
13.3 Usingthe Convol|ltior
Integralto Find
an oljtp|lt Signat 535
13.4 Usingthe Transfer
Function
to Find
the Steady-State
Sinusoidal
Response539
Chaptor
14
14.1 Designing
Filtet 574
a Low-Pass
14.2 Designing
a SeriesfC Low-Pass
titter 575
14.3 0esigning
a SeriesRl High-Pass
tillet 579
Chapter
16
16.1 Findingthe FourierSeriesof a Triangutar
Waveform
with NoSymmetry660
16.2 Findingthe FourierSeriesof an odd Function
with Symmetry667
16.3 Catcutating
Formsof the Trigonometric
Folrier
Sedesfor Periodic
Voltage 669
16.4 Calcutating
Average
Powerfor a Circlit
with a Periodic
VoltageSo[.e 677
16.5 Estirnating
the rmsValueof a Periodic
tun(tion 679
16.6 Findingthe Exponential
Formof the Fourier
Series 681
Slnusoidal
Steady-state
Response716
17.3 ApplyingParevaltIheoren 719
17.4 ApptyingParsevalt
Theorem
to an Ideal
Eandpass
tilter 720
17.5 AppMngParseval's
fheoremto a Low-pass
Fiatet 721
Measurements
235
18.3 Findingt parameters
fromMeasurements
ard Table18.1 738
18,4 Anatyzing
a Terminated
Two-port
Circuit 246
18.5 Analyzing
Cascaded
Two-port
Circuits Z5O
Variables
Circuit
1.1 EtectricatEngineering:
An overviewp. 3
1.2 TheInternationatSystemof Units p. 8
1.3 CircuitAnatysisrAn overviewp. 10
1.4 Voltageand (urrent p. 1?
1.5 TheIdeal BasicCircuitEtementp. 12
1.6 Powerand Energyp. 14
1 Unde6tand
andbe abteto use5I unitsandthe
standadprefi:esfor powersof 10.
2 Knowandbe abteto useihe dennitionsof
3 Knowafd be abt€to us€ihe defifitionsof
4 Beabteto usethe passive
siqnconvention
to
caLcutaie
the powerfor af idealbasiccjrcujt
€tementgivenits vottageandcunent.
ElectricaI engineering is an excitingand challcngingprofcssion
for anvone who has a genuine interest in, and aptitude lor,
applied scienceand mathematics.Over the past century and a
hall, electrical engineershave played a dominant rolc in thc
developnrentof svstemsthat have cbangedthe v',aypeople live
and work. Satellitecommu cation links,telephones,digital com
puters, tcicvisions,diagnosticand surgical medical equipment,
assembiylinerobots,and eleclricalpower tools are representative componentsof systemsthat define a modern technological
socicty.Asan clcctricalengincer.you canparticipatein this ongoing technologicalrevolution by improvilg and refining these
existingsystemsand by discovcdngand dcvclopingnew systems
to meet the needsot our ever-changing
socjety.
As ]'ou embark on the study of circuit analysis,you nccd to
gain a feel for where tlris study fits into the hierarchyof topics
that conlprisean introductionto electricalengineering.Hencewe
begin by presentingan oveNiew of electricalelElineering,some
ideasabout an engineeringpoint of view as jt relatesto circuit
analysis.and a reviervof the internationalsystemof units.
Wc then describegenerallywhat circuit analysisentails.Next,
we intloducethe conceplsolvollage and current.Welbllow these
conceptswith discussiolof an ideal basicelementand the necd
for a polarity reference system.We conclude the chapter by
describinghow current and voltagerclatc to powcr and cncrgy.
An0verview
1.1 EtectncatEngin€€nng:
An Overview
Engineering:
1.1 Etectrical
Electrical engineeringis the profession concetned with systemsthat
oroduce.transmit, analmeasureelectric signals.Elect cal engineeing
;ombines the pbysicist'smodels of natural phenomena ith the mathematician'stools lor manipulatingthose modelsto producesystemsthat
meet practicalneeds.Electiical systemspervadeour lives;theyare found
ir homes, schools,workplaces, and transportation vehicles everywhere
We beginby presentinga few examplesfrom eachofthe five major classificationsof electricalsystems:
. commu catlonsyslems
' computel systems
t control systems
. power sysEms
. signal-processing
systems
Then we describehow electrical engineersanallze and designsuchsystems
Conrmunication systems are electrical systems that generate' transmit. and distribute information. Well-known examples include television
equipment, such as cameras,transmitters, receivers,and VCRS; radio telescopes,used to explore the univeise; satellite systems,which return images
of other planets and our own: radar systems,used to coordinateplane
fl ights;ard telephonesystems.
Figure 1.1depictsthe major componentsof a modern t€lephonesystem. Sta ing at the left of the figure,inside a telephone,a microphone
turns sou d waves into electric signals.These signalsare caded to a
s$ritchingcenter where they are combined with the sigmls from tens,hundreds, or thousands of othei telephones.The combined signals leave the
switchingcenter;tleir form dependson the distancethey must travel.In
our example,tley are sent througl wires in underground coaxial cablesto
a microwavetransmissionstation.Here, the signalsare transformedinto
microwavefrequenciesand broadcastfrom a tansmission antena through
air and space,via a communicationssatellite,to a rcceivingantenna.The
microwave receiving station translates the microwave signals into a folm
suitablefor furtler transmission,perhapsas pulsesof light to be sent
through fiber-opticcable.On arival at the secondswitchingcenter,the
combinealsignalsare separated,and each is routed to the approprrate
telephone,where an earphoneacts as a speakerto convert the leceived
electricsignalsback hto soundwaves.At eachstageof the process,electric ci.cuits operate on tlle signals. Imagine the challenge involved in
designing, building, and operating each circuit in a way that guarantees
that all of the hundredsof thousandsof simultaneouscallshave high quality
Computer syslemsuse electric signalsto processinformation mrging from word processingto mathematicalcomputations.Systemsrange
in size analpower from pocket calculato$ to personal computers to
supercomputersthal perform suchcomplextasksas plocessingweather
data and modelingchemicalinteractionsof complexorganicmolecules.
Thesesystemsincludenetworksof microcircuits,or iniegratedcircuitssYstem.
postage-stampsized
assembliesof hundreds,thousands,or millions of tigure1,1 A Atetephone
Circuit
Vanabbs
electricalcomponentsthat often operateatspeedsandpowerlevelsclose
to fundamenlal physical limits, including th.3speedol light and the thermo
dynamic laws,
Control systemsuse electricsignalsto regulateprocesses,
Examples
include the control of tempemtures, pressures,and flow mtes in an oil
refinery;the fuel-airmixlure in a fuel-injectedautomobileengine;mechanismssuchasthe molors,doors,andlightsin elevators;andthe locksin the
Panama Canal.The autopilot and autolanding systemsthat help to fly and
land airyIanes aie also familiar control systems
Powersyst€msgenerateand distributeelectricpower.Electricpower,
which is the foundatior of our technology-based
society,usuallyis generated in large quartities by nuclear, hydroelectric, and thermal (coal-, oil-,
or gas-fired)generato$.Poweris distributedby a grid of conductorsthat
crissooss tlle country. A major challenge in designing and operating such
a systemis to provide sufficient redurdancy and control so that failure of
any piece of equipment does not leave a city, state, or region completely
figure1.2 a A tT scanofan adLrtt
head.
Figure1.3 A Anajrplane-
Signal-processingsystens act on electric signals that represent information. They translorm the signalsand the informarion contained in tlem
into a more suitable form. There are many different ways to process the
signalsand their information. For example,image-processing
systems
gathermassivequantitiesof data from orbiting weathersatellites,reduce
the amount of data to a manageablelevel,and transform the remaining
data into a video imagefor the eveningnewsbroadcast.A computerized
tomography(CT) scanis another€xampleofan image-processing
system.
It takessignalsgenemtedby a specialX-ray machineand transformsthem
into an imagesuchas the one in Fig. 1.2.Although the original X-ray signals are of little useto a physician,once they are processedinto a recognizableimagethe informationthey containcanbe usedin the diagnosisof
djseaseand injury.
Considerable interaction takes place among the engineering disci
plines involved in designhg and operating these five classesol systems.
Thus conmunications engineersuse digital computeis to control the flow
of information,computers contain control systems!arrd control systems
contail computers,Po\ver systemsrequire extensive communications sys
temsto coordinatesafelyand reliably the operationof components,
which
may be spread acrossa continent. A signal-processingsystemmay involve
a conmunicationslink, a computer.and a control system.
A good exampleof the interaction among systemsis a commercial
airplane,such as the one showl in Fig. 1.3.A sophisticatedcommunicatiom systemenablesthe pilot and the air traffic controllerto monitor the
plane'slocation.permittingthe air traffic controllerto designa safeflight
path for a of the nearbyaircraftand enablingthe pilot to keep the plane
on its designatedpath. On the newestcommercialairplanes,an onboard
compulersystemis usedfor managingenginetunctions,implementingthe
navigation and flight control systems,and gererating video information
scrcensin the cockpit.A complexcontrol systemusescockpit commands
to adjustthe positionand speedofthe airplane,producingthe appropriate
signals to the engines and the control surfac€s (such as the wing flaps,
ailerons, and rudder) to ensure the plane rcmains safely airbome and on
the desiredflight path.Theplane must have its own power systemto stay
aloft and to provide and distributethe electricpower neededto keep the
1.1
cabin lights on, make the coffee, and show the movie. Signal-processing
systems reduce the noise in air taffic communications and tlansform
information about the plane's location into the more meaningful form of a
video display in the cockpit. Engineering challengesabound in t}le design
of each of these systemsand their integration into a coherent whole. For
example,these systemsmust operate il widely varying and unpredictable
environmental conditions. Perhaps the most importalt engineering challenge is to guarantee that sufficient redundanry is incorporated in the
designsto ensurethat passenge$arive safelyand on time a! their desired
destinations.
Although electrical engineersmay be interestedprimarily in one
area, they must also be knowledgeable il other areas that interact with
this area of interest. This intemction is part of what makes electrical engi
neering a challenging and exciting profession.The emphasisin engineer
ing is on makingthingswork,soan engineelis free to acquireand useany
technique,ftom anyfield, tlat helpsto get t}lejob done
CircuitTheory
In a field asdiverseas electricalengineering,you might well askwhether
all of its brancheshave anythingin common.The answeris yes rlectric
circuits.An elect'riccircuit is a mathematicalmodel that approximates
the behaviorof an actualelectricalsystem.Assuch,itprovidesan important foun{:lationfor leaming-in your later coursesand as a practicing
engineer-the details of how to design and opefate systemssuch as tnose
iust described.The models,the mathematical techniques,and the language
of circuit theory will form the intellectual framework for your future engi
neering endeavors.
Note that the term electtic circuit is commonly used to reter to an
actual electrical system as well as to the model that represents it. In this
text. when we talk about an electdc circuit, we always mean a model,
unlessotherwisestated.Itis the modelingaspectof circuit theorythat has
broad applications acrossengineering disciplhes.
Circuit theory is a specialcaseof elestromagneticfield theory: the study
of static and moving electdc charges.Altlough generalized Iield theorl
might seemto be an approp ate starting point for investigating electric signals, its application is not only cumbersomebut also requires the use of
advanced mathematics. Consequendy,a course in electromagnetic field
theory is not a prerequisite to urderstanding the material in this book. We
do, ho ever, assumethat you have had an introductory physics course ilr
which electdcal and magneticphenomenawere discussed.
Three basicassumptionspermit us to use circuit theory,rather than
electromagnetic field theory to study a physical systemrepresented by an
electric circuit. These assumptionsare as follows:
7. Electrical effects happen instantaneously throughout a system.We
can make this assumption because we know that electric signals
travel at or near the speed of light. Thus, if the system is physica y
small, elecfic signals move through it so quickly that we can consider them to affect every point in the systemsimultaneously A system that is small enoughso that we call make this assumptionis
system.
calleda lumped-paramet€r
An0verview
Engjneenng:
EL€ciricat
2. The net charge on ewty component in the system is always zero.
Thus no component can collect a net excess of charge, although
some components,as you will learn latei, can hold equal but oppositeseparatedcharges.
3. Therc is no magnetic co pling betteeenthe componentsin a system.
As we demonstrate later, magnetic coupling can occtrr within a
componenL
Thafs it; there are no other assumptions.Using circuit theory provides
simple solutions (of sufficient accuracy) to problems that would become
hopelesslycomplicatedif we were to use electromagneticfield theory
These belefits are so great that engineeN sometimes specifically design
electical systemsto ensurethat theseassumptioNare met. The importance of assumptions 2 and 3 becomes apparent after we inlroduce the
basic circrlit elementsaltd the rules for analyzing interconnected elements.
However,we needto take a closerlook at assumption1.The question
r\ "How smalldoesa pb).icalsyslemhare ro be ro quatifyas a lumped
parameter system?"We can get a quantitative handle on the question by
noting that electdc signals propagate by wave phenomena. If the wavelength of the signal is large compared to the physical dimensions of the system, we have a lumped-parameter system.The wavelength I is the velocity
dividedby the repetitionrate,or frequency,of the signal;thatis,,\ = c/f.
The frequency/ is measuredin hertz (Hz). For example,power systems
in the United States operate at 60 Hz. If we use t]le speed oI light
103m s) as lhe velocir) ot propagation.
1r - 3
lhe walelenglhis
5
l0'm. If rbe po\rer(ystemot inrere\ric pbysically
smatterlhan $is
wavelength,we can representit as a lumped-parametersystemand use
circuit theory to analyzeits behavior.How do we definemarel? A good
N\e 1sthe rule of 1/10ri: if rhe dimensionoI the systemis 1/10th (or
smaller.)of the dimension of the wavelength,you have a lumped-parameler
system.fhus,aslong asthe physicaldimensionof the power systemis less
than 5 x 10)m,we cantreat it as a lump€d-parameter
system.
On the other hand, the propagation frequency of mdio signalsis or tlle
order of 10vIIz. Thus the wavelength is 0.3 m. Using the rule of 1/10ttr, rtre
relevant dimerNionsof a connunication syslemthat sendsor receivesmdio
signalsmust be less tian 3 cm to qualify as a lumped-parameter system.
menever any of the pertinent physical dimensionsof a systemurder study
approachesthe wavelength of its signals,we must use electromagneticfield
theory to aralyze that system. Thioughout this book we study circuits
de ved ftom lumped-parameter systems.
ProblemSotving
As a practicingengineer,you will not be askedto solve problemsthat
have already been solved. Wlether you are trying to improve the perfomance of an existing systemor qeating a rew system,you will be working on unsolved problems.As a student, however, you witl devote much ol
your attention to the discussionof problemsalready solved.By reading
about and discussinghow theseproblemswere solvedin the past,and by
solving related homework and eram problems on your own, you w.ill
begin to develop the skills to successfully attack the unsolved problems
you'll faceas a practicingengineei.
AnoveMew
Engineering:
1.1 Etectrical
Some general problem-solving prccedures are presented here Many
perlainlo lhinkingaboutand organLing)our solulionslr0legy
rhem
ol
proceedingwith
calculations.
6elor€
1.. Identify what'sSiven and what's to be found h problem solving,you
need to know your destilation before you can selecta route tbr getting tlere. What is the problem asking you 10 solve or find?
Sometimesthe goal of tle prcblem is obvious; other times you may
need to paraphrase or make lists or tables of known and unknowr
inJormation to seeyour objective.
The problem statement may contain extraneous information
tlat you need to weed out betore proceeding.On the other hand' it
may offer incomplete information or more complexities thar can be
handled given the solution methods at your disposal. In that case,
you'll need to make assumptionsto fill in the missinginformation or
simpliJy the problem context. Be prepared to circle back and reconsider supposedlyextraneousinformatior and/or your assumptionsif
youl calculationsget boggeddown or produce an answerthat doesn't
seemto make sense.
2. Sketcha cicuil diagram or other risual model. Translathg a verbal
problem description into a visual model is often a useful step in the
solution process.If a circuit diagam is aheady provided' you may
need to add inlormation to it, such as labels, values' or leference
directions.You may also want to redraw the circuit in a simpler, but
equivalent, form. Later il this text you vrill Ieam the methods for
developing such simplified equivalent clrcuits
3. Think of several solution methoh and decide on a w6Jrof choosing
among theru'f\is colnse will help you build a collection of analytical tool$ several of which may work on a given pioblem. But one
method may pioduce fewer equations to be solved than another, or
it may require only algebra bstead of calculus to reach a solulion.
Suchefficiencies,if you can anticipate them, can sfeamline your cal_
culations consideiably. Having an alternative method in mind also
givesyou a path to pursue iJ your first solution attempt bogs down
4. Calculate a solution. Your planning up to tltis point should have
helped you identify a good analytical method and the correct equationstor lhe probtem.Not\ comesrhe\olulionof lhoseequalions.
Paper-and-pencil, calculator, and computer metlods are all available for performing t]le actual calculations of circuit analysis
Efficiency and your insxuctor's prefeiences will dictate which tools
you shoulduse.
5. UseWw creatirit!.ltyou suspectthat yolu answeris off baseor if the
calculationsseemto go on and on without moving you toward a solu_
tion, you shotlld pause and consider altematives.You may need to
revisit your assmlptions or selecta diffeient solution method. Or, you
may need to take a less-conventionalproblem_solvingapproach'such
asworking backwad from a solution.This text provides answen to all
of the AssessmentPrcblems and many of the Chapter Problems so
that you may wo* backward when you get stuck. ln the real world,
you wonl be given answersin advance,but you may have a desired
prcblem outcomein mind from which you can work backward.Other
seative approachesinclude allowing yourself to see parallels witl
CircuiiVariabtes
other t!?es of probleris you've successfullysolved, following your
intuition or hunches about how to proceed, and simply setting the
prcblem asidetemporadly and coming back to it later.
6. Test your sol tion. Ask yourself whether the solution you've
obtained makes serlse.Does the magnitlde of the answer seemiea
sonable?Is the solution physically realizable? You may want to go
furtler and rework tle problem via an alternative method. Doing
so will not only test the validity of your odghal answer,but will also
help you develop your intuition about the most efficient solution
methodsfor various kinds of problems.In the real \qorld, safetycdtical designsare always checked by several iDdependent means.
Getting into the habit ol checking your answeis will benefit you as
a studentand as a practicingengheer.
Theseproblem-solving stepscannot be used asa recipe to solve every prob
lem in this or any other cource.You may need to skip, changethe order of,
oI elaborate on certail stepsto solve a particular problem, Use thesesleps
as a guideline to develop a problem-solving style that works for you.
1.2 TheInternational
Systemof Units
Engineen comparc theoretical results to experimental results and compare competingengine€rhgdesignsusingquantitativemeasures.
Modern
engineering is a multidisciplinary profession in which teams of engineers
work together on projects, and they can connnunicate their results in a
meaningful way only if they all use the same units of measure. The
Intemational System of Units (abbieviated SI) is used by aI the major
engheeringsocietiesand most engifleersthroughoutthe world;hencewe
useit in this book.
Th€ SI units are basedon sevend€lired quantities:
. length
' fime
. elect c current
. thermod]'namic tempemture
. amount of substance
. luminousintensity
:rI4*l:lrlillrri
Qumtily
BasicUnit
Length
Mass
kg
Tine
Thermodynamic temperature
AmouDt of substane
K
1.2
System
ofUnits
TheIntenationat
Thesequantities,along rvith the basicunit and s]'mbo1for each,are
lisled in Tablc 1.1.Although not sldclly SI units,the familiar time unils of
minute (60 s), hour (3600s), and so on arc often usedin engineeringcal
culalions.In addirion.defined quanlitiesarc combinedto form deriv€d
units.Some,suchas force,energl',power,and elcctriccharge.you already
krow through previousphysicscou$es.Table 1.2lists the derived units
usedin this book.
Ir manycases,
the SI unit is either ioo smallor too largeto useconveprefixcs
niently. Standard
correspondingto polve$ of 10, as ljsted in
Table 1.3,are then appliedto the basicunit. All o{ lhescprefixesare correct.but engineenoftcn useonly the onesfor powersdivisibleb]' 3;thus
centi,deci,deka,and hectoare usedrare\,.Also, engineersoflen selectthe
prefix lhat places the base number in the range between I and 1000.
Supposethat a time calculationyieldsa result of 10 ' s,that js,0.00001s.
Most engineers would dcscribe this quantity as 10 /is. that is.
j
ps.
10 = 10 x 10j s,mlher than as0.01ms or 10,000,000
prefiresfor powersof 10
objective1-lJnderstandand be abteto useSI units and the standard
1.1
1.2
How many dollarsper millisecondwould the
federalgovemmenthaveto collectto retire a
delicit of$100 billion in one year?
If a signalcantmvelin a cableat 80%of the
speedof light,whatlengthof cable,ir inches,
rcpreserts1 ns?
Answer:9.45'.
Answ€r: $3.17/ms.
NOTE: Abo try ChaptetPrcblems1.1,1.3,and L6.
Pr€fixesto signify
TABLE1.3 Standardized
Prefu
Symbol
Pol€r
t0 r3
TABLE
1.2 DerivedUnits in SI
Qufllity
UnitNrne (Slnbol)
pico
p
hetv (Hz)
106
newtoD(N)
joulc (J)
N m
nilli
1 0 3
t0 l
J
coulonb (c)
Electricpotential
vo1l(V)
d
h
C/V
Magnetic llux
lrenry (H)
1o'
t0l
kilo
sienens(s)
l0'
l|)
I/C
ohm (o)
Electricconduclance
10 r'
10v
M
106
G
10e
T
l0''
10
1.3 CircuitAnalysis:An 0verview
modet
forelechcatenqitiguret.4 A A conceptuaL
needng
design.
Before becoming involved in the details of circuit analysis, we need to
take a broad look at engineeing design,specilicallythe designof electric
circuits. The purpose of this oveNiew is to prcvide you with a perspective
on where circuit analysis fits within the whole of circuit design. Even
though this book focuseson circuil analysis,w€ try to provide opportudties for circuit design where appropriate.
A11engineeringdesignsbegin with a need,as shownin Fig. 1.4.This
need may come from the desire to improve on an existing desig , or it may
be something brand-new A careful assessmentof the need results in
design specifications,which are measumble characteristics of a proposed
desigl. Once a designis proposed,the designsp€cificationsallow us to
assesswhether or not the design actually meets the need.
A concept for the designcomesnext. fhe concept derives ftom a complete understandingof the designspeciJicationscoupled with ar insight into
t]!e need,which comesfrom education and experience.The concept may be
realized as a sketch, as a written description, or ir some other fom. Often
t]le next step is to translate tle concept into a mathematical model. A commoily usedmathematicalmodel for electrical systemsis a circuit model.
lhe elements fhat comprise the circuit model are called ideal circuit
components. An ideal circuit component is a mathematical model of an
actualelecldcalcomponent,like a battery or a light bulb. It is important
for the ideal circuit component used in a circuit model to iepresent the
behavior of the actual electrical component to an acceptable degree of
accuracy.The tools of circuit snalysis, the focus of tlis book, are then
applied to ttre circuit. Circuit analysisis basedon mathematical techniques
and is used to predict the behavior of the circuit model and its ideal circuit
components.A comparison between the desired behavior, from t}le design
specifications,and the predicted behavior, from circuit analysis,may lead
to rcfinements ir the c cuit model and its ideal circuit elemerts. Once the
desiredandpredictedbehavioraie in agreement,a physicalprototypecan
be constructed.
The physical proroqpe is an actual electrical system,constnrcted ftom
actualelectricalcomponentsMeasuiementtechniquesare usedto determine the actual,quantitativebehaviorof the physjcalsystem.This actual
behavior is compared vrith the desired behavior from the designspecifications and the predicted behavior from circuit analysis.The comparisons
may result ir rcfhements to the physicalprototype,the circuit model,or
botb Eventually, this iterative process,ir which models, components,and
systems are coflthually rcfined, may produce a design that accurately
matches the designspecifications and ttrus meets the need.
From this desciption, it is clear that circuit analysis plays a very
importantrole in the designprocess.Becauseciicuit analysisis appliedto
circuit models,practicingengineerstry to use mature circuit models so
that the resultingdesignswill meet the designspecificationsin the filst
iteration.In this book. we use modelsthat have been lestedfor between
20 and 100 years; you can assume that they are matuie. The ability to
model actual electrical systemsvrith ideal circuit elements makes circuit
theory extremely useful to engineers.
Sayingthat the interconnectionof ideal circuit elementscan be used
to quantitativety predict th€ behavior of a system implies that we can
1.4
Voltage
andCLrent
describethe interconnectionwith mathematicalequation$For the mathematicalequationsto be useful,wemustwrite them in termsof measurable
quantities.Inthe caseof circuits.thesequantitiesare voltageand current,
whjch we discussin Section 1.4.The study of circuit analysisinvolves
understandingthe behavior of eachideal circuit elementin terms of its
voltage and current and understandingthe constraintsimposedon the
voltageand currentas a resultof interconncctingthe ideal elcmcnts-
1.4 VottageandCurrent
The conceptof electricchargeis the basisfor describingall electricalpheof electriccharge.
nomena.Let's review someimportant characteristics
. The charge is bipolar, meaDingthat electrical effects are described in
termsof positiveand negativecharges.
. The electric chargeexistsin discretequanrities,which are integral
multiplesof the electroniccharge,1.6022x 10-1eC.
. Electical effectsare attributcdto both the sepafationof chargeand
chargesin motion.
ln circuit theory, the separation of charge creates an electric force (vol!
age),andthe motion of chargecreatesan electricfluid (cunenr).
The conceptsof voltageand cu ent are usetul from an enginee.ing
point ol view becausethey can be expressedquantitatively.Whenever
positiveand negativechargesare separated,
energyis expended.Voltage
is the energyper unit chargecreatedby the separation.We expressthis
ratio in differential form as
(r.1)
4 Definitionof voltage
0 = the voltagein volts,
1, = the energyinjoules,
4 = the charge ir coulombs.
The elect cal effectscausedby chargesin motion dependon the rate
of charge flow The rate of charge flow is klown as the electric currcnl,
which is expressedas
,=#,
(1.2) < Definitionof current
j : the currentin amperes,
q = the chargein coulombs,
I : the time in seconds.
Equations1.1and1.2are definitionsfor the magnitudeofvoltage and
cu ent,rcspectively.The
bipolarnatureof electdcchargerequircsthat we
assignpolarity referencesto thesevariables.we will do so in Section1.5.
11
t2
Although current is made up of discrete,moving elechons, w€ do not
need to considerthem hdividualy becauseof the enormousnumber of
them. Rather, we can think of electrons and thei corresponding charge as
one smoothly flowing entity. Thus, i is heated as a continuous variable.
One advantageof usingcircuit modelsis that we canmodel a compo,
rent stricUy in terms of tie voltage and current at its terminals. Thus two
physically different components could have the same relationship
betweenthe teminal voltage and terminal current. lf they do, for purposesof circuit analysis,they are identical.Once we know how a component behaves at its terminals, we can anal'ze its behavior in a circuit.
However,when developingcircuit models,we are interestedin a component's intemal behavior.We might want to know, for example,whether
charge conduction is taking place because of free electrons moving
through the dystal lattice structureof a metal or whetherit is becauseof
electronsmovingwithin the covalentbondsof a semiconductormaterial.
However, these concerns are beyond the rcalm of circuit theory In this
book we usec cuit modelsthat have alreadvbeen develoDed:
we do not
di.cu\\ho$ compoDent
modelsarede\eloped.
1.5 TheIdealBasicCircuitElement
Figure
1.5A Anid€albask
circuit
ehment.
An ideal basiccircuit el€menthasthree attributes:(1) it hasonly two terminals,which are points of connection to other circuit components;(2) it is
describedmathematicallyin terms of current and/or voltage;and (3) it
cannotbe subdividedinto other elements.
We usethe word deal to imply
that a basiccircuit elementdoesnot exist as a realizablephysicalcomponent. However,aswe discussed
in Seclion1.3.ideal elementscan be connected in order to model actual devices and systems We use the word
Dartcto imply that the circuit elementcannotbe further reducedor subdivided into other elements.Thus the basic circuit elemelts form the bui]ding blocks for constructing circuit models, but they themselvescannot be
modeledwith any other type of element.
Figur€ 1.5is a rcpresentation of an ideal basic circuit element.The box
is blank becausewe are making IIo commitment at this time asto the type of
circuitelementit is.In Fig.1.5,the voltageacrossthe teminals of the box is
denoted by o, and t}le cunent in tlle circuit element is denoted by i. The
polaiity reference for the voltage is indicated by the plus and minus signs,
and the reference direction for the curent is shown by the arrow placed
alongsidethe current. The interprctation of these rcIercnces given positive
or negativenumerical valuesof o and i is sunmarized in Table 1.4.Note that
.u voltagedrop from terminal1 to terminal2
i
voltage rise from terminal 1 ro terminal 2
vollageise ftomterminal2to terminall
voltagedrop from terminal2 to terminal1
positivechargeflowing from terminal1 ro terminal2
positive charge tlowing from terminal 2 to terminal I
negativechargeilowing ftom terminal2 io tenninal1
negative charge flowi.g fiom terminal 1 to termilal2
1.5
TheldeatEasicCiftuitEtement
the notion ofpositivechargeflowingin one directionis equiv
algebraically
aleni lo the notion ofnegativechargeflowingin the oppositedircction
Thc assignmcntsof the rcferencepolarily for voliage and the reftr
encedircction for curreni irc entirely arbitrary.Howevel.olrccyou have
assignedthe Iefercnces,you must wrilc all subsequentequations to
agr;e \vith the choscnreferencesThe most widely Lrscdsign convention
applied lo theseretcrencesis called the passivesign conv€niion,which
we use throughoul thjs book The passivesign conventioncan bc staled
Whcneverthe referencedirectionIor the currentin an elementis in
the directionof the re{ercncevoltagcdrop acrossthe elemcnt(asin
Fig.1.5).usea positivesignjn any exprcssionthat relateslhc voltage
ro Lhccuff<nt.O,hef$iri. u.e a n(galr\e \igrr'
-l Passivesign convention
We applythis signconvertion in all the analysesthat follo$r Ourpurpose for inlroducirg it even before we havc introduccd the dilTereni
lypcsofbasic circuit elenentsis !o impresson )routhe Iact that the selection of polaity referenccsalong with the adoption ot the passivcsign
conventior is rol a function of the basicelementsnor thc type oI interconncctionsnadc with thc basic elements.We present thc applicatlon
ard interpretalionof the passivesignconventionin power calculationsin
Seclion1.6.
Objective2-Know and be abteto usethe definitionsot voltageandcu ent
1.3
The currentat the terminalsof thc clement
Fig.1.5is
The expressionfor the chargeeDtedngthe
upper t€rminalofFjg.1.5 is
1
i :20e 5ooo'A, r > 0.
Calculatethe total charge(in microcoulombs)
enteringthe elementat its upper terninal-
Answer: 4000/..C.
NOTE: Also try Chapter Ptoblem 1.9
- ( . r . , le'" c.
Find lhe maximumvalueofthe curlenl entering the terminalifa = 0.03679sr.
Answer: l0 A.
14
1.6 PowerandEnergy
Power and energy calculations also are important in circuit analysis.One
reason is that although voltage and current are useful variables in the
analysis and design of electrically based systems,the useful output of the
systemoften is nonelectrical, and this output is conveniently expressedin
terms of power or energy.Another reason is that all practical deviceshave
Iimitations on the amount of power that they can handle. In the design
process,therefore, voltage and current calculations by tlemselves are not
sufficient.
We now relate power and ene.gy to voltage and curent and at the
sametime use the power calculation to illustrate ttre passivesEn convention. Recall from basicphysicsthat power is the time rate of expending or
absorbing energy. (A water pump rated 75 kW can deliver more lite$ per
secord than ore rated 7.5 kW.) Mathematically, energy per unit time is
er-Dressedin the form of a derivative. or
,:'H,.,
Definitionof power>
(1.3)
the power rn watts,
the energyin joules,
the time in s€conds.
Thus1W is equivalent
to 1 J/s.
The powei associatedwith the flow of chargefollows directly from
thedefinitionof voltageandcurrentin Eqs.1.1and
1.2,or
aw
/ aw\( aq\
dt
\ dq l\dt J'
Power
equation>
(1.4)
p : t}le power in watts,
1)= the voltage in volts,
i : the curent in amperes,
1.6
Equation 1.4 shows that the pow€r associatedwith a basic ckcuit element
is simply the pmduct of the current in the element and the voltage adoss
the element. Therefore, power is a quaDtity associatedwith a pair of tei_
minals, and we have to be able to tell from our calculation whether power
is being delivered to the pair of teminals or extracted from it. This information comes ftom the corect application and interpretation of tle passive sign convention.
If we usethe passivesign convention, Eq. 1.4 is correct if the rcference
alirection for the current is in the direction of the rcference voltage drop
acrossthe terminals. Otherwise, Eq. 1.4 must be written with a minus sign.
In other words, if the current reference is il the direction of a refererce
voltage rise acrossthe teminals, the expressionfor the power is
Power
andEneqy
+
(b)p = -?)i
(1.5)
'The algebraic sign of power is based on charge movement tbiough
voltage drcps and rises.As positive chargesmove ttrrough a drop in voltage,they lose energy,and as they move through a rise in voltage, they gain
energy.Figure 1.6 summarizesthe relationship between the polarity rcferencesfor voltage and current and the expressiol for powerWe can now state the rule for interprcting the algebraic sign of power:
If the power is positive (that is, if p > 0), power is being deiivered to
the circuit iNide the box. If the power is negative (that is, iJ p < 0),
power is being extracted from the circuit inside the box.
For example, suppose that we have selected the polarity refeiences
shown h Fig. 1.6(b). Assume further that our calculations for the curent
and voltage yield ttre following numeical results:
j=4A
and D:
1 0V .
Then the power associatedwith the terminal pair 1,2 is
? = _(_10x4): 40w.
Thus t}Ie circuit inside the box is absoibing 40 W.
To take this analysis one step furthel, assumethat a colleague is solv_
ing the same problem but has chosen the reference polarities shown in
Fig.l.6tcl.Theresuldngnumerical\aluesdre
i:-4A,
0:10V,
and
p:40w.
Note that interprcting these results in terms of this reference system gives
the same conclusions that we proviously obtained-namely, that the circuit inside the box is absorbing 40 W In fact, any of the reference systems
il Fig. 1.6 yields tlis sameresult.
,
I;:]
l - - : l
*-113
(c)p=-d
41.-_stl}-l
.
L - - - I
*----f:--]
1
(d)p = ,i
andth€expr€sion
rcferences
figure1.6a Potadty
signof power
atgebraic
{ Interpreting
76
objective3-Know andusethe definitionsofpowerandenergyt
Objective
4-Be abteto usethe passive
sign
1.5
Assumethat a 20 V voltagedrop occursacross
an clcmcnt from tcrminal2 to lcrninal l and
that a curcnt of4 A entcrstcminal2
a) SpeciJythe valuesof d and t for the polarity
ret<fence.
shoqnin rig. 1.6(r)(J .
b) Statewhetherthe circuit insidethe box is
absorbhgor deliveringpo$'er.
c) How much power is the circuit absorbing'?
Calculatcthc total cncrgy (injoulet dclivered
lo the circuit clement.
Answer: 20 J.
1.7
Answen (a) Circuit1.6(a):r,= 20V,t:
4A;
circuii1.6(b):1]:
20V,i = 4A;
c 20V,t = 4A:
c i r c u j1t . 6 ( c ) : =
circuit1.6(d):0: 20 V. t : 4 A;
(b) absorbing;
(c) 80w:
1,6
A higi-voltagedirecl curenl (dc) lransmissioll
line betweenCelilo,Oregonand Sylmar,
C ,litufDiI i. opcrrtirgJt 800kV andcarr\ i19
1800A, asshowr. Calculalelhe power (in
megawalts)at the Oregonend ofthe line and
statethe directionofpower flow
Assumethat the voltageat the terminalsofthe
elementin Fig.1.5correspondingto the currcnt
in AssessnentProblem1.3is
kv,
1]: 10e-s00e
I > 0
Answ€r: 1,140MW Celilo to Sylmar.
NOTE: Also try ChapterProblems1.12,1.17,1.24,
and 1.26.
Srimnrary
'lhe
InternationalS]'stemof tjnits (SI) enablesengjneers
to communicatc in a mcaningful way about quanlilalive
rcsults.Tablc 1.1summarizcsthc bascSl unils;Table1.2
prcsentssomcuscill dcrivcdSI units,(Seepages8 and9.)
Circuit analysisis basedon the variablesofvoltage and
current.(Seepaser1.)
Vollageis the erergy per unit chargecreatedbt' charge
sepantion and has the SI unit of \olt Q) = dlNldll).
(Seepage11.)
Cunenl is the rate ofchargetlow and hasthc SI unit of
ampere(, : d4ldr). (Sccpagc I l.)
The id€albasiccircrit elemenlis a two-terminalcompoDert that cannot be subdivided;ii car be described
malheDralically
iD tenns ofits terminalvoltageard currenl. (Seepage12.)
The passivesign conventionusesa positivesign in the
expressionthal relales lhe voltagc and cu cnt at thc
le niials oI an elemenl when thc relcrcncc dircction
Ior the cu enl lhroughthe elemenlisin thc dircclionof
the relerence vollage drop across the element. (Scc
page13.)
Pofler is energy per unit of iime and is equal to the
productofthe terminalvoltageand curren! it hasthe SI
unit of$'att (t : dflrlll = ot). (Seepage1a-)
The algebraicsignof power is interpretedasfollows:
. If p > 0, power is being deliveredto the circuit or
cfcurt component,
. If r, < 0, power is beingextracledtuomthe circuil or
circuii component.(Seepage15.)
17
Problems
Seclion 12
L1 There are apprcximately 250 million passenger
vehicles rcgistered ir the United States. Assume
that the batteiy in the average vehicle stores
440 watt-hours (Wh) of energy. Estimate (in
gigawatt-houis) tlle total energy stored in U.S.passengei vehicles,
1.2 The line desuibed in AssessmentProblem 1.7 is
845 mi in lengtl. The line contains lour conductors,
eachweighing2526lb per 1000ft. How many kilo
$ams of conductorare in the line?
L3 The 4 giga-blte (GB : 10ebytes) flash memory chip
for an MP3 player is 32 rnm by 24 rnn by 2.1mm.This
memory chip holds 1000thee-minute songs
a) How many seconds oI music fit into a cube
whosesidesarc 1 mm?
b) How many bytes of memory are stored in a cube
whosesidesaie 100pm?
1,4 A hand-held video player displays 320 x 240 picture
elements(pixels)in eachframe of the video.Each
pixel requires 2 bytes of memory. Videos are dis'
played at a mte of 30 frames per second.How many
minutes of video will fit in a 30 gigabyte memory?
1,5 Some species of bamboo can grow 250 nn/day.
Assume individual cells in the plant are 10 pm long.
a) How long, on average,does it take a bamboo
stalkto $ow 1 cell length?
b) How many cells are added in one week, on
average?
1.6 One liter (L) oI paint covers approximately 10 m2
of wall. How thick is the layer before it dries? (/{tnr:
1L=1x106mm3.)
35 /rA curent, due to the flow of electrons.What is
the averagenumber of elechonsper secondthat
flow past a fixed reference qoss section that is perpendicular to the direction of flow?
1,9 The current enteriry tlle upper terminal of Fig. 1.5is
i : 24 cos4000tA.
Assume t}le charge at t}le upper terminal is zerc at
the instant the current is passingthrough its maximum value.Find the expressionfor q(t).
L10 How much eneryy is extracted ftom an electron as
it flows through a 6 V battery from tlle positive to
tbe negarivererminal: F\pfess )ouf answef iD
attojoules.
Sections1.F1.6
1,11 Ore 9 V battery supplies 100 rA to a camping
flashlight. How much energy does the battery supply in 5 h?
1,12 Two electiic circuits, repiesented by boxes A and
B, are connectedas shown in Fig. P1.12.Therefer
ence dhection foi ttre current i h the inteiconnection and the reference poladty for the voltage ?.r
across the inter connectiotr are as shown in the figure.For eachof the following setsof nume cal values,calculate the power in ttre interconnection and
state whether the power is flowing ftom A to B or
a) i:5A,
b) i:
8A,
c) i:16A,
d) t:
u: l20Y
r:25OY
t,: -1s0v
10A, 0:
480V
figuruP1.12
Section 1.4
A curent of 1200A exists in a copper wfue,wrth a
circular cioss-section (iadius = 1.5 nrm). The current is due to free electrons moving through the
wfue at aIt average velocity of o meten/second If
the concentntion of free electrons is 1de elecrrons
per cubic meter and iJ they are uniformly dispersed
tbroughout the wire, then what is the averagevelocity of an electon?
1.8 In electronic circuits it is not unusual to encounter
currents in the micioampere iange. Assume a
1.13 The references for the voltage and cu ent at the
terminal oI a circuit element are as shown in
Fig.1.6(d).ThenumedcalvaluesIor o andt are 40V
and 10A
a) Calculate the power at the teminals and state
whether the power is being absorbedor deliveredby the elementin rhe box.
18
va'iabtes
Circuit
b) Given that the current is due to electronflow,
statewhetherthe electronsare enteringor leaving terminal2.
c) Do the electronsgain or loseenergyasthey pass
thmugh the elementin the box?
1.14 RepeatProblem1.13with a voltageof
60V.
1.15 Wlen a car hasa dead battery it can often be started
by connectingthe battery from another car acrossits
terminals. The posilive terminals are connected
together as are the negative terminals.The connection is illustrated in Fig. P1.l5. Assume the crment i
in Fig.P1.15is measuedandfound to be 30A.
a) Which car hasthe deadbattery?
b) If this connectionis maintainedfor 1 min, how
much energyis transferredto the deadbattery'?
a) Find tlre power absorbedby the element at
1=10ms.
b) Find the total energyabsorbedby the element.
1,19 The voltage and current at the terminals of the circuit elementin Fig.1.5are shownin Fig.Pl.19.
a) Sketchthe power versuslplot for 0 < t < 50 s.
b) Calculalethe energydeliveredto the circuit ele
ment at t : 4, 12, 36,ard 50 s"
fiqureP1.19
1](v)
t0
Figure
P1.15
2
0
2
8 1
8
10
1,16 The manufacturer of a 9V diy-cell flashlight battery
saystlat the battery will deliver 20 mA for 8u conrinuoushours.During thal time lhe vollageqill
drop tromq v lo o V Assumethe drop in !ollageis
linear witl time.How much energydoesthe battery
deliver in tlis 80 h interval?
1,17 The voltage and current at th€ t€rminals of the cir
cuit elementin Fig.1.5are zerolor , < 0. For , > 0
they are
?) = e
s00,
i : 30
?-1500! v,
]0
Il.4
0
2024'2832364044
s2 s6 r(s)
-1.0
mA
40e56'+ 10e-1s00i
a) Find the power at r - 1 ms.
b) How much energyis deliveredto the circuit element between0 and 1 ms?
c) Find the total energydeliveredto the element
1.18 The voltage and curent at the terminals of fte cir
6trG cuit elementin Fig.1.5aie zerofor I < 0.Fort > 0
they are
t' = 400eroo'sin 200t V,
j = 5e-1m'sin200tA.
1.20 lhe voltage and cu ent at the terminals of t}le cir
6pre cuit elementin Fig-1.5are zerofor 1 < 0. For I > 0
r):75 - 75a1oauv,
j = 50e rm/mA.
a) Find the maximumvalue ofthe power delivered
to the circuit.
b) Find the total energydeliveredto the element.
Probt€ms19
1:1 The voltage and current at the teminals of the eleP*r€ ment inFig.1.5 are
36 sin 200rt V,
r':
1,23 The voltage and currcnt at t}Ie teminals of the ciitsi'c crit element in Fig. 1.5 are zero for t < 0. For t > 0
they are
i = 25 cos 2007t A.
1)- (16,000r+ 20)e-eo'V,
a) Find the maximum value of the power being
delivered to the element.
b) Find t]le maximum value of the power beirg
erlracted frcm the element.
c) Find the aveiage value of p iJl the interval
0<t<5ms,
d) Ftnd the average value oI p in tlle htelval
0<l<6.25ms.
L22 The voltage and cu:rent at the teminals of an auto6PIcrmobile battery du ng a charge cycle are shown in
fi$.P1.22.
a) Calculate tne total charge tmnsfered to the
battery
b) Calculate the total energy transferred to the
battery.
Figure
P1.22
t,(v)
12
10
t = (128r + 0.16)?{0' A.
a) At what instant of time is ma-\imum powel
delivered to the element?
b) Find the maximum power in watts.
c) Flnd the total energy delivered to tlle element in
millijoules.
1.24 The voltage and clment at the terminals of tlte cir6flc cuit elementin Fig.1.5are zerofor t < 0andt > 3s
are
In rhejniervalbelween0 and 3 s lbe er?ressjons
b:t(3
t)V, 0<t<3s;
i=6-4tr,4.,0<t<3s.
a) At what instant of time is the power being delivered to the c cuit element maximum?
b) What is the power at the time found in pa (a)?
c) At what bstant of time is the power behg
extracled frcm tle c cuit element maximum?
d) Wlat is the power at the time found in part (c)?
e) Calculate the net energy delivered to the circuit
at 0,1,2 and3 s.
6
4
z
12
lo
20
l(ks)
(A)
16
14
12
10
8
6
4
2
1,2
t6
20 r(kt
1.25 The voltage and curent at the terminals of the cirE Ic cuit element ir Fig. 1.5 are zero for t < 0. For t > 0
mey are
+ s)eroo'V,
?)= (10,000r
1> 0;
t = (40r + 0.05)e4orA,
r > 0.
a) Fird the time (in milliseconds) when the power
deli\eredlo lhe ciJcuitelemeDlis maximum.
b) Find the maximumvalue olp in milliwatts.
c) Find the total energy delivered to the circuit element in millijoules.
20
'.26 The numerical values for the currents and voltages
in the circuit in Fig.P1.26are givenin Table P1.26.
Find the total power developed in the circuit
P1.26
Figure
5.0
2.0
3.0
-5.0
150
250
200
400
1.0
50
350
400
I
-2.0
g
350
6-0
P1.26
TABI.E
Etemenl
Currenr (A)
Yoltage (n\7)
150
150
0.6
100
250
300
-0_8
-0.8
rigureP1,28
2.0
7.2
300
f
1.28 The numerical values of the voltages and currents
in the interconn€ction seenin Fig. P1 28 are grven in
Table P1.28. Does tlle interconnection satisfy t]le
Powercheck?
4 fl3
r27
Assumeyou are an engineerin chargeof a project
and one of your subordinate engineen reports that
the interconnection in Fig. P1.27 does noi pass the
power check. The data for the interconnection are
givenin TableP1.27.
a) Is the subordinate corect? Explain your answer'
b) If the subordinate is corect, can you find the
error in the data?
P1.27
Figure
,
+
Elem€nt
.
L
l
+
Volt.ge (\,)
36
250
-250
2a
t
108
32
+
f
g
48
h
j
80
80
100
150
350
200
- 150
-300
129 One method of checkirg calculationsinvolving
interconnectedcircuit elementsis to see that the
total power deiivered equals the total power
absorbed (conservation-of-energy principle). Witl
this thought in mind, check the interconnection in
Fig. P1.29and statewhether it satisfiesthis power
check.The current and voltagevaluesfor eachelement arc given in Table P1.29.
1.30 a) In the circuit shown in Fig. P1.30, identity
which elemenls are'absoibing power and which
are detilefiDgpower. usi0g the pa$ive sign
conventron.
b) The numerical values of tle currents ard voltagesfor each element are given in Table P1.30.
How much total power is absorbed and how
much is delivered in this circuit?
Figure
P1.29
figureP1.30
!
,t" "l
1.6
2.6
1,.2
1.8
f
g
j
80
60
50
2t)
1.8
3-6
3.2
30
40
30
-20
2.4
30
L
+
Elemenl
volraee(mv)
b
300
-100
200
2M
350
I
g
h
200
250
50
!
c
+
Curenl (irA)
25
10
15
35
25
10
35
10
Elements
Circu'it
Thereare five ideal basic circuit elements:voltascsourccs.
2.1 VoLtage
and CunentSourcesp. 24
2.2 ELectric;L
Resistance
(Ohmt Law) p. 28
2.3 Con5trudionof a Circuitl4odetp. 32
2.4 Kir€hhoff'slaws p. .t6
2.5 Anatysisof a CircuitContainingDependent
Undertandth€ symbots
for afd the behaviorof
the fottowifgideatbasiccircujtel€ments;
jndependent
voltag€andcuir€ntsources,
d€pefdertvotiageaid cuiientsourc€s,
and
Beabteto stat€0hmt taw,Kirchhoffscurrent
ta& and (irchhoffsvottagetaw andbe abteto
usetheselawsto anatyze
simpt€cjrcujtr.
Knowhowto caLculate
th€ powerfor €ach
elementjf a simpl€circujtandbe abteto
detefmjiewhetheror not the powerbatarces
22
cunent sources,r'esistors,
inductols.and capacito$.In this chaptcr wc discussthc chaaactcristicsof voltage sources,current
Altlrough this ma,yseemlike a small num
sources,and r-esistors.
pracber of elementswith which to beginanalyzingcircuits,mal1y
tical systemscan be modclcdwith iust sourcesand resistors.They
are alsoa LlseftLl
startingpoint becauseo[ their relativesimplicity;
thc mathcmaticalrclationshipsbctwcen voltage and current ill
sourcesand resistorsare algebraic.Thusyou will bc ablc to bcE:in
leaning the basic techniquesoI cjrcuit analysiswith only algebraic manipulations.
We will postponeintroducinginductols and capacitorsuntil
Chapter6, becausetheir userequiresthal you solveintegral and
However. the basicanalyticaltechniqucs
differential ecluations.
for solvingcircuitswith induclols and capacitorsare the sameas
those introduced in this chapter.So, by the time you need tcr
begirlmanipulatingmore difficult equations,you shouldbe vcry
familiar with the methodsof writirg them.
PracticaI
Perspective
Etectrical.
Safety
"Danger-High
VoLtage."
Thiscommonty
seenwarning
is misteading.
Attformsof energy,
including
etectricaL
energy,
can
gut ifs not ontythe vottagethat harms.
be hazardous.
The
staticetectriciry
shockyou receive
whenyou watkacross
a
carp€t
andtoucha doorknob
is annoying
butdoesnotinjure.
Yetthat sparkis caused
by a vottagehundr€ds
or thousands
oftimeslargerthanthevoltages
that cancause
harm.
Theetectricat
energy
that canactuatty
cause
injuryis due
to etectricai
currentandhowit ftowsthroughthe body.Why,
then,doesthesignwarnof highvottage?
Eecause
ofthe way
poweris produced
eLectrical
anddjstributed,
it is easierto
determjrevoLtages
than cuffents.Atso, most etectdcaL
produce
sources
constant,
specified
voltages.
So the signs
warnaboutwhatis easyto measure.
Determining
whether
andunderwhatcondjtjons
a sourcecansupptypotentiaLty
dangerous
currents
is moredifficutt,asthisrequjres
anunderstanding
of etectrical
engineering.
Before
wecanexamine
thisaspect
of eLectricaL
safety,
we
haveto learnhowvoltages
andcurrents
areproduced
andthe
reLationship
between
them.Theetectrical
behavior
of objects,
suchas the humanbody,is quite comptexand often beyond
comptetecomprehension.
To atLowus to predjctand controt
phenomena,
eLectfical
we usesimpLiryjng
modelsjn whjchsimple mathematjcaL
retationships
betweenvottageand current
jn reaLobjects.Suchmodapproximate
the actuaL
retationships
elsandanalyticaL
methods
formthe coreofthe electrjcaL
engineering
technjques
that wjLlattow
usto understand
attetectrjcaL
oheromena.
incldding oseretdLirgto electdcaI
safety.
At the end of this chapter,we witt usea sjmpteetectric
circuitmodelto describehow andwhy peopleareinjuredby
etectriccurrents,Eventhoughwe rnayneverdeveLop
a compteteand accurateexptanation
of the etectricaL
behaviorof
the humanbody,we can obtaina ctoseapproximation
usjng
simptecjrcuit modeLs
to assessand improvethe safetyof
etectrjcalsystemsand devices.DeveLoping
modetsthat pro,
videan understanding
that h jmperfectbut adequate
for solvinq practicalprobtems
lies at the heartof engineering.
I4uch
of the at of etectricat
engineering,
whichyou wil[ Learnwith
js in knowingwhen and how to sotvedifficutt
experience,
probtems
by usingsimptiryingmodets.
23
24
2.1 VoltageandCurrentSources
ideal voltageand curent sources,we need to consider
Befoie aliscussing
the general nature of electncal sources.An elechical sourc€ is a devjce
that is capable oI converting nonelectric en€rgy to electric energy ard vice
versa.A discharging battery converts chemical energy to electric energy,
whereas a battery being charged converts electric energy to chemical
energy.A dynamo is a machhe that conveits mechanical energy to electdc
mode,it is
energyandvice versa.If operalingin the mechanical-to-electic
energy,
it is
to
mechanical
ftom
electric
called a generator. If transforming
about
these
to
remember
thing
referred to as a motor. The impotant
sourcesis tlat theY can eitler deliver or absorb elect c power, generally
maintaining either voltage or current. This behavior is of particular
interest for circuit analysisand led to the creation of the ideal voltage
sourceand the ideal current sourceas basiccircuit elements.Thechallenge is to model practical sourcesin telms of the ideal basic circuii
An ideal voltage source is a circuit element that maintains a prescibed voltage across its terminals regardless of tle curent flowing ilt
those terminals. Similarly, an idesl cuEent source is a circuil element that
maintains a presoibed current thmugh its terminals regardlessof the volf
age aooss those teminals. These circuit elements do not exist as practical
devices-they are idealizedmodelsof actualvoltageand currentsources.
Usingan idealmodelior currenlandvollagerourcesplacesan important restriction on how we may descdbe them mathematically. Becausean
ideal voltagesourceprovidesa steadyvoltage,even if the current in the
it is impossibleto speciry'the current in an ideal voltage
elementchanges,
source as a function of its voltage. Likewise, iJ t}le only inlormation you
have about an ideal ctment sourceis the value of curent supplied,it is
impossibleto determinethe voltageaooss that current sourceWe have
sacrificed our ability to relate voltage and current in a practical source for
the simplicity of using ideal sourcesin circuit analysis.
Ideal voltageand curenl sourcescan be further describedas either
independent sourcesor dependent sources An independent sowc€ estab
lishes a voltage or current in a circuit without relying on voltages or cur
rents elsewherein the circuil. The value of the voltage or current supplied
is specifiedby the value of the independentsourcealore. In contrast'a
avoltageor curent whosevaluedependson
dep€ndenlsouce establishes
tie value of a voltage or current elsewherein the circuit. You cannot spec,A
( - )
ify the value of a dependent source unlessyou know the value of the volt
ageor currenron whichil depends.
The circuit symbolsfor the ideal independentsourcesale shown rn
Fig.2.1.Note that a circle is usedto rcpresentan independentsourceTo
completely specify an ideal independent voltage source in a circuit, you
(b)
G)
must include the value of the supplied voltage and the reference polarity,
ideal independforG)anideatinde- as shownin Fig.2.1(a).Similarly,to completelyspecifyan
Figure2.1^ Ihecircuiisymbob
pendent
vothge
source
and(b)anjdeatindependent ent cuirent source,youmust includeth€ valueofthe suppliedcurrentand
its rcferencedirection,asshownfuFig.2.1(b).
Y
L
(b
Y
2-1 Vottag€
andCunent
Sourc€s 25
The circuit symbolsIor the ideal dependentsourcesare shown in
Fig. 2.2.A diamond is used to representa dependert source.Both the
depeDdentcurrent sourceand the dependentvoltagesourcemay be controlled by either a voltage ot a current elsewherein the circuit, so there
are a total of four variations,as indicated by the symbolsin Fig. 2.2.
Dependent sourcesare sometimescalled controlled sources.
To completely specify an ideal dependent voltage-controlled voltage
source!you must identify the controlling voltage,the equationthat permits you to compute the suppliedvoltage ftom the controlling voltage,
and the referencepoladty for the suppliedvoltage.ln Fig.2.2(a),tlrecontrollirg voltage is named o,, the equation that determinesthe supplied
I, = Itrx.
? t
G)
( c)
?
and the referencepolarity for ?r is as indicated.Note that ri is a multiplying constantthat is dimensionless.
Similar requircments exist for completely specifying the other ideal
(b)
(d)
dependentsources.InFig.2.2(b),the controllingcurrcnt is i' the equation
Figure2.2 A Thecircujtsymbols
for G)an ideat
for the supplied voltage ," is
Ds = p1\,
the relerencepolarity is as show4 and the multiplyingconstantp hasthe
dimensionvolts per ampere.In Fig. 2.2(c),the conrroling volrageis o-,
the equationfor the suppliedcurrentrsis
rs:
dl)r,
the referencedirectior is asshown,and themultiplying constantd hasthe
dimensionamperesper volr. In Fig.2.2(d),the controling currentis j,, the
equation for the supplied current i, is
i, = Bi,,
the referencedirection is as shown,and the multipllng constantp is
dimensionless.
Finally, in our discussionoI ideal souices,we note that they are
examplesof activecircuit elements.Anactiveelementis one that models
a device capableof generatingelectdc energy.Passiveetementsmodel
physicaldevicesthat cannot generateelectric energy Resistors,inductorE and capacitors are examples of passivecircuit elements.Examples 2.1
and 2.2 illustrate how the characteristicsof ideal indepefldent and
dependentsourceslimit the t'?es ofpermissibleinterconnectionsofthe
dependent
voltage-controthd
vottas€
(b) an
source,
jd€atdependent
current-controtted
vothgesoufte,(c)an
jdeatdependent
vottage-conhoLled
currcnisource,
and
(d) anjdeatd€pendent
cunent{ontrolLed
cunentsource.
26
C cuitEtements
TestingInterconnections
of ldealSources
Using t}le defidtions ofthe ideal independentvoltage and current sources,state which interconnections in Fig. 2.3 are pemissible and which violate
tlle constraints imposed by t}le ideal sources.
10v
10v
Solution
Conneclion(a) is valid. Each sourcesuppliesvolt
age aooss the samepair of terminals,marked a,b.
This requires that each sourcesupply the samevoltagewiti the samepolarity,which they do.
Connection (b) is valid. Each source supplies
current tbrough the same pair of terminals, marked
a,b.This requiresthat eachsourcesupplythe same
currentin the samedirection,which they do.
Conrection (c) is not permissible. Each source
supplies voltage across the same pair of terminals,
marked a,b.This requires that each sourcesupply
the samevoltage with the samepolarity, which they
(b)
+\
LOV
/+
5V
G)
Connection(d) is not permissible.Each source
suppliescurrenr througl the samepair oI teminals,
rnarked a,b.This requiresthat each sourcesupply
the same current in the same direction, which they
Connection(e) is valid.The voltagesourcesupplies voltage acrossthe pair of terminalsmarked
a,b.The current sourcesupplies curent through the
same pair of terminals.Becausean ideal voltage
sourcesuppliesthe samevoltageregardlessof the
current, and an ideal curent sourcesuppliesthe
sane currentregardlessof the voltage,thisis a permissibleconnection.
fortuampte
2.1.
Figur€2.3a Thec'rcuits
(d)
2.1
Voltage
andCurentSources
TegtingInterconnections
of IdeatIndependent
andDependent
Sources
Using the definitions of the ideal independent and
depende0r
sourcer.
sralewhichiDterconneclrons
in
Fig. 2.4 are valid and which violate the consuamN
imposedby the ideal sources.
T\\
|),=3r,,
Solution
b
Connection (a) is invalid. Both the independenr
souce and t])e dependent souice supply voltage
across the same pair of terminals, Iabeled a,b.This
requircs that each source supply the same voltage
with the samepolaity.The independent sourcesupplies 5 Y but the dependentsouce supplies 15V
Connectiotr (b) is valid. The independent volragesource suppliesvoltage acioss the pair of terminals maiked a,b. The dependent current source
suppliescurent through the sarnepair of terminals.
Becausean ideal voltagesourcesuppliesthe same
voltage regardless of cuirent, and an ideal currert
source supplies the samecurent regardlessof voltage,this is an allowable contrection.
Connection(c) is valid. The independenrcurrent source supplies crment tlrough the pait of termhals marked a,b.The dependentvoltage source
supplies voltage aqoss the same pafu of terminals.
Because an ideal cufient source supplies the same
current regardless of voltage, and an ideal voltage
source supplies the same voltage regaidless of currcnt, this is an allowable connection.
Connection(d) is invalid. Both the irdependent sourceand the dependent sourcesupply cunent
through ttre samepair of terminals, labeled a,b.This
requkes that each source supply the same curent
in the same reference dircction. The independent
souce supplies 2 A, but the dependent source suppfies 6 A iII the opposite direction.
(.)
r ) /", ="3", , ( t
+\
b
(b)
t)
'.=4r.
G)
I
T
i,
b
(d)
Figure2.4lr Thecircuitstor
Examph
2.2.
28
Objective1-understand ideal basiccircuit etements
2.7
2.2
For the circuit shown,
a) what vdl e ofzr is rcquired in order for the
interconnectionto bc valid?
b' for lhi. valu<ol "s.finLllle po$err"ociatedwith lhe 8 A source.
For lhc circtlil shown,
a) What valueof d is requiled in order for the
interconncctionto be valid?
b) For the valueof a calcularedin part (a), fhd
with the 25 V sonrcethe power associated
Answer; (a) 0.6 A/Vl
(b) 375 w (37sw absolbed)
Answer: (a) -2 V;
(b) 16 w (16 w delivered).
NOTE: Abo try ChuptetProblens 2 2 dnd2 3.
fiesista$ee{$hm's!-aw}
2.2 El.cctfiea{
+\\.-
R
a resktorhavinga
figur€ 2.5 :" Thecjrcuitsynrbotfor
Resistrnc€is lhe capacityof mateials to impede the llow ot curcnt or'
thc flou of clectdc charge.ll]c circuil clement usc'l to
more specifically,
model ihis behavioris the resistor-Figure2 5 showsthe circuit symboltor
Lhcresistor,with R denotingthc resislancevalueof the resistor'
Conceptually,we can undcrstandresistarceif we think aboul the
moving electronsthat nake up clectriccurrentiltcracting with and being
resistedby the atonic structurcof the rnaterialthrough which they are
noving. In lhe cou6c of thesc interactions,somc amount of electric
eDergyis conr,ertedto thcrmal encrgyand dissipaledin the [om1 of heat'
Tlis cfTectmay be undeslrable.However,many uselu] electdcaldevices
take advantageot resislancehealing.includingstoves.toasters,rrons.and
Most marerialsexhibit measumbleresislanceto current The amouDl
of resistancedcpendson the matcrial Metals such as copper and alu
minlrm have smal1values of resistance,making then good choiceslbr
wiriig usedto conductelcctriccurcnl ln fact' when rcpresenledin a cir
cuit diagram.copperor aluminumwiring isn't usuallymodeledas a resN
tor; thc resistanccof the wire is so small compared1()the resistanceoI
other elementsilr the circuit that we can neglcctthe lYiringresistanceto
simplify
'For the diagran.
purposesof circuil analysis.we musl rcference the currcnt in
the resistorio the tcrminal voltage.We can do so in two waysieilher in
2.2
ELectricat
Resktance
(ohmtLaw)
29
the direction of the voltage drop acrossthe resistor or in the direction
of the voltage rise acrossthe resistor,as shown in Fig. 2.6.If we choose
the former, the rclationship betweenthe voltage and current is
?r- iR,
(2.1) .{ 0hm'staw
il
r : the voltagein volts,
^ 'l
i = the curent in amperes,
r=iR
R : the resistancein ohms,
r=
,R
Figure2.6.q Twopossibte
r€ference
choices
forthe
drrentandvottage
atthe ternrinals
of a resistor,
and
If we choosethe secondmethod,we must wnte
u = -iR,
:1
\2.2)
where r,, j, and R are, as before,measuredin volts,amperes,and ohms,
respectively.
The algebmicsignsusediII Eqs.2.1 and2.2are a directconsequerceof the passivesignconvention,which we introducedin Chapter1.
Equations 2.1 and 2.2 are known as Ohm's law after Georg Simon
Ohm, a German physicistwho establishedits validity early in the nineteenth century Ohm's law is the algebraic relationship between voltage
8r}
and current for a resistor. In SI units, resistanceis measured in ohms.The
Grcek letter omega (O) is the standard symbol for an ohm. The circuit
Flgure2.7d. Thecircuit
symboLfor
an8 O rcsjstor.
diagramsymbolfor an B O resistoris shownin Fig.2.7.
Ohm's law expressesthe voltage as a function of the current. However,
expressingthe curcnt as a function of the voltage also is convenient.Thus,
frcm Eo.2.1.
(2.3)
or,fromEq.2.2,
12.4)
Thereciprocal
of theresistance
is rcferredto asconductance,
is sym(S).Thus
in siemens
bolizedby theletterG, andis measured
" = * "
\2.5)
An 8 O resjstor has a conductancevalue of 0.125S In much of tie profes
sionaliiterature, the ur t usedfor conductanceis the mho (ohm spelledbackward), which is symbolized by an inverted omega ((J). Therefore we may
also descdbean 8 O resistor ashaving a conductanceof 0.125nho, ((').
30
Circuit
Etements
We use ideal resistors in circuit analysis to model the behavior of
physical devices.Using the qualifier ideal rem'inds us that the resistor
model makes several simplifying assumptions about the behavior of
actualresistivedevices.lhe most impo ant ofthese simplifyingassumptions is that the iesistanceof the ideal resistoris constantand its value
aloesnot vary over time. Most actual resistive devicesdo rot have constant
resistance. and their resistance does vary over time. The ideal rcsistor
modelcanbe usedto representa physicaldevicewhoseresistancedoesn't
vary much from some constant value over the time period of interest in
the circuit analysis.Inthis book we assumethat the simplifyhg assumptions about resistanc€devicesare valid, ard we thus use ideal resistors in
circuit analysis.
We may calculate the powel at the termhals of a resistor in several
ways.The first approach is to use tle defining equation and simply calculate the product of the terminal voltage and curent. For the refeience systemsshoM in Fis.2.6.we x'dte
\2.6)
w h e n o= i R a n d
(2.7)
when1' = -iR.
A secondmethodof expressingthe power at the terminalsol a resistor expresses power in terms of the currcnt arld the resistance.
SubstitutirgEq. 2.1into Eq. 2.6,weobtain
p=ui=(iR)i
Power
in a resistorin termsof currentb'
P=i2R
(2.8)
Likewise,substitutingEq.2.2 into Eq.2.7,we have
p = -1)i : -(-i
R)i = i2R.
(2.e)
Equations2.8 and 2.9 are identicaland demonstrateclearly that, regard_
lessof voltagepolarity and currentdirection,the powel at the terminalsof
a resistor is positive. Therefore, a resistor absorbspower from the circuit.
A tlird methodof expressingthe power at the teminals of a iesistor
is in terms of the voltage and resistance The expressionis independent of
the Dolaritv references.so
Power
in a resistorin tennsof vottageF
p
R
(2.10)
?.2
Etectricat
(0hm'sLaw)
Resistance
31
Sometimesa resistor'svaluewlll be er?rcssedasa conductance
rather
than as a resistance.Usirg tlle relationship betweer resistance and conductancegivenin Eq.2.5,we may also wrire Eqs.2.9 and 2.10in lerms of
the conductance,
or
i,
(2.11)
l?.12)
Equations2.G2.12providea vadety ofmethodsfor calculatingthe power
absorbedby a resistor. Each yields the sameanswer.In analyzing a circuit,
look at the informationprovidedand choosethe power equationthatuses
lbal inlormaliondirecll).
Example2.3 illustratesthe applicationof Ohm,s law in coniunction
with an ideal sourceand a resistor.Powercalculationsat the terminalsofa
iesistor also aie illustrated.
Cal.cutating
Vottage,
Current,andPou/er
for a SimpteResistive
Circuit
Ir eachcircuitin Fig.2.8,eittrerthe valueof z or, ts
I ) 1 A 1 , , ,8 o
50v
rb= (50x0.2)- 10A.
The voltager). in Fig.2.8(c)is a dse in rhe direction of the current in the rcsistor. Hence
(a)
f ) r a " .: o o
0.2s
The currenti, in the resistorwith a conductance
of 0.2 S in Fig. 2.8(b) is ir the direction of rhe
voltagedrop acrossthe resistor.Thus
(b)
(
)50v
(c)
0c= -(1)(20)= 20v.
25n
'.1
(d)
tigure2.8A Thecircuih
forExampL€
?.3.
a) Calculatethe valuesof 0 and i.
b) Determineth€ power dissipatedin eachresistor.
The curent id in the 25 O resisrorin Fig.2.8(d)
is in the direction of the voltagedse affoss the
resistor.Therefoie
. 5 0
' 2 5
b) The power dissipatedin eachof the four resistorsis
r8l ( 1 ) 1 8=) 8 w .
r s o= ;
= 500w,
pors: (s0),(0.2)
Solution
a) The voltageoain Fig.2.8(a)is a drcp in the direction of the curent in the resistor.Therefore,
,o : (1X8)
I 7n\)
= 20w,
r:on= _fr = t1t(20)
{50)2
= ( 2)12s)- 100w.
PL'a= i-
of ideal sourcesand resisHaving introducedthe generalcharacteristics
lors,wc next showhow to setheseelementsto build the circuit model of
a practicalsystem-
objective2-Be abteto state and useohm's Law. . .
2.3
For the circuit shown.
For the circuit shown,
a) Ifos = 1 kV and ls = 5 mA, find the vahre
of R and lhe power absorbed by tlre resistor
b) Ifis : 75 mA and thc powcr dcliveredby
the voltagesourceis 3W,find os,n, and th€
power absorbedby the rcsistor,
c) If R - 300 O and the power absorbedby R
is 480 mW. find ts and z,g.
a) ff is = 0.5A and G : 50 mS,find ?s and
t}repower deliveredby the curent source
b\ lr ds - l5 V rnd lhe poqcr Lleli\efed
lo rhe
conductoris 9 W.find the conductanc(G
and the sourcecurrentis.
c) If G : 200pS and thc power deliveredto
the conductanceis 8 W find is ard z's.
r)
G
Answer: (a) 10V5 W;
(b) 40mS,0.6A;
(c) 40mA,200v
Answer: (a) 200kO,5 W:
(b) 40v s33.330,3 W;
(c) 40mA,i2V
2.6and2.8.
NOTE: Alsotty ChapterProblems
2.3 eonstruction
cf a eireuitiv!*dei
We have alreadyslatedlhat onc reasonfor an interestin the basiccircuit
elementsis thal they can be uscd 1o constructcircuit modelsof practical
systens.Thcskill rcqujred1odevelopa circuitmodel ofa deviceorsysten
is as conplex as thc skill rcquired to solvethe derived circuit.Although
thc skills requiredto solvecircuits,I'oualsowill need
this text emphasizes
other skills in the practiceof clcctricalengineering,and one of the most
impoilanl is modeling.
We deveiopcircuit modelsin the next two cxamples.ln Exampie2.1
we construcla circuit modei basedon a knowledgeofthc bchaviorof the
system'scomponentsand how the componenlsare intcrconncctcd.ln
Example2.5we createa circuitmodelbymeasuringthc tcrminalbehavior
2.3
Consiiuction
ofa Cncuitllodet
Constructing
a CircuitModetof a Ftashtight
Consrructa circuitmodel oia flashlight.
SoLution
We chosethe flashlighl to illustrate a practical system
becauseits components are so familar. Figure 2.9
shows a photograph of a widely available flashlight.
When a flashlight is regarded as an electrical
system,the componentsof primary interestare the
batteries, the lamp, the connector, the case,and the
switch. We now consider the circuit model for each
A dry-cell battery maintahs a reasonably constart teminal voltage if the current demand is rot
excessive.Thus if the dry-cell battery is opemting
witlin its intended limitq we can model it wrth an
ideal voltage source.The prescribed voltage ther is
constant and equal to the sum of two dry-cell value$
The ultimate output of the lamp is light energy,
which is achieved by heating the filament in the
lamp to a tempeiaturehigh enoughto causeradia
tion in the visible range. We can model tlte lamp
with an ideal resistoi. Note in this casethat although
the resistor accounts for the amount of electric
enelgy converted 10 thefitlal energy,it does not predict how much of the thermal energy is converted to
liglt eneigy.The rcsistor used to represent the lamp
does predict the steady current dmin on the batteiieE a charactedsticof the systemthat also is of inteiest.In thismodel,Rr symbolizestle lamp resistance.
The connector used in the flashlight se es a
dual role. First, it provides an electdcal conductive
patll between the dry cells and the case.Second,it is
formed into a springy coil so that it also can apply
mechanical pressure to the contact between the
batteries and the lamp.The puipose of this mechanical pressureis to maintain contact between the two
dry ce s and between the dry cels and the lamp.
Hence, in choosing the $rirc lor the connector, we
may find that its mechanicalprope ies are more
important than its electrical properties for the
flashlight design. Electdcally, we can model the
€onnector with an ideal resistoi, labeled Rr.
The case also se es both a mecharical and an
electdcal purpose Mechanicaly, it crntains all the
other componentsand provides a $ip for the peNon
usingit. El€ctrically,it provides a connectionbetween
Flglre 2.9 A A flashtight
canbeviewed
asanetectricat
system.
other elementsin the flashlight. If the caseis metal,it
conductscurent betweenthe batteries and the lamp.
If it is plastiq a metal strip inside the case connects
the coiled connector to the switch. Erther way, an
ideal resjstor,which we denote &, models the electri,
cal connectionprovided by the case.
The final component is the switch. Electically,
the switch is a two-statedevice.It is either oN or
oFF.An ideal switch offen no resistanceto the current when it is in the oN state, but it offers iDfinite
resistance to curent when it is in the oFF state.
Thesetwo statesrepresentthe limiting valuesof a
resislor:
rharic,rheO\ slalecoffespoDds
ro a resi,
tor with a numedcal value of zero, and the oF! state
correspolds to a resistor with a numerica] value of
infinity. The two extreme values have the descriptive names sho circuit (R = 0) and open circuit
(R : m).Figure 2.10(a)and (b) showttregraphical
representation of a short circuit and an open circuit,
respectively.
The symbolshown in Fig.2.10(c)represents the fact that a switch can be either a shofi
circuitoran open circoit,dependingon the position
oI its contacts,
34
Etements
Circuit
(b)
OFF
*/-
ON
G)
(b)openci(uit.
cncuit.
Figure2.10A Circuit
synbols.
G)Short
G)swiich.
We now construct the circuit model of the
flashlight.Staiting with the dry-cell balteries,the
positive terminal of the first cell is connectedto
the negativeterminalof the secondcell,asshown1n
Fig.2.11.Thepositiveterminal of the secondcell is
connected to one terminal of the lamp. The oth€r
teminal ofthe lamp makescontactwith one sideof
lhe .sirch,and lhe other.ide of lhe s\\ilchis connect€dto the metalcase.Themetal caseis then con
nectedto the negativeteminal of the fust dry cell
by mcans of the metal spring.Note that the ele
ments form a closed path or circuit. You can seefte
closed path formed by the connected elements in
Fig.2.11.F$re 2.12showsa circuit model for the
flashlight.
offtashtight
components.
Figure2,11 ,s Thearangement
a flashtight.
Figure2.12 A A ciftuitmodettor
We can make some generalobservationsabout modeling hom our
flashlight example:First. in developing a circuit model, the eli?cr'icdlbehavior of each physical component is of primary interest. In the flashlight
model, three very diJferent physical components a lamp, a coiled wire,
and a metal case-are all representedby the samecircuit element (a resistor), becausethe electrical phenomenon taking place in each is the same
Each js presenting resistanc€10the current flowing through the circuit
Second.circuit modelsmay need to accountfor undesiredas well as
desiredelectricaleffects.For example,the heat resultingftom the resist
ancein the lamp producesthe light, a desiredeffect.However,the heat
resultingfrom the resistancein the caseand coil representsan unwanted
or parasiticeffect.It drainsthe dry cellsand producesno usefuloutput
Suchparasiticeffectsmust be consideredor the resultingmodel may not
adequalelyrepresentthe system.
And finally,modelingrequiresapproximation.Even for the basicsystem representedby the flashlight,w.3made simplifying assumptionsir
developjngthe circuit model.For example,we assumedan ideal switch,
2.3
Construction
ofa Ciftuit!1odet
35
but in practicalswitches,contactresistancemay be high enoughto interfere with proper operationofthe system.Our model doesnot predicttlis
behavior. We also assumedthat the coiled connector exerts enough pressureto eliminateany contactiesistancebetweenthe dry cels. Our model
does not predict the eflect of hadequate pressure.Our use of an ideal
voltage source ignores any internal dissipation of energy in the dry cells,
which might be due to the parasiticheatingjust mentioned.We could
account foi this by adding an ideal resistor between the source and the
lamp resistor.Our model assumesthe internal lossto be negligible.
In modelhg the flashlightasa c cuit,we had a basicunderstandingof
and accessto the internal components of the system.However, sometimes
we know only the terminal behavior of a device and must use this infor
mation in constructingthe model.Example2.5 exploressucha modeling
problem.
Construding
a Cir.uitModelBasedon Terminat
Measurements
The voltage and current arc measued at the tenninals of the device illushated in Flg. 2.13(a),and the
values oI o! and ir are tabulated in Frg. 2.13(b).
Constructa circuit model of the deviceiiside t}le box.
,, (v) t (A)
-40
-10
20
-5
0
20
40
(")
5
10
Solution
Plotting the voltage as a function of the cunent
yieldsthegraphsho*nir Fig.2.14(a).The
equation
of tle lille in this figue illustratesthat the termhal
voltageis directlyproportionallo the terminalcurrcnt, t'! = 4i. In tems of Ohm's law, the device
insidethebox behaveslile a 4 ( I resistor.Therefore,
lhe cfcuir modelfor lhe de\iceiniide rhe bo-\is a
4 O resislor,
asseenh Fig.2.14(b).
Wecomebackto this techniqueof usingterminal characteristics
to constmcta circuit modelafter
i n t f o d u c i n g K i r c b h o f fs l a $ s a D d c i r c u i r a n a l y s i s .
(b)
Figure2.13 A lhe (a)device
and(b)datator Exampte
2.5.
r,,(v)
,10
20
20
40
G.)
(b)
Flgure2.14 ^ (a)Thevatues
ofor versus
t, forthedevicejnFig.2.13.(b)Ihe cncLrit
modet
forthedevjce
in Fi9.2.13.
NOTE:
Assessyow undetstunAing ofthk
efumpk by tyine Chapter Prcbkms 2.10 and2.11.
36
Cncuit
ttements
2.4 Kirchhoff'sLaws
a R 1
b
R "
Flgure2.15l Circuitmodetofthe
flashtight
with
assigned
voltage
andcurcntvariabtes.
A circuit is said to be solved when tle voltage across and the current in
everyelementhavebeendetemhed. Ohm\ law is an important equation
for deriving such solutions. However, Ohm's law may no1 be enough to
provide a complote solution. As we shall see in trying to solve th€ flash
light circuit from Example2.4,we need to usetwo more important alge
braic relationships, knom as Kirchhoffs laws,to solve most circuits.
We begh by redrawing the circuit as shown in Fig. 2.15, with the
switch in the oN state.Nole that we have also labeled the current and volt
agevadablesassociated
with eachresistorand the currentassociated
with
the voltage source. Labeling includes reference polarities, as always.For
convenience,we attach the same subscripl to the voltage and current
labelsaswe do to the resistorlabels.InFig.2.15,wealsoremovedsomeof
the terminal dots of Fig. 2.12and have insertednodes.Terminaldots ate
the start and end pointsofan individualcircuit element.A node is a point
wheretwo or more circuit elementsmeet.It is necessary
to identify nodes
in order to use K chhoff's currert law. as we will see in a moment. In
Fig.2.15,the nodesare labeleda, b, c, and d. Node d connectsthe battery
and the lamp and in essencestrelches all the way acrossthe top of t]le diagrum, though we label a single point for convenience.The dots on eithet
side of the switch irdicate its terminals, but only one is needed to representa node,so only one is labelednode c.
For the circuit shown in Fig.2.15,we can idenlify sevenurftnowns:
t,, t1,i., i, 1J1,
z'.,and ?,r.Recall that o" is a known voltage,as it represents
the sum of the terminal voltagesof the two dry cells,a constantvoltage
of 3 V The problem is to find the sevenunknown va ables.From algebra,you know that to find r unkno$,nquantitiesyou must solve,?simultaneousindependentequations.From cur discussionof Ohm's law in
Section2.2,you know that three of the necessaryequationsare
or : irRr,
(2.13)
\2.14)
12.15)
What about the otler four equations?
The interconnection of circuit elements imposes consttaints on the
relationshipbetweenthe teminal voltagesand current$Theseconstraints
are rcferred to as Kirchhoff's laws, after Gustav Kirchhoff, after Gustav
Kirchhoff, who first stated them in a paper published in 1848.The two laws
that state the constiaints in mathematical form are known as Kirchloff's
current law and Kirchhoff's voltage law.
We can now state Kfuchhoffs cunent law
Kirchhoff's
currenttaw(KCL)>
The algebraic sum of all the cufients at any node in a circuit
equalszero,
To use Kirchhoff's curent law, an algebraic sign co[esponding to a
reference direction must be assigned to every current at the node.
Assigning a positive sign to a current leaving a node requires assigninga
negative sign to a cu[ent entering a node. Convenely, giving a negative
sign to a current leaving a node requires giving a positive sign to a current
enteringa node.
Applying Kirchhoff's curent law to the four nodes in the circuit
shownin Fig. 2.15,usingthe conventionthat currentsleavinga node are
considered positive, yields four equations:
j"-i1 :0,
(2.16)
(?.17)
-ic-r:t),
q
\=t)'
(2.18)
(2.1e)
Note that Eqs.2.1G2.19are not an independentset,becauseany one
of the four can be dedved hom the ottrer tbree. In any cilcuit witl r nodes.
l'
1 independent crment equations cail be dedved from Kirchhoffs
current law.r Let's disiegardEq. 2.19 so that we have six independent
equations,
namely,Eqs.2.13-2.18.
We needone more,which we cande ve
from K cbloff's voltage law.
Before we can state K chhoffs voltage law, we must define a dosed
psth or loop. Starting at an arbitra ly selected node, we trace a closed
path in a circuit througl selected basic circuit elements ard returr to the
original node without passingtlrough any intermediare node more tlan
once.The circuit shown in Fig. 2.15 has only one closed pat}l or loop. For
example, choosing node a as the starting point and tracing tlte circuit
clockwise,we form the closed path by moving through nodes d, c, b, and
back to node a.We can now state Kirchhoff's voltase law:
The algebraicsum of all the voltagesaround ary closedpath in a circuit
equalszero.
To use Kirchhoffs voltage law,we must assignan algebnic sign (ieferencedirection) to eachvoltage ir the loop. As we tiace a closedpath, a voltagewill appeai either asa rise or a drcp in the tracing direction. Assigning a
positive sign to a voltage rise requiies assignirg a negative sign to a voltage
drop. Conversely,giving a negative sign to a voltage rise rcquires givhg a
positive sign to a voltage drop.
We now applyKirchhoffs voltagelaw to the circuitshowllin Fig.2.15.
We elect to trace the closed path clockwise, assignhg a posilive algebraic
signto voltagedropr Startingat node d leadsto the expression
11-1)c+1)1
I We say more about lhis obseNalion in Chapier ,1,
.=U.
(2.20)
{ Kirchhoff'svottageLaw(RVL)
38
CircuitEtemenis
which represents the severth independent equation needed to find the
sevenunknown circuit variables mentioned earlier'
The thought of having to solve seven simultaneous equations to find
the curent delivered by a pair oI dry cells to a flashlight lamp is not very
appealing. Thus in the coming chapte$ we introduce you to analytical
techniques that will enable you to solve a simple one-loop circuit by writing a single equation. However, before moving on to a discussionof these
circuit techniques, we need to make several observations about the
aletailedanalysisof the flashlight circuit ln general,these observations are
true and therefore are important to the discussionsin subsequentchapters. They also suppo the contention that the flashlight circuit can be
solvedby defininga singleunknown
Firsi, note that if you know tlte curent ill a resistor, you also know the
voltage across the resistor, because current and voltage are directly
relatod through Ohm's law. Thus you can associateone unknown variable
with eachresistor.eithel the current or the voltage Choose,say,the current as the unknovn vadable. Then, once you solve for ttre unknown current in the resistor, you cal! find the voltage acrossthe resistor. In general,
if you know the current in a passive element, you can find the voltage
across it. greatly reducing the number of simultaneous equatrons to be
solved. For example,in ttre flashlight circuit, we eliminate the voltages ?.,
ot, and ,l as unknowns.Thus at the outset we leduce the analytical task to
solving four simultaneous equations rather than seven
of conThe secondgeneralobser\alion felaleslo rhe coDsequences
curto
Kiichhoff's
nectingonly two elementsto form a node.According
you
know
the
if
to
a
node,
rent law, when only two elementsconnect
element.
the
second
you
know
it
in
also
current in one of the elements,
In other words,you need define only one unlnown current for the two
elements.Whenjust two elementsconnectat a singlenode,the elements
are said to be in series.The importance of ttris second observationis
obviouswhen you note that each node in the circuit shown in Fig 2.15
Thus) ou needto delineonl) one unknown
invol!eronly rqo elemenls.
current.The reasonis that Eqs.2.16-2.18lead directly to
\2.2r)
which states that if you know any one of th€ element currents,you
know them all. For example,choosingto use ir as the unknown elimi
nates4, i., and ir.The problem is reducedto determiningone unknown,
Examples 2.6 and 2.7 illustrate how to write circuit equations based
on Kirchhoffs laws. Example 2.8 illustrates how to use Kirchhoff's laws
and Ohm's law to find an unknown current. Example 2 9 er?ands on the
technique prcsented in Example 2 5 for constructing a circuit mod€l ibr a
device whose terminal characteristics are known.
39
UsingKirchhoff's
CurrentLaw
Sum the curents at each node in the circuit shown
in Fig.2.16.Note that there is no connectiondot (.)
in the center of the diagram, where the 4 {) branch
crossesthe branch containing the ideal current
Sotution
In wdting the equations,we use a positive sign for a
crment leavirlg a node.The four equations are
i\ + ia
n o d eb
i2+i3
i2
is: 0,
Figure2.16A Thecjrcuit
forExanrpte
2.6.
i 1- i 6 - t , = 0 ,
i3 - i4 - i.:0,
ib
node d
4+ta+ic:0.
UsingKirchhoff's
VottageLaw
Sum t}le voltages around each designated path ln
the circuit shown in Fig- 2.17.
Solution
In writing the equations,we use a positive sign for a
voltage drop. The four equations are
-r'1 + ti2 + ?r4- 2rb
path a
60
parh b
rJa+?]3+zrs:0,
path c
palh d
z'3: 0,
ab-14
-r.
- ,r
ir.
uc
?,6
q , rr7
?)5:0,
,J - 0.
Figure2.u A lhecircuit
forExampte
2.7.
40
CncuitEtements
Current
Lawsto Findan Unknown
Applyingohm'sLawandKirchhoff's
a) Use Kirchhoff'slawsand Ohm\ law to find i, in
the circuit shownin Fig.2.18.
10()
-
5oa
)tzov
We obtain the secondequation from Kirchhoff's
vollage law in combination with Ohm's law.
Nolinglrom Ohm c law lhrt ,- ic Int. and d' is
504,we sum the voltagesaroundthe closedpath
(t
forExanrpte
?.8.
Figur€2.18A lhecircuit
b) Tcst the solution for i, by verirying that the total
power generatedequalsthe total power dissipated.
Solution
120+10t,+50ir=0.
In writing this equation,we assigneda positive
sign to voltage drops in the clockwise direc
tion. Solving these two equations for i, and
it yields
t.:
3A
and
ir=3A.
in rh( 50 Q re.i.lorrs
b) fte po$ef dirsipaled
a) We begh by redrawingthe circuit and assigning
an unknown current to the 50 O resistor and
unknown voltages across the 10 O and 50 O
resisto$.Figure2.19showsthe circuit.The nodes
arelabeleda, h, andc lo rid the discussion.
p5oo= (3)'(50) : 4s0 w.
The power dissipatedin the l0 O resistoris
21oo=( 3),(10)=90W.
.
10() ,,,
The power deliveredto the 120V sourceis
- )rzov
= -120i" =
p121,y
jn Fig.2.18,withth€
shown
Figure2,19A Thecircuit
i1,0., andrr d€fined.
uiknowns
Becausei, also is the current in the 120V
source, we have two unknown currents and
lherefore must deive two simultaneousequa
tions involving l, and 4. We obtain one of the
equatioN by applying Kirchhoff's current law to
eithernodeb or c. Summingthe currentsatnode
b and assigning a positive sign to the cunents
leavingthe node gives
il
io-6:0.
120(-3) : 360 W
The power deliveredto the 6 A sourceis
pra =
u(6),
P6a =
but
tr = 50t1: 150 V
-1s0(6) : -900 w'
The 6 A source is delivering 900 W, and the
120 V source is absorbing 360 W. The 1otal
power absorbed is 360 + 450 + 90: 900 W
Therefore,the solution verifies that the power
deliveredequalsthe power absorbed.
41
a CircuitModelBasedon TerminalMeasurements
Construding
The terminal voltage and terminal current were
measuredon the deviceshown in Fig. 2.20(a),and
the valuesof tr and i ar€ tabulatedin Fig.2.20(b).
:r
-
,
,, (A)
+
t Device
' l
I
I
|
:
-
I
30v
10o
Figlre2.20A (a)Devjce
and(b)datatorExampLe
2.9.
a) Construct a circuit model of the device inside
b) Using this circuit model, predict the power tlis
devicewill deliverto a 10 O resistor.
(b)
jn
t/ torihe devic€
Figurc2,21 i (a)Thegftphof r, versus
r i s .2 . / 0 ( a )(.b i I 5 p' e \ Lf , n gc i ' c i' L - o d e t t r ! \ e d e r c e ' .
to a 10 O r€sjstor.
Fig.2.20(a),
connected
Solution
a) Plotting the voltage as a function of the current
yieldsthe giaph shownin Fig.2.21(a).Theequation of the line plotted is
u':30
5'!
Now we treedto identiJy the componentsof a circuit model t}Iat will produce the same rclationship between voltage and current. Kirchhoffs
voltage law tells us tlat the voltage drcps aqoss
two components in sedes add. From the equa
tion, one of those components produc€s a 30 V
drop regardless oI t}le current. This component
can be modeled as an ideal independent voltage
source.The other component producesa positive
voltage drop in the direction of t}le current ir.
Becausethe voltage drop is propo ional to the
current, Ohm's law tells us that this component
canbe modeledas an ideal resistorwith a value
of 5 (r. The resulting circuit model is depicted in
the dashedbox in Fig.2.21(b).
b) Now we attach a 10 O rcsistor to the deuce m
Fig. 2.21(b) to complete the circuit. Kirchhofs
current law tells us that t}le current in the 10 f}
resistor is the sameast]Ie current in the 5 {) resistor. Using Kirchhoffs voltage law and Ohm's law,
we can write the equation for the voltage drops
around the circuit, stafiing at the voltage source
and proceeding clockwise:
-30+5i+10i:0.
Solving for t, we get
i=24..
Because this is the value of currcnt flowmg in
the 10 O resistor,we canusethe powerequation
p : j2R to computethe power d€liveredto this
rcsistor:
Proo - (2)'(10) = 40 w
42
0bjective2-Be ableto stateanduseohm'slawandKirchhoff's
currentandvoltagelaws
For the circuit shown,calculate(a) i5;(b) "r;
(c) ?,r;(d) z,\;and (e) thc power deliveredby
the 24V sorrce.
2.5
2.7
Answ€r: (a) 2 A;
(b) 4Y
(c) 6 V;
(d) 14v;
(e)a8w
30
21V
a) The terminalvoltageand terminalcurrcnt
weremeasuredon the deviceshown,The
valuesof 2,,and j, are providedin the table.
Using thesevalues,createthe st(aightlil1e
ptot of 4 versusir-Computethe equalionof
the lire and usethe equarionto constructa
cfcuit model for the deviceusjngan ideal
voltagesourceand a resistor.
b) Use the model constructedh (a) to predicr
the power that the devicewill deliverto a
25 O resistor.
Answer: (a) A 25 V sourcein serics$iith a 100O
resistor;
( b )1 w
1A
21)
2.6
Use Ohm's law and Kirchhoffs iawsto find the
valueoIn in the circuit shown-
Answer:R = 4().
200v
(lt)
2.8
120\:24!)
8r)
(b)
RepeatAssessment
Problem2.7 bul usethe
equationofrhe graphedline to constructa circuit model containingan ideal currentsource
and a resistor,
Answer: (a) A 0.25A cunert sourceconnected
betweenthe terminalsof a 100O rcsistor;
(b) 1w.
\OTf:
A L , ' ! r j ( h d p l e r P t u b | , n 5 2 . 1 1 . 2 . 1 ' , 2 . 1 8 . d n J) . 1 u .
2.5 A*nlysisqf e eircuitCcmt*iming
Sependent
5o$rees
We concludethis introductionto clenenlary circuit analysiswith a discus
sion of a circuitthat containsa depeident source.asdepictedin Fig.2.22.
5i,r
We want to use Kirchholls laws and Ohm\ la$'to find o, in this cir
cuit. Before writing equations,il is good practiceto examinethe circuit
diagramclosel_v.
This rvill help us identify the information that is known
aDdthe informationrje musl calculate.It may also help us dcvisca slrat
Figure2.22Z A circuitwitha dependent
source.
egyfor solvingthe circuitusingonly a few calculations.
2.5 Anatysis
of a Cjrcujt
Contajning
Depend€nt
Sources
A look at the ckcuit in Fig.2.22revealsthat
. Oncewe know to,we can calculateoousingOhm's law
. Once we know i^, we also know the cu enl supplied by the dependent
. The curr€ntin the 500V sourceis i^.
Thereare thus two unknowncurents, iA and io.Welleed to constructand
solvetwo independentequationsinvolving thesetwo currerts to produce
From the circuit,noticethe closedpath containingthe voltagesource,
the 5 O resistor, and the 20 O resistor. We can apply Kircbloff's voltage
law aro nd this closed path. The resulting equation containsthe two
500: 5ia + 20t,.
\?.?2)
Nowweneed to generatea secondequationcontainingthesetwo cur,
rents. Consider the closed path formed by the 20 O resistor and the
dependentcurent souice.If we attemptto apply Kirchhoff'svoltagelaw
to thjs loop,we fail to developa usefulequation,becausewe don't know
the value of the voltageaooss the dependentcurrent source.In faclj the
voltageaooss the dependentsourceis oo,whichis the volragewe ate trying to compute.Writing an equationfor this loop does not advanceus
towarda solution.For tlis samereason,wedo not usetie closedpath conlainhg the voltagesource,the 5 O resistor,and the dependentsource.
Thereare three nodesin the circuit.so we turn to Kirchhoff'scurrent
law to generatethe secondequation.Node a connectsthe voltagesource
andthe 5 O resistor;aswe havealreadyobserved.the currentin thesetwo
elementsis the same.Eithernode b or node ccan be usedto constructthe
secofldequatior from Kirchhoff'scurrentlaw We selectnode b and producethe following equation:
lo=iA+5tA=6ta
\2.23)
SolvingEq$2.22 and 2.23for the currents,we get
j^:4A'
in:A
4,.
12.24)
Using Eq.2.24 and Ohm's law for the 20 O resistot,we can solvefor the
voltage tro:
?)o= 20to : 480 V'
Think about a circuit analysisslrategy before beginning to write equation$As we havedemonstrated,
no1everyclosedpath providesan oppor
tunity to write a useful equation based on Kirchhoff's voltage law. Not
every node provides for a useful applicatior of Kirchhoff's current law
Somepreliminary thinking about the problem can help in selectingthe
most fruitful approachand the most usefulanalysistools for a particular
Ciruit Ehments
problem. Choosing a good approach and the appropriate tools will usually
reduce the numbei and complexity of equations to be solved.Example 2.10
illustrates another application of Ohm\ law and Kirchhoff's laws to a circuit with a dependentsource.Example2.11involvesa muchmore complicated circuit, but with a careful choice of analysistools, the analysjsis
relatively uncomplicated.
Vottage
Lawsto Findan Unknown
Apptyingohm'sLawandKirchhoff's
a) Use Kirchhoff's laws and Ohm's law to find the
voltage,r, asshowninFig.2.23.
b) Show that your solution is consistent with the
constraintthat the total power developedjn the
circuit equalsthe total power dissipated.
2A
Applying Ohm's law to the 3 O resistor g.Ives
the desiredvoltage:
?)o: 3to - 3V
b) To computethe power deliveredto the voltage
sources,we use the power equationin the form
p : ol. The powei deliveredto the independent
volragesourcels
: -16.7W.
p : (10x 1.67)
Eramph
2.10.
Figure
2.23A Thecircuittor
The power delivered to the dependent voltage
p = (3r"x-i"): (5)( 1) = -5 w.
Solution
a) Acloselook at the circuitin Fig.2.23revealsthat:
' There are two closedpaths,the one on the
left with the curent ir and the one on the
right with the curent i,.
. Once i, is known, we can compute 2),.
We need two equations for the two currents.
Becausethere are two closedpaths and both have
voltage sources,we can apply Kirchhoft's voltage
law to eachto give the following equations:
10 _ 6i'
3i"=2i.+3i..
Solvingfor the curents yields
i" : 1.67A,
Both sourcesare developing power, and th€
total developedpower is 21-7W
lo compurelhe powerdeli\eredlo LheIe.i<
ton, we use the po er equation in the form
p - lTR.Thepoqer deliveredro rheb Q resisloris
p = (1.61)2n: M.1w.
The power deliveredto the 2 O resistoris
P : (1)'(2:)= 2w '
The power d€liveredto the 3 O iesjstoris
P = (1F(3):3w.
The resiston ail dissipatepower, and the total
power dissipatedis 21.7 W equal to lhe total
power developed in the sources.
2.5
Anatysjs
ofa CncujtContaining
DependentSources
Apptying0hm'sLawandKirchhoff's
Lawin an AmptifierCircuit
The circuitin Fig.2.24representsa commonconfiguration encounteredin the analysisand designof
transistor amplifien. Assume that the values of all
the circuit elements'_R1,R2,Rc. RE,Ucc,and U$a) Developthe equationsneededto determinethe
current in each element of this circuit.
b) From these equations, devise a formula for compufing is in terms of the circuit element values.
(3) iE- iE- ic-0.
A fourth equation iesults ftom imposing the
constraintpresentedby the sedesconnectionof
Rc and the dependentsource:
(4) ic = BiB
We tum to Kirchhoff's voltage law in de ving the remaining two equalions We need to
selecttwo closedpaths in order to useKirchhoffs
voltage law. Note that the voltage across the
dependentcurrent souice is unknown, and that it
cannot be detemined from the source current
Bis. Thereforc, we must select two closed patbs
that do not contain tlis dependentcurrent source.
We choose the patls bcdb and badb and
specifyvoltagedrops aspositiveto yield
(s) Uo+
(6)
Figure2.24A Thecircuit
forExampte
2.11.
Solution
A careful examination of the cfcuit reveals a total
of six unknown currents,designatedir, rj, is, ic, ir,
and jcc. In definingthesesix unlnowr curents, we
usedrheobservalion
thal rheresisror
R( i\ in series
with the dependentcurent source Bir. We now
must derive six independentequationsinvolving
thesesix unknowrs.
a) We can derive three equations by applying
Kirchhoff's curent law to any thiree of the nodes
a, b,c, and d. Let's usenodesa, b, arld c and label
thecurrenlsaqay from lhe 0odesasposili!e:
(1) i1 + tc - icc: 0,
(2) iB+ i2
it=0,
iERE
i2R2: 0,
- 4& + Ucc
i2R2: 0.
b) To ger a single equation for i, in terms oI t}le
known circuit variables,you can follow thesesteps:
. Solve Eq. (6) for t1,and substitutethis solurion for jl into Eq. (2).
. Solvethe transfomed Eq. (2) for tr, and sub
stitute tiis solution for i, into Eq. (5).
' Solvethe transformedEq. (5) for ir, and sub
stitute this solution for iE into Eq. (3). Use
Eq. (4) to eliminateic in Eq. (3).
. Solve the transformed Eq. (3) for is, and
rearange the termsto yield
(uccR)/(&+ R2\-uo
(nlRt/(Rr + n,) + (1 + p)RE
(2.25)
Problem2.27asksyou to verify thesesteps.Note
that once we know ir, we can easilyobtain the
remanlng currents.
CncuitElements
pow6rfor eachetement
objective3-Know howto calcutate
in a simplecircuit
2.9
For the circuit shownfind (a) the currentir in
(b) thc voltageo in volls,(c) the
microamperes,
iotal power gcncratcd,and (d) the total power
absorbed.
c) ihe power deliveredby the independentcurrenl source!
d) the power deliveredby the contlolledcurrent source,
€) the lolal power dissipaledin the lrvo rcsislors.
Answer: (a) 25 /..A;
(b) 2 v:
(c) 61s0/,w;
(d)61s0/..w.
Answer: (a) 70 V;
(b)210w
(c) 300w;
(d)40w;
(e)130w
2,10 The currentt{ in the circuitsho*n is 2A.
Calculate
b) Lhepower absoibedby the independent
vollagesource,
NOTE: Also tt| ChapterPrcblems2.21und 2.28.
PracticalPerspective
ttectrical
Safety
At thebeginning
of thischaptetwesaidthat current
throughthebodycan
caus€
injury.Lefsexamine
thisaspect
of etectricaI
safety.
jnjuryis dueto burrs.However,
Youmightthjnkthat etectrjcaI
thatis
jnjury
js
notthe case,Themostcommon
etectricat
to the n€rvous
system.
Nerves
useetectrochernical
signats,
andetectrjc
currents
candisruptthose
signats.
Whenthe currentpathjnctudes
onLysketetaI
musctes,
the effects
paratysis
(cessation
caninctude
temporary
of neruous
signats)
or involuntarymuscle
contractions,
whicharegeneralty
notfifethreatening.
Howevet
jncLudes
whenthe currentpath
nerves
andmuscLes
that controL
th€ supply
js muchmorese ous.Temporary
paral
of oxygen
to thebrain,the problem
ysisof thesemusctes
muscanstopa pe6onfrombreathing,
anda sudden
cLecontraction
candisruDt
thesiona[s
that reoutate
heartb€at.
Theresuttis
47
deathin a few
btoodto the brain,causing
a hattin theflowof oxyqenated
js
gjven
a range
immediatety.
Tabte
2.1shows
emergency
aid
minutes
untess
in this
reactlons
to variouscurlentlevets.Thenumbers
of physiotogicat
froman anatysis
of accidents
they are obtajned
tableare approxinate;
experiments
on
it is not ethicatto perform
eLectricaL
obviousty,
because,
orless
people.
witllimitcurrenttoa fewmilLiamperes
Good
etectrjcal
design
conditions.
underal[ possible
PhylioLoglcal
Reaction
p€rcephbLe
BarcLy
Heat stoppage
3 5 5 0m A
5 0 - 7 0m A
500mA
rvot* Datatakenfrcm w F.coop€t,Ehcttico!SaJetyEngtnging, 2d ed. (London:ButteNodh,
N.Y.:l,4arcel
Dekket1988).
P@cttditledr?dlsdrtry(Monti@tlo,
1936);andc. D,Winburn,
eLectrjcal
modelof the humanbody.The
Nowwe devetop
a sinpLified
so
a
reasonable
starhngpointis to
of current,
bodyactsas a conductor
potentiatly
dangerous
Figurc2.25showsa
modelthe body0singresistors.
onearmandoneleg of a
A voLtage
difference
existsbetween
situahon.
model
ofthehunanbodyjn
2.25(b)
shows
aneLectricaL
human
being.Figure
eachhave
Thearms,tegs,neck,andtrunk(chestandabdomen)
Fig.2.25(a).
path
is through
the
Notethatthe
of thecurrent
a characterjstic
resistance.
deadLy
arnngement.
trunk,whichcontains
the heart,a potentjatty
bodywjtha vottase
Figure2.25 A (a)A human
oneam andoneteg.(b)A simdjfference
between
bysoling Chapter ptifiedmodetofihe humanbodywitha vottasedifyourunderstonding
of thePradicalPerspective
NO|E: Assess
oneatmandoneteg.
ference
bebveen
2.34-2.38.
Prcblens
Summary
The circuit elementsintroduced in this chapter are voltagesources,
curent sources,and rcsisto$:
. An ideal voltage sourc€ maintains a prescribed voltage regardlessof the current in the device.An ideal
q|rletrt sourcemaintains a prescdbed cuffent regardless of the voltage acrossthe device.Voltage and current sources are eitler indepetrd€nt, that is, Dot
influenced by any other curent or voltage in the ckcuit;or d€petrdent,that js! determined by some other
current or voltage ir the circuit. (Seepages24 and 25.)
. A resistor constrains its voltage and current to be
proportional to each other. The value of the proportional constant relating voltage and current in a resistor is called its resislance and is measured in ohms
(Seepage28.)
Ohm's law establishes the proportionality of voltage
and current in a resistor. Specifically,
\2.?6)
4S
Circuit
Etements
if the current flow in the resistor is in the directiol of
the voltagedrcp acrossit, or
n=-iR
12.27)
if the curent flow in the resistoris in the direction of
the voltagedse aooss it. (Seepage29.)
By combidng the equation for power, p : ot, with
Ohm's law,we can determinetle power absorbedby a
p=izR-uzlR.
(2.28)
through connecting elementq starting and ending at the
samenode and encounteringintermediatenodeso y
onceeach.(Seepages36 38.)
fhe voltagesand cullenfs of interconnectedcircuit elementsobey Kirchhoffs laws:
. I(irchhoffs current law states that the algebraic sum
of all the currentsat anynode in a circuitequalszero.
(Seepage36.)
. KirchhofPs voltage law statesthat the algebraic sum
of all the voltages aiound any closed path in a circuit
equalszero.(Seepage37.)
(Seepage30.)
. Circuits are described by nodes and closed paths. A
node is a point wheretwo or more circuit elementsjoin.
When just two elements connect to form a node, t]!ey
are said to be in seri€s.A clos€dpath is a loop traced
A circuit is solvedwhen the voltageacrossand the cur
rent in every elementhave been detemined. By combining an undentandingof independentand dependent
sources,Ohm's laq and Kirchhoff's laws, we can solve
many simple circuits.
Problems
Section2.1
figlre P2.2
2.1 a) Is the interconnection of ideal sourcesin the circuil in Fig.P2.1valid?Explain.
b) Idenrify which sourcesaie developing
and which sourcesare absorbingpower.
c) Verify that the total power developedin thecircuit equalsthe total power absorbed.
d) Repeat (a)-(c), reversing the polarity of the
10V source.
figureP2.1
10v
2.3 If the interconnectionir Fig. P2.3is valid, find rhe
total power developedbythe voltagesources.Ifthe
interconnectionis not valid,explainwhy.
FigureP2,3
30v
10v
2.2 Tf rhe inlerconneclion
in Fig.P2.2is valid.find rhe
power developed by the current sources.If the
interconnectionis not valid,explainwhy.
Probtems49
2.4 If the interconnection il Fig. P2.4 is valid, find ttre
total power developed in the circuit. If the inteiconnection is not valid, explain why.
a) Is the interconnection in Fig. P2.7valid? Er?lain.
b) Can you find the total energy developed in the
circuit? Explain.
Flgurc
P2.4
Figurc
P2.7
10v
20v
5A
The inteiconnection of ideal sourcescan lead 10 an
indeterminate solution. With this thought in mind,
explail why the solutions foi ot and t'2 in the circuif
in Fig. P2.5 are not unique.
FlgueP2.5
100v
l)
, o A (I
i)
troov(
cnv{
t
3ir
If the interconnection in Fig- Y2.8 is valid, find the
total powerdevelopedin the circuit.Il the irterconnection is not valid, explain why.
FlgureP2.8
50v
t0A
MA
10A
25 If the interconnection in Fig. P2.6 is valid, find the
total power developed ir the circuit. ff the interconnectionis not valid,explainwhy.
Figure
P2.6
6V
2.9 Find the total pov/er developedir the circuit m
Fig.P2.9it a. = 100V andis : 12 A.
tigureP2.9
50
Etenenb
Circuit
FlgurcP2.12
Seclions2.2-2.3
r (A) p 0v)
2,10 The teminal voltage and terminal current were
measuredon the deviceshownin Fig.P2.10(a).The
values of ? and i are given in the table of
Fig. P2.10(b).Use tlle valuesin the table to construct a circuit model for the device consisting of a
singlercsistor.
z
8
10
12
Figure
P?.10
t (mA) ?,(v)
-20
-160
-10
80
10
80
20
160
u0
30
<!:
- ?
--E
100
400
900
1600
2500
3600
(b)
(b)
2.13 A pair of automotive headlamps is connected to a
2.tl A variety of voltage source values were applied to
l2 V barrer) via tbe arfangementsshown in
Fig. P2.13.In the figure, the tdangular symbol V is
used to hdicate that the terminal is coturected
directly to the metal ftame of the car-
the device shown in Fig. P2.11(a).The power
absorbed by the device for each value of voltage is
recordedin the table givenill Fig.P2.11(b).Usethe
values in the table to construct a circuit model for
ttre device consisting of a single resistor.
a) Construct a circuit model using resisto$ and an
independent voltage source.
Flgure
P2.11
b) Identify the corespondencebetweenthe ideal
circuit element and the s]'mbol component that
it rcpresents.
--E
! (v)
10
---__l
/A
(
lDevie I
l- f - rl l
Y
G)
),,
' (mwl
25.0
6.25
6.Zs
15
t0
25.0
56.25
20
100
(b)
2.12 A variety of curent source values were applied to
tlle device shown in Fig. P2.12(a). The power
absorbed by tle device for each value of curent is
recorded in the table given in Fig. P2.12(b).Use t})e
values in the table to construct a circuit model for
the device consisting of a single resistor.
tigurcP2.r3
51
2.14 Tho voltage atrd cunent werc measured at the terminals of ttre device shown in Fig. P2.14(a). The
resultsare tabulatedin Fig.P2.14(b).
a) Construct a circuit model for this device using
an ideal current source and a resistor,
b) Use ttre model to predict the amount of power
the device will deliver to a 5 O resistor.
Figwe P2.74
i, (A)
100
180
260
340
420
0
8
12
16
mhals of the device shown in Fig. P2.15(a). The
resultsare tabulatedin Flg.P2.15(b).
a) Construcl a circuil model for lhis device usiDg
an ideal voltage source and a resistor,
b) Use the model to predict the value of ii when or
Figure
P2.15
?,',(V)
r ----:
:L
l
t
;
G)
40
35
30
25
18
8
0
10
20
30
40
50
55
(b)
L15 The voltage and curent were measured at the ter-
l r
figureP2.16
(b)
(")
fr--l
-
r)
Use your circuit model to predict the open-circuit
voltagg of the current souice.
Wlat is the actual open-circuit voltage?
E)plain why the answersto (d) and (e) are not
,, (A)
0
50
58
66
'74
z
6
8
82
90
IO
(b)
(b)
LM The table in Fie. P2.16(a) gives the rclationship
betweenthe teminal current afid voltage ot the Factical constantcunent sourceshoM in Fig.P2.16(b).
a) Plot i, versus ,'".
b) Construct a circuit model of this current source
that is valid for 0 < z'" < 30 V, based on the
equationof tle line plotted in (a).
c) Use your circuit model to predict the currcnt
delivered to a 3 kO resistor.
2.17 The table in Fig. P2.17(a) gives the ielationship
betweenthe terminal voltage and current of the practical coDstantvoltage sourcesho$n in Fig.P2.17(b).
a) Plot ,J versus ir.
b) Construct a circuit model of ttre piactical source
that is valid for 0 < i" < 225 mA, based on the
equation of the line plotted in (a). (Use an ideal
voltage source in serieswitl an ideal rcsistor.)
c) Use your circuit model to predict the curent
delivered to a 400 O resistor connected to the
teminals of tle practical souce.
d) Use your circuit model to predict the orrrent
delivered to a short circuit connected to the terminals of t}Ie practical source.
e) What is the actoal short-circuit current?
f) Explain why the a$wers to (d) and (e) are not
Figurc
P2.r7
'75
0
60
45
30
20
10
75
150
225
300
400
500
(a)
(b)
52
Circuit
ELements
Section 2.4
FigureP2.21
2,18 Given the circuit shownin Fig.P2.18,tind
"""
a) the valueofi.,
b) the value of 4,
c) t]le value of o,,
d) the power dissipated in each resistor,
e) the power deliveiedby rhe 50 V source.
250
10()
5()
180V
Flgure
P2.18
2.22 For rhe circuit shown in Fig. P2.22,find (a) R and
6trc (b) the power supplied by tle 125V source.
50v
r,ilzo
o
800
2,19 a) Find the currents il and t, in t}le circuir in
Fig.P2.19.
b) Fitrd tlle voltage !'o.
c) Verify that t}le total power developed equalsrhe
total power dissipated.
ligurcP2.22
R
125V
30f,r
2.23 The variable rcsistor R in rhe circuir in Fig. p2.23 is
EEc adjusted until z'dequals 60 V Find the value of R.
Flgure
P2.19
300
Fig(reP2.23
450
80()
90f,)
180r)
18()
2.20 The current ia in the cftcuir shown h Fig. p2.20 is
Errd 2 mA. Find (a) io (b) is; and (c) rhe power deljvered
by the independent current source.
FlgueP2,20
1k()
'.Jizro
4k()
3ko
221 The cunenl i, in the circuit in Fig. p2.21 is 4 A.
"""
a) Find i1.
b) Find the power dissipatedin eachrcsisror.
c) Vedfy thar the rotal power dissipated in rhe circuit equalsthe power developedby the 180V
2.24 The voltage across ttre 15 kO resistor in the circuit
6nd in Fig. P2.24is 500V positive at rhe upper terminal.
a) Find the power dissipated in each resistor.
b) Find the power suppliedby the 100 rLA ideal
cufient source,
c) Verify tiat the power supplied equals t}le total
power dissipated.
li$rc ?2-24
2r5 The cuffents ia and ib in the circuil in Fig. P2.25are
6fl4 4 A and 2 A, respectively.
a) Fitrd is.
b) Find the power dissipated in each resistor.
c) Find 0s.
d) Show that the power delivered by the current
source is equal to the power absorbed by all the
other elements,
Figu€P2.25
2.28 a) Find tie voltage o, in the circuit in Fig. P2.28.
"""
b) Show that t]le total power genemted in t}le circuit equalsthe total power absorbed.
tigureP2,28
15.2
V
72o'
4(}
24o"
2r9 Find (a) t,, (b) ir, and (c) i, in the circuitin Eg. P2.29.
zzn The curents
i1 and i2 in the circuit in Fig. P2.26are
10 A and 25 A, respectively.
a) Find the power supplied by each voltage source.
b) Show that the total power supplied equals the
total power dissipated in the resistors
Flgurc
P2.29
1ko
5000
Figure
P2.26
100r}
2.30 Find rJ1and os in the circuit shown in Fig. P2.30
6xa when oo equals250 mV (Hirtr start at tle dght end
of the circuit and work back toward os.)
S€dion 2.5
227 Derfte Eq.2.25. Hint: Use Eqs (3) and (4) fiorD
Example 2.11 to expressiE as a function of is. Solve
Eq. (2) for i, and substitute the result into both
Eqs. (5) and (6). Solve the "new" Eq. (6) for tr and
subshtute this rcsult into tle "new" Eq. (5). Replace
tr in the "new" Eq. (5) and solvefor is. Note that
becausercc appearsonly in Eq. (1), the solution for
is involves the manipulation of only five equatiofls.
FiguleP2.30
10()
12.5I)
54
cncuitEtements
2,31 Foi thecircuitshownin Fig.P2.31,calculate(a) ia and
Ede o" andO) showdratthe powerdevelopedequalsthe
powerabsorbed,
rigure
P2.31
in excessof two, Show how one four-way switch
plus two three-way switches can be connected
between a and b in Iig. P2.33(c) ro control the
lamp from three locations. (Hirr The four-way
switch is placed between the three-way switches.)
Flgure
P2.33
tat]l
t| \\ l l
232 For the circult shoxn n Fig.2.24,Rr = 20kO,
Emr R, : 80ko, Rc : 5000, RE = 100(), vcc = 15v,
y0 : 200mY and B : 39.Calculateis, ic, ;8, 2,3d,
,bd,,2,4,oab,
icc,ando13.(NoterIn the doublesubscript notation on voltagevariables,the first subscript is positive with respect to tlle second
subscript.
SeeFig.P2.32.)
figureP2.32
3
Il +
R,i ,,,,
tt z. r3 l1
Position 2
11---21
f1
|1 l l 1
\ V / lI
n l
l
?
-t]
9
1 3 4 l
Position2
Position 1
(b)
1
d
Sectiono2.1-2.5
2.33 It is often desirable h designing an electric wiring
pffiTil| systemto be able to contrcl a single appliance from
two or more locationE for example, to control a
Iighting fixtue from both the top and bottom of a
stairwell. In home wfting systems,this type of control is implemented \irith three-way and four-way
switches.A tbree-way switch is a three-terminal,
two-positior switch, and a four-way svdlch is a fourteminal, two-position switch.The switchesare shown
schematica[y in Fig. P2.33(a), which illustrates a
three-way switch, and P2.33(b),which illustmtes
a four-way switch.
a) Show how two three-way switches can be connected between a ard b in the circuit ill
Fig. P2.33(c) so that the lamp I can be tumed oN
or orT lrom two locations
b) If the lamp (appliance) is to be controlled hom
more than two locationE four-way switches are
used in conjunction with two thrce'way switches.
one four-way switch is required for eachlocation
tc/
234 Suppose t]le power company installs some equiptigl;ii! ment that could provide a 250 V shock to a human
being.Is the cuffent that results dalrgerous enough
to warmnt posting a warning sign ard taking otlei
piecautions to prevent such a shock? Assume that
if the souce is 250 Y the resistance of the arm is
400 O, the resistance of the trunk is 50 O, and the
rcsistance of the leg is 200 O. Use ttre model given
inFig.2.25(b).
2.35 Basedon the model and circuit sho$n in Fig.2.25,
pll|;l#; draw a circuit model ol the path of current tlrougl
tlle humar body for a person touching a voltage
source with both hands who has both feet at the
samepotential as the negative teminal of the volt-
55
2.36 a) Using the values of resistance for arm, leg, and
""'"'''"' lrunl Drovidedin Pfoblem).14,calculatelhe
po.", di.ripuredin the arm.leg.andlr unJ.
b) The specificheat of water is 4.18 x 10rJ/kg'C,
so a massof water M (in kilograms)heatedby a
power P (in watts)undergoesa rise in temperalure ar a mre glven oy
dT
i=
2.39 . 10-aP
'^^' 1 "
M
Assuming that the mass of an arm is 4 kg, the
massof a leg is 10 kg, and t]le massof a trunl is
25 kg, and that the humanbody is mostlywater,
how many secondsdoesit take the alm,leg, and
living tissue?
trunk to dsethe 5'C that endangers
c) How do the valuesyou computedin (b) compare vrith the few minutes it takes for oxygen
sta ation to injure the brain?
2.37 A person accidently $abs conducton connected to
each hand
?lltfff*Jreach end of a dc voltage source,one in
a) UsiDg the resistancevalues for the human body
provided in Problem 2.34, what is the mhimum
source voltage that can produce electdcal shock
sufficient to causepamlysis, preventing the person ftom letting go of the conducton?
b) Is there a significant dsk of this type ol accident
occurring while servicing a pe$otal computer,
which typically has 5 V and 12 V souces l
2.38 To unde$tand why the voltage level is not the sole
nI oI polentialiniur] due to elecrrical
,lli,l*l,
'' _' delelmina
shock
sbock,con.iderlhe caseol a .laticeleclricily
mentioned in the Practjcal Perspective at the start
of this chapter.W}IeIl you shuffle youi feet acrossa
carpet, your body becomes charged The effect of
this charye is that your entire body iepresents a
vollage potential.w]len you touch a metal doorknob, a voltage difference is created between you
and the doorknob, and curent flows-but the conductionmaterialis air, not your body!
Supposethe model of the spacebetweenyour
hand and the doorknob is a 1 MO resistancewhat
voltage potential exists between your hand and
the doorknob ifthe current causingthe mild shock
is 3 mA?
Circuits
SimpleResistive
3.1 Resistorsin Seri€sp. 58
3.2 Resistorsin Parattetp. 59
and Current-Divider
3.3 TheVoltage-Divider
Circlits p. 62
3.4 Vottag€Divrsionand CurrentDivisionp. 65
3.5 li'leasuring
Voltageand Cunent p. 68
3.6 t'leasuringResistance-TheWheatstone
Bndge p. 71
(Pi-to-Tee)Equivatent
3.7 DeLta-to-Wye
Circuitsp. 7.t
in
connected
I 8e ableto rccogfjzeresistors
and useth€ rutesfor
s€riesandin paratLet
serjesconiected
resjstolsand
combining
paraiteL-connected
resistors
to yietdequivatent
and
2 Knowhowto designsjmptevottage-divjd€r
cunent-divid€r
circuits.
3 Beabt€to usevoltagedivjsionandcur€it
to sotvesjmpi€cjrcuits.
djvisionapprcpriatety
the readingof an ammeter
4 Beabteto determine
curfenLbe
whenaddedto a circuitto measure
abteto deteminethe readjngof a voLtmeter
vottage.
whenadd€dto a cjrcujtto measure
bfidgeis usedto
howa Wheatstone
5 Understand
6 Knowwhenandhowio used€Lta-to-wye
equival€itcitcuitsto sotvesimpb cncuits.
0ur analytica[ toolbox now containsOhm's law and Kirchhofls
laws.In Chapter2 we usedthesetools in solvingsimple circuits
In this chapterwe continue applying these tools,but on more_
complexcircuits.The greatercomplexitylies in a greaternumbelThis chapoI elementswith more complicatedinterconnections.
ter focuseson reducingsuchcircuitsinto simpler,equivalentcircuits.We continLreto focus on telatively simple circuits for two
reasons:(1) It givesus a chanccto acquaintourselvesthoroughly
with the laws underlyirg more sophisticatedmethods,and (2) it
allowsus to be introducedto some circuits that have important
application..
engineering
The sourcesin the circuits discussedin this chapterare limited to voltageand curent sourcesthat generateeither constant
voltagesor curents; that is,voltagesand cunents that are invatant with time. Constantsourcesale often called dc sources.The
dc standsfordirectcurleat,a dcscriptionthat hasa historicalbasis
but can seemmislcadingnow. Historically,a direct current was
definedas a currentproducedby a constantvoltage.Therefore,a
coNtant voltagebecameknown asa direct cunent, or dc,voltage
The use of dc for conrldtf stuck,aDdthe terms dc cuftent arr'ddc
volldg?arc now universallyacceptedin sciencealld engineering
to meanconstantcurrentand constantvoltage.
PracticaI
Perspective
A RearWindow
Defroster
js an examTherearwindowdefrostergrid on ar automobite
pteof a resjstivecircuitthat pedormsa usefulfunction.one
suchgridstructureis shownon the teft of the figurehere.The
gid conductors
can be modeledwjth resistors,as shownon
the ight of the figure.Thenumberof hoizontaIconductors
varieswith the makeandmodeL
of the carbut typicattyranges
from9 to 16.
Howdoesthis gridworkto defrostthe rearwindow?How
are the propertiesof the g d determined?
We witt answer
thesequestions
in the Practical
Perspective
at the endofthis
chapter.The circuitanatysisrequiredto answ€rthesequestiors adsesflom the goaLof havinguniformdefrostingin
bothth€ horizontalandverticaL
directions.
57
58
Simpte
Resistjve
Circuih
3.1 Resistors
in Series
'ii
In Chapter 2, we said that when just two elementsconnectat a single
node.lheyare said to be in series.Seri€\-connected
circuit eternentscanr
lhe samecuffenr.Theresi.rorsin rhe circuilshowniD frg.3.t are connecledin series.
\ e canshowlhal lhe.e fesistorc
cafry lh; .amecurrerr
b\ appllingKirchhoffscurrenllaw to cacbnodein rh; circuir.The
scrics
interconnectionin Fig.3.1requiresthat
i
\t'i
) "
R,
RO
Rr
n ; c - i r - i "
jn sedes.
figurc3.1 A Resjstors
connected
R,
RO
h
g
is=4=
RS
f
e
i2=ir=ia:
is= -i6= i. ,
(3.1)
which statesthat if we know any one ofthe sevencurrents,we know them
all.Thus we can redrawFig.3.1as shownin Fig.3.2,retainhg rhe idenrity
of the singlecurent i".
To find i., we apply Kirchhoff's voltage law around the single closed
toop.Delinirg thevoltageacrosseachresistorasa drop jn the directionof
rr grves
Figure3.2 A Series
resistors
witha sjngteunkno\i/n
-ol + i"Rr + irn2 + irR3 + j.Rj + lrR5 + trn6 +
irRT:0,
(3.2)
z ' .= t " ( R+1R 2 + & + & + R s + R 6 + R ? ) .
(3.3)
The significanceof Eq.3.3 for calculatingir is that the sevenresistorscan
be replacedby a singleresistorwhosenumeiical value is the sum of the
individualresistors.
that is.
R e q= R 1+ R 2 + R 3 + R 4 + R 5 + R 6 + R 7
\3.4)
and
(3.5)
h
Figure3.3 A A simplified
version
of the cjrcujtshown
in Fig.3.2.
Combining
resistocin series>
Thus we can redrawFig.3.2assho$n in Fig.3.3.
In general,if tr resistorsa.reconnectedin series,the equivalenlsircle
re.islorhasa resrstance
equallo lhe sumot lhe & re,i.rancir.or
(3.6)
Note that the resistanc€of the equivalentresistoris alwayslarset than
!harot lhe largcslresi.torin rheseriesconneclion.
3.?
in PaEttet
Resisto6
59
Another way to think aboutthis conceptof an equivalentresistanccis
ro visualizethe string of resistorsas beinginsidea black box. (An elcctri
+
cal engineerusesthe term blackbor to imply an opaquecontainer;that is,
the conterts are hidden from view The engineeris theD challengedto
modei ihe contentsof the box by studyingthe relationshipbetweenthe
h R7 R6 .It5
voltageand current at ih termillals.)Determiningwhether the box concircuit
tains k resistorsor a single equivalentresistoris impossible.Figure 3.4 Figure3.4 A Theblackboxequjvahniofthe
in fi9. 3.2.
shown
illustratesthis methodofsttdying the circuit shownin Fig.3.2.
in Faratlel
3.2 Resistors
When two elemerts connectat a singlenode pair, they are said to be in
parallel.Parall€l-cotrnecled
circuit elementshavethe samevoltageacross
their terminals.Thecircuitshom in Fig.3.5illust.aleslesistorsconnected
in parallel.Don't make the mistake of assumingthal two elementsare
parallelconnectedmerelybecausethey are lined up in parallelin a circuit
elementsis that
diagran.The definingcharactedsticof parallel-connected
jn parattel.
they have the samevoltageacrosstheir terminals.In Fig.3.6,you can see Figure3.5 A Resistors
belwecntheir respecthat Rr and R3 are not parallelconnectedbecause,
R2
rive termhals.anotherresistordissipatessomeofthe vollag.3.
Resislorsin parallel can be reducedto a single equivalentresistor
usingKirchhoff'scurrent law and Ohm's law. as we now demonstmte.In
the circuit shownin Fig.3.5,we lel the curenis ir, tr. ii. and ia b.3the curWe also let the positive
rentsin the fesistorsRr througl Ra,respectively.
referencedirectionfor eachresislorcurrentbe down firough th.3resistor, Figrre3.6 { Nonparalhtresistors.
that is,from node a to node b.From Kirchhoff'scunent la\
i r : t 1+ t r + i r + i 4 .
(3.7)
The parallelconnectionot the resistorsmeansthat the voltageacrosseach
resistormust be the same.Hence,fromOhm'slaw,
rrR,: irRz= 4Rr = raRr- r,
(3.8)
Therelore,
Rr'
:t
l,
:I
R1
and
(3.e)
60
simpte
Resktjve
Circuits
Substitutirg
Eq.3.9into Eq.3.7yields
,. " : " \[^r, '
1
1
1 \
&, k. R,l
(3.10)
from which
i
r
1
1
1
1
1
,,=R*:4-&-&-&
(3.11)
Equation3.11is whatwe setout to show:that the four resistorsin the circuit shownin Fig.3.5canbe replacedby a singleequivalentresistor.The
r€sisto$shown circuit shownin Fig. 3.? illustratesthe substitution.For k resistorsconFlgorc3.7 A RepLacing
th€ foff palaLtet
nectedir parallel,Eq.3.11becomes
in Fiq.3.5witha singleequivat€nt
resistor.
combining
resistors
in patallel>
13.\2)
Note that the resistanceof the equivalent resistor is alwayssmaller than the
rcsistance of the smallest rcsistor in the parallel comection. Sometimes,
usirg conductancewhen dealing witl resistorsconnectedin parallel is more
cotrvenient,In that case.Eo. 3.12b€comes
'-Fl'
jn palaLteL.
Figure3.8,| Tworcsistols
connecied
+G2+
G " q => G i : G l
+Gk
(3.13)
Many times only two resistors are connected in pamllel Figure 3.8
illustrates this special case We calculate tle equivalent resistance ftom
Eq.3.121
1
1
1
R*=&-&:
n2+ R1
'
Rt&
R.R"
(3.14)
(3.15)
Thus for just two resistorc h parallel the equivalent resistarce equals
the product of the resistaDcesdivided by the sum of the resistance$
Remember that you can only use this result in the special caseofjust two
rcsistors in Darallel. Examole 3.1 illustrates the usefulnessof these results.
3.2
Relisto6inPanttet
61
ApplyingSeries-Parall.et
simpLifi
cation
Find i,. rr . aod i2 in lhe ci-rcuilshownin Fig.3.s.
40
t20v
+
ttl:18o ',{ 6 0
Solution
We begin by noting that the 3 f,) resistor is in senes
wirh rbe6 Q resistor.weLherelorereplacelhis series
combination with a 9 O resistor, reducing the circuit
to the one shown in Fig. 3.10(a).We now can replace
the parallel combhation of the 9 O and 18 O resistors with a singleresistanceof (18 x 9)/(18 + 9), or
6 O. Egure 3.i0(b) shows t]Iis turtler reduction of
the circuit.The nodesx and y marked on all diagmms
facilitate tracing through t]le reduction of the circuit.
From Fig. 3.10(b) you can veiify that iJ equals
120/10,or 12 A. Figure3.11showsthe iesult at this
point in the analysis.We added the voltage ?)1to
Using Ohm's
helpcladfy the subsequentdiscussion.
law we compute the value of ,1:
'72v.
1\ = (12)(6)-
1)1
:72
9
v
Figur€3.9 a Th€crrcujtfor E,dmph3.1.
4f)
r,l$rso
r,,f9 f )
l20v
y
G)
40
120V
x
+\ '"
60
(3.16)
But t 1is the voltage drop fmm node x to node y, so
we can return to the circuit shown in Fig. 3.10(a)
andagainuseOhm'slaw to calculatetr and ir. Thus,
- 72 : 4 A '
18 1 8
3()
v
(b)
jn Fjg.3.9.
Figue3,10a A sinpLification
ofthecjrcuitshown
(3.17)
4f)
+\ 12A
9
:8A.
120V
(3.18)
We have found t}le thrce specified currents by using
series-parallel reductions in combination with
Ohm'slaw'
6()
v
figuru3.r1 A lhecircuit
of Fig.3.10(b)
showins
thenumedcaL
Belore leaving Example 3.1, we suggest that you take the tlme to
show that the solution satisfies Kirchloffs curent law at every node and
Kirchloff's voltage law around every closed path. (Note that there are
thee closed paths that can be tested.) Showing t]Iat the power delivered
by the voltage souce equalsthe total power dissipatedin the rcsistors also
is informative. fsee Prcblems 3.3 and 3.4.)
62
SimphRsistiveCircuits
objective1-Be abteto recognize
resistors
conneded
in seriesandin parattel
3.1
Fortbecircuitshown,find (a) thevollageo,
(b) thepowerdeliveredto thecircuilby the
cufent source,
and(c) thepowerdissipated
in
the 10() resistor.
7.211
101)
Answer: (a) 60Vr
(b) 300w;
(c) s7.6w
NOTE: Alsotry Chapter
Problems
3.1,3.2,3.5,
and3.6.
3.3 The\foltage-Sivider
andeurrent-Sivider
eireuits
(")
(b)
Figure3.12a (a)A votiase-divider
circuitand(b)the
vottagsdivids
circuitwithcrrent ; indicaied.
Al timss especiallyin eleclronic circuils developingmore than one
voltagelevel from a singlevollagesupplyis necessary.
One way of doing
this is by usinga voltsge-divid€rcircuil,suchasthe one in Fig.3.l2.
We analyze this circuit by directly applying Ohm's la$' and
Kirchhoff'slaws.Toaid the analysis,
we introducethe curent i asshownin
Fig.3.12(b).From Kilchhoff's current la\ Rr and R, carry the samecurrel1l.Applyilg Kilchhoff's voltagelaw aroundthe closedloop yields
r.-lRr+iRz,
Rr+R,
(3.20)
Nowwc canuseOhm'slawto calculale
ul and,2:
R
r
01 =rr(1 : ?rnr
+ R2,
(3.21)
R,
.0 2= u < 2 : x s R t
+ R,
13.22)
Equations3.21and 3.22show that ,r and o, are fractionsof o,- Each
ftaction is the ratio of the resistanceacrosswhich rhe divided voltageis
definedto the sum ofthe two resistances.
Becausethis ratio is alwayslcss
than 1.0.the divided voltagesor and ?J2are alwaysiess than the source
If you desirea particularvaluc of ?)2,and o, is specified,an infinile
numberofcombinarionsof Rr and R, yicld thc propcr ratio.For example,
supposethat !. equals15 V and 1J2
is to bc 5 V Thon r2/o, = i and,liom
3.3
TheVottage'Divider
andCurrcntDjvider
Circuiis
Eq.3.22,wefind that this mtio is satisfiedwhereverR, : ;R1.Other factors that may enter into the selection of R1, and hence R2, include the
powei lossesthat occur in dividing the source voltage and the effects of
connecting the voltage-divider circuit to other circuit components.
Consider connecting a resistor Rz fu parallel witl R2, as shown in
Fig. 3.13.The resistorR, acts as a load on the voltage-djvidercircuit.A
load on any circuit consistsof one or more circuit elementsthat draw
power from the circuit. With the load Rr connected,the expressionfor the
output voltagebecomes
-'
'"
Rr + R.q
\3.23)
Figure3.13A A voLtage
djvjder
connected
to a toadR..
RtRr
R2+ RL
(3.21)
Substituting
Eq.3.24into Eq.3.23yields
R2
Rlt1 + (,R/R')l + R"'
Note that Eq. 3.25 reduces to F'q.3./2 as Rz+c!,
as it should.
Equation 3.25showsthat, as long as R, >> R2,the voltageratio r,/?i" is
essentially undisturb€d by t})e addition of the load on the divider.
Atrother chaructedstic of the voltage-divider circuit of interest is the
sensitivity of the divider to tie tolerances of the resistors.By roler.ar?ce
we
mean a range of possible values.The resistancesof commercially available
resistorsalways vary within some percentageof theh stated value.
Example3.2illustratesthe effectof resistortolerancesin a voltage-divider
circuit.
Anatyzing
the Voltage-Divider
Circuit
The resistorsused in the voltage-dividercircuit
shownin Fig.3.14have a toleranceof t10%. Find
the maximumand minimum valueofo,.
25kO
t00v
Sotution
From Eq. 3.22,the maximum value of r)ooccurswhen
R is l0oohighandRr is l0'o lo\\.and(beminimum
value of t', occurswhen R2 is 10% low and Rr is
10% high.Thereforc
,lma\) =
,,(Illrn) =
100kI)
Figur€3,14 A Thecircujtfor Exampi€
3.2.
(100v110)
= 83.02
v.
1101-(100x90)
e0 +--
: 76.60v.
Thus,in makingthe decisionto use10% resistorsin
this voltage divider, we recognize that the noload
orltput voltage will lie between 76.60and 83.02V
64
Simpte
R€sistive
Circuits
TheCurrent-Divider
Circuit
t
o , * 1,, '*
rigure3.15A Ihedrnent-divider
circuit.
The curent-divider cfucuit showr in Fig- 3.15consistsof two tesistors connectedin paralleladoss a current source.The current divider is designed
to divide the current i. between R1 and R2. We find t]le relationship
between ttre curent i, a.ndthe current in each resistor (that is,i1 and ir) by
directly applying Ohm's law and Kirchhoffs current law. The voltage
aoossthe parallelresistomis
(3.26)
Ffom Eq.3.26,
,"'. =
R,
,
R, + R7-"'
13.27)
,"'. =
R'
,
R, + Rz-''
(3.28)
Equations3.27and 3.28showthat the current dividesbetweentwo resis
tors in parallel such that the curent in one resistor equals the curent
entering the parallel pail multiplied by the other resistanceand divided by
the sum of the rcsistors.Example 3.3 i ustratesthe use of the cuflentdivider equation.
Analyzing
a Cufient-Divider
Circuit
Find the power dissipatedin the 6 f) resistor shown
inFig.3.16.
and the power dissipated in the 6 O resistor is
p=(3.2)'(6)=61.44w.
Solution
1.6r}
First,we mustfind the curent in the resistorby simplifying the circuit with series-parallelrcductions
ltus, the circuit shownir Fig. 3.16rcducesto the
one shownin Fig. 3.17.We find the qrllent i, by
usingthe formulafor currentdivision:
16
j" = _--(10)
tb+ 4
frro 3on
figure 3.16 A Theci,cuitforErampte
l.l.
:8 A.
Note that i, is the cunert in the 1.6() iesistor in
Fig. 3.16.We now car turther divide i, between the
6 O and 4 O resistols.The current in the 6 O resistoris
rd: _(8)
t)
: 3.2A,
10A
I )
160
4{)}
Figure3.17 A A sjmptjfication
oftheciftuitshown
in Fig.3.16.
3.4
Vottage
Division
andCrrentDivision
0bjective2*(now how to designsimplevottage-dividerand cuff€nt-divid€rcircuits
3.2
^) Find the no-loadvalueof?J,in the
Find n, when nr is 150kO.
c) How muchpower is dissipatedin thc 25 kO
resistorifthe load terminalsare accidentally
shon-circuited?
d) What is the maximumpower dissipatedjn
the ?5 kO resistor?
3,3
a) Find the valueof R that will cause4 A of
currentto flow throughthe 80Q resistorin
the circuit shown.
b) How muchpower will lhe resistorR from
part (a) nced to dissipale?
c) How much power will the currentsource
genemtelor the valueof R froln pafi (a)?
60()
100
20A
It
800
(a)150v;
(b)133.33
V
(c)1.6W
(d)0.3w
Answ€r: (a) 30O;
(b) 7680w;
(c) 33,600
W.
NOTE: Also tty ChaptetPnblems 3.13,3.15,antl3.21.
3.4 tfqlltaqefiivisisn
a$deurrentllivision
We canrow generalizethe resrlts from analyzingthc voltagedivider circuit in Fig.3.12and tlre currentdivider circuil in Fig.3.15.Thegeneralizations will yield two addiiional and very usctul circuit analysislechniques
known asvoltagedivisionand currentdivision.Considerthe circuil shown
inFig.3.18.
The box on tlre left can cortain a singlevoltagesourceor any other
combinationof basiccircuitelementsthat resultsin thc vollag€?)shownin
the tigure.To the right of the box are r resistorsconncctcdin series.We
division.
are interestedin finding the voltagedrop ol acrossan arbitraryresistorRl figure3.18t Cjrcuitusd to ittustEievottage
in terms of the voltager. We start by usingOhm's law !o calculatei, fte
currentthrough all of the resistorsin serics,in lerms of the current o and
'
n l + R 2 + - + R , , R.q
(3.2e)
SjmphResjstive
Cjrcuitr
The equivalentresistance,
Rcq,is the sum ol the r resistorvaluesbecause
the resistorsare in series,as shownin Eq. 3.6.We apply Ohm's law a second time to calculatethe voltagedrop o/ acrossthe resistorRj. usingthe
cureni t calculatedin Eq.3.29:
Vottaqe-division
equation>
ur:
iRt :
RI
=-1)
(3.30)
Note that we used Eq.3.29 to obtain the right-hand side oI Eq.3.30.
Equation 3.30 is the voltage division equation.It saysthat the voltage
drop ri acrossa singleresistorRi ftom a collectionof se es-connected
resistorsis pioportionalto the total voltagedrop , acrossthe setofseriesconnected resislors.The coDstantof propor'tionality is the mtio of the single resistanceto the equivalert resistanceof the seriescomected set of
resistors,
or R'/Red.
Now considerthe c cuit showr in Fig. 3.19.The box on the left can
contain a single curent source or any other combination of basic circuit
elemenls that results in the current i shown in the figure. To tlre right of
the box are r resistorsconnectedin para el. We are inrerestedin finding
the curent ii throughan arbitraryresistorRr in terms of the current i. We
stafi by usingOhm'slaw to calculatez),thevoltagedrop acrosseachof the
resistorsin parallel.in termsof the current; and the r resistors:
0 = (R1 R2 ... iln,) : rReq.
(3.31)
Theequivalent
resistance
ofn resistors
in parallel,Rcq,canbe calculated
usirgEq.3.12.We
applyOhm'slawa second
timeto calculate
thecurrent
l, throughtheresistorR;,usingthevoltageo calculated
in Eq.3.31:
Current-division
equationt
n
R,
Note that we used Eq.3.31 to obtain the right-hand side of Eq- 3.32.
Equation 3.32is the current division equation.It saysthat the current i
through a singleresistorRr {rom a collectior of parallel-connected
resisto$ is proportionalto the total current i srpplied to the set of parallelconnectedresistors.The constantof proportionality is the ratio of the
tigure3.19A Circuii
used
to ittustrate
cunentdivjsion.
3.4
Vottage
Divisjon
andCurrent
Division
6l
equivalentrcsislanceof the parallel-connecled
serofresistorsto the siDgle
resrstance.
or Rcq/&. Note that the constanlof proportionaliryin thc cul'
rent divisionequationis rhe inverseof rhc constantof proportionalilyil1
the voltagedivisionequationl
Examplc 3.4 usesvoltage divislon and current division to solve for
voltagesand culrenlsin a circuit.
UsingVottageDivisionandCurrentDivisionto Solvea Circuit
[Jsecurrenr divisionto find thc cullett i, and use
voltagedivisionto find the voltager,, for the circuit
]n Fig.3.20.
Solution
We san use Eq. 3.32if we can find the equivalcnl
tcsislanceof the four parallel branchescontaining
rcsislors.Symbolically,
Figur€3.20 ,",ThecircuitforExamph
3.4.
Req= (36 + 44) l0 (40 + 10 + 30) 24
= 8 01 0 8 0 2 4 -
I
* 1 * 1
lJ0 r o s 0
t
=60.
-t
Applyhg Eq.3.32.
i, =
6
t(8A)
= 2A.
We can usc Ohm's law to find thc \,ottagedrop
acrossthe 24 () resistor:
This is also tlle voltage drop acrossthe branch con
tainingthe 40 Q, thc 10 O, and the 30 0 resistorsin
.<ri(sWe c.n rhcl U.c\ ollxg<di\ isionro de.cr1riir
the voltagedrop ,. acrossthe 30 () resistorgi\,cn
l l J L $ < k n o q t h e r o l t r g i o r u p r c f o . . r l ' ei c r i c \
connectedresistors,using Eq. 3.30.To do this, wc
recognize that the cquivalenl resisranceof the
series-connected
rcsislorsis 40 + 10 + 30 = 80 O:
,,=l!,**u,: ,av.
1]: (24)(2) = 48 V.
obiective3-ge abteto usevottageandcurrentdivisionto sotvesimplecircuits
3.4
a) Use voltage division to determine the
voltagc ?),acrossthe 40 O resistorin lhe
circuit shown.
b) Use o,,from part (a) to determincrhe current throughthe 40 J) resistor,and usethis
currcnt and currelt divisionto calcu]atethe
currentin the 30 O resistor.
c) How much poweris absorbedby lhe 50 O
400
',o,,1
60v
rrl
+
70o
Answer: (a) 20V;
(b) 166.67I1lA;
(c) 347.22mw.
NOTE: Alxo ttr Chapter Problens 3.22 and 3.23.
50(:}
,0,,
68
citcujts
Simpte
Resistive
Vottageand€urrent
3.5 Measuring
When working with actual circuits, you will oJten need to measue voltagesand cullents.we will spendsometime discussingseveralmeasuring
deviceshere and il the next section,becausethey are relativelysimpleto
analyze and offer practical examples of the curent- and voltage-divider
configurations we have just studied.
An .nmet€r is an hsuument designedto measue curent;it is placed
in seies with the circuit element whose cuuent is being measured A
the
connected
to measure
Figurc3.21A Anammeter
voltrnet€ris an instument d€signedto measurevoltage;itis placedin par'
the
connected
to measure
curentin Rr. anda voLtmeter
allel with the elementwhosevoltageis beingmeasured.Anideal ammeter
or voltmeter has no effect on the circuit variable it is designedto measure
That is, an ideal ammeler has an equivalent resistanc€ of 0 o and functions as a short circuit in sedes with the element whose current is being
measured.An ideal voltmeter has an i inite equivalent resistance and
thus functions as an open circuit in parallel with ttre element whose voll
age is being measu.ed.The configurations fol an ammeter used to measure the current in R1and for a voltmeter used to measurethe voltage irl 1R2
are depictedin Fig.3.21.The ideal modelsfor thesemetersin the samecilmodet
forthejdeatamme cuit aie shownin Fi9.3.22.
Figue3.22A A short-circujt
jdeatvoLtmeter.
modeLtotthe
ier,andanopen'circuit
There arc two broad categoriesof meters used to measurecontiluous
voltages and currents: digital mete$ and analog meters. Digital meters
measde the continuous voltage or current siglal at disclete points in
time, called the sampling times.The signal is thus conve ed from an analog sigml, which is continuous ir time, to a digital signal, which exists only
at discrete hstants in time. A more detailed e4lanation of th€ workings
of digital meters is beyond the scopeof this text ard coufte. However, you
are lilely to seeand use digital meten in lab settingsbecausethey olfer
several advantagesover analog meters.They introduc€ less resistanceinto
the circuit to which they are connected,they are easier to co rect, and the
precisior of the measurement is greater due to the nature of the rcadout
diagram
0f a dl\rsonvat
Figue3.23A A sch€matjc
tigue 3.24A A dcammeter
crrujt.
Figue3.25a A dcvottmeter
cncuit.
Analog m€t€rs are based on the d'Arsonval meter movement which
implements the readout mechanism.A d'A$onval meter movement consistsof a movable coil placedin the field of a pemanent magnet wlen current flows in the coit, it creafesa torque on the coil, causingit to rotate and
move a pointer across a calibrated scale.By design, the deflection of the
pointer is directly proportional to the current in t]le movable coil- The coil is
chaiactefzed by both a voltage rating ard a curent rating. For example,
one commercially available meter movement is rated at 50 mV and 1 rnA.
This meansthat when tlle coil is carrying 1 ntA, the voltage drop affoss the
coil is 50 mV and the pointer is deflected to its full-scale position. A
schematrcillustration of a d'Arsonval meter movement is shown in Fig.3.23
An analog ammeter consists of a d'A$onval movement in parallel
with a rcsistor, as shown in Fig. 3.24.The purpose of the pamllel resistor is
to limit the amount of curent in the movement's coil by shunting some of
it ttrrough R,4.An analog voltmeter consistsof a d'Arsonval movement in
series with a resistor, as showtr in Fig. 3.25. Here, the iesistor is used to
limit ttre voltage drop across tlle meter's coil. In both meteis, the added
rhefull-scale
feadingof rhemelermo\emenl.
resistorderermines
From thesedescriptionswe seethat an actualmeter is nonideal;bottr
the added resistor and tlle meter movement i troduce resistance in the
3.5 Measudns
vottage
andCureni
69
circuit to which the meter is attached.In fact, any instrument used to make
physical measu{ements extracts energy from the system while making
measurements.The more energy extracted by the instnments, the more
severcIy the measurement is disturbed, A real ammeter has an equivalent
resistanceUrat is not zero, and it thus effectively adds resistanceto the circuit itr series with the element whose current the ammeter is readins. A
real\ohmelerbasan equivalenr
resistance
lhal is not inLinite,
so it eFectively adds resistance to the circuit in pamllel with the element whose
voltage is being read.
How much these meters disturb the circuit beinA measured deDends
on LheeltecliveresistaDce
of the meter.compared;lh the resista;ce
in
t]le circuit. For example,using the rule of 1/10th, the eflective resistanceof
an ammetershould be no moie than 1/10th of the value of the smallest
iesistance in the circuit to be sure that the current being measured is
nearly t]le samewitl or without the ammeter. But in an analoe meter. the
vaiueoI resistanceis determinedby lhe desiredtull-scalefead;g \re wisb
to maLe, and it cannot be arbitarily selected. The following examples
illustrate the calculations involved in detemining the rcsistanceneeded it
an analog ammeter or voltmeler. The examplesalso consider the resulting
effeclive resistanceof the meter when it is insefied in a circuit.
lJsinga d'Arsonval
Amrreter
a) A 50 nV 1 mA d'Arsonval movement is |o o€
used in ar ammeter with a full-scale reading of
10 mA.Determine R,a.
b) Repeat(a) for a full-scalereadingof 1A.
b) Wlen the full-scale deflection of the ammeter is
1A, R,amust cary 999mA when the movement
carries1mA.In tlis case,then,
999x10-3R,{=50x10-3,
c) How much resistance is added to the or€urt
when the 10 mA ammeter is inserted to mcasurc
R,{ : 50/999 ! 50.05mO.
d) Repeat(c) for the 1A arnmeter.
c) Let R- rcpresert the equivalelt resistanceof the
ammetei. For the 10 mA alnmeter,
50mV
10mA
Solution
or, alternatively,
a) From the statement of the problem, we know
that when the current at the terminals of the
ammeter is 10 mA, 1 mA is flowing rlrough the
meter coil, which means that 9 mA must be
dive ed through R4. We also know that when
the movement caries 1 mA, the drop acrossits
terminalsis 50mV Ohm'slaw requiresthat
9x103Rn-50x103
R1 : 50/9 : 5.555O.
r50)r50l9
|
xm:sn+(50/g=so.
d) For the 1 A arnmeter
50mV
or, alternatively,
(5oxso/eee)
:
().
0.050
50+ (s0/eee)
70
SimpLe
Resistjve
Cncuits
Usinqa d Arsonval
Vottmeter
a) A 50 mV, 1 nA d'Arsonval movementls to be
usedin a voltmeter in which the full-scalercad
ing is 150V Determinenr.
b) Repeat(a) for a lUll scalereadingoI5V
c) How much resistancedoes the 150 V meter
insertinto the circuit?
d) Repeat(c) for the s V neter.
b) Fora tul] scalereadingof5 Y
50x103:nJso(s),
R, : 4950O.
c) If we let R", representthe equivaleniresistance
R- =
Solution
t0
or, altematively,
a) Full-scaledeflectionrequjrcs 50 mV acrossthe
meier movement.and the movementhasa resisl
'lterefore
anceof 50 O.
we apply Eq. 3.22with
= 50nV.
Rr - RtR, : 50,D.= 150.and1)2
-
150.v=
,l,.
r5o.ooo
'A
n,,, - 149.950+ 50 - 150,000O.
d) Then,
5rl
= ,ooo
^-' : l]L
,,.
t0'A
s0. 10':=-(1s0).
K/ + )ll
Solvingfor Rr gives
or, alternatively.
R- : 4e50+ 5U= 50U0A.
R. = 149,950O.
objective4-Be abteto determine
the readingof ammet€rs
andvottmeterc
3.5
a) Find the currentin the circuitshown.
b) lfthe ammeterin Example3.5(a)is usedto
measurethe cullent, what will it read?
3.6
a) Find the voltage1,acrossthe 75 k0 resistor
in the circuit shown.
b) lfthe 150V voltmeterof Example3.6(a)is
usedto measue l]re vollage,whatwill be
ttre reading?
75kO
Answer: (a) 10mA;
(b) 9.524mA.
NO'.E: Abotry Chapter
Probtems
3.30a
Answer: (a) 50Y
(b) 46.15v
3.33.
3.6
I'leasuingResistance-TheWheatstoneBddqe
71
3.6 Measuring
ResistanceTheWheatstone
Bridge
Many difrerent circuit conJigurationsare used to measue resistance,Heie
we will locus on just one, the mealstone bddge. The Wheatstone brialge
circuit is used to precisely measureresistancesof medium values,that is.itr
the range of 1 O to 1MO. In commercial models of tie Wheatstone
bridge, accuracieson the order of +0.1oloare possible.The bridee circuit
consistsof four resislorra dc voltagesource.anl a delector.The;sisraoce
of one of the four rcsiston can be varied, wbich is indicated in Fis_3.26 bv
lhe arrow tlu-oughRr. Tbe dc \ohage sourceis usuallya bartery,whicb is
indicated by the battery symbol for the voltage source z, in Fig. 3.26.The
detector is generally a d'Arsonval movement in the micioamD ranqe and is
bidqecircuit.
caueda galvanometef.Hgure 3.2bsho\rsthe circuil aflansemenlot lhe Figure3.25^ TheWheatstone
resislaoces
balteryanddetectofq bereRt. R.. andR3areL;owo resislors
and R, is the unknown resistoi.
To find the value of R,, we adjust the variable resistoi R3until theie is
Iro cunetrt in thg galvanometer, We then calculate the unknown rcsistor
from the simole exDiession
&:
-R.- R r .
The derivalion of Eq. 3.33 follows directly from ths apptication of
Kirchhoffs laws to the bddge circuit. We redmw the bridsi circuit as
Eg. 1.27ro sho\r (be curenrsappropriateto lhe derilationof Eq. .J.l3.
W}len is is zero, that is, when the bddge is balanced, Kirchhoff\ curent
law requires that
Q3q
tz = tt.
(3.35)
Now, becauseis is zerc, there is no voltage drop acrossthe detector, and
thercfore points a and b are at the samepotential. Thus wher the bddge is
balanced,Kirchhoff's voltage law requires that
i:Rs : i,R ,
(3.36)
iRI : i2R2.
(3.37)
CombiningEqs.3.34and 3.35with Eq.3.36gives
i1R3 : 12R".
(3.38)
WeobtainEq.3.33by firsrdividing8q.3.38by Eq.3.37andr}tensolving
the resultingexpressionfor Rr:
R:
Rr
R,
Rz'
(3.39)
a
R, -'/
Figure
3.27l A batanced
wheatstone
bfidse
(ls = 0).
Simpte
Resistive
Circuitr
R.
ft,: *n:.
(3.40)
Now that we have voriticd rhe validity of Eq.3.33,severalconments
aboul lhe r'esullare in ordcr. First.note that ifihe ratio R /R1is unily.the
unknown resistorR, cquals Rr. ln this case,the bridge resislorn3 must
vary over a rangcthat includesthe valueR.. For example,iI lhe unknown
lvere1000O and Rl couldbevariedfrom0 to 100(), the bridge
resistance
couldneverbe balanced.Thus to covera wide rangeof unknownresislors.
we must be able to vary the ralio R2/Rr. In a commercialWheatstone
bridge,Rr aDd 112consisloI dccimal values of resistancesthat can be
switched into the bridge circuil. Normall,v,the decimal values are
1, 10,100,and 1000O so |hal the ratio R2/Rrcan be varied fron 0.001to
1000in decimalsteps.ThevariablcrcsistorRr is usuall]'adjustableiD iDle
gral valuesofresislancefrom :l to 11,000O.
Altlough Eq.3.33 inplies that R, can vary from zero to infinit),.the
practicalrarge ofR, is approxinalely I () to 1 MO. Lower resistances
are
whealslone
because
of
thermodifficrlt to measureor a standard
bridgc
eleciric voltages generatedat the juncliorls of dissimilar metals and
bccauseof thermalheatingeffects lhat is,i'R eIiecls. Highcr rcsjstances
arc ditTicultto measureaccuratelybecauseof leakagecurrcnts.In other
words.if R, is large,the currentieakagein lhe eleclricalinsulationmay be
conparableto the curfent in the branchesoI lhe b dgc circuit.
. obiectives-lJnderstandhowa Wheatstone
bridoeis usedto measure
resistance
The bridgecircuit shownis balancedwher
R1: 100O../1,- 1000O,and Rj - 150O.
Thc bridgc is energizedfrom a 5V dc source.
a) What is the valueofR,?
b) Supposceachbridge resistoris capableof
dissipating250mW Can the bridgebe left
in thc balancedstatewithout exceedingthe
power dissipatingcapacityofthe resistors,
therebydamagingthc brjdgcl
NOTE: AIso tr! Chupter Problem 3.4ti.
Answer: (a) 1500{);
(b) yes.
3.7
DeLta
to'Wye(Pi-to-lee)
EquivaL€ni
Cncuiis
3.7 Detta-to-Wye
(Pi-to-Tee)
Equivalent
Circuits
+
The bridge configuratior in Fig. 3.26 introduces afl interconnectiofl of
resistancesthat warrantsfurther discussion.If we reDlacethe salvanomeler$1lhrtsequi\alenrresisrance
R,. qe candraq rle circuir.houn in
Fig. 3.28.We cannot reduce the interconrected rcsiston of this circuit to a
singleequival€ntresistance
acrossthe teminals ofthe battervif reslricted
to t]le simple seriesor parallel equivalent circuits introduced earlier in this
chapter. The hterconnected resistors can be reduced to a single equivalent resistorby meansof a delta-to-r,ye(A{o-Y) or pi-to tee (,7-ro-T)
equivalentcircuit.l
The rcsiston Rl, R2,and R- (or R3,n- and R,) in the circuit shown
in Fig.3.28 are referred to as a delta (d) intercorn€ction becausethe
interconnectionlooks like the creek letter A. It also is refened to as a
pi int€rconnectionbecausethe A can be shapedinto a ?' without disturbing the electricalequivalenceof the two conJigurations.
The electrical equivalencebetween the A and z interconnectionsis aDDarentin
Fig.J.2,r.
The resistorsRl, R-, and R3 (or Rr, R- and R,) in the circuit shown in
Fig. 3.28are referred to as a rrye (Y) interconnectionbecausethe interconnectior can be shapedto look ljke the letter Y It is easierto seethe y
shapewhen the interconnection is drawn asin Fig. 3.30.The y configuration
also is rcferred to as a tee (T) int€rconnection becausethe Y structure can
be shapedinto a T structure without disturbing the electdcal equivalenceof
the two structures.The electrical equivalenceof the y and t}le T configurationsis apparentfrom Fig.3.30.
Figure3.31illustratesthe A-to-Y (or r {o-T) equivalentcircuit tiansformation. Note that we cannot transfom t}le A interconrection into the
Y interconnection simply by changing the shape of the interconnections.
Sayingthe A-connectedcircuit is equivalentto the Y-connectedcircuit
meansthat the A configuration can be replaced witi a Y configuration to
mal(e the termhal behavior of the two configurationsidentical.Thus if
each circuit is placedin a black box, we can't tell by extemal measuremelts whetherthe box containsa set of A-coinectedresistorsor a set of
Y-comectedresistors.This
conditionis true only if the resistaflce
between
correspondingterminal pails is the samefor each box. For elample,the
re'istance
belweenterminal.a and b mu.l be lhe ."me $herher$e use
the A-connectedset or the Y connecteds€t.For eachpair of terminalsin
r A andY sLuctnrcsare presenrin a variely of usefulcircuits,rcr jusr resislivc
nelworks
Hencethe Aro-Y transfomationis a helptulioot in circuit analysis.
Figure3.28A A resistjve
netwo*generated
bya
Wheaistone
bridsecircuit.
F i g u r e 3 . 2 9 L configuration
AA
vjewed
asa r
Figlre3.304 A Y niuctureviewed
asa T stiuciure.
Figure3.31A TheA to-Y transformation.
'14
Simple
Resktive
Cncuits
the A-connected circuit, the equivalent rcsistance can be computed using
seriesand pamllel simplifications to yield
Rat :
R.(R" + nb) - R 1
+R2,
&+R6+R.
(3.11)
R.(R, + &) = R : + R : ,
R,+nb+R.
(3.42)
Rr(R. + R,) = R 1
+R}
(3.43)
Ra+Rb+R.
StaightJorward algebnic manipulation of Eqs. 3.41 3.43 gives values
for the Y-connected resistors in terms of the A-connected rcsistors
reouired for the A-to-Y eouivaletrt circuit:
RaR.
-Rd+ R, + &'
n.&
R,+R6+R.'
&Ra
&+Rr+R.
11.11)
(3.45)
13.46)
Revening tle A-to-Y trarsfomation also is possible.fhat iq we can start
with the Y shucture aDd replace it with an equivalent A structure. The
expressionsfor the three A-connected resistors as functions of the three
Y-connected rcsistors are
n1R2+R2R3+R3R1
R1
n1R2+R2R3+R3Rr
R2
n1R2+n2R3+R3R1
\3.47)
(3.48)
(3.1e)
Example3.7illustiatesthe useof a A{o-Y hansfomation to simplify
tlle analysisof a circuit.
3.7
(ncrirs
(Pi-ro-TeF)
Detta-to-Wye
Equivdlert
Applyinga Delta-to-Wye
Transform
Find the current and powei supplied by the 40 V
souce in the circuit shownin Fig.3.32.
resistance
acrossthe terminalsof the 40V sourceby
se es-parallel simplifications:
(501450)
: 55 + -_= 80 ().
R,"
-'
I ll(,
The final step is to note that the circuit reduces to
an 80 f) resistor across a 40 V souice, as shown ln
Fig. 3.35, from which it is apparent t]lat the 40 V
source delivers 0-5A and 20 W to the circuit.
Figure3.32A lhecircuit
forExampte
3.7.
Solution
We are interestedoDly in the current and power
drain on the 40 V source,so the problem hasbeen
solved once we obtain the equivalent rcsistance
acrossttre terminals of the source.We can find this
equivalenl resistanceeasily after replacingeilher
the upper A (100,125,25 O) or the lower A (40,
25, 37.5O) with its equivalent Y We choose to
replacethe upper A- We then computethe threeY
resistances,
defined in Fig. 3.33,from Eqs.3.44 to
3.46.Thus,
Figure3.33A TheequivalentY
rcsistor.
37.50
100x 125= 5 0 r ) ,
250
125 25
250
-^_^
100 25
250
.^ ^
Substituting the Y-rcsistoN into t]le circuit
shown iD Fig. l.J2 produce( the ciJcuit sholn in
Fig.3.34-From Fig.3.34,wecan easilycalculatethe
Flgure3.34t Atrandom€d
version
ofthecncuitshown
in
Fjg.3.32.
,J'
jn thesimptification
Flgure3.35 ^ Thefinatstep
of thecjrdrit
s h o wjnn F i g . 3 . 3 2 .
75
76
Simph
Resistive
Cncuits
0bjective6-Xnow wh€nandhowto usedetta-to-wye
equivaterit
circuits
3.8
Use aY-to-A tansforinaticjnto find the voltage
, in Lhecircuit.horrn.
105{}
Answer:J5\.
NOTE: Abo try Chapter
Prcbletns
3.52,3.53,
and3.51.
PracticalPerspective
A ReaIWindow
Defroster
A nodel of a defrostergridis shownin Fig.3.36,wherer and], denotethe
horizontalandverticaL
spacingof the grid etemerts.Giventhe dimensions
of the grid, we needto find expressions
fof eachresistorjn the gfid such
that the powerdissipatedper unit tengthjs the saniejn eachconductor.
ThiswjlLensureuniformheatjrg of the rearwindowin both the .v and l
directjons.
Thuswe needto find vatuesfor the gridresistors
that satis! the
foUowinqretationshiDs:
:,,(+)
=(+):-(+)
"(+)
-(+)
(3.50)
'(+),(+)
(+)='(+):,'(+):,
(3.51)
tigur€3,36e Modetof
a deftustagrid.
(3.52)
,,(?)=',(*)
(3.53)
Webegjnthe anatysis
of the grid by takingadvantage
of its structure.
portion
(i.e.,
Notethat if we disconnect
the lowef
of the circuit
the resjstor
i1, iz, il, andib areunaffected.
Thus,instead
&, R,/,Ra,andR5),the curr€nts
of analyzjng
the circuitin Fig.3.36, we can analyzethe simplercircujtin
PmcticaLP€rspectjve
77
Fig.3.37.NotefurtherthatafterfindingR1,R2,R3,Ra,andR, in the circuit
in Fig.3.37,we havealio foundthe vaLues
for the remaining
rcsjstors,
since
Rt -
R1
Rz,
(3.54)
R" : RA,
,
R
.
Ra - R,.
Beginanalysisof the simpLified
grid circuit in Fig. 3.37 by writing
expressions
for the currentsi1, i2, i?, andib. Tofind ir, descrjbe
the equivaFigure
3.37a A sjmptified
modeL
ofthe
lent resistance
in paratlelwith R3:
-
^^
x r ( R 1+ 2 & )
_ (& + 2R.)(R2+ 2R) + 2R2Rb
(Rt I R, 2RJ
Forconvenience,
definethe numerator
of Eq.3.55as
D = (\
+ 2R)(R2 + 2R) + 2R2Rb,
(3.56)
andtherefore
(Rr+R,+2R")
(3.57)
It foLtows
directtythat
'.=&
-'
R.
Ydc(Rr+Rr+2&)
D
(1 58)
Expressjons
for i1 and 12can be found djrectlyfrom ib usjngcurrent
division.Hence
'
iiR1
%,R'
R--R, r'R.,
D
and
.
ib(R] + 2R4)
(Rt' R, 2R
vd.(Rr + 2&)
D
Theexpression
fori3 js simpLy
.
''
v,t
R3
(3.61)
Sinrpte
Resistive
Circuits
in Eqs.3.50-3.52to deriveexpressions
for
Nowwe usethe constraints
R., nb, R2,andR3 asfunctionsof R1.FromEq.3.51,
R,
RI
R^=LR,=,rR,.
13.62)
Thenfrom Eq.3.50we have
., =(1)'^,.
(3.63)
Theratio(ir/ir) is obtained
directty
fromEqs.3.59and3.60:
Rz
tJRz
:
+
+
i2 Rr 2R" Rr zoRl
(3.64)
WhenEq.3.64js substitutedinto Eq.3.63,we obtain,aftersomeaLgebraic
(seeProbiem
maniputation
3.69),
Rr=(t+2o)' Rr.
(3.65)
Theexpression
for R6 asa fundjon of R1js derivedfromthe constraint
imposedby Eq.3.52,nametythat
/: \2
n o :l r l n .
(3.66)
\tb/
Theratio (i/i/,) js derivedfrom Eqs.3.58and3.59.Thus,
n
\_
(.\ + R2 + 2R")
ib
(3.67)
WhenEq.3.67is substituted
into Eq.3.66,weobtain,aftersomeatgebraic
manipuLation
3.69),
GeeProbtem
^
^'-
(L + 2c)2oR1
4(1j "r,
(3.68)
FinaLly,
the expression
for R3canbe obtainedfromthe constnintgiven
jn Eq.3.50,or
o.: i1)'",,
(3.6e)
sLrmmary 79
RzR:
D
i3
onceaqain,aftersomeatgebraic
(seeProbLem
maniputation
3.70),the
expression
for R1canbereduced
to
(t + 2o)zoR1
4(I + o)'
(1 + 2o\1
(I+o)'
R 'l : - R , .
(3.70)
Theresutts
of ouranaLysis
aresumnarized
in Tabte
3.1.
Rz
(1 + 2o\!R1
I
+ 2d\a
a *'F"'
R3
yourundeRtanding
NqTE:Assess
of thePrddicalPetspective
bytryingChopter
Problens
3.71-3.73.
5ummary
. Sedes resistorccan be combined to obtain a single
equivalentresistanceaccordingto the equation
R"q=)Rr:Rr+R2+.
. When currelt is divided betweenpara]lel resistors,as
shown in the figure, the curent Urrougheach resistor
canbe found accordingto the equations
+R&.
.
(Seepage58.)
''
. Pamll€l resistors can be combined to obtain a single
equivalentresistanceaccordingto the equation
|
R.q
: r
e
Ri
I
I
r
Rr
R2
Rfr
-'
Whenjust two resistorsare in parallel,the equationlbr
equi\dlcnlresisrance
canhe simplifiedro gi!e
,."q
.R R,
Rr + R2.
(Seepages59-60.)
. When voltage is divided between sedes resistors,as
shownin the figure,the voltageacrosseachresistorcan
be found accordingto the equations
", ,'-
Rl
,,
Rr
Rr+ Rr''
R:
*="r*&-'"
(Seepage62.)
ir"
R
l
2
Rr + tRr'"
R1
l)
' r l n, nl
Rr + Rr'"
(Seepage64.)
Voltagedivisionis a ciicdt analysistool thar is usedto
find the voltagedrop acrossa singleresjstancefrom a
collectionof series-coDnected
resistances
wheDthe vol!
asedroDacrossthe collectionis known:
Rj
t '] ' = - ? ) '
Kcq
where oi is the voltage drop acrossthe resistanceRl
and o is the voltage drop acrossthe se es-connected
resistanceswhose equivalent resistanceis R.q. (See
page66.)
80
R€sistjve
Circuits
Simpt€
Curr€trt division is a circuit analysis tool that is used to
find the cunent through a single resistanceftom a collection of paiallefconnected resistanceswhen t]Ie curient into ttre collection is kno$n:
R""
,j_
Digital met€rs and analog meterc have intemal resisG
ance,which influences ttre value of the circuit variable
beins measured.Meters based on the d'Arsonval meter
mov;ment detbemtely include internal rcsistance as a
way to limit t]le current in the movemenf's coil. (See
page68.)
p.,
where ir is the curent thmugh the iesistance Rj and i is
the current into the parallel-connected resistances
whose equivalent resistanceis Req.(See page 66 )
A voltmeter measuresvoltage and must be placed in parallel with the voltage being measuled.An ideal vollmeter
hasinfinite inlemal resistanceand tlus doesnot alter tlle
voltage being measured.(Seepage 68.)
. An ammeter measurcs current and must be placed in
serieswith the curent beingmeasured.Anideal ammeter has zero internal rcsistance and thus does not alt€r
the currentbeingmeasured.(Seepage68.)
The Wh€atstone bddg€ circuit is used to male precise
measuementsof a rcsistoi's value usingfour resistors,a dc
voltage source,and a galvanometer'A Wheatstonebiidge
js balancedwhen the resjstorsobey Eq. 3 33, resulting in
a galvanometerreading of 0 A. (Seepage 71.)
. A circuit with three resistors connected in a A configuration (or a ?]'configuiation) can be transformed into an
equivalent circuit in which the ttrree resistors are Y connected(or T connected).The A-to-Y transformationis
given by Eqs.3.44-3.46;the Y-to-A hansformationis
givenby Eqs.3.47-3.49.(Seepage74.)
Problems
Sections3.1-32
b) simplify the circuit by replacing ttre seriesconnectedresistonwith equivalentresistors.
3,1 For each of the circuits shown,
a) identify the resistors connected in serieq
P3.1
Figure
240{l
5ko
3ko 8ko
300()
1800
1400
10v
200()
400
soo :ov( )
60f)
450
3oo
81
3.2 For eachof tle cftcuitsshowr in Eg. P3.2,
a) identify the rcsistorsconnectedin parallel,
b) simplify tle circuit by replacing the parallelconnectedrcsisto$ with equivalentrcsistorc,
3,3 a) Find the powerdissipatedir eachresistoriII the
cLcuit showr)in Fig.3.9.
b) Fhd tle powerdeiiveredby the 120V source.
c) Showthat the powerdelivetedequalsthe power
dissipated.
3,4 a) Showthat the solution of the circuit in Fig. 3.9
(seeExample3.1) satisfiesKirchhoff's current
law at junctionsx andy.
b) Show that the solution of the circuit in Fig.3.9
satisfies Kirchhoff's voltage law around every
closed loop.
3,5 Find lhe equivale0rresisrance
seenby the sourcein
each of the circuits of Problem 3-1.
3.6 Find the equivalentresistanceseenb' lhe sourcein
each of the circuits ol Problem 3.2.
3J Eind the equivalent resistance R"b for each of the
circuitsin Fi9.I,3.7.
3.8 Find the equivalent resistance Rabfor each of the
circuits in Fig. P3.8.
Figure
P3.2
Flgure
P3.7
2A
10()
7ko
14()
Figure
P3.8
300
140
12[r
30
500
20a ' a
4{O
300
\1
2{l
G)
z7a 24o"
120
82
Simple
Resisiive
Cncuitr
3.9 a) In the circuitsin Fig.P3.9(a)-(c),find the equivalentresistance-Rab,
b) For eachcircuit find the powerdeliveredby the
S€€{ioE3.}3.4
c) Using tle results of (b), destn a resistive network with an equilaieDlresislaDce
o[ using
1 kO resistois.
d) Using tle rcsufts of (b), design a resistive network with atr equivalent resistanceof 5.5kO
using 2 kO resistors.
3.13 a) Calculate the noload voltage t, foi the voltage-
3.10 Find the powerdissipatedin the 30 O resislorin the
EnG c cuir ir Fig.P3 10
!!99!!1
FigueP3.10
3,11 For the circuit in Fig. P3.11calculate
divider circuit shoM in Fig. P3.13.
b) Calculate the power dissipated in & and R2.
c) Assume tlat only 0.5 W resisto$ are available.
The noload voltageis to be the sameas in (a).
Specify the smallest ohmic values of R1 and Rr.
tigur€P3.13
t60v
b) the power dissipatedin the 12 O resistor.
c) the power developed by the curent souice.
Figure
P3.U
10()
3,14 In the voltage-divider circuit shown in Fig. P3.14,
12A
the no-Ioad value of oo is 4 V Wlen the load resisf
ance Rr is attached acrossthe teminals a and b, ?r,
drops to 3 V Find Rr.
Figure
P3,14
40()
1800
-t*l-?
3.12 a) Find an expressionfor the equivalent resistance
oI two resistois of value R in parallel.
b) Find an expressionfor the equivalent resistance
of ,?resistomof valueR in parallel.
-r
(:)
tigurcP3.9
14()
144v
16O
18fI
100
14o
10()
.l]]]
3.,,'
lR.
l_ *___l
Probtems83
3.1S The no-load voltage in the voltage-divider circuit
shown ir Fig. P3.15is 20 V The smallestload resistor
!ryq!!11
that is ever connecled 1ot]le divider is 48 kO. When
the divider is loaded, r', is not to drop below 16V
a) Design the divider circuit to meet the speciJications just mentioned. Specify the numerical value
of R1and R2.
b) Assume the powei ratings of commercially
availableresistorsare 1/16, 118,114, 1, aad2 W.
Wlat power rating would you specify?
Figure
P3.15
n1
3.18 There is often a need to produce more than one
voltage using a voltage divider. For example, the
memory components of many penonal computen
require voltagesof -12 V, 5 V and +12 V, all with
respect to a common reference terminal. s€lect the
valuesof R1,R2,ard R3in the circuitin Fig.P3.18to
meet the followitrgdesignrequirements:
a) The total power supplied to the divider circuit
by the 24 V source is 80 W when the divider is
unloaded.
b) The three voltages,all measuied with respect to
the commonreferenceterminal,are ,r = 12 V,
?)2-5V,andt'3= 12V.
+
100v
Fig(reP3.18
3.16 Assume tie voltage divider in Fig. P3.15 has been
constructed from 0.15 W rcsistors, How small can
R, be before ore of tlle resistorsin the divider is
operating at its dissipation limit?
3.17 a) The voltage divider in Fig. P3.17(a)is loaded
with the voltagedivider shownin Fig.P3.17(b);
that iq ais connectedto a', and b is connectedto
b'. Find !',.
b) Now assumethe voltagedivider in Fig.P3.17(b)
is connected to the voltage divider in
Fig. P3.17(a)by meansof a current-controlled
voltagesourceasshownin Fig.P3.17(c).Findr..
c) What effect does adding the dependent-voltage
source have on the opention of the voltage
divider that is connectedto the 480V source?
3,11)A voltage divider like that in Fig. 3.13 is to be
designedso that o. : /,.r'"at no load (R, : oo) and
?, = aor at full load (Rr = R,). Note that by defia) Showthat
figuleP3.17
20ko
L
R, = -R"
60ko
o
AR
and
k - d
^
^
^'qt _ rf-
20ko
80,000
i 180ko
b) Specify the numedcal values of R1 and R2 iI
& : 0.85,a = 0.80,andRo : 34 kO.
c) If x'" = 60 V, specify the maximum power that
will be dissipatedin R1and R2.
d) Assume the load resistoi is accidentally short
circuited. How much power is dissipated in R1
and Rr?
84
Simpte
Resjrtiv€
Cjrcuits
3.20 a) Show that the current in the ,tth branch of the
6BG circuit in Fig. P3.20(a)is equal to tle source current
is times the conductance of the *th branch divided
by the sum of the conductanceqthat is,
iFr
b) Use the result derived in (a) to calculate the current in the 6.25 f,) resistor in t]le circuit in
Fig.P3.20(b).
Figuru
P3.20
b) Using your result from (a), find tle current flowiog i0 the 240 O resisrorlrom lei lo right.
c) Using your result form (b), use curent division
to find the current in the 140 O resistor.
324 a) Find the voltage ?J,in the circuit in Fig. P3.24.
"""
b) Replacethe 30V sourcewith a generalvoltage
source equal to 14.Assume % is positive at the
upper teminal Find o, as a function of %.
hg,rrcP3.24
5ko
60ko
1ko
15 k{)
30v
3.2S Find ?1 and ?J2in the circuit itr Fig. P3.25EPr'r Figure
P3,25
120
50()
321 Specify tle iesistorc ir the circuit in Fig. P3.21
meet the following design criteda:
is=5mA;0s:lvti=4i21
i2 : 84; and i3 : 5la.
FigueP3.21
3.26 Find o, in the cfcuit in Fig. P3.26.
'5"!'
tigureP3.26
R. i.
Look at the circuit in Fig.P3.1(a).
a) Use current division to fhd the current flowing
from top to bottom ir the 10 kO resistor.
b) Using your result from (a), find the voltage drop
affoss the 10 kO resistor, positive at the top.
c) Using your result ftom (b), use voltage diusron
to find the voltage drop acrossthe 6 kf,) iesistor,
positive at t])e top.
d) Using 'our resuit lrom (c). use !oltage division
to find the voltage drop acrossthe 5 kO resistor,
positive at the left.
Look at the circuit in Fig. P3.1(b).
a) Use voltage division to find the voltage drop
acrossthe 240 O resistor, positive at tle left.
10ko
3ko
2kO +ir,- 4kO
15kO
12kO
327 Find i. and is in tlle circuit in Fig. P3.27.
"""
Figure
tl,zz
10()
6'75
V
3.28 For the cftcuit in Fig. P3.28, calculate (a) t, and
tsdc (b) the power dissipated in ttre 15 O resistor.
Figure
P3.31
60f)
figuruP3.28
4()
2ll
329 The curent in the 12 O resistor in the circuit in
Fig.P3.29is 1A, asshoM.
a) Find 0s.
b) Find tlle power dissipated in the 20 O resistoi.
The ammetei described in Problem 3.31 is used to
measruethe current jo in the circuit in Fig.P3.32.What
is the percetrtageof elror ir the measuredvalue?
figureP3.32
tigureP3.29
20()
1.20
5f)
20mA
2tl
t)
3,n
l1A
I-/
t2f}
3()
6f)
Section3.5
A d'Arsonvalvoltmoteris shownin Fig.P3.33.
Find
t}le value of Ro lor eachof the follovrhg full-scale
(a) 100v, (b) 5V,and(c) 100mV
readinss:
figureP3.33
3.30 a) Showfor tle ammetercircuit in Fig. P3.30that
the current in the d'Arsonval movement is
always1/25thof thecurent beingmeasured.
b) Whatwould tle fractionbe if the 100mV 2 mA
movementwereusedin a 5 A ammeter?
c) Would you expect a unifom scale on a dc
.
d'Arsonvalannneter?
FigunP3.30
100mV.2 mA
(25/n)a
3.31 The ammeter in the circuit in Fig. P3.31has a resistanceof 0.5 O. What is the percentageof error in the
reading of this ammeter if
^.
,/oerror:
measwedvalue "\
/ ___________
_^^.r | \ IUU/
I
varue
rrue
\
/
a
3.34 Suppose tlle d'Arsonval voltmeter described in
Problem3.33is usedto measurethe voltageaqoss
the 24 O resistorin Fig.P3.32.
a) What wil tle voltmeter read?
b) Usinglhe definilion of the percentageoferror in
a meter reading found ir Problem 3.31, what is
the percentageof eror in tle voltmeter rcading?
A shunt resistor and a 50 mV, 1 mA d'Arsonval
movement are used to build a 10 A ammeter.A
resistanceof 0.015O is placedacrossthe terminals
of tle ammeter. Wlat is the new full-scale range of
the ammeter?
86
5impte
Resjstive
Circuits
3.36 A d'Arsonval movement is rated at 1 mA and
50 nV Assume0.5 W precisionresistorsare available to use as shunt$ What is the largest full-scalereadingammelerlhal canbe designedlEyplain.
A d'Anonval ammeter is shown in Fig. P3.37.
Design a set of d'Arsonval ammeten to rcad the following tull-scale cunent readings: (a) 5 A, (b) 2 A,
(c) 1 A, and (d) 50 mA. Specify the shunt resisror
R,afor each ammeter-
c) What will the voltmeter read if it is placed aooss
the 20 kO resistor?
d) Will the voltmeterreadingsobtainedin parts(b)
and (c) add to the readingrecordedin pal1 (a)?
Explain why or why not.
Figure
P3.39
Figurc
P3.37
3.40 You have been told that tlle dc voltage of a power
3.3E 1!e elementsin the circuit h Frg.2.Z havethe folowadc ing values:R1 = 20 kO, R, - 80 kO, rRc= 0.82kO,
Rr: 0.2kO, Vcc= 7.5V,% : 0.6V,andB = 39
a) Calculatethe valueof/s inmicroamperes.
b) Assume that a digital multimeter, when used asa
dc ammeter,has a resistanceof 1kO. If the
meter is inseted between terminals b and 2 to
measurethe current is, what will the meter read?
c) Using the calculated value of iB in (a) as the correct value, what is the percentage of elror in the
measurement?
l.Jg The vohage-dividercircufl shown in Fig. PJ.Jq is
designedso that the noload output vollage is
7/9ths of the irput voltage.A d'Arsonval voltmeter
having a sensitivity of 100 O/V and a full-scale rating of 200 V is used to check the operation of tlte
a) What will the voltmeter read if it is placed aqoss
t]le 180V source?
b) What will the voltmeler rcad if it is placed acioss
the 70 kO iesistor?
supply is about 500V Wren you go to the instrument
Ioom to get a dc voltmeter to measue t]Ie power
supply voltage, you find that there are only two dc
voltmeters available.The voltmeters are rated 400 V
full scaleand have a sensitivity of 1000 O/V.
a) How can you use the two voltmeters to check
the power supplyvoltage?
b) What is the maximum voltage that can be
measured?
c) If ttre power supply voltage is 504 Y what will
each voltmeter read?
3.41 Assume that in addition to the two voltmeters
describedin Problem3.40.a 50 kQ precisjonresistor is alsoavailable.The 50 kO rcsistoris connected
in sedeswith the sedes-colnectedvoltmeters,fhis
circuit is then connected acrcssthe terminals of the
power supply. The reading on the voltmete$ is
328V Wlat is the voltage of the power supply?
3.42 Th€ voltmeter shown in Fig. P3.42(a) has a fu[scale rcading of 800 V The meter movement is
rated 10OmV and 1 mA. Wlat is the percentage oI
error in the meter readingif it is used to measure
the voltageo in t}le circuit of Fig.P3.42(b)?
Prcbl€ms87
tigurcP3.42
-----800V
3.44 Assume in designing the multLange voltmetet
shown in Fig. P3.44that you ignore the rcsistanceof
the meter movement,
r----.
R-:
I
100mv/ /\
t 'A\-/
T
a) Specify the values of R1,R2,and R]].
3.5mA
b) For eachof the three ranges,calculatethe percent
ageof er:rorthat this desig[ strategyprcduces.
co*'on
tiglre P3.44
(b)
50v
3.43 A 600 kO resistor is connected from the 200 V ter'
minal to the common termhal of a dual-scile volt'
meter, as shown iD Fi& P3.43(a). This modified
voltrneter is then used to measurethe voltage across
the 360 k(! resistor in the circuit ir Fig. P3.43(b).
20v
2V
a) Wltat is the reading on lhe 500 V scale of
the meter?
b) What is the percentage of ellor in the measued
voltage?
3.,15Design a d'Alsonval volfieter
that will have the
three voltage mnges shown in Fig. P3.45.
a) Specify the values of R1,R2,and R3.
FigureP3.43
| 500v
b) Assume that a 500kO resistor is connected
between the 100 V terminal and the common
terminal. The voltmeter is then connected to an
unknown voltage using the common teminal
and the 200 V teminal. The voltmeter reads
188V What is the unknown voltage?
c) W}lat js the maximum voltage the voltmeter h (b)
tigureP3.45
r-- -- -l
500v|
I
I
Modified
360ko
(b)
88
5jmple
Resistiv€
Circuits
3.46 The circuit model of a dc voltage souce is shown in
Flg. P3.46.The following voltage measurementsare
made at 1Ie terminals of the source: (1) With the
terminals of the source open, the voltage is measured at 80 mY ard (2) with a 10 MO resistorconnected to the terminals, the voltage is measured at
72 mV All measurements are made with a digital
voltmeterthat hasa meter resistanceof 10 MO,
a) What is the internal voltageof the source(or) in
millivolts?
b) Wlat is the internal resistanceof the source (Rr)
ir kilo-ohms?
figureP3.46
R"
3,50 Find tle power dissipatedin the 18 O rcsistorir lhe
6r.r circuit in Fig P3.50.
figureP3.50
3.51 In the Wleatstone bddge circuit shown in Fig. 3.26,
the ratio R2/Rr catr be set to the following values:
0.001,0.01,0.1,1, 10, 100,and 1000.The resistorR3
can be varied from 1to 11,110O, in inffements ol
1 (1. An unknown resistor is knoim to lie between
4 and 5 O. Wlat should be the setting of the R/Rl
ratio so that the unknown resistor can be measured
to foui signiJicantfigures?
352 Use a A-to-Y transformation to find the voltages ,1
and t,2in the circuit in Fig. P3.52.
Sectiotrs3.6-3.7
3.47 Assume the ideal voltage source in Fig. 3.26 is
replaced by an ideal cuirent source. Show tlat
Eq.3.33is still valid.
3.48 The bddge ctucuitshown in Fig.3.26 is energized
6@ ftom a 21 V dc source.The bddge is balanced when
Rl = 800o, R' = 1200o,and R3 : 600o.
a) What is the value of &?
b) How much curent (in miltamperes)doesthe dc
source supply?
c) Wlich resistor ir t]le circuit absorbs the most
power? How much power does it absorb?
d) Which resistor absorbs the least power? How
much power does it absorb?
Figure
P3.52
28f}
16|l
3.53 a) Find the equivalent rcsistance Rabin the chcuit
in Fig. P3.53by using a d-to-Y transfomation
invol!iDgrberesislo^Rr. Rr.and Ra.
b) Repeat (a) ushg a Y{o-A transfonnatior
involving resiston -R2,Ra,and R5.
c) Give two additional A-to-Y or Y-to-A transformationsthat could be usedto find Rab.
figuruP3.53
3.49 Find the detector curent i,i in the unbalanced
Edc bridge in Fig. P3.49 if the vollage drop across the
detector is negligible.
Figure
P3,49
60f}
20o"
Prcblems89
3,54 Find the equivalent resistance R.b in the circuit in
rigureP3.54
15()
100
fia
3.58 a) Find the resistance seen by the ideal voltage
source in the circuit in Fig. P3.58.
b) If oab equals 600 Y how much power is dissipatedin the 15 O resistor?
Figure
P3.58
a
20
60J)
80()
140f)
38()
15(}
60J)
3.55 In tlle circuit in Fig. P3.55(a) the device labeled D
P5E.Erepresentsa component that has the equivalelt circuit shown in Fig. P3.55(b).The labels on the terminals of D show how the device is connected to the
circuit. Find ?,xand the power absorbedby the device.
Figure
P3.55
3.59 Use a Y-to-A transformation to find (a) i,; (b) irt
6dc (c) 4; and (d) the power delivered by the ideal current sourceir the circuit in Fig. P3.59.
tigureP3.59
320{))
3.56 Find id and the power dissipated in t}le 140 O resisEtrG tor in the circuit ir Fig. P3 56.
3.60 For the circuit showrl in Fig. P3.60,find (a) ,1, 0) o,
k) t , and (d) the power supplied by tle voltage
Flgure
P3.55
224
20 tL
Figure
P3.60
300
240Y
) "n-'..-. Itogrruon
10()
182()
80
120
3.57 Find nabin the circuit in Fig. P3.57.
Figure
P3.57
1.8kO
3.61 Derive Eqs.3.44-3.49from Eqs.3.41 3.43.Thefollovrhg lwo hints should help you get stated in the
dglt dhectioni
a) To find Rl as a tunction of &, R6, and Ra,filst
subtractEq. 3.42ftom Eq.3.43and then add this
result to Eq. 3.39.Use similar manipulationsto
find R2 and R3 as functions of R,, R6,and R..
90
Simpte
Resistjve
Cncuits
b) To fitrd Rb as a function of n1, R2, and R3,take
advantage of the derivations obtained by hint
(1),namely,Eqs 3.44-3.46.
Note that theseequations can be divided to obtait
R2
R^
= ;,
^b
R.
R, = ;Rb,
or
b
and
xr
Ro
&:&.
!--Rr-
t--..1 d
Attenuator
R2^
. ' ^^ , : x ^ '
3.64 a) The fixed-attenuator pad showl in Fig. P3.64 is
Now usetheseratios in Eq.3.43to eliminate&
and &. Use similarmanipulationsto find Raand
& as functions ol R1,R2,and R3.
Show that the expressions for A conductances as
lunctions of the three Y conductancesare
called a brillgel tee. Use a Y{o-A transformation to show that Rab : RL if R = RL.
b) Showthat when R : RL,the voltageratio o,/o,
equals0.50.
FlgueP3.54
GzGz
Gr+G2+G3'
GGz
Gt+GziGt'
GtGz
G\+G2+G3'
"
i________,_
pad
Fixed-attenuator
^
r, :
1
^
o.
r. :
1
&.
erc.
I '
The designequationsfor the bddged-teeattenuator
circuit in Fig. P3.65are
2RRi
Seclions3.1-3.7
Resistor networks are sometimes used as volume
control c cuits. In this application, they are
refened to as resr.rtdrceatknu&tors or pads. A typrcal fixed-attenuator pad is shown in Fig. P3.63. In
designiDg
an drlenuado0pad.lbe circuildesigner
will select the values of R1 and R2 so that the iatio
of r,/oi and the resistanceseenby the input voltage
souice R"6 both have a speciJiedvalue.
a) ShowthatifRab: RL,then
R i : 4 R r ( R 1+ R ) ,
R2
2R1 -r R2 + RL
b) Select the values of R1 and R2 so that
Rab- RL = 600O and 0,/?,r : 0.6.
3n-RL
3R + RL'
when R2hasthe valuejust given.
a) Designa fixed attenuatorso tlat oi : 3oowhen
RL:600o
b) Assume the voltage applied to the input of the
pad desigtred in (a) is 180 V Which iesistor in
the pad dissipates
t}le most power?
c) How much power is dissipatedin the resistorin
part (b)?
d) Which rcsistor in the pad dissipatesthe least
e) How much power is dissipated ill the resistor in
part (d)?
Probtems91
Figure
P3.65
show that the percent elror in the approximation of o, in koblem3.66 is
/A P\P.
o/oenor = -::il:i:. 100.
(fi, + &r)ll4
b) Calculate the perce error in od,using the values
h Problem 3.66(b).
3.68 Assume the eror in t'd in the bridge circuit in
3.66 a) For the circuit shown in Fig. P3.66 rhe bridge is
balancedwhen AR = 0. Showthat if A-R << R,
the bridge output voltage is approximately
-An&
U"^ = - r ) , -
( R , + R 4 r-
Fig.P3.66is not to exceed0.5%.Whatis tlle largest
percent change in Ro that can be tolerated?
3.69 a) Derive Eq.3.65.
b) DedveEq.3.68.
,:lNfl[ii,
3.70 DeriveEq.3.70.
b) Given R, : 1 kO, R3 : 500O, n4 = 5 kO, and
oin - 6 V, what is the approximate bridge output voltageif AR is 3oloof Ro?
c) Find the actual value of 1'" in part (b).
Figure
P3.66
3,fl Suppose
the grid structurein Fig.3.36is 1 m wide
and the veitical displacementof fie five hodzontal
gdd linesis 0.025m. Specifythe numericalvaluesof
R1 R5and R, - Rd to achievea unilorm power
dissipationof 120rym, usinga 12V powersupply.
(Hrrri Calculate
(' filsr,thenR3,R1,Ra,Rr, andR2
in thatorder.)
3,72 Checkt}le solutionro Problem3.71by sho$ringthat
J#;lHlEthe total power dissipatedequalsthe powerdevele;ai- opedby the 12V sowce.
3.73 a) Design a defroster grid in Fig. 3.36 having five
horizonral conductors lo meel
the following
T!ryla!
specifications:The gdd is to be 1.25m wide, the
ry,ryL. ve ical sepamtion between conductors is to be
3.67 a) llperce
Ifpercent eror is definedas
approxrmale value
r] , roo
0.05 m, and the power dissipation is to be
150Wm when the supplyvoltageis 12V
b) Check your solution and make sure it meets the
design specifications.
Techniques
of
Circuit
Analysis
Sofar, we haveanalyzedretativetysimpleresistivecircuits
4.1 T€rminotogyp. 94
4.2 Introductionto the Node-Vottage
4.3 The Node-VoLtigc
Methodand Dependent
4.4 The Noile-Vottage
Method:SomeSpeciaL
Casesp. ?01
4.5 tntroductionto the l4esh-[urrent
l4ethodp. 105
4.6 Thel'lesh-Current
l4ethodand Depend€nt
4.7 Tte li,lesh-Current
l,lethod:Som€Spe.iat
t.ses p. 109
4.8 Th€Node-Voltage
Merhod
V€rsus
the
M€sh-Current
ethodp. 112
4.9 Source
lransformations
p. 116
4.10Thevenin
andNortonEguivalents
p. 119
4.11 ltlor€on Derivinga Th6venin
EquivaGnt p. 1?3
4.12 i\,laximunPowerTransferp. 126
4.13 Superpositionp. ?29
1 Understaid
andbe abteto us; ihe iode-voltage
nrethod
to soLve
a circuit2 Understand
andbe abiet0 us€the mash-cunent
methodio solvea circuit.
3 Beableto decidewhetherth€node-vottage
methodor the m€sh-cunent
methodis the
preturred
approach
to soLvjng
a partjcutar
circuit.
4 Understand
sourcetransforflatiorandbe abLe
to us€it io sotvea circuit.
5 Und€Etand
the conceptofthe Th6venin
ard
Noton equjvaLent
circujtsafd be abteto
construct
a Theverjnor Nortonequivatent
for a
6 Knowtlr€ conditionfor maximum
powertransfer
to a resistivetoadandbe ableto calcuLate
the
vatueofthe Loadresistorthatsatisfiesthis
g2
by applyingKirchhoff's laws in combinationwith Ohm's law We
can use this approachfor all circuits,but as they becomestructurally more complicatedand involve more and more elements,
this direct methodsoonbecomescumbersome.Inthis chapterwe
introduce two powerful techniquesof circuit analysisthat aid in
the analysis of complex circuit structures: the node-voltage
method and the mesh-curent method.Thesetechniquesgive us
two systematicmethodsof describingcitcuitswith the minimum
Ilumberof simultaneousequations.
In additioll to these two generalanalyticalmethods,in this
chapterwe also discussother techniquesfor simplifyingcircuits.
We have alreadydemonstratedhow to use sedes-parallelreduc
tions aIId A-to-Y transformatjonsto simplify a circuit'sstructure.
We now add sourceffansfor'rnationsand Thdvenin and Norton
equivalentcircuitsto thosetechniques.
We also considertwo other topics that play a role in circuit
analysis.One,maximumpower transfer,considersthe conditions
necessary
to ensurethat the power deliveredto a resistiveload by
a sourceis maximized.Th6venin equivalentcircuits are usedin
establishingthe ma,\imum power transfer conditions.The final
topic in this chapter,superposition,looks at the anatysisol circuitswith more than one independentsource.
Perspective
Practicat
Resistors
Circuits
withReatistic
the effectofimprecise
In the tastchapterwebeganto expLore
on
of a circuit;specificaLty,
resistorvatueson the peformance
are manufac
the perfomanceof a voltagedivider Resistors
vaLues,
andanygiven
turedfor ontya smatlnumberof discrete
witl varyfiom its statedvalue
resistorfroma batchof resistors
wjth tighter toterance,say
within sometoteBnce.Resistors
than resjstors
wrth greatertolerance,
1ol0,
are moreexpensive
it
in a circuitthat usesmanyresjstors,
say 100/".Therefore,
whichresisto/svatuehas
woutdbe importantto understand
perfornrance
ofthe circljt.
on the expected
the grcatestimpact
In otherwords,we wouldtiketo predjctthe effectof varying
vatueon the outputof the circuit.If we know
eachresistor's
resistormustbe veryctoseto its statedvatue
that a palticul"ar
for the crrcuitto functioncofiectly,we can then decldeto
to achieve
a tr'ghtertoLerance
spendth€ €xtramoneynecessary
on that resisto/svatue.
valueon the
ExpLorirg
the effectof a circuitcomponenfs
0ncewe have
cifcujfsoutputisknownassensitivityanaLysis.
the topic of
presented
technjques,
additionatcircuitanatysis
wL[be examjned.
sensitivjtyanatysis
93
94
ofCncuitAnatysis
Techniques
4.1 Ternninology
To discussthe more involvedmethodsof circuit analysis,we must define
a few basic terms. So far. all the circuits presentedhave been planar
circuits-that is, those circuits that can be drawn on a plane with no
oossing branches.A circuit that is drawn witb crossingbranchesstill is
consideredplanar if it can be redrawn with no crossoverbranches.For
example,the circuit shown in Fig. 4.1(a) can be redrawn as Fig 41(b);
the circuits are equivalentbecauseall the node connectionshave been
maintained.Therefore,Fig.4.1(a) is a plarar circuit becauseit can be
redmwn as one. Figure 4.2 shows a nonplanar circuit-it cannot be
redra$n in sucha way that all the node connectionsare maintainedand
no branchesoverlap.The node-voltagemethodis applicableto both pla_
nai and nonplanarcircuits,whereasthe mesh-currentmethod is limited
to planar circuits.
(b)Ihesame
cncuii
rigurc4.1A (a)A planar
cncuit.
rednwn
to venry
thatit is ptanar
RI
R2
RS
Fiqlre 4.2 d A nonplanarcircuit.
Describing
a Circuit-Thevocabulary
In Section1.5we defined an ideal basiccircuit element.Whenbasiccircuit elementsare interconnectedto form a circuit, ihe resulting interconnectionis describedin terms of nodes,paths,branches,loops,and
meshes.We defined both a node and a closed path, or loop, jn
Section2.4.Here we restalethosedefinitions and then define the terms
ail of thesedefinitionsare
path,blanch, ardmesh.Foi your convenience,
presentedin Table 4.1.Table 4.1 also includesexamplesof eachdefinition taken fromthe circuit in Fig.4.3(seepage95.).whichare developed
in Example4.1.
Exanple Fron Fig. (3
A point wheretwo or more circuit elenentsjoin
A nodeehere three or more circuitelemenrsjoin
path
;
A traceof adjoiningbasicelementsvilh no
elementsincludedmore than once
A path that connectst\vo nodes
A path {hich connectstwo esential nodeswithout
pasiDg throughan essentialnode
Apath whoselastnodeis the sameasthestartingnode
Aloop that doesnotencloseany olher loops
A circuir that can be drawn on a plane lvith Do
,r Rr
2r-Rr
R5 R6- Ra ,?
Rj R,
?r-Rr-nj
Fis.4.3is aplanarcncuit
circuit
Fig.4.2is a nonplanar
4.1 Teminotogy 95
IdentifyingNode,Eranch,MeshandLoopin a circuit
For the circuit in Fig 4.3,identify
a) a[ nodes.
b) all essentialnodes.
c) all brarches.
d) all ess€trtialbranches.
e) all mesh€s.
0 two paths that are not loop6 or essentialbranches.
g) two loops that are trot meshes.
Sotution
a) The nodes are a, b, c, d, e, f, and g.
b) The essentialnodesare b, c, e,and g.
c) The branchesare ?J1,tb, Rt, R2,R3, R4,R5, R6,
R?,and 1.
d) The essential branches are
Rz - Rl
2,2 R4,R5,R6,R7,and I
e) The meshes are \ - R1- R5 - R3 - R2,
u 2 R 2 R 3- R 6 - R 4 , R 5 - R 7 - R 6 , a n d
& r .
2
)
R,
R z d R 3
",(
\ )
jttustnting
nodes,
bmnches,
nreshes,
Figure4.3 r A circujt
D R1 ns R6 is a path, but it is not a loop
(becauseit does not have the same starting and
ending nodes), nor is it an essentialbmnch
(becauseit doesnot connect two essentialnodes),
o, - R2 is also a path but is neither a loop nor an
essentialbranch,for the samereasons.
g) ?J1 Rl
R5 - Re - Rr - ?'2 is a loop but is
not a mesh,becausethere are two loops wilhin it.
/
R5 Rois alsoa loop but ool a mesh.
NOTE: Assessyour understandingof this motelial bt ttting Chopter Prcblettus4.2 snd 4.3
Equations-HowMany?
Simultaneous
The number of unktrown currents in a circuit equals the number of
blaltrche$ ,, where the current is not known. For exadple, the circuit
showtr itr Fig. 4.3 has nitre branches in which the cu[ent is uoknown.
Recal that we must have D itrdependent equations to solve a circuit with
b unknown currenls If we bt ,r rcpresent the number of nodesin the circuit,
we can derive ,l
1 independent equations by applying Kirchhoffs current law to any set of'' - 1 nodes.(Application of the current law to the
,rth node doesnot generatear iDdependentequation,becausethis equation can be derivedfrom the previousrl - 1 equation$SeeProblem4,5.)
Becausewe need b equationsto describea giveDcircuit and becausewe
can obtain n - 1 of these equations from Kirchhoff's current law, we must
apply Kirchhoff'svoltagelaw to loops or meshesto obtain the remaining
b (/r
l) equations.
Thus by counting nodeE meshes,and brancheswhere the cuuent
is unknoM, we have establisheda systematicmethod for writing the
necessarynumber of equationsto solve a circuit. Specifically,we apply
1 nod€s and Kirchhoff's voltase law to
Kirchhoff's curent law to n
96
Techniques
of CncuitAnatysh
b - (n - 1) loops (or neshes). These obseryations also are valid in terms
of essential nodes and essential branches.Thus iI we lel ne represent the
number of essentialnodes atrd 4 the number of essential brancheswhere
the current js unlno*n, we car apply Kirchhoffs current law at n. - I
oodes and Kirchhoffs voltage law around 4 - (n - 1) loops or meshes.
In circuits, the Dumber oI essential nodes is less than or eoual to the number of nodeq and the number of essential bmnches is less than or eoual to
the number ot branches.
Th us it is ofletr convenient to useessentialnodes
and essential branches when anallzing a circuit, because they produce
fewer independent equations to solve.
A circuit may consist of disconnectedparts.An example of such a circuit is examinedin Problem4.3.The statementspe ainingto the number
of equationsthat can be derivedfrom Kirchhoffs curent law,r - 1, and
(n - l), apply to connectedcircuits.If a circuit has
voftage laqwib
,?nodesand b branchesand is made up of r parts,the current law can be
appliedn s times,and the voltagelaw 6 - /, + r times.Anytwo separate partscanbe connectedby a singleconductor.This connectionalways
causestwo nodesto fofin one node,Moreover.nocu[ent existsin the single conductor,so any circuit made up of s disconnectedparts can always
be reducedto a connectedcircuit.
TheSystematic
Approach-AnIllustration
/-a
R5
)
R z a R :
r
i
R,c
:
' r ! R6
\
i
J
n4
f
We now illustmte this systematic approach by using the circuit shown in
Fig. 4.4. We write the equations on the basis of essential nodes and
branches.
Th€ circuit hasfour essentialnodesand six essentialbranches,
denoted ir - ,;, for which the current is unhown.
We derive thrce of the six simultaneous equations needed by applying
Kirchloffs current law to any three of the four essentialnodes.We use the
nodesb,c,and e to get
-
Fi$re 4.4a ne circuitshown
in Fig.4.3withrix
unknou,n
bEnchcunents
defined.
it+i2+i6-I:0,
ir-ir
i5
4 + i4
i2 = 0.
0.
(4.1)
We derive the remaining ttrree equations by applying Kirchhoff's voltage
law around three meshes.Becausethe circuit has four meshes,we need to
dismiss_one
mesh.Wechooseri? 1, becausewe don't know the voltage
across1,,
Using the otler threemeshesgives
R1il + R5i, + 6(R, + Rt - 1,r : 0,
-i3(n, + Rl) + iaR6+ i5na
t2 = 0,
-irR5+i6R7-loR6=Q.
' W. saymole aboutthisdcchionin Section4.7.
(4.2)
a.2
hiroduction
to theNod*VoLtage
Method
91
RearrangiDgEqs.4.l
and4.2ro tacilitare
lheir.olulionyieldslheser
-ir + i + 0t3+ 0i4+ 0i5+ t6 = 1,
i1 + Oi2 i) + ola i5 + 0t6:0,
qir- i2+ h + i4 + 0i5 + 0i6 = 0,
R1i1+R5i2+(Rr+R3)t1 + ota+ ots+ 0t6= r,1,
0i1+ 0i' - (R' + R3)i3+ R6i4+ Rai5+ 0i6= or,
0i1
R5i2 + 0i3
R6ia + Ois + R?i6 = 0.
\4.3)
Note that summing the curent at t}le nth node (g in this example) gives
is
ia
i6+I:0.
(4.4)
Equation 4.4 is not independent, because we can deiive it by summhg
Eqs.4.1and tlen multiplyingtle sum by -1. Thus Eq.4.4 is a linear combination of Eqs. 4.1 and theiefore is not itrdependent of them. We now
cally the procedureone stepfurther.By introducingnew vadables,we can
des€ribea circuit with just,1 - 1 equationsor jusr b - (n - 1) equations.
Therefore these new variables allow us to obtain a solurion by manipula!
ing fewer equationq a desirable goal even if a computer is to be used to
obtain a nume cal solution.
The new variables are known as rode voltages and mesh curents.The
rode-voltage method enables us to describe a circuit in tems of r?
1
equations;tle mesh-currentmethod enablesus to desffibe a circuit in
tems of be (na 1) equations.We begin in Section 4.2 $dth the nodevoltagemethod.
NOTE: Assessyow unde$tunlling of this mateflil by trying Chaptel
Prcblens 1.1 an.l 1.4
4.2 Introductionto the
Node-Vottage
Method
We introducethe node-voltagemetlod by usingthe essentialnodesof the
circuit. The first step is to make a neat layout of the circuit so that ro
branchesclossover and to mark clearlythe essentialnodeson the circuit
dia$am, as in Fig.4.5. This circuit has three essentialnodes (/,": 3);
therefore, we need two (r?" - 1) node-voltage equations to descdbe the
circuit. fhe next step is to select one of the three essential nodes as a rcferencenode. Although theoretically the choice is arbihary, pmctically the
choice for the reference node often is obvious.For examDle.the Dodewith
10v
Figlre 4.5 A A ciruit usedto ittustntethe node-vottage
method
of circujtanatysk.
98
Techniques
of CncLrit
AnaLysis
1f,)
^
lq
jn Fig.4.5witha
Figure4.6 A Thecncuitshown
rcference
nodeandthe nodevoltages.
10v
Figure
4.7A Computation
ofthebranch
currcnt
t.
the most branchesis usually a good choice.The optimum choice of the reference node (if one exists) wi become apparent after you have gained
someexpefience
usinglhi\ method.In rhe circuirsboqDiD fig.4.5.rbe
lo\rernodeconnect5
rhemoslbranchelso $e useir asLhereference
node.
We flag the chosenreferencenode with the symbolV,as in Fig.4.6.
Atter selectingthe referencenode,we definethe nodevoltageson the
circuit diagram. A nod€ volfage is defined as the voltage rise from the reference node to a nonreference node, For tlis circuit, we must define two
node voltages,which are denotedol and 2'2in Fig-4.6.
We are now ready to generate the node-voltage equations.We do so by
first writing the cufient leaving each branch connected to a nonreference
node as a function of the node voltagesand then summinb thesecunents to
zem in accordancewith Kirchhoff's current law. For the circuit in Fis. 4.6.
lhe currenl awayfrom node I lhroughthe I O res;\rori( lhe rohag;rop
acrosst]le resistor divided by the resistance(Obm's law). The voltage drop
acrossthe resistor, in the direction of the curent away hom the node, is
r)r - 10.Thereforet]le cunent in the 1 J) resistoris (o1 10)/1.Figure4.7
depictstheseobsenations.Itshowsthe 10 V-1 O branch,with the appropriate voltages and culrent.
This samereaso ng yields the curent in every branch where the current is unknown.Thus the current away ftom node 1 thtough the 5 O
resistoriso/5, and the cunent awayfrom node l thrcughthe 2 O resistor
is (or ?)r)/2.The sum of the three curents leavingnode 1 must equal
zero;thereforethe rode-voltageequationderivedat node 1is
r'r - 10
1
r'1
5
2
'
14.5)
The node-voltageequationderivedat node 2 is
2
1
0
\4.6)
Note that the first terminEq.4.6 is tie current awayfrom node 2 through
the 2 O rcsistor, the second term is the current away ftom node 2 tlrough
the 10 O resistor,ard the third term is the cu[ent away from node 2
through the current source,
Equations 4.5 and 4.6 are t]te two simultaneous equations that
descdbethe circuit shownin Fig.4.6in tems of the node voltagesor and
?)2. Solving for r'1 and ?,l2yields
.l-
r_
r,-
10n
-ll
1 t-o
tl
= 9.09V
- 1 0 . 9 v1.
Oncethe node voltagesare knowi\ all the branchcurents canbe calculated.Once theseare knoM, the branchvoltagesand powen can be
calculated.Example4.2illustratesthe useofthe node-voltagemethod.
4.2
llethod
to the NodeVottage
Introduction
99
Method
Usingthe Node-Voltage
a) Use the node-vollagemethod of circuit analysis
to fhd the branchcurrentsi., ib,and ic in ihe cir
cuit shownin Fi9.4.8.
with eachsource,and
b) Find the powcr associated
statewhetherthe sourceis deliveringor absorbing power.
5o
50v
1 ) ' | , n n , , $ r o( ot
42
Figur€4.8 .i, ThecircuitforEranrph
SoIution
a) We beginby notingthatthe circuithastwo essential nodes;thirs we need to write a singlenodeWe selectthe lower node as
voltageexpression.
the rcference node and define the unknown node
voltage as r,1.Figure 4.9 illustratesthese deci'
sions.Surming the currenls away fiom node I
generatestlle node-vohageequation
50
0,-
s
,1
+i]+i
1
0
0l
4
-3:0
0
l-./Tl'<po\re' a*uciatcdt\irh Ih. <U\ \ourccis
-100w (delivedng).
15ov= 50ta=
Thepo$er a..ocialcduillr Ihe r A .oLfceic
gives
Solvingfor ?J1
= ) A
; : ' '
'"
in Fig.48 witha rcference
shown
figure4.9 r. Thecircuit
or.
node
voliage
nodeandtheunknown
10
40
' 4 0
p3A=
3?,r:
3(40)=
120w (deh'ering).
We check thesecalculationsby noiing that the
total deliveredpower is 220W.The total power
absorbedby the three resisrorsis 4(5) + 16(10)
+ 1(40), or 220W, as we calculalcd and as it
method
andb€ableto usethe node-voltag€
1-Understand
Obiective
l)
For the circuit sho\ln, use the node-voltage
methodto lind r,1,r,2,ancli1.
How much power is deliveredto the cfcuit
by the 15A source'l
c) Repeat(b) for the5 A source.
Answ€r:(a) 60V 10Y 10A;
(b)eoow;
(c) s0w.
5r}
FL
--
)rro, |uon *"n
I
+
}rn , (l
100
Iechniqu€5
ofCncLrit
Anay5is
4.3 TheNode-Voltage
Method
andDependent
Sources
Ifthe circuitcontahs dependentsources,
the node-voltageequationsmust
be supplementedwith the constraintequationsimposedby the presence
of the dependentsources.Example ,1.3illustratesthe applicationof the
node-voltage method to a circuit containing a dependent source.
Usingthe Node-Voltage
Methodwith Dependent
Sources
Use the node-voltage metiod to find the power dis(ipatedin lhe 5 Q resisrorin lhe circuilsho$n in
Fig.4.10.
5f)
+
20o-
8b
10(l
Summing the curents away from node 2 yields
01- x1
u2
Ir2- 8lp
^
As \raritten,these two node-voltage equations con,
tain tluee unknowns, namely, 01,?r2,and id. To elim
inate id we must expressthis controlling current in
terms of the node voltages,or
Figure4.10A Thecircuit
forExampte
4.3.
Substitutingthis relationshipinto the node2 equation simplifies
thetwonode-voltage
equations
to
SoLution
We begin by noting t]Iat the circuit has three essential nodes,Hence we need two node-voltage equations to describethe circuit. Four branchestemtnare
on the lower node, so we select it as the rcfcrcrce
node. The two unknown node voltages are defined
on t}te circuit sho$,nin Fig. 4.11. Summing the currents away ftom node 1 generatesthe equation
o-, ' 2 0
o,
+ j +
2
2
s
"
br+16u2=0.
Solving for ?J1atd zJ2gives
q:16v
and
u-, - t )' , = n
0
0.75u1-0.2u2- 10,
'
t'2 = 10V.
4.4
l,lethod:
Some
speciatcases 101
TheNodeVotiage
2{)
16
51)
1
2{t
10
lsr) : (1.14)(5): 7.2w.
A good exercise10build your problen solving
intuition is to reconsiderthis example,usingnodc 2
as the rcferencenode. Does it make the analysis
tigure 4.11 ;1lThecircuitshownin Fig.4.10,\i/itha refelen.e
nodeandthe nodevoLtaqes.
method
objedive l-ljnderstand and be able to usethe node-vottage
4.3
a) Usc the node-voltagemcthod to find the
power associated
with eachsourcein ihe
3it
6f)
b) Statewhetherthe sourceis delive ng power
2A
to the circuitor extractingpower from the
Answen(a)p5ov: 150W,11= -144W.
zre = -80 w;
50v
80
4 ' , (
(b) all sourcesare deliveing power to the
clrcurt.
NOTE: Also try ChaptetProbkms 1.19and 424.
f4ctltsd:
4.4 TheFlodc-tfo{tage
5*rn*Special{ias*s
I i0{] 2
When a voltagesourccis the onll' elemenlbctweentwo essentialnodcs,
the node-voltagemethod is simplified.As an example,look at the circuit
in Fig. 4-12.There are lhree cssentialnodesin this circuit. which means
100v
that two simultaneousequationsare needed.From thesethree essential
nodes,a rcferencenode hasbccn chosenard two olher nodeshave been
latreled.But the 100V sourceconstrainsthe voltagebetwcennode 1 aDd
the retercncenode to 100V This meansrhat there is only one unknown
nodc voltage(rr). Solutionoflhis circuitthus involvesonl]' a singlenode figurc4.123 A circuit\vitha knownnodevottaqe.
vollageequationat node2i
,
10
2
.
,
50
,
14.7)
But rr = 100V. so Eq.4.7 canbe solvedlbr tr.
1)2= 125V.
(4.8)
Knowing riz,we cancalculatethe curent in everybranch.Youshouldverify that the current into node 1 in lbc branchcontainingthe independent
voltagesourceis 1.5A.
102
Iechniques
of CircuitAnatysis
Figufe4.13A A cjrcuitwitha dependent
votiage
source
connected
betw€en
nodes.
In generul,when you use the node-voltagemethod to solvecircuits
that havevoltagesourcesconnecteddirccrly betweenessentialnodeqthe
number of unknown node vollages is reduced. The reason is that, whenever a voltage sourceconnectstwo essentialnodes.it constraiN tle differencebetweenthe rode voltagesat thesenodesto equalthe voltag€of the
source.Taking the time to seeif you can ieduce the number of unknowns
in this way will simplily circuit analysis.
Supposethat the circuit shownit Fig.4.13is to be analyzedusingthe
node-voltagemethod.The circuit containsfour essertial nodes.so we
anticipatewriting three flode-voltageequations.However,two essential
nocles arc connected by an independent voltage source, and two other
essentialnodesare connectedby a current-controlleddependentvoltage
source,Hence,there actuallyis only one unknownnode voltage,
Chooring$hich node lo useas rhe reterencenode in\olvesseveral
possibililies.Either node on each side of the deDendentvoltase source
look. atbacrivebecause.
iJ cho<en.
oDe of lhe ;ode volldger;ould be
known to be eitler +10id oeft node is rhe rcference) or 10td(righr node
is the reference)The lower node looksevenbetter becauseone nodevoltage is irnmediately known (50 V) aIId five branches terminate there. We
tierefore opt for the lower node as the reference.
Figure 4.14showsthe redrawn circuit, with tie reference node flagged
and the rode voltages defined. Also, we intrcduce the current i becausewe
cannot express the cufent in the dependent voltage souce branch as a
functior of the node voltageso2and 2)3.Thus,at node 2
5
jn fig.4.13.wiihthe
rigure4.14a Thecncuit
shown
setected
node
voliages
defined.
5
0
'
"
14.e)
'
aIId at node 3
fr-; +:o
(4.10)
We eliminatei simplyby addingEqs.4.9and 4.10ro get
I)
Ur
5
i
'I)1
5(J too
-
"'
(4.1L)
TheConcept
of a Supernode
4A
Figure4.15A Consideing
nodes
2 and3 to bea
Equation 4.11 may be written directly, without resorting to the intermediaie steprepresented
by Eqs.4.9and4.10.Todo so,we considernodes2 and
3 10 be a single node and simply sum the currents away from the node in
terms of the node voltages02 and ?J3.
Figure 4.15illustrates this approach.
When a voltage sourceis betweentwo essentialnodes.we can combine
those nodes to form a supenode. Obviously, Kirchhoffs current law must
hold for the supemode.Ir Fig.4.15,srartingwirh rhe 5 O bmnchand moving counterclockwise
aroundthe supernode,wegeneratethe equation
b) -
5
Di
'01
50
u1
100
-
"'
(4.r2)
SpecjaL
Cases 103
l'4ethod:
Some
4.4 TheNode_Vottage
which is identical to Eq. 4.11. Creating a supemode at nodes 2 and 3 has
madethe task of analyzingthis circuit easier'It is tlerefore alwaysworth taking the time to look for this tlpe of shortcut before writing any equations.
After Eq.4.12hasbeendedved,thenext stepis to reducethe expression to a single unknown node voltage. FiISt we eliminate ,1 from the
equation becausewe know that ,r = 50 V. Next we expressD3as a func_
tion of 02:
,3=12+10i0.
(4.13)
We now expressthe current controlling the dependent voltage source as a
functionof the node voltases:
u2 50
14.14)
UsingEq$4.13and4.14andor = 50VreducesEq4.12to
( 1
, , \ s o I- 5 -# . # )
:10+4+1,
r2(025)= Is,
FromEqs.4.13
and4.14:
60
50
5
24,
in
circuitshown
Figsre4,16A Thetnnsistoramptifier
69.2.?4.
?r3:60+20=80V.
Analysisof the AmplifierCircuit
Node-Voltage
Let's use tlle node-voltagemethod to analyzethe circuit ftst introduced
in Section2.5and shownagah in Fig.4 16
wlten \le used the branch-current metlod of analysis in Section 2.5,
we faced the task oI writing and solving six simultaneous equations Herc
we will show how nodal analysiscan simplify our task.
The ciicuit hasfour essentialnodes:Nodes a and d are connectedby
an independentvoltagesourceas are nodesb and c Thereforethe prcblem reduces to finding a single unknown node voltage, because
(n" 1) - 2 = 1 Using d as the referencenode,combinenodesb and c
the voltagedrop acrossR2astb, andlabel the volf
hto a supernode,label
agedrop acrossRE asoc,asshowllinFig.4.17 Then,
*
,,fn
uo
=u" r*-Bi":0.
,'-, '-'
,-\
f i g u r e4 . 1 7A l h ei r c u ' t+ o w r : r F 9 . 4 . 1 6w. i r h
rdentified
vottages
andthe lupernode
104
Techniques
of Cncuit
Analysis
We now eliminateboth tc and iB from Eq.4.15by noting rhar
.,.
(in
Brr)Rr.
\4.16)
'Dc:n6-Ur'
14.17)
Substituting
Eqs.4.16
and4.17inroEq.4.15yields
- l
R
' r
R
.
rl
_
B ) R ,I
l
R
- n o
rI I P,R/
'-'
SolvingEq.4.18for ob yields
yc.n (1 + B)Rr + %R1R,
RrR2
+ ( 1+ B ) R E ( R I + R )
(4.1e)
Using the node-voltagemethod to analyzethis c cuit reducesthe prob
lem bom manipulatingsix simultaneousequationsGee Problem2.27)ro
manipulatingthree simultaneousequations.You shouldverify rhat,when
Eq.4.19is combinedwith Eqs.4.16and 4.17,the solutionfor is is identical
to Eq.2.25.(SeeProblem4.30.)
Objective
l-t
ndectandandbeabt€to usethe node-voltag€
method
4.4
U s e t b e n o d e - v o l l a g em e t h o d l o l i n d , . i n r h e
cifcuil .bo\ n.
4,5
U(e (henocle-voltage
merhodlo find , in lhe
circuitsbo$D.
Answer:8V
Use lhe node{ollaqe melhod to find L, in lh,
NOTE: Also try ChapterProbkms 4.21,426,and 1.27.
30r)
4.5
llethod
Introduction
to the Mesh-Curent
105
4.5 Introductionto the
Method
Mesh-Current
As stated in Section 4.1, t}]e mesh-currentmethod of circuit analysis
(/,e 1) equations.Recall
enablesus to describe a circuit in terms of r"
tlat a mesh is a loop with no otler loops inside it. The circuit in Fig.4.1(b)
t+
is shown agai in Fig. 4.18,with cu ert arrows inside each loop to distinguish it. Recall also that the mesh-curert method is applicable only to
planar circuits. The circuit in Fig. 4.18 contahs seven essential branches
where the current is unknown and four essentialnodes,Therefore, to solve
(4
1)] mesh- Figue4.18A Thecjrcujtshown
in Fjg.1.1(b),
withthe
it via the mesh-curent method, we must *.dte four [7
A mesh cunent is the current that exists only il the penmeter of a
mesh. On a circuit diagram it appears as eitier a closed solid line or an
almost-closed solid line that follows tlte pedmeter of the apprcpriate
mesh.An arrowhead on the solid line indicates the reference direction for
the mesh current. Figuie 4.18 shows the four mesh ctments that describe
tle circuir in Fig. 4.1(b). Note that by definition, mesh currents automatically satis{y Kirchhoff's curent law. That is, at any node in the circuit, a
given mesh current both enten and leaves the node.
Figure 4.18 also shows that identifying a mesh current il terms of a
bmnch current is not always possible. For example,the mesh curent t2 is
not equal to any branch cutent, whereasmesh currents i1, 4, and i4 can be
identified with branch currenls. Thus measudng a mesh curent is not
always possible; note that tlere is no place where an ammeter can be
insefted to measure t]]e mesh current i2. The fact that a mesh current can
be a fictitious quantity doesn't mean that it is a uselessconcept. On the
contrary, the mesh-curent mefiod of circuit analysis evolves quite naturally ftom the bmnch-current equations.
We can usethe circuit in Fig.4.19to showthe evolutionof the mesh_
current technique.We begin by using the branch currents (ir, i2, and i3) to
formulate the set of independent equations. For this circuit, 4 = 3 and
}?, = 2, We can write only one independentcurent equation,so we need
two independent voltage equations Applying Kirchhoff's curent law to
the upper node and Kirchhoff's voltage law around the two meshesgeneidevetopnrent
to itLusirate
tiglre4,19A A circuiiused
ates the followins set of equations:
analysis.
method
of circuit
0fthemesh-cunent
(1.20)
iI: i2 + i3,
?)r = i1R1 + i3R3,
-D2 = izR2
(4.21)
\4.22)
iR}
We rcduce this set of three equations to a set of two equations by solving
Eq.4.20for i3 and then substitutingthis expressioninto Eqs.4.21and4.22:
?,r = ir(R1 + R3)
-r2:
4nj,
i7R3+ tr(R, + R3).
\4.23)
(4.24)
we can solve Eqs 4.23 and 4.24 for i1 and t, to replace the solution of
three simultaneous equations \.ith the solution of two simultaneous equa1 curent
tions.We derived Eqs.4.23 and 4.24 by substitutingthe 'l"
106
Techniques
of CncuitAnaLysis
equations into the 6d (,?" - 1) voltage equations.Thevalue of t}le meshcurrent method is that, by defining mesh curents, we automatically eliminate the ,1e- 1 curent equations. Thus the mesh-curent method is
equivalent to a systematicsubstitution of the /,a 1 curent equations into
the b" - (n" 1) voltageequations.The
meshclrllentsir Fig.4.19that are
equivalent to eliminating the branch current i3 frolxl Eqs.4.21 al:td4.22 arc
shownin Fig.4.20.We now apply Kirchhoff'svoltagelaw aroundthe two
mesheq expresshg all voltages acrossrcsistois ilr tems of the mesh cuirents,to get the equations
?r1: ianl + (ia tb)R3,
-?,, - (rb tJn3 + i6R .
tigure4.20A Me5h
cuirents
ia andtb.
\4.25)
(4.26)
Collecting tlle coefficienls of ir and ib in Eqs.4.25 and 4.26 gives
u1 = ia(n1 + R3) ibR3,
-1tz:
ia& + tb(n, + Rr).
\4.?7)
(4.28)
Note ttrat Eqs.4-27and 4.28and Eqs.4.23and 4.24are identicalin form,
wilh the mesh currentsia and ib replacingthe branch curents 4 and i2.
Note alsothat the branchcurrentsshownin Fig.4.19can be expressedin
termsofthe meshcurents shownin Fig.4.20,or
\4.29)
(1.30)
(4.31)
The ability to write Eqs.4.294.31 by inspectionis crucial to the meshcurent methodof circuit analysis.Onceyou know the meshcurrents,you
also know the branchculrenls.And once you know the bmnch currents,
you cancomputeanyvoltagesor powersolinterest.
Example4.4 illustrateshow the mesh-currentmethod is usedto find
souce Dowersand a branchvoltape.
Usingthe Mesh-Cunent
Method
a) Use the mesh-currentmethod to determinethe
power associated
with eachvollagesourcein the
circuitshownin Fig.4.21.
b) Calculatethe voltageo, acrossthe 8 f) rcsistor.
SoLution
a) To calculate the power associatedwitl each
so[rcej we need to know the culent in each
source. The circuit indicates that these source
curents will be identicalto meshcurrents.Also,
note that the circuit has sevenbrancheswhere
Figure4.21A Thecjrcujtfor
Lyampt€
4.4.
the current is unknown and five nodes.Therefore
(,? 1) : 7 - (5 i)]
we need three [b
mesh current equations to descdbe the circuit.
Figure 4.22showsthe three mesh currentsused
to descdbethe ctucuitin Fig. 4.21.If we assume
4.6
llre l,4esh-Cutrent
l4ethod
andDependent
sources 107
that the voltage drops are positive,the three mesh
+\
,!
, 8 a ? ',
7'
I
4 0 + 2 t a+ 8 ( i . t b ) : 0 ,
+6ir+
ll('r,
lb) + 4i. + 20 = 0.
6(,.
(4.32)
Your calculatorcan probably solvetheseequations, or you can use a computer tool. Cramer's
method is a useful tool when solvirrg tlree or
more simultaneous equatioDsby hald. You cdr
review this important tool in Appendi\ A.
ReofgJntingtqs.4.32 in anriciparion
of using
your calculator,a computer program, or Cramer's
methodgives
10i, - Stb+ 0i. =
8i. + 20ib
oir
)
/+
? 6 ! 'r \
I
figure4.22,\ Thethr€€mesh
to anatyze
the
cu enisused
circuit
shown
in Fi9.4.21.
Thc mcshcnrrent i" is identicalwith the branch
'n
crrrrcnr rl'c 4rr\ source..orheposer ds\ociatcd with this sourceis
prov = -40i, : -224 W.
The minussignmcansthat this sourceis delivering power to the nelwork. Thc currcnt in the
20 V sourceis identical 10 thc mcsh current ic:
6;c =
6lb + 101.=
.io
60
2ll
prov - 20i" = -16 W.
(4.rr)
The 20 V soulcealso is delivcringpower to the
The threemeshcurrentsare
b) The branch curfent in the 8 () resistor in the
direction of the voltage drop 0, is i" - tb.
ib:20A,
r. :
0.80A.
,,:
8(1"
lb)
objective2-llnderstand and be able to usethe mesh-current
method
4.7
Use lhe mesh-currentmethodto find (a) the
power deliveredby the 80V sourceto the circuit shownand (b) the power dissipaledin the
8 O resistor.
Answer:(a)400W;
(b)50w
30(:}
80v
NOTE: Abo try Chaptet Problens 4.31 snd 1.32.
Stethed
4.6 TiieMesh-eurrent
amdmependent
So|'srees
II lhe circuilconlainsdependenlsources.the nesh-currenlequalionslnusl
be supplenenledby lhe approprialeconslraiil equalions.Exanple 4.5
illustralesthe applicationof the mesh curenl method when the circuit
includesa dependenlsource,
ti - 28.8V.
108
lechniques
of CircuitAnatysis
Usingthe Mesh-Current
Methodwith Dependent
Sources
Use lhe mesh-curentmethod of c cuit analysisto
determinethe power dissipatedin the 4 O resistor
in the circuitshowninFig.4.23.
15 t6
Figure4.23 A Thecjtcujtfor Exampte
4.5.
Solution
This circuit has six branches where the current is
unknown and four nodes.Thereforewe need three
mesh curents to describe the circuit. They are
delined or the circuit shownin Fig.4.24.Thethree
meshcurent equationsare
so
l')
qo
''';),,
15 i,L
50: 5(i1- r') + 20(rr ir),
0 = 5(i,
0 = 20(b
Figure4.24 a, Thecjrcuitshown
in Fig.4.23
withthethrce
11)+ 1i2+ 4(iz - h),
Q + 4(i
i) + 15i/".\4.34)
We now er?ress the branch currcnt controlling the
dependentvoltage source in terms of the mesh
Becausewe are calcuiatingthe power dissipatedin
the 4 O resistor,we computethe mesh currentsi2
and i3:
i2=26A,
iO: iI
i3,
which is the supplementaleq ation imposedby the
presenceof the dependent source. Substituting
Eq. 4.35 into Eqs. 4.34 and collectingthe coefficientsofi1, i, and i3 in eachequationgenerates
50:25ir-5i2-20h,
0:
5i1+ 10i2 4i3,
0:
54
4iz+9h.
j3:28A
11.35)
The curent in the 4 f) resistor o ented from left
to right is i3 - i2 , or 2 A. Therefore the power
dissipatedis
p4n= (4
i2)2(4:): (2)'(4) = 16 w.
What if you had not been told to use the meshcurrent method? Would you have chosentlre nodevoltage method? It reduces tlre problem to finding
one u*nown nodevoltagebecauseof the presence
oftwo voltagesourcesbetweenessentialnodes.We
presentmore about makingsuchchoiceslater.
4 . 7 T h eI l e s h - C L r rMr eent ht oS
do
: mS
e p e c i a t i a s e s109
objective2-lJnderstand
andb€ableto usethe mesh-current
method
4,8
a) Determinethe numberofmesh-cunent
equationsneededto soh'ethe circult shown.
Answert (a) 3;
(b) 36]V.
b) Use the mesh-currentmethodto find how
much power is being delivered to t}le
vollJgesource.
dependenl
4,9
Usethemesh-current
methodto lind ?r,in lhe
circuit sho$n.
Answer: 16V
NOTE: Also try ChapterPnblems 4.37dnd 4.38.
4.7 TheMesh-eurrent
Method:
Son"re
SpeeiaI
eases
l0 f)
W})ena branchincludesa currentsource,thcmcsh currcntmethodrcquircs
Thc circuit shownin Fig. 4.25depictsthe
somc additionalmanjpulations.
natureof the problem.
We havedefinedthe meshcurrenlsia,ib, and ic.as well as the voltage
acrossthe 5 A curent sowce,to aid the discussion.
Nole thal the circuil
conlainslive essentialbrancheslvhere the cur'renlis uDknownand four
l00v
essenlialnodes.Hence we need 10write lwo 15 (4 1)] nesh'current
equationsto soivethe circuit.Thepresenceof the curreni sourcereduces
the three unknown mesh cufents to tlvo such currents.becauseit corjttustrating
4.25a A circujt
nresh
anatysis
when
strainsrhe differencebetweeni" and t. to equal5 A. Hence,if we know 1.. Figure
a
branch
contains
an
independent
cLrrrent
source.
we krow i., and vice versa.
However,whenwe attemptto sum the voltagesaroundeitler mesha
or rneshc- we must inlr'oduceinlo the equationsthe unknown voltage
acrossthe 5 A currentsoulce,fhui for rnesha:
r00=3(i"-lb)+t,+6ia,
50=4ic
1]+2(i.
ib)
(4.36)
\4.37)
110
Techniquesof
Ci'cuitAnaLysis
We now add Eqs.4.36and 4.37to eliminate? and obtaitr
50=9i"-5ib+6r..
(4.38)
Summingvoltagesaroundmeshb gves
0 = 3(ib i") + 10rb+ 2(ib - iJ.
(4.3e)
We rcduceEqs.4.38and4.39to two equationsand two unknownsby using
the constraint that
(4.40)
We leave to you tlle verification that, when Eq. 4.210is combined witl
Eqs.4.38 and 4.39,the solutions for the tbree mesh currents a-re
h = 1.75A,
ib : 1.254.,
and
t" = 6.75A
TheConcept
of a Supermesh
10(}
l:t)
100v
60
50v
4lL
shown
in Fiq.4.25,
iltustrattigure4.25,| Thecircuit
ingtheconc€pt
of a supemesh.
We can derive Eq. 4.38 without introducing tle unknown voltage o by
using the concept of a supemesh. To qeate a supermesh, we mentally
rsmove the current sourceftom the circuit by simply avoiding tlis branch
when lTiting the mesh-crrrent equations.We expressthe voltages around
the supermesh in terms of ttre original mesh currents. Figure 4.26 illusfiates tho supermesh concept. When we sum the voltages around t]le
supemesh (denoted by the dashedline), we obtair the equation
-100 + 3(i. - ib) + 2(tc
,b) + 50+4i"+
6ia=0,
(4.4\)
which reduces to
50=9ia-sib+66.
o, i,
a'f,,)
\4.42)
Note that Eqs.4.42 and 4.38 are identical. Thus the supemesh has eliminated the need for introducing the unknown voltage aooss the curent
sourc€.once agaio, taking time to look carefully at a circuit to identify a
shortcut such as this provides a big payoff in sirnplifyhg t]le analysis.
Mesh-Cunent
Analysisof the AmptifierCircuit
in Fig.2.24with
the
tigure4,27a Thecircujtshown
mesh
cun€nts
i", ib, andt".
We canusethe circuitfirst introducedin Section2.5 (Fig.2.24)to illustnte
how the mesh-cu ent method works when a bmnch contains a dependent
current source.Figure 4.27showsthat crrcuit, with the three mesh currents
denoted t,, ib, and i". This circuit has four essentialnodes and five essential
a.7
Tle !1esh-(unent
l4erhod:
Some
Special
Cases 111
bmnches wherc the current is unlmown. Therefore we know that the citcuit can be analyzed in terms of two t5 - (4 - 1)] mesh-curent equations.Although we defined three mesh cunefis in Fig. 4.27,the dependent
crurent source forces a constaint between mesh currents ia and ic, so we
have only two unlnown mesh currents, Using tle concept of the supermesh,we iedmw the circuit as show[ in Fig. 4.28.
We now sum the voltages arormd the supermeshin terms of the mesh
cunents i!, ib, and L to obtait
Rri^ + 'Dcc + RE(tc
ib) - Vo = 0.
\4.43)
The mesh b equation is
R2tb+Yo+nr(ib-i")=0.
Figure4.28A Th€circuitshownjn Fig.4.27,
depicting
(4.44) the sup€nnesh
crcated
bythepresence
ofthe dependent
cuftentsourc€.
Theconstaint imposedby the dependentcunent sourceis
piB=i,-i..
14.45)
The bmnch current contiollhg tle dependent curent source, expressed
as a function of the mesh currents! is
(4.46)
FromBqs.4.45
and4.46,
jc=(1 r B)'a- Ir,b.
14,47)
WenowuseEq.4.47to eliminateicfrcm Eqs.4.43and4.44:
fn1 + (1 + p)RElia- (1 + B)REib=V)-Ucc,
(t
B\Rd" - [R/ + fr - B)Rr]t"- -Yo.
(4.48)
14.4eJ
You should veriiy that the solution of Eqs. 4.48 and 4.49 for ia and ib gives
.
h:
uoR2- uccR2- ucc(1 + B)RE
RrRtr (1 -r B)R CR,+ R, '
-uo& - (1+ B)REUcc
R1R'.r (1 + B)R,(R1+ R )
\4.50)
(4.51)
We also leave you to verify that, when Eqs. 4.50 and 4.51 are used to fhd
iB, the resultis the sameas that givenby Eq.2.25.
112
Iechniques
of Circuit
Analysis
objective2-Understand and be abteto usethe nesh-currentmethod
4.10
Use the mesh-curent method to find the power
dissipatedin the 2 O resistor in the circuit shown.
30
Answer 72 W:
4.11 Use the mesh-curlentmethodto find the mesh
currert ia in the circuit shown.
i0A
Answer:15A.
4.12 Use the mesh-currentmethodto find the
power dissipatedin the 1 O resistorin tlle ci!cuit shown.
10v
NOTE: AIso try ChsptetPrcblems1.41,4.12,4.17,and 4.5A.
4.8 TheF{ode-VoltaEe
MethodVersus
the Mesh-eecrremt
Methsd
The greatestadvantageofboth the node-voltagcand mesh-currentmethodsis that they reducethe numberofsimultancousequationsthatmust be
manipulated.Thcyalsorequirethe analystto be quite systematicin terms
oforganizingand wdling theseequations.Itis naturalto ask,lhen,"When
is the node-voltagemethod preferred to the mesh-currentmelhod and
vice versa?"As you might suspect.there is no clear-cutanswer.Askinga
number of questions,however,may help you identify the more efficient
methodbelbre plunginginto the solutionprocess:
. Does one of the melhods result in fewer simultaneousequations
. Does the circuil contain supernodes?If so, using the node-voltage
method will permil you to reduce the number of equalions to
4.8 Th€Node
Voltase
Method
Versus
th€l,lesh
Current
Method
. Does the circuit contain supermeshes?If so, using the mesh-current
method will p€rmit you to reduce the number of equations to
be solved,
. Will solving some portion of the c cuit give the requested solution?
If so, which method is most efficient for solving just the pertinent
portion of the circuit?
Perhaps the most important obse ation is that, for any situation, some
time spentthinking aboutthe problemin relationto the variousanalytical
apprcachesavailable is time well spent.Examples 4.6 and 4.7 ilustrate the
processol decidingbetweenthe node-voltageand mesh-curentmethods
Understanding
the Node-Vottage
MethodVersus
Mesh-Current
l{ethod
Find the powei dissipaled in the 300 O resistor in
the circuit showl in Fig.4.29.
3000 l!
150{}
\
1000
500
5000
\50t{
t256v ,000 (
)
rooo "uu(
Figure
4.29A Thecircuit
forBGnrpte
4.6.
dependentvoltage source between two essentral
nodes,we have to sum the currents at only two
nodes.Hencethe prcblem is reducedto writing two
node-voltageequationsand a constraintequation.
Becausethe node-voltagemethod r€quires only
three simullaneousequations,it is the more altractive approach.
Once the decision to use the node-voltage
method has been made,the next step is to selecta
refercnce node,Tlvo essentialnodes ir the circuit in
Fig. 4.29 merit consideration. The fiist is the referencenode in Fig.4.31.Ifthis nodeis selected,one of
the unknown node voltagesis the voltage acrossthe
300() resistor,namely,?2 in Fig. 4.31. Once we
know this voltage, we calculate the power in the
300O resistorby usingthe expression
prmo : oji 100.
Solution
To find tlre power dissipated in the 300 O resistor,
we need to find either the curent in the resistor or
tlle voltage across it. The mesh-current method
yields tlle curent in ttre resistor; this approach
iequires solving five simultaneous mesh equations,
as depicted in Fig. 4.30. In writhg the five equations,we must includethe comtraint iA :
r,.
Before going furtler, let\ also look at the cir
cuit in termsof the node-voltagemethod.Note that,
once we know the node voltages, we can calculate
eitler the current in the 300 O rcsistor or the voll
age acrossit. The circuit has four essentialnodes,
and therefore only tlree node-voltage equations
are requiredto descibe the circuit.Becauseof the
1.28V
jn Fig.4.29,withthenve
Figure4.30A Thecircuit
shown
114
Techniques
ofCircuii
Analysis
we compare thesetwo possiblercference nodes
by mea]ls of the lollovring setsof equations.The first
set pertains to the sircuit shown in Fig. 4.31,and the
secondsetis basedon the circuitshownin Fig.4.32.
300{)
100_O
I ir'r250_O
set 1 (Fig4.31)
At thesupernode,
Llr_
r)
jn Fjg.4.29,wjtha
Figur€4.31 a Th€circujtshown
r00
Note that, in addition to selecting the reterence
node,we definedthe threenode voltages"1, ?,r,and
?.,3and indicated that nodes 1 and 3 form a supernode, becausethey are cormected by a dependent
voltage source.It is understood that a node voltage is
a risefrom the referencenode;tlerefore,in Fig.4.31,
we have not placed tlle node voltage polarity refererces on the circuit diagram.
The second Dode that merits considemtion as
the reference node is the lower node in the circuit,
as shoM h Fig. 4.32.It is attractivebecausert has
the most branchesconnectedto it, and the nodevoltage equations are thus easierto $'rite. However,
to find eitlei the current in the 300 f,, resistor or
the voltage acioss it requies an additional calculation oncewe know the node voltagesoaand oc.For
example, the curent in the 300 O iesistor is
(.,. - r,,)/300.$hereactbe voltageacfo.<the re<is
250
200
,r\ + 256
150
^
\
rol]o
"b
250o ,.:q!o
'*o
n(
"
l2s6v moo(
)'''
\
128)
- + 1r2+ 128
u3
D l
500
From the supemode,the constraint equatron$
t)t :
)UrI :
Ut
Ltl
-
. Set2 (Fig4.32)
0" - 256
1)"
2c0-
,c
1500
-
500
.100
250
'
4oo
-ij
3000
400
i jb : : : +X -) - D - + D-) 300
o -(D
D
ot
150
u. + 128
5oo
-
-
,a - 0b
1oo
1).- rb
250
-
-
oa
0c
3oo
3oo
=''
-'
From the suDemode.the constraint equation is
128V
Figure4.32A Thecircujtshown
in Fis.4.29
withan
attemative
r€ference
node.
50(r"
ub: 50ia: -ff
,")
6
You should veify that the solution of either set
leadsto a power calculationof 16.57W dissipated
in the 300 O rcsistor.
4.8
TheNode-Vottage
Meihod
Versus
thel4esh-Current
l,4ethod ,15
Conparing
the Node-Voltage
andMesh-Current
Methods
fiDd rhevoltage,, in rhecjrcuitsho$nin Fig.4.33.
Sotution
At firsl glance,the node-voltagemethod tooks
appealing,becausewe may define the unknown
voltage as a node voltage by choosing the lov/er terminal of the dependent curent source as the reference node, The circuit has four essential nodes and
two voltage-contmlleddependentsources,so the
node-voltage method requires manipulation of
three node-voltage equations and t\ro constraint
equatiorN.
Let's now turn to the mesh-currentmethodfor
finding oo. The circuit contains three meshes,and
we call use the leftmost one to calculatezr.,If we
let i. denote the leftmost mesh curent, then
oo = 193 10ia.The presenceof the two curent
sourcesreducesthe problem to manipulatinga single supermeshequation and two constmint equations.Hence the mesh-curent method is the more
attractive technique here.
Figure4.35A Thecircujtshown
in Fig.4.33
withnode
vottages.
To help you compare the two approaches,we
summaize both methods.The mesh-curent equations arebasedonthe circuitshownin Fig.4.34,and
the node-voltageequationsare basedon the circuit
shownin Fig.4.35.ThesupermeshequationIS
193 : 10i, + 10tb+ 10tc+ 0.80d,
alld t]re constraint equations are
ta : 0.4?,r:0.8tc;
lb
1)d=
7.5id and
ic - ib = 05
We usethe coNtraint equationsto write the supermeshequationirl termsofla:
4f)
2:4
160 : 80i,, or ia : 2 A,
1). = 193 - 20 : 173 Y.
2A
The node-voltageequationsare
D.
6J)
7.5()
193
10
80
o.+,tr+\f
: o,
o,-(z'b+0.80,
tigure4.33A Ihecncuit
forExampte
1.7.
10
0b+0.80d-r,"
++o.s+
10
The constraintequationsare
- (16+ 0.8r'd)
J^
U r = , 0 ,, O = l &
60
7.5{)
8r)
jn Fig.4.33wjthth€threemeshcurrents.
Figun4.34A Thecncujtshown
10
l-
We usethe constraintequationsto reducethe nodevoltage equations to three simultaneous equations
involving o,, oa, and ob. You should vedfy that the
node-vollage
approach
alsogi\esd.
173V.
116
Techniques
of CncujtAnatysis
objective3-Decidingbetw€en
the node-vottage
andmesh-current
methods
4.13 Find the power deliveredby the 2A current
sourcein the circuit showr.
4.14 Find the power deliveredby the 4A current
sourcein the circuitshown.
15()
2A
25V
Answer: 40W:
NOTE: AIso try ChapterProblens 4.54and 1.56.
4.9 S*ure* Tyaslsf*r$fi
a"iioris
G)
rj
Even though t1renode-voltageand mesh-currentmethodsare powcrful
tcchniqueslor solvingcilcuits,weare still irterestedin methodsthat canbe
usedto simplifycircuits.Sedes-parallel
reductionsand I{o-Y transfbrma
tions are alreadyon our list of simplifyingtechniques.
We begincxpandnrg
this lisl with source transformations. A solrlc€ transformation. shown in
Fig.4.36,allows
avoltagesourcein serieswitha resistorto be replacedbya
curent sourcein parallelwith the sameresistoror vicc vcrsa.Thedouble
headedarrow emphasizes
tiai a sourcetransformationis bilateral that is,
lve canstartwith eitherconfigurationand dedvethe other.
We needto find the relationshipbctwcorlo, and ir thal guamnteesthe
two configurationsin Fig. 4.36are equivalentwith respect1()iodes a,b.
Equivalenceis achievedif an_vresistorR, experiencesthe samecurrent
flo\l:,and thus the samevoltagedrop,whetherconnectedbetweennodes
a,bin Fig.4.36(a)
or Fig.4.36(b).
SupposeR, is connecledbetlveennodes a,b in Fig. 4.36(a).Using
Ohm\ law.thc currentin n, is
Figure
4.36r. Source
transtormatioN.
14.52)
Now supposethc sameresistorR, is connectedbetweennodesa,b in
Fig.4.36(b).Usingcurrent division,lhe cunent in R, is
If the lrvo circuits in Fig. 4.36are equivalent.these resistor currents must be
the same.Equatingihe righlhand sidesofEqs.4.52and:1.53
andsirnplitying,
,'=R
14.54)
4.9
SourceT6nsformations
717
When Eq.4.54 is satisfied for the circuits in Flg.4.36, tle curenl itr Rr is
the same for both circuits in tle figure for all values of Rr. If the current
through R. is the samein botl circuits5then the voltage drop acrossR, is
the sameir both circuitq and tle circuits are equivalent at nodes a.b.
lf lhe polarily of d. is reversed.rhe orie0rarionolir mu<tbe reversed
to mahtaitr equivalence,
Example 4.8 illustmtes the usefulnessof making source ftamformations to simplify a circuit-analysis problem.
UsingSource
Transformations
to Solvea Circuit
a) For the circuil showr in Fig.4.37,find the power
associated
wifi the 6V souce.
b) State whether the 6 V sourceis absorbitrgor
delivedngthe powercalculatedin (a).
4{}
+'
60
30o
5{}
t20o
40v
10{}
Flgure4.37A Thecircujtfor E\ampte
4.8.
Solution
a) ff we study the circuit show! ill Fig. 4.37,knowitrg
rhat lhe power asqociated
witb fte o V sourceis
of interest, several approachescome to mind, The
circuit has four essential nodes and six essential
brancheswhere the curent is unlnown, Tnus we
can find the current in the brarch containitrg the
6 V source by solving either thrce 16 (4 - 1)l
mesh-currert equations ot three [4 1] nodevoltage equations, Choosing the mesh-curent
approach involves solving for the mesh current
that cor:respondsto the branch current in the 6 V
source. Choosing the trode-voltage approach
involves solving for the voltage adoss t}le 30 O
resistor, from which the branch current in the 6 V
source can be calculated.But by focusing on just
one branch culrent, we can first simptily tlle circuit by using sourcetransformations,
We must reduce the circuit in a way tlDt
preserves
$e idenriryof lhe braDchconlatning
the 6 V source.We have no leasol to preserve
the identity of the branch cotrtaining the 40 V
source.Beginnhg witl tlis btanch, we can transform the 40 V sourcein serieswith the 5 f,) reslstor into an 8 A curent sourc€ in parallel witl a
5 O resistor,asshownin Fig.4.38(a).
(a) First step
(b) secondstep
(c) Thnd step
(d) Fourth step
Flgur€4.38 Siep-by-step
sjmplification
of ih€ circujtshown
in Fig.4.37.
118
Techniques
of Circuit
Anatysis
Next, we can rcplace the pamllel combination of
the 20O and 5O resistorswith a 4O resistor.
This 4 O resistor is in parallel with the 8 A source
and tlerefore can be replaced with a 32 V source
in series with d 4Q resisror, as sho\rn in
Fig. 4.38(b).The 32 v sourceis in seiies with 20 O
of resistanceand,hence,can be rcplaced by a currcnt sourceof 1.6A in parallel with 20 l), asshown
in Fig.4.18(c).Tbe
20 A and l0 0 parauelle\i.
tors can be reduced to a single 12 O resistoi. The
parallel combination of the 1.6A cu:rent souice
t
t
,zi\
(
Y
-
),, tR"
I
L -,,----a---
b
/n
(
P6v= (0 82s)(6)= 4 9s w'
b) The voltagesourceis absorbingpower.
A question that arisesftom use of the souce transformation depicted
in Fig.4.38is,"Wlat happensif thereis a resistanceR, ir parallelwith the
voltage source or a rcsistance Rr in se es with the current source?" In
both cases,the resistancehas no effect on the equivalent circuit that prc_
dicts behavior with rcspect to terminals a,b. Figure 4.39 summarizes this
l
Y
and the 12O resistortiansfoms hto a voltage
sourceof 19.2V in serieswidt 12o. Figure4.38(d)
showsthe result of this last transformation.The
curent in the directionof the voltagedrop across
the 6 V sourceis (19.2- 6)/16, or 0.825A.
Theiefore the power associatedwith the 6 V
)u,
+b
-r--*T1."- (r-]
f),;fn
'"
The two circuitsdepictedin Fig.4.39(a)are equivalentwith iespectto
terminals a,b becausethey produce the same voltage and cuirent in any
resistorRr insertedbetweennodesa,b.The samecan be saidfor $e circuits in Fig.4.39(b).Example4.9 illustratesan applicationof tlle equivalent circuitsdepictedin Fig.4.39.
-r__L....-,
r_____l_,
(t)i
fn
(D)
a
Figure
circuits
containing
4.39 Equivatent
otin series
in paratteL
witha voltage
source
rcsistance
@
Techniques
sour€e
T]ansformation
Usingspecial
I
I a) Use sourcetransfomations to find the voltage
ooin the circuit shownin Fig.4.40.
I
by rhe25uv lolrage
ut nnd rnepo*eraeleloped
I
source.
I
c) Find the power developedby the 8 A current
I
source.
I
I
1250
250V
{ J8A
1000
150
100
forErampL€
4.9.
tigure4.40A Thecjrcuit
I Solution
25o"
I
I d) Webeeinbv removinstbe I25 Q and l0 Q fesisror". l-.*ut. the l2'5O re\islor is connecled
I
the 250V \oltagesourc€and the l0 Q
across
I
resisror
is conDected
i0 rerieswjlh lhe R A curI
rent source.we also combinethe .eries-conI
nectedresistorsinto a singleresistanceof 20 O.
I
Figure 4.41 showsthe simplified circuit.
I
5f)
25A
250V
1)
(f)ro '.|,oo
20o"
F i g u e4 . 4 1A A q i_ p t ' F evde l s ' oonf l h e , c u t s ' r o r .i l
Fig.4.40.
4.10 Th-;venin
andNofton
Equivatents119
We now use a source transformation to
replacethe 250V sourceand25 O resistorwith a
10A sourcein parallelwith the 25 O rcsistor,as
shoqnin Fig.4.4).Wecdnnoq simpliJylhe circuit showr in Fig. 4.42 by using Kircbhoff's current law to combinethe parallelcurent sources
into a singlesource.The pamllel resistorscombine into a singleresistor Figure4.43showsthe
Iesull.Hencer), = 20 V.
b) The currentsuppliedby the 250V sourceequals
the currentin the 125O resistorplus the current
in the 25 O resistor-Thus
''
250
125
250 20
25
.-^
"'
-
souce, posilive at the upper terminal of the
D J+ 8 ( 1 0 ) = r o = 2 0 . o t
.N.:
and the polver developedby the 8 A sourcers
480W Nole that the 125O and 10 O resistors
do not alfect the value of ,". but do affect the
power calculations.
f ) r o aJ z : o( { ) r , r , } r , , r , o
Figure4.42A Thecircuiishownin Fig.4-41aftera
source
Thereforethe power developedby the vottage
: (250)01.2) = 2800W.
/25ov(developed)
c) To find &e power developed by the 8 A current
source,we fi$t find the voltage acrossthe source.
It we let o" represent the voltage aqoss the
t)
lUO
Figurc4.43d Thecircuitshown
in Fig.4.42aftercombjning
objective4-llndersiandsouicetransformation
4.15 a) Usea seriesdf sdurceti:;sfonilalions to
find thevoltageo in thecircoifshown.
b) How mtch powerdoesthe 120V source
deliverto ttre circuit?
(a)asV
(b)374.4w
NOTE:
AIso tty Chaptel Problens 1.59 and 4.62.
4.10 Th6venin
andNortonfrquival,ents
At timesin circuit analysis,we wanl 10concentrateon what happensat a
speciJicpair ofterminals.For example,whenwe plug a toasterinto an out
let, we are interestedprimarily in the voltageand currentat the tcrminals
of the toaster.Wehavelittle or no interestin the effectthat connectingthe
toasterhason vollagesor curents elsewherein the circujt supplyingthe
outlet.We can expandthis irterest in terminal behaviorto a set of appli,
ances,eachrequiringa different amountof powcr.We then are interested
in how the voltage and current delivered at thc oullel changeas we
changeappliances.Irotherwords,wewant to focuson thc bchaviorol the
circuitsupplyingthe outlet,but only at the ortlet terminals.
1.6r}
8()
120
Techniques
of Cncuit
Anatysk
Th6venin and Norton equivalents aie circuit simplification techniques
that focus or termiml behaviorand thus are extremelvvaluableaids in
"ndllsi..Althoughbere$e drscuss
them a. theypenainro re.isri\eciF
nelqork @ntaining
cuitE Th6venin and Norton equivalent circuits may be used to represent
independcnrand
any circuitmadeup of lineat elements.
We can best describe a Th6venin equivalent circuit by relercnce to
Fig. 4.44,which representsany circuit made up of sources(both hdepend,
(a)
(b)
ent and dependent) and rcsisto$. The letters a and b denote the pair of ter,
minals of interest.Egure 4.44(b)showsthe Th6venjnequivalent.Thus,a
Flgure4.44a (a)A generatcircuit.
(b)TheThdvenjn
Th€ve n €quivalent circuit is an independent voltage souce yrh in series
witl a resistor Rll, , which replacesan interconnection of sourcesand rcsis,
tors.Tfus seriescombination of yrh and Rr, is equivalent to the orighal circuit in the sensethat,if we connectthe sameload acrossthe tetminalsa,b
of each circuit, we get the samevoltage and current at the terminals of the
load-l]!is equivalenceholds for all possiblevalues of load resistanc€.
To represent the original circuit by its Thivenin equivalent, we must
be able to delermine the Th6venin voltage /rl1 and the Thevenin resistance Rr,. First, we note that if the load resistanceis infinitely large,we
have an open-circuil condition. The open-circuit voltag€ al the terminals
a,b in the circuit shownin Fig.4.44(b)is V16.By hlpothesis,this must be
the sameasthe open-circuitvoltageat the terminalsa,bin the o ginal circuit.Therefore,lo calculatethe Th6veninvoltageyrh, we simplycalculate
the open circuit voltageir the originalcircuit.
rigure4.45.4A circuit
used
io ittunrate
aThevenjn
Reducingt}le load resistance
to zerogivesus a short-circuitcondition.
If we placea short circuit acrossthe terminalsa,b of the Th6veninequiva
lcnt circuit, the shortcircuit current directed ftom a to b is
By h!?othesis,this shorlcircuit current must be identical to the short-circuit
currentthat existsin a short circuit placedacrossthe terminalsa,b of the
originalnetwork.From Eq.4.55,
Rrr, =
figure4.46A Thecjrcuitshown
in Fig.4.45
with
ierminaba andb shot cncuit€d.
8r)
\4.56)
Thusthe Th6veninresistanceis the ratio ofthe open-circuitvoltageto the
sho -circuit curent.
Finding
a Thdvenin
Equivalent
To find the Th6venin equivalent of the circuit shown in Fig. 4.,15,we first
calculatethe open ciicuit voltageof o,b.Note that when the teminals a,b
are open,thereis no currentin the 4 O resistor.Therefore
the open,circuit
voltageoabis idcnticalto the voltageacrossthe 3 A curent source,Iabeled
We find the voltage by solvinga singlenode-voltageequation.Choosing
the lower node as the referencenode,weget
-[*
3 2 V[ )
zrr - 25
I-
Figue4,47A TheTh6venin
equivatenr
ofrhecircuit
shown
in Fig.4.45.
vn
s
- 6D1-
(4.51)
1 1 : 3 2v .
(4.58)
solvingfor or vields
4.10 Th6v€nin
andNorton
Equivatentr121
Hence ttre Th6venin voltage for the circuit is 32 V
The trext step is to place a sholt circuit affoss the terminals and calculate the rcsulting short-cftcuit culrent. Figue 4.46 shows the circuit vrith
the short in place. Note that the shot-circuit curent is in tle direction of
the opetr-circuit voltage drop auoss the terminals a,b. If the shon-circuit
curent is in tlle direction of the open-circuit voltage rise aqoss the temioals.a miDussign must be insenedin Eq.4.5o.
The short-circuit curent (isc) is foutd easily once ,2 is known.
Therefore tle problem reducesto finding 1,2with the short in place.Agaiq
iJ we usethe lower node asthe reforence node. the equation for ,, becomes
,,-25
5 2
0
-
\4.59)
SolvingEq.4.59for ?]2gives
u2:16Y'
(4.60)
Step1:
Source transformation
HeDce,the shofi-circuitcurrentis
.
1
6
| s _ / =
(1.61)
Wenow find theTh6veninresistanceby substitutingthe numericalresults
fromEqs.4.58
and4.61inroEo.4.56:
^*=*:?='n
Step2:
14.62)
Figure 4-47showsthe Th6venin equivalent for the circuit showtrin Fig. 4.45.
You should verify that, if a 24 O resistor is connected across the terminals a,b in Fig. 4.45, the voltage acrossthe resistor will be 24 V and the
current in t}le resistor will be 1 A, as would be the casewith the Thdvenit
circuit iD Fig. 4.47.This same equivalence between the circuit in Figs.4.45
alrld4.47holds for any resistor value cornected between rodes a,b.
TheNortonEquivalent
Step3:
Souce kansformationj series
resistorscombined,producing
the Thevenin equivalent circuit
A Norton equirelent chcuit consists of an independent current source in
parallel with the Norton equivalent rcsistance.We call dedve it from a
Thevenin equivalent circuit simply by making a source tansfomation.
Thus the Norton curont equals the shorlcircuit current at tlle terminals of
interest, and the Norton resistanceis identical to the Th6venin resistanc€.
UsingSourceTransformations
Step1:
Sometimeswe can make effective use of source transformations to derive
Souce traDsfornation, producing
the Norton equivalent circuii
a Th6venin or Norton equivalent circuit. For example,we can derive the
Th6venin and Nortor equivalents of tle circuit shown in Fig.4.45 by making tle series of source tansformations shov,n in Fig. 4.48.This technique
is most useful when the network cotrtains only independent sources,The
presenceof dependent sourcestequires retainhg the identity of the controllhg voltages and/or curents, and this constraint usually prohibits
conthued reduction of the circuit by source tmnsformations We discuss
tlle problem of finding the Th6venin equivalent when a circuit contains Figure4.48A Step-by-step
denvation
of theIh€venin
dependentsouces in ExamDle4,10.
jn Fjg.1.45.
andNoironequivatenis
ofthecjrcujtshown
122
Anatysis
Techniques
of Circuit
Source
of a Circuitwith a Dependent
Equivalent
Findingthe Thdvenin
Find the Th6venin equivalent for the circuit con
taining dependentsourcesshownin Fig-'1.49.
2ko
a
in Fig.4.49wiihtermjnaLs
Figure4.50 s Thecircuitshown
eqLrilatent
a Th6venin
Figure4.49 a A circuitus€dto itLustraie
sources.
dependent
whenthecircuitcontajns
As the voltagecontrollingthe dependentvoltage sourcehas been reduced to zero, the current
controlling the dependent current source is
Sotution
The first slep i]l analyzingthe circuit in Fig 4'19 js
to recognizethat the current labcled i. must be
zero. (Note the abscnceof a retum path for ir to
enter the leffhand portion of the cjrcuit.) The
open circuit, or Th6venin.voltage will be the voll
ageacrossthe 25 O rcsistor.Withi" : {),
Yrh=oob:(
2 0 t ( 2 s )=
500i'
5
2000
Combiningthesetwo equationsyieldsa shortcircuit
i," =
20(2s) =
Fromi,. and Yrhwe gel
The current i is
-
':
5-30
:
2ooo
5
50 mA.
x 10e= 100O.
3yft
,(no '
In writing t}le equation for i, wc recognize that the
Th6veninvoltageis identicalto the control voltage
wllen we combinethesetwo equalions,weobtarl
Figure 4.51illustratesthe Th6veninequivalent
Note that the ref
for the circuit shownin Fig.4.,19.
voltage
the
Th6venin
erence polarity marks on
,{.51
preceding
equathe
agreewitb
sourcein Fig.
tion for Yft.
Yrh=-5v
To calculatethe short-circuitcurent, we place
a short circuil acrossa,b.Whenthe terminalsa,b are
shortedtogelher,thecontrol voltage? is reduccdto
zero.Therelore,with the short in place,the circuit
shor\n in Fig. 4.4,rbecomerlhe one \ho$n in
Fig. 4.50.WitI the sho.t circuit shuntingthe 25 O
resistor.all the curent from the dependentcunent
sourceappearsin the short,so
100()
('
in
cncuitshown
equjvahntforihe
Figure4.51',. TheTh6venin
Fil.4.49.
4.11 r,4ore
on 0eriving
a Th6venin
EquivaLent123
objective5-Understand
Th6venin
andNortonequival€nts
4,16 FindtheTh6veninequivalentcircuitwith respect
to the termimls a,b Ior the circuil shown.
7 2l l
Answer: Yab= Yrh = 64-8V, Rft : 6 O.
4,17 Find thc No on equivalentcircuirwith respeci
to the teminals a,b tbr dre circuit sho!v!.
15A
Answe.: 1N = 6 A (directedtoward a), RN : 7.5 O.
4.18 A voltmeterwirh an inlernalresistance
of
100kO is usedto measurethe voltage?ABin the
circuil shown.Whalis the volrmeterreading?
12kO
Answer 120V
NOTE: Aho try ChaptetProblems4.63,4.66,and4.67.
4.11 More*n $erivi*g* Th$v*n-in
Hquivx[ent
The lechniquefor determiningnrh rhat wc discussedand iltustraredin
Scclion4.10is not ahvaysthc easiestmerhodavailable.Two other meth
odsgenerallyare simplerto use.Thefirst is usefulif thc netlvorkcontains 2 5 V
only indepcndentsourccs.To calculatonft for such a nerwork,wc first
deactivatcall independcnlsourcesand theDcalculatethe resistanceseen
looking inlo lhe network allhe designaredrerminatpair.A voltagcsource
is deactivaledby replacingit $'ith a shorl c cuit.A curent sourceis deac- Figure4.52 .r A cjrcuitusedto itlustctea Th€venin
tivatedby replacingit wirh an open circuil.Fof example,considerrhe cir,
cuit shown in Fig. 4.52.Deactivatingthe independcnlsourcessimplifies
the circuit to the one shownin Fig-4.53.The resisranceseenlooking into
5('
4(:}
the terminalsa.bis denotcdR"b,which consistsofthc 4 O resistorin se es
with ihe parallelcombinalioDsof thc 5 and 20 Q rcsislors.Thus.
R
R
,
.
r2o_8,,.
(4.63)
Nole that thc derivation of nTn with Eq.,1.63is much simpler than the
sane derivationwith Eqs.,l-57+.62.
figure4.534 Thecircuitshown
in Fig.4.52afterdeac
tjvatjonof thejndependent
soLrrces.
124
lechniques0fti,cuitAnatysis
If the circuit or network contains dependent sources,an alternative
procedure for finding the Th6venin resistance Rrr is as follows. We fi$t
deactivate all independent souces, and we then apply either a test voltage
sourceoi a test culrent sourceto the Th6venin terminals a,b.The Th6venin
resistanceequals tle ratio of the voltage acmss the test source to t}Ie current delivered by the test source.Example 4.11 illustrates this alternative
procedure for fhding Rrh, using the sameciicuit as Example 4.10.
Equivalent
Usinga TestSource
Findingthe ThEvenin
Find the Thdvenin resistance Rrh for the circuit in
Fig.4.49,usingthe alternativemethoddescdbed.
5olution
We filst deactivate the independent voltage source
ftom t}Ie circuit and then excite the circuit ftom the
teminals a,b with either a test voltage souice or a
test current soruc€.If we apply a test voltage source,
we will know t}Ie voltage of the dependent voltage
sourceard hencethe controlling curlent i. Therefoie
we opt for t]le test voltage soluce.Figure 4.54shows
the circuit for mmputirg the Th6venin resista[ce.
The externally applied test voltage source N
denoted o?, and the current that i[ delivers to the
ciicuit is labeledir. To find the Thdveninresistance,
we simplysolvethe circuitshownin Fig.4.54for the
ratio of the voltage to the current at the test source;
that is,Rrh = Dr/ir. From Fig.4.54,
i,=fr+zoi,
-30"
i=;'nA.
(1.64)
\4.65)
We then substituteEq.4.65 into Eq.4.64 and solve
the resulthg equation for the mtio t r/i?l
b.r
25
60t)r
2000'
i.f,7
_ 6
ur
25 200
(4.66)
50
5000
1
100
(4.61)
From Eqs.4.66and 4.67,
Figure4.54
Anatternatjve
method
for compuiing
the
nrn=3=tooo.
(4.68)
In general,thesecomputationsare easierthan thoseinvolvedin computing the shofcircuit current. Moreover, in a netwoik containing only
resiston and dependent sources,you must use the alternative method,
because the ratio of the Th6venin voltage to the shorFcircuit current is
indeteminate.That is.it is the ratio 0/0.
4.11 Moreon Deriving
a Th€venin
Equivatent 125
objective5-Understand Th6veninand Nortonequivatents
4.19 Find the Th6veninequivalenrcircuitwilh respect
to the lerminals a,b lor the circuit shown.
- ?rab
= 8V,Rrh:
Answeri Y1.h
1O.
4.2O Find thc Th6venin equivalent circuit with
respectto the terminalsa.b for the circuit
showr. (girtr Define the voltage at the left
most node as?),and wrile two nodal equations
with ynr asthe righr node voltage.)
Answer: Y1]r: oab= 30 V. Rft : 10 O.
160;r
NOTE: Ako trt Chaptet Ploblems 1.71 dnd 4.77.
Usingthe Th€venin
Equivalent
in the AmplifierCircuit
At trmeswe canuseaThivenin equivalcnlto reduceone portion of a cir
cuir ro greatll'simplify analysisof the larger network.Let,s reiurn to the
circuil first introduced in Section 2.5 and subsequentlyanalyzed in
Sections4.4and 4.7.To aid our discussion.
we redrcw the c cuit and identified the branchcurrentsofinterest,asshor\ltlin Fig.4.55.
A. L'urprerrou\analvsi\hds\no$n./s i, Inc ke) ro tindingrhcorlrer ' r
branchcurrents.We redrawrhe circuil as shownin Fig.4.56to prcpareto
replacethe subcircuilto rhe left of % wirh its Th6veninequivatcnt.you
shouldbe able to determinethat lhis modificalion has no etlect on the
branchcurrentsi1,ir, iB, and iE.
h
Now we replace the circuit made up oI y.t., R1, and R2 with a
Thdveninequivalent,with respectto the terminalsb,d.TheThdvenir volr
ir ,
ageand reslslanceare
VccRz
Rr+R2
-\€
-\p.€.'
4.55,
apolica
io- o d
eq na a
li9u.,e
r4.6e,
'
r no r ( u ra
r narvsE.
Rrn:
R1 + Rr'
14.10)
of CircuitAnaty5is
126 Te.hniqLres
with the Th6veninequivalent,the circuit in Fig. 4.56 becomesthe one
shownhFig.4.57.
We now derive an equation for iB simply by surnming t]le voltages
around the left mesh. ln writing this mesh equation, we recognize that
i, : (1 + B)t,. Thus,
vn,: R*i, +va+ RE(1+ B)iB,
\1.71)
irom which
i,,
Vm
d
sho$,n
Figure4.56a A modified
veEionofthe circLrit
in Fig.4.t5.
V,t
Rrh + (1 + B)R,
\4.72)
Wiren we substituteEqs. 4.69 and 4.70 into Eq. 4.72,we get the same
er?ressionoblainedin Eq.2.25.Notelhatwhen we haveincoiporatedthe
Th6veninequivalentinto the original circuit,we can obtain the solution
for is by wdting a singleequation.
,r,
jn Pig.4.56modified
tigurc4.57A Theciftuitshown
equivatent.
bya Th€venin
power
maximum
figure4.58A A circuitdescribing
PowerTransfer
4,12 Maxircrum
Circuit analysisplaysan importantrole in the analysisof systemsdesigned
to transferpower lrom a sourceto a load,We discusspower transferin
terms of two basic types of systems.The first emphasizesthe efficiency of
the power transfer. Power utility systemsare a good example of this type
becausethey are concernedwith the generation,tmnsmission,and distribution of large quantitiesof electricpower. If a power utility systemis
inefficient,a large peicentageof the power generatedis lost in the trans
and thus wasted.
missionand distributionpiocesses,
The secondbasictype of systememphasizesthe amount of power
transfeired. Communication and instrumentation systemsare good examples becausein the transmissionof information,or data,via electdcsignals,the power availableat the transmitteror detector is limited. Thus,
transmittingas much of this power as possibleto the receiver,or load, is
desilrable.
In such applicationsthe amount ol power being lransferredis
small,so the efficienryof transferis not a primary concen. We now considermaximumpowertransfeiin systemsthai canbe modeledby a purely
resistivecircuit.
Ma-\imum power transfer can best be described wilh t}le aid of the circuit shown in Eg. 4.58.We assumea resistivenetwork containingirdependentand dependentsourcesand a designatedpair of terminals,a,b,to
problemis to determinethe value
which a load,Rr, is to be connected.The
4.12 l,4aximum
Power
Transfer 127
ol Rr lhat permiLsmaximum power deliver) ro Rr. The first slep in thjs
processis to fecogniTelhat a fesisrivenerworkcan aiwaysbe feplacedby
its Thdvenin equivalent.Therefore, we redraw the circuit sho$n ir Fig. 4.58
as the one shown.in Fig. 4.59. Replacing the original network by its
Thdvenin equivalent geatly simpiifies the task of finding Rr. Derivation of
Rr rcquires expressing the power dissipated in R, as a function of the
three circuit parameters yTh,Rft, and Rr, Thus
rigur64,59 A circuitus€d
to d€t€rmine
thevalue
of
nr for maximum
power
transfer.
p=PRL:(#E)""
\4.73)
Next, we rc€ognize that for a given circuiq yft and Rrn will be fixed.
Therefore the power dissipated is a function of the single variable Rr. To
find the value of R, that mMi;izes the power, we use elementary calculus.
We begin by writing an equation for the dedvative ofp with rcspect to Rr:
^:'^l
{^Trlr + ^r)_ - KL. Z\Krh + ltL)
(Rft+Rf
l
\4.74)
The dedvative is zerc and p is maximized when
(^Tb -
r(L.,-,
1l(r(Kft
-
Rrl
\4.75)
SolvingEq.4.75ields
(476) < Condition
for m.ximumponertransfer
Thusmaximumpower,hansferoccurswhenihe load resistanceR, equals
lheTheveDi0
resisraoce
Rrb.To findthema"\ilnum
povr'er
de[!eredto Rr.
we simplysubstituteEq. 4;76]JJtoEq.4.731
^ =v'*a"
"*
enD'
4R.
\4,77)
The analysisof a circuit wheii the Ioad resistotis adiustedfor maximum
powerhansferis illustratedin Example4.12.
128
Techniqu€s
ofCircuit
Anatysh
Cal.cutating
the Condition
for MaxinumPower
Transfer
a) For the circuit shownin Fig. 4.60,find the value
of Rr_ that results in maximum power being
transferredto Rt.
25o"
300v
R.
300
360V
-
jn Fig.4.60by
Figure4.61 3. Reduction
ofthe circuitshown
means
of a Thdvenin
equivateni.
) rsoo
b
Figure4.50a Thecjrcuitfor
ExampLe
4.12.
b) Calculate the maximum power tlat can be deliveredto Rr,
c) When R._is adjustedfor maximumpower tiansfer, what percentageof the power deliveredby
the 360V sourcereachesRr?
b) The maximumpowerthat can be deliveredto
Rr is
,*"=(S)i,a: 'o*.
c) When Rr equals25 O, the voltage?,b is
/too\
j(2s) : 150v.
,"h : l=
\rtr/
Solution
a) The Th6veninvoltagefor the circuitto the left of
t]reterminalsa,b is
v* :
= :oo,r.
'*141:eo;
The Th6venin resistanceis
(150V30)
Replacingthe circuit to the left of the terminals
d.bwilh ilsThevenin
givesusrhecirequivalenr
cuit shown ir Fig. 4.61,which indicatesthat Rr_
must equal25 J) for maximumpower transfer.
From Fig. ,1.60,when o"b equals150V the current in the voltagesourcein the dnecdon ofthe
voltagerise acrossthe sourceis
(=
360 150- 3210
0 = / _A
:ro
Therefore,
thesource
is delivering2520W
to the
p, = t,(360)= 2s20w.
The percentageofthe sourcepowerdeliveredto
the load is
900
^ 100:35.71%.
25?]J
a.1l
Supe,positio1
n29
obiective6-Know the conditionfor andcatcutate
powertransferto resistiveload
maximum
4.21 a) Find the valueofI that enablesthe circuit
shownto delivermaximumpower to the
terminalsa.b,
b) Find the maximumpower deliveredto R.
4.22 A..u.nelhdl rhccircuuin Assessmenr
Problenr4.21isdeliveringmaximumpower to
the load resistorR.
a) How much powcr is the 100V sourc€delivering to the network?
b) Repeat(a) for the dependentvoltage
c) Wlul percentageof the total power gcner
atedby theseiwo sourcesis deliveredto the
load resistor-R?
Answen(a)3 O;
(b) 1.2kw
Answer: (a) 3000W:
(b) 800w;
(c) 31.s8%.
NOTE: AIsott! ChaptetPtublems
1.79and4.80.
4.13 5up*rg:**iti*n
A linear systcn obeys the prirciple of superposilion.which statesthat
whenevcra linear systemis excited,or driven,by more than one indc
pendentsoulceoI energy,the total responscis the sum of the individual
responsos.
An individual responseis thc resull of an independentsource
acting alone. Becausewe are dcaling lrilh circuits made up of inter
connccledliDear-circuitelements.wc canapplythe prirciple ofsupcrposi
tion direclly to the analysisof suchcjrcuilswhen they are driven by lnore
thanore independentenerg,\'sourcc.Atpresent,werestrictthediscussion
10 simple resistivenetworks:however.the principle is applicable1()any
Superposition
is appliedin both the analysisand thc designoI circuiis
In ana\zing a complcxcircuil with multiple independcntvoltageand current sources.therc are oflen fewer,simpler equations1{)solvewhen the
effectsof the indepcndcnlsourcesare consideredone at a 1ime.Applying
superyositioncan thus simplify circuit analysis.Be aware.though,that
somedmes
applyingsrpcrpositionacluallycomplicates
thc analysis,producrl1glnore equationsto soh'cthanwith an alternativemethod.Superposition
is requiredonly if thc jndepeDdentsourcesin a circuit are lriDdamental]y
different.In thcsccarly chapters,
all independentsourcesare dc sources,
so
superposition
is nol required.We introducesuperposilionhere ir anticipaiion ot latcr chaple$ir whichcircuiiswill rcquie i!.
130 Iechniques
of cncuiiAnatysk
Superpositionis applied in design to synthesizea desired circuit
responsetlat could not be achievedin a circuit with a singlesource.Ifthe
desiredcircuitresponsecan be written as a sum of two or more terms,the
response can be realized by including one hdependent source for each
term ol the rcsponse.This approach to the desigt of circuits with complex
responsesallows a designei to consider several simple designsinstead of
one complexdesign,
We demonstrate t}le superposition pdnciple by using it to find the
bmnch curents in the circuit shoMr in Fig. 4.62.We begin by finding the
bmnch
curents resulting from the 120V voltage source.We derote those
120V
12A
currents with a pdme. Replacing the ideal current sourcewith an oper circuit deactivatesiq Fig. 4.63 shows this. The branch currents ir this circuit
tigure4.62A A circuitusedto iLLustrate
supeposition. are the resultof only the voltagesource.
We can easily find the bmnch curents in the circuit in Fig. 4.63 once
we know the node voltageacrosst}le 3 O resistor.Denoting this voltage
q
120V
120
6
r . r -
11.t8)
from which
01 :30 v.
jn Fjg.4.62u/jththe
Figue4.53 Th€circuitshown
curentsource
deactivat€d.
(4.79)
Now we can wite the expressionsfor the branch currents ti - ti directly:
ii = 14-lq
= 15A,
=;=10A,
.
724
jn Fig.4.6?wjththe
Figure4.64A Thecircuiishown
voltage
source
deactivaied.
6{)
2tl
3
(4.81)
0
-
(4.82)
To find the componentol the bnnch cu:rents resulting from the curent
source,we deactivatetle ideal vollage sourceand solve the circuit shown in
Fig. 4.64.The double-prime flotation for the currents hdicates they are the
componentsof the total cunent resulting from the ideal cuffent source.
We determinethe branchcurrentsin the c cuit showr h Fig.4.64by
first solving for the node voltages across the 3 and 4 O resisto$, respectively. Figurc 4.65 shows tie two node voltages. The two rode-voltage
equationsthat describethe circuit are
t2 A.
3
Figlre4.55-r Thecircujtshown
in Fig.4.64showjng
the nodevottag€s
rr and?a.
(4.80)
U4
6
0l
2
,4
.^
0,
14.83)
\4.84)
1.13 tupeeosition131
Sol\iDgEqs.4.83and 4.84for ,r aDdd.. we get
a3 : -72 V,
(4.85)
ot=-24V
(1.86)
Now we can write the branch currents ii tlrough ii directly in tems of the
node voltages?r3and 04:
ii: =:::2
A,
\4.87)
il=l=
aA.
(4.88)
3
l=
3
-12 + 24
(4.8e)
ii=i:i=-sa.
(4.e0)
To furd the branch currents in the original citcuit, that is, the curents
iI, i2, 4, and ia n FIE 4.62, we simpty add rhe cunents given by
Eqs.4.87-4.90
to the currentsgivenby Eqs.4.804.821
ir:i\+ii:15+2:17A,
\4.91)
i2=ii+i5=10-4=6A,
14.e2)
it= i\+ ii:5 + 6:11 A,
(1.93)
i + = i L + i i =5 - 6 :
(4.e4)
1A.
You shouldveriry that tle curents givenby Eqs,4.914.94 aretlle corrcct
valuesfor the bmnchcurrentsir the circuit shownin Fig.4.62.
Whenapplyingsuperpositionto linear circuitscontainingboth independentand dependentsources,you must recognizethat the dependent
sourcesare neverdeactivated.Example4.13illustratesthe applicationof
superpositionwhen a circuit containsboth dependert and indeDendent
132
]echniqles
of CjrcujtAnatysjs
UsingSuperposition
to Solvea Circuit
Use the principleof superpositionto find o, m the
circuit shownin Fig.4.66.
when the 10 V souce is deactivated,
the circuit
reducesto the one shownin Fig. 4.68.We have
addeda reference
nodeandthe nodedesignations
a, b, andc to aid the discussion.
Summingthe curientsawayfromnodea yields
_2
2
-!
0
5
O.du';- 0. or rr:
B.i
u.
Summingthe currentsawayfrom nodeb gives
Figure4.66 A ThecjtcujttofExamph
4.13.
0 . 4_0+( -
4D'\+ Db
Sotution
5=0.or
l0
2iL:
50.
I)b=21L+ll' L
We begin by linding the component of oo resulting
from the 10 V source.Figure4.67showsthe circuit.
With the 5 A source deaclivated,r,l, nust equal
( 0.ari!X10).Hence,oi must be zero,tlre branch
containingthe two dependentsourcesis open,and
to find the valuefor 0i. Thus,
5ui = 50, or l)l = 10 V.
From the node a equation,
4:2*@) : Bv.
soii = 80, or 06 = 16 V.
The value of o, is the sum of z'i,and o;, or 24 V
10v
Figure4.57A Thecircuit
shown
in Fig.4.66withthes A
tigure 4.68 A Thecjrcuitshown
in Fig.4.66wjththe 10V
NOTE: Assessyour understanding of this matetial by trting Chapter Problems 1.91 and 4.92.
PEctjcatPersp€ctive
133
PracticalPerspective
Circuits
with Realistic
Resistors
It is notpossibte
to fabricate
identicaI
etectricaL
components.
Forexampte,
produced
resis(ors
fromihe )amemanu%ctuing
process
car varyin vatue
byasmuchas20%.Therefore,
in creatjng
anelectrjcal
systern
thedesigner
mustconsider
the impactthat component
variation
wilt haveon the per
fornance
of the system.
onewayto evaluate
thisjmpactis by performing
sensitiviry
permitsthe designer
anatysis.
Sensitiviw
anatysis
to calcutate
theimpactof variations
in the component
vatues
onthe outputofthe system.WewitLseehowthis information
enabtes
a designer
to speci! an
acceptable
component
vatuetoterance
foreachofthesystem's
conponents,
Considerthe
circuitshown
in Fig.4.69.
Toittustrate
sensitiviw
anatysjs,
wewjt[investigate
thesensitivity
ofthenodevottages
rl and?2to changes
jn theresistor
Rt. Usingnodalanatysis
wecanderivetheexpressjons
fof r,1
and22asfunctions
of the circuitresistors
andsource
currents.
Theresults
aregivenin Eqi.4.95and4.96:
- lRz(Rr+ R4)+ n.Ral/rr}
Rr{RrRy's,
(Rl+nr(R3+R4)+R3n4
-'
\4.95)
R3R4[(R1+ R)Ir2-
RJ 91]
( R 1+ R ) ( R 3 + R 4 ) + R 1 & '
(1.e6)
Ihe sensitivjtyof ?,1with respectto Rr is foundby differentiating
Eq.4.95
with respectto R1, andsimitarty
the sensitivityof zJ2with respectto R1 js
foundby differentiating
Eq.4.96with respect
to R1.Weget
drr
dRr
[R3& + Rr(R3+ n4)]{R3R4rs, lR3R4+ R (R3+ Ri)11s1}
I(R1 + R )(R3+ R4) + R3R4l2
\4.97)
d?,r, R3R4{R3R418,
lRr(R3 + R4) + R3R4U81}
d&
l(Rr + R)(R3+ R4) + R3R4l'
Figure4.69A Cjrcuitusedto introduce
sensiiivity
ana\6is.
(4.e8)
134
ofCircuitAnalysis
Techniques
to itLustrate
vatues
withactualcomponent
anexampLe
Wenowconsider
and
4.98.
theuseof Eqs.4.97
EXAMPLT
in Fig.4.69are:
in thecircuit
vatues
ofthecomponents
Assume
thenominat
R r = 2 5 0 o ; R z = 5 O ; R 3 = 5 0 O ; R a = 7 5 O ; I s 1= 1 2 A a n d
to predictthe vatuesof 01 and t'2 if
1, = 16A. ljse sensitivityanatysis
from
its nominalvatue.
10%
the vatueof Rr is differentby
Solution
of 01andtr2.Thus
vaLues
FromEqs.4.95and4.96wefindthe nominal
-'
) 5 { 2 7 5 0 t l o r 5 ( 1 2 5-, J 7 5 0l 2 l
30(125)+ 37s0
and
- 25(12)l
37s0130(16)
1-',=
3odt+3dj=eov
(4.100)
to
NowfromEqs.4.97and4.98we canfind the sensitivityof t)1and?J2
in Rr. Hence
changes
- 137s0
+ s(12s)112}
+ 5(125)l- {37s0O6)
d?-'1
_ 137s0
dRr
l(30x125)+ 37sol'
-
(4.101)
ivla,
and
duz 37s0{3750(16) [5(12s)f 37s0]12]l
LtR
(?soo/
: 0.5v/o.
(4.102)
Assume
that
givenby Eqs.4.101and4.102?
Howdoweusethe resutts
:
22.5
O.
Then
is,
Rr
value,
that
its
noninat
Rl is 10% lessthan
A?)1wittbe
ARr = -2.5 11un, ,0. ,.101prcdicts
t 7 \
aa - l';Jf-25)
14581v'
predicts
vatue,ouranatysis
if Rr is 107olessthanits norfiinal
Therefore,
that01witLbe
v.
or : 25 - 1.4583: 23.5417
(4.103)
P'acticatPe,spe.iive135
Simjtarty
tor Eq.4.102wehave
A?],: 0.5(-2.5): 1.25V,
?,,: 90 1.25= 88.75V.
\4.104)
jn Eqs.4.103and4.104by substituting
Weattemptto confirmthe resuLts
thevatueR1 : 22.5O intoEqs.4.95and4.96.When
wedo,theresutts
are
r"r = 234780V,
02= 88.6960
V.
(4_105)
(4.106)
Whyis therea difference
betweenthe valuespredjctedfromthe sensitivity
anaLysis
andthe exactvatuescomputedby substitutingfor Rr in the equationsfor or and?'2?Wecar seefromEqs.4.97and4.98that the s€rsjtiviry
of 01 andz)2with respectto Rr is a functjonof Rr, because
R1 appears
in
the denominatorof both Eqs.4.97 and 4.9a. Thjs meansthat as Rl
changes,
the sensitivities
changeandhencewe cannotexpecttqs,4,97and
4.98 to give exactresuttsfor largechangesin R1. Notethat for a 10%
change1n11, the percenterrorbetweenthe predjctedandexactvatuesof
js smatl.Specjficatty,
0?)1
and0o2
the percenterrorin Dt = 0-2713"/0
aft the
errorin ,2 : 0.06'764/".
P€rcent
Frornthis exampte,
we can seethat a tremendous
amountof workis
jn the
invotved
ifwe areto determjne
the sensitivjtyof?r and02to changes
remajning
component
vaiues,nametyR2,R3,-R4,/sr, and1s2.FortunateLy,
Pspicehasa sensitivityfonctionthat witl pedormsensitivityanatysjs
for us.
Thesensitivityfunctionin Pspicecatculates
two typesofsensitjvjty.
Thefirst
is knownas the one-unjtsensitiviry,and the secondis knownas the 1ol.
sensitivity.In the examptecircuit,a one-unitchangeir a resistorwould
changeits vatueby I O and a one-unitchangein a currentsourcewoutd
changeits vaLue
by 1A. In contEst,1% sensitivityanatysis
determines
the
effectof changjng resjsto$oI sourcesby I o/oof th€ir nominaI vaLues.
Theresultof PspicesensjtivityanaLysis
of the circujtin Fig.4.69 js
shownin Table4.2. Because
we are analyzinga Unearcircuit,we can use
superpositjon
to predictvaluesof zJland1)2if morethan onecomponent's
vaLuechanges.Forexanpte,L€tus assumeR1 d€creases
to 24O and R2
decreases
to 4 O. FromTabte4.2 we cancombinethe unit sensitjvityof or
to changes
in Rr andR2 to get
A1)'
ln,
+
Az,,
*
:oss::
- 541'/:
4.8337Y/a.
Simjtarty,
Ar)
an'
Ar, 05- 65
tln
-ov
o
Thusif bothRr andR, decreased
predict
by 1 O wewouLd
Dt=25 + 4.8227= 29.8331v,
12: 90 '7 = 83Y.
136
Technjques
of cncuitAnatysis
TABLE
4.2
Etement
Pspice
Sensitivity
Analysis
Res ts
Elenent
.!:r......_ - rillJ
ElenentSensttlvlty
orinaliz€dSensitivity
09tt"/,u1q,.,.,...,...,....(-v-9!!:1l9TJl!l
(a) DCSensitivities
oJNodeVoLtase
V1
R1
R
R3
R4
IG1
!G2
2
25
5
50
75
72
16
0.5833
-5.477
0.45
0.2
,74.58
o.7458
-o.27Q8
0.225
0.15
-\,15
2
(b) Sensitiities of 1utput Vz
R1
R
R3
R1
IG1
IG2
25
2
5
50
75
72
76
0.5
6,5
0.51
4.24
-72.5
15
0.125
0.325
Q,27
0.18
?.4
If wesubstitute
Rr = 24O andn, : 4 O jnto Eqs.4.95and4.96weget
ut = 29'793Y'
1tz= 82759Y'
In bothcases
ourprcdictions
arewithjna fractjon
of a vottoftheactualnode
vottage
vatues.
Circujtdesigners
usethe resuttsof sensitivity
anaLysis
to determine
whi€hcomponent
vatuevarjatjon
hasthe greatest
impactonthe outputof
the circuit.Aswecanseefromthe Pspice
sensitivity
anaLysis
in Tabte
4.2,
q and,2 aremuchmoresensitive
the nodevoLtages
to changes
in R2than
to changes
in Rr. Specificatty,
r\ is (5.417
/0.5833)or approximatety
9 timesmoresensitive
to changes
in R2thanto changes
in R1 andtt2js
(6.5/0.5)or 13timesmoresensitive
to changes
in R2thanto changes
in
Rl. Hence
in theexampte
circuit,thetolerance
on R2mustbemorestringentthanthetolerance
on R1if it js important
to keepo1ando2closeto
theirnominal
vatues.
yau understanding
ll0TE: Assess
af thisPradicolPe5pective
byttyingChapter
Problems
4,1054.107.
Summary
137
5ummary
For the topics in tlis chapter,mastery of somebasictems,
and the conceptsthey represent,is necessary.
Thoseterms
are nod€, ess€ntisl node, path, brsnch, essentialbmnch,
m€sh, and planor circuit. Table 4.1 piovides definitions
and examplesof theseterm$ (Seepage 94.)
Two new circuit anaiysjstechniqueswere introducedin
tlis chapter:
. The node-voltage method works with both planai
and nonplanar circuits. A reference node is chosen
ftom among the essentialnodes.Voltage variables
are assignedat the remaining essentialnodes,and
Kirchhoff'scurrentlaw is usedto write one equation
per voltage variable.The number of equations is
,?"- 1, where r, is the number of essentialnodes.
(see pagee7.)
. The mesh-current method works or y with planar
circuits.Mesh cunents are assignedto each mesh,
and Kirchhoff's voltage law is used to write one
equation per mesh. The rumber of equations is
6 - (, - 1), where b is the number of branchesin
which the curent is unknown,andn is the numberof
nodes.The mesh currents are used to find the bralch
curents. (Seepage105.)
. Th6v€nin equiElenrs and Norton equivalents allow
us to simplifya circuitcompris€dof sourcesand resistors into an equivalent circuit consisting oI a voltage
source and a series resistor (Th€venin) or a current
souce and a parallel resistor (Norton). The simplified
circuit and the original circuit must be equivalent in
terms of their terminal voltage and curent. Thus
keep in mind that (1) the Thdvenin voltage (yrJ is
the open-cjrcuit voltage aooss lhe teminals of the
odginal cjrcuit, (2) the Th6venin resistaflce (Rn) is
the ratio of the Th6venin voltage to the shortcircuit
curent across the terminals of t]re oiginal circuit;
and (3) the Nonon equivalent is obtained by performing a source transformation on a Th6venin
equivalent.(Seepage119.)
Maximum power transfer is a technique for calculating
the maximumvalueofp that canbe deliveredto a load,
Rr. Maximum power transferoccu$ when Rl = Rrh,
thc Th6venin resistanceas seen from the resistor Rr,
The equationfor the maximumpower tmnsferredis
v+d
'
Several rew circuit simplification techniques were
introducedin this chapter:
. SourcetraNformationsallow us to exchangea voltage source (o") and a sedesresisror (R) for a current
qouce (1.) and u paralielfesisror(R) and \ icever.a.
The combirations must be equivalent in terms oI
tleir terminal voltage and current- Tefininal equiva,
lenceholdsprovidedthat
(Seepage116.)
4Rt
(Seepase 126.)
In a circuit witl m ltiple independent sources,
superyositionallowsus to activateone sourceat a timc
and sum the resultingvoltagesand curreDtsto det€r'
mine the voltagesand currentsthat existwh€n all inde
pendent sources are active, Dependent sources are
never deactivatedwhen applying superposition.(See
pase 129.)
138
Techniques
of Cjrcujt
Anatysis
Problems
Sectiotr4.1
41 Assune ttre current is in the circuit in Fig. P4.1 is
known.TheresistorsR1 Rs are alsoknown,
a) How many unknoM currents are there?
b) How many independent equations can be witten usingKirchhoff'scurent law (KCL)?
c) Wdte an indeperdent set of KCL equations.
d) How many independent equations can be
derived from Kirchhoff's voltage law (KVL)?
c) How many branches are theie?
d) Assume tlat tle lower node in each part of the
circuit is joined by a single conductor. Repeat
fie calculationsin (a)-(c).
Flgufe
P4.3
nr
l)
jn, r'J
e) Write a set of hdependent K\aL equations.
Figur€
P4.1
{) pr,
R5
4.4 a) If only the essential nodes and bnnches are
42 For the circuit shown in Fig. P4.2,state the trumerical
value of the number of (a) hanches, (b) brarches
wherc the current is urkno\rn, (c) essentialbmnches,
(d) essentialbrancheswherc the cunent is unknown,
(e) nodes,(f) essentialnodes,and (g) meshes.
FigureP4.2
identified in t}le circuit in Fig. P4.2, how many
simultaneousequationsare neodedto descdbe
ttre circuit?
How many of these equationscan be derived
using Kirchhoff's current law?
c) How many must be derived using Kirchhoff's
voltagelaw?
d) Wbarrso meshes
shouldbe dvoidedin dpplying
the voltage law?
4.5 A current leaving a node is defined as positive.
a) Sum the currents at each node in the circuit
showr in Fig. P4.5.
b) Show ttrat any one of the equations in (a) can be
derived from tlle remaining two equation$
43 a) How many separate parts does the circuit in
Fig. P4.3 have?
b) How many nodes?
i)
,,ll^, ,,llo, ili
3
PrcbL€ms
139
Section 42
FiglreP4.10
18{))
46 Use t}le node-voltage method to find 2,. in the cir6rc cuit in Fig.P4.6.
128V
FigureP4.6
a7 ^) Find
t]le power developed by tle 3A current
source in the circuit in Fig. P4.6.
Find the power developed by the 60 V voltage
source in the circuit in Fig. P4.6.
c) Verify that tle total power developed equalsthe
total power dissipated.
411 The circuit shoM in Fig. P4.11is a dc model of a
Atrft rcsidential power distdbution circuit.
a) Use the node-voltage method to find rhe branch
currentsi1 i6,
b) Test your solutior for the bianch currents by
showingthat the total powei dissipatedequals
the total power developed.
Figure
P4.11
1{)
125V
4.8 A 10 O resistoris connectedin serieswith the 3A
6flft current source in the circuit in Fig. P4.6.
a) Find ?,,.
b) Find the power developed by the 3,{ currenr
125V
c) Find the power developedby the 60 V voltage
d) Ved$' thatthe totalpower developedequalsthe
total power dissipated.
e) What effect wil any finite resistance connected
in serieswith the 3,{ current source have on the
valueof o.?
t3
4.12 Use the node-voltage method to find ?,1and 02 in
am the circuit in Fig P4.12.
tigureP4.12
4f}
80|l
4.9 Use the node-voltage method to find ?J1atd z)2in
the circuit sho$n in Fig. P4.9.
FlglreP4,9
25tl
4.13 Use the node-voltagemethod to find how much
E!'c power the 2A sourceextracts ftom the circuit in
Fig.P4.13.
Figure
P4.13
410 a) Use the node-voltage method to find the bmnch
curreirts ir
;€ in the circuit shom in Fig.P4.10.
b) Find the total power developed in the circuit.
140
Iechniques
ofCjrcuit
Anatysis
4.14 a) Use the node-voltagemethodto find t r, 02,and
t'3 in ttre circuit in Fig. P4.14.
b) How much power does the 640V voltage source
deliver to t}le circuit?
Figure
P4.17
20o'
figureP4.14
3A
2.54
+
640V
+
, , J ' 0 o "j,' o ( 1
l)
2A
12.8A
4,18 a) Find the node voltagestt, ,2, and r'3in the circuit in Fig.P4.18.
2.5A
4.15 Use the node-voltagemethod to find the total power
tstrG dissipatedin the circuit in Fig. P4.15.
b) Find the totalpower dissipatedin the circuit.
tigur€P4.18
Figure
P4.15
+ l
30v
)
25ll
31.25()
50()
5oo
(
500
4,16 a) Use the node-vollage method to show that the
output voltage o, in the circuit in Fig. P4.16 is
equal to the averagevalue of the sourcevoltages.
b) Fhd o, if 2,1-150V,
r'2 = 200V, and
03 = -50 V-
20()
5r}
1 ) 2 , "" i r o o or | : o o o ^ {
\_-,/
15()
25o"
+
(:
38.5V
5 i,,
4.19 Use the node-voltagemethod to calculate the
Fr't po$ef delivered
vollagesourcein
by rhedependenl
the circuit in Fig.P4-19.
figurcP4.19
5()
10J)
'jlsoo
80v
tigureP4.16
150
420 a) Use the node-voltagemethod to find the total
power developed in the circuit in Fig. P4.20.
b) Check your answerby finding the total power
absorbedin the circuit.
Seclion 4.3
4.17 a) Use the node-voltage method to find o, in the
circuitin Fig.P4.17.
b) Find the power absorbedby the dependentsource.
c) Find the total power developed by the independent sources.
tisul€P4.20
5()
300
-
t)
Jtso J:oo Jruo
5ir
Prcbtems14X
S€ction 4.4
FigureP4.24
500()
421 Use the node-voltage method to find the value of o,
Atrc ir the circuit in Fig. P4.21.
Figure
P4.21
800J)
10v
4r5 Use the node'voltage method to find the value of o,
EnG in the circuit in Fig.P4.25.
Figure
P4.25
4.22 Use the trode-voltage method to find i, in rhe cir6nc cuit in Fig. P4.22.
Figure
P4.22
10A
423 a) Use the node-voltagemethod to find the power
+
426 Use the node-voltage method to find r', and the
poser deliveredb) lhe 40 V voltagesoufcein the
circuitin Fig.P4.26.
dissipated in the 5 O resistor in the circuit in
Fig. P4.23.
b) Find the power supplied by the 500 V source.
rigureP4.25
40v
figureP4.23
417 Use the node-voltage method to find o, in the cir6trc cuit in Fig.P4.27.
Figure
P4.27
424 a) Usethe node-voltagemethodto find the bmnch
currentsil, j2,and i3in the circuit in Fig.P4.24.
b) Checkyoursolutionfor tr, tr, andi3by showing
that the power dissipatedin the circuit equals
thepowerdeveloped.
100
50v
rals
| , .
l ' v l
10o
: r oo
+
i3oo,,
142
Technique5
of Cncuit
AnatJ6k
428 Assume you are a project engineer and one of your
Bfl(! staff is assigned to analyze the circuit shown in
Eg. P4.28.The referencenode and node numbers
given on the figurc were assigned by the analyst.
Her solution gives the values of D3and oa as 235 V
and 222V, respectively.
Test these values by checking the total power
developed in the circuit against t}le total power dissipated. Do you agee vrith the solution submitted
by the analyst?
Figure
P4.28
(30)ro
FigureP4.31
3 r)
40v
+
2A
4A
r'
,,1+*'
+ 64V
1.5o
4.32 a) Use t]le mesh-currcnt method to find tlte total
powei developed in the circuit itr Fig. P4.32.
b) Check your answer by shovring that the total
power developed equals the total power
dissipated.
Figurc
P4.32
2tl
I
1()
3
I+ ti
')
25ov(
zoa
4r}
F-
10{}
40r}
4
q
10r)
3.2ra
Da
110V
429 Use the node-voltagemethod to find the power develBnc oped by the 20 V sourceh the circuit in Fig. P4.29.
2A
4.33 SolveProblem4.10usingthemesh-current
method.
Figur€
P4.29
35ia
434 SolveProblem4.11usingttremesh-currentmetlod.
4.35 SolvePrcblem4.22usingthe mesh'curent metlod.
3.125rA
436 SolveProblem4.23usingthe mesh-currentmethod.
Sectior 4.6
430 Showthat when Eqs.4.16,4.17,
and 4.19are solved
for is, the rcsult is identicalto Eq.2.25
Section45
431 a) Use the mesh-current method to find the branch
currentsi, lr, and i. in the circuitin Fig.P4.31.
b) Repeat (a) if the polarity of the 64 V sourceis
revelsed.
4.37 Use the mesh-current medod to fitrd the power dis"
tsrG sipated in the 8 O resistor in the circuit ir Fig. P4.37.
Figure
P4,37
16f}
4A
Probt€ms
143
438 Use the mesh-current method to find the power
6dc delivered by the dependent voltage source in the
chcuit seenin Fig.P4.38.
FigureP4.41
tigureP4.38
4.42 a) Use t]le mesh-curent method to solve foi iA in
the circuit in Fig. P4.42.
b) Find the power deliveied by the independent
439 Use the mesh-curent metlod to fhd the power
si,G developedin 1hedeperdent voltage sourcein the
circuit ir Fig. P4.39.
c) Fhd the power deliveredby the dependentvol!
age soulce,
FiglreP4.42
Flgure
P4.39
9800
1.8kO
2.65u!
' , 1 * ,o, n
15()
200ir
4.7kA
250
4,43 Use tle mesh-currentmethod to find the total power
6dc developedin the circuit in Fig. P4.43.
125V
FigneP4.43
4.40 a) Use the mesh-curentmetlod to find ,l, in the
circuit in Fig.P4.40.
b) Findthepowerdeliveiedby thedependent
souce
FigureP4.40
2A
0.5 ,I
12o"
165V
4.,|4 Use the mesh-cuffent method to fhd the total power
6dc developedin tlle circuit in Fig. P4.44.
S€ction 4,7
441 a) Use the mesh-curent method to find how much
power the 12 A current souce delivers to the
circuir in Fig. P4.41.
b) Find the total power delivered to tle circuit.
c) Check your calculations by shovring that the
total power developedir the circuit equalsthe
total power dissipated.
FigureP4.44
25{}
20()
144
Technjques
of Cncuit
Anatysis
445 a) Use the mesh-currentmethodto find the power
defiveredto the 2 O resistor in the ckcuit in
Fig. P4.45.
b) What percentage of the total power developed
in the circuit is delivered to the 2 {) resistor?
b) Repeat(a) if the 3 A curent sourceis replaced
by a short circuit.
c) Explah why the answersto (a) and (b) are
the same,
4.50 a) Use the mesh-clrlrent method to find the branch
curents in i.
i" il the circuit in Fig. P4.50.
b) Check your solution by showingthat the total
power developedin t}le circuit equalsthe total
power dissipated.
tigur€P4.45
L Zr ^
Figure
P4,50
10v
100f}
4.46 a) Use the mesh-curent method to determine
which sourcesil the circuit in Fig. P4-46are genemting power.
b) Find the total power dissipated it the circuit.
figureP4.46
2A
4.51 a) Find the branch currents ia - ie for the circurt
5 ()
shownin Fig.P4.51.
Check your answersby showingthat the total
power generated equals the total power
dissipated.
FigureP4.51
1 5d
*tc
4.47 Use the mesh-currcnt method to find the total
xnc power dissipatedin the circuit in Fig.P4.47_
FigureP4.47
3{)
9{))
21,48Assume the 18 V sourcein the circuit in Fig. P4.47is
""c increasedto 100V Find the total power dissipated
in tlle circuit.
4.49 a) Assume t}le 18V sourcein the circuil inFig. P4.47
is changedto -10 V. Find the total power dissipated in the circuit.
S€clion 4.8
452 The circuit in Fig. P4.52is a directcurrent ve$ion
6rt! of a typical tbree-wire distdbution system. fhe
resistoisR", Rb,and R" representthe rcsistances
of
the three conductorsthat connectthe three loads
Rr, R2,and R3to the 110/220Vvoltage supply.The
resistoft Rl and R2 representloads connectedto
Prcblens145
the 110V circdts, and R3 tepresentsa load con,
nectedto the 220V circuit.
a) What circuit analysis method will you use
and why?
b) Calculateq, 2,2,and q.
c) Calculatethe power deliveredto Rr, R2,and R3.
d) What percentageof the total power developed
by the sourcesis deliveredto the loads?
e) The Rb branch representsthe neutral conductor
in the distribution circuit. What adverse effect
occurs if the neutml conductor is opened? (Hir?r:
Calculateot arld r'2and note that appliancesor
loads designedfor use in this circuit would have
a nominalvoltagerating of 110V)
Rr=18o
Rr=54625O
1t0v
A 4k O resjstor is placed in parallel with the 10 nA
curent source in the circuit in Fig. P4.54.Assume
you have been askedto calculatethe power developedby the curent source,
a) Which methodof circuit analysiswould you rccommend?Explain why
b) Find tlre power developedby the currentsource.
4.56 a) Would you usethe node-voltage or mesh-current
method to lind the power absorbed by the
10 V source in the circuit in Fig. P4.56?Explain
your choice.
b) Use the method you selectedin (a) to find
the power.
Figure
P4.52
R. = 0.1O
110V
Flgure
P4.54
Figure
P4.56
r,E.
R, = 110.5
o
1qv
4.53 Show that whenever R1 : R, in the circuit in
Fig. P4.52,the curent in the neutral conductoris
zero. (,r1/rti Solve for t}le neutral conductor curent
as atunction of Rr and -R).
d54 Assume you have been askedto filld the power dissiFc, pdledin lhe I kO resi.rorin rhecircuirin Fig.P4.54.
a) Which method of circuit aflalysis would you rcc
onrmend? Explain why.
b) Use your rccommendedmethod of analysisto
find tle power dissipated in the 1 kO resistor.
c) Would you changeyour recommendationif the
problem had been to find the powei developed
by lhe l0 mA currenrsource?
Lxplain.
d) Find the power delivered by the 10 mA current source.
10A
2i,
4.57 The vaiiable dc current source in the ciicuit in
Fig. P4.57is adjustedso that the power developed
by the 15A currentsourceis 3750WFind the value
Figure
P4.57
15A
\-./
'7.2
{l
420y
)
,/
15[)
200
40()
' n (
146
Iechniques
of CjrcuitAnatysis
4.58 The variable dc voltage source in the circuit in
Anc Fig.P4.58is adjustedso that i, is zero.
a) Find the value of Ydc.
b) Check youi solution by showing the power
developedequalsthe power dissipated-
4.61 a) Use source transfomations to find o, in the cir,
cuit in Fig. P4.61.
b) Find the power developed by the 340V source.
c) Find the power developedby the 5 A current
d) Vedfy that the total power developed equals the
total power dissipated.
figureP4.58
figureP4.61
5{)
340V
15()
+
20()
10r)
Section 4.9
459 a) Use a sedes of source transformations to find
the cunent i, in tle circuit in Fig. P4.59.
b) Verify your solution by using the node-volrage
method to find i,.
4.62 a) Use a seriesof sourcetransformationsto find i,
in tlre circuit in Fig. P4.62.
b) Verify your solution by using the mesh-cufient
m€thodto find i,.
Figure
P4.62
l0A
Figure
P4,59
2.7kO
l)
$ 2 . : r oJ r r . o( f
4.60 a) Find t})e current in the 10 kO resistor in the circuit in Fig. P4.60 by making a succession of
appropriate source tmnsformations.
b) Using the rcsult obtained in (a), work back
through the circuit to lind the power developed
by the 100V source.
figurcP4,60
20ko
Seclion4.10
4.63 Find the Th6venin equivalent with respect to rhe
5r.r telminals a,b for the circuit in Fig. P4.63.
FigureP4.63
rko
5ko
* ) r o o v$ s o r (ot ) , : . o f u o o o
1ko
10(}
10ko
80
Probtems
147
4.64 Fnd the Thdvenin equivalentwith respectto the
5rft terminalsa,b for the circuit in Fig.P4.64.
Figure
P4.67
4ko
3ko
FigueP4.54
't2{r
2a
468 a) Find the Th6venin equivalent with respect to the
terminals a,b for the circuit in Fig. P4.68by find'
ing the open-circuit voltage and the shortc cuit
1,2v
4.65 Find the Th6venin equivalent with iespect
\c - !erminals
a.bfor lhe ciJcuitin Fig.P4.o5.
b) Solle for rheThdleninre.isrance
b! removing
the independent sources.Compare your result
to the Th6veninresistancefound in (a).
FiglreP4.68
Figure
P4,65
20o"
4.66 Find the Norton equivalentwith rcspectto the terErft minalsa,bin thecircuith Fig.P4.66.
FiglreP4.65
t5 ko
4.69 An automobilebatterywhenconnected
to a car
radio,provides
12.5Vto theradio.When
connected
provides11.7Vto theheadto a setofheadlights,it
lighfs.Assumethe mdio canbe modeledasa 6.25O
iesistoi and the headlightscan be modeledas a
WhataretheTh6veninandNorton
0.65{) resistor.
equivalents
for thebatlery?
470 DetermineL ando, in the circuitshownin Fig.P4.70
tsr.r whenR, is 0,2,4,10,15,20,30,50,60,
and70O.
6()
4.67 A voltmeterwith a resistanceof 100kO is used to
amE measrre the voltage ,"b in the circuit in Fig. P4.67.
a) What is tlte voltmeterreading?
b) mat is the percentage of eiror in the voltmeter
r€ading if the percentage of error is defined as
l(measured actual)/actuallx 100?
60()
148
4fl
*rft
Techniques
ofcncuitAnatysis
Determinethe Theveninequivalentwith respectto
the terminals a,b for the circuit shown in Fig. P4.71.
liglureP4.74
Figure
P4.71
9800
16(l
5 x l0 5.u,
1140
4.72 Find the Th6venin equivalent with respect to the
EEe teninals a,b for the circuirseenin Fig.p4.72.
ligtrrcP4.72
30ra
-\.;.-
zko
5ko
) ,,120ko
4,75 A Thdvenh equivalent can also be determined
trom measurements
made at the pair of terminals
of interest,Assume the following measurements
were made at the terminals a,b in the circuit ill
Fi+.P4.75.
When a 15 kO resistoris connectedto the terminals a,b,the voltage o,b is measuredand found
ro be 45V
10ko
:50ko
96{)
40ko
When a 5 kO resistoris conrected to the terminals a,b, the voltage is measuredand found to
be25v'
Find the Th6venin equivalent of the network
with respectto the teminals a,b.
4.73 Wlten a voltmeter is used to measure the voltage zre
En.r in Fig.P4.73,itreads7.5V
a) Wha[ is the resistanceofthe voltmeter?
b) What is t}le percentageof erroi in the voltage
measurement?
tigureP4.75
Figure
P4.73
_^
0.4v
4.74 When an ammeteris usedto measurethe currenttd
6tr( in the circuit shown in Fig. P4.74,ir reads 10 A.
a.) Wlat is the rcsistance of the ammete(
b) What is the percentage of error in the current
measurement?
4.76 The Wheatstone bddge in the circuit shown in
tsda Fig.P4.76is balancedwhenR3equals1200O.If tlle
galvanometerhas a rcsistanceof 30 O how much
cuuent will tlte galvanometer detect when the
bridge is unbalancedby setting R3 to 1204O?
(ftrf Find the Th€veninequivalentwilh respecrto
the galvanometerterminals when R3 = 1204O.
Note that once we have found rhis Th€venin equivalent, it is easy to find the amount of unbalanced
curent in the galvanometer branch for different
garvanometer
movements.)
Probt€ms
149
Figure
P4.76
Figure
P4.79
50()
----,,
1200C)
120V
60f,l
€qoo
200(:}
100v
+)60i
4.80 The variable resistor (R, in tle circuit in Fig. P4.80
6dc is adjusted for maximum power hansfer to RL.
a) Find the numericalvalueof RL.
b) Find the maximumpouer tmnsferredto RL.
S€ctiotr4.11
4.77 Find the Th6venin equivalent with respect to the
6trc terminalsa,b in the circuitin Fig.P4.77.
Figure
P4,80
tigureP4.77
10()
120
4.81 The vaiable resistor in lie circuit in Fig. P4.81 is
6flc adjustedfor maximumpower traNfer to R,.
a) Find the valueof R,.
b) Find the maimum power that can be delivered
to R,.
4.78 Fird the Th6venin equivalent with respect to the
ade terminalsa,b for the circuit seenin Fig.P4.78.
FigureP4.81
4k{)
1.25
ko
Figure
P4.78
10{)
120
4.82 What percentageof th€ total power developedin
Er( the circuitin Fig P4.81is deliveredto R. when R"is
set tbr maximumpower transfer?
Section 4.12
4.79 The variable resistor (&) in the circuit in Fig. P4.79
Eflc is adjusted until the power dissipatedin the resistor
is 1.5W. Find the values of R" that satisrythis condition,
4.83 A variable resistor R, is coinected across the teram minals a,b in the circuit in Fig. P4.?2.The vadable
resistoris adjusteduntil maximum power is translerred to R,.
a) Find the value of Ro.
b) Find the maxjmumpower deliveredto Ro.
c) Find the percentage of the total power devel
oped in the circuit that is delivered to R,.
150
lechniquesof
ftcojtAnatysjs
4.84 a) Calculate the power delivered for each value of
n, usedin Problem4.70.
b) Plot the power delivercdto R, ve$us the resistc) Ar what value of R, is the power delivered to Ro
a maximum?
b) Find the maximum power.
c) Find the percentage of the total power developed in the circuit that is delivered to R,.
tiguleP4.88
2A
40
4,85 The variable resistor (Ro) in tle circuit in Fig. p4.85
6trc! is adjusted for maximum powet transfer to R,.
What percentageof the total power developedin
the ciicuit is delivered to R,?
5fI
Figure
P4.85
4.89 The vadable rcsistor in the circuit in Fig. P4.89 is
6flc adjusted for maximum power transfer to _R,.
a) Find the numedcalvalueof &.
b) Find the maximum power delivered to R,.
c) How much power does ttre 280 V source detiver
to the circuit when R, is adjustedto the value
found in (a)?
FlgureP4.89
486 The vadable resistor (R") in the circuit in Fig. p4.86
6trc is adjusted for maximum power transfer to R .
a) Find the value of &.
b) Find t}le maximum power that can be delivered
to R,.
50 i4
10f)
so
20a
0.5725
a^
FigureP4.86
14ia
4.90 a) Find the value of the variable resistor R, in the
circuilin Fig.P4.S0lhal wijl fesullin ma\imum
power dissipation in the 6 O resistor. (I*nr:
Hasty conclusions could be hazardous to your
career.)
b) Wlat is the maximum power rhar can be delivered to the 6 O resistor?
4.87 w}lat percentageof the total power developedin
Atrc the circuit in Fig. P4.86is delivered to R ?
4.88 The va able rcsisror (R,) in t})e circuit in Fig. p4.88
tsdd is adjusted ulltil it absorbs maximum power ftom
the circuit.
a) Find rhe value of Rd.
figureP4.90
Probtems
151
Section4,13
Figure
P4.94
491 a) Use the principle of superpositionto find the
voltageo in tle circuitof Fig.P4.91.
b) Findthepo\aerdissipaled
in lhe20n resistoFiguleP4.91
4.95 Use superposition to solve for i, and z), in the cir6nrr cuit in Fig.P4.95.
FigureP4.95
400
30()
7 . 5A
20tr
492 Use the principle of superpositionto find the vottagex' in the circuit of Fig. P4.92.
I
Figore
P4,92
4()
180V
:60 O
80f}
250
496 Use the principle of superposition to find the curBflc rent L in the circuit shown in Fig. P4.96.
r-
Figure
P4.95
1.8()
10r)
50v
A 12o"
I ,)4.s
4.93 Use the principle of superposition to find
Erra rent L in the circuit in Fig. P4.93.
Flgurc
P4.93
5()
10J)
f):oetr+o
) s ov
497 a) In the ctucuit in Fig. P4.97,before the 10 mA current source is attached to ttre terminals a,b, the
current io is c?lculated and found to be 1.5 mA.
Use superyosition to find the value of i, after
the current source is attached.
b) Verily your solution by finding i, when all three
sourcesare acting simultaneously,
figuruP4.97
4.94 Use the principle of superposition to fhd r,. in the
6da ckcuit in Fig. P4.94.
15{)
mv
l)
r,J$
roro r,Jfrsr,o
rsro (t
Jroro
152
Techniques
ofCjrcuitAnatysis
Sections4.1-4.13
4.98 Laboratory measurements on a dc voltage source
yield a teminal voltageof 75 V with no load connectedto the soutceand 60 V when load€dwiti a
20 O,rcsistor.
a) What is the Th6veninequivalentwirh respectto
the terminals of the dc voltage source?
b) Show tlat the Th6ve n resistanceof the source
is given by the expression
/ t:-"
Rrh=[-1
\ r,,
\
1JRr.
/
4.100 Assume your supervisorhas askedyou to detemine
the power developedby the 16V sourcein the circuit
in Fig.P4.100.Before calculatingthe power developed
by the 16V source,the supervisorasksyou to submit a
proposal describirg how you plan to attack the problem. Futhemore, he asks you to explain why you
have chosenyour prcposedmethod of solution.
a) Describe your plan of attack, explainingyour
reasoning.
b) Use rhe merhod you have outlined in (a) to Iind
t}le power developed by the 16 V source_
Figure
P4.100
16V
q'i = the Th6venin voltage,
?, = the terminal voltage correspondiflg
to the load resistanceRL.
4,99 Tko ideal dc voltage souces are connected by electdcal conductors that have a resistanceof / O/m, as
shown in Eg. P4.99.A Ioad having a resistance of
R O moves between the two voltage soucen Let .r
equal the distancebetween the load and the source
?)1,and let .L equal t}re distancebetween the souices
a) Show that
8r,
4.101 Find the power absorbedby the 2 A current source
Edc in the circuit in Fig.P4.101
Figure
P4.101
30
8{)
UrR.+R(U, O)J
u=--.
RL + 2rLx
2rr'
b) Show that the voltage ?,will be minimum when
"Y:-l
-0r +
c) Find i when a = 16km,?)1: 1000V, u2 :
1200V, R : 3.9 O, and r' : 5 x 10 s O/m.
d) What is the minimum value of o for the circuit
of part (c)?
Figure
P4.99
4102 Find 01, r)2,and r,3in the circuit in Fig. P4.102.
figureP4,102
0.20
0.2A
110V
180
110V
| < - l - _
Probtems
153
4.103 Find i in the circuit in Fig. P4.103.
4,105 Assume the nominal values for the components in
t!61!q,v!
tigureP4.103
the circuirin Fig.4.60are: R. = 25 Q: R2 5 O:
R3 - 50 O;Ra : 75 O;1s1= 12A;and1s2= 16A.
Predict the values of q and D2if 1s1deqeases to
11A and all other componerts stay at thefu nominal
value$ Check your predictions using a tool like
Pspice or MATLAB.
4.106 Repeat Problem 4.105 if 1s2increasesto 17 A, and
Jrliflfii! all otler components stay a1 t]reir nominal values.
5;;i Check your predictions using a tool like Pspice or
MATLAB-
4.1o7 Repeathoblem 4.105if Is1 deqeasesto 11A ard
1()
t! !t!!!v!
/d2increase.
ro l7 A. Checkyour prediclions
using
a tool lile Pspice or MATLAB.
4108 Use the results given in Table 4.2 to predict tle val-
4.t04 For the circuit ir Fig.4.69 derive the expressionslor
the sensitivity of ?r and 1,2to changesin the source
currents1"j and1"r.
ues of 1)1and q if Rt and n3 increaseto 10oloabove
their norDinal values and R2 and Ra deqease to
l0o. belovr'
lbeiJnomiDai!alues./st and /s2remajn
at theh nominal values.Compare your predicted
values of or and ?2 with their actual values.
The0perationaI
Amplifier
Theelectroniccircuitknownas an operationaI
amptifierhas
5.1 operationalAmptifierT€rminalsp. 156
5.2 TerminatVottagesand Currentsp. 156
5.3 TheInverting-AmptifietCitctit p. 161
5.4 TheSumming-AnpLifiet
Citcuit p. 163
5.5 TheNoninverting-AnptifletCircu:tp- 164
5.6 TheDifference-Amplifier
Circuit p. 165
5.7 A MoreReatisticl4odetforthe operational
Beabteto nam€th€ five op ampteminalsand
describe
and usethe vottageandcurreft
constrairtsandthe r€sultingsimptjficatjons
t h e yL e a tdo i n a r i d e a t o pa m p .
Beableto anaLyze
simptecircujt' coitajning
ideatop anrps,and'ecosniz€the fottowjngop
ampcjrcujc:irveftingampfifier,summing
amptifier,nonjnvertirsampLjfi
er, anddiffererce
Understand
the morereaListjc
modetfor ar op
ampandbe abteto usethis modetto anatyze
simpt€cjrcujtscontainingop amps.
becomeincreasinglyimpofiant- However, a detailed analysisof
this circuit requircs an understandingof electronicdcvicessuch
as diodcs and transisloN.You may wondel, then, why we are
introducirlgthe circuit bcforc discussingthe circuit's electronic
componcnts.There are severalreasons,First,you can developan
appreciationfor how the opcrationalanlplifiel cat1be usedas a
circuit buildirg block by focusingon its terminal bchavior.At an
introductory level,you need not fully understandthe operation
of the electronic componentsthat govem termiDal behavior.
Second,the circuit model of lbe operationalamplificr requires
thc useof a dependelt source.Thusyou havea cbanceto usethis
type oI sourceill a practicalcircuit rather than as an abstractcircuit component.Third, you can combinc thc operationalamplificr with resistorsto pedorm son,cvery usefuLlhDctions,suchas
scaling,summing,sign changing,and subtracting.Finally, after
iltroducing inductorsand capacitorsin Chapter 6, we cal1show
you how to use the opcrational ampliiier 1()designintegrating
and dillercntiatingcircuits.
Our iocuson the termiral bchaviol of the operationalampli
fiel irnpliestaking a black box approachto its operation;thar is,
we arc not interestedin the internal structureof the ampliliernor
in the cuuents and voltagesthat existill this structure.Thc important thing to remernberis that the intemal belraviorof the anplifier accourtsfor the voltageand cu ent constraintsimposedat
the terminals.(For now,we ask that you acccpttheseconstrairlts
on liith.)
PracticalPerspective
StrainGages
jn a metaL
Howcou[dyoumeasure
theamount
of bending
bar
jn
physjcaLty
suchasthe oneshown the fiqurewjthout
contactjngthebar?onemethod
woutdbeto usea straingage.A
straingagejs a typeof transducer.
A transducer
is a devjce
that neasures
a quantityby convefingjt into a moreconvenientform.Thequantiwwe wishto measure
in the metal
baris the bending
angte,but measuring
the angtedjrectty
is
quitedifficuttand coutdevenbe dangerous.
lnstead,we
attacha strajngage(shown
in the Ljredrawing
here)tothe
netatbar.A straingageis a gridof thjn wjreswhoseresjstancechanges
whenthewiresaretengthened
or shortened:
\t
AR = 2R=
L
whereR is the resistance
of the gageat rest, AI/a js the
fractionatlengthening
of the gage(whichis the definitionof
"strajn"),the constant2 is typicalofthe manufacture/s
gage
factor,andAR is the changejn reshtance
dueto the bending
pairs of stEin gagesare attachedto
of the bal, TypicaLty,
oppos'tesidesof a bar.Whenthe ba-is bent,the wiresin one
pajr of qagesget longerand thinner,increasjngthe resistance,whjlethe wjresin the otherpajf of gagesget shorter
andthjcker,decreasing
the resistance.
But howcanthe changein resistance
be measured?
one
way woutdbe to usean ohmmeteLHowever,
the changejn
resistance
experienced
by the strain gaqejs typjcattynruch
smatterthan couldbe accuratety
measured
by an ohmmeter,
pairs
gages
Usuatlythe
of strain
are connectedto form a
Wheatstone
bridge,and the vottagedjfferencebetweentwo
legsof the bridgeis measured.
In orderto fiake an accurate
fieasurement
ofthe voltagedifference,
we usean operationat
ampufiercircujt to amptit/, or increas€,the voltagedifference.After we intfoducethe operationalampLifier
and some
of the impotant circujtsthat emptoythesed€vjces,we witL
presentthe circujt usedtogetherwith the strain qaqesfor
rreasudng
rheamounr
of beldingin a metalb:,,.
Theoperationat
amptifiercircuitfirst cameinto existence
asa basicbujLding
btockin analogcomputers.
It wasreferred
to asopetutional
beca$eit wa5usedto impLement
the mathematicalopentionsof integation, differentiation,addition,
sign changing,and scaLing.In recentyears,the rangeof
hasbroad€ned
appLication
beyondimptem€rtingmathenratjcaIoperations;
however,
the orjginaInamefor the cjrcuitper
sists,Engineers
andtechnjcians
havea penchantfor creating
technicaljafgon; hencethe operationalampLifieris wideLy
knownasthe op amp.
155
156
TheoperationatAmptifier
5.1 Sperationa[
.{rnptifierTerrnina[s
Becausewc are stressingthe temhal behaviorof the opemtionalamplifier (op amp), we begin by discussingthe terminalson a commercialty
availabledevice.In 1968,FairchildSemiconductorinlroducedan op amp
that hasfound widespreadacceptance:the
/.A741. (The /rA prefix is used
by Fairchiidto irdicate a microcircuitfabricationof the amplifier.)This
amplifieris availablejn severaldi{ferentpackages.
For our discussion,
we
assumean eighllead DIP.I Figure 5.1 showsa top view of thc package,
with the terminal designalionsgiven alongsidethe terminals.The rerminalsof primary inierestare
. invertinginput
' noninvertingi put
. output
. positjvepower supply(r+)
. negativepower supply(Y )
The remahing thrce teminals are of little or no concern.The ofliet null terminals may be used in an auxiliary circuit to compensatefor a degadation
Figure5.1 .,Ihe eishtteadDIPpackase
(topview).
in performance becauseof aging and imperfectiors. Howcver, the degradalion in most casesis negligible.so the offset terminals often are unused and
P o s i r n Lp o $ e rs u p p l )
Nonir\LLIrns
play a secondaryrole in circuit analysis.Terminal B is of no interest simply
F\.t
Inpur l.+ t\\
becauseit is an unusedterminal;NCstandsfor no connection,
whichmeans
I
>-outDur
nrverrrng-F J/
that the terminal is not connectedto the amplifier circuit.
i n'D u t l - - 4 .
Figure5.2 showsa widelyusedcircuit symbolfbr an op amp thar con
rcgrrre nowersupprv
tainsthe five teminals ofprimary interest.Using word labelsfor thc terFigure5.2 n Thecircuitsymbottor
anopeEtional
minals is jnconvenientin circuit diagams, so we simplify the rerminal
designationsin the following way. The roninve ing input terminal is
labeledplus (+), and thc inverting input terminal is labeled minus (-).
The power supplyterminals,which are alwaysdralvnoutsidethc triangle,
are marked y- and y_. The terminal at the apex of the triangularbox is
alwaysunderstoodto be the output terminal.Figure5.3 summarizesthese
sirnplifieddesignations.
Figure
5.3rr A simptified
circuit
symbotforan
opanrp.
5.2 TerminaI
Voltages
andeurrents
We are now readyto introducethe terminal voltagesand curtentsusedto
descdbethe behaviorof the op amp.The voltagevadablesare measured
from a commonrefercncenode.2 Figure 5.4 showsthe voltagcvariables
with lheir referencepolarities.
Conmonnode
tigure5.4 ,g lemjnatvottage
larjabtes.
t DIP is an abbeviation lor .fual d lte pd.nda..'Ihis neans thar rhe temrinals
on eachside of
tlre packageare in linc, and that lhe temnrah on opFsite sidesof the packagelho line up.
I l]1ecotunor nodc is extemai to the op anp It h the reterene temnral
ol the circuir in whicl
the op anp is embcdded.
5.2 TeminalVottages
andturrents 157
All voltages arc considered as voltage sesfrom the common node.
This conventionis the sameas that used in tle node-voltagemettrodof
analysis.A positive supply voltago (ycc) is connectedbetween l^ and
the common node. A negative supply voltage (-ycc) is connected
between y- and the common node. fhe voltage betweetr the irverting
input terminal and the commonnode is denotedo,. The voltagebetween
the noninvertinginput termhal and the common node is designatedas
o,. The voltage betweenthe output lermiltal and the common node is
Figue 5.5 shows the curent variables with their reference dircctions,
Note that all the current rcference d[ections are into t]le terminals of the
operational amptJier: i, is the curent into the inverting input terminal;i?
is the curent into the noninveting input terminal; io is the current into
the output terminal;i.* is the curent irto the positive power supply terminal and i" is the curent itrto the negative power supply terminal.
The teminal behavior of the op amp as a linear circuit element is
characterized by constaints on the input voltages and the input currents.
The voltage constraint is dedved from the voltage transfer characteristic
of the op ampintegratedcircuit and is picturedin Fig.5-6.
The voltage transfer characteristic descdbes how the output voltage
varies as a function of the input voltages;that is, how voltage is tmnsferred
ftom the input to the ouQut. Note that for the op amp, the output voltage
is a function of the difference between the input voltages,o, - o,,. The
eouationfor the voltasetransfercharactedstic
is
\:$
- "^)
A(Dt,-D")<-Ucc.
- Vcc = A(up rn) < +Ucc,
A(np n") > +Ucc.
Figure5.5 A Ie,minatcu,reni
vadabe5
(5.1)
tigure5.6A Thevoliage
transfer
chancteristic
ofan
We seeftom Fig. 5.6 atrd Eq. 5.1 that the op amp has three distinct
iegions of opemtiotr. When the magnitude of the input voltage difference
( o, - o, ) is small, t]Ie op amp behaves as a linear device, as the output
voltage is a Linearfunction of tlle input voltages.Outside this linear region,
t]Ie output of the op amp satumtes,and the op amp behavesas a nonlinear
device, becausetlle output voltage is no longer a Lhear functiol of the
input voltages.When it is opemting linearly, the op amp's output voltage is
equal to the difference in its input voltages times the multiplying constant,
or gaitr.,4.
wllen we conline the op amp to its linear operaling region, a corstraint is imposed on the input voltages,o? and r'" . The constraint is based
on tpical numerical values for ycc and ,4 in Eq. 5.1.For most op amps,t]le
recommendeddc power supply voltages seldom exceed20 V and t]Ie gain,
,4, is mrely lessthan 10,000,or 104.We seefrom bottr Fig.5.6 and Eq.5.1
that in the linear region, the magnitude of the input voltage differerce
( l r ? , , I m u s l b el e s st b a o2 0 l 0 " . o r 2 m V
The0p€rabonat
Amplifi
er
Typically, node voltages in th€ circuits w€ study are much larger than
2 mV,so a voltagedifferenceoflessthan 2 mV meansthe two voltagesare
essentiallyequal.Thus,when an op amp is constrainedto its linear operal
ingregion andthe nodevoltagesarernuchlargerthan2 mV the constraint
on the input voltagesof the op amp is
Inputvottageconstraint
for idealop ampts
(5.2)
Note tlat Eq.5.2 characterizes
the relationshipbetweenthe input voltages
for an ideal op amp;thatiq an op ampwhosevalueof.4 is infinite.
The input voltage constrajnt jn Eq.5.2 is called the yiltual short
condition at the input of the op amp.It is natural to ask how the virtual
short is maintained at the input of the op amp when the op amp is
embeddedin a circuit, thus ensudnglinear operation.Theansweris that
a signalis fed back from the output terminal to the inverting inpul ter
minal. This configuration is known as negative feedback becausethe
signal fed back ftom the output subtiacts from the input signal.The
negative feedback causes the input voltage difference to declease.
Becausethe oulput voltage is proportional to the input voltage difference,the outpul voltage is also decreased,and the op amp operatesin
its linear region,
If a circuit containing ar op amp does not provide a negative leedback
path from lhe op amp output to the invertinginput, then the op amp will
nonnally saturate.The difference in the input signals must be extremely
smallto preventsaturationwith no negativefeedback.B t evenif the circuit providesa negativefeedbackpath for t}le op amp,linearoperationis
not ensured. So how do we know whetler the op amp is operating in its
linear region?
The aNwer is,we don't!We deal with this djlemmaby assuminglin
ear operation, peforming the circuit analysis,and then checking our
results for coltradictions. For example,supposewe assumethat an op
amp in a circuit is operating in its linear region, and we compute the
output voltage of the op amp to be 10 V On examiningthe circuit, we
discover that ycc is 6 Y resulting in a contradiction, becausethe op
amp's output voltage can be no larger than Vcc, Thus oul assumption
of linear operation was invalid, and the op amp output must be saturated at 6 V
We have identified a constrainton the input voltagesthat is based
on the voltage transfer characteristicof the op amp integrated circuit,
the assumptionthat the op amp is restricted to its linear operating
rcgion and to typical valuesfor ycc and.4. Equation 5.2 representsthe
voltage constmint for an ideal op amp, that is, wittr a value of,4 that is
infinite.
We now tum oul attention to the constraint on the input currents.
Analysis of the op amp integrated circuit reveals that the equivalent resistanceseenby the input termimls of tlre op amp is very large,tpically 1 M O
5.2 Te,minatVottages
andcu'renis 159
or more. Ideally, the equivalent input resistanceis infinite, resulting in the
(5.3) < Inputcurrentconstraint
for ideatop amp
Note ttrat the curent constraint is nor based on assuming the op amp is
confined to its linear operating region as was t}te voltage constraht.
Together, Eqs. 5.2 and 5.3 folm the constaints on terminal behavior that
define our ideal op amp model.
From Kirchhoffs current law we know that the sum of the currents
entering the operational amplfier is zero, or
ip+ir-
ia
i,
r. =u.
(5.4)
Substituting the constraint given by Eq.5.3 into Eq.5.4 gives
io:
Q , .+ i . ) .
The significance of Eq. 5.5 is that, even though the curent at the input terminals is negligible, there may still be appreciable curent at the output
terminal.
Beforc we starl analyzingcircuils containing op ampElet's further simplify tlle circuit symbol.w1len we know that the ampliEer is operaringwithin
its linear rcgion, the dc voltages+Vcc do not enter into t}Ie circuit equations.
In this case,we can remove the power supply teminals from tle s]'rnbol
and the dc power suppliesfrom the circuit, as showt ir Fig. 5.7.A word of
caution: Becausethe power supply teminals have been omitted, therc is a
dangerof inJening from the symbol that i, + in + i, = 0. We have already
noted that suchis not the case;that is,i, + i, + i, + i., + i.- : 0. In other
words! the ideal op amp model conshaint that i, : i, : 0 does no1 imply
that i" : 0.
tigure5.7 A The0pampsyrbotwiththe powersuppty
Note that tle positive and negative power supply voltages do not
have to be equal ir magnitude. In the linear operating region, z, must lie
between the two supply voltages.For example, if V+=15V
and
y = -10V,then-10V< D. < 15V.Be awarealsothatthevalueof.4
is not constant under all operating conditions. For now, however, we
assumet]tat it is. A discussion of how and why the value of .4 can change
must be delayed until after you have studied the electronic devices and
componentsused to fab cate an amplifier.
Example 5.1 ilustrates t}le judicious applicationof Eqs.5.2 ard 5.3.
When we use these equations to predict the behavior of a circuit containing an op amp, in effect we are using an ideal model of the device.
160
lhe operationaL
Amptifier
Anatyzing
an 0p AmpCircuit
Theop ampin thecircuitshownin Fig.5.8is ideal.
a) Calculate 1,, if ?J!- 1 V and ,b = 0 Vb) Repeat(a) for 0o : 1 V and ob : 2 V.
c) Ifo, = 1.5V, specifythe rangeof rb that avoids
amplifiersaturation.
t ! ! 1 0 0k O
The current constraint requires i, = 0.
Substiluting the values for the three currerts
into the node-voltageequation,weobtain
7
25
1).
100
''
Hence, t, is 4V, Notc that because,, lies
between j 10V, the op amp is in its linear
region of opemtion.
b) Using the sameprocessas in (a),we get
ap:ah:un=)y,
'"
25
1-2 :
25
100
ff-"'
Figure5.8 ..' Theci,cuitforEMmpte
5.1.
,25 :
1
25.^,
,1tl(]
Sotution
a) Becausea legative feedback path exists from the
op amp's output to its inverting irput througl the
100 kO resistor, let's assumethe op amp is confined to its linear operating region. We can write
a rode-voltage equation at the inverting input
terminal.The voltage at tlle irverting input terminal is 0, as r'r, : x'b = 0 from the connectedvoltage source, and uh:1)p from the voltage
constraintEq. 5.2.The node'voltageequationat
Therefore, ?,o= 6 V. Again, ?, lies within + 10 V.
c) As beforqo, = op = ob,andi25= itoo.Because
?]a:15V,
1 , 5- 0 b
25
0o-0t
100
Solving for ob as a function of o, gives
I.'b=i(o+"i
From Ohm'slaw,
,,r = ru. ,,,)/25:
l
-
mA.
= (?Jd 0,)/100:0,/100 mA.
4o1r
Now, if the amplfier is to be within the linear
regior of operation, -10 V < 0, < 10V.
Substituting these limits on o, into the expression lor ?b,we seethat ob is limited to
-0.8V<'Db<3.2V.
r ceu, i t 1 6 1
5 . 3 T h eI n v e d i n g - A m p (t ji F
objective1-Use voitageand curent constraintsin an ideat op amp
5.1
AqsumerbarIhe op amp i0 rhecircuir,hown
is ideil.
80ko
a) ( alcuialedolbr rhe louowingvaluc\oi, :
-0.6. 1.6,and -2.,1V.
0.4,2.0,3.5,
h) SpeciJ)
lhe rang.-ol d, requiredlo a\ord
amplifier saturation.
A n s w e r :( a ) 7 . 1 0 . - 1 5 . 3 . 8 . a nld0 \ ;
rbr 2\
,.1l\.
''IOl L: Also lry Charpr Ptublen\ 5.1-5.3.
5.3 TheXnvefri*lg-Amp['!fier
efrcsit
We are now readyto discussthe operationof someimportant op amp cir,
cuils,usirg Eqs.5.2 and 5.3 to model the behavior of the device itseli
Figure5.9showsan inverting-amplifiercircuit.We assumethat thc op amp
is operatingin jts linear regior Note that, in addition to thc op amp,the
circuit consistsof lwo resistors(Rl and R,), a voltagesignalsource(1,,),
anda shortcircuitconnectedbetweenthe noninvefiinginput terninal and
We row analyze this circuit, assumingan ideal op amp.The goal is ro
obtainan expressionfor the output voltage,.ro,as a functionof the source
voltage,or.Wc employa sirgle node voltageequationar the invertingterminal ofthe op amp,givenas
Figurc5.9 ). AnjnvedjngimpLifier
circuit.
is + if: in.
(5.6)
The vollageconstraintof Eq. 5.2 setsihe voltageat un = 0, becausethe
voltageat tlp = 0. Therefore,
(5.7)
'
.
J
1)a
D
..t
(5.8)
Now we invoke the constraintstatedin Eq.5.3,namely.
(5.e)
SubstitutingEqs.5.7-5.9into Eq.5.6 yieldsthe soughrafter resuit:
-R,
(5.10) { Inverting-amptifierequation
162
TheoperationatAmptifier
Note that the ouQut voltage is an inverted, scaledreplica of the input. The
sign revercalfrom input to ouQut is!of course,the reasonfor refening to tlle
circuit asan invertnS amplifier.The scalingfactor, or gain,is the ratio -Rf/Xs.
The result given by Eq. 5.10is valid only if the op amp shownin the
circuit h Fig.5.9is ideal;that is,iL4 is infinite and the input resistanceis
infinite. For a practicalop amp, Eq.5.10 is an approximation,usually a
good one. (We say more about this later.) Equation 5.10 is important
becauseit tellsus thatifthe op amp gain,4 is large,we canspecifythegain
ofthc invertingamplifierwith the externalresistorsRf and Rr.The upper
limflon rhegain.R/ R..is derermined
b) rhepot\er
supplyvoltagesand the value of the signalvoltageo,. Ifwe assumeequal
y- : fcc, we get
power supplyvoltages,thatis,y" :
ro < Vcc'
lnr I
=
l4*l u-'
R.
p
- lY-l
t,"l
(5.11)
For example,if ycc : 15 V and o" : 10 mV, the ralio Rl /R, must be less
than 1500.
In the inverting amplifier circuit sho*'n in Fig. 5.9,the resistor Rf pro
videsthe negativefeedbackcornection.That is,it connectsthe output ter
minal to the inveting input termiml. If Rf is removed, the feedback pattr
is openedand the amplifier is saidto be operatingopen loop.Fi9].lle5.10
showsthe open-loopoperation.
Opedng the feedbackpath drasticallychangestlle behavior of the
circuit.First,the output voltageis now
Figure5.10.d AninleidngampLifier
operating
(5.12)
assuming
asbetorethat V* : V : ycc; th€nlo,l < VccfA lor llnear
operatjon.
Because
theinve ing inputcurent is almostzero,thevoltage
drcpaoossR"is alnostzero,andtheinve inginputvoltagenearlyequals
thesignalvoltage,
o"; thatis,on - os, Hence,theop ampcanoperateopen
loopin thelinearmodeonlyif ltr, < ycc/A.If ?)"> ycc/A, theop amp
simplysatuates.In particular,
if tr, < ycc/A, the op ampsatumtesat
+ycc, and if tr, > ycc/A, the op a1npsaturates
at -ycc. Becausethe
path,the
relationship
shownin Eq.5.12occu$whenthereis no feedback
gainof the op amp.
valueofA is oftencalledtheopen-loop
0bjedive2-Be abteto analyze
sinpl; circuitscontaining
id€al.dpamps
5.2
\ollcgeu. in thecircuitin
lhesource
Assessment
Problem.s,lis-640 mV.The
6UkIt IeeoDacx
resrslor
rsreplaceo
b) a \anableresislorR. . Whar.angeot R, alloqslhe
NOTE: Abo try Chapter Problems 5.8 and 5.9.
,
inverting amilifier to opeiate in its lheaf
region?
Answer: 0 < Rl < 250ko
5.4 lhesumnins
Anrptjfier
cjrcujt 163
5.4 TheSumming-Amplitier
Circuit
The output voltageof a summingamplifier is an inveited,scaledsum of
the voltagesappliedto the input of the amplifier.Figure5.11showsa sum
ming amplifierwith three input voltages.
We obtain the relationshipbetweentlte output voltage o. and the
threeinput voltages,zra,?h,and o., by summingthecurrentsawayfrom the
invertinginput termiMll
&
*--*
'r1o
tn - uc
un
uo
1-+--+t":o
(5.13) Fisure
5.11.{ A summins
amptifier.
Assumingan ideal op amp,we cannsethe voltageand curenl constraints
together with thc ground imposed at z), by the circuit to see that
rn= DD: o zndin = 0.ThisreducesEq.5.13ro
"=
(\
\e*
Rt
+4'b+
Rr\
&"(/'
(5.14) r€ Inverting-sumningamplifier equation
Equation5.14statesthal the output voltageis an inverted,scaledsum of
the thrce input voltages.
If R" : Rb = Rc : &, then Eq.5.14reducesto
(5.15)
Finaily,if we make Rt = Rr, the output voltageisjust the irverted sum of
the input voltages.
That is,
1,, =
(oa+ 4+
0c).
Although we illustratedthe summingamplifier wjth just three input
signals,the number oI input voltagescan be inffeased asneeded.For example, you might wish to sum 16 individually recorded audio signalsto form a
single audio signal.The summing amplifier conJigurarionin Fig. 5.11could
irclude 16 djfferent input resistorvaluesso that cach of the input audio
tracks appears in the output signal with a differeni amplification factor.
The summhg amplifier thus plays the role of an audio mixer. As with
invefiing amplifier circuits, the scaling factors in summirg-amplifier circuitsare deternired by the exlemalresistorsRr, na, Rb,R., . . . , R,.
164
TheopeationatAmptifier
ideatop amps
0biective2*Be abteto anatyze
simptecircuitscontaining
5.3
a) Find r, jr the circuit shownifo. : 0.1V
and tb = 0.25V.
b) If tl6 = 0.25V, how large canoabe before
the op amp saturates?
c) lf x'" - 0.10V, how largecan 1]l)be before
I he op ampsarurares?
d) Repeat(a), (b), and (c.)wjth the polaity of
(c) 0.5v:
(d) -2.s,0.2s,and2v
'!_bIeVCned.
Answer: (a) -?.5 v;
(b) 0.15v;
NOTE: Alsotty ChaptetProblems
5.12,5.1i,dnd5.15.
5.5 Thef{oninverting-Amptitr'er
Cireuit
Figure 5.12depiclsa nonirveriing amplifier cilcuit. The signalsourceis
representedby us in serieswith the resistorRg.In derivhg the expression
Ior the oulput voltageas a function of the souce voltage,we assumeaD
ideal op amp operatingwilhin its linear region.Thus,as belore, lve use
Eqs.5.2 and 5.3 as t}le basisfor the dedvation.Becausethe op amp iDput
curenl is zero,we can write a, = 1" o"4,1to- tq. 5.2,2,, = u! as well
Now,becausethe input currert is zero (i, : ,/, : 0), the resistorsRf and
R, form an unloadedvoltagedivider acrossD..Therefore,
u,R,
Figurc5.12A A noninverling
amplifier.
xJ+R/
(5.17)
SolvingEq.5.17for o,,givcsusthesoughtafterexprcssion:
Noninverting-amptifi
er equationv
,,,:
R"+Rl
R,
0s.
(5.18)
Opcrationin the linear regionrequireslhat
o,*or.ly..
&
lr"
Note again that, becauseof the ideal op amp assumption,we can
expressthe output voltageas a functionoftheinput voltageandthe external resistors in this case,R" and Rt.
5.6
TheDiffeence-Amptifie,
tncuit
objective2-Be abteto analyze
simptecircdtscontaining
ideatop anps
5.4
Assumethat the op ampin the circuirshown
is ideal.
a) Find the output voltage when tlle vadable
resistoris set to 60 kO.
b) How large can n" be before the amplifier
saturates?
Answer (a) 4.8V;
(b)7sko.
NOTE: Abo try ChapterPnblems 5.17and 5.18.
5.6 The$ifference-"4r'nptifier
efreuit
Thc output voltageof a differenceamplifier is proportionalto rhe dificr
enccbetweentlle two inpul voltages.Todemonstrate,
we analyzethc diJfcrence-amplifiercircuit showDin Fig.5.13, assumiDgan ideal op amp
operatingin its linear region.We derive thc relationshipbetweeno, and
lhe two input voltagesDaand ?rbby summingthe currentsawayfrom the
inve ing input nodc:
Dn -
Xa
'nr
R.
1)a
(5.1e)
fir
Becausethl3op amp is ideal,weusethe voltageand currentconstraintsro
\5.20)
",r = 0e =
Rd
RJ
Rd
0Lr.
\5,2r)
Conbining Eqs.5.19,5.20,and 5.21 gives the desired relarionship:
'
1id(R"+ Rb)
R,(R.+ t?,r)"
Rr,
n"
"'
\5.??)
Equation5.22showsthar the output voltageisproportionalto the diflerencebetweenascaledreplicaofob and a scaledreplicaoI?i".Ingeneral
the scalingfactor applied to ob is not the samc as that applied to oa.
Howcver,Lhescaiingfactor applied to each input voltagecan be made
cqualby setting
R.
n"
Rr
Ra
\5.?3)
Figurc
5.13i,\ A differcnce
amptjner.
165
166
The0pe6tjonat
Amplifier
When 8q.5.23 is satisfied, the expiession for the output voltage reducesto
simptifieddifference-amp[iIier
equation>
15.24)
Equation 5.24 indicatesthat the output voltage can be made a scaled
replica of the difference between the input voltages ob and oa.As in the
previous ideal amplifier circuits, t]le scaling is controlled by the external
resisto$, Furthermore, tlte relationship between tle output voltage and
the input voltages is not affected by cornecting a nonzeio load resistance
acrossthe outDut of the amDlfier.
TheDifference
Amplifier-Another Perspective
We can examine t]te behavior of a difference amplifier more closely if we
redefine its inputs in tems of two othei voltages. The fbst is the
difiercnlial mod€ input, which is the difference between the two input
voltagesin Fig.5.13:
(5.25)
The second is tlre commotr mode input, which is the average of the two
input voltagesin Fig.5.13:
r.
= (o^+ Di/2.
\5.?6)
5.6
TheDjfference-Amptjfi€r
Circujt 167
UsingEqs.5.25and5.26,we cannowrepresentthe originalinput voltages,
oaand 0b,in tems of the differential modeatrdcommonmodevoltases.
lta:
D6
I
\5.27)
t|)dn,
I
rb:1)d+l:dn
(5.28)
SubstitutirgEqs.5.27and5.28into Eq.5.22givesrhe outpurof the differenceamplifier in terms of the differertial mode atrd commonmode
voltaees:
, t t"n"Rd - RbRc
R"(n. + Rd)
.I
Rd(Ra+R6)+Rb(R"+Rd)l
= ?4!nod
,RJRJ RJ
+ ,4dm0dm,
j'd-'
15.?9)
(5.30)
where Ad is t}le common mode gain and .4dn is tlle differential mode
gain. Now, substitute Rc = Ra and Rd : Rb, which are possible values for
Rcand Rd rharsatisfyEq.5.23,into Eq.5.29:
/nt\
\4/**
Thuq an ideal dilference amptifier has /d = 0, amplifies only rhe differential mode portion of the input voltagg and eliminates the cormon mode
po ion of the input voltage.Figure 5.14 shows a difference-amplifier
circuit lvith differential mode and common mode input voltagesin place of
1!4
2
3ss
Equation 5.30 provides an important pe$pective on t}le function of
2
the difference amplfier, since in many applicatioDs it is the differential
mode signal that contains tle infomation of hterest, whereas the common mode signal is the noise found in all electric signals.For example,an
electrocardiograph electrode measures the voltages produced by your
body to regulateyour heartbeat.Thesevoltageshave very small magni- rigure5.14A A differcnce
ampLifierwith
common
anddjff€r€ntiat
mode
inputvottages.
tudes compared with the electdcal noise that the electrode picks up from mode
sourcessuch as lights and electrical equipment. The noise appears as the
common mode poftion of the measured voltagg whereas tle hea rate
voltages compdse the differential mode portion. Thus an ideal difference
amplifier would amplify only the voltage of interest and would suppress
the noise.
168
TheoperationatAmptjfi€r
Measuring
Difference-Amplifi
er PerformanceTheCommon
ModeRejectionRatio
Ar ideal difference amplifier has zero conrmon mode gain and nonzerc
(and usually large) differe tial mode gain. Two factors have an influence
on tle ideal common mode gain-resistance mismatches(that is,Eq. [5.23]
is not satisfied)or a nonidealop amp (that is,Eq. [5.20]is not sarisfied).
We
locus here on the effect of resistancemismatcheson the performance of a
difference amplifier.
Supposethat resistorvaluesare chosenthat do not preciselysatisfy
Eq. 5.23.Instead,the relationshipamongthe resistorsR,, Rb,R", and Rdis
R"
R^, , '
-€)R
i;:(1
X" = (1 - €)R.
and
Rr = Ra,
(1 - €)Rb
and
R, = R",
Rd:
(5.32)
where € is a very small number.We car seethe effect of this resistarce
mismatchon the commonmode gain ofthe differenceamplifierby substituting Eq.5.33into Eq.5.29and simplifyingthe expressionfor ,4..,:
R,(1 -€)Rb-RaRb
n,[R"+(1 - €)Rb]
15.34)
-exr
Ra+(i-€)Rb
-eRr
-- R , . &
(5.36)
€ is verysmall,
Wecanmakethe approximation
to giveEq.5.36because
andthercforc(1 €) is approximately
1 in the denominator
of Eq.5.35.
Note that, when the resiston in the differenceamplifier satisfyEq. 5.23,
€ : 0 andEq.5.36givesAcn = 0.
Now calculatethe effectof the resistancemismatchon the differential
mode gain by substitutingEq. 5.33into Eq. 5.29and simplityingthe
expression
for ,4dm:
(r - e)n(n" + Rb) + R6[R"+ (1 - €)nb]
2RalRa+ (1 €)Rbl
'l
(e/2)Ra
n.I
R"L R , . ( 1 " ) & ]
(5.37)
(5.38)
5.6 lhe Differcnce-Amptifier
Circuir 169
= R,[4L'
(€/2)xiI
(5.re)
R "+ R ' ]
We usethe samerationalefor the approximationin Eq.5.39asin the com_
putation of r cm.Wlen the resistors in the difference amDtifier salisfv
Fq.5.)J.c = 0 and Eq 5.lq gi!ec?4o,- Ro RI
The conrmon mode r€jection ratio (CMRR) can be used to measure
how nearly ideal a difference amplifier is. It is defined as rhe rario of rhe
diflerenrial
modcgajnlo rhecommonmodesrin:
a. )
(5.10)
The higherthe CMRR,the more nearlyideal rhe differenceamplifier.We
can seethe effect of resistancemismatchon the CMRR bv substitutins
Eqs.5.36and 5.39into Eq.5.40:
* *",1
lltt ,o,+r,,,^,
C M R R- l - '
-l
-€fb/(Rr
+ Rb)
\5.41)
R"(1 €/2)+ Rbl
-&
1 1 +R b / R] "
I
\5.4?)
15.13)
From Eq. 5.43,if the resistorsin the differenceamDlifier are matched.
.
0 rnd CVRR - rc. L\en it rhe iesislorsare mi.marched.
ue can
minimize the impact of the mismatch by making the differential mode
gain(Rb/Ra)very large,therebymaking the CMRRlarge.
We saidat the outsetthat anotherreasonfor nonzerocornrnonmode
gain is a nonideal op amp. Note that the op amp is itself a difference
amplifier,becausein the linear operatingregion,its output is proporrional
to tlle difference of its irputs; that is, r,, : ,4(rp , o,,). The outpur of a
nonidealop amp is not strictlypropofiional to the differencebetweenthe
inputs (the differentialmode input) but also is compdsealof a common
modesignal.Internalmismatchesin the componentsofthe intesratealcircuil make(hebeha\ior ot rheop amp Donideal.
in rhe.ameua."y
rharrhe
resistor mismatchesin the difference-amplifier circuit make its behavior
nonideal.Even though a discussionof nonideal op amDsis bevond the
.copeol lhiste)\r.you ma! notelha(lhe CM RR i, orrenu:edin i.se*ing
how neaily ideal an op amp'sbehavioris.In fact,it is one ofthe main wavs
of mting op amps in pnctice.
NOTE: Assesstour undentanding of thb materiat by trying Chapter
Prcblems
5.32and5.33.
170
lheopentionat
tunptifier
5.7 A MoreReatisticModetfor the
0perational
Amptifier
We now consider a more realistic model tlat piedicts the perfomance of
an op amp in its linear region of operation. Such a model includes three
modifications to the ideal op amp: (1) a finite inpul iesistance, Rj; (2) a
finite oper-loop gain,,4;and (3) a noMero output iesistance,Ro.The circuit shown in Fig. 5.15illustr4tes the more realistic model.
A \rp - 1,),
Wheneverwe use the equivalentcircuit shownin Fig.5.15,we disregard the assumprionsthat t^ = Dp (Eq. 5.2) and ti = ,e = 0 (Eq. 5.3).
Furthermore,Eq. 5.1 is no longer valid becauseof the presenceof the
nonzero output resistance, Ro. Arother way to undentand the circuit
That is,we canseethat
shownin Fig.5.15is to reverseour thoughtprocess.
the circuitreducesto the idealmodelwhen R; + co, 14- m , and Rd+ 0.
For th€ /rA741 op amp,the tlpical valuesof &, ,4, and Ro are 2 MO, 10,,
and 75 O, respectively.
Figure5.15A Anequivalent
cncuitforanopeEtional
Although the pres€nceof Ri and Ro makes the analysisof circuits containing op amps more cumbeNome,such analysisremains straightforward.
To illustmte, we anal'ze both an invefiing and a noninverting amplifier,
ushg the equivalent circuit shown in Fig- 5.15.We begin with tle inverting
amplifier.
Analysisof an Inverting-Amplifier
€ircuit using
the MoreReatistic0p AmpModet
If we use the op amp circuit showl in Fig. 5.15,the circuit for the inverting
amplifier is the one depicted in Fig. 5.16.As before, oul goal is to express
ttre output voltage, o,, as a function of the souice voltage, r'r, We obtain
the desiredexpressionby writing the two node-voltageequationsthat
describe the circuit and tlen solving the rcsulting set of equations foi tro.
In Fig.5.16,the rwo nodesare labeleda and b. Also note that 2,, : 0 by
vi ue of the extemal short-c cuit connection at the noninverthg input
terminal. The two node-voltaee eouatioN are as follows:
node a:
Dn-bs
un , bn
R,
nooe o:
-
+
uo
15.11)
xr
D" - A(-D,\
(5.15)
(ncuit.
tigure5.16A Aninveiring-amptifier
We rearrangeEqs.5.44and 5.45so that the solution for o, by Cramer's
methodbecomesaDoarent:
lt
r
r\
1
1
\a*^,*4/" 4%-A.,"'
(5.40)
(*-+)...
(+.;)":"
15.47)
5.7 A li4orc
ReaListic
l4odeL
torihe 0perationat
Amptifrer 171
Solving for 0" yields
A + (R"/R)
".*).
(t.').*"
+('.
(5.48)
Note that Eq.5.48reducesto Eq.5.10asn, '0, &+ co, and "4 + oo.
I{ the inve ing amplifier shown in Fig.5.16 wero loaded at its output
term;nal(wi$ a loadre.istaoce
ot R. ohms.rhe relaliooship
belwee;,.
and o" would become
A + (R./Rf)
D.
t + A
--
.t.*).("r"X'
-4)-4
R,\
1,..(5.49)
&
Analysisof a Noninverting-Amplifier
€ircuitUsing
the MoreRealistic
0p AmpModel
w}Ien we use the equivalent circuit shown in Fig.5.15 to analyze a roninve ing amplifier, we obtain the circuir depicred in Fig. 5.17.Helq the voliage source os, in se es with tle resistatce Rs, represents tlle signal
source.The resistor Rr denotes the Ioad on the amplifiet. Our analvsis
consislsof derivingaDexpre<(ioDtor ,, as a flrncrionol 0". We do so by
writingrhenode-\ollage
equadoDs
ar nodesa andb. Ar nodea,
un
on-xg
Dn-Do
^,
&-Rr+R,'
= "^
(5.501
and at node b,
Figure
5.17A A noninverting
amptifrer
cncuit.
Becausethe currentin Rs is the sameasin Rr, we have
rp-Dg
I"-
Dg
RI+R"
15.52)
172
Theop€ntjonal
Amptifi€r
We use Eq. 5.52 to eliminate o/ from Eq. 5.51, giving a pair oi equations
itrvolving the unknown voltages on ard 2,. fhis algebraic manipulation
leadsto
lt
r
r\
/r\
a\n.' n"-n,'
\)
/
t
\
"\E)-%\R".R,/
(55r)
(t
en,
tl
I
r\
" lf n , r n- ," , - 4 l ' ".\ & - & ' & /
f
en. l
- ',Ln
rn,* n,rl
(5'51i
Soivjngfor lJ, yields
[(R/ + x") + (R,R"lARt\]Ds
4 * f1, * *.; +
^,,' :
Rc
&
Rr
&
RtR"+ (Rrl f"x^r + Rs)
R,
{5.55)
R1R. r R1R,- RrR.
RB,
&
Note that Eq.5.55 reducesto Eq.5.18 when R,+0, ,4+oo, and
Ri+ co. For the unloaded (Rz = c.) noninverting amplifier,Eq. 5.55
simplifies to
'
(nr+&)+
- . |n . ( "I .
R.
;\L
R"R"/ARID|
^ |
n , +" n , \r .. ^ R"
R,
)'
A:,R-,IR,
(R,
4XRi
R!)l
(5.56)
Note that, in the derivationof Eq.5.56 from Eq.5.55,1(, reducesto
(& + Rs)/n,.
Pra.ti.atPe6pe.tive 173
objective3-Understandthe morerealisticmodelfor anop amD
5.6
The invertingamplifierin the circuit shownhas
an input resislanceof500 kO, an output resisf
ance()15kO, and an openloop gain of300,000.
Assumcthat the amplifier is operatingin its
linear region.
Answer: (a)
19.9985:
(b) 69.995pV;
(c) 5000.3s
o;
(d) 20,0pV,5ko
a) Calcnlatethe volragegain (r,,/us) of the
anlpnner.
b) Calculatethe valueof z',,ir rticrovoltswhcll
D c= 1 V '
c) Calculatethe resistanceseenby thc signal
source(0s).
d) Rcpeat(a)-(c) usingthe ideal modetfor the
op amp.
NOTE: AIso tly Chaptcr PtobLems5.42dnd 5.43.
PracticalPerspective
StrainGages
Changes
in the shapeof etasticsotidsareof greatimpotanceto enqjr€ers
whodesjgrstructures
that twjst, stretch,or bendwhensubjected
to exter,
nalforces.An ajrcraftframeis a primeexampte
of a structurein whichengi
neersmusttaker'ntoconsideration
etasticstrain.ThejnteLtigent
apptjcatjon
of strain gagesrequiresinformationabout the physicatstructureof the
gage,methodsof bondingthe gageto the surfaceofthe structure,andthe
orjentationof the gageretativeto the forcesexertedon the structure.our
purposehereis to pointout that straingagemeasurements
areimportantjn
engjneering
apptications,
anda knowledge
of eLectric
cjrcujtsis germane
to
their oroDeruse.
Thecircuitshownin Fjg.5.18provid€s
onewayto measure
the change
in resistanceexperienced
by strain gagesjr appLicatjons
Ike the one
F+ARa
fR-tR
Figure5.18n Anopampcircujtusedfor measuing
thechange
in nraingage
r00ko
174
Th€0pe,aiionaLAmplifier
in thebeginning
ofthischapterAswewit[see,thiscitcuitis the
described
thetwo
familiardifference
amptifier,
wrththe strainqagebridgeproviding
Thepairof strajngagesthat are
whosedjfference
is amptified.
vottages
R + An in the bridge
lengthened
oncethe baris benthavethe vatues
that are
amp$fier,
whereasthe pairof straingages
feedingthe difference
thiscircuitto discover
havethevaL0es
-R- AR. Wewittanatyze
shortened
jn resistthe outputvottage,
o, andthe change
the retationship
between
bythestrainqages.
ance,AR experienced
at
thattheopampis ideal.wdtingtheXcLequations
Tobeqin,assume
inputterminat5
ofthe opampwesee
theinvedingandnoninverting
R+AR
R A R
:R
L)n
,
AR'
Dn
Ua
&
=R+AR-&
'
(5.58)
for the vottageat the roninEq.5.58to get an expression
Nowrearrange
vertingtermjnaIof the op amp:
(5.5e)
* *,(-!-al*.a)
in its Linear
regjon,
so
Asusuat,
wewitlassume
thattheopampis operating
bethe expression
for ?, in Eq.5.59mustaLso
r', = o, andtheexpression
theright-hand
sideof Eq.5.59in ptaceof zJ,
for r',. Wecanthussubstitute
maniputation,
in Eq.5.57andsotveforr,. AftersomeaLgebraic
r' =
Rr(2AR)
F'
r1;1't'"t
(5.60)
js verysmalL,
by straingages
in resistance
experienced
Because
thechange
(AR)'<< R2,soR2 (AR)' - R' and Eq.5.60becomes
Rf
(5.61)
where6 : AR/R.
NOTE: Assets,ou anderstandinq of this Practical Perspectiveby
tlying Chapter Prcblem 5.48.
summary175
Surnnrary
The equatior that definesthe voltage transfercharac
tedsticofan ideal op amp is
A(t;o - r") < V66,
l-Ur.,
ro = \A\rp D), -Ucc = A(up - u") < + Vcc,
A(ur 1),1> + ucc,
I + t..,
where .{ is a propofiionaUty constantknoM as the
open-ioopgain, and ycc representsthe power supply
voltages.(Seepage157.)
An inverting amplifier is an op amp circuit producing
an output voltagethat is an inverted,scaledreplica of
the input. (Seepage161.)
A sunming amplifier is an op amp circuit producing an
outp tvoltage thatis ascaledsum olthe input voltages.
(Seepage163..)
A feedback path between an op amp's output and
its nrverting inpnt can constrain the op amp to its
linear operatingregion where ,', = ,4(?,r o,.). (See
page157.)
AnoDinvertingamplifieris an op amp circuitproducing
an output voltage that is a scaledrcplica of the input
voltage.(Seepage164.)
A voltageconstmintexistswhen the op ampis confined
lo its linear operatingregion due to typical valuesof
Ucc and A. If the ideal modeling assumprionsare
made-meaning A is assumedto be infinite the ideal
op amp modelis characterizedbythe voltageconstmint
A difference amplifier is ar op amp circuil producing an
oulput voltagethat is a scaledreplicaof the input voitagedifference.(Seepage165.)
rP:
0"
(Seepage158.)
. A currenl constraini further characterizes the ideal op
amp model,becausethe ideal input resistanceof the op
ampintegratedcircuitis infinite.This curent constraint
is givenby
iP: i" = 0'
The two voltageinputs to a differenceamplifier can be
used to calculate the common mode and difference
mode voltageinputs,om and ud.. The oulput from the
difference amplitier can be written in the form
r,-/m0cm+r
dnrid,i,
where ,4cnis the common mode gain, and ,4dmis the
differentialmode gain.(Seepage167.)
(Seepage159.)
. We consideredboth a simple,ideal op amp model and a
more reaiistic model in this chapter.The differences
betweenthe two nodels are asfollows:
Simpli6cdModel
More Reslislic Model
Infinite input resislance
Finite lnput resistrnce
Infinite openloop gain
Piniteopenloop gain
Zefo output resistance
NoMero oueur resislancc
(Seepage170.)
Tn an idealdiflerencedmplifierA.n,= 0. To mca(ure
how nearly ideal a differenceamplifier is. we usc rhe
commonmode rejectionratio:
C M R :R l + 1 .
l^.ml
An ideal difierence amplifier has an infinite CMRR.
(Seepage169.)
176
lheopentionat
Amptifier
Probtems
S€ctions5.1-5.2
5.3 Find i, in t}te circuit in Fig. P5.3if the op amp is ideal.
5,1 The op amp in the circuit in Fig. P5.1 is ideal.
tsPItta) Label
the five op amp term.inalswitl their names.
Figurc
P5.3
9ko
b) W}tat ideal op amp constraht determines the
value of i,? What is this value?
c) W}lat ideal op amp constrairt detemines the
valueof (rr - 0,)?What is this value?
d) Calculate t'o.
figureP5.1
8ko
5,4 A voltmeterwith a tull-scalereadingof 10V is used
tsP!.rto measurc the output voltage in the circuit in
Fig. P5.4.What is tlle reading of the voltmeter?
Assurnethe op amp is ideal.
Fig!reP5.4
3.3MO
The op ampin the circuit in Fig.P5.2is ideal.
a) Calculate2,.if o" : 1.5V and?rb= 0 V
b) Calculate 1,oif z,! : 3 V and ?,b : 0 V.
c) Calculate ,, if o. : 1 V and z'o : 2 Y.
d) Calculate0, if ou : 4 V and ,b = 2V.
e) Calculatet, if o, = 6 V and ,b - 8 V.
f) If ob = 4.5V, specifythe range of o, such that
the amplifierdoesnot saturate.
Figuru
P5.2
55 The op amp in tlle circuit
Ed.r Calculate the folowinC:
in Fig. P5.5 is ideal.
b) 0,
")h
d) i,
Fig!reP5.5
160ko
20ko
9V
z0 ko
u;60kO
Prcblems177
5.6 Find ir (in microamperes)in the circuirin Fig.P5.6.
5ko
5.9 a) Design an inverting amplifier using an ideal op
amp t]tat hasa gain of 4. Use only 10 kO resisto$.
b) Il,ou wirb lo amplill a 2.5V inputsignalu.ing
the circuil you designedin part (a), what are the
smallest power supply signals you car use?
5.10 a) The op amp in the circuit sho\an in Fig. P5.10 is
ideal. The adjustable rcsistor R^ has a ma-ximum
valueof 120kO, and a is rcstrictedto the range
of 0.25 = d < 0.8. Calculatethe range of o, if
5.7 A circuit desigrer claims t}le circuit in Fig. P5.7 will
an output voltage that will vary berween
pl$flg;Eproduce
_j9
as os variesbetween0 and 6 V Assumethe op
6mi
amp is ideal.
a) DIaw a graph of the output voltage od as a functionof theinputvoltage?s foro < zs < 6V
b) Do you agree with the designer'sclaim?
b) If d is not restricted, at what value of a will tle
op amp satumte?
figureP5.10
Figur€P5,7
15 k()
5.11 The op amp in the circuit in Eg. P5.11is ideal.
a) Find t]le rarge of values for (' in which the op
amp does not saturate.
b) Find i, (in microamperes)when d :
Section 5J
5"8 The op amp in the circuit oI Fig. P5.8 is ideal.
"""
a) What op amp circuit conliguration is tlis?
b) Calculate0d.
tiguleP5.8
100ko
figureP5.11
30k{}
178
Theoperationat
Amptjfi€r
5.15 Designan in\ erlrngsummingamplilier50thal
Section5.4
5,12 The op ampin Fig.P5.12is ideal.
"""
a) What circuit configuration is sholl,n in this figure?
b) Find r, if ou : 0.5V, ?]b = 1.5V, and
?''=-25V
c) The voltages0aand ,'bremainat 0.5V and 1.5V,
respectively.w}lat are the limits on ?c if the op
amp operateswithin its Linearregion?
Figure
P5,12
180ko
1!9!€ll
u":
-(3r, + sxb+ 4Dc+ 2Di.
If the feedbackresistor(RJ) is chosento be 60 k0,
draw a circuit dia$am of the amplifier and speciJy
the valuesofRa, Rb,&, and Rd.
5.16 Refer to t}le circuit in Fig. 5.11, where the op amp
Efl( is assumedto be ideal. Given that R. = 4kO,
Rb : 5 kO, & = 20 kO, ?,, - 200mV,
,b : 150mV, zrc= 400mV, and ycc - a6 V, specify the range of Rl for which tlle op amp operates
within its linear region.
SecfionSj
5.17 The op amp in the circuit of Fig.P5.17is ideal.
a) What op amp circuit configumtion is this?
b) Find ?. in termsof ?".
c) Find the range of values for os such that z,odoes
not saturateand the op amp remainsinits linear
regionof opention.
Figure
P5.17
32kO
5.13 a) The op amp in Fig. P5.13 is ideal. Find 1,oif
?]a= 16V, ub: 12V, 1J.= -6V,and,Ud= 10V.
b) Assume th ,'c, and ?rdretain their values as given
in (a). Specifythe range of ?,bsuchthat the op
amp operateswithin its linear region.
Figure
P5.13
330ko
33kO
5.14 The 330kO feedback resistor in the circuit in
Afln Fig.P5.13is ieplacedby a variableresistorRJ. The
voltagesoa- od have the samevalues as given in
Problem5.13(a).
a) Whal value of .lRfwill causethe op amp to saturate?Note that 0 < nf < co.
b) Wlen Rt hasthe valuefourd in (a), what is the
current (in microamperes)into the output ter
minal of the op amp?
5.18 The op amp jn the circuit shown in Fig. P5.18is ideal.
"""
a) Calculate o, when z'sequals 3 V
b) Specifythe range of valuesof ts so that the op
amp operatesin a linear mode,
c) Assumethat z)sequals5 V and that the 48 kO
resistor is replac€d with a variable resistor.What
value of the variable resistor will cause the op
amp to saturate?
Figure
P5.18
.18kO
hobtens 179
5.19 The op amp in the circuit of Fig. P5.19is ideal.
""" a)
What op amp circuit configurarion is this?
b) Find r]" in terms of 0r.
c) Fird the range of values for Dr such that o, does
not saturateand the op ampremainsin its linear
regionof opemtion.
Flgure
P5,21
Figure
P5,19
40ko
5.20 The op amp in the circuit shown in Fig. P5.20 is
Erft ideal.Thesignalvoltageso" aIId ob are 500mV and
1200mY respectively.
a) Wlat circuit conJigurationis showl in tlle figure?
b) Calculate r', in volts.
c) Find i, and 4 in microamperes.
d) wllat are the weighting factors associatedwith
?aand ,b?
Figure
P5.20
The circuit in Fig. P5.22is a noninverting summing
amplifier.Assume the op amp is ideal. Design the
lr9r!!1
circuit so that
ra=4ud+1)b+2u.,
a) Specifythe numericalvaluesofRb, R., and Rr.
b) Using the values found in part (a) for R/, RD,and
R., calculate (in microamperes)i,, ib, and tc
when ?)a= 0.75V, 0b : 1.0V, and r,. - 1 5 V-
Figure
P5.22
72kO
5,1kO
4.7kO
5.21 The op amp in tbe noninverting summing amplifier
Erie of Fig.P5 21is ideal.
a) Specifythe valuesof &, Rb,and Rcso thar
0o:60a+3rb+41)".
b) Usingthe valuesfound in part (a) for R/, Rb,and
n., find (in microamperes)ia, ib, i., ie, and is
when ?,'a= 0.5V, ?b : 2.5 V, and ?. : 1 V.
Section 5.6
5.23 a) Use the principle of superpositionfo clerive
Eq.5.22.
b) Derive Eqs.5.23and 5.24.
180
Ihe0perationat
AmpLjfier
524 The op amp in the circuit of Fig. P5.24is ideal. What
value of Rf will give the equation
uo: 15 zDd,
for this circuit.
Design the difference-amplfier circuit in Fig. P5.27
so that ?,, : 7.5(r,b 0,), and the voltagesourceob
!rg|!r!.
seesan jnput resistanceof 170kO. Specifythe values ot Ra,Rb,and Rf. Use the ideal model for tle
op amp.
Figure
P5.24
Figure
P5.27
15kO
5.25 The op amp in the adder-subtracter c cuit shown in
6trc Fig. P5.25is ideal.
a) Find oDwhen oa = 0.4V, ?b : 0.8V, ?rc: 0.2V,
and?')d:06V'
b) ff o,, r'., and 2d are held constant,what values of
0b will not saturate the op amp?
5.28 Select the values of Rb and Rr in the circuit in
*fJfljr Fis. Ps.28 so t}lat
,r, : 4000(ib
r").
The op amp is ideal.
figureP5.25
375kO
Flgufe
P5.28
5.29 Designa differenceamplifier(Fig.5.13)to meet tne
5.26 The resistors in the difference amplifier shown in
Ma Fig.5.13are R" = 20 kO, Rr : 80 kO, R" = 47 kO
and Rd = 33 kO. The signal voltages ?raand ,6 are
0.45and 0.9 V respectively,and ycc : 19 V.
a) Find 1)".
b) W[at is the resistance seen by the signal
c) What is the resistance seen by the signal
following cdteda: I)a :zDb - 5a^.The resistance
seen by the signal source rJbis 600kO, and tlle
resistanceseen by the signal source o, is 18kO
when the output voltage o, is zero. Specilf the valuesof Ra, Rb, nc, and Rd.
to,r_!ll1
The op amp in the circuit of Fig. P5.30is ideal.
a) Plot Daversusa when RJ : 4Rr and DE= 2V.
Use inuements of 0.1 and note by hypothesis
that0 < a < 1.0.
pmbtems181
b) Wdte atr equation for the staight lire you plotted in (a). How are the slope and intercept of the
line relatedto os and the ratio Rr/R1?
c) Using the results from (b), choose values for ?is
and tllreratio Rr/R1 suchthata" = -6a + 4.
5.33 In tle differerce amplifier shom in Fig. P5.33,what
rangeofvalues of Rr yieldsa CMRR > 750?
Figure
P5.33
47kA
Figure
P5.30
Rr
-10v
531 The rcsistor Rr in the circuit in Fig. P5.31is adjusted
EtrG until the ideal op arnp saturates.Specill' lRrin kilohms.
Sectiotrs5.1-5.6
5.34 a) Show that when the ideal op amp in Fig. P5.34is
operating in its Linearregion,
Flgure
P5.31
.
3r,
, a : ^ .
b) Show that the ideal op amp will saturate when
R(tUcc - zr,E)
T0 the dif{erence ampUfier sbosD in fig. P5.J).
compure (a) the differential mode gain, (b) the
commonmode gain,and (c) rhe CMRR.
Flgufe
P5,32
100ko
;
tigur€P5.34
182
lhe 0peBtionat
Anptifier
5.35 The c cuit insidethe shadedareain Fig.P5.35is a
arc constant current source for a limited range of val
uesof Rl,
a) Find tle valueofi, for R, : 2.5kO.
b) Find the maximum value for Rr for which i_ wiil
have the valuein (a).
c) Assumethat RL = 6.5kO. Explainthe operation
of the circuit. You can assumethat in : i? - 0
under all operatingconditions.
d) Skeichir versusR' for 0 < R, < 6.5kO.
5,37 Assume that the ideal op amp in the circuit in
Ers Fig.P5.37is operatingin its linear region.
a) Calculate the power delivered to the 16kO
b) Repeat (a) witi the op amp removed from the
circuit,that is,with the 16 kO resistorconnected
in the serieswith t}le voltage source and the
48 kO resistor.
c) Find the ratio of the power found in (a) to that
found in (b).
d) Does the insertionof the op amp betweenthe
source and the load serve a useful purpose?
Explain.
FigurcP5.35
Figure
P5,37
48kO
320mV
itit
s.38 Assumethat the ideal op amp in the circuit seenir
Fig.P5.38is opemtingin its lineai region.
5.36 The op ampsin the circuitin Fig.P5.36aie ideal.
""" a) Find 1".
b) Find the value of the right sourcevoltage for
which i, = 0.
Figure
P5.36
500mV
a) Showthatr',: [(Rr + Rtlnr]?r.
b) What happensif Rl + oo and R2+ 0?
c) Explain why this circuit is referred to as a vol!
agefollower when -R1: co and R, = 0.
Probt€ms
183
figureP5,38
Figur€
P5.40
2ko
18kO
3.9kO
5.39 The two op ampsin the circuit in Fig. P5.39are
Edc ideal. Calculate 2,1and 1,,2.
5.41 The voltageos shownin Fig. P5.41(a)is appliedto
6ntr the inverting amplifier shown in Fig. P5.41(b).
Sketchrroversust, assumingfie op ampis ideal.
Figure
P5.39
FigureP5.41
2Y
' \
/
\
/
0.s l q 1.5 1.0 2.s 3.q 3.s y'.o /G
4.7kO
2
(a)
75tO
5.40 Th€ signalvoltage o! in the circuit shownin Fig.P5.4.0
Fa., is describedby the followingequations:
r<0,
?,s= 4 sin(srl3)rV,
8.2kO
0 < t < oo.
Sketch?r,ve$us ,, assumingtlte op ampis ideal.
(b)
184
The0peGtionat
AmpLifier
Section 5.7
Figure
P5.44
240ko
5.42 Repeat Assessment Problem 5.6, given tlat the
Efld irverting amplifier is loaded with a 500 O resistor.
5.,(} The op amp in t}le noninverting amplifier circuit of
xrc Fig. P5.43has an itrput resistanceof 440 kO, an output resistance of 5 kO, and an openloop gain of
100,000.Assume that t}le op amp is operating in its
linear region,
a) Calculatethe voltage gain(1,"/D).
b) Find the inve ing and noninverting input voltageson and o/ (in millivolts) if lls = 1 V.
c) Calculate the difference (?i? on) in microvolts
when t,s = 1V.
d) Fhd the current drain in picoamperes on rhe
signal source rrswhen l)s = 1 V.
e) Repeat(a)-(d) assumhgan ideal op amp.
2lmko
J
5.45 Repeat Problem 5.44 assumingan ideal op amp.
5.216Assume rbe inpur resislanceof the op amp in
qr+ Fig.P5.46is infinireandilc outputresi.rance
i. /ero.
a) Find o, as a function of ?s and the open-loop
galn,{.
b) What is the value of t,o if os : 0.5V and
/ : 150?
c) What is the value of 1), if o" = 65Y ,tt.
d) How largedoes.4haveto be so that o,is 98% of
its valuein (c)?
FigureP5.46
150ko
s.44 a) Find the Th6venin equivalent circuit vdth rcspect
lo lbe ourpul lermjnals a.b lor the inverrjng
ampliJier of Fig. P5.44.The dc signal souice has a
value of 150 mV The op amp has an itrput resist
ance of 500 kO, aJI output resistance of 750 O
and an open-loop gain of 50,000.
$4lat is rbe ourpurresistance
of the in\efling
amplifier?
c) lvhat is the re.isrance
fin ohms)seenb) thesig,
nal source ?rswhen the load at the teminals a,b
is 150O?
5.47 DedveEq.5.60.
P'obtems185
S€ctions5.1-5.7
5.50 a) IfperceDterroris deliredas
5.48 Supposethe stmin gagesin the bridge in Fig. 5.18
pPRi.Hthave the value 120 O I 1%. The power supplies to
the op amp are I 15 V, and the reference voltage,
OEr,is taken from the positive power supply.
a) Calculate tie value of Rt so that when the strain
gage that is lengtlening reaches its maximum
Iengtl, the output voltage is 5 V
b) Supposethat we can accurately measure 50 mV
changes in the output voltage. What change
in strain gage resistance can be detected in
milliohms?
r!:r!!lv!
approximate value
-'l"'*,
show that the percent eroi in the approximation of I'o in Problem 5.49is
AR (R + Rf)
.
%eror=
=
-:vlot,.
^ t^-z^t
b) Calculatethe percent erloi in ?r,for Problem 5.49.
5.49 a) For the circuit showr ir Fig. P5.49,show rhar if
AR << R, the output voltage of the op amp is
,ll|;iH;,
approximately
iiiiii
do !;;
-R,rR + n.l
, -
.(-^R)2,,0.
b) Find ?,,if Rr = 470kO, R = 10 kO, AR : 95 O,
and !'in = 15 V.
c) Find tie actual value of oo in (b).
Figure
P5.49
5.51 Assume the percent error in the approximation of
pRflH;lod in the cncuit in Fig. P5.49 is not to exceed 1ol".
Fna Wlat is the largest percent change in 1Rthat can be
toleiated?
5.52 Assume the resistor in the variable branch of tlle
p'#flI#hbridge circuit in Fig. P5.49is n - AR.
Ma; a) What is the expressionfor o, if AR << R?
b) What is the expression for the percent erroi in
0o as a function of R, Rr, and AR?
c) Assume the resistance in the variable arm of
the bridge circuit in Fig.P5.49is 9810O and the
vallresof R. &. and ," are the same as in
Problem5.49(b).Whatis the approximatevalue
What is the percenteror in the approximation
ot ro \rhen the variable arm resi.ranccj(
9810()?
Inductance,
Capacitance
andMutual
Inductance
6.1 TheInductor p 188
6.2IheCapaeitot p 195
6.3 Series-Parattet
Combinations
of Inductance
p ?00
and Capacitance
6.4 Mutuatlnductan€ep 203
6.5 A CtoserLookat llutuallndtdan.e p 2A7
Knowandbe abteto usethe equations
for
voLtage,
curcnt, pow€r,anden€rgyjr an
jnductor;understafdho\aan inductorbehaves
jn th€ presence
of cofstait cLrrrent,
andthe
requnement
that the cunentbe continuous
in
Knowandb€ abt€to usethe equatjorsfor
voltage,cunent.power,andeneryyin a
capacitor;
undestandhowa capacjtorbehaves
in the presence
of consiantvottag€,andthe
rcquirement
tlrat th€ vottagebe continuous
in a
Beabteto combineinductoBwith iiitiat
condjtions
in sen€sandin panltetto forma
singt€equivatent
inductorwith an jfiiiat
coidjtion;be abteto combir€capacjtoBwjth
jfitial coidjtionsjn senesandjf pamttetlo
forma singLe
equivatent
capacitorwith an
Urde6tafdthe basiccofceptof mutual
inductance
andbe abteto write m€sh-cufleit
equatjoisfor a citruit containingmaqneticatly
couptedcoitsusjfs the dot conveniion
186
We begin this chapter by introducingthe last two ideal circuit
elementsmentioneclin Chaptet 2, namely.inductorsand capacjtors.Be assuredthat the circuit analysistechniquesintroducedin
Clrapters3 and 4 applyto circuitscontaininginductorsand capac
itols. Thercfolc, once you Lrlde$fand the tcrminal bebavior oI
these elementsin terms of current and vollage, vou can usc
KirchholT'slaws to desc be anv intetconnectionswith the other
basicclcments.Like other compoflents.inductorsand capacitols
are easier 1()describeirl terms of circuil variablesrather than
electromaSnetic
ficld variables.T-Iowever,
before we focuson the
circuit desc ptions,a blief reviervof thc field conceptsul1der-lying thesebasicelcmcntsis in order'.
An incluctor is ar1electdcal component that opposesany
changein electrical current. It is composedoI a coil of wire
wound ar-ounda supportingcore wbosematerial may bc lnagnetic or nonmagnetic.'l'he
behaviorof inducto$ is baseclon phenomena associatedwith magnetic fields. The source of thc
magnetictleld is chargein motioD,or current. If thc culrent is
varvingwith tjrne.the magneticfield is varyingwith time.A timevaryiig ma[iletic licld inducesa voltagein any conduciorlinked
by ihe ticld. The cir-cuitparameter of ind[ctance relates the
inducedvoltage1()the cufrent.We discussthis quanlilativerela
tiorNhipin Section6.1A capacitoris an electricalcomponenl that colsists of two
corrductorsseparatcdby an insulator or dielectricmatcrial.The
capacitoris the only device other than a battery ihat can store
electricalcbarge.Thebehaviorofcapacitorsis basedon phel1omena associated
with electricficlds.Thesourceof the electricficld
is separationof charge,or voltage.lf the voltagc is varying with
time,the electliclield is varyingwith time.A time-varyingelec ic
field producesa displacementcurrel1lin the spaccoccupiedby
thc field. The circuit parameterof capacitancelelates the dis
placementcurrentto thc voltage,wherethe displacementcurcnt
is equal to the conductioncurrent at the terminalsof thc capacitor. Wc discusslhis quantitative rclationship in Section 6.2.
PracticaI
Perspective
ProxinitySwitches
Theetectricaldeviceswe usein our dailytjvescontainmany
switches.
l4ostswitchesaremechanicat,
sucha5the oneused
in the ftashlightintroduced
in Chapter2. l4echanical
switches
usean actuatorthat is pushed,putLed,
stid,or rotated,causjng two piecesof conductingfiretalto touch and createa
shot circuit. Sometjmesdesignerspr€fef to use switches
withoutmovingparts,to r'ncrease
the safety,retjabitity,convenience,or novettyof their products.Suchswitchesare
caLtedproxjmityswitches.Proxjmityswitchescan empioya
varietyof sensortechnotogies.
For exarnpl€,
someetevator
doorsstayopenwhenever
a Ljghtb€amis obstructed.
Anothersensortechnotogyused in proximityswitches
detectspeopleby responding
to the djsruptionthey causein
etectrjcfields.Thistype of proximityswjtchis usedin some
desklampsthat turn on andofFwhentouchedandin etevator
buttonswith no movingparts(as shownin the figure).The
switchjs basedon a capacitorAsyouareaboutto dtscover
in
this chapter,a capacitoris a cifcuit €Lement
whoseterminaL
characterjstjcs
are deteminedby etectricfietds.!!hen you
proximjtyswitch,you producea changein
toucha capacitjve
the vatueofa capacitor,
causinga voltagechange,
whichacti
vatesthe switch.Thedesignof a capacjtjvetouch-sensjtjve
swjtchis the topic ofth€ Practicat
Perspective
exanrple
at the
endof this chapter.
187
188
and14utuat
lnductance
Inductance,
Capacjtance,
Section6.3describestechniquesusedto simplifycircuitswith seriesor
parallelcombinationsof capacitorsor inducto6.
Energy can be stored in both magnetic and electic field$ Hence you
should not be too surprisedto learn that inducto$ and capacitorsare
capableof storingenergyFor example,energycan be storedin an inductor and then releasedtofirea sparkplug.Energycanbe storedin acapac_
itor andthenrcleasedto fire a flashbulb.Inidealinductorsand capacitors,
or y as much energycan be extractedas hasbeen stored.Becausehductors and capacitoiscannotgenerateenergy,they arc classifiedas passive
In Sections6.4and 6.5we consideithe situationin which two circuits
are Linked by a magretic field and thus are said to be magnetically cou_
pled.In this case,the vollageinducedin the secondcircuit canbe related
to the time-varyilg cment in the first circuit by a parameterknofir as
mutual inductance. The practical significance of magnetic coupling
and
unfoldsaswe study1ie relationshipsbetweencurrent,voltage,power,
We introducethese
severalnew parametersspecificto mutual inductancerelationshipshereand then describetheir ulility in a devicecalleda transformer in Chapters 9 and 10-
6.1 TheInductor
Inductanceis the circuit parameter usedto describean inductor. Inductance
is synbolized by tlle letter l,, is measuredin henrys (H), and is represented
graphicatly as a coiled wire a reminder that inductance is a consequence
of a conductorlinking a magneticfield. Figure 6.1(a)showsan inductor.
Assignhg the rcfererce direction of the current in lhe direction of the volf
agedrop acmsstheteminals ofthe inductor,asshownh Fig.6.1(b),ields
- t equation
Theinductorr
ts
,di
(6.1)
where o is measuredin volts,L in henrys,i in amperes,and I in second$
Equation6.1reflectst}le passivesignconventionshownin Fig.6 1(b);that
(a)
is, the current reference is in the diroction of the voltage drop acrossthe
inductor.If the current referenceis in the direction of the voltagerise,
L
F-,ffi<
Eq.6.1is written witl a minussign.
Note fromEq.6.1 tlat the voltageaoossthe termjnalsofan inductor
i
is proportional to ttre time rute of change of the current in the inductor.
(Dl
We canmake two important obsenationshere.First,if the currentis con'
stant, the voltage acoss the ideal inductor is zeio. Thus the inductor
Figure6.1 A (a)Thegraphic
symbotfor
aninductor
behaves
as a short circuit ir the presence of a constant, or dc, current.
(b)Assigning
of /, henrys.
refer€nc€
withaninductance
in an inductor;that is,the
Second,
cu
en[ cannotchangeinstantaneously
fottowjng
the pasvottage
andcmenttotheinductor,
curent cannotchangeby a finite amountin zero time. Equation 6.1 tells
us that this changewould require an irfinite voltage, and infinite voltages
are noi possjble.For example,when someoneopens the switch on an
L
. f f i -
6.1
Thelnductor 189
inductive circuit in an actual system,the current irutially contlnues to flow
in the ah across the switch, a phenomenon called a/cr'r& The arc across
the switch prevents the current ftom drcpping to zero instantaneously.
Switching inductive circuits is an important engineering prcblem, because
arcing and voltage surgesmust be controlled to prevent equipment damage.The first step to urderstandbg the natue of this problem is to master
tle hfoductory material presented in this and tle following two chapteG
Example 6.1 illusirates the application of Eq. 6.1 to a simple circuit.
Determining
the Voltage,
Giventhe Current.at the Terninalsof an Inductor
The independent current soruce in the circuit
shownin Fig. 6,2-generates
zerc curent for t < 0
anda pulse10te"A, for t > 0.
t=0,
r<0
100mH
i=10te 5lA, ,>0
c:)u = Ldild.t = (0.1)10€5!(1-5, =e "(1
5 l ) v , l > 0 ; ? ,: 0 , t < 0 .
d) Figure 6.4 shows the voltage waveform.
e) No;the voltageis proportionalto dt/dt, not i.
f) At 0.2 s,which coresponds to the moment when
/t/d/ is pa$ing rhrougbzero and changingsign.
g) YeE at t : 0. Note that the voltage can change
instantaneouslyacross the teminals of an
inductor,
Fisure5.2r Th€cjrcujt
forb(ampl€
6.1.
a) Sketch the curent waveform.
b) At what instant of time is the current maximum?
c) Er?ress the voltage adoss the lemrinals of the
100mH irductor as a function of time.
d) Sketch tlle voltage waveform.
e) Are the voltage and the clrlrsnt at a maximum at
t])e same time?
f) At what instant of time does the voltage change
polarity?
g) Is therc ever an instantaneouschange in voltage
aqossthe inductor?If so,at what time?
r (A)
Flgure5.31Thecunentwaveform
forExample
6.1.
?r(v)
Sotution
a) Figue 6.3showsthe curent waveform.
b) dildt = lo(-ste-st + s 1= 10e 5'(1
- sr) Als;dildt = 0 whent = s. (SeeFis.6.3.)
l
Figure6.4
ThevoLtage
waveform
fo' E\ample
6.L
190
Inductance,Capacitance,and
Mutuatlnductance
Currentin an Inductorin Termsof the Voltaae
Acrossthe Inductor
Equation6.1expresses
the voltageaoossthe terminalsofan inductorasa
function oI the curent in the inductor. Also desirable is the ability to
expressthe curent as a function of the voltage.To find i as a function of o,
we startby multiplyingboth sidesofEq.6-l by a differentialtime dt:
=,(*),,
(6.2)
Multiplying the rate at which i vadeswith tby a differenrialchangein rime
generatesa differentialchangein t, so we $,riteEq.6.2 as
adt = Ldi.
{6.3)
We next integrateboth sidesof Eq. 6.3.For convenience,
we interchange
the two sidesofthe equationand write
"l*,'o'=1,"'
' o'
(6.4)
Nole that we useI and i as the variablesof irtearation. whereasI and I
becomelimirsonlhe integral..Thcn.
trom Lq.0.4.
Theinductori - z' equationb
1 r
4.t) : i I 1'!1r + j(ro),
(6.5)
where i(r) is the current coresponding to 1,and i(r0) is the value of the
inductor curent when we initiate the integration,namely,t0. In many
practicalapplications,to
is zero and Eq- 6.5becomes
iI'.*.
(0).
(6.6)
Equations6.1 and 6.5 both give the relationshipbetweenrhe volrage
and current at the terminalsof an inductor. Equation 6_1exDresses
the
\olrageasa funcrionot currenr.u hereasLq.o.5i\pfesse\lhe curfcnlasa
tunction of voltage.In bolh equationsthe referencedirectionfor the current is in the directionof the voltagedrop acrossthe teminals. Note that
i(lo) cardesits own algebraicsign.If the initial cu enr is in the samedircc
tion as the reference direction for i, it is a positive quantity. If the initial
current is in the oppositedirection,it is a negativequantity.Example 6.2
illustratesthe applicationof Eq. 6.5.
6.1 TheInductor 191
Determining
the Current,Giventhe Voltage.
at the Terminals
of an Inductor
The voltage pulse applied to the 100 mI{ inductor
shown in Fig. 6.5 is 0 for t < 0 and is given by the
expfession
r=0,
100mH
1)= z\te-|tuV,t>0
a(t) = 20tu ntV
Figure6.5
foit > 0. Also assumei : 0 for I < 0.
lhe circuitfor Exanpte
6.2.
a) Sketch the voltage as a function of time.
b) Find the inductor currenaas a function of time.
c) Sketch the current as a functiotr of time.
Solution
a) The voltage as a lunction of time is shorm ln
Fig.6.6.
b) Thecunentin thehductor is 0 a = 0.Therefore,
thecurrentfor t > 0is
1= t
2
I
I t0
10te10'- s 10)A,
Flgure6.6 A ThevoLtage
wavefo,n
fortuample
6.2
r (A)
| 2 r . x et c d t + o
t ",o'
ll,
= 2 0 0 11 m ( 1 0 +
r 1 )l l .
=2(1-
r<0
r> 0
c) Figure 6.7 showsthe cment asa function of time.
Figure5.7 Thecun€ntwavetorm
forExampte
6.2.
Note ir Example 6.2 that i approaches a constant value of 2 A as t
increases.We say more about this rcsult after discussingth€ energy stored
in an hductor,
PowerandEnergyin the Inductor
The power and energy relafionships for an inductor can be derived
directly ftom the current al1d voltage relationships. ff the curent reference is in rhe dircction of t]le voltage drop across the terminals of the
inductor.the Doweris
(6.7)
192
Inductance,
CaDacitance,
andlluiuatInductanc€
Reqgmbe-Ilhlt pqwg.r.istu wa!t!, voltageis in voltq atrd aurrent is in
amperes.If we expressthe inductorvoltageas a function of tlte hductor
currcnt,Eq. 6.7becomes
'i:lt:',;1.. :: ':
:'i'i:Li:i!..
':.
Power
in aninductor
>
..
\t dt .:i.
(6.8)
We can also expressthe curent it terms of the voltage:
o:,1!1,"',*.
r,"t)
(6.e)
Equation 6.8 is useful in expressing the energy stored in the inductor.
Power is the time mte of expending energy,so
d.w
,.di
(6.10)
Multiplying both sidesof Eq.6.10 by a differential time gives the differential relationship
dw : Li .1i.
(6.11)
Both sidesof Eq.6.11 are integiatedwith the understandingthat the refeietrce for zerc energy correspotrdsto zero curent h the inductor. Thus
I
Energy
in anindudor>
dx=Llyd)J,
$u,iffi
(6.12)
As before,we use different s]'mbolsof integration to avoid confusion
with the limits placedon the integrals.In Ed: 6.12,the energyis itr joules,
inductanceis in henrys,and curent is in amperes.
To illustrate the applicatiotrof Eqs.6.7and 6.12,we return to Examples6.1and6.2by meansof
Example6.3.
6.1 lheInductor 193
Determining
the Current,
Voltage,
Power.
andEnergy
for an Inductor
a) Por Example6.1,plot i, o,p, and x, versustime.
Line up the plots veftically to allow easy assess
ment of each variable's behavior.
b) In what time interval is eneryy being stoied in
the inductor?
c) In what time interval is energy being extmcted
from the inductoi?
d) What is the maximum energy stored in the
hduclor?
e) Evaluate tl]e integrals
f"
J" oo'
and
e) Fiom Example6.1,
i : 10t?-5'A
and
? , : e 5 ' ( 1- 5 4 v .
Therefore,
p = 1)i = Tote 1or- 5or2e 1orw.
/.l",oot,
0.6
and comment on thet significance.
Repeat (a) (c) for Example 6.2.
0.8
1'(v)
In Example6.2,why is there a sustainedcurent
ir the inductor asthe voltageapproaches
zero?
Solution
a) The plots of i, o, p, and ?.'follow direcdy from the
er?ressions for i and t, obtained in Example 6.1
and are .hown^in I ig. 6.8.ln particulaf.p - ,t.
andl/, - (;)Lt.
b) An increasing energy cun'e indicates that energy
is being stored.Thus erergy is being stored in the
time intenal 0 to 0.2 $ Note that this correspondsto the irterval when p > 0.
c) A decreasingenergy curve indicates that energy
is being extracted.Thus energy is being extracted
in rherimeinrerval0.2s lo oo.Nole thatrhiscorrespondsto t}le inte al when p < 0.
d) From Eq.6.12 we seethat eneigy is at a naxlmum
when current is at a maximum; glarcing at the
gmphscontirms this.From Example 6.1,ma)omum
curent : 0.736A. Therefore,unax : 2'7.07mJ.
u (nJ.)
figure 6.8 A Thevarjabtes
t, r, p, andu venusr for
Example
6.1.
194
Induchnce,
Capacitance,
aid l,lutuat
Inductance
Thus
ta.2
f,tu
lo.z
rat : tul*t-to' - tt
l"
/
'{t#. *[#r-,,-
g) The application of tle voltage pulse stores
ene4y in the inductor. Because the inductor is
ideal,this energycannotdissipateafter the vottage subsidesto zero, Therefore, a sustained current circulates in the circuit. A losslessinductor
obviously is an ideal circuit element. Practical
inductors rcquire a rcsistor in the circuit model.
(More about this laler.)
',]):'
= 0.2e) : 27.0'7tJ,l,
, (v)
/-
f "1at
lm
= IoLlmt
to, t,
Jate',t
]0.,
'{3#..e[#,
*'-
"]);
i (A)
: ^0.2e 2 = -21.M mJ.
Based on the definition of p, the area undei the
plot of p ve^us r represents the energy
erpeDded over the intenal of inregration.
Hence tlte integmtion of t}le power between
0 and 0.2 s representsthe energystored in the
inductor during tiis time interval. The inteFal
of p ovei the interval 0.2 s co is the energy
extracted. Note that in this time interval, all the
energyodginallystoredis removed;thatis,after
the cment peak haspassed,no energyis stoied
in the inductor.
f) The plots of o, 4p, and 1, follow dircctly from the
er?ressions for ?,and i given in Example 6.2 and
are shown in Fig. 6.9. Note that in this case the
po$er is alwals posilire.and henceenergyis
alwaysbeing stored during the volrage pulse.
? (-w)
u (mr)
0.4 0.5 0.6
Figure5.9 l. Theva abtes
1',t, p, and?ove6usr for
Exampte
6.2.
'
6.2
TlreCapacjtor 1 9 5
for vottage,
current,power,andenergyin aninductor
objective1-Know andbeab[€to usethe equations
6.1
The curent sourcein the circuit showngeneratesthe curent pulse
lr(t)=O,
t<0,
is(l) = 8€ 3m'- 8d 12oo'A,
r > o.
Fnrd (a) u(0); (b) the iDstantof time,greater
than zero,when the voltageo passesthrough
zerc;(c) the expressionfor the power deiivered
to the inductor:(d) the instantwhen the poNer
deliveredto the inductoris maximum;(e) the
naximum power;(t the instantof time when
the storederergy is maximum;and(g) the maximum energystoredin the inductor.
An5wer (a) 28.8V
(b) 1.s4ms;
(c) 76.8e-6ooi
+ 384e15oo'
- 307.2e-2{otw.
r > 0;
(d) 411.0s
i!s;
(e) 32./2w;
(f) i.54rns;
(g) 28.s7mJ.
NOTE: Also try Chaptet Pnbkns 6.1 and 6.3.
6.2 Thefapacitor
The circuit parameteroI capacitanceis representedby the leller C. is
measuredin farads(F), and is synbolizedgraphicallyby tlvo short paral
-lel conductiveplates,as shown in Fig.6.10(a).Becausethe farad is an
(a)
extrernelylargequartity ofcapacitance.practicalcapacitorvaluesusually
lic in the picofarad(pF) to microtarad(,.rF)range.
C
Thc graphic symbol for a capacitoris a reminder that capacitance
occurswheneverelectricalcoDduclorsarc scparatedby a dielectric,or
insulating,material. This condition implics that electric charge is nol
lransportedthrough the capacitor.Although applying a voltage to the
(b)
terminalsofthe capacitorcannotmove a chargethroughthe dielectric.it
a capacitor.
can displacetI chargc within the dielectric.As lhe voltage varies with Figure6.10;]: (a)Ihecircuiisymbotfor
(b)Assignjnq
vottage
and
cu
entto
the
rcfercnce
time, the displacementot chargealso varies with lillle, causingwhat is
thep$sivesignconvention.
capacitor,
totLowing
knowr as the displecem€ntcunent.
fiom a
At the terminals.Lhedisplacementcurrentis indistinguishablc
conductioncurcnt. The current is propolioDal 1()thc ralc at which the
voltageacrossthc capacitorrarieswith time, or, malhematicall-v,
t =r # ,
(6.13) n; capacitori
whcre i is measuredin amperes,C in larads.?rin volts,and tin seconds.
t, equation
196
Inductance,
capacitance,
andtlutuat
Inductance
Equation 6.13reflects the passivesign convention shown in Fig. 6.10(b);
that is,the current referenceis in the direction of the voltaqe drop acrossthe
capacilor.It tbe currenl referencei\ in fie direclion oirhe volraserise.
Eq.o.lJ i\ wrirrenwitha mjDu.srgn.
Two importantobsenationsfollow from Eq. 6.13.Iirst.voltagecarmot
change instantaneouslyacrossthe terminals of a capacitor. Equation 6.13
indicares
tbrl sucha changewouldproduceintinirecurrenr,a phrsical
impossjbiiir).second.
it the vollageacrosslhe lermtnal.is con.ranr,rhe
capacrtorcurrent is zero.The reason is that a conduction current cannot be
establishedin the dielectdc male al of the capacitor. Only a time_varying
voltage can produce a displac€mentcurrent, Thus a capacitor behavesas an
open circuit in the presenceof a constant voltage.
Equation 6.13givestle capacitorcurent as a function of the capacitor voltage.Expressingthe voltageas a lunction of the cu ent is alsouseful. To do so,we multiply both sidesof Eq. 6.13by a differcnriattime dr
and then integrate the resulting differentials:
id! = Cdo
l))'*=il,"''*
Carrying out the integralion of the left-hand side of the secondequa,
tlon grves
Capacitor?)- i equationb
,r,t=ll',0,*,<^t.
{6.14)
In many practicalapplicarionsof Eq. 6.14,rhe initial rime is zero; thar is,
lo = 0. ThusEq.6.14becomes
,(t) :
Il,',0,,^o,
(6.15)
We can easily derive the power and energy relationships lor the capacitor.
From the definitionof power,
powerequationts
Capacitor
(6.16)
,='11f,""*.
^,",]
Combiningthe definitionof energywith Eq.6.16lelds
du : C1)d't),
16.r7)
6.2 TheCapacitor197
from which
I d.x: C I jdy,
(6.18) .{ Capacitorenergyequation
In the derivationof Eq.6.18,the referencefor zeroenergycorresponds
to
Examples 6.4 and 6.5 ilustrate the application of the current, voltage,
power, and energy relationships for a capacitor.
Determining
CurrentVottage,
Power,
andEnergy
for a Capacitor
The voltage pulse described by the following equations is impressed auoss the terminals of a 0.5 /rF
Solution
a) FromEq.6.13,
{li,,"
1)0):
t < 0s;
0s<1<1s;
( (0.sx 10 6)( 4a(i r)) :
a) Derive the expressionsfor the capacitor current,
powei, and energy,
b) Sketch the voltage, current, power, and eneigy as
functions of time. Ljne up the plots ve ically.
c) SpeciJytle interval of time when energy is being
stored in the capacitor.
d) Specify the interval of time when energy is being
deliveredby the capacitor.
e) Evaluate tlle integrals
I'
pdt
and
* 10-9(0)= 0,
[
{
o
s
, : 1 ( o . s1x0 6 \ ( :=) 2 p " A ,
l* oo,
and comment on their significance.
1< 0s;
0s<r<1s;
ze (t r) LLA, t > rs.
Theexprcssion
for the poweris derivedfrom
Eq.6.16:
: ,,,*,
1,oiuo,
1) = -8e 2( 1)
(4e (' 1)(-2rt
[
/rW,
t < 0s;
0s=1<1s;
, > 1s.
The energy expressionfollows directly fmm
Eq.6.18:
t < 0s;
0s<t<1s:
r)
r)
:
(1(o.s)t0""t'
4e2(' pJ, t > 1 s ,
: o,',r,
{ l,to.a,u,'
198
Inductance,
Capaciiancq
andMutuat
Inductance
b) Figure 6.11 shows the voltage, clurentj power,
and eneryy as functions of time.
z'(v)
c) Energy is being stored in the capacitor whenevei
the power is positive. Hence energy is being
storedin the illteival0-1s.
d) Energy is being delivered by the capacitor whenever the power is negative.Thus energy is being
deliveredfor all t geater than 1 s.
e) The integral of p d/ is the energy associatedvrith
the time inte al cofiesponding to the limits on
the integral.Thus the first integralrepresentsthe
eneryy stored in the capacitor between 0 and 1 s,
whereas the second integral represents the
energy returned, or delivered, by the capacilor in
the interval 1 s to co:
2
I
0
1
rG)
z
p @w)
8
tr
tt
,lr
t a r = , 8 t d t: 4 i l : 4 p I .
I
,,0
J0
rG)
l0
u (pJ)
/N
I
J
tF
"
e a, = , r-8c
')dt = (-8)
- 2r,-lJlN
,It
il
z
= -a /,J
ll
The voltage applied to the capacitor rctums to
zero astime inoeases without limit, so the energy
returned by this ideal capacitor must equal the
eneigy storcd.
!@!|
|
||
I
Figure5.11 .A Thevariabt€s
x,,i, p, and?ove6us,for
Exanple
6.4.
Finding,tl,p and ?, Inducedby a Triangularcurrent Pulsefor a capacitor
I An uncharged0.2/,F capacitoris ddvenby a tlianI gularcurrentpulse.Thecurrentpulseis desciibedby
I
rG)
tn
, <o;
s000/A.
0. r-20ss:
",". -. - J
s
o
o
o
,
a
2
, 0 < r< 4 0 p s ;
l|o . . z
r - 4ops
to
I a) Deri!e lhe er"ression.for the capacilorvollage.
Wwer, ana enereVfor eachof the four time rnterI
I
valsneededto descdbethe current.
b) .P]glt:
y lersusI AligotheplotsasspecI
Ineornilflr"l
rneprevlous
exampres,
I
c)
Why
does
a
voltage
rcmain
onthecapacitor
aJter
I
I
thecurrentrcturnsto zero?
Solution
4
Flli
==
arr
are
zero'
?':r;iii"D
o=5x106 l1sornlar+0=12.5x10er' v,
Jo'
p
.,i 025 lorrl'w'
w:1Ctl
= 15.625 < 101'14J.
For20ps< , < 40ps,
a: S x tOu/ (0.2- S000r)dr+ 5.
Jzo^
6.2
(Notcthal 5 V is lhe voliageon thecapacitor
at
theendof thc prccedirgiDlerval.)
Then,
Thetapa.to
199
t (mA)
| = (106r- 12.5x 10er2 10)V.
t (i.t
- (62.5x 10tr13 7.5)<10et,+ 2.5x 105r- 2)w,
,,(v)
. =lc,t,
= 05.625x 1or'ia 2.5 x 10qrr+ 0-125x 106t2
I (,..s)
? + lo 5)J.
p ('nw)
500
400
300
200
100
a = 10V,
1r =
1
^
tclr
= 10pJ.
0
b) The excitationcurrcnt and the resullingvoltage,
po\\,er.and cncrgy are ploned in Fig.6.12.
c) Note thal thc power is alwayspositive for the
duralion of the cullenl pulse,which meansthat
encrgyis conlinuouslybehg storedin the capacitor. When the curreit returDsto zero,the stored
energy is lrapped becausethe ideal capacitor
oflers no meansfor dissipatingenergy lhus a
vollageremainson the capacitorafter i rclurns
r0
20
30
40
50
60
10
20
30 ,10 50
60
r (pt
" 0,1)
l0
8
6
2
0
I (!t
Figure6.126.Thevariabhs
i, r, p, andu versus
r for
txamph
6.5.
obl'edive2-Know andbeableto usethe eqqations
for voltage,current,power,andenergyin a capacitor
6,2
Thc vollageal t11etermimls ofthe 0.6pF
capacilorshowDin the figure is 0 for I < 0 and
40e rs'0mrsin
30.000tV tor t > 0. Find (a) i(0);
power
(b) the
deliveredto the capacitorat
r = 7/80 ms; and (c) the energystoredin lhe
capacitorafi : d80 ms.
0.6pF
NOTE: Also rty ChapterProblems6.14aru16.t5.
Answer: (a) 0.72A;
(b) -649.2 mw;
(c) 126.13pt.
6.3
Thc curcnt in the capacitorofAssessment
Problenr6.2is 0lbr r < 0 and 3 cos50.0001A
for r > 0. Find (a) ?)0)i (b) the ma"{imum po}'er
delivered to the capacitor at any one ilNtant of
time;and (c) the maximumenergystoredin the
capacitor at any one instant of time.
Answer: (a) 100sin 50,000tV, / > 0;
(b) 150w;(c) 3 mJ.
200
inductance,Capacitance,andl4utuatlnductance
6.3 Series-Paratlel
Combinations
of Inductance
andCapacitance
Lt
+ ? L
Lz
+ L 2
L3
+ r 3
Just as serieepaiallel combinalions of resistois can be reduced to a sinAle
equivalenl
reristor.
series-parallel
combinarion5
ot inducrors
or capacir6n
car be reduced to a single inductor or capacitor. Figure 6.13 shows inductors in series,Here, the inducto$ are forced to cat:rythe samecurent;thus
we define only one curent for the s€ries combination. The voltage drops
acrossthe hdividual inductors are
=i'.11<.1T*J
i
Figure
6.13A Inductors
in sedes.
,
Lt
Lz
d i, .
- t t. a
0z-Li
di
and
dt
u -t.d,.
L1
@
The voltageacrossthe seriesconnectionis
i(to)
1l
i
Leq: L1+ L2+ L3
01-tLt+ut
L,l;t.
i'g'e
(,0)
Flgure
6.14 Anequivatent
circujt
forinductors
in
series
canying
aninitialcuri€nt
t(d.
from which it should be apparent that the equivalent inductance ofseriesconnected inductors is the sum of the individual irductances. For ', induc-
Combining
inductors
in series>
f,.t,;
(6.19)
If the original inductors carry an initial curtent of i(to), the equivalent
inductor carries the same initial curent. Figure 6.14 shows the equivalent
circuit for se es inductors carrlng an initial culrent.
Inductorsin parallelhavethe sameterminalvoltage.Inthe equivalent
circuit, the current in each inductor is a function of tle terminal voltage
and the initial curent in the inductor. For the tlree inductors in pamllel
shown in Fig. 6.15,the currents for the individual inducton arc
jnductors
PiguE6.15A Three
in parattet.
;
1 . f
I r a I + ir(ro),
1 r
1 r
+ i2Gd,
+ /3(to).
(6.20)
6.3
Series-ParattetCombinatjonsof
Inductance
andCapaciiance201
The current at the terminals of the tlree parallel inductors is the sumot
the inductor currents:
i=ir+i2+h.
\6.?1)
Subsrituring
Eq.6.20into Eq.6.21ields
I
\.r
l\
f'
- + ; - l l u d r + ir(ro)+ iz(ro)+ 4(ro).
(6.22)
Now we can interpret Eq. 6.22in telms of a single inductor; that is,
. 1 r
t:
I
\6.23)
lltar+
CompadngEq.6.23with (6.22)yields
- 1: - + 1
-+
L"o
L,
1
Lt
1
Lt
;(ro)= 400)+,,G0)+ 4(ro).
(6.24)
L=1+
Lcq
\6.25)
L1
L+!
L2
L3
j(ro.)= ir(,o)+ t (rJ + i(ro)
Figue 6.16shows the equivalent circuit for the three parallel inductoE in
Fig.6.15.
The results er:pressed in Eqs. 6.24 and 6.25 can be exterded to
, hductors in pamllel:
Leq
figure 6.16 A An equjvatent
cjrcujtfor ihreeinductoE
in parattet.
(6.26) < Combining
inductors
in paratlel
(6.27) a Equivateni
inductance
initiaI current
Capacitois connected in sedes can be reduced to a single equivaletrt
capacitor.The reciprocal of tle equivalent capacitanceis equal to the sum
of t]le rcciprocals of the individual capacitances.If each capacitor cades
its own initial voltage, the initial voltage on the equivalent capacitor is tle
202 Inductancq
Capacitance,
andltlutuat
Inductance
algebmic sum of the initial voltageson tlle individual capacirors,Figure 6.17
and t]Ie following equations summadzetheseobservations:
Cornbining
capacitors
in seri€s>
(6.28)
fquivatentcapacitance
iniiial voltage>
(6.2e)
We leave tlle dedvation of the equivalent circuit for sedes-connected
capacitors as an exercise.(SeePrcblem 6.30.)
The equivalent capacitanceof capacitors connected in parallel is simply the sum of the capacitancesof the individual capacitors, as Fig. 6.18
and the follouring equation show:
Co|nbining
capacitors
in paraltet>
(6.30)
Capacito$ connected in parallel must carry the samevoltage.Therefore, if
fiere is an initial voltage acrossthe origbal pamllel capacitors,this same
idtial voltage appeaN acrossthe equivalent capacitance Ceo.The derivation of the equivalent circuit for pamllel capacitors is left ai an exercise.
(Seehoblem 6.31.)
We saymore about se es-parallel equivalent circuits of inductors and
capaciton in Chapter 7, where we interpret results based on their use.
;--r-r''T
+
cl
c1
q (ro)
IT
l +
"^i"t^s
(a)
.
- _),
;
1
_)"- T
1
;1
l
(a)
_('.q
L=!+ L+...*!
cr c
c"
cea,
1,(d = ?,(d + !(h) + " + 1'"(ro)
(b)
C e 1 , =C r + C 2 + - + C n
Figure
5,17A Anequivatent
cjrcuittor
capacitors
connected
in
(d)Ihesedes
series.
caDacirors.
(b)Tteeqriva.ent
crcur.
(b)
figure6.18"| Anequjvatent
cjrcuitfor capacitors
connected
in
palattet.
(a) CapacjtoB
(b)Ihe equivatent
in paratteL.
cncuii.
6.4
l4utuatlnductance203
objective3-Be ableto combine
inductors
or capacitors
in seriesandin paral(elto forma singLe
equivaLent
indudor
6.4
The initial valuesoflr and l2 in the circldt
showl are + 3 A and 5 A, respective]y.
The
voltageat the terminalsofthe paralielinductorsfort > 0is 30e"mV.
a) Il lhc paralleliDductorsare replaccdby a
singicinductor',what is its irduclancc?
b) Whal is Lheinitial currentand ils leference
directior in the equivalcn1inductor?
c) Use the equivalcntinduclor to find l(r).
d ) I n d ; { , 1 a n Lr lt ) . \ e r i r } r r J . r r e . o l u r ' o n .
fbr i1G),ir(,), and ,0) satisfyKirchhofl's
An5wer: (a) ,18mH;
(b) 2 A,|p;
(c) 0.125e5, 2.125A,1 > 0;
( d )r r 0 ) = o l e s r + 2 . 9 4 . r > o ,
jy'r) = o.o25e5' 5.025A, r > o.
6.5
Thc currenl ai the terminalsof the two capaci
lors showDis 240? "'l.A for I > 0. Thc inilial
valuesofur andq are 10Vand 5V.
respectivellCalcuiatethe total cncrgy trapped
in the capacitoKasr + N. (H,irij Don'lcombine the capacitorsin series lind thc energy
lrappedin each,and then add.)
240mH
Answer: 20AJ.
NOTE: ALto try ChapterProbler8 6.21,6.22,6.26,
urul6.27.
6.4 t4util;linducta*se
The magneticijcld we consideredin orn studyof iiduolors in Secrion6.1
wasrestrictedto a singlecircuit.We saidthat induclancei! the parameter
that relatcsa voilage 1()a time-\,aryingcuflcnt in the samecircuit;thus,
jnductanccis nore preciselyreferredto as scll inductance.
We now considerthe situationin which lwo circuitsare linkcd by a
- + v V - t , _
magnclic[e]d.In this case.the voltageinducedin rhe secondcircuil can
bc rclaled lo the time-varyingcurrcnl in the first circuir bv a paramcter
kno\n asnutual inductance.
Thc circuil showr in Fig.6.i9 reprcscnlstlvo
magnclicallycoupled coils.The scll nrductancesof the tr,o coils are
labeled/,1 ard lr, and the mutual inductanccis labeledM. The doublc
headedarrow adjacentto M indicatesthc pair oI coitswith this valuc of Fig!ie6.19 i. Twomagneticatty
coupted
coiLs.
mutualinducQnce.Tlisrotation is necdcdparlicularlyin circ its conrrin
ing morc than oDepair ofrnagneticallycouPledcoils.
The easiestway to analvzecircuitsconlainhg murualinducranceis 1{)
uscmeshcu ents.Theproblemis 1owdle thecircuitequationsthat describe
lhe circuitin termsof thc coil currents.First,choosethc rclcrencedirection
lbr eachcoil current.Figurc6-20lholvs arbitrarilysclcctcdrelcrencecurrerls.After choosingthc rcfcrcncedirectionsfor i and ;2.sum the voltages
aroundeachclosedpath.Becauscol lhe mutualinductanceM. thcrc will be Figure6.20i Coitcurients
tLandl2 usedto describe
two vollagesacrosseachcoil.namcll a self-hducedvoltageand a nutually tlrecncuitshownin Fiq.6.19.
.,o ',1[,', i^'
204
Inductance,Capacitance,andllutualInduchnc€
r------.d -r.
.l'rra
|
I
*Qrl.,l ["? *"
t . l
Figurc5.21A Thecjrcuitof Fig.6.20withdotradded
toth€ coitsindicating
the potarity
ofthe muiuatly
Dotconvention
for mutuattycoupl€dcoits>
induced voltage.The self induced voltage is the product of the self,
inductance of the coil and the 6rs1derivative of th€ cunent in tftat coil. The
mutually induced voltage is the pioduct of tlrc mutual inductanceof the coils
and the fust derivative of the curent in the other coil. Consider the coil on
the left in Flg. 6.20whose selJ-itrductarcehasthe value la. The selJ-induced
voltage adoss this coil is ar(dildt) and the mutualy hduced voltage aooss
this coit is M(di/dr). But what about the polariries of rheserwo voltages?
Using the passivesign convention, the self-induced voltage is a voltage
drop in the direction of the curent producing the voltage. But the poladty
of the mutually induced voltage dependson the way t]le coils are wound in
relation to tte reference directior of coil clrllents. In general, showing ihe
details of mutually coupled windings is very cumbersome.Instead,we keep
track of the polarities by a method known as t}Ie ilot convention,in which a
dot is placed on one lerminal of each whding, as shown in Fig.6.21.These
dots carry the sign informatior and allow us to draw the coils schematically
rather than showing how they wrap around a core structure.
The rule for using the dot convention to determine the polarity of
mutually induced voltage can be summaized as follows:
When the reference direction for a curent enters the dotted teminal ol a coil, t]le reference polarity of the voltage that it induces in
the othei coil is positive at its dotted terminal.
Or, stated alternatively,
Dotconvention
for mutuattycoupted
coits
(alternate)>
W}len tie reference dircction for a current leaves the dotled terminal of a coil, the reference pola ty of the voltage that it induces in
the other coil is negative at its dotted terminal.
For the most part, dot markings will be provided for you in the circuit
diagrams in this text. The important skill is to be able to wdte the appropriate circuit equations given your underctanding of mutual inductance
and the dot convention. Figu ng out where to place the polarity dots if
they are not given may be possible by examining tie physical configura,
tion of an actual circuil or by testing it in fhe laboratory. We will discuss
these procedures afrer \i'e discussthe use of dot markings.
In Fig. 6.21,the dot convention rule indicates that the reference polarity for the voltage induced in coil 1 by the current i2 is negative at tie do!
ted terminal of coil 1.This loltaEe (Mdi2/dt) is a voltage dse with respecr
to 4. The voltage i duced in coil 2 by the curent ilis M dh/.Lt, ^nd its rcferencepolarity is positive at the dotted terminal of coil 2. This voltage is a
voltage ise in the direction of 4. Figue 6.22 sho*s t}le self- and mulually
inducedvoltagesaoosscoils 1 and 2 alongwith their polarity maiks.
M
+
Ll
tiguru6.22a Thes€lfandmutuattyinduced
vottases
appearjng
across
ihecoits
shown
in Fig.6.21.
6.4 Muiuatlnductance
205
Now let\ look at the sum of the voltagesaroundeachclosedloop.In
Eqs.6.31and6.32,voltageises in the referencedircctioll ofa currentare
negative:
-u" - i,R, t t,d"
"
d
t
t,*, * t,*
d
, o ' t. - - o .
(6.31)
= o.
\6.3?)
,*
TheProcedure
for Determining
DotMarkings
We shift now to two methods of determining dot markings. The first
assumesthat we know the physicalarangement of the two coils and the
mode of eachwindingin a magneticallycoupledcircuir.The following six
steps,applied
here to Fig.6.23,determinea set ofdot markirgs:
,l
a) Arbitradly select one terminal say,the D terminal- of one coil and
mark it with a dot.
b) Assigna currentinto the dotted termiml and labelit rD.
?)
c) Use the ight-hand iule1 to determinethe direction of the magneric
lsrel
field establishedby iD iruide the coupl€dcoilsand label this field dD.
rA.rbitrarily
d) Arbitrarily pick one terminal of the secondcoil-say, terminal A and
assigna curent into this terminal, showirg the current as iA.
(Step1)
e) Use the ght-handrule to detemine t}le direction of the flux established by ta ir ide the coupled coils and label this flux dA.
Figure6.23A A setof coilsshowing
a method
for
d€t€rmining
a
set
ofdot
nrarkings.
f) Compare the direcrions of the two fluxes dD and dA. If the flrlxes have
the samercference direction, place a dot on the terminal of the second
coil wherethe test current (iA) enters.(In Fig.6.23,the fluxesdD and
dA havethe samereferencedirection,and thereforea dot goeson terminal A.) If the fluxes have different reference directioni place a dot
on the teminal of the secordcoil wheretie testcurrentleaves.
The relativepolaritiesof magneticallycoupledcoilscan alsobe determhed experimentally. This capability is important becausein some situations,determininghow the coilsare wound on the core is impossible.One
experimentalm€thodis to connecta dcvoltagesouice,a resistor,a switch,
and a dc voltmetei to the pair oI coiiE as shownin Fig.6.24-The shaded
box coverjng the coils implies tlat physical inspection of the coils is nol
possible.The
resistorRLimitsthe magnitudeof tie currentsuppu€dby the
dc voltagesource,
Flgure
6.24A AnexperimentaL
setup
tordeteririning
The coil terminal connected to the positive terminal of tle dc source
via the switch and limiting resistor receives a polarity mark, as show, in
Fig.6.24.Wlen the switchis closed,the voltmeterdeflectionis observed.If
the momentarydeflectionis upscale,the coil teminal connectedto the
positiveterminalof the voltmeterreceivesthe poladty mark.ffthe deflection is dowNcale, the coil terminal connected to the negative telmhal of
ure voltmeter ieceives the polarity mark.
Example6.6 showshow to usethe dot markingsto fomulate a set of
circuitequationsin a circuit containingmagneticalycoupl€dcoils.
-'
s".ti*ion
otru.uauyttawonpage207.
206
Inductance,Capacitance,andMutualInductance
FindingMesh-Current
Equations
for a Circuitwith Magenticatty
Coupted
Coits
a) Write a set of mesh-crment equations that
descdbe the circuit in Eg. 6.25 in tems of the
currents i1 and ,2.
b) Verify that if there is no energy stored in tlle clrcuit at r - 0 and if ic = 16 - 16e-5'A, tie solutions lbr i1 and i2 are
i 1= 4 + 6 4 a 5 t - 6 8 e 4 t A ,
i1(0)=4+64-68=0,
i,(0)=1
i2= 7 - 52a5t+ 51e4tA..
t.)
8H
b) To check the validity of i1 and i2, we begin by
testhg 1}leinitial and final values of i1 and i2.We
know by h}?ot}lesis that t1(0) = ir(0) = 0. From
the giver solutions we have
20f,l
reu f i'-i aoo
5 2 + s 1= 0 .
Now we observe that as l approachesinlinity the
source current (is) approachesa constatrt va.lue
of 16 A, and ttrerefore the magnetically coupled
coils behave as shon circuils.Hence al I - m
lhe circuil feducesro rhat (howo in Fg. b.26.
From Fig. 6.26 we see that at I : co the thrce
rcsistors are in pamllel aqoss the 16 A source.
The equivalent resistanceis 3.75 O and thus the
voltageacrossthe 16 A curent sourceis 60V It
tbllows that
4(co)
Figure6.25 Thecircuit
forExample
6.6.
60
20
60
60
t(,:o): r @ = t o
Solution
a) Summingtle voltagesaroundthe il meshyields
4d
- I3dtlis t2) .20(i -i)
l5(;
-r")=0.
Thesevaluesagreewith the final valuespredicted
by t}Ie solutions for 4 and i2.
Finally we check the solutions by seeing if
they satisfy the differential equatiom dedved in
(a). We will leave this fhal check to the readei
via Problem6.37.
The i mesh equation is
20lir
)
l,) | 60i, + lb:{/,
dt-
i"r
'
si,
8"1' - ".
dt
Note that the voltage aqoss the 4 H coil due to
the current (is - i), that is,gd(is - i)/dt,1s a
voltage drop in the d ection of tt. The voltage
induced in the 16 H coil by tle curent i1, that is,
8dtldl, is a voltage rise in tle direction of i2.
50
20a
Figure5.26A lhecircuiiof Lumpte
6.6when
r = co.
6.5
A Ctoser
Lookat l4utual
Inductance 207
0bjective4-Use the dot convention
to writemesh-current
equations
for mutuallycoupted
coils
6.6
a) Wite a set of mesh-curtentequationslor
the circuit in Example6.6if the dot on the
4 H inductor is at the righl-hand telminal,
the referelce djrectionof is is reversed,alrd
the 60 O resistoris increasedro 780 ().
b) Verify that ifthere is no energysroredin rhe
circuitat r : 0, andif is - 1.96 1.96e4tA,
the solutionsto the differentialequations
derived in (a) of this AssessmentProDremare
Ans$,er:(a) 4(dirldr) + 25ir + a(rlizlrh) 20i2
: sis s(disldt)
8(di/dt) - 20ir + 16(rli2/ dt) + 800i2
- 16(di:r/dt);
(b) vedfication.
ir = -0.,1 11.6e{'+ 12€-5'A,
/2 -
0.Ul
0'1o,""
,' 5 A
NOTE: Also try ChaptetPrcblen 6.34.
6.5 A fl[oserLookat Mutual3*duetaalee
In order to fully explainthe circuit paramercrmutual inductance,and to
examinethc limilationsandassunptionsmadein the qualitativediscussion
presentedin Section6.4,wc beginwith a more quanritativcdescdptionof
self-inductance
thanwasprcliousiy providcd
A Reviewof Setf-Inductance
:lhe conceptofinductancecan be tracedto Michael Faraday.
who did pio
neeringwork in this area in t]re early 1800s.Faradaypostulatedthat a
magnelicfield consistsof lines of force surroundingthe current-carrying
conductor.Visualizetheselines of lbrce as energystoring elastjcbands
that closeon themselves.As
the currentincreasesand decreases.
the clastic bands(that is.thc liDesof force) spreadand collapseaboutrhe conductor.The voltagejnducedin the conductorisproportionalto rhe numberoI
lines that collapseinto, or cut. thc conductor.This imageof inducedvolf
ageis expresscdby what is callcdFaradays law;lnar N,
,:n'
(6.33)
where,\ is rcferredto asthc flux linkageand is neasuredin weber{urns.
Howdowegetfrom Faraday'slaw to the definitionofinduclaDcepresentedin Scction6-1?Wecanbeginto draw thisconnectionusingFig.6.27 Figure6.27A Represeniation
of a magnetjc
fieldtink
208
Inductanc€,
CaDaciiance,
andlllutuat
Inductance
The linestbreadingthe N tums and labeledd representthe magnetic
lin€s of foice that mal(e up the magnetic field. The shength of th€ mag'
netic field dependson the strengttrof the current,and the spatialorientation of ihe field depelds on the diection of the current. The right-hand
rule rclates the odentation of the field to the direction of the cment:
w}len the fingers of tlle right hard are rrrapped around the coil so that the
fingers point in the directior of tlle curent, tle thumb points in the direction of that potion of the magneticfieldinsidethe coil.Theflux linkageis
the product of the magneticfield (0), measuredin webers(Wb), and the
numberof turns linted bv the field fN\:
16.34)
The magnitude of the flu\, d, is rclated to the magniiude of the coil
currentby the relationship
O: sNt,
(6.35)
whereN is the numberof tums on the coil, and O is ttrepermeanceofthe
space occupied by the flux. Permeance is a quantity that describes the
magneticpropertiesof this space,and assuch,a detaileddiscussionofpermeancejs outsidethe scopeof this text. Here,we need only obsene that,
when the spacecontainhg the flux is made up of magnetic materials (such
as iron, nickel, and cobatt),the permeancevarieswith the flux, giving a
nonlinearrelationshipbetween4 and t. But when the spacecontaining
the
flux is comprised of noimagnetic materialsr the permeance is constant,
giving a linear relationship between d and i. Note from Eq. 6.35 that the
flux is alsopropoitional to the numberof tums on tle coil.
Here, we assumethat the core material-the spacecontaining the flux
is notunagnetic.Then, substituting Eqs.6.34 and 6.35into Eq. 6.33yields
d^
= N;:
d(No)
N;tENrl
= Nza!! _ t!!
(6.36)
which showsthat self-inductance is propoitional 10 tlle square of the number of tums on the coil.We make useof this observationlater.
The polarity of the induced voltage in tie circuit in Fig. 6.27reflects the
reaction of the field to the cwent creating the field. For example,when i is
increasing, dt/dl is positive and o is positive. Thus energy is required to
establishthe magnetic field. The product ?i givestlre rate at which energy is
stored in the field. Wlen the field collapses,di/dt is negative,and again the
polarity of the induced voltage is in opposition to the change.As the field
collapsesabout the coil, energy is returned to the circuit.
Keeping in mind this furtler insight into the concept of sef-inductance,
we now turn back to mutual inductance,
6.5 A ctoser
Look
at l{utuat
Inductance209
TheConcept
of MutualInductance
Figure 6.28 shows two magnetically coupled coils.You should verily that
the dot markings on the two coils agreewith the direction of the coil windings and curents shown.The number of turns on each coil are N, and N".
respeclively.Coii I is energizedb) a trme+arling cunenr tour.. rlai
establishesthe current ir itr the Nl turns. Coil 2 is not etrergized and is
open. The coils a-rewound on a ronmagnetic core. The flux produced by
the cment il can be divided into two components,labeled du and d21.
The flu\ component d11is tlle flux produced by ir that links only the N1 Figure
6.28A Twonragneticatly
coupted
coib.
turns.The component 421is t}le flrlx produced by i1 that links the N2 turns
and the N1 tums. The tint digit in the subscript to the flux gives the coil
number, and the second digit refers to the coil current. Thus dj r is a flux
linking coil 1 and produced by a current in coil 1,whereasd2r is a flux lhking coil2 and producedby a currentin coil 1.
The total flux linling coil 1 is dl, tle sum of d11and d\:
.lt + 6*
et:
(6.37)
The flux dr and its components dx and d2r are related to the coil curent
i1 asfolows:
6t = 94rir
0i
(6.38)
: 9l ,Ni1,
(6.3e)
(6.40)
where 9r is the permeance of the spaceoccupied by the flux d1, 9rr is the
permeanceof the spaceoccupied by the flux d1r, and 921is the permeance
of t}tespaceoccupiedby the flux d21.SubsrirutingEqs.6.38,6.39,
and 6.40
into Eq. 6.37 yields the telatioNhip between the permealce of the space
occupied by the total llux dr and tle pemeances of rhe spacesoccupied
by its componentsd11andd2r:
91=gr+@21.
(6.41)
We use Faraday's1awto derive erpressions for r)r and o2:
: N,*@u,
,' =# =o'ry;fu
o,')
: N]g.11+s^\*
= *p,*
= ",*
(6.42)
and
d(Nzdzt)
dt
= N,w,g,,\
=u.frg,,N,i,1
(6.43)
210
Induchnce,Capacitance,andl,4utuallnduciance
The coefficient of dt/dt i]) Eq. 6.42 is the self-inductance of coil 1. The
coefficientof dtldl in Eq. 6.43is the mutual inductancebetweencoils 1
and 2.Thus
\6.44)
The subscript on M specifiesan irductance that relates the voltage induced
in coil2 to the curent in coil 1.
The coefficient of mutual inductance sives
D2= M2r
di.
dt.
(6.45)
Note that tle dot convention is used to assignthe polarity rcference to ?)2
in Fig.6.28.
For the coupled coils in Fig. 6.28, exciting coil 2 from a time-varying
current source (i2) and Ieaving coil 1 open produces the circuit arrangement shown h Fig. 6.29. Again, the polarity reference assignedto or is
basedon the dot convention.
The total flux linking coil2 is
Figure5.29A Themagnetkatly
coupted
coil50f
Fig.6.28,withcoit2 exciied
andcoill open.
6z:6zz+4o.
(6.46)
The flux d, and its components d22afil dD are related to the coil curent
i2 as follows:
O2= g2N2i2,
\6.47)
4n : qrNzrz,
(6.48)
0D: grzN2i2.
\6.49)
The voltagesu2and ur afe
u=ff=uZo,ft=1,ff,
(6.50)
- Nfi2{to*.
,,:ff =fiw,q-l
(6.51)
The coefficient of mutual inductancethat relates the voltage induced
in coil 1 to the time-varyingcurrent in coil 2 is the coefficient of dizldt
in Eq.6.51:
Mp = N1N20p.
(6.52)
For nonmagnetic materials, the pemeances O12and @r are equal, and
rherefore
MD:
M2r = M.
(6.53)
Hence for linear circuits with just two magnetically coupled coils, attaching subscriptsto the coefficientofmutual inductanceis not necessary.
6.5 A Ctos€r
Loolat l,iutuat
Inductance211
MutualInductance
in Termsof Self-Inductance
The value of mutual inductance is a furction of the seiJ-inductances.We
derivetlis relationshipasfollows.From Eqs.6.42atrd6.50,
Lr = Nlgr,
(6.54)
L2 = N1!t2,
(6.55)
respectively.
FromEqs.6.54
and6.55,
LrL2- N?N1!trs2.
(6.56)
We now use Eq.6.41 and the corresponding expressionfor @2to write
L[2 = N1Ni(gr + @2)(922+ gn).
(6.57)
But for a linearsystem,
O21: 9p, soEq.6.57becomes
Lr4= (N1N2s1)2(1.
#)('.
#)
=*(,H)('.H)
(6.58)
Replacing the two terms involving permeances by a single consIant
expresses
Eq.6-58in a more meaningfulform:
i=("H)(.H)
(6.5e)
Substituting8q.6.59 into Eq.6.58yields
tf
= kzLrLz
(6.60) < Retating
setf-indudances
andmutuar
inductance
usingcouptingcoeffiaient
wherc the constantt is called t}le coefficientof coupling.Ac€ordingto
Eq.6.59,1/fr'musrbe$eater than1,whichmeansthat ,t musrbetessthan1.
In fact,the coefficientof couplingmustlie between0 and 1,or
0<k<1.
(6.61)
The coefficient of coupling is 0 when the two coils have no conmon
flux; that is, when dD = drl = 0. This condition implies that 912 : 0, and
Eq. 6.59 indicates that Yle - cr', or tr : 0. If there is ro flux linkaqe
betweenthe coils,obviouslyM is 0.
212
IndLrctance,
Capacitance,
andl4uiuat
Inductance
The coefficient of coupling is equal to 1 when 411and 422are 0. This
condition implies that aI the flux that finks coil 1 also links coil 2. In terms
of Eq. 6.59,gr1 = @z : 0, which obviouslyrepresentsan ideal state;in
ieality, winding two coils so that tley share precisely the sameflux is physica y impossible. Magnetrc matedals (such as alloys of iron, cobalt, and
nickel) createa spacewith high permeanceand are usedto establishcoefficients of coupling that approach unity. (We say more aboul this important quality ofmagneticmaterialsin Chapter9.)
NOTE: Assessyou undeqtanding of this material b! tying Chapter
Problems6.12and6.43.
EnergyCalculations
Figur.5.30
A Thecircuit
used
io denve
thebasic
We concludeour first look ai mutual inductancewith a discussionof the
total energy stoied in magnetically coupled coils. In doing so, we will
confirm two observationsmade earlier: For lirear magnericcoupling,
(1\ MD: M^ = M,^nd(2) M: krZE, where0 < t < 1.
We use the circuit shown ir Fig. 6.30 to dedve the expression for the
total energy stored in the magtetic fields associaled$rith a pair of iinearly
coupled coils. We begin by assumirg tiat the currents ir and t2 are zero
and that this zero-current state corresponds to zero energy stored in the
coils.Then we let i1 increasefrom zero to some arbitrary value 1r and compute the eneigy stored when ir = 1r. Becausei:0,
t]le total power
input into the pair of coilsis ?,r4,and t}le energystoredis
I.
d1t,= Lt.la idir
1
W:;L1Ii.
^
(6.62)
Now we hold ir constant at 11 ard increasei, from zero to some arbitrary
value 12. During this time interval, ttre voltage induced in coil 2 by il is
zero because1r is constant.The voltage induced h coil1by i2 is M Driizldt.
powefinpullo thepairof coilsis
Therefore.lhe
di.
p=IrMn;+i2o2.
The total energystoredin the pair of coilswhen i2 : 12is
tw
J*
o*:
tt,
J"
ItMldi,+
th
J)
L,idi,,
w =w+ rJ,MD+ tk4.
:)r,4 +lkt1+ t,r,u,,.
(6.63)
6.5 A Ctoser
Look
at l4utuat
Inductance
If we reverse the procedure-that is, if we fkst increase 4 from zero to 12
and then inuease i1 from zero to 11-the total energy stored is
w =l,r?,1b4*r,t,u,,.
16.64)
Equations 6.63 and 6.64 express the total energy stored in a pair oI lin,
early coupled coils as a function of t}le coil curents, the selJ-inductances,
and t]le mutual inductance. Note that the only diflerence between rhese
equations is the coefficient of the cu ent product 1rI2. We use Eq. 6.63 if
i1 is establishedfirst and Eq.6.64ifi, is estabtished
first.
When the coupling medium is linear, the total energy stored is the
sameregardlessofthe order usedto establishlr and 12.The reasonis that
in a linear coupling, tlle resultant magnetic flux depends only on the final
valuesof 4 and i2, not on how the clllrents reached their final values.If the
resultant flux is the same,the stored energy is t}le same.Therefore, for linear coupling, M12 : Mz1.Also, because11arrd 12are arbitrary values of i1
and i2, respectively,we rcpresert the coil currents by their instantaneous
values 4 and i2. Thus, at any instant of time, the total energy stored it the
couDledcoils is
-Ol=7,& +lr,i1+ui,i,.
(6.65)
We de ved Eq. 6.65 by assuming that both coil curents entered
polarity-marked terminals. We leave it 10 you to vedfy that, if one current
enters a polarity-marked terminal while the other leaves such a teminal,
the algebraic sign of the telm Mi1i2 reve6es Thus, in general,
*1t7: ,\,i? +
f,r,i} r ui,i,
(6.66)
We use Eq. 6.66 to show that M cartrtot exceed \/LJ;. The magnetically coupled coils are passive elements, so the total energy stored can
never be negative.If ?r(r) can never be negative,Eq. 6.66indicates that the
quanlrty
t
,Li'
-
l
)Lri)
- tt|.t,
must be greatertlan or equalto zero when it and i2 are eilher both positive or both negative.The Limitiry value of M corresponds to setting the
quantityequalto zero:
I,t*)"a-
Mi1i2:0.
(6.67)
To find the Limiting value of M we ^dd and subtract the telm
ilirvall2 to tle left-handsideofEq.6.67.Doing so genetatesa term that
I E;. \!7'
/ _
\
+ iiz\.t/LrL2 - M
):0.
(6.68)
214
I n d u c t a nC
. ea, p a . j t a n . € ,M
a nudt u a t h d u ( t a n ( e
The squaredlerm in Eq. 6.68can never be negative,but it can be zero.
Thereforcu(t) > 0 only if
^/Lrt2 > M,
(6.69)
which is anofter way ofsayingihat
M = k^/L;L; (o=&<1).
We derivedEq.6.69 by assumingthat 4 and i2 arc either both positiyeor
both negative.However,we get the sameresult if il and i2 are of opposile
sigD,becausein this casewe obtainthe limiting valucof Mby selectingthe
plus signin Eq.6.66.
NOTE: Assessfoar unde^tanding of thi,Jmaterial by tJing Chapter
Problems6.17and6.18.
PracticaI
Perspective
ProxinitySwitches
At the b€qinningof thjs chapterwe jntroducedthe capacitiveproxjnity
swjtch.Therearetwo fonis of this switch.Theoneusedjn tableLarnps
js basedon a sinqle-electrode
switch.It is Leftto your jnvestigationin
Probtem6.50. In the exampleher€,we considefthe two-etectrode
swjtch
usedin elevatorcatlbuttons.
EXA14PLE
Th€eLevatof
call buttonis a snrallcupinto whichthe fingeris inserted,as
shownin Fig.6.31.Thecupis nadeof a metatring eLectrode
anda circutar
ptateetedrodethat areinsutated
fromeachother Sometimes
two concentric
ptasticare usedinstead.The eiectrodes
rings embedd€d
in insuLating
are
coveredwith an insutatingtayerto preventdirectcontactwith the metat.
Theresuttingdevicecanbe modeted
asa capacitor,
asshownin Fig.6.32,
cl
.
Figlre 6,31"'. Anetevator
cattbutton.
G) Fronivieu (b)Sideview.
-i--.
Figure5.32 ^4.A capacitor
modeL
ofthetwo-eteclrode
prcximity
switchusedjn etevator
caLlbuttons.
PractjcatPe6p€ctive
215
Untikemostcapacitors,
proximjty
thecapacitive
switchpermits
youto
cl
insertanobject,suchasa finger,between
yourfintheetectrodes.
Because
geris muchmoreconductive
thanthe insutating
covering
surrounding
the
etectrodes,
the circuitresponds
asthoughanother
electrodg
connected
to
ground,hasbeenadded.
Theresuttis a three,terminal
circuitcontainjng
threecapacitors,
asshownin Fig.6.33.
Theactualvatues
of the capacitors
in Figs.6.32and6.33arejn the
rangeof 10to 50pP,depending
ontheexactgeometry
of theswitch,how Figure6.33 A A circujtmodeLofa
capacitive
the fingeris jnserted,
switchactjvated
gtoves,
whether
the person
byfingert0uch.
is weadng
andso forth. proxjmiry
Forthefotlowing
problems,
assume
that all capacjtors
havethesamevalue
of 25pF. Atsoassume
the elevator
catlbuttonis ptaced
in the capacjtive
€quivatent
of a vottage-divider
circuit,asshown
in Fjg.6.34.
a) Catcutate
theoutputvoltage
wjthnofingerpresent.
b) Catcutate
theoutputvottage
whena fingertouches
thebutton.
tt:r
-.,Tr.,-
Solution
a) Beginby redrawing
thecircujtin Fig.6.34withthecatlbuttonreplaced
by its capacitive
modet
catlbuttoncncuit.
fromFig.6.32.TheresuLting
circuitis shownjn Figure6.34 A Anelevator
Fig.6.35.Writethecurrent
equation
at thesingLe
node:
".
d(u +)
* c'. dI4 : o .
t1l
(6.70)
Rearrange
this equationto produce
a differentialequationforthe output
vottageo(t):
dn _
CI
dt"
dt
Cr+C2 rlt'
(6.71)
Finatty,
integrateEq.6.71to find the outputvoLtagei
,{ti:-
C.
u , ( r )+ u ( 0 ) .
Figure6.35 A A modet
ofthe etevaior
caltbutton
cjrtujtwithnofingerpreseni.
(6.12)
Theresuttin tq. 6.72shows
thatthesedes
capacitor
circuitin Fig.6.15
formsa voLtage
dividerjust
asthes€ries
resistor
cjrcujtdjdin Chapter
3.
In bothvottage-divider
circuits,
theoutputvoltage
doesnotdepend
on
thecomponent
vatues
butontyonthejrratio.Her€,C1= C2 = 25 pF,
sothecapacjtor
ratjois Cr/C, : 1. Thustheoutputvoltage
is
0(t)=0.5r,(r)+"(0).
(6.73)
Theconstant
termin Eq.6.73is dueto theinitiaL
charge
onthecapacitor,
Wecanassume
thato(0) : 0 y, because
thecircuitthatsenses
theoutputvoltage
etiminates
theeffectoftheinitiatcapacjtor
charge.
Therefore,
thes€nsed
outoutvoltaoe
is
0(i) = 0.5,,(1).
16.74)
216
Inductance.
CaDacitance.
andMuiuatlnductance
b) NowwerepLace
the catLbltton of Fig.6.34withthe modet
of theactivatedswitchin Fig.6.33.Theresultisshown
in Fig.6.36.Again,wecal
cutatethecurrents
leaving
theoutplt node:
d ( u - u- s- c) )^ddr D
- e t d^ -d-u0 .
c,1,
(6.15)
Reafianging
for?,(r)resutts
to writea differential
equation
in
dD
du"
Ct
A: cn c. + qn'
(6.76)
FinatLy,
sotving
the differentiai
equation
in tq. 6.76,wesee
rlt):
?,"(r)+ o(0).
c 1+ c 2 + c r
\6.77)
If Cr = C2 = Cz = 25pF.
?'(t) = 03331'r(t)+ t'(0)'
(6.78)
As before,the sensing
circujtetiminates
?r(0),so the sensed
output
voltage
is
o(t) = 0.3330"(r).
(6.7e)
Comparing
Eqs.6.74and6.79,weseethat whenthe buttonis pushed,
the outputis onethird of the inputvoltage.Whenthe b0ttonis not
pushed,
theoutputvoLtage
is onehaLfoftheinputvoLtage.
Anydropin
outputvottage
is detected
by the etevato/s
controlcomputef
anduttimateLy
resuLts
in theelevator
a ivingat theappropriate
floor.
yourundestanding
NoTE:Assess
of thisProctical
Perspective
by tryingChaptel
Problens
6.49ond6.51,
Figure6.36d A modet
oftheei€vator
cattbutton
cncuitwhen
activated
byfinger
touch.
Sumnrary2 7 7
Surnmary
Inductanc€is a linear circuit parameterthat relalesthe
voltageinducedby a time-varyingmagrericfield to the
currentproducingthe field. (Seepage188_)
Capacifance
is a linear circuit pammeterthat relatesthe
cment induced by a time-varying electric lield to the
voltageproducingthe field. (Seepage195.)
Inductorsand capacitorsare passiveelemenls;theycan
store and releaseenergy,but they cannot generateor
dissipateenergy.(Seepage188.)
The instantaDeous
poweral the terminalsof an inductor
or capacitorcan be posilive or negative,dependingon
whetherenergyis beirg deliveredto or extractedfrom
.
and Capacitors
(v)
,=L*
i=iJ,,,a"+4t,,1
p:Iii-Li#
(w)
(r)
.o=ilia,+x1to;
(v)
(A)
. doesnot permit an instantaneous
changein its termi. does permit an inslantaneouschangein fusteminal
voltage
. behavesasa shortcircuitin the presenceofaconsrant
terminalcurrent(Seepages188-189.)
p:ri=CD*
(w)
w:icf
(r)
TABLE
6.2 Equationsfor Series-and Pahllet-Connected
Inducto15.nd Capacitors
. doesnot permit an instantaneous
changein its teminal voltage
. cloespemit an instantaneous
changein ils terminal
. behavesas aII open circuil in t]re presenceof a con
stantterminalvoltage(Seepage196.)
Equationsfor voltage,curent, power, and encrgy in
ideal inductorsand capacitorsare givenin Table6.1.
Inductorsin seriesor in parallel can be repiacedby an
equivalenlinductor.Capacitorsin seriesor in parallel
can be replacedby an equivalentcapacitor-The equations are summaizedin Table6.2.SeeSection6.3 for a
discussionon how to handle the initial conditionsfor
seriesand parallel equivaientcircuitsinvolving inductols ano capacrton.
r..:zr+l+. +f
ceq: c|+ c?+ ,,, + c,
Mulual inductance, M, is the circuit parameter relating
r h c v o l l a g ei n d u c c d i n o n e c i r c u i t t o a l i m e - v a r y i n gc u r .
rent in another circuil. Specifically,
dri
r,t=Lri+Mi;
Jt.
_. ,tiL
- dit
+ L )u) = tulr.
dt
,lt
218
Inductance
Capacitance,andl4utuatlnductance
where ?)1and i1 are the voltage and cment in circuit 1,
and ?)2and i2 are the voltage and current in circuit 2. For
coils wound on notunagneticcore<.Mp - M1] - V
(Seepage210.)
. The dot conv€ntion establishesthe polarity of mutually
inducedvoltages:
When ths referencedir€ction for a curent enters
the dotted teminal of a coil, tle referencepolarity of the voltage that it inducesh the other coil
is positive at its dotted terminal.
Or, altematively,
When the rcferencedirection for a curent leaves
t]Ie dotted teminal of a coil, the refererce polarity of the voltage that it inducesin the other coil
is negativeat its dotled teminal.
the ielationship between the self-hducfance of each
winding and the mutual inductance between windhgs is
M : k^/LJ4.
The coefffcient of coupling, &,is a measureof the degiee
of magnetic coupling. By definition, 0 < I < 1. (See
page 211.)
The energy stored in magnetically coupled coils in a linear medium is related to the coil currents and inductancesby tlre relationship
*:LS,ii *l,i?+ uri,.
(seepaee213.)
(Seepaee20a.)
Probtems
Figure
P6.2
Secrion6,1
6.1 The triangular current pulse shoM in Fig. P6.1 is
""" appliedto a 375mH inductor.
a) Write the expressionsthat desc be i(t) ill the
foul intelvalsl < 0,0 = t = 25ms,25ms< l <
50ms,andt > 50 ms.
De
ve the expressionsfor the inductor voltb)
age,power, and energ]. Use the passivesign
convention.
fr
0
(a)
1 2
(b)
6.3 The current in the 4 mH inductor in Fig. P6.3 is
Figure
P6,1
i (nA)
400
known to be 2.5 A for I < 0. The inductor voltage
for t > 0 is giver by the expression
Ir(t) : 30e3rmv, 0* <t < oo
Sketchr'.(t) and tr(,) for 0 < t < oc'.
0
t(ns)
25
50
,(m0
6.2 The voltage at the terminals of the 300 /-.H hductor
""" in Fig.P6.2(a)is shownin Fig.P6.2(b).The hductor
cu[ent i is known to be zerofor I < 0,
a) Derive tlre expressionsfor i for t > 0.
b) Sketchtversuslloro< I < @.
figureP6.3
iL(.4
u,@
,1mH
prcbrems219
The curent in a 100/rH inductor is known to be
iL = zotelt A Ior t>0.
a) Find the voltage aqoss the inductor for t > 0.
(Assume the passivesign convention.)
b) Find the power (in microwarts) at the rerminals
of the inductor when r : 100ms.
c) Is the inductor absorbing or delivering powei at
100ms?
d) Find the energy (in microjoules) srored in the
inductor at 100ms.
e) Find t}le maximum energy (in midojoules)
slorediD rhe inducloraDdthe time (in micro
seconds)when it occurs.
The current in and $e voltage acrossa 2.5 H inductor are known to be zero for t = 0. The voltage
acrosslhe inductofis gjren bv the graph in Fig.po.5
lort > 0.
a) Derive the expression for the curent as a
function of tim; in the intenals 0 = t < 2s,
2 s < r < 6 s , 6 s < r = 1 0 s ,1 0 s< , < 1 2 s ,
a n d 1 2 s= t < c o .
b) For r > 0, what is the current in the ilductor
when the voltage is zero?
c) Sketchiversus,foro< t < c.o.
figureP6.5
, (v)
6.7 a) Find the inductor curent in tle circuit in
Fig. P6.7 if t) = 250 sin 10001V,l, = 50 mH, and
(0) : -5 A.
b) Sketch r', i, p, and tl) veNus r. In makhg rhese
sketches,usethe fomat usedin Fig. 6.8.plot over
one complete cycle of the voltage waveform.
c) Describe tle subintervals in the lime inte al
between 0 ard 27 ms when power is being
abso$ed by the inductor.Repeat for the subintervals when power is being delivered by the
inductor.
Figure
P6,7
fl
6.8 The current ir a 15 mH inductor is known to be
t<0;
i : Ale-zm, .+ A2e-3,m,A,
t > 0.
The voltage affoss the inductor (passive sign convention) is 60 V at I = 0.
a) Find the e4)ression for the voltage across the
inductor for r > 0.
b) Fitrd the time, greater than zero, when the power
at t]le terminals of the inductor is zero.
6.9 Assume in Problem 6.8 that the value of the voltage
asoss the inductor aff = 0is 300 V insteadof 60V
a) Find the nume cal expressions for j and o for
r>0.
b) Specify the time itrtervals when the inductor js
storing eneryy and the time intenals when the
inductor is delivedng eneryy.
.) Show that the total energy extracted from the
inductor is equal to t}le total energy stored.
6.10 The current in a 2 H inductor is
6,6 The curent in a 20 mH inductor is known ro oe
7 + (15 sin 140r- 35 cos1zl0r)€20'mA for I > 0.
Assume the passivesign convention.
a) At what instant of time is tle voltage acrossthe
ilductor maximum?
b) What is the maximum voltage?
i =254,
t=0;
i = (81 cos 51 + 82 sir sl)e-,A,
I > 0.
The voltage adoss the inductor (passive sign qjnvention) is 100V a : 0. Crlculate the power at the
terminals of the hductor at t = 0.5 s. State whether
the inductor is absorbingor deliverhg power.
220
Inductance,capacjtance,andMutuatlnductance
6.11 Initially there was no energy stored in the 20 H
inductor in the circuit in Fig. P6-11when it was
placed acioss the terminals of the voltmeter. At
t = 0 the inductor was switched instantuneously to
positionb whereitiemainedlor 1.2s beforereturning instantaneously10 position a. The d'Arsonval
voltmeterhasa full-scalereadingof25V and a sensitivity of 1000 O/V. W}lal will the reading of the
voltmeter be at the instant the switch returns to
position a if the inertia of 1ie d'Arsonval movement is negligible?
Section6,
6.14 A 0.5pF capacitoris subjeciedto a voltagepulse
having a duration of 2 s.The pulseis describedby
the following equations:
( qot'v.
o=1<1s;
, . ( t ) = 1 a 0 ( 24 ' v , 1 s < r < 2 s i
elsewher(.
t0
Sketchthe cuuent pulsethat existsin the capacitor
duringthe2sinterval.
Figure
P5.11
6.12 Evaluate the integral
1,"oo,
for Example 6.2. Cornment on the significance oI
the result.
6.13 The expressionsfor voltage, power, and energy
derived in Example 6.5 involved both integntion
and manipulation of algebraic expressions.As an
engineer, you cannot accept such results on faith
alone.That is, you should develop the habit of asking yourself,"Do theseresultsmake sensein tems
of the known behavior of the c cuit ttrey porport 1I)
describe?" With tlese tloughts in mind, test the
expressionsof Example6.5 by performingthe following checks:
a) Check the expressionsto see whether the voltageis continuous ir passhg from one time interval to the next.
b) Check the power expression in each interval by
selecting a time within the inteflal and seehg
whether it gives the same result as the corresponding product of o and i. For example,test at
10 and 30 ps.
c) Check the eneryy expressionwithin eachinterval
by selectinga time within the interval and seeing
whether,the-eneigy equation gives the same
rcsultas;Co'. Use 10 and30 ps astestpoints.
6.15 The rectangular-shapedcurrent pulse shown in
End Fig-P6-15is applied to a 0.2IrF capacifor.The initial voltage on the capacitoris a 40 V drop in the
relerenco direction of the currert. Derive the
expression for the capacitor voltage for the time
intervalsin (a)-(c).
a) 0 < t < 100ps;
b) 100ps < t < 300l..s;
c) 300ps < r < oo;
d) Sketch r(l) over the interval -1001.s < t<
500ps.
Flgure
P6.15
r(/d
6.16 The voltage at the terminals of the capacitorin
Eflc Fig.6.10is known to be
"
r - rr:
[ --rov.
1m,(4
+ sin3000r)
cos3000r
v I > 0.
lto ro"
AssumeC:0.5/rF.
for I < 0.
a) Findthecurent in thecapacitor
b) Fnrdtlre curent in the capacitortbr r > 0.
Prcbtems
221
c) Is there an instantanoous change in the vottage
aooss tle capacitor at I : 0?
d) Is tlere an instantaneous change in the current
in the capacitor at t = 0?
e) How much energy (in mioojoules) is stored in
the capacitorat, = co?
6,17 The curent shown in Fig. P6.17 is applied to a
6flft 0.25pF capacitot. The idtial voltage on the capacitor is zero.
a) Fird the charge on the capacitor at t = 30 ps.
b) Find the voltage on the capacitor a1r = 50 ps.
c) How much energy is stored in the capacitor by
this culrent?
figureP6.17
6.19 The voltage across the terminals of a 0.4/r,F
t
fzsv,
r5m'+,42€ ]soev. I > 0 .
[erre
The initial qrllent ir the capacitor is 90 mA. Assume
t}Ie passivesign convention.
a) What is the initial eneigy srored ir the capacitor?
b) Evaluatethe coefficients,4land.42.
c) What is the expiessionfor the capacitor current?
Section 6.3
6.20 Assume tlat the initial energy stored in the inductors of Fig. P6.20is zero. Find the equivalent inductance with respect to the terminals a,b.
figureP6.20
21H
6.21 Assume that the initial energy stored in the inducton of Fig. P6.21is zero Find rhe equivalent iDductancewith respectto the terminalsa,b.
6.18 The initial voltage on the 0.2 rlF capacitor shown in
6tre Hg. P6.18(a)is -60.6 V. The capacitorcurrent has
the wavefom shownin Fig.P6.18(b).
a) How much energy, in miffojoules, is stored in
the capacitorat t = 250ps?
b) Repeat(a) for t = oo.
Flgure
P6.18
O.2pF
--
-60.6V
(a)
Flgure
P5,21
12H
5H
14H
222
Inductance,
Capacitance
andl4utuat
Inductance
Figurc
P6.23
622 The two parallel inductors in Fig. P6.22 are con-
necledacrosslhe terminaL of a black bo),al / - 0,
The resulting voltage u for t > 0 is known to be
1800€ m' V. It is also known that ir(o) = 4 A and
ir(0) = -16 4.
a) Replacerhe originalinducror.with an equivalent inductor and find i(r) for I > 0.
b) Find ir(t) for t > 0.
c) Find i2(t) foi t > 0.
d) How much energy is delivered to the black box
in the time intenal0 < 1 < oo?
e) How much energy was initially stored in the par'
allel inductors?
f) How much energy is tiapped in ttre ideal
inductoft?
g) Show that your solutions for tr ard t, a$ee with
the answerobtainedin (0.
32H
624 For the circuit shown in Fig. P6.23,how many milliseconds after the switch is opened is the energy
delivercdto the black box 80o/oof the total energy
delivercd?
625 Find t}le equivalent capacitancewit}l respect to the
terminals a,b for tlle circuit shown in Fig. P6.25.
Figure
P6.25
12p.F
8V
20 pF
FigureP6.22
--rts?1PF
- L
28 pF
24 p.F
5V
;,oi10H,?0)l
2Y
6.:26 Find the equivalent capacitancewith respect to the
terminals a,b for the circuit shown in Fig. P6.26.
figu|eP6.26
8nF
6.23 The thee inducton in tlle circuit in Fig. P6.23 are
30v +
conrected across the termimls of a black box at
t = 0. The resulting voltage for I > 0 is knorn to be
u. : 7250e25'Y
If tr(0) = 10 A and i(0) : -s A, ftrd
a),,(o);
b) ',(r),r > 0;
18 DF
1 5 V+
5.61F, //
40 Dr
b.-
+ 5V
6.n
l0v +
c) il(r),r > 0;
d)ir(r),r>0;
The two series connected capacitors in Fig. P6.27
are connected to the terminals of a black box at
I = 0. The resultingcurrent l(t) for t > 0 is known
pA.
to be 900e 15oot
e) the initial energy stored in the three inducto$;
f) the total energydeliveredto the blackbox;and
g) the energy tmpped in tlre ideal inductors
a) Replace'he origjnalcapacirors
wirh an equi!dlent capacitor and find z"(r) for / > 0.
b) Find 01(t)for t > 0.
ProbLems
223
c) Find r'r(r) for r > 0.
d) How much energyis delivercdro the black box
in the time interval0 < r < oo?
e) How much energy was initially stored in the
seriescapacitors?
f) How much energy is trapped in the ideat
capacitors?
g) Showt]tat the solutionsfor or and o, agreewith
the answerobtainedin (1).
Figure
P6.27
45V
15V
+
6.?2 The four capacitorsin the circuit jn Fig.P6.28are connected acrossthe tenninals of a black box at 1 = 0.
The resulting current i, for a > 0 is kllown to be
d) the percentageof the initial energystoredthat is
deliveredto the black box| and
e) the time, in miliiseconds, it takes to deliver 5 /..J
to the blackbox.
6.30 Derive the equivalert circuit Ior a seriesconnection
of ideal capaciton.Assumethat eachcapacitorhas
its o$,ninitial voltage.Denote theseinitial voltages
as rr(r0), o2(r0),and so on. (Ht rr Sum rhe voltages
acrossthe stdng of capacitors,recognjzingthat the
seriesconnectionforcesthe current in cachcapacitor to be the same.)
6.31 Derive the equivalentcircuit for a parallel connection of ideal capacitors.
Assumethat the initial volf
age acrossthe paralleledcapacitorsis u(lo). (dir?rr
Sum the currentsinto th€ string of capaciton,recognizing that the parallel connection forces the
voltageacrosseachcapacitorto be the same.)
Sections6.1-{,3
6.32 The cment ir the circuit in Fig. P6.32is knowl to be
i, : 5043o0'(cos60001+ 2sin60001.1mA
for / > 0+.Find ?,r(0+)and z,l0+).
Figure
P6.32
it = 50e-^a'tt A.
If o"(0) : is v, t.(0) = 45 v, and z'd(o): 40 V,
tind the following for / > 0: (a) olr, (b) ?,,0),
320Q
(c)o.(1),(d)trlr, (e)n(0, and(f) ,,(,).
Figure
P6.28
6.33 At t = 0, a seriesconnectedcapacitotand inductoi
are placedacrossthe terminalsof a black box, as
shownin Fig.P6.33.For r > 0, ir is known that
j, :
e-30'sin6ot A
If0c(0) =
tigureP6.33
6.29 For the circuitin Fig.P6.28,calculate
a) the initial eneigystoredin the capacitors;
b) the final energystoredin the capacitors;
c) the total energydeliveredto the black box;
300V lind ", for I > 0.
224
Inductance.
Calacitance.
andlllutuat
Inductance
Section 6.4
6.34 There is no energy srored in the circuit in Fig. P6.34
at the time the switchis opened.
a) Derive the dilferential equation that govems
the behaviorot 12 if Lr - l0 H. L, - 40 H.
M:5H,andRd:90O.
b) show that when ie = 10;' - 10 A, t > 0, tlle
differential equation derived in (a) is satisfied
when i2 = at - 5e 225tA, t > lJ.
c) Find the expressionfor the voltage ?1 acrossthe
current source.
d) what is the initial value of x'r? Does tlis make
sensein terms of known circuit behavior?
Figure
P6.34
l),
LI
6.35 Let o. represent the voltage acrossthe 16 H inductoi in the circuit in Fig. 6.25.Assume D, is positive at
thedot.AsinExample6.6,i"= 16 - t'n5' O.
a) Can you find 2o without having to differentiate
the expressionsfor the currents? Explain.
6.37 a) Showthat the differential equationsdedvedin
(a) of Example6.6canberearangedasfollows:
di.
4;
di"
+ 2sit - 8i
di,
8dt
- 2Oi2= si. - 8
di"
dt:
di,
di.
20r, I t6-i + 80i,- t6dt
dt
b) Show that the solutions for it, and 12given in (b)
of Example6.6satisfythe differentialequations
given in pa (a) of this problem.
6,38 The physical construction of four pairs of magneticaly coupled coils is shown in Fig. P6.38. (See
page 225.) Assume that the magnetic flux is confined to the core material ir each structure. Show
two possible locations for the dot markings on each
pair of coils.
6.39 The polarity markings on two coils are to be determhed expeimentally. The experimental setup is
shoM in Fig. P6.39.Assumethat tho terminal connected to the negative termhal of tle battery has
been given a poladty mark as shown. When the
switch is close4 the dc voltmeter kicks upscale.
Where should the polarity mark be placed on the
coil connected to the voltmeter?
rigureP6.39
b) Derive tle expressionfor r',.
c) Check your answer in (b) using the approp ate
current derivatives and inductances,
6.36 I-et DsreFesentthevoltageacrossthe currentsource
in the circuitin Fig.6.25.Thereferencefor ?,sis positive at theupperteminal of the currentsource.
a) Find-ts as a Iunctionof time whents = 16b) wlat is the initial valueof os?
c) Fitrd t])e expressionfor the powerdevelopedby
the currentsource,
d) How muchpoweris the currentsourcedeveloping whenI is infinite?
e) Calculatethe power dissipatedin eachresistor
wherlt is infinite.
6.40 a) Show that the two coupled coils in Fig. P6.40can
be replaced by a single coil having an inductance
of aab : a1 + L2 + 2M. (Hint: Express o^bas a
tunction of tab.)
b) Show that if the connectionsto the terminals
of the coil labeled a2 are reveNed,
Ldb= Lr + Ia
2M.
Figure
P6.40
M
a5:r._._ri'^_'b
Pmblems225
rigureP6.38
0)
G)
6.41 a) Show that the two magneticaly coupled coits in
Fig.P6.41canbe feplaced
by a singlecoithaving
a.ninductance of
(d)
b) Show that if the magnetic polarity of coil 2 is
reversed,then
Lrlo - M'
M2
hlo
Lr+ L2 2M
Lr+ L2+2M
Fi$re P6.41
(Airi Let it and i2 be clockwise mesh curents in
the left anddght "whdows" of Fig.P6.41,respectively. Sum t]te voltages around the two meshes.
In mesh 1 let ?,abbe t})e unspecified applied voltage.Solve for d4/dt as a function of oab.)
226
Inductanc€,
Capacitance,
andl{otuat
Inductance
S€.tion 6.5
6,42 TWo magnetically coupled coils are wound on a
nonmagneticcore.The se[-inductanceof coil 1 is
250mH,the mutualinductanceis 100mH, the coef
ficient of coupling is 0.5, ard tlle physical structure
ofthe coilsis suchthat 91r = 922.
a) Find 12 and the turns ratio N/N2.
b) If Nr = 1000,what is the valueofgr and 0r?
6.43 The self-inductances of two magnetically coupled
coils are l,r = 400 riH and 12 : 900 &H. The coupling medium is nonmagnetic. If coil t has 250 tums
and coil 2 has 500 tums, find 9x and 921 (in
nanowebersper ampere) when lie coefficientof
couplingis 0.75.
6.,14 Two magneticallycoupled coils have self-inductances
of 52 mH and 13 mH, respectiveryThe mutual induc
tance betweenthe coils is 19.5mlL
a) What is the coefficient of coupling?
b) For these two coils,what is the largest value that
,14can have?
c) Assume that the physical structure of these coupled coils is such that Or : 92. What is the turns
ratio N/N, if Nl is the number of tums on the
52 mH coil?
6.45 The self-inductances of two magtetically coupled
coils are 288 mH and 162 InH, respectively. The
2B8 mH coil has 1000 tums, and the coefficient of
coupling between the coils is Z. The coupling
medium is nonmaglelic. When coil 1 is excited with
coil2 open,thefluxlinking only coil 1 is 0.5aslarge
as the flux linkhg coil 2.
a) How manyturff doescoil2 have?
b) What is the value of 02 in nanowebersper
c) What is the value of 9n in nanowebersper
ampere?
d) What is the ratio (drxldrr?
6.47 The self-inductances of thl3 coils in Fig. 6.30 are
L1 : 25 mH andL2 : 100mH. If the coefficientof
couplingis 0.8, calculatethe energystored in the
systemin millijouleswhen (a) t1 : 10 A, 4 : 15 A:
(b) i1:
10A, i=
15A; (c) tr = -10A,
'2 = 15 A; and (d) t1 = 10 A, i2 : -15 A.
6.48 The coefficient of coupling in Prcblem 6.47 is
increasedto 1.0.
a) If ,r equals10 A, what value of i resultsir zero
storedenergy?
b) Is there any physicallyrealizablevalue oft2 that
canmake the storedenergynegative?
Sections6.1-6.5
6.49 Rework the Practical Penpective example, except
pll$lf?i! that this time, put the button on the hottom of tie
divider circuit, as shown in Fig. P6.49.Calculate tlre
output voltage o(t) when a finger is present.
Figuru
P6.49
t"(t)
6.50 Somelamps are made to tum on or off when the
,l$llJjhbaseis touched.Theseusea one-tenDinalvariation
of the capacitive s$itch circuit discussed in the
PracticalPerspective.
Figue P6.50showsa circuit
model ol such a lamp. Calculate the change in the
voltage t'(t) when a person touches the lamp.
A(.umeall capacirorr
are i0iliallyJischarged.
Figure
P6,50
10pF
6,46 a) Startingwith Eq.6.59,show Oratthe coefficient
of couplingcanalsobe expressedas
' J(#X#)
b) On the basis of the fracnons 62\/6r and.OdQ,
explainwhy k is lessthan 1.0.
..{-
=ti
Lamp Person10pF
Probtems
227
6.51 In the Practical Perspectiveexample,we calculated the output voltagewhen the elevarorbufton
pPEIifiIHLE
is the upper capacitor in a voltage divider. In
Problem 6.49,we calculatedthe voltage when the
button is the bottom capacitorin the divider, and
we got the same tesult! You may wonder if this
will be true for all suchvoltagedividers.Calculate
the voltagedifference(fingerversustro finger) Ior
the circuits in Figs. P6.51(a) and (b), which use
two identicalvoltage sources.
FigucP6.51
Fixed
_p5 pF +
)(t)
,"(r)
;
\
""(,)
-f
No finger
?5pF ri^"a
capacitor
E5 pF +
?,(r) Finger
Response
of First-0rder
Rl andRCCircuits
7.1 TheNaturatResponse
of an RL[itcuirp.230
p.236
7.? TheNaturatResponse
of anRCaircuit
7.3 TheSt€pResponse
of f,l and
RCCitcuits p. 24A
7.4 A GeneratSoLutionlor Stepand NaturaL
p. 248
Responses
7.5 SequentiatSwitchingp- 254
p. 258
7.6 UnboundedResponse
7.7 TheIntegnting Amplifier p. 26,
1 Beableto deiermireihe faturatfesponse
of
botlrRLaid RCcncuits.
2 Beabteto determine
the steprcsponse
of both
3 Knowhowto anatyze
circuitswith sequentjat
swiiching.
4 B€abt€to analyze
op ampcncujtscontaining
rcsistors
anda sjngl€capacitof.
228
In Chapter6, we notedthatanimportanlalt buteo[ inductois
and capacitors
is their ability to storeenergy.We are now in a
positionto determinethc currcntsandvoltagesthat alisewhen
cnergyis either releasedor acquiredby an inductor or capacitor
ln responseto all abrupt changein a dc voltaE:cor curent source,
In this chaptcr.we will focus on circuits that consistonly of
sources,resiston,and either (but nol both) inducton or capaci'
tors. For brevity, such configurationsare called lla (.esistorinductor) andRC (resistor-capacitor)
circuits.
Our analysisof R, and RC circuitswill be dividedinto three
phases.II1 the first phase,wc considcrthe curents and voltages
that arisewhen storederergy in an inductor or capacitoris suddenly releasedto a resistivenetwork. This happcnswhcn thc
inductor or capacitoris abruptlydisconnectedfronr its dc source.
Thuswe canreducethe circrlit to one ofthe two equivalenttbrms
shownin Fig.7.1on page230.Thecurents and voltagesthat arise
in this configurationare leferred to asthe natuml tesponseof the
circuit,to emphasizethat the natureof the circuit itsclf,not cxtcr
nal sourcesof excitation.doterminesits behavior.
Tn tbe secondphaseof our analysis,we considerthe cunents
and voltagesthat arisewhenenergyis beingacquircdby an inductor or capacitordue to the suddenapplicationof a dc voltageor
cunent source.This responseis referred to as the step rcsponse.
The proccssfor finding both the natural and stepresponses
is the
same;thus.jnthe third phaseof our analysis,
we developa general
method that can be usedto find thc responseof RL and nC cjrcuitsto any abruptchangein a dc voltageor cuflent soulce.
Figure7.2on page230showsthe lour possibilitiesfo. the gen
eral configurationol RZ and RC circuits.Note that when there
are no indcpendentsourcesin the circuit.the Th6veninvoltageor
Norlon cunent is zero, and the circuit reducesto one of those
shownin Fig.7.1;thatis,we havea natural-response
problem.
na and RC circuits are also known as first-order circuits,
becausetheir voltagesalrd culrents are describedby first ordcr
ditferential cquations.No matter how complex a cjrcuit may
PracticaI
Perspective
A Fl.ashing
LightCircuit
Youcan probablythink of rnanydjfferentapplications
that
requirea flashingtight.A stitLcamerausedto takepicturesjn
low light conditionsemptoysa brightftashof Lightto ittuminatethe sceneforjust tongenoughto recordthe inrageon
fitm. GereralLy,
the cameracannottakeanotherpjctureuntjl
the circuitthat createsthe ftashof Lighthas'te-charged."
otherapplicatjons
usefLashing
tightsas warnjfgfor hazards,such as tat[ antennatowers,constructionsites, and
secureareas.In designingcircuitsto producea flashof tight
the engineermustkrow the requir€nr€nts
of the apptication.
Forexampte,
the designengineerhas to knowwheth€rthe
flashis controtLed
manuatLy
by operatinga switch(asin the
caseof a camela)of if the flashis to repeatitsetfautomaticattyat a predetermined
rate.Theengineeratsohasto knowif
the flashinglight js a permanent
fixture(ason ar antenna)or
a temporary
instatlation(as at a construction
sjte). Another
qre\fionrhathas[o beanswFred'.
power
wherLrer
a
sorrcei.
r€adiLy
avaitabte.
Manyofthe cjrcuitsthat areusedtodayto controlflashjng
Lightsare basedon €tectroniccircujtsthat are beyondthe
scopeof this text. Neverthetess
we can get a feel for the
jnvotvedin designing
thoughtprocess
a fLashing
tight circuit
by anatyzjng
a circuitconsisting
of a dc voLtage
source,
a reststor, a capacr'tor,
and a lampthat is designed
to discharge
a
jn
flashoftight at a criticatvottage.
Sucha cjrcuitisshown the
figure.Weshatldjscuss
this cjrcuitat the endofthe chapter.
229
230
Response
of Fi6t-0rder
fi andRrCircuits
c.,r
(a)
(b)
FiguE7.t A Thetu/oforms
ofthecncuits
fornatuGt
(a)Rrcncujt.
(b)nrcircuit.
rcsponse.
appear,if it can be reduced to a Th6venir or Norton equivalent connected
to the terminals of an equivalent inductor or capacitor, it is a first-order
circuit. (Note that if multiple inductors or capacitors exist in the original
circuit, they must be intercoinected so that they can be replaced by a single equivalentelement.)
Aftei introducing tlle techniques for analyzing the natural and step
rcsponsesof fi$t-order circuits, we discusssome special casesof interest.
The first is that of sequentialswitching,involving circuits in which switching
can take place at two or more hstants in time. Next is the unbounded
rcsponse.Finally, we anallze a useful circuit called the integrating amplifier.
Rrh
I
l
-
fl"*
Y
7.1 TheNaturat
Response
of an RLCircuit
.j,
I
G)
vrh
t
L
(b)
Yrh
Rft
The natural responseof an Ra circuit can best be desc bed in terms of the
circuit shown in Fig. 7.3.We assumethat the independent current source
generatesa constantcurent of 1"A, and that t]re switch has been in a
closed position for a long time. We define the phrase a long time morc
accumtely later in this section.For now it meansttrat all currents and voltageshave rcached a constant value.Thus oDly constant, or dc, currents can
exist in the circuit just prior to t]le switch's being opened, and therefore
the inductor appears as a shot c cuit (adi/dl : 0) prior to the release of
the stored energy.
Becausethe inductor appears as a short circuit, the voltage acrossthe
inductive branch is zero, and there €an be no current in either Ro or R.
Therefore, all the source current 1, appea-rsin the inductive branch.
Fhding the natuial response requires finding the voltage and curent at
the terminals of the resistor after tlle switch has been opened, that is, after
the source has been disconrected and the inductor begins releasing
energy.If we let t = 0 denotethe instantwhen the switchis opened,the
problem becomesone offinding ?(r) and t(t) for I > 0. For t > 0, the cir,
cuit shownin Fig.7.3reducesto tlle one showr in fig.7.4 on the next page.
Derivingthe Expression
for the Current
(d)
tigure7,2 A Fourpossible
fi'st-order
circuits.
(a)Aninductor
connected
t0 a Th6venin
equjvateni.
(b)Aninductor
connected
to a Norionequjvatent.
connected
to a Th6venin
equjvatent.
k) A capacitor
(d)A capacitor
connect€d
to a Norton
equivaLent.
Figure
7.3l' AnRtcircuit.
To find i(t), we use Kirchhoff's voltage Iaw to obtain an expressioninvolving i,R, and a. Summingthe voltagesaroundthe closedloop gives
t4n^r=o,
(7.1)
wherewe uselhepassivesigncorvention.Equation7.1is known asa firs!
oider ordinaly differential equation, becauseit contains terms involvhg
the ordinary derivative of the unknown, that is, dy'dt. The highest order
derivativeappeaing in the equationis 1;hencethe telm fitst-otdet.
We can go one step fu her in descdbing this equation. The coeffi
cients in the equation, R and a, are constants; that is, ttrey are not ftrnctionsof either the dependentvariableior the independentvariablet.Thus
ttre equation can also be described as an ordinary differcntial equation
witl constant coefficients.
7.1
To solve Eq. 7.1,we divide by a, transposerhe rerm involving i ro the
right-hand side, and then multiply both sidesby a differenrial time dr. The
resuttrs
di,
-4tot
TheNatumt
RespoNe
ofanil Circuit 231
-,=.'8,
1 7 . 2 ) F i g u r7e, 4A t t p , , . 1 ' ts o w -i - t g . / . ? .f o ., - 0 .
Next,we recognizethe lelrhard sideofEq.7.2 as a differertial changein
the current i, that is, di. We now divide through by j, gettitg
di
i
R,
= -70,
(7.3)
We obtair an explicit expressionfor i as a function of r by integrating both
sidesofEq.7.3. Usingr and I asvariablesof integiationyields
-:1,"'*,
li,,'+=
17.4)
in which i(lo) is the cu[ent corespondingto time r0,and j(1) is the current
correspondingto time t. Here, r0 = 0. Therefore,carrying out the indicaledintegrationgives
' ( 0
(0)
R
L"
(7.5)
Basedon the definitionofthe naturallogarithm,
i\t) = i\o)e tR Ltt
(7.6)
Recall from Chapter 6 that an instantaneous change of current cannot
occur in an inductor. Therefore, in the first instant after the switch has
beenopened,the curent in the inductor remainsunchanged.Ifwe use0
to denotethe timejustprior to switching,and 0+for the time immediately Figure
7.5A Thecurrent
response
forthecncuitshown
Iollowing switching, then
jn Fjg.7.4.
,(0-):(0+)=10,
d Initial inductorcurrent
where,asin Fig.7.1,10denotesthe initial curent in the inductor.The initial
current ir the inductor is oriented in the same directiorl as t]re reference
directionof i. HenceEq.7.6 becomes
itJ\:
r^" lR/L)t
(7.7) { Naturalresponse
of annt circuit
which showsthat the currentstartsfrom an initial value10and decreases
exponentiallytowardzerc as r increases.
Figure7.5showsthis response.
We derivethe vollageacrossthe rcsistorin Fig.7.4ftom a direct appli
cationof Ohm'slaw:
r:
iR=
IoRe-\RtL)t, t >0+.
(7.8)
232
Response
of Fnst-0rd€r
RtandRrCjrcuits
Note that in contrust to the expressiotrfor the curent shown in Eq. 7.7,
the voltage is defined only for I > 0, not at t = 0. The reason is that a step
change occurs in the voltage at zero. Note that for r < 0, the derivative of
the current is zerq so the voltage is also zero. (This result follows hom
1)=Ldi/dt-\.Jmus
"(0):0,
(7.e)
t (0-) = 1oR,
(7.10)
wherer{0+) is obtainedfrom Eq.7.8 withr = 0+.1Withthis stepchangeat
an hstant in time, the value of the voltage at 1 = 0 is unknowr. Thus we
use I > 0- in defining the rcgion of validity for these solutions.
We dedve the power dissipated in the resistor ftom any of the follow-
p = izR,
(7.11)
Whicheverform is used,theresultingexpressioncan be reducedto
p
lzRe-2(RtL)t,
t > 0+,
17.12)
The energy delivered to the rcsistor during any intenal of time after the
swilchhasbeenopenedis
-= | ,,": fo'na",,'',,a,
- "u'*'",
:4fu13*1t
= L*tZrt - e-2tR/Lr),t >0.
(7.13)
Note ftom Eq. 7.13 tlat as t becomesiniinite, the energy dissipated in the
iesistor approachesthe initial energy stored in t]te inductor.
TheSignificance
of the TimeConstant
The e\pr-essions
lof j(1)(tq 7.7iand d tEq. 7.8)includea termot rhe
"'.
"
plorm
The coefLicienrot l-namel]. n L determinesrbe rate al
which the cwent or voltage approacheszero. The reciprocal of this ratio
is the time constsntof the c cuit. denoted
Timeconstant
for Bl circuit>
\7.11)
t we can de6ne the er?ressio.s 0 and 0+ more fomaly. The exprcsion r(0 refeB to the
)
linit or the variable x as t +0 fron tne left. oi lron negalive tine. The expresion r(0+)
relers to the limil of ihe rariable x as r + 0 fron the right, or fton posilive tine
7.1 TheNatumt
Response
ofanfit Circuii 233
Using the time-constantconcept,we write the expressionsfor curtent,
voltage, power, and energy as
ilt)= Ioea',
t>0,
1 ) ( t )= I o R e - ' / 1 ,
p = I1Re2th,
*=
t
-
rut^r,
t>0+,
t>0+,
e-o,,t. t>0.
(7.15)
(7.16)
\7.17)
(7.18)
The time constant is an important parametet fot firsForder circuits, so
mentioning several of its charactedstics is worthwhile. First, it is convenient to fhink of the time elapsedafter switching in terms of integral multiples of r. thus one time constant after the inductor has begun to release
its storedeneigy to the resistor,the curent has been reducedto tl, or
apprcximately 0.37of its initial value.
Table 7.1 gives the value of ?-t" for integal multiples of r from 1 to
10. Note that when the elapsed time exceedsfive time constants,the current is less than 1ol" of its initial value. Thus we sometimes say that five
time constants after switching has occuned, the currents and voltages
have, for most practical purposes,rcached tlle final values. For single
time-constant circuits (first-order circuits) with 1% accuacy, the phnse a
/org dme implies that tive or more time constants have elapsed.Thus the
existenceof current in the R-L circuit shown in Fig.7.1(a) is a momentary
event and is refered to as the tramient r€sponse of the circuit. The
response that exists a long time after the switching has taken place is
called the steady-stateresponse.The phrase d long ,tne then also means
the time it takes the circuit to reach its steady-statevalue.
Any first-oder circuit is characterized,in part, by the value of its time
constant.If we have no method for calculating the time constant of such a
circuit (perhaps becausewe don't know the values of its components), we
candetermineits valuefmm a plot of the circuit'snaturalresponse.
That's
becauseanother important characteristic of the time constant is that it
givesthe time required for the current to reach its final value if the curent
continues to change at its initial rate. To illustmLe, we evalrratedi/ dt at 0*
and assumethat the cunent continuesto changeat this mte:
firot=-lr,: lt
2r
3r
5r
3.6788x
1.3534x
4.9'787x
x
1.8316
10_1
10 1
10n
10-,
6.7379\ 10 3
2.4i88x 10 3
9.1188x 10+
8r 3.3546x 1or
9r 1.2341x 1or
10t 4.5400x 10 5
6r
'7r
(7.19')
Now, if i starts as 10 and decreasesat a constant rate of 1o/r amperesper
second.the exDressionfor i becomes
Io
11.zoJ
Equation 7.20 indicates that i would reach its fhal value of zero in
r seconds.Figue 7.6 showshow this graphic interpretation is useful in esti.
mating tle time conslant of a circuit from a plot of its natural rcsponse.
Such a plot could be generated on an oscilloscopemeasuring output currcnt. Drawing the tangent to the nalural responseplot at t : 0 and readhg Figure7,6 A A g€phicjntenretatjon
of thetimecor
the value at which the tangent intersects the time axis givesthe value of r.
stantoftheRt cjrcuitshown
in Fig.7.1.
234
Response
of Fnst-oder
Rlandfc Cncuits
Calculating the natural responseof an RL circuit can be summarized
Calculating
the naturalresponse
of
Rl circuitts
1. Find the initial current,10,throughthe inductor.
2. Find the time constant o{ the circuit, r : a/F
3. Use Eq. 7.15,Ioe '/', to genemtet(l) from 10and i
A
other calculations of interest follow ftom knowing i(1).
Examples7.1 and 7.2 illustratethe numericalcalculatioNassociated
with
the natuml responseofan Rt circuit.
Determining
the NaturalResponse
of anil Circuit
The switch in th€ circuit shown in Fig. 7.7 has
been closedfor a long time before it is opened at
t : 0. Find
a) iro) for t > 0,
b) i,(t) for t > 0*,
c) ?,o(,)for r > 0+,
d) the percentage of the total energy slored tn the
jn the 10 O resistor.
2 H inductorthat is dissipated
2Q
0 . 1 o, i , l 2 H
)'"
b) We find the current in the 40 O resistor most
easilyby usingcurrent division;lhatis,
'"
'"10
10
+ 40
Note that this er?ression is valid for I > 0*
becausejo = 0att:0
. The inductorbehavesas
a shor circuit prior to the s$ritch behg opened,
producing ar instantaleous changein the cufent
i". Then,
t,(r)= -4d5rA,
100
40f)
Figure7.7 4 ThecircuiiforExampte
7.1.
r>0*.
c) We find the voltage r, by direct applicationof
Ohm's law:
D.(t) = 40i": -160e 5'v,
r > 0'.
d) The power dissipatedin the 10 O resistorls
Solution
a) The switchhasbeenclosedfor a long time p or
to r :0, so we know the voltage across the
inductor must be zero at r : 0 . Therefore the
initial cuirent in the inductor is 20 A at r = 0 .
Hence,iz(O-) also is 20 A. becausean instantaneou.changein rbe cuffenrcannotoccufiD an
inductor. We replace the resisrive circuit connectedto the terminalsof the inductor wu a
singleresistorof 10 ():
R.q=2+(40 10)=10O.
The time constantof the c cuit is I/Rcq, or 0.2s,
giving the expressionfor the inductor current as
,r(r) - 20e-5,A, r>0.
,?
D N ' u r = ? - 2 5 b 0 lf0 W .
t(l
/'0'
The total energydissipatedin the 10 O resistoris
u )^ " ( t \ :
/I 2 5 o 0 P r u ' d: l - 2 5 b 1 .
.lo
The initial energystoredin the 2 H inductorrs
l
_
1
ru(O)= -Zi'(0)::(2)(4n0)
= 40uJ.
Thereforethe percentageoI energydissipatedin
the 10 O resistoris
=o.to..
ffirroot
7.1 TheNatumt
ResDons€
ofanNlCircuit 235
Determining
the NaturalResponse
of anBl.Ci]cuitwith ParallelInducto]s
In the circuit shown in Fig. 7.8,the initial currents in
inducto$ 1,1 and Z2 have been established by
sourcesnot shown.Theswitchis openedat I = 0.
:
7.6
t > 0,
= 5 . 7 o zP A . / > 0 + .
i,: --
a) Find i1,i2,and ir for I > 0.
b) Calculate the inilial erergy stored in the paiallel
2.4e2' A,
Nole that the expressionsfor the inductor cunents
it and ,2 are valid lor r > 0, whereasthe expression for the resistorcurent i3 is valid for t > 0*.
c) Determine how nuch energy is stored in tl]e
inductorsast+co.
d) Sbowthar$e roralenergydelivefedro lhe re.istive network equals the dilference between the
resultsobtainedin (b) and (c).
,o1
8()
Solution
jn Fig.7.8.
ofthecircuit
tisure7.9 A simptification
shown
a) The key to finding curents ir, i2, and i3 lies in
knowing the voltage o(t). We can easily find o(r)
if we reduce the circuit shown in Fig. 7.8 to the
equivalent folm shown in Fig. 7.9. The parallel
inductors simplify to an equivalent inductance of
4 H. carryingan i0ilialcurrentof l2 A.The resi*
tive network reduces to a single rcsistance of
8 O. Hence the initial value of i0) is 12 A and
rhelimeconslanr
is 4/8.or n.5s. Iherelore
b) The initialenerg)sroredin rheinduclor(h
1
1
d: -(5X6a) + -(20)(16):320J.
i(t)=12e)1A,
j1'1.6A
c)As t+co,
and i21.6A.
Thereforc, a long time after the switch has been
opened, the energy stored in the two inductors is
t
_
t
d
J2J.
^(5Xl.o) -:(20X l.ol
t>0.
d) we obtainlbe toralenergydeliveredlo the resisrive network by integrating the expression for
powerfromTerolo infinir):
lhe ioslcnraneous
Now z'(l) is simply tlle prcduct 81,so
.,(t) = 96e2'Y,
t>o+.
u::
The circuit showsthat o(l) : 0 at t = 0 , so the
expressionfor D(r) is valid for r > 0*. After
obtaining o(,), we can calculatei1,i, and 4:
"-4t )"
ln
i,' =: I 96e2'dr - 8
:Jo
=1.6-9.6e2A, t>0,
i,:j l'sa"^a',t
4{}
(5H)
,ln
= 1152:,l
-+
1 I l
L1
I pdt:
lo I r = 0
r2QoH)
1
Figurc
7.8A Thecncuit
forExamph
7.2.
40{))
,rrn
"l
/.m
I 1rs2e4,dt
.la
: 288J.
This result is the dillerence between the initially
stored eneryy (320 J) and the energy trapped in
the paftllel induclors (32 J). The equivalent
inductor for the parallel inductorc (which predicts the terminal behavior of the parallel combination) has an initial energy of 288 J; that is,
the energy stored in the equivalent inductor representsthe amount of energythat will be delivered to the rcsistive netwoik at the terminals of
the original inductors.
10r)
236
Response
of First-order
8l andRrCircuits
objective1-Be ableto determine
the naturaL
response
of both [l andRCcircuits
7.1
The swirchin thc circuil shorvnhasbeenclosed
for a lo g time and is openedat t - 0.
a) Calcular.3
the initial valueof;.
b) Calcularethe inirial cncrgv sloredin the
7,2
c) w}lal is the time corstant of the circuit lbr
t>0'l
d) Wlat is thc nunericat expressionfor ;0) for
e) What percentageof the initial energystored
hasbeendissipatedin the 2 r) resislor5 ms
afler the switchhasbeenopcned?
120V
Answer: (a) -12.5 Ar
(b) 62snJ;
(c) a nis:
(d) -12.5a1sorA.r > 0r
(e) e1.8o/..
At r : 0, the s*itch in the circuit shownnoves
instantaneously
from posiliona to posilionb.
a) Calculate,. for t > 0'.
b) What percentageof the initial cncrgyslored
in the inclucroris eventuallydissipaledin
the ,1O resistor'?
8c "'V. t > 0:
(b) 80%.
(a)
NOTE: Also try ChapterPrabkn$7.1 7.3.
ResBrnne
7 . 2 l-heN*tsrraI
*'i'e* R{ eirer"rit
F i g r r e7 . 1 0 , 1 A nf C c i r c u i i .
rigurc7,11;! Thecir.uitshownin
As mcntionedin Section7.1,the naturalresponscot an RC circuil is aral,
ogousto thal ofan Rl, circuit.Consequentl],.
wc don't treal lhe RC circuit
in thc sane detail aswe did the nl- circuit.
The Daturalresponseof an RC circuil is dcvclopedfroD the circuit
shownin Fig.7.10.Webeginb]' assumingthal lhc ss,ilchhasbeenin position a lbr a long time,allowilrgthe loop nade up of thc dc vollagesource
ys.the resistorn1. aDdthe capacitorC to reacha steadystalc condilion.
Recall fionr Chapler6 that a capacitorbehavesas an opcn circuil il1 the
presenccof a constantvoltage.'lhusdre voltage sourcccaniot suslaiDa
currcnt,and so lhe sourcevoltageappearsacrossthe capacitortenninals.
In Scction7.3,we will discusshow thc capacitorvollageactuallybuildsto
the sleady-state
value ofthc dc voltagesource,but for now the important
pojnl is that when the switch is moved from posiiioDa to position b (at
i = 0), the voltageon the capaciloris y-. Becausethere can be no instantancouschangein the voltageat the tcrminalsoI a oapacitor.the problem
rcduces1lrsolvingthe circujtshownin Fig.7.l1.
7.2 TheNaturaL
Response
of annCCircuit 237
Derivingthe Expression
for the Voltage
We can easily find tlle voltage ?(t) by thinldng in tems of node voltages.
Using the lower junction between R and C as the reference node and summing tlle curents away from the upper junction between R and C gives
c4 * {:0.
(7.21)
Comparing Eq. 7.21 with Eq.7.1 shows that tlte same mathematical techniques can be used to obtair the solutior for z(t). We leave it to you to
showthat
D(i) = u(o\e1tRc, t-0.
\7.22)
As we have already roted, t]le initial voltage on tle capacitor equals the
voltage sourcevoltage %, or
(7.23) < Initial capacitor
vottage
where yo denotesthe ioitial voltage on the capacitor.The time constant for
the nC circuit equals the product of the resistarce and capacitance,
namely,
i:rli,;#!l
(7.24) < Tine constant
for BCcircuit
SubstitutingEq$ 7.23arrd'7.24n\b Eq.1.22yields
(7.25) < NalulalresDonse
of annc circuii
which indicates that tle natural rcsponse of an RC circuit is an exponential decay of the initial voltage. The time constant RC governs tle mte of
decay.Figure 7.12 shows the plot of Eq. 7.25 and the $aphic interpretation of the time constant.
After determiningD(l), we can easilydedve the expressionsfor i,p,
and 1,:
44 = !!
=\-t1,
, - ()*,
vo
1,(t,= ve r'
a(t) :vo
17.26J
Flgure
7.12 ThenaiuraLrcsponse
ofanif circuit.
p-Di
= 2 R e - 2 ' , , t. > O + .
:?,a,,
17.27)
17.28)
238
Response
of Fi6t-0rder
Rrandfr Cjrcujts
Calculating the natural responseol an RC circuit car be summarized
as followsi
Catcutating
the naturalresponse
of an
fC circuit>
l . Find the initial voltage, y0, acrossthe capacitor.
2. Find the time constant of the circuit, r = RC.
3. Use Eq.7.25,?(t) : U& 't', to Eerleifre1)(1)from yo and r.
Al1 other calculations of interest follow from knowing t)0).
Examples7.3 and 7.4 illustratethe numericalcalculationsassociated
rvith
the naturalresponseofan RCcircuit.
Determining
the NaturalResponse
of an fC Circuit
The switchin the circuit shownin Fig.7.13hasbeen
in position r for a long time. At I : 0, the swltch
moves instantaneously to position y. Find
a) oc(t) for l > 0,
b) 0,(t) for r > 0*,
c) to(4 for r > 0*, and
d) the total energydissipatedin the 60 kO resistor.
b) The easiestway to find o,(r) is to note that the
resistive circuit forms a voltage divider across
the teminals of the capacitor. Thus
, r')
48
- b0. -_
z-'v.
s;rcfl)
r> u .
This expression for ,"(t) is valid for / > 0+
becausetro(0-)is zero.Thuswe havean instantaneouschang€in the voltageacrossthe 240kO
c) We find the current i,(t) ftom Ohm's law:
Figure7.13A Thecircujt
forExampte
7.3.
i"(t\ :
r,"(t) : e 2 5 r
mA,
60 x 103
t>0+.
Solution
a) Becausethe switch has been in positionr for a
long time, the 0.5/rF capacitor will charge to
100V and be positive at the upper terminal. We
can replace ttre resistive network conlected to
the capacitorat a : 0' with an equivalentresist,
ance of 80 kO. Hence the time constantof rhe
circuiris (0.5 x 10-1(80 x 1d) or 40 ms.Then,
"c(t) = 100e' 5'V, / > 0.
d) Thepowerdissipatedin the 60kf,l tesistoris
p60ko0): t3(r)(60x 103)- 60e-50,mw, r > 0-.
The total energydissipatedis
7.2 ThelaructRespolse
o'ar nr('(uit
239
Determining
the NaturalResponse
of an 8CCircuitwith SeriesCapacitors
The initial voltages on capacitors C1 and C2 io the
circuit shown in Fig. 7.14 have been established by
sourcesnot shown.The switch is closed at t : 0.
a) Find o(t, or(t), and o(t) for / > 0 and t(t) for
r>0-.
b) Calculare
rbe inirialenergysroredin lhe capacitors C1and Cr.
c) Determine how much energy is stored in tlle
capaqlonast+oo,
d) Show tlat the total energy delivered to the
250kO resistor is ttre difference between the
resultsobtainedir (b) and (c).
r(,
Cl(5pF) "(r)
+
1'(r_)250ko
+
24V
c2Q0 t/F) nzl)
Figure7.14A Thecjrcuit
forE\ampte
7.4.
Solution
l(4
a) Once we know o(t), we can obtain the cunent t(t)
from Ohm's law. After determining i(l), we can
calculateq(1) and ?rr(l) becausethe voltage aooss
a capacitor is a function of the capacitor cunent.
To find ?0), we roplace the series-connected
capacitoN with an equrvalent capacitor. It has a
capacitanceof 4 pF and is chargedto a voltage of
20 V Therefore, the circuit shoM ir Fig. 7.14
reduces to the one shown in Fig. 7.15, which
revealsthat the initial value of o(t is 20V and that
the time constantof the circuit is (4)(250) x 10 r,
or 1 s.Thus the exprcssionfor o(r) is
r(t) = 20e-'V,
250ko
Figure7.15 A A simphfication
of theckcuitshown
in Fjg.7.14.
b) Theinitial energystoredir C1is
1
t,j - :(5 \ 10n)(16) = a0 sJ.
The initial energy stored in C, is
t>0.
w2:
The current i(t) is
,/,1
i ( r l= : * : 8 o e i
I)U,WU
!A.
/>o+.
- (16e-'
i.'6
, r ( r )= - =
20) V,
4
t > 0,
tt
I 80 . to6e'dr + U
&'o : 40 + 5760 : 5800pJ.
rr+
20V
and u2+ +20V.
Ttere[orelhe energysloredin rbe two capaci
fols is
.\t.ta
= ( 4 e . + 2 0 ) V ,r > 0 .
x 10 )(s76) = 5760pJ.
The loralenergysroredin rherwocapacilors
is
Knowing i(t), we calculate the expressionsfor
0(t) and or0):
r0b /'
r,(t) = --| 80 . l0-6e "dx
5 ./o
;Q0
rr& =
1
,
+ 20) x 10-"(400): 5000pJ.
;(5
240
Response
of First-order
RLandRCCircujts
d) The totalenergydelivcredto the250kO resistoris
f- 4onu 2'
r l t- R t . ' r u J
lpd!
l^^^
/
./ r Z)lr.(rlru
Comparingthe results oblained in (b) and (c)
800riJ : (5800
Thc enelgy storedin the equivalcntcapacitorin
Fig.7.l5 is +(,1x 10 6)(400),or 800/..J.Because
this capacilorpredictsthe terminal bchavior of
the original series-connectedcapacilorsjthe
encrgyslored in the equivalentcapacitolis the
encrgydeliveredto the 250kO rcsislor.
5000)/1J.
objective1-Be abteto determin€
the naturalresponse
of bothnt andnCcircuits
7.3
The switchin the circuit shownhasbeen closed
for a long rime and is openedat r : 0. Find
a) the initial valueof?r(/),
b) the time constantlol r > 0c) the numericalexptessionfor ?r(l)after the
switchbasbeenopened.
d) thc inilial energystoredirl the capacitor,and
e) the lengthoI timc requiredto dissipate757o
of ihe initially storedenergy
7,4
The switchin the circuit shownhasbee! croseo
for a long time beforebejngopenedat r = 0.
a) Find 1r,,(l)for r > 0.
b) Whal percentageofthe initjal encrgystored
in thc circuit hasbeendissipatedafter lhe
svilch hasbeen open for 60 rns?
50kJ)
rsv )r,
Answen (a) 200Y
(b) 20 ns;
(c) 200€ 5o'V, r > 0j
(d) 8 mJ;
''uloo*,
Answer: (a) 8e '5' + 4e Iu v. L > 0;
(b) 81.05o/o.
(c) 13.86ms.
NOTE: ALto try ChapterPrcbleft 7.21und7.21.
7.3 TheStenll{esponse*f ffA
ai'dfff e!rcu'its
We arc now readyto discussthc problemof firding the currentsand vol!
agcsgeneratedin first-ordcrna or RC circuitswhen either dc vollageor
currenl soulcesare suddcnlyapplied.Theresponseof a circuit to the sudden applicationofaconstantvollageorcurrent sourccis referredto asthe
7.3
lhe SteDResDome
offt andRCCncuits 241
steprcsponseof ttre circuit.In presentingthe stepresponse,
we showhow
the circuit rcspotrdswhenenergyis beingstorediI) the inductoror capacitor. Webeginwith the stepresponseof an RL circuit.
TheStepResponse
of an nl Circuit
To begin, we modiry ttre firsrorder circuit shown in Fig. 7.2(a) by adding a
switch.We usethe resultingcircuit, shownin Fig.7.16,in developingthe
stepiesponseof an Ra circuit.Energy storcdin the inductor at the time
the switch is closed is given in terms of a nonzero initial current i(0). The
task is 10find t]le eeressions for the current in the c cuit and for the voltage acrossthe inductor after the s$ritch has been closed.The procedure is
the sameas that usedin Section7.1;we usecircuit analysisto derive the
differential equation that descdbes the cftcuit in terms of the variable of
intercst,and then we useelementarycalculusto solvethe equation.
AJter the switchin Fig.7.16has been closed,Kirchhoffs voltagelaw
rcquires that
v":Ri+L+,
11.2e)
which can be solved for the current by separating the variables t and r ard
then integrating.The first stepin this approachis to solveEq.7.29for the
deivative di/ dtl
Ri+u"
dt
, * )
L
(7.30)
Next, we multiply both sides of Eq. 7.30by a differential time dl. This step
reduc€s the leffhand side ot the equation to a differential change in the
ctrllent, Thus
.h*
a \
,
R/.
dr-
L\i-
- ; )v"\
0,,
v.\
R)/lt
We now separatethe variablesin Eq.7.31to get
di
-R,
t-tu/R)
L
"''
Figure
7.15A A circujt
used
to ittustrate
ihestep
tesponse
ofantst-order
ru cncuit.
\7.31)
242
Response
of Fiui-order
Rl andRrCircuiis
and then integate both sidesof Eq. 7.32.Using .r and y asvariables for th€
intesration. we obtain
[i(t)
dx
-R
f .
1 , " , - , v 1 4 =7 1 , ' t '
(7,33)
where 10 is the crment at I = O and j(t) is the curent at any t > 0.
Performing tlle integration called for in Eq.7.33 generatesthe exprcssion
.
n
i(t) - (4/R)
hlvr$
=
-R
Lr'
\7.34)
from which
i(t) (ulR)
I o - ( WR )
Stepresponse
of & circuit>
(7.35)
When the initial energy in the inductor is zero, 10 is zeio. Thus Eq. 7.35
i0 =f; \^"t*ro'
(7.36)
Equation 7.36 indicates that after tlle switch has been closed,the currcnt hcreases exponentially from zero to a final value of y,/R. The time
constalt of the circuit. Z/R. detemines the mte of hclease. One time
constant after the switch has been closed. the cunent will have reached
approximately 6370of its final value, or
u,':\-\^"'=o.o:zr*.
17.37)
If ihe currentwereto continueto increaseat its initial rate.it wouldreach
its final valueat t = 7: that is.because
di
clt
-
4( '\.,,,
u -,,.
R \ , , /
L
(7.38)
7.3
TheStepResponse
ofRt andffCCjrcuits
the initial rate at which i(1) inoeases is
_u
L'
firot
\7.39)
If the current were to continue to increaseat this rate, the erpression for i
would be
'=l''
(7.40)
from which, at t = 7,
. uL
u
I, R
R.
\1.41)
Equations7.36and ?.40are plottedin Fig.7.1?.The valuesgivenby
Eqs.7.37
and7.41arealsoshownin thisfigure.
anitrductoris ady'dt,sofromEq.?.35,forI > 0-,
Thevoltageacross
TX.E), r"r,r,
R)'
= (v" - ry!.\RtL,.
Q.4z)
The voltage across the inductor is zero before the switch is closed.
Equation 7.42 indicates fiat the inductor voltage jumps to I{ - 1oRat the
instant the switch is closed and then decaysexponentially to zeio.
Doesthe valueof o at t = 0* make sense?Becausethe initial current
is 10 and the inductor prevents an instantaneous change i[ curent, the Figure7.17.. Thest€presponse
0fthenLcircujt
current is 10 in the instant after the switch has been closed.The voltage shownin Fiq.7.16when10 = 0.
drop aooss the resistor is 1oR,and t]le voltage impressedacrossthe inductor is the souce voltage minus t]le voltage drop, that is, ys 10R.
When the initial inductorcurent is zero,Eq.7.42sjmplifiesto
r : Vse IRILY
17.13)
If the initial cuirent is zerq the voltage acrossthe inductor jumps to 14.We
also expect the inductor voltage to approachzero as1inoeases,becausethe
curent in the circuit is approachhg the constantvalue of VJ-R.Figure 7.18
showsthe plot of Eq.7.43 and the relationship between the time constant
and the initial rate at which the hductor voltage is deueasing.
If there is an initial current in the inductor, Eq. 7.35 gives the solution
for it. The algebraic sign of 10is positive if the initial current is in ttre same
0 ? 2 t 3 r 4 r
directionasi;otherwise,lo carriesa negativesign.Example7.5 illustrates
ve66 tjme.
the application of Eq.7.35 to a specfic circuit.
figure7.18A Inductorvottag€
244 Response
of Fjrst-OrderRL
andfr Circujts
Determining
the StepResponse
of an Rl Circuit
The switchin the c cuit shownin Fig.7.19hasbeen
in position a for a long time. At I = 0, the swltcn
movesfrom positiona to positionb.The switchis a
make-before-break
type; that is, the connectionat
position b is establishedbefore the conneclionat
position a is broken,so there is no interruption of
currentthroughthe inductor.
b) The voltageacrossthe induclor is
a) Find the expressionfor t0) for, > 0.
b) What is the initial volrageacrossthe irducrorjust
afterthe switchhasbeenmovedto positionb?
c) How many milliseconds after the switch has been
moveddoesthe inductorvoltageequat24V?
d) Does this initial voltagemake sensein terms or
circuitbehavior?
e) Plot both t(r) and ,l(r) versusr.
The initial inductorvoltageis
dt
7 ' = L ,
: 0.2(200e-10,)
:40e lo'V, t>0+.
2(0-.) = 40 V.
c) Yes; in the instant after the switch has beefl
movedto positionb, the inductor sustainsa current of 8 A counterclockwisearound the newly
formed closedpath. This current causesa 16 V
drop acrossthe 2 O resistor.This voltage drop
rddc ro lhe dfop acros.lhe.ource.producinga
40 V drop acrossthe inductor.
d) We find the time at which the inductor voltage
€quals24V by solvirg the expression
24 : 4De tol
tor t:
':
1 , 4 0
1o-t
= 51.08x 10 l
- 51.08ms.
figure 7.19 A ThecircuitforExamph
7.5.
Solution
a) The switchhasbeenin positiona for a long time,
so the 200 mH inductor is a short circuit across
the 8 A current source.Therefore,the irductoi
carriesan initial curent of 8 A. This currenr is
orientedoppositeto the referencedirectionfor i;
Jhus10is -8 A. w}len the switchis in posiuonD,
the final valueof i will be 24l2,or 12A.T\e time
constaDtof the circuit is 200/2, or 100 ms.
Substitutingthesevaluesinto Eq.7.35gives
r = 12 + (-8
=12
e) Figure7.20showsthe graphsoIt0) and r,(r) versus/. Note that the instantof time when the cur
rert equalszero correspondsto the instant of
time when the inductorvoltageequalsthesource
voltageof24V aspredicredby Kirchhoff,svolrage law
1J(v)t(A)
40
32
21
1(r
8
, (nt
12)€4'rr
2 0 et n A , t > 0 .
figure7.20.d Thecurrent
andvottaqe
wavetorns
for
7.3
TheSiepRespoNe
of{L andRfCircujts 245
of bothnl' andnCcircuits
obj€ctive2-Be abteto determine
ihe stepresponse
7-5
\O7 f:
Assumethal the switchin the circuit showDro
Fig.?.19hasbeenin positionb for a long timeand at r = 0 ir moves10positiona.Find
(a);(o*); (tr) z(o-); (c) r, t > 0: (d) i(r), r > 0;
and (e) ,l0), t > 0+.
Al,o rt Chaplrt Prohlrn'
Answer: (a) l2A:
(b) 200v:
(c) 20ms;
(d) -8 + 20?-' A, I >
(e) -200eso'V,I> 0 .
-..1.1-.13.
We can also describethc voltage?J(r)acrossthe inductor in Fig.7.16
directly,Dotjusl in lemrsol lhe circuit current.Webeginby noling thal thc
voltage acrossthe resistoris the ditferencebetweenthe sourcevollage
and the inductorvollage.Wc$.rite
11.44)
$ h c r cv i . a c o n \ t d n r . D i r t e f e n l i rbtoj nt h! r i d c \ u i r hc \ t e c lr o r i ' n e ) i e l d \
dt
Rdt
(7.45)
Theq iI we mulliply eachside of Eq. 7.15by the induclalce a, wc gct an
expressionfor thc voltageacrossthe inductor on the lefl hand sidc,or
(7.46)
PuttingEq.7.46into standardIoIm yiclds
d x R
(7.47)
You shouidvedly (in Problen17.40)that the solutionto Eq.7.47i! idcnti
cal io that givenin Eq.7.42.
At this poinl, a generalobservationabout the step responseoI an
Rl, circuil is pertincnt.(This observationwill prove helplul latcr.) When
we derivedlhe dificrcntial equatior for the inductorcurrenl,we obtained
Eq.7.29.Wenow rcwritc Eq-7.29as
(li
A
R.
t
L
V"
L
\1.48)
Obscrvc that Eqs.7.47 and 7.48 have Lhesame form. Specifically,each
equatcsthc sum ofthe firsi derivativeoI lhe variablcand a constanttimes
the variableto a constantvalue.In Eq.7.47,thcconstanton the right-haDd
sidehappensto be zero;hencethisequationtakcson the sameform asthe
246
Response
of First-order
Rl andfftCi(uits
natuml rcsponseequationsin Section7.1.Ir bolh Eq.7.47 a$d,Es.7.48.
rbe consla0lmultiplyinglhe dependent
!ariableis the rec;procal
of tbe
time constant, that is, R/a = 1/r. We encounter a similar situation iII the
de vationsfor the step responseof an RC circuit.In Section7.4.we will
use these observations to develop a general approach to finding the llatural and step responsesof Ra and RC circuits.
TheStepResponse
of an nCCircuit
We can find the step responseof a firsforder RC c cuit by analyziry the
circuit shownin Fig.7.21.For mathematicalconvenience,
we choosethe
Norton equivalent of the retwork comected to the equivalett capacitor.
SummingrhecuJrenFa\rayfiom tbe lop nodein Fig.r.:t generiresrhe
differcntial equation
Figure
7.21AA circuit
used
to iltustnterhe
step
response
ofa first-oder
Rrcjrcuit.
,!; ,3 =,"
(7.49)
Division of Eq.7.49by C gives
dt
-
RC=V
(7.50)
CompadngEq. 7.50vrith Eq.7.48rcvealsthat the form of the solutionfor
oc is the same as that for the curent in the inductive circuit. namelv.
Fq. 7.15.Tberelore.
b) simpl)substitlrling
rhe appropriare
variabtes
anl
coefficients, we can wdte the solutiotr for ?c directly. The translation
requircs that 1, replace 14,C replace a, 1/R reptace R, and y0 replace 10.
We get
Stepresponse
of anfC circuit>
(7.51)
A similar derivation for the curent in the capacitor yields the differential
equation
d
dt
i
l
RC'
^
''
\7.52)
Equation ?.52has the sameform as Eq. ?-47,hencethe solution for j is
obtained by usjng the same translations used for the solution of
Eq.7.50.Thus
i-(r.-\\t,'^",,-6.,
\7.53)
where y0 is the initial value of oc, the voltage acrossthe capacitor.
We obtainedEqs.7.51and 7.53by usinga mathematicalamlogy to
the solution for the step rcsponse of the inductive circuit. Let's see
whether these solutions for the RC circuit make sensein tems of known
circuit behavior.From Eq.7.51, note tlat the initial voltage affoss the
capacitoris yo,the final voltageacrossthe capacitoris IrR, and t]te time
ofil andRCCncuits 247
7.3 TheStepR€sponse
constantof the circuit is RC.Also note that the solutionfor oc is valid for
t > 0. fhese obse ationsare consistentwith the behaviorof a capacitor
in parallelwith a resistorwhendrivenby a constantcurrentsource.
Equation7.53predictsthat the curent in the capacitorat t - 0- is
1" - yo/R.Thispredictionmakessense
because
thecapacitor
voltagecannot changehstantaneously,
andthereforethe initial curent in the rcsistor
is yo/R,The capacitorbranchcurent changeshstantaneouslyftom zero
att = 0 tols - yo/Rat I : 0+.Thecapacitorcuirentis zeroat I : l)o.
Also rote that the final valueof 1, = 1"R.
Example?.6illustrates
howto useEqs.7.51and7.53to fhd the step
resDonse
of a first-orderRC circuit.
Determining
the StepResponse
of annCCircuit
The switch in t}Ie circuit shown in Fig.7.22 has been
in position 1 for a long time. At t = 0, t})e switch
moves to position 2. Find
a) 0.(t) for / > 0and
b) i"(r) for t > 0*.
The value of the Norton current sourceis the
ratio of the open-circuit voltage to the Th6venin
resistance,or 60/(40 x 10) : -1.5 mA. The
resullrng
Norronequivalenlcifcuit is sboqn in
Fig. 7.23. From Fig.'7.23, I,R = -60V and
RC:10ms. We have already noted tiat
o,(0) : 30 V, so the solutionfor z'. is
.t). =
=
Figule
7.22A lhecircuiitor
Exampte
7.6.
vo,
io :
160x 103
(-7s) = -60 v
(40+ 160)< 103
( 60)]rlm,
60 + 90e-1mrv, ,>0.
b) We write the solution for i, directly from
and
Eq. 7.53 by noting -that 1, = -15mA
:
(30/40)
x
0.7s
mA:
10-r,
or
U/R
Solution
a) The switch has been in position 1 for a lorg time,
so the initial value of ?, is 40(60/80),or 30 V To
take advantageof Eqs.7.51and 7.53,wefind the
Norton equivalent with respect to the terminals
of the capacitorfoi I > 0.To do this,we beginby
computing the open'circuit voltage, which is
given by the 75 V source divided across the
40kO and 160kO resistors:
60+ [30
2.25ermt mA, t > o+.
We check the consistencyof the solutions foi o,
and L by noting that
t,,
C?
(0.25
t11
=
l0 6)(-o000er'"',
2.25elau mA.
Becausedp.,(D-)/dt:0, the expressionfor i,
clearly is valid only for t > 0-.
Ner1, we calculate the fh6venin iesistance, as
seento the right of the capacitor, by shoning t}le
75V source ard making sedes ard pamllel
combinatioN of tlle resistors:
RTl' : 8000 + 40,000lL160,000= 40 kO
Figlre 7.23 a Theeqoivatent
circuiifort > 0 forthecircuit
s h o wjnn F j g . 7 . 2 2 .
248
Response
of Fnsi,order
Rl andffCircuih
objective2-8e abteto determine
the stepresponse
of bothnl andfC circuits
7.6
a) Find the expressionfor the voltagcacross
the 160kO resistorin the circuirshownm
Fig.7.22.Let this volragebe denoleuui, anu
assumethat the reference polarity for the
voltageis posjtivea1the upper termillalof
rhe 160kf) resistor.
b) Specifythe interval of time for which the
expressionobtainedin (a) is valid.
Answer: (a) -60 + 72e 1t)rtyl
( b )I > 0 ' .
NOTE: Also try ChaptetPtublems7.50und 7.51.
7.4 A SeneralSolutiog'l
for $tep
ftndNsturelResponses
arr
, - l
)
V* .J
/G\
G)
The generalapproachLo finding cither the natural responseor the step
responseof the first order Rl, and RC circuitssbownin Fig.7.24is based
on their differential equationshaving the sanrclorm (compareEq.7.48
and Eq.7.50).Togeneralizethe solutionofthesc lour possiblecircuits,we
let r(r) representthe unknown quanriry,givingr(,) four possiblevatues.Ir
canrepresentthe currentor voltageat the lcrminalsof an inductoror rhe
current or voltage at the terminalsof a capacitor.From Eqs.7.47.7.4tj.
7.50.aDd7.52.wc knolv that thc differentialcquariondescribingany one
of rhe four circuilsin Fig.7.24rakesthe fornr
i/p
L
\7.51)
dt-;
(b)
wherethe value ofthe constant1( canbe zero.Becauscthe sourcesin rhe
circuit are conslantvoltagcsaDd/orcurrcnls,the final value of .r will be
conslantlthal js,Lhefinal valuemust satisfyEq. 7.54,and.rvhenx reaches
its firralvaluc.lhe derivativedr/dl must be zero.Hencc
r, = Kr.
where.r/ represeDts
the final valueofthe variable.
We soivcEq.7.54by separatingthevariabtes,beginnil1gby solvingtor
thc first derivative:
frr
(x
..
Figure7.24a Fourpo$jbhfirt,ordercircuits.
connected
to a Th6venin
equivatent.
{a)Anjnductor
(b)Aninductorconnected
to a Nodonequivateni.
connected
to a Th6venjn
equivateni.
G)A capacjtor
(d)A capacitor
connected
to a No,tonequjvahnt.
Kr)
-1x
-r1)
(1.56)
In w ling Eq.7.56,we usedEq. 7.55to subsriruter, for Kr. We now mul
tiply both sidesofEq.7.56 by dr and divideby.r
.r/ to olrtain
o'
:1at.
17.51)
7.4
A GenentSotuiion
forSteDandNaturat
ResDon5es249
Next,we integrateEq. 7.57.To obtair asgenerala solutionaspossible,we
usetime lo asthe lorqerlimit and I asthe upperlimit. Time t0 corresponds
to the time of the switchingor other change.Previouslywe assumedthat
r0 : 0, but t}Iis changeallows the switchingto take place af any time.
Using& and o asslanbolsof integration,we get
f't') a"
t
J'1';u-t7
1' f
r du.
1J"
(7.58)
Canying out the integiation called for in Eq.7.58 gives
(7.59) < ceneralsolutionfor naturalandsteo
resDonses
of f[ andfC cirauits
The importanc€of Eq.7.59becomesappaientif we write it out in words:
the unknown
the final
variable as a - value of the
variable
function of time
tlre idtial
thefinaLl
value of the - vaiueof rhe x
I
variable
variablel
€
rlne$tulrd)
(7.60)
In many cases,the time of switching that is, t0 is zero,
W}len computing the step and natural rcsponses of circuits, it may
help to follow thesesteps:
1. Identify the va able of interest for t}le circuit.For RCcircuits,it is
most convenient to choose the capacitive voltage;for -Ra circuits,
it is best to chooset}le inductive cuffent
2. Detemine tlle initial value of the variable, which is its value at 10.
Note that if you choosecapacitive voltage or inductive current as
your variable of interest, it is not recessary to distinguish berween
t : l; and I : td.2This is becausethey both are continuous variables.If you choose another variable, you need to remember that
its initial value is defined at , = 16.
3. Calculatetle fhal value of the variable,which is its value asI - m.
4. Calculate the time constant for the circuit.
With these quantitieE you can use Eq. 7.60 to produce an equation
describing the variable of interest as a function of time. You can then find
equations lor other circuit variables using the circuit analysis techniques
introduced in Chapters 3 alld 4 or by rcpeating the preceding stepsfor the
other vadables.
Examples 7.7-7.9 illustrate how to use Eq. 7.60 to find the step
respoff e of an -RCor Rl, circuit.
'Tne expressions
r; andrf areanalogous
to (' and0'. ttus r(rt) is the linit of r(r) asr + ,0
lron the left, andrG is the linit of r(t) ast - rofron the rieh!.
< Calcutating
the naturalor steprcsponse
of Rl or nCcircuits
250
Response
of Fi6t order8r andfC (i(uits
Usingthe General
SolutionMethodto Findannf CircuifsStepResponse
The switchin the circuit showr in Fig.7.25hasbeen
in position a for a long time. At r = 0 the switch is
moved to position b.
a) What is the initial valueof 0c?
b) What is the final valueof 0c?
c) What is the time constantofthe circuit when the
switchis in positionb?
d) Wlat is the expressionfor z,c(r) when I > 0?
e) Wrat is the expressionfor i(r) when r > 0+?
t) How long after the switch is in position b does
the capacitorvoltageequalzero?
g) Plot "c(1) and i(r) venus r.
Solution
a) The switch has been in position a for a long time,
so the capacitor looks lite an open circuit.
Thercfore the voltage acioss the capacitor is the
voltage acmsstle 60 (} resistor.From tie voltage,
divider rule, the voltageacrossthe 60 O resistor
is a0 x [60/(60 + 20)], or 30 v As rhe refer
ence for oc is positiveat the upper lerminal of
the capacitor,wehavez,c(o): 30 V.
b) After the switchhasbeenin positionb for a long
time,the capacitorwill look like an open circuit
in terms of the 90 V source.Thus the final value
ofthe capacitoivoltageis + 90V.
c) The time constantis
T:RC
= (400x 103x0.5
x 10{)
= 0.2s.
d) Substitutingthe appropriatevaluesfor z/, t'(0),
and r irto Eq.7.60yields
0c(t)=90+( 30- 90)e{'
= 90 120ts'V, I > 0.
e) Here the value for r doesn't chaflge.Thus we
need to find only the initial and final values for
the curent in the capacitor.When obtainingthe
initial value, we must get t}le value of t(0+),
becausethe currett in the capacitorcan change
instantaneously.
This current is equal to the current in the iesistor, which from Ohm,s law is
[90 ( 30)]/(400 x 103)= 300/,A. Nore that
whenapplyingOhm'slawwe recognizedrharthe
Figure7.25A Thecircuit
forExample
7.7.
capacitorvoltagecannotchangeinstantaneously.
The fiMl value of i(t) - 0, so
i0)-0+(300-0)e-5.
= 300e-srpA, r>0+.
We could have obtahed this solution by diffbrentiating tlle solution in (d) and mulriplying by
the capacitance.You may want to do so for yourseli Note that this altemative approachto finding
r(r)alsopfedrclc
lhedisconlinu] dl I = 0.
f) To find how long the switch musr be in position b
before the capacitorvoltage becomeszero, we
solve the equation derived in (d) for rhe rime
when 0c(t) : 0:
120t5r = 90 or
-
,.
720
90'
, =1r,-/\43\ /
= 57.54ms.
Note that when ,c = 0, i = 225pA and the
voltagedropacross
the400kO resistoris 90V
g) Figure7.26showsthe graphsof oco) and i0)
, (i.A) t)cry)
300
250
200
150
100
50
Figure7.26a Thedrr€ntandvottage
waveforms
tor
Exanrpte
7.7.
7.4 A CeneratSotutjon
torStepandNaturalResponses
251
Usingthe General
SolutionMethodwith ZeroInitiaLConditions
The switch in t]le circuit shown in Fig.7.27 has been
open for a long time. The initial charge on the
capacitor is zero. At 1 = 0, the swirch is closed.Find
the expressior for
Substitutingtlese valuesirto Eq.7.60genemtes
(, =o+(3 -o\ett5tn:
a) i(r) for t > 0- and
: 3 e 2 o o ' m Ai,> 0 + .
b) ,0) when t > 0-.
30ko
b) To find ?,(1),we note ftom the circuit that it
equalsthe sum of the voltage acrossthe capacitor and the voltageacrossthe 30 kO resistor.To
find the capacilor volrage (which is a drop in
the direcrion of the current), we note that its
inirial value is zero and its final value is
(7.5)(20),or 150VThe time constantis the same
as before, or 5 ms. Therefore we use Eq. 7.60
to wnte
Figure7.27A Thecircujttor
ExampLe
7.8.
z'c(l)- 150+ (0
SoLution
150)e'mr
= (150- 150e,oo,)v,
a) Becausethe initial voltage on the capacitoris
zero,at the instantwhen the switchis closedthe
currentin the 30 kO branchwill be
(7.5X20)
(0) =
I > 0.
Hencethe expression
for thevoltagez'(,)is
z(t) = 153- lsOeioo'+ (30x3)eioo'
50
: 050
6oa'rir) v,
I > 0*.
= 3 mA.
The final value of the capacitor current will be
zero because the capacitor eventually will
appear as an open circuit in terms of dc cwent.
Thus if : 0. The time constant of the circuit will
equal the product of the Th6venin resistance(as
seen ftom the capacitor) and the capacitance.
Thercforer = (20 + 30)10'(0.1)x 10 " : 5ms.
As one checkon this expression,
note that rt
predictsthe hitial valueof the voltageacross
the 20O resistoras 150 60, or 90 V The
instant t}re switch is closed,the curent in the
20ko resisroris (7.5)(30/50),
or 4.5 nA. This
curent producesa 90V drop acrossthe 20kl)
resistor,confirming the value predictedby the
solutiofl,
252
andRCCircuits
Response
0f First-0rderft
SolutionMethodto Findan RLCircuifsStepResponse
Usingthe General
The switchin the circuit shownin Fig.7.28hasbeen
open for a long time.At I : 0 the switchis closed.
Find the expressionfor
constantis 80/1,or 80 ms.we useEq. 7.60to
foi 0(t):
wite theexpression
oO=0+(15
o)e'l3orrr'
a) "0) whenI > 0- and
: 15t12stV,
I > 0+.
b) i(t) when r > 0.
b) We have alreadynoted that tlle initial value of
the inductor cuirent is 5 A. After the switchhas
been closedfor a loDgtime, the inductor current
reaches20/1,or 20A. The circuittime constantis
80 mq so the expressionlor i(t) is
i(t) : 20 + (s - 20)e r25t
= (20 - 15?-125,)
A,
r > 0.
7.9.
Figure7.28 A ThecjrcujtforExampLe
We determine t}lal the solutions for t'(l) and i(l)
agree by noting that
Solution
a) The switch has been open for a long time, so the
initial current i]l th.3 inductor is 5 A, onented
lrom top to bottom. Immediately after the swrtch
closes,the current still is 5 A, and tierefore the
initial voltage across the inductor bemmes
20 5(1),or 15V The fhal valueofthe inductor
voltageis 0 V WitI the switch closed,the time
NOTE:
1' 5i]
:80 x 10 3[15(12.5)e
: 15e l'r'v,
r > ot.
Assessyour ande$anding of the seneral solution method b) trrng Chapter Problems 7 53 a d 7-54-
Example7.10showsthat Eq.7.60 can even be used lo find th€ step
responseof somecircuitscontainingmagneticallycoupledcoils
7.4
A GenentSotution
forStepandNaturaL
Responses253
Determining
StepResponse
of a Circuitwith Magnetical.ty
Coupted
Coils
There is no energystoredin the circuit in Fig.7.29
at the time the switchis closed.
a) Find the solutions for i., u., ir and i2.
b) Show that the solutions obtained in (a) make
sensein termsoI known circuit behavior,
'.3 H
120V
Figure7.29a Thecircuit
forExampte
7.10.
Solution
a) For the circuit in Fig.7.29,the magneticallycoupled coils can be replacedby a singleinductor
having an inductance of
LrL2 - Ml
Lr+ L2- 2M
(SeeProblem6.41.)It follows thal the circuit in
Fig.7.29canbe simplifiedasshownin Fig.7.30.
By hypothesisOleinitial valueof i, is zero.
From Fig. 7.30we seethat the final value of i.
will be 120/7.5or 16A. The time constantof the
circuit is 1.5/7.5or 0.2 s.It follows directly from
Eq.7.60that
io=16
1 6 e - s t A ,t > 0 .
fhe voltage r,, follows from Kirchloff's
voltage law.Thus,
D.: 120
7 5i"
- 12(le-5tV.t>O+.
To find i1 and i2 we first note from
Fig.7.29that
^air
.diz
. dir
dt
at
dt
di
i :
.1t
IT
j j
"-diz
/11
d;
n,tt-n
Therefore
Figure7.30,d ThecjrcuitinFjg.7.29wjththe magneiicaLty
coupLed
coitsreptaced
byanequivatent
cojt.
Because i2(0) is zero we have
a=
.ltj
-
because
ale " dx
8+8€s'A,
r>0.
Using Kircbioffs current law we get
iI:24
2 4 es 'A ,
t > 0.
b) First we observethat i,(0), ir(0), and ir(0) are all
zero, which is consistent with the statement that
no energyis stored in the circuit at the instant
the switch is closed.
Next we obsene r',(0-) = 120V, which is
consistentwith the fact that i,(0) = 0.
Now we observethe solutions for ir and
4 are consistentwith the solution for o, by
observing
dt
dt
'
.lt
follows from Fig. ?.29 tlat
+ iz,
1.5H
t1l
At
= 36oe st - 24oe sl
: 720e-sty, t>O*,
di.
u-:6:+15::
di"
: '720e 5t - 600e 5l
: 120e-5tv,
t > 0'
254
Response
of Fnst-order
Rl andnr Cncuitr
The final values of 4 and i2 can be checked
using flux linlages. The flux linking the 3 H coil
(11) must be equal to the flux linldng the 15 H
coil (12),because
The tinal valueoflr is
n(co)
andthetioalvalueofi is
', : dn'
ir(oo) =
The consistencybetween these final values
for i1 and i2 and the final value of the flux Linkage can be seenfrom the expressions:
3i1 + 6i2 Wb-turns
^r(co) = 3i,(@) + 6i(,.o)
= 3(A) + 6( 8) : z wb{urm,
6i1 + 15i, Wb-tums.
Regardless of which expression we use, we
obtain
= 6Qa) + E(
,\1 = i2 : 24 - 24ts' Wb-tums.
Note the solution for It or 12 is consistent with
the solution for rro.
The final value of the flux linking either
coil 1 or coil 2 is 24 Wb-tums, that iE
I(co) = r,(oo) : 24 wb-tums.
^,(o.) : 6i1(oo)+ 15ir(,tro)
8) : 24 wb-tums.
It is worth noling that the final values of 11
ard i2 can only be checked via flu\ linlage
becauseat t : oo the two coils are ideal short
circuits. The division of curent between ideal
short circuits cannot be found from Ohm's law.
NOTE: Assessyour understanding of this material bt using the genersl solution method to sotve Chaptet
froDlems /.ol and /.o /.
7.5 Sequentiat
Switching
Whenever switching occuN more that once in a circuit, we have sequential
switching.For example,a single,two-position switch may be switched back
and fo ]\ or multiple switchesmay be opened or closed in sequerce.The
time reference for all switchings cannot be I : 0. We detemine tlle voltages and cu ents generated by a switching sequenceby using tle techniques described prcviously in this chapter. We derive the er:pressionsfor
o(t) and i(, for a given position of the switch or switches and rhen use
these solutions to determine the initial conditions for the next position of
the switah or switches.
With sequential switching problems,a premium is placed on obtaining
the inilial value r0o). Recall that an)'thing but inductive curents and
capacitive voltages can change instantaneously at the time of switching.
Thus solving fi$t for itductive currents and capacitive voltages is even
more pertinent in seque ial switching problems. Drawing the circuit that
pefiains to each time irterval in such a problem is often helpful in the
Effmples 7.11 and 7.12 illustrate the analysis techniques for circuits
with sequential svritching.The fircl is a natual responseproblem with two
switching times,and the secondis a step responseproblem.
SequentjatSwjtchjng
255
7.5
Analyzing
an ff Circuitthat hasSequential.
Switching
The two switches in the circuit shown in Fig. 7.31
havebeenclosedfor a long time.At r - 0, switch1
is opened.Then,35ms later,switch2 is opened.
a) Find ir(l) for 0 < , < 35 ms.
b) Find iz for t > 35 ms.
c) Wlat percentage of the initial energy stored in
the 150 mH inductor is dissipatedin the 18 O
r€sistor?
d) Repeat(c) for the 3 O resistor.
e) Repeat(c) lor the 6 O resistor.
F i g u r € 7 . 3 2 l^t e L i ' L i . s h o w l i n - i g . 7 . l l . r o '0r .'
150mH
,:35ms
40
\
)60v
-1i-
18()
Figure
7.33A Thecncujtshown
in Fig.7.31,
for0 < r =35 ms.
3r}
2
: 1 , 2 A 1 6ll
tt. 150inH
18()
b) When t:35ms,
the value of the inductor
ir = 6e rL = 1.48A
Figure7.31 A Thecircujtfor Exampte
7.11.
Thus, when switch 2 is opened, the circuit
rcducesto the one shown in Fig. 7.34,and the
time constant changesto (150/9) X 10-3, or
16.67ms The expressionfor i. becomes
Solution
oo3s)
iL : 1.48e4411
A,
a) For t < 0 both switches are closed, causing the
150nH induclor ro shorr-flrcurlthe l8 O resistoi. The equivalent circuit is shown in Eg. 7.32.We
determine the initial current in the inductor by
solving for ir(o-) ir the circuit showr in Fig.7.32.
AJter making several soutc€ transformations,we
find 4_(0-) to be 6 A. For 0 < t < 35 ms, switch 1
is open (switch 2 is closed),which disconnectsthe
60 V voltage souice and the 4 O and 12 O resis'
torc from the circuit. The inductor is no longer
behaving as a short circuit (becausethe dc source
is no longer in the circuit), so t}Ie 18 O rcsistor is
no lorger shortcircuited. The equivalent circuit is
shown in Fig.7.33.Note tlat t}le equivalent rcsistance across the teminals of the inductor is the
parallel combhation of 9 O and 18 O, or 6 O.
The time constantof the circuit is (150/6) x 10 3,
or 25 ms,Theieforc the expressionfor ir is
r > 35 ms.
Note that the exponentialfunction is shiftedin
time by 35 ms.
150mH
l, (0.035)= 1.48A
tigure7.34A Thecjrcuitshown
in Fig.7.31,
forr > 35ms.
c) The 18 O resistor is in the circuit only during the
first 35 ms of the switching sequence.During tlfs
intelval, tle voltage auoss the rcsistor is
)
u, = o.-ts:(6Pro')
tll
iL = 6e Nt A,
0 < , < 35 ms.
=-36eNtY,
0<r<35ms.
256
Response
of Fi'st-o'der8land8r CkcLriis
fhe powerdissipated
in the 18I) r€sistoris
'D
.'?
-2-j2?
3 0\ \ .
Iti
o
Hence the energydissipatedin the 3 O resistorfoi
I > 35msis
r. ismr.
[:,
Henc€ the energy dissipated is
fi,
r0,035
'72e!o'dt
I
.lo
w:
""
3(36)e2 3e 12o\t o o35)dt
lo.or:
,o'
i4o/, 3,tt
N
--120(r-0035)
1^
rzu
l00rs
: 1 0 8 P' 3
lo
: 0.9(1 e{)
t0a __
- 845.27nL
Theinitial energystoredh the 150mH inductoris
The total energydissipatedin the 3 O resistoris
: 563.51+ 54.73
?{)3o(otal)
' r', : -){ 0 . 1 5 ) { 3=6 )2 . 7J : 2 7 0 0 m J .
: 618.24
mJ.
Theretore (845.2'7
/2'700)x 100, or 31.31ol"of
the initial eneryystoredin the 150mH inductor
is dissipated
in the 18o resistor.
d) For 0 < t < 35ms, the voltageacrossthe 3 O
The percentageof the initial energystoredis
6 tJl )/
==
tA t
leslstof1s
I uL\.^.
,r!!=l;l(r)
\ r I
1
=
u3tt -
e) Becausethe 6 O resistoris in serieswith the 3 O
resistor,the energydissipatedand the percentageof the initial energystored$rillbe twice that
ofthe3Orcsistor
u6o(rotal) = 1236.48mJ,
12e"Y.
Thefefore
lheenergy
dissipated
in lheI O resis
tor in the fiNt 35msis
*,
/o.n"t++"
-----dr _
|
r
,n\
:0.6(1
? ' 3)
- 563.51
mJ.
lou : 22.90",".
and the p€rcentageof the initial energystoredis
45.80%.Wecheckthesecalculationsby observing that
1236.48+ 618.24 + a45.27 : 2699.99ml
and
31.31+ 22.90+ 45.80:100.010/..
For t > 35ms,the cunent iII the 3 O resistoris
1a)e{o('oo3s)
A.
ha= iL= (6e
The small disqepanci€s in the summations are
the result of ioundoff errors,
7.5 Sequ€ntiat
Switching 257
Anal.yzing
anRf Circuitthat hasSequentiat
Switching
The uncharyed capacitor in the circuit showt in
Fig. 7.35 is iritialy switched to terminal a of the
throe-positionswitch.At t = 0, the switchis moved
to position b, where it rcmains lor 15 ms.After the
15 ms delay,the switch is moved to position c, where
it remains indefinitely.
a) Derive tle numerical expression for the voltage
acrossthe capacitor,
b) Plot the capacilor voltage versus time.
100ko b
Figure7.35a lhecircujt
fortxampte
7.12.
c) When will the voltage on the capacitor equal
200v?
Solution
a) At the instant the switch is moved 10 position b,
the initial voltage on the capacitor is zero. If tlle
switch were to remain in position b, the capacitor
would eventually charge to 400 V The time constant of the circuit when the switch is in position b
is 10 m$ Therefore we can use Eq. 7.59 with
t0 : 0 to write the expression for the capacitor
vorEge:
u:400
+ (0
: (400
In wiiting the expression for o, we recognized
that lo : 15 ms and tlat this expression is valid
only for I > 15 ms.
b) Figure 7.36 showsthe plot of o versus r.
c) The plot in Fig. 7.36 rcveals that the capacitor
voltage will equal 200 V at two different times:
once in the inte al between 0 and 15 ms and
once after 15 ms.We find the fiISl time by solving
200=400-400tlmr',
400)ermr
400erm) v,
0 < , < 15 ms.
Note thal. becauserhe swirch remainsin position b for only 15 ms,this expressionis valid or y
for t]le time interval from 0 to 15 ms.Aftcr ure
switch has beer in this position for 15 ms, the
voltage on the capacitor will be
?r(15mt = 400 - 400e-rs: 310.75V.
Therefore,when the switch is moved to position c,
the initial voltage on the capacitor is 310.75V
Wit}l the switch in position c, the final value of
the capacitor volaageis zero, and the time constantis 5 m$Again, we useEq.7.59to write the
er?ression for the capacitor voltage:
which yields 11: 6.93ms. We find the second
time by solving the expression
-oors)
200 : 31O;75etno(t
In this case,t2: 17.20ms.
r,(v)
300
200
/:310.75.
ao(/ od:J
100
?, = 0 + (310.75- o)€ 20r(!ools)
- 310.75?-200(r-0015)
V,
15ms < r.
Flgure
7.35^ Thecapacitor
votiage
fortxampte
7.12.
258
Response
of Fifst-Order
nl andRCCjrcujis
objective3-Know howto anatyze
circuitswith sequential
switching
7.7
In the circuit sho$'n.switchI hasbeenclosed
r n d . $ i r c h) h ,\ . . c n o p c l o r : rl o ' r gl i r n eA
.l
t - 0. s$,itch1 ls opcncd.Ther l0 ms later,
switch2 is closcd.Find
S$ir(hJ iI rheci-cuir.h^qn hl. bccnop<ntul
a loDgtine, and slvirchb hasbccn closedfor a
long time.Slvitcha is closedat I = 0 and,after
remainingclosedfor 1 s.is openedagait.
Switchb is opcncdsimultaneously.
and both
switchesrenain open indcfiriitely.Delermine
lh. e\pfe..ion
f o r l h er r r d u c r (oLr r r ( n l/ r h , l i .
validwhen(a) 0 = I < I sand(tr)r > 1s.
7,8
a) r.(r) for 0 < r < 0.01s.
b) ,tJ.(/)for r = 0.01s.
c) dre total cncrgydissipatedin the 25 kQ
d) rhe lolal energydissipatedin thc 100ko
20
AFFrI
b
- 1
0.8o
9ll
l f ok ( l
\
i3o
r:(l
2H
ii 3o
60
Answer: (a) 80.rrrrlv;
oorrVl
(b) 53.63e5o1r
Answer: (a) (3 3c t'5')A,0 < , < 1s;
(b) (-4.8 + 5.98r r'25(/r) A. r > I s.
(c) 2.e1nJ;
(d) 0.2enrJ.
NOTE: Aha iry ChapterProbl ts7.72and7.76.
7.6 i,inbcu*ti{:d
RrsFs*se
A circuil responsemay grox'.rather lhan decay.exponentiallywith timc.
This ttpe of responsc.
calledan unbomded response.is possiblcif rhe cir
cuit containsdcpcndcn! soulces.Ir that case,the Th6vcnin cquivalenl
resistance
with rcspccl10the lerminalsofeither an inductoror acapacilor
nav be negaiivc.This negadveresistancegeneratcsa ncgativctine con
stant.and thc rcsullingcurerts and voltagesincreascwithoul limi1.Tn aD
actual circuit. thc rcsponseeventuallyreachesa limiting valuc whcn a
componentbreaksdo\\n or. goesinto a saturationstate,prohibiling rur
ther increasesin voltagcor cult'enl.
wlren we considcrunboundedresponses,
the conceptof a tinal laluc
is confusirg.Hcncc,ralher lhan usingthe step responsosolutiongi!en in
Eq.7.59.wedcrivc lhe differentialequationthat describcsthc circuil con,
taining thc negalile resistanceand then solvc it using the separariorof
variablcs lcchnique.Example 7.13 prcscnts an exponenlialll,growing
responscin termsoI lhe vollageacrossa capacitor-
Response 259
Unbounded
7.6
Response
in an Rccircuit
Findingthe Unbounded
a) When the switch is closed in the circuit shown m
Fig.7.37.lhe \ollage on the capaciloris l0 V
Find the expressionfor ?),for t > 0.
b) Assume that the capacitor short-cilcuits when
its terminal voltage reaches 150 V How many
millisecondselapsebefore the capacitorshorG
circuits?
metfod
r e d t of r d R n
F i g u r €7 . 3 8^ T f er e s r - l o i ( e
-5 ko
7.13.
Figure7.37 l Th€crrcujtfor Exampt€
jn
ofthecjrcujt
shown
Fig!rc7.39l A simptjfication
fig.7.31.
Sotution
a) To find the Th6venin equivalent resistance with
respectto the capacitor teminals, we use the tes!
souce method describedin Chapter 4-Figure 7 38
shows the {esulting circuit, where ?,r is the test
voltage ald ir is the test current. For th expressed
in volts,we obtail
.
rr:
UT
10
- -/l )rL2 0 t +
1)T
,
For , > 0, the ditferential equation describing
the circuit shownin Fig.7.39is
_3
15 1s_oyy_:g
111
,a
r0_r: o.
Dividing by the coefficient of the first derivative
yields
2-mA.
4
Sol\ingfof tbe rario ,r"/ir yieldsthe fie\enin
Rrr, :
5
= -5 kO
d
t
,'r,u^=o.
"
We now usethe separationof variablestechnique
to find 0"(r):
q(t):10e4utY,
with this Th6veniil- resistance,we car simplify
lo lheone\hownin
thecircuirsho\rnin Fig.7.37
Fig.7.39.
t > 0.
b) o, : 150V when?au'- 15.Therefore,4ot= ln 15,
and/ = 67.70ms.
NOTE: AssessyoLr understandingof thk material by trying Chapter Probkms 7 86 and 7.87.
The fact that hterconnected circuit elements may lead to everhcreasing currents and voltages is impofant to engineels. If such inter:
connections are unhtended, the resulting ckcuit may experienc€
unexpected,and potentially dangerous,component failures'
260
Response
of Fi6t-0rder
fi andRrCircuiis
7.7 TheIntegratingAmplifier
tigurc7.40e Animeg,aLing
anplifier.
Recall from the introduction to Chapter 5 tlat one reason for our interest
in t})e operational ampMier is its use as an integrating amplifier. We are
now ready to analyze an integrating-ampiifier circuit, which is shown in
Fig. 7.40. The purpose of such a circuit is to generate an output voltage
proportionalto the integal of the input voltage.InFig.7.40,we addedthe
branch curents it and i,, along with the node voltages o, and o?, to aid
our analysis.
We assumetlat the operational amplifier is ideal- Thus we take
advantageof the constraints
\7.61)
\7.62)
Becauseu? : 0.
(7.63)
dD^
(7.64)
Hence,from Eqs.7.61,7.63, and'7.64,
du.
n:
1
R{r,,"
\7.65)
Multipling both sidesof Eq.7.65 by a differentialtime dl and thetrintegrating from lo to I generatesthe equation
1
R.
t"'
,,r d) + ?,o(10).
(7.66)
In Eq. 7.66,t0representstlle instantin time whenii'ebeginthe inte$ation.
Thus ?r,(to)is the value of the output voltage at that time. Also, because
bn: tp: 0, t"00) is identical to the idtial volrage on tlre feedback
caDacitorC,,
Equation 7.66 states that the ouQut voltage of an inregrating amplifier equalsthe initial value of the voltage on the capacitor plus an inverted
(minus sign), scaled (1/&Cl) replica of tle inregral of the input voltage. ff
no energyis storedin the capacitorwhenintegation commences,
Eq.7.66
ol0 -
1 l
nq J,"o'ot
\7.67)
7.7
TheIntegBiingAmptifier 261
If 0Jis a step changeifl a dc voltage level, the output voltage will vary linearly with time. For example, assumetiat the input voltage is the rectangular voltagepulseshownin Fig.7.41.Assume
alsothat the initial valueof
,"(1) is zero at the hstant or steps from 0 to y,. A direct application of
Eo.7.66vields
."=
l-,v^t
+ 0, 0<r<r,.
(7.6s)
tigure7.41,i, Aninputvothgesjgnal.
W})en t Liesbetween 11ar'd2t1,
-#,'*'
#1"'"-'*
u^
fi"cr
'
2uR,Cr'"
t1<t<2tr
(7.6e)
Figuie 7.42 shows a sketch of o,(t) versus r. Clearly, the output voltage is
an inverted, scaledreplica oI the integal of the input voltageThe output voltageis propo ional to the integralof the input voltage
only ifthe op amp operateswithin its linear range,that iE i{ it doesn'tsa!
urate.Examples7.14 and 7.15 fuitler i ustratethe analysisof the inte- Figure7.42A Theoutputvottage
of anjniegcting
gratn€ amplfier.
Analyzing
an IntegratingAnplifier
Assume that the numerical values for the signal
voltage shown in Fig. 7.41 are y-:50mV
and
11: 1s. This signal voltage is applied to the
integrating-amplifier circuit sho\\.n in Fig. 7.40.The
circuit parametersof the amplifier are R" = 100kO,
Cf = 0.1 pF, and Vcc : 6 V. The initial voltage on
the capacitor is zerc.
Forl<t<2s,
o, = (5r - 10)V.
b) Figure7.43sho*sa plot ofo.(t) ve^usL
a) CalculatezJ,(t).
b) Plot r',0) venus ,.
Solution
a)For0<r=1s,
(1oox 10)(0.1x 1oi)
=
5l V,
0<t<1s.
50x103r+0
Figure7.43 A Theoutputvottage
for Exampte
7.14.
262
Response
of Fnst'0rder
il andRCCircuits
Analyzing
an IntegratingAmptifierthat hasSequentialSwitching
At the instant th€ switch makes cofltact with terminal a in th€ circuitshownin Fi9.7.44,the voltageon
the 0.1/rF capacitoris 5 V The switch remainsat
terminala for 9 ms and then movesinstantaneously
to terminal b. How many millisecondsafter making
contactwith terminalb doesthe opemtionalampli
fier saturate?
Thus,9 msaftertheswitchmakescontactwith termiDala,theoutputvoltageis 5 + 9. or 4V
Theexpression
for theoutputvoltageafterthe
switchmovesto terminalb is
D' ^ : 4
.
+ 5V
-4
0.1pF
I
t, ^.
nd\J
I
t0 . ../,r
^ o.
-
800(1 9x10-3)
: (11.2- 800' V.
Duringthistimeinterval,thevoltageis deoeas
ing,and the operational
amplifiereventuallysatu
mtesat 6 V. Thereforewe setthe expression
for oo
equalro 6 \ lo obldinrheraruradon
rimer_
tigure 7.44 A Thecjrcuitfor Examph
7.15.
11.2
Solution
800r, : -6,
The expression foi the output voltage during
time the switchis at terminala is
,": -s- ft;l'etota:,
= (-s + 1000tv.
t' = 215 ms
Thus the integrating amplifier saturates21.5 ms
after makirg contact with terminal b.
From the examples,we seethat t}le integrating amplifier can perform
the integration function very well, but oniy within specified limits that
avoid salurating the op amp. The op amp saturates due to the accumulation of chargeon the feedbackcapacitor.We canpreventit from satuat
ing by placing a rcsistor in parallel with the feedback capacitor.We
examine such a circuit in Chapter 8.
Note that we can convert the integrating amplifier to a differentiating
amplifierby interchangingthe input resistanceR" and the feedbackcapacitor Cr.Then
tln"
u. = -R'Ct
clt
\7.70)
We leavethe derivationof Eq.7.70asan exercisefor you.The differentiating amplifieris seldomusedbecausein practiceit is a sourceofunwanted
or noisysignals,
Finally, we can desigr botl integrating- and differentiating-amplifier
circuitsby usingan inductor insteadof a capacitor.However,fabricating
capaciton for inlegmted-circuitdevicesis much easier,so inductorsare
rarely used in integrating amplifiers.
Pa.trct Pespe.tive 263
obiective4-Be abteto analyze
op ampcircuitscontaining
resistors
anda singtecapacitor
7.9
Thereis no energystoredin rhe capacitorat
the lime the switchin the circuit nakes contact
wilh termitul a.The switchrcnrainsaLposition
a lor 32 ms and then movcsjnstantaneously
to
positionb. How nany millisecondsafter making contactwith terminal a doesthe op amp
0-2r.F
7.10 a) Whcn the switchslosesin the circuir
shown.there is no energystoredin thc
capacitor.How long doesit rakc to saturale
lhe op amp?
b) Repeat(a) with an initialvollage on the
capacitorof l Ypositivc at the upper
lermlnal.
t0 ko
.10ko
0.8to
Answer:262ms-
Answer:(a) 1.l1ms;
(b) 1.76ns.
NOTE: Aho rry Cha?@ Ptublems7.92and 7.93.
PracticaI
Perspective
A Ftashing
LightCircuit
Wearenowreadyto anatyze
the ftashjnglight cjrcuitintroduced
at the stat
of thjs chapt€rand shownin Fig.7.45. Thetampjn this circuitstats to
conductwhenever
the lampvoltagereaches
a valuey-i,. Durjngthe time
the larnpconducts,
it canbe modeledas a resistorwhoseresistance
is X,.
TheLampwiL[contjnueto corductuntiIthe Lanrp
voLtaqe
dropsto the value
y, i. Whenthe [ampis not conducting,
it behaves
asan opencjrcuit.
Beforewe devetop
the analyticaL
expressions
that descrjbe
the behavior Figurc7.45 A fla!hingtight. ,cuit.
of the circuit,let us developa feelfor howthe circuitworksby notingthe
fotlowing.Fi6t, whenthe lampb€haves
as an opencjrcuit,the dc voLtage
sourcewiL[chargethe capacitor
viathe resistor,l?towarda valueof14voLts.
However,
oncethe tampvottagereachesy,,a,,it stats condLrctjng
andthe
capacitorwittstartto discharge
towardthe Th6veninvottageseenfromth€
t€rrnjnats
of the capacitor.But oncethe capacjtorvottagereaches
the cutoff vottageof the lanp (y",i,,),th€ Lampwjlt act as an opencircujtandthe
capacitorwit[ stat to recharg€.
Thiscycleof chargjrganddischarqing
the
capacitor
is summarized
in the sketchshownin Fig.7.46.
In drawingFig.7.46we havechoserr = 0 at th€ instantthe capacitor
startsto charge.Thetjrner, represents
the instantthe Lampstartsto con- Figlre7.46; LampvoLtage
versus
tine for the
duct,andr. is the endof a comptetecycte.WeshouldaLsomentionthat jn c i f c u iitn f i g . 7 . 4 5 .
Respome
of Fjrsi-Orderfl
andRf Circuih
constructing
Fig.7.46we haveassumed
the circuithasreachedthe repeti
tive stageof its operation.0ur desjgnof the ftashingtjght circuitrequires
we deveLop
the equationfor r'r(t) asa functjonof V-*, y.,i., %, R, C, and
R, for the intervats0 to t, andt, to t".
Tobeginthe anaLysis,
we assume
that the circuithasbeenin operation
for a long time. Let t : 0 at the jnstantwhenthe Lampstopsconducting.
Thus,at / = 0, the tampjs modeled
asan opencircuit,andthe voLtage
drop
acrossthe tampjs yni,,,as shownin Fig.7.47.
Frofirthe circuit,wefind
Dr(c') = %'
r,.(0) : ynh,
Egure7.47A Theftashing
tjghtcircuit
att = 0,
when
ihetamp
is notconducting.
r _ RC,
Thus,whenthe Lamp
is not conductrng,
"r(t) : 14 + (Y-r"
he-iac.
Howlongdoesit takebeforethe Lampis readyto conduct?
Wecanfind this
tjme by settingthe expression
for or(r) equaL
to y."" andsoLvjng
for t. If
we callthis vatuet., then
^^-
''
RL
'-
U-it
\
t -""
{'
Whenthe [ampbeginsconductjng,
it canbe modeLed
asa resistance
Rz,
as seenin Fig.7.48. In orderto find the expression
for the vottagedrop
acrossthe capacitorjn this circuit,we needto find the Th€venin
equivatent
as seenby the capacitor.
Weleaveto you to show in Problem7.106,that
whenthe Lampis conducting,
Figure7.48A Ihe ftashjng
tjghtcjrcuiiai, = ro,
whenthelampis conductjng.
DL(.t)= vrh+ (y.*
yrh)e ('-,t"
and
'
RRLC
n+R/'
Wecandetermine
howlongth€ lampconductsby settjngthe aboveexpres'o'
s on lor r/ (/) to l-_. andso-ving
tr. r,J,giving
RR1C . %'^ - v,.
'n y.n
yrh
R Rr
your understanding
NqTE: Assess
ofthis PradicdLPerspediveby ttying Chdpter
Prcblems
7.103-/,105.
265
5ummary
A iirst'order circuit may be reducedto a Th6venin(or
Norton) equivale.rtconrectedto either a singleequivalent inductoror capacitor.(Seepage230.)
The natursl r€sponse is the currents and voltages that
exist when stored energy is releasedto a circuit that
containsno independentsources.(Seepage228.)
The time conslant of an Xa circuit equals the equivalert inductance divided by the Thevenin resistalc€ as
viewed from t1leterminalsof the equivalentinductor.
(Seepage232.)
The tim€ constentof an RC circuit equalsthe equivalent capacitancetimes the Thdvenin resistance as
viewedfrom the terminalsof the equivalentcapacitor.
(Seepase237.)
The step r€spons€ is the cuuents and voltages that
resultfiom abrupt changesin dc sourcesconnectedto a
circuit.Storedenergymay or may not be presentat the
time the abrupt changestake place.(Seepage240.)
The solutionfor either the natural or step responseof
boti R-Land RC circuitsinvolvesfinding the initial and
final value of the curent or voltage of interest and the
time constantof the circuit. Equations 7.59 and 7.60
summarizetlis approach.(Seepage249.)
Sequential switching in first-order circuits is analyzed
by dividing the analysisinto time intervalscorrespon
ding to specificswitchpositions.Initial valuesfor a particular interval are determined ftom the solution
correspondingto the immediatelyprecedjnginte al.
(Seepage25,1.)
An unbound€d r€sponseoccurs when the Th6venir
resistanceis negative, which is possible when the
first order circuit contains dependent sources.(See
page258.)
An integratingamplifier consistsof an ideal op amp,a
capacitorin the negativefeedbackbianch,and a resis,
tor in serieswith the signalsource.It outputs the inte
gral of the signal source,within specifiedlimits that
avoidsaturatingthe op anp. (Seepage260.)
Prob[erns
Sectiotr7,1
7.1 In the circuit shownin Fig. P7-1,the switchmakes
contactwith positionb just beforebreakingcontact
with positiona.As alreadymentioned,thisis known
as a make'belbrebreak switch and is dcsignedso
that the switchdoesnot interrupt the cment in an
inductive circuit. The inte al of lime between
"making" and "breaking"is assumedto be negligible.Theswilchhasbeenin the a positionfor a long
time.A1 t = 0 the switchis thrown from positiona
to posilioDb.
a) Determincthe initial currert in the inductor.
b) Determine thl9time constantof the circuit for
r>0.
c) Find i. 1)r,and 02for 1 > 0.
d) What percentag€of the initial energystoredin
the irductor is dissipatedir the 20 O resistor
12 ms after the switchis tbrown fiom positiona
to positionb?
figurcP7.1
5o
7.2 The switchin the circuit ir Fig.P7.2hasbeenclosed
EDGfor a long lime before openingat 1 = 0.
a.) Find t1(0 ) and i,(r).
b) Find 4(0+) and,,(0+).
c) Find tio) for r > 0.
d) Find tr(1) for r > 0*.
e) Explain why t (0 ) * tr(o-).
figureP7.2
266
RlandRfCircuits
Response
of Fnst-oder
The switch shown in Fig. P7.3 has been open a long
time before clositrgat, = 0.
a) Find i,(0 ).
Figure
P7.5
b) Findtr(o-).
10A
c) Find i"(0*).
d) Find i,(0-).
e) Find t,(c.).
f) Find tr(,ro).
g) Wdte tlle expressionfor ir(,) for t > 0.
h) Find z'r(0 ).
i) Find r'r(0*).
j) Fird ?,.(co).
k) Wite tlle expressionfor !'r(t) for t > 0'.
l) Wdte the expressionfor i"(l) for I > 0*.
rent source, o represent tlre fracrion of initial
energy srored ir the inductor that is dissipated in to
seconds,ard a represent the inductance.
a) Show that
L1LI1|0 - o)l
2t.
b) Test the expression derived in (a) by using it to
find the value of R in Problem 7.5.
Figure
P7.3
50(l
7.6 In the €ircuit in Fig. P7.5,Iet 1s represent tlle dc cur-
100()
2000
7,4 ln the circuit in Fig. P7.4, the voltage and current
expressionsare
o = 100t30!v,
r >0+;
i = 4e4'A,,
t>0.
Find
a) R.
b) r (in miltiseconds).
c) L.
d) tle idtial energy stoied in the inductor.
e) the time (in milliseconds)it takes to dissipate
80% of the initial stored energy.
Figure
P7.4
7,7 The switch ir ttre circuit in Fig. P7.7 has been open
6fl( for a long time. At , = 0 t]le switch is closed.
a) Determinei,(0") and i,(m).
b) Determinei,(t) for t > 0*.
c) How many millisecondsafter the switch has been
closedwill t]le curent in the switch equal 3.8A?
figureP7.7
12o"
160
80v
'(r=0
+\
20nH
40
|
8()
7,8 The switch in the circuit in Fig. P7.8hasbeen closeda
6dc long time At I : 0 it is opened.Find "o(r) for l > 0.
FigueP7.8
15O
\
)80v
50O -r
i s oo
lf}
0.2H o.a:60r)
20o"
2A
The switch in tlle circuit seen in Fig. P7.5 has been
in posirion 1 for a long time. At r : 0, the svritch
moves instantaneously to position 2. Find the value
of R so tlar 50ol. of the initial energy stored in the
20 In}I inductoris dissipatedin R in 10lrs.
7.9 Assume that the switch in the circuit in Fie. P7.8has
been open for one time constant. At this instant,
what percentage of t}le total energy stored in the
0.2 H inductor has been dissipated in ttre 20 O
Prcbtems
?67
7.10 In the circuit shown in Fig. P7.10,the switch has
6fl( beenin positiona for a longtime.AtI :0,itmoves
instantaneously
from a to b.
a) Find r"(t) for I > 0.
b) W}lat is the total energydeliveredto the 1 kO
rcsistor?
c) How manytime constantsdoesit taketo deliver
95%of the energyfound in (b)?
7.14 The switch in the circuit in Fig. P7.14 has been
6tr4 closed for a long time before openirg at t = 0. Find
?o(t) for r > 0'.
tigur€P7,14
50nH
tigureP7.10
7.11 The switch in the circuit in Frg. P7.11 has been in
6nc position 1 for a lory time. At t = 0, the switch moves
instantareously to position 2. Find ?o(t) for t > 0*.
figlre P7.11
7A
1
7.15 The switchin Fig.P7.15hasbeenclosedfor a long
time beforeopeningat t : 0. Find
a)iL(t),t>o.
b)rL$),t>o-.
c)iA(r),r>0-.
Figure
P7.15
3r)
50
96mH
')s.rv
2
15()
5(}
50 ia
4.5f,)
T<{2--l*!l
i,. 100o.
:9o
2oomH:
:'
20()
l;
200():
7,16 Wlat percentage of the initial energy stored in tlle
inductor in the circuit in Fig. P7.15is dissipated by
t}Ie cunetrt-contiolled voltage soulce?
7.12 For the circuit of Fig. P7.11,what percenlage of the
initial efergy stored in the inductor is eventually
dissipated in t]te 20 O resistor?
.
7.13 In the circuit in Fig. P7.13, the switch has been
closed for a long time before opening at I = 0.
a) Find the value of a so that r'"(t) equals0.25 ?,,(0-)
whenr:5ms.
b) Find the percentage of the stored energy tlat
has been dissipatedin the 50O re stor when
t - 5ms.
7.17 The two switchesin tle circuit seenin Fig. P7.17are
synchronized.The switches have been closed for a
long time before opening at t = 0.
a) How many microsecondsafter the switches are
open is the eneigy dissipatedin the 60 kO resistor 257oof the initial energy stored in tho 200mH
hduclor?
b) At the time calculated in (a), what percentageof
the total energy stored itr the inductor has been
dissipated?
Fig!reP7.17
Figure
P7.r3
200nH
)',."
40ko
20ko
:60ko
30ko
268
R€sponse
of First'order
Rl andRrCircuits
7.18 The 220 V 1 O source in the circuit in Fig. P7.18is
6nc inadvertendy short-circuited at its terminals a,b.At
the time the fault occurq the circuit has beer in
operation for a long time.
a) Wlat is the initial value of the current iabin the
short-circuit connectionbetween terminals a,b?
b) Wlat is the final value of the current iab?
c) How many microseconds after the short circuit
has occurred is the current in the short equal
ro 210A?
S€cfion 7.2
721 The svritch in the circuit in Fig. P7.21 has been in
position a for a long time and 1)2= 0V. At t : 0,
the switchis thmwn to positionb. Calculate
a) i,?r, and ?r,for, > 0*.
b) the energy storcd in the capacitor aft = 0.
c) rhe energy happed in the circuit and the total
energy dissipated h the 5 ko resistor if the
switch remains in position b indefinitely.
Figure
P7.21
FigueP7.18
4.'7kA
1()
a
b
5kO
+tj.
12{r
60(])
2nH
15nrH
220V
7.22 In the circuit in Fig. P7.22 the voltage and current
exPressionsare
?r:100e-1m0lv, r > 0;
b
7.19 The two switches shom in the circuit in Fig. P7.19
6tG operate simultaneously,Prior to t : 0 each switch
has been in its indicated position for a long time. A1
I : 0 the two switches move instantaneously to
their new positions. Find
a) 0"0), t > 0-.
b) ,o(r),I> 0.
FigueP7.r9
j - 5e-1000!
mA,
I > 0+-
Find
a) R.
b) c.
c) ' (in minisecondt.
d) the initial €nergystoredin the capacitor.
e) how many microsecondsit takes to dissipate
80o/, of the initial eneryy stored in the capacitor.
Figj'Je ?7.22
80 mH "o
i^
7.23 The switch in the circuit in Fig- P7.23 is closed at
ED.r I = 0 aft€r beingopen for a long time.
7.m For the circuit seenin Fig. P7.19,find
a) the total energydissipatedin the2.5 kO resistor.
b) the energy trapped in t]le ideal inductors
a) Find tr(r) and i(0 ).
u) nna 4(o+)ana;z(o+).
c) Explahwhyil(o) = t1(0*).
Probtems
269
d) Explain why i(0 ) + i(0+).
e) Findil(t) for t > 0.
l) End t (t) for t > 0*.
tigure
P7.23
726 In tllle circuit sho*n in Fig. W.26, both switches
operate together; tlat is, they either open or close at
the same time. The switches are closed a Iong time
before opening a1I - 0.
a) How many microjoules of energy have been
dissipatedin the 12 kO resistor 2 ms after the
switchesopeD?
b) How long does it take to dissipate95% of rhe
initialy stored erergy?
Figur€
P7,26
7.24 The switch in the circuit in Fig. P7.24 has been in
position a for a long time. At r : 0, the swifch is
thown to positionb.
a) Find t,(t for t > 0+.
b) What percentage of the initial energy stored in
the capacitoris dissipatedin the 4 kO resistor
250 /,csafter the switch has been tfuown?
7.25 Both switches in the ckcuil in Fig. P7.25have been
sflc closed for a long time.At t = 0, both switchesopen
simultaneously.
a) Find roc) lor t > 0+.
b) Find ,"(1) for , > 0.
c) Calculate tle energy (in microjoules) trapped in
the circuit.
7.27 The switch in the circuit in Fig. P7.27has beenin
En@posirion 1 for a long time before moving to position 2 a - 0. Findl,(t) foi t > 0t.
figtr'e P7.21
7.28 The switch in the circuit seen in Fig. P7.28has been
in position x for a long time. At r : 0, the switch
moves instantaneously to posirion y.
a) Find d so that the time constantfor I > 0is1ms.
b) For the a found in (a), find t,a.
Flgure
P7.28
20ko
Figure
P7.25
10ko
270
Response
of Fnst-0rder
Rtandrr Cinuits
7.29 a) In Problem 7.28, how many microjoules of
energy !!re generated by the dependent curent
source during the time the capacitor discharyes
to 0V?
b) Show that for t > 0 the total energy stored and
genemted in the capacitive circuit equals t]Ie
total energy dissipated.
730 After the circuit in Fig. P7.30has beer in operation
tsd.r for a long time! a screwdriver is inadvertently connected aqoss the terminals a,b. Assume the resis!
ance of ttle screwdriver is negligible.
a) Find the current in the screwdriver at I : 0+ and
b) Derive the expression foi the curent ir the
screwdriver for t > 0+.
7.32 At the time the switch is closed in the circuit in
Fig. P7.32,the voltage acrosstlle paralleled capacitors is 30 V and the voltage on the 200 nF capacitor
is 10V
a) Wlat percentage of the initial energy stored in
the three capacitors is dissipated in the 25 kO
resistor?
b) Repeat(a) for the 625O and 15 kO resistois.
c) What porcentage of the initial energy is tapped
in the capacitors?
tigureP7,32
l0 nF
tigureP7.30
7.33 The switch in the circuit seen in Fig. P7.33has beer
arft closed for a long time. The switch opens at 1 = 0.
Fhd the numerical expressions tor i,(t) and oo(t)
when t > 0*.
Figure
P7.33
7,31 At the time the switch is closed in the circuit shown
in Fig. P7.31,the capacitors are charged as shown.
a) Find ?J,(t)for t > 0'.
60ko
,J4
b) Wlat percentage of the total ene4y initially
stored ir the three capacitors is dissipated in the
25kO rcsistor?
0 Find 2,1(tfor t > 0.
d ) Find ?r20)for t > 0.
e) End the energy (in milijoules) trapped itr the
idealcapacitois.
Figure
P7,31
+
+
0.6pF "l
" 25kO
Section 7.3
7,34 The switch ir the circuit sho$n in Fig. P7.34 has
Eflft been closed foi a long time before opening at I : 0.
a) Find tlle numerical expressions for i.(t) and
,"(l)fort > 0.
b) Find tlre numedcal values of 0r(0+) and o,(0+).
FlgureP7.34
Probtems
271
7.35 The switch in tle ckcuit shown in Fig. P7.35 has
Btrc been in position a for a long time. At r:0, the
switch moves instantaneously to position b.
a) Find the numerical e4ression for io(t) when
Figur€
P7.38
b) Find the numerical expression for o,(r) for
l>0*.
FiqueP7,35
7.36 After the switch in the circuit of Fig. P7.36has been
open for a long time, it is closed at I : 0. Calculate
(a) the initial value of i; (b) the final value of i;
(c) the time constant for t > 0; and (d) the nunedcal expressionfor i(t) when I > 0.
Flgur€
P7.36
5kO
4kO
75mH
7.39 The switch in the circuit iII Fig. P7.39 has been
closed for a long time. A student abruptly opens the
switch and repo s to her instructor that when the
switch opened, an electdc arc with noticeable per,
sistence was established across the switch, and at
the same time the voltmeter placed aqoss the coil
was damaged. On the basis of your aralysis of the
circuit in koblem 7.38,can you explainto the student why this happened?
Pigure
P7.39
20kO
737 The currert and voltage at the teminals of tlle
inductorin the circuit in Fig.7.16are
(D : (10- 10r5m')
A, r>0;
u(t\ : 20oesm'v,
l>0*.
a) Specify the numerical values of %, R, 1,, and I.
b) How many miliseconds after the switch has
beencloseddoesthe energystoredin tlle inducior reach 25% of its final value?
?.38 The svritch in the circuit shown in Fig. P7.38 has
been closed for a long time. The switch opens at
l:0.Foit>0*i
a) Find I'o0) as a function of 1s, Rr, R2,and a.
b) Explain what happers to ?,(t) as R2 gets larger
and larger.
c) Find osw as a function of 1s, Rt, R2,ard L.
d) Explain what happens to osw as R2 gets larger
and larger.
7.4O a) Derive Eq.'7.4'7byf st conveting the Th6venin
equivalent in Fig. 7.16 to a Norton equivalent
and then summing the curents away from the
upper [ode, using the inductor voltage o as the
variable of interest.
b) Use the separation of variables technique to find
the solution to Eq. 7.47.Verily that your solution
agees with the solution given in Eq- 7.42.
7.41. The swilch in tle circuit in Fig. P7.41has been open
6ntr a long time before closing ar r = 0. Find ,"G) for
t>0.
Figure
P7.41
87.2mH
20()
10O
09?]i
40O
272
Rtandlr Circujts
Response
of Fitst-od€r
7.42 The swilch in the circuit in Fig. P7.42has been open
6xa a long time before closhg at I = 0. Filld o,(1) for
r>0-.
P7.42
Figure
100
),,o
150
5 0 " i2mH
:
l'*(
,n ot^(
7.43 There is no energy stored in the inductors a1 and az
at the time the switch is opened in the circuit shown
in Fig. P7.43.
a) Derive t]le exp.essionsfor the curents i1(t) and
i20)forl > 0.
b) Use the expressionsderived in (a) to find i1(x')
and i2(oo).
7,46 The make-before-break switch in the circuit of
6r,t! Fig.P7.46hasbeen in positiol a for a long time At
t = 0, tle switch moves instantaneously to position b. Find
a)?,,(,),r>0*.
b)t10),t>0.
c)4(t),t>0.
Figure
?7.46
15o r,+
?.47 Theswitchir thecircuitin Fig P747hasbeenopen
6ire a long time before closingat t = 0. Find o,(t) for
FiglreP7.43
u
iz,.l)l
figureP7.47
t=0
40 mH
5rI
10A
7.44 The switch in the circuit in Fig. P7.44 has been in
6xd position 1 for a long time. At I = 0 it moves instantaleously to position 2. How many milliseconds
after the switch opentes does ?t"equal 80 V?
60 nH
ligve P7.44
7.48 The switch iII the circuit in Fig. P7.48 has been in
tsdG position i for a long time. The initial charge on the
15 nF capacitor is zero At t : 0, the switch moves
iNfantaneouslyto positiony.
a) Find 0.0) for t > 0".
b) Fhd 1)(, for t > 0.
tigur€P7.48
?,45 For the circuit in Fig. P7.44,find (in joules):
a) the total energy dissipated in [he 80 O resisfor;
b) the energy trapped in the inductors;
c) tle idtial energy stored in tlle inductors.
7.49 For the circuit in Fig. P7.48,find (ia microjoules)
a) the energydeliveredto the 200kO resistor;
b) the energytmpped in the capacito$;
c) the initial energy stored in the capacitors.
7.50 The switch in tle circuit shown in Fig. P7.50 has
antr been closed a long time before opening at r = 0.
For, > 0+,find
a) ro\t).
b) i,(r).
c) 4(r).
d) i2O.
e),r(0-).
figueP7,50
19
4ko
Figure
P7.53
1g.ko"
120V
4;
6125tO
,:0
150ko
40nF
7.54 The switch in the circuit of Fig. P7.54 has been in
position a for a Iong time. At t - 0 the switch is
movedto positionb. Calculate(a) the initial voltage
on tlre capacito! (b) the fhal voltage on the capacitor; (c) tlre time constant (in micmseconds) for
t > 0; and (d) tle length of time (in microseconds)
required for the capacitor voltage to reach zero
after the switch is moved to position b.
Figure
P7.54
20ko
7.51 Thecircuit in Fig.P7.51hasbeenin operationfor a
6trc Iong time.At t = 0, the voltagesouce dropsfrom
100V to 25V andthe currentsourcereversesd ection. Find r,,(r) for , > 0.
Figrre
P7.51
5ko
15kO
75V
10k()
2ko
100v
7.52 The curent and voltage at le terminals of the
capacitor in the ckcuit in Fig. 7.21 are
i(r) : 50e-'sm'mA'
v.
,{r} - (80 - 80P25m')
t > 0t;
r > r.
a) Specify t}le numerical values of 1", U, R, C,
and r,
b) How many midoseconds after the switch has
been closed does the energy stored in the capacitor reach64% of its final value?
7.53 Assume that the switch in the circuit of Fig. P7.53
has been in position a for a Iong time and that at
1 = 0 it is moved to position b. Find (a) oc(o+);
(b) ?,c(oo);(c) r for I > 0; (d) (0*); (e) oc, r > 0;
and (f) ,, t > 0+.
755 The switch in the circuit shownin Fig. P7.55has
6rc beencloseda long time beforeopeningat I : 0.
a) Whatis the initial valueof i,(l)?
b) Whatis thefinalvalueof i,(tx
c) Whatis tlle time constantof the circuitfor t > 0?
d) What is the numericalexpressionfor i.(t) when
r>0-?
e) What is the numericalexpressionforu,,(t) when
figureP7.55
6.8kO
75V
274
Response
of First-ord€r
nt andfr Circujts
7.56 The switch in the circuit seenin Eg. P7.56hasbeen in
Edc position a for a long time. At t : 0, the swrtchmoves
instantaneouslyto position h For t > 0*, find
b),,(r).
c) r.lt).
7.59 The switch in the circuit shown in Fig. P7.59 has
EdG been in the oFF position for a long time. At 1 = 0,
the svdtch moves instantaneously to the oN position. Find o,(l) fo. t > 0.
Figure
P7.59
30 x ldia
d) ,g(0-).
7,60 Assume that the switch in the circuit of Fig. P7.59
6na has been in ttre oN position for a long time before
switching instantaneously to the oFF position at
t = 0. Find ?r,(t) for 1 > 0.
7.57 The switch in the circuit seenin Fig. P7.5?has been
6tra in position a lor a long time. At t = 0, the switch
moves instantaneously to position b. Find oo(r) and
i,(t)forr>0'.
Figure
P7.57
.i,r,r-!
,,,,"
7.61 a) Derive Eq.7.52 by tust converting the Norton
equivalent circuil shown in Eg. 7.21to a Th6venin
equivalent and then summingthe voltagesaround
the closedlooA usilg the capacitorcurrent i asthe
relevant vadable,
b) Use the separalion of variables technique to find
the solution to Eq. 7.52.Verify that your solution
agees with that of Eq.7.53.
7.62 There is no energy stored in tlle capacitors C1 and
C2at the time the switch is closed in the circuit seen
in Fig.P7.62.
a) Dedve the expressions for o(t and zJr(t) for
t>0.
b) Use t})e expressionsderived in (a) to find o(oo)
and 02(').).
Figure
P7.62
,r(t)
758 The switch in the circuit shown in Fig. P7.58opensat
mPi.r , : 0 alter behg closed for a long time. How many
milliseconds after the switch opens is the energy
stored in the capacitor 90% of its final value?
rig'lle P7.58
?,n,
7.63 The switch in the circuit of Fig. P7.63 has been in
sP,m position a for a long time. At t : 0, it moves instantaneously to position b. For t > 0*, find
a) polt).
b)
c)
d)
e)
i,(r).
?,(0.
,10.
theenergyfappedin thecapacitonast + oo
Figure
P7.63
7.67 There is no energy stored in the circuit in Fig. P7.67
EPj.Eat the time the switch is closed.
a) Find i,0) for t > 0.
b) Ftnd u"0) for r > 0*.
c) Fird ir(,) for r > 0.
d) Find i(D for I > 0.
e) Do your answen make sensein terms of knowlr
circuit behavior?
100v
I
Figure
P7.57
-+i"
7.64 The switch in the circuit in Fig. P7.64 has been in
6Ee position a for a Iong time. At I = 0, the switch
moves instantaneously to position b. At the instant
the switch makes contact $rith terminal b, switch 2
opens.Find o,(r) for t > 0.
t+
.
10v
7.68 There is no energy stored in the circuit in Fig. P7.68
Edc at the time the switch is closed.
a) Find (i) for I > 0.
b) Fhd r,(t for t > 0*.
c) Find r,r(r) for 1 > 0.
d) Do your answe$ make sensein terms of known
circuit behavior?
Figure
P7.68
Section 7,4
7.65 Therc is no energy stored in t}le circuit of Fig. P7.65
at the time the switch is closed.
a) Fird io(t) for I > 0.
b) Find 0o(t)for t > 0*.
c) Find t1(t for I > 0.
d) Find irQ) for t > 0.
e) Do your answersmake sensein terms of known
circuit behavior?
FiWleP7.55
,/10riH\.
200v
8H
d'
7.69 Repeat Problem 7.68 if the dot on the 8 H coil is at
6nc the top of the coil.
Section 7.5
7.70 In the circdt in Fig. P7.70,switch A has been open
E ic and switch B has been closed for a long time. At
t = 0, switch A closes One second after switch A
closes,switchB opens.Find tr(t) for t > 0.
Figure
P7.70
7,66 Repeat (a) and (b) in Example 7.10if the mutual
inductance is reduced to zero,
l0 nH
276
Response
ofFirst-0rderfl
andRCCncujts
7.71 There is no energy stored in the capacitor in the cir
6trr! cuit ir Fig. P7.71when switch 1 closesat r :0.
Three microsecondslater, switch 2 closes.Find od(t)
fort > 0.
ligurc P7.74
tigureP7,71
r:0 + 50/rs
+
7.72 The action of the two switchesin the circdt seenin
*rI( Fig. It.72 is asfollows. For t < 0, switch 1 is in position a and switch 2 is open.This state has existed for
a long time. At t = 0, switch 1 moves instantaneously frcm position a to position b, while switch 2
remains open. Tivo hundred fifty microseconds
after switcb I operales.switch 2 closes.remain(
closed for 400 ps, and then opens.Find o,(t) 1 ms
after switch 1 moves to position b.
7.75 The switch in the circuit shom in Fig. P7.75has
EPicbeenin positiona for a long time.At I :0, the
switchis movedto positionb, whereit remainsfor
100ps.The switchis then movedto positionc,
whereit remainsindefinitely.Find
a) (0).
b) t(25ps).
c) i(200ps).
d) 0(100 i.is).
e) r(100+ps).
FigureP7.75
IrgurcP7.72
500v
7.73 For the circuit in Fig.P7.72,how manymilliseconds
after switch 1 movesto position b is the energy
sroredin the inductor4% of its initial value?
+
J
96()
480f)
20nH
276 The suritch in the cftcuit in Fig. P7.76 has been in
Edc position a for a long time. At I = 0, it moves instantaneouslyto positionb, whereit remainsfor 250ms
before moving instantaneouslyto position c. Find
I,otbrr > 0.
tigureP7.76
7.74 The capacitor in the circuir seen in Fig. P7.74 has
Efld been charged to 494.6mV Aft = 0, switch 1 closes,
causingthe capacitor to discharye hto the resistive
network. Switch 2 closes50 &s after switch 1 closes.
Find t])e magnitude and direction of the current in
the second switch 100 ps after switch 1 closes.
120()
4{}
50-A
7.77 In the ckcuit itr Fig. P7.77,switch t has been in posisPI.r tion a and switch 2 has been closed for a long time.
At t - 0, switch 1 moves instantareously to position b. Four hundred microseconds later, switcb 2
opens,rcmahs open for 1ms, and then recloses.
Find ,, 1.6 ms after switch 1 makes contact with
teminal b.
0+40Ops
35kO
Figure
P7.80
I J sko
7.81 The voltage waveform shown in Fig. P7.81(a) is
Erla applied to the circuit of Fig. P7.81(b). The initial
voltage on the capacitor is zero.
a) Calculate0o(.).
b) Make a sk€tchof r',(t) ve$us t
Figure
P7.81
7.78 For the circuit in Fig. P7.77,what percentage of the
initial energy stored in the 50 nF capacitor is dissipatedin the 10kO resistor?
7.79 The voltage wavefonn shown in Fig. P7.79(a) is
am applied to the circuil of Fig. P7.79(b). The initial
curent in the inductor is zero.
a) Calculate0"(l).
b) Make a sketch of ?,,(t) venus L
c) Find io at I : 4 &s.
FlgueP7.79
?,"(v)
20
.---.1
1t)
6
(a)
t(nd
(b)
7.82 The voltage signalsourceh dre cncuit in Fig.P7.82(a)
is geneiatingthe signalsholvl in Fig.P7.82(b).Thereis
no stored energyat t = 0.
a) Derive tlre expressionsfor o,(t) that apply in the
i n r e r l a lrs. 0 : 0 = t < l 0 m s ; 1 0 m s < r <
20ms;and20ms < t <oq
b) Sketch oo and ?r"on the samecoordinate axes.
c) Repeal(a) and (b) with R reducedto 10 kO.
20ko
tigureP7.82
R=50kO
G)
(b)
7.80 The cunent sourcein the circuit in Fig. P7.80(a)
6r,c generales
the currentpulseshownh Fig.P7.80(b).
Thereis no energystoredat t : 0.
a) Derive the numedcalexpressionsfor o,(t) for
the time inteivalst < 0, 0 < 1< 50ps, and
5 0 p s< , < o o .
b) Calculale
,, (50 Fs)aod,, (50 p./
c) Calculatei" (50- /rs) andi. (50' l..s).
278
ResDonse
of First-0rder
RlandRCCncuiis
7.83 The cunent souce in the circuit h Fig. P7.83(a)
6r,( generatesthe current pulse shown in Fig. P7.83(b).
There is no energy stored at I - 0.
a) Derive the erpressions for i,(r) and oo(t for the
time inte als t<0;
0<t<0.5rns; and
0 . 5 m s< 1 < o o . b) Calculate l,(0 );
?5x ldra
25V
i"(0+); io(0.0005-); and
,,(0.000s).
c) Calculate o,(0 ); o,(0+); 1),(0.0005
); and
o,(0.0005*).
d) Sketch t"0) veNus I for rhe inrerval
-2ns <, < 2ms.
e) Sketch Oo(t) veisus r for tle inteival
-2ms < t < 2ms,
Flgure
P7.83
k (nd)
2ko
tigureP7.85
7,86 The gap in the cfcuit seenin Fig. P7.86will arc over
Btr( whenever the voltage aqoss the gap reaches36 kV
The initial curent in the inductor is zero. The value
of P is adjusted so tle Th6venir resistance with
respect to tle terminals ol tle inductor is -3 kO.
a) WIat is the value of B?
b) How many microseconds after the switch has
been closed will the gap arc over?
tigureP7.86
20
0.1rrF
0.sr(mt
(a)
(b)
Section 7.6
7.84 The inductor curent in the circuit in Fig. P7.84 is
AnG 2; mA at the instant the suritch is opened. The
inductor will malfunctior whenever the magnitude
of the inductor current equals or exceeds 12 A.
How long after the switch is opened does the inductor malfunction?
7,87 The svritch in the circuit in Fig. P7.87 has been
tsflft closed for a long time. The maximum voltage rating
of the 16 nF capacitoi is 930 V How long after the
switch is opened does the voltage acrosstle capacitor reach the maximum voltage mtirg?
Figure
P7.87
4ko
^ . ^ r _ ,_ .
Figurc
P7.84
2ko
tjo
rxn n,r
7.85 The capacitor in the cLcuit shol*il in Fig. P7.85 is
Eqc charged to 25 V at tle time tle switch is closed. If
the capacitor ruptures when its terminal voltage
equalsor exceeds50 kV how long does it take to
rupture the capacitor?
7,88 The circuit sho$,n in Fig. P7.88 is used to close the
switch between a and b for a predetermined Iength
of time. The electdc relay holds its contact arms
down as long as the voltage aooss the relay coil
exceeds5 V When tle coil voltage equals 5 V the
relay contacts return to their initial position by a
mechanical spring action. The switch between a and
b is initially closed by momentarily pressing the
push button. Assume that the capacitor is fully
charged when the push button is first pushed dov.n.
,robtems279
The rcsistance of t]te relay coil is 25 kO, and the
inductance of the coil is negligible.
a) How long wi[ the switch between a and b
remaitrclosed?
b) Write the numerical expression for i ftom the
time tlle relay contacts first open to the time the
capacitor is completely charged.
c) How many milliseconds (after the circuit
between a and b is interrupted) does it take t])e
capacitor to reach 85o/oof its final value?
7.91 Therc is no energy stoied in the capacitofi in the
Aflc circuit shown in Fig. P7.91 at the instant the two
switchesclose.
a) Find x', as a funclion of ?r,,,b, R, and C.
b) On the basis of the result obtained in (a),
describerhe operaLionof lhe circuit.
c) How long will it take to saturatethe amplifier
if t,a = 10mv; ?)6 = 60mv; R = 40ko;
C - 25 n\ and Ucc: 72Y?
Figure
P7.88
Flgure
P7.91
S€ction 7,7
7,89 The eneryy stored in the capacitor in the circuit
6trc shown in Fig. It.89 is zero at the instant the switch
is closed. The ideal operational amplifier ieaches
saturation in 3 ms Wtat is the numerical value of R
in kilo-ohms?
tiguruP7.89
7.92 At the instant the switch of Fig. P7.92is closed, the
6r,a voltage on the capacitor is 16 v. Assume an ideal
operational amplifier. How many milliseconds
after the switchis closedwill t}le output voltageoo
equalzero?
60 nF
Figurc
P7.92
+ 16V
10k[]
7.90 At the instant the switch is closed iI} the circuit of
Fig. P7.89,t}te capacitor is charyed to 5 Y positive at
the ight-hand terminal. If the ideal operational
amplifier satumtes in 8 ms.what is the value of n?
280
Response
of First-order
Rl andnrCircuits
7.93 At the time the double-pole switch in the circuit
tsdG shown in Fig. It.93 is closed,the initial voltages on
the capacitors are 45 mV and 15 mV as shown.Find
the numedcal expressionsfor r'"(t), olt, ard ,10)
that are applicableaslong asthe idealop ampoper
atesin ils linear range.
Rgure
P7.93
0
()ko
-
12.5nF
!r(t) +
')
-6V
4OkO +
7.95 Repeat Problem 7.94 with a 4 MO resistor placed
EpI( acrossthe 50 nF feedback capacitor.
7.96 fhe voltagesourcein the circuit in Fig. P7.96(a)is
Erc generating the triangular waveform shown it
Fie. P7.96(b).Assume the energy stored in the
capacitoris zerc at t = 0.
a) Derive the numerical expressions for r',(r) for
t h e f o l l o $ i n g L i m e i n l e r v a l ' :0 r r 5/rs:
5ps < t < 15ps;and15/..s= t < 20Fs.
b) Sketchthe output wavelorm between0 and 20,.s.
c) ff the triangular input voltage continuesto repeat
itsef for r > 20 /-.s,what would you expect the
ouFut voltage to be? Er?lain.
Figure
P7.95
10mV e0) 15
25nF
7.94 The voltage pulse shown in Fig. P?.94(a) is applied
Edc to the ideal integrating amplifier shown in
Fig.P7.94(b).Derive the nume cal expressions
for
o,(t) when t,(0) : 0 for the time intenals
a)t<0.
b) 0 < r < 50 rns.
c)50ms<t<100ms.
d)100ms<t<co.
Figule
P7.94
ljs (mv)
50 nIF
Sections7.1-7,7
7.97 The circuit showr in Fig. P7.97 is known as an
astable multivibrutor ar].dfinds wide application in
pulse circuits The purpose of this problem is to
relate tle charging and discharying of the capacitors to the operation of the circuit. The key to analyzing the circuit is to undentand the behavior of
the ideal transistor switch€sT1 and T2.The circuit is
designedso that the switchesautomaticallyalternate between oN and oFF.When Tt is oFF,T2 is oN
Probtems
281
and vice versa.Thus in the analysisof tltis circuit, we
asstmlea switch is either ONor OFF,We also assumc
ttrat the ideal transistor switch can change its state
instantaneously. In other words! it can snap from
oFFto ONand vice velsa,When a transistor switch is
oN, (1) the base curent lb is geater than zcl{r,
(2) the teminal voltage ob€is zero, and (3) the terminal voltage o@ is zeio. Thus, when a transistor
switch is oN, it presents a sho circuit between the
terminals b,e and c,e. When a transistor switch is
oFF,(1) the terminal voltage obeis negative, (2) the
basecurent is zero, and (3) there is an open circuit
between the termhals c,e, Thus when a hansistor
switch is oFF, it presents an open circuit between
tle teminals b,e and c,e.Assume that T2 has been
on and hasjust snapped oFFrwhile T1 has been oFF
and has just snapped oN. You may assumethat at
this instance, C2 is charged to tle supply voltage
ycc, and the charge on Cr is zero. Also assume
C1 = C2and R1 : R2 = 10Rr.
a) Derive t]le expression for ob!2during t]le interval that T2 is OFF.
b) Derive tlle expression for o@2during the interval that T2 is oFF.
c) Find the length of time T, is oFF.
d) End the value of r'""2 at the end of the interval
that T2 is oFF.
e) Derive the expressionfor ib1duritrg the interval
that T2 is oFF.
I) Find the value of ib1 at the end of the interval
that T2 is oFF.
g) Sketch ?r."2versus t during the interval that T2
is oFF,
b) Stelch tbt versusr during tbe inter!al lhal T2
figureP7.97
7.98 The component values in the circuit of Fig. P7.97
are Ucc : 9V; Rr : 3 kO; Cl - C2 : 2 nF; and
R 1: R 2 : 1 8 k O .
a) How Iong is T2 in the oFFstate during one cycle
of opemtion?
b) How long is T2 in the oN state du ng one cycle
of operation?
Repeat(a) forT1.
a) Repeat(b) forT1.
At the first instant after T1 turns oN, what is the
value of ib1?
At the instant just before Tr tums oFF,what is
tle value of ib1?
g) What is the value of ?ce2at the instant just
before T2 tums oN?
7.99 Repeat Problem 7.98 with C1 :3 nF and
C2:2.8nF. All other component values are
unchanged.
7.100 The astable multivihator cftcuit iII Fig. P7.97 is to
satisly tlte lollowing criteda: (1) One traNistor
switch is to be oN for 48/rs and oFF for 36ps for
(3) ucc: 5v;
each cycle; (2) RL:2k0t
(4) R1 : Rr; and (5) 6R, < R1 < 50-Rr.wllat are
the limjting values for the capacitors C1 and C2?
7.101 The circuit shom in Flg. P7.101 is ktrown as a
m onostabIe multfuib r ator. -lhe adlectlve monostlib k
is used to descdbe the fact that t]le circuit has one
stable stale. That is, if left alone, the electronic
switch T2 will be oN, and Tr $rill be oFF.(The operation of the ideal hansistor switch is descibed in
Problem 7.97.) T2 can be turned oFF by momentarily closing the s$ritch S. After ,t returns to its open
position, T2 will rctum 1oits oN state.
a) Show that if T, is oN, Tr is oFF and will stay oFF.
b) Explain why T, is turned oFFwhen S is momentadly closed.
c) Show that T2 will stay oFFfor RC ln 2 s.
242
Response
of Fjrst-order
flLandRf Circuits
FigureP7.101
continue to conduct until the lamp voltage drops to
5 V Wlen the Iamp is not conducting, it appears as
an open circuit. 14 = 40 V; R : 800kO; and
+
+
a) How many tim€s per minute will the lamp
turn onl
b) The 800 kO rcsistor is replaced with a variable
rcsistor R. The resistance is adjusted until the
lamp flashes12 times per minute. What is the
valueof n?
7.105 In t]le flashinglight circuit shown in Fig.7.45,the
lanp can be modeledas a 1.3kO resjstorwhen it is
pillfffii,
_
;itrii conducting.The lamp triggers at 900 V and cuts off
at 300V
7.102 The parametervaluesin the circuit in Fig. P7.101
a) If % : 1000V, R : 3.7kO, and C : 250pF,
are Ucc = 6 V;
R1 : 5.0kO;
Rr = 20 kO;
how many timesper minute wil the light flash?
C = 250pF; and R : 23,083O.
b) What is the averagecurrent in milliampsdeUv
a) Sketch 0@2versus t, assuming that after .t is
ered by t]le source?
momentarilyclosed,it remains open until the
c) Assume the flashinglight is operated24 houn
circuit has rcachedits stablestate.AssumeS is
per day If the cost of power is 5 cenKper kilowatt
closedat I : 0. Make your sketchtor the interhour, how much doesit cost to operatethe light
val 5<r<10ps.
per yeafl
b) Repeat(a) for ib, versusr.
7.106 a) Show tlat ttre exprcssion for the voltage drop
across the capacitor while tle lamp is conduct
,:l|;l,i;,
7.103 Supposethe circuit in Fig. 7.45models a portable
ing in the flashing light circuit in Fig. 7.48 is
flashing light circuit. Assume tlat four 1.5V battergrvenby
e?lifl,#iLr
jes power the chcuit, and that the capaciforvalueis
10pF. Assume that the lamp conductswhen its
or0) = Yn + (v^*
Vr)e \' '")t'
voltage ieaches 4 V and stops conducting when its
voltagedrops below l VThe lamp hasa resistance
of 20 kO when it is conducting and has ar infinite
resistancewhenit is not conducting.
R-,
V'- '" =
V
a) Supposewe don't want to wait more than 10 s in
x+xi "
betweenflashes.What value of resistancen is
requiredto meet this time constraint?
RRLC
R+Rr
b) For the value of resistance from (a), how long
doesthe flashof light last?
b) Show that the expressionfor the time the lamp
conducts in the flashing light circuit in FiC.7.48
is givenby
7.104 In the circuit of Fig. 7.45,the lamp starts to conduct
whenever
the
lamp
voltage
reaches
15
V
During
,:l#11;r
v-", - vrh
RRrC
iiai the time whenthe lamp conducts,it canbe modeled
(I. - r,, :
Ln ymn
as a 10kO resistor.Once the lamp conducts,it will
R + Rr
%rr'
hoblerTrs
283
7.107 The relay shol*n in Fig. P7.107connects rhe 30 V dc
p:Sfi*iiEgenentor to t}le dc bus as long as the relay cment
is greater tlan 0.4 A. If the relay current drops to
0.4 A or less,the sprhg-loadedrelay immediarely
connectsthe dc busto the 30V standbybatteryThe
resistanceof the rclay winding is 60 O. The inductanceof the relay windingis to be detemined.
a) Assume t]le pdme motor ddving the 30 V dc
genemtor abruptly slows down, causingthe generated voltage to drop suddenly to 21 V What
value of a will assure that the standby battery
will be coinected to the dc bus in 0.5 seconds?
b) Using the value oI L determined in (a), state
how long it will take the relay to ope{ateif the
generatedvoltagesuddenlydropsto zero.
FigureP7.107
30v
gen
DC loads
Natural
andStep
Responses
of RICCircu'its
In this chapter,discussion
of the ratural rcsponseand step
8.1 Introduction
to th€ NaturatResponse
of a
P.ratLeL
flc Circuitp. 286
8-2 TheForms
of th€ NaturatResponse
of a
PaElbtnlc Circuitp. ?91
8.3 TheStepResponse
of a ParalLeL
RLCCitcuit p. 3a1
8.4 TheNaturalandStepResponse
of a Series
RLCffrctljt p- 3A8
8.5 A CircuitwithTvoIntegrating
p. 31?
Amptifi€rs
1 B€abteto determine
the naturalrespoiseand
the stepresponse
of paratlet
RtCcncuits.
2 Beabteto determine
the natuat response
and
the st€prcsponse
of seriesRLfcjrcuits.
284
responseof circuits containingboth inductors and capaciton is
limited to two simplesffucturesrthe parallelRZC circuit and the
seriesnLC circuit.Findingthe naturalresponseof a parallelRaC
circuit consistsof finding the voltagecreatedacrossthe parallel
branchesby the releaseof elergy storedin the inductoror capacitor or both. The task is defined in terms of the circuit shown in
Fig.8.1on pagc286.Theinitial voltageon the capacitor,V6,representsthe initial energystoredin the capacitor.Theinitial current
throughthe inductor,10,representsthe initial energystoredin the
illductor.If the individualbranchcuuents are of interest,you can
find them after detemining the terminalvoltage.
We derivethe stcpresponseofa parallelRl,Ccircuit by using
Fig.8.2on page286.Weare interestedin the voltagethat appears
acrossthe parallelbranchesas a result of the suddenapplication
of a dc current source.Energy may or may not be stored in the
circuit when the currellt sourceis applied.
Finding the natural responseof a seriesRLC circuit consists
of finding the current generatedin the se esconnected
elements
by the releaseof initially storedenergyin the inductor,capacitor,
or both.lhe task is defined by the circuit shown in Fig.8.3 on
page286.As before,the initial inductorcurrent,lo, and the initial
capacitorvoltage,yo,representthe initially stored energy.If any
of the individual element voltagesare of interest,you can find
them after determi ng the current.
We describethe stepresponseof a seriesRZCcircuit in terms
of the circuit shown in Fig. 8.4.We are interestedin the curent
resultingfrom the suddenapplicationof the dc voltage source.
Energy may or may not be stored in the circuit when the switch
is closed.
If yolr havenot studiedordinarydifferentialequations,de vation of the natural and step responsesof parallel and se.iesRaC
circuitsmay be a bit difficult to follow. However,the resultsare
important elough to warant presentationat tbis time.We begin
with thc natural responseof a parallel RZC circuit and cover
Perspective
Practical
AnIgnitionCiJCuit
weinkoduce
the stepresponse
of anRlf cirIn thjschapter
ignitioncircuitis based
onthetransient
cuit.Anautomobile
operresponse
of anRlCcircuit.In sucha circuit,a switching
in the currentin an jnductive
ationcauses
a rapidchange
windingknownasanignitioncoit.Theignjtjoncoilconsists
in series.
This
coupted
coitsconnected
of two magneticatly
The
is alsoknownas an autotransfomer.
seriesconnection
to the batteryis referred
to as the primary
coil connected
to the sparkptugis referrcd
wjndingandthe coilconnected
in
winding.TherapidLy
chanqing
current
to asthesecondary
(mutuai
prjmary
via magnetic
coup$ng
the
windinginduces
in the secondary
winding.
inductance)
a veryhighvoLtage
Thisvottage,
whichpeaksat frorn20 to 40 kV is usedto
ignitea sparkacross
the gapof the spafkp[ug.Thespark
ignitesthefueL-air
mixture
in thecytinder.
A schematic
diagramshowingthe basiccomponents
of an
jgnition systemis shownin the accompanying
figure. In
(as opposedto mechanicat)
todays automobite,eLectronic
switchingis usedto causethe rapidchangein the primary
ofthe electronicswitching
windingcurrent.An undefstanding
that is
circuitrequiresa knowtedge
of etectroniccomponents
beyondthe scopeof thjs text. However,
an anatysisof the
ignjtion circujtwjt[ seweas an introdlcoLder,ionventionaL
in the designof a
encountered
tion to the typesof probtems
usefulcircuit.
Capacitor .
245
286
NatuEtand
StepResponses
of RrcCircuits
C
Figure
8.1A A cjrcuitused
to itlunrate
ihenatumt
response
of a paraLtet
Rr circuit.
tlis material over two sections: ofle to discuss the solution of the
differertial equation that describesthe cicuit and one to prcsent the
three distinct Iorms that the solution can take. After introducine these
lhreelorms.$ e shos lhat rhe(dmetormsapplyro the srepre.pon.eof a
parallel RLCciicuit as well as to the natural and step responsesof sedes
RIC circuits.
8.1 Introductionto the Natural
Response
of a Paraltel
RLCCircuit
Figure
8.2A A circujt
used
t0 jltustmte
thestep
rcspon*
ofa panLlet
tu circuit-
va
The fint stepin findingthe naturalresponseof t}le circuitshownirl Fig.8.1
is to derive the dilferential equationthat the voltage ?i must satisfy.We
chooseto find the voltagef st.becauseit is the samefor eachcomponent.
After that, a branch current car be tound by using the currenfvoltage
relationshipfor the branch compoDenr.
We easilyobtain the differential
equation for the voltage by summing the curents away from the top node,
whereeachcurrentis expressedas a function ofthe unknown voltage?:
u
Figure
8.3A A circuit
used
io jltustrate
thenaturat
r€sponse
of a sedes
rur cjrcujt.
1f '
n'i.lnoo'*
4*cfi=0.
We eliminate the htegral in Eq.8.1 by ditrerertiating once with respect to I,
and,because10is a constant,weget
na-i
Figure
8.4A A circuit
used
i0 jltustctethe
step
response
of a s€ries
SLfcjrcuit.
(e.r)
(8.2)
We now divide through Eq. 8.2 by the capacitanceC and arrange the
deivatives in descendingorder:
d2r
1 du
't)
^
(8.3)
ComparingEq. 8.3 vith the differential equarionsderived h Chaprer?
reveals that tley differ by the presenceof the term involving the second
deivative. Equation8.3is an ordinary,second-orderdiffeietrtialequation
with constantcoefficients Circuits in this chaptercontain botll inducto$ and
capacitors,so the diJferential equation describingthesecircuits is of the second order.Therefore,
we sometimescall suchcircuitssecond-ord€r
circuits.
TheGeneral
Sotutionof the Second-0rder
Differential
Equation
We can't solveEq. 8.3 by separatingthe variablesand integmting as we
were able to do with the firstorder equationsin Chapter7. The classical
approachto solvingEq.8.3 is to assumethat the solutionis of exponential
folm, that is. to assumethat the voltase is of the form
r, = Ae',.
whereA ands aie unknownconstants,
(8.4)
8.1 Intrcduction
totheNaturat
Resoonse
ofaParaltetRlC
Circuit 297
Before showhg how this assumptionleadsto tle solutionof Eq. 8.3,
we needto showthat it is mtional.Thestrcngestargumentwe canmakein
favorof Eq.8.4is to noteIrcm Eq.8.3that the seconddeivativeof the
plusa constant
plusa constant
solution,
timesthefint derivative,
timesthe
solutionitself,mustsum to zero for all valuesof L This can occur only if
higher order derivativesof the solutionhavethe sameform as the solu
tion.The exponentialfunction satisfiesthis cdterion.A secondargument
in favor of Eq. 8.4is that the solutionsof all the firsl-order equationswe
derivedin Chapter7 wereexponential.It
seems
reasonable
to assume
that
the solution of the second-orderequationalso involvesthe exponential
function.
If Eq.8.4is a solutionof Eq.8.3,it mustsatisfyEq.8.3for all valuesof t.
Substituting
Eq.8.4into Eq.8.3generates
theexpression
er'e'rfi*rff=0,
-
,
1 \
RC-i)
(8.5)
which can be satisfied for aI values of t oDly iJ ,4 is zero or the parenthetical term is zero,becauseer' f 0 for ary finite valuesofrt. We cannotuse
,4. = 0 as a geneml solution becauseto do so implies tlat the voltage is
zero for all time-a physical impossibility if energy is stored in either the
inductor or capacitor. Therefore, in order for Eq. 8.4 to be a solution of
Eq.8.3,theparentheticalterm in Eq.8.5 must be zero,or
(8.6) < Characteristic
equation,parattel
ruCcircuit
Equation 8.6 is called the characteristic equation of the differential equation becausethe roots of this quadmtic equationdetemine the mathematicalcharacterof t'O).
The two roots of Eq.8.6 are
'
I
2RC
2RC
(8.7)
(8.8)
288
Natunt
andStepRespons€s
of fr Cjrcuits
If either root is substitutedinto Eq.8.4, the assumedsolutionsatisfiesthe
giver differential equation,that is, Eq.8.3. Nore fiom Eq.8.5 thaf this
resultholdsregardlessof the valueof,4. Therefore,both
u = Ard,' ar.d
D = A2ett
satisfyEq.8.3.Denotingthesetwo solutionsz,iand o2,respectively,
we can
show that their sum also is a solulion. Specifically,if we le1
'D = z)1+ D2 - Aret
(8.e)
+ A2erI,
du=
(8.10)
Arfie\' + A2sae,'
(8.11)
SubstitutingEqs.8.9-8.11into Eq.8.3 gives
/
4/"('i
t
- - rn )\ o n
R;,
I
*
I
\': ^.,
' . ,rJ -\ o
re.rzr
But each parenthetical term is zero because by definition 11 and 12 are
roots of the characteristicequation.Hence the natural iesponseof the
parallel RaC circuit shown in Fig. 8.1 is of the form
0:
Ares\t + A2e5,r.
(8.13)
Equatior 8.13 is a repeat of the assumptionmade in Eq.8.9. We have
shown that z'r is a solution, ?)2is a solution, and tJ1+ ,2 is a solution.
Therefore,the geneml solutionof Eq. 8.3 has the form givenin Eq. 8.13.
The roots ofthe characteristic
equatjon(st and sr) are determiDedby the
circuitparametersR, a, and C The initial conditionsdeterminethe values
ofthe comtants/1and,42. Note that the form of Eq.8.13must be modified if the two rcots sr and r, are equal. We discussthis modification when
we turn to the criticallydampedvoltageresponsein Section8.2.
The behaviorofo(t) dependson the valuesof sl and 12.Thereforethe
first step in finding the natural response is to determine the roots of the
characteristicequation.We retun to Eqs.8.7 and 8.8 ard rewrite them
t u s i D gnao l a r i o\n i d e l yu s e d i nl h e k l e r a l u r e :
\=-d+l&4,
"r= -'
\/&
(8.14)
"1,
(8.15)
8 . t - r r ' o d r r i o l . o h el a r u r a l R e 5 D oo,tsaeP a n l t eRrl t ( i l ( r i r
(8.16) < Nep€rfrequenqr,
paralletRICcircuit
(8.17) < Resonant
parattel
radianfrequency,
Rlf circuit
These results are summarized in Table 8.1.
The eeonent of e must be dimensionless,
so both sr and r, (and
hence a and o0) must have the dimension of the reciprocal of time, or frequency.To distinguishamongthe ftequeflciess1,12,a, and oo, we usethe
fotlowingterminology:sr and s2are refeired to asmmplexftequencies,a
is called the neper frequenct', and oo is the resonant m.li&n ftequenq.'I]Je
full significance of this teminology unfolds as we move t}lrough the
remainingchaptersof this book. All theseftequencieshave the dimensior of angular frequency per time. For complex frequencies, the neper
frequency, and the rcsonant radian frequency, we specify values using the
nni,tndilns per second(rad/s).The nature of the roots 11and s2depends
on the ^valuesof c and o0. There are three possible outcomes. First, if
oii < a'. both roots will be real and distirct. For reasonsto be discussed
late^r,the ^voltageresponseis said to be overdamped in this case.Second,
if 06 > l, both s1and s2will be complexand,in addition,will be conjugatesof each other. In this situation,ttre voltage rcsponseis said to be
underdamped.The third possibleoutcomeis that oB = a2.In this case,rl
and 12will be real and equal.Here the voltage rcsponseis said to be
critically damped. As we shall see, damping affects the way the voltage
responseleachesits final (or steady-state)value.We discusseach case
separatelyin Section8.2.
Example 8.1 illustmtes how the numerical values of 11 and 12 are
detemhed by the valuesofR, a,and c-
Terminology
rj-
a+Vd'
s1:-d1
Resonant radian frequency
"'
2RC
I
.vzr
va-
06
onl
289
290
Naturat
andStepRespons€s
ofnr Cjrcuits
Findingthe Rootsof the Characteristic
Equation
of a PaEttetfl"CCircuit
a) Find the roors of the chamcteristic equation that
govems the transient behavior of the voltage
shownin Fig.8.5if R : 200 O, L = 50 mH, and
C = 0.2 p.F.
b) Will tle responsebe overdamped,underdamped,
or critically damped?
C
Figure8.5l A cjrcujt
used
to illustrat€
th€natuntr€sponse
of
a palattet
Rrrcircuit.
c) Repeat(a) ard (b) for R : 312.5O.
d) What value of R causesthe responseto be criticaly damped?
c) ForR = 312.5O,
106
= 8000rad/s,
d = ..:-.^
(oz)){u.rl
SoIution
& : 64 x 106: 0.64 x 103rad'ls'.
a) For the given values of -R,a, and C,
As ofr remains at 103rad2/s2,
-
2RC
1
,"
_r,{.,na_Cs.
(400)(0.2)
r, = -8000 - ./6000rad/s.
/ rnl ! 1n6r
(In electrical engineering,the imaginary number
V-1 is representedby the letter i, becausethe
letter j representscu[ert.)
In this case,t}le voltage responseis underdamped since ofr > L
d) For critical damping, a' : (,3, so
FrcmEqs.8.14
and8.15,
\ - -1.25^ td + r"ftsox
= -12,500+ 7500:
to3
to3
5000rad/s,
\2= -r.25^ ro'- lffirsox ro' rd
:
12,500 7500:
rr = -8000 + i6000 radls,
20,000
rad/s.
b) The voltageresponseis overdampedbecause
oZ < a).
(#)'=#='.''
#='*'
106
(2 x 10)(0.2)
: 250().
Circuit 291
ofa PaEttetRlC
s.2 TheForms
ofthe NatuBtResDonse
of parattelruCcircuits
obj€dive 1-Be abteto determinethe naturalresponseandth€ st€presponse
8.1
The resistanceand inductanceofthe circuit in
Fig.8.5are 100O and 20 mH, rcspectively.
a) Find the valueof Clhat makesthe voltage
responsecriticallydamPed.
b) If Cis adjustedto give a neperfrequencyof
5 krad/s,find the valueof C and the roots of
the charactedsticequation.
c) If C is adjustedto give a resonantlrequcncy
of20 krad/s,find the value of C and the
equation,
roots of the characteristic
Ansu,er:(a) 500nF;
(b)c:1pF,
11: 5000+ j5000rad/s,
r, = 5000- i5000rad/s;
(c) C - r2snF,
'!r - -5359rad/s,
s2: -14,647radls
NOTE: Also by ChapterPtublem8.1.
of th* !$aturai$tesB*nse
8.2 TlteForsns
8l{ Ciieuit
ef a Far*tte!.
Sofar we have seenthat the behavior of a second-orderR LC circuit depends
on the valuesof rt and 12,which in turn dependon ihe circuit paramete$ R,
l-. and C.Thercfore, the first step ill finding the natural responseis to calcu
late thesevalucsand,relatedly,dctcrmine whetherthe rcsponseis over ,
under-,or criticallydamped.
Completing the description of the natural resPonserequires finding two
untnown coetficientqsuchas A1and,4, in Eq.8.13.Themcthodusedto do
this is basedon rnatchingthe solulion fbr the natural rcsponscto the initial
conditionsimposedby the circuil, which are ihe idtial value of the cufent (or
voltage) and the initial value of lhe Iirsl derivative of the omenl (or voltage).
Note that thesesamejnitial conditions,plus thc final value of the variable,wil
of a second-order
circuit
alsobe neededwhenfindingthe stepresponsc
In this section, wc analyze the natuml responseform {or each of the
three types of damping,beginningwith the overdampedresponse.As we will
see, the response equations, as well as lhe equations for evaluating the
unknown coefficients,are slightly different for each of the three dampjng
configurations.This is why we want to determine al thc outsetof the Problcm
whether the responseis over . under-.or critically damped.
VoltageResponse
Theoverdamped
equationarereal anddistinct,thevoltWhen lhe rootsof the charactenstic
solu
ageresponseof a parallelRLC circuitis saidto be overdamped.The
tion for the voltage is of the lorm
r-A1e\r+A2et1t,
(8.18)'j! voLtage
naturaIresponse-ovetdarnped
parattetnlf circuit
NatuntandSiepResponses
0f flr Circuitjs
where J1and s2are the roots of the characteristic equation, The constants
,41 and ,42 are detemined by the initial conditiorq speciiically from the
vaiuesof r(0 ) andd0\0 ) dr, wbich in rurn are determinedtr;m |he initial voltage on the capacitor, y0, and the initial current in tle inductor. 1n.
\ext. we showhow to uselhe inilial\okageon lhe capacilordodthe
initial current in the inductor to find ,41and ,42.First we note from Eq.8.18
rharA and,42.FirslwenoleftomEq.8.l8rhal
z'(0'):Ar+,4r,
dr(o+)
: srAr + szA2.
(8.1e)
(8.20)
wilh rr and12known.lhe taskot tinding/4| and,42reduces
to finding
u(0 ) and /rfo ) dI. The valueot r{0-) is rhe inirial vokageon lbe capacitor y0.We get the initiat value of dr/dt by first finding the currenr i; the
capacitor branch at I - 0+.Then,
dr'(0*)
,.(0.)
C
(8.21)
We use Kirchloffs current law to fhd the initial current in the capac_
itor bmnch. We ktrow that the sum of the three branch currents at r : 0+
musl be zerc. The currcnt in the resistive branch at r : 0+ is the initial
voltage % divided by the resistance, and the curent in the inductive
branch is 10.Using the reference system depicted in Fig. 8.5,we obtain
-%
ic(o)
\8.2?)
After finding the numericalvalueoft6(0+),we useEq.8.21to fhd rhe jniIial \ante of da/dt .
We can summarize the processfor finding the overdamped response,
o(l), asfollows:
1. Find the ioots of the characteristic equation, st and s2,using the valuesof R, Z, and C
2. Find r'(0+) and do(O+)/dl using circuit analysis.
3. Find the values of "{r and ,,t2 by sotving Eqs. 8.23 and 8.24
simultaneously:
u(0-):A1+4,
db(o+)
,c(o-)
C
(8.23)
s1A7+ s2Az.
\8.24)
4. Substitutethe valuesfor sl, 12,,41,and,42 into Eq.8.18 to determinerheexpression
for ,(r) tor r > 0.
Examples 8.2 and 8.3 illustrate how to find the overdamped responseof a
pamllel RIC circuit.
Rr Cncuit 293
Reslonse
of a Parattet
8.2 TheForms
oftheNatunt
NaturalResponse
of a Parattet
8tCCircuit
Findingthe overdamped
For the circuit in Fig. 8.6, o(0"):12V.
ir(0') : 30 mA.
and
a) Find the initial current in each branch of the
b) Find the initial value ofdt'ldr.
c) Find the expressionfor 0(t).
r
d) skelch,(r) in rheinrerval0
Examph
8.2.
Figure8.6 a Thecircuittor
250ss.
Solution
a) The inductor preventsan instantaneouschange
in its current,so the initial value oI the inductor
currentis 30 mA:
ir(0 ) : tr(0.) - i-(0-) : 30mA.
The capacitorholdsthe initial voltageacrossthe
parallel elementsto 12V Thus t}le initial current
in tbe resistive branch, ir(O-), is 12/200, or
60 mA. Kirchhof|s currentlaw requircsthe sum
of the currents leaving the top node to equal
zero at everYinstant-Hence
rc(O) = rr(0) iR(o)
- 90mA.
Note that if we assumedthe inductor current and
capacitor voltage had reached their dc values at
the instant that energy begins to be released,
tc(o-) = 0. In other words,there is an instantaneouschangein the capacitor current at I : 0.
b) Becauseic : C(.tu/dt),
du('')
dt
0.2pF
the form ofEq.8.i8.We find the co-efficientsA1
and ,4, from Eqs.8.23 and 8.24.We've already
determinedrr, r,, r'(0+),and dt'(0*)/dr, so
12: Ar+ A2,
450 x 1d:
5000/r - 20,000/r.
We solve two equations for ,4r and ,4, to obtajn
Ar :
1.4y and A2 - 26 V. substitutingthese
valuesinto Eq.8.18ields the overdampedvolt
ageresponse:
'"' t \. r . ..
( t p-smolt- 2bp)
,(/)
As a check on these calculations,we note that
the solution yields o(0) : 12 v and da(o*)/dt
:
4s0,000v/s.
d) Figure 8.7 showsa plot of o(l) versus1 over the
interval0 < r < 250ps.
,(, (v)
12
10
9 0 . r 0 r: 4 5 0 k v / s .
0.2 x 10-6
c) The roots of the characteristicequation come
from the values of R, Z, and C. For the values
specifiedand from Eqs.8.14and 8.15alongwith
8.16and It.17,
. t'T tF
c, - -t.25 . t0" \,{5= -12,500+ 7500= 5000rad/s,
r.25' l0'- \,/i5b25 r0' u'
12
: -12,500- 7500: 20,000
rad/s.
Because
therootsarerealanddistinct,weknow
is overdamped
andhencehas
thattheresponse
rcsponse
for Example
8.2.
Figure8.7 A Thevottage
294
Naturaland
StepRespomes
ofR|rCircuits
Catcutating
BranchCurrents
in the NaturalResponse
of a parattet
flc Circuit
Derivc the expressionsthat describe the rnrce
branch currents tx, ir, and r. in Examplc 8.2
(Fig-Il.6) during thc lime the storcdenergyis bcing
A second approach is to find rhe current in the
capacitivcbranch first and then use the tacl that
iR + it. + ic = 0. Let's usc this approach.The current in thc capacitivebranchis
,.(r)=
Solution
5r{r0' 520,000ar0,000/
: 0.2 x 101j(70.000e
)
We know the voltage acrossthe thrcc branches
frollr the solLrtionin Example8.2,narncly,
solro'
+ 26e 2oooo')
v,
44 = ( L1e
= (14e tooo/ 104e'oooo/)mA,
. > 0.
The currcntin the resistivebranchis theD
i"tt\:
:I , 70. s000,
+ I tn, :0000r
\, ,m, ,r A . ' :
200
u.
'I}lcre
rre two waysLofind the curent in thc inductive branch.Onc wav is to usc the integralrelationship lhat existsbclweenthe currentand rhc voltage
allhe terminalsoI an inductor:
it(t) =
c+
lI',uo*
t1o
Notc thal tc(0*) = 90 mA, which agreeswirh th.rcsull in Examplei1.2.
Now we obtain the iDductivebranch curreni
&om the relationship
iL(t) = -iRQ)
ic(t)
= (56d 5ooor 26d 2o,ooo,)
mA,
Use the integralrelationshipbelweeni, and ?
to find rhce\pfes.i.nfor r, in fig.U.o.
Answer: tr(1) = (56e taht
8.3
26e 20m,1 m,r, r - U.
The elcmentvaluesin the circuit showtrare
R - 2kO, Z = 250mH, and C : 10 nF. The
iDitialcurrent10in the inductoris 4 A, and
the initjal voltageon the capacitoris 0V The
output signalis the voltage2,.Find (a) t,r(o+);
(.h)ic(1'); (c) dr(c')/.Lt, (d) ,41:(e),4r;ard
(f) ?J(1)
when I > 0.
NOTE: AIso try ChapterPrcblens 8.2,8.5,an(t8.19.
r > rr.
We leave it to you, in AssessmentProbtem8.2,10
show that the integral relation alluded to leads to
the saDe result. Note that thc expressionfor iz
agreeswith thc initial inductor curreDt,as it musl.
obiedive 1-Be abteto determinethe naturalresponseand the stepresponse
of parattet
f,lf circuits
8.2
I > 0+
(a) o;
(b),rA;
(c) ,l x 103v/s;
(d) 13.333
V;
(e) -13,333V;
- e rn,mo
(0 13,333(e-10.0''0'
) v.
nrcircuit
ofih€Naturat
Response
ofa Paiatkt
8.2 rheFonns
295
TheUnderdamped
VoltageResponse
Whetr06 > l. $e roots of lhe characrerislicequadoDare complex.and
the response is underdamped, For convenience, we express the rcots sl
and J2as
:
a + j\/;{-i
(8.25)
(8.26)
(8.27) < Damped
radianfrequency
(8.28)
(8.?9)
+ i(/r
- A)snt adt).
< vottagenatoralresponse-undedamped
DarattelruCcircuits
296
N a t u € t a nSdt e pR e s p o 6 e sRoLf t c i ( u i t s
At this point in the transitionfrom Eq. 8.18to 8.28,replacet}le arbitiary
constants,4r + ,4, and t(A1 - At with new arbitary constantsde oted
81 and B, to get
o = € "'(Br cos,rdt + 82 sin.dd,
: B1e "t cosadt + B2e-"tsin.odt.
The constants-Brand 82 are real,not complex,becausethe voltageis
a real tunction.Don't be misledby the fact that B, : j(,4r - ,4r). In this
and thus81 and 82
underdampedcase,,4rand ,42are complexconjugates,
are real.(SeeProblems8.13and 8.14.)Thereasonfor definingthe underdampedresponsein terms of the coefficients81 and 82 is that it yieldsa
simpler expression for the voltage, ,. We determine B1 and B, by the initial energystoredin the circuit,in the sameway that we found,4r and ,42
for the overdamped response:by evaluating ?Jat t : 0' and its derivative
at I : 0*. As with sl and sr, a and ro2are fixed by the circuit pammeters R,
L,ancl C.
For the underdamped response,the two simultaneous equations that
determifle 81 and B2 are
(8.30)
u(0'):uo-81,
at'(o) _ i.(o) :
d
t
C
dBr + odBz.
(8.31)
Fint,
Let's look at the generalnatue of the underdampedresponse.
the trigonometric functioff indicate tlat this responseis oscillatory; that
is, the voltage alternates between positive and negative values.The rate at
which the voltage oscillates is fixed by .dd.Second,the amplitude of the
oscillation decreaseser?onentially. The iate at which the amplitude falls
off is detemined by a. Becaused detemines how quickly the oscillations
subside,it is also referred to as the damping factor or dsmping coeflicient.
That explains why (1),?
is called the damped radian frequency. If there is no
damping, d = 0 and th€ frequency of oscillation is o0. Whenever there is a
dissipativeelemenqR, in the circuit, a is not zero and the ftequencyof
oscillation, ar, is less than o0. Thus when .Yis not zeio, the frequency of
oscillationis saidto be damped.
The oscillatory behavior is possiblebecauseof the two Opes of energystomge elementsin the circuil the inductor and the capacitor. (A mechanical analogy of this electric circuit is that of a masssuspendedon a spring,
where oscillation is possible because erergy can be stored in both the
spring and the moving mass.)We saymore about the characteistics of the
underdamped responsefollowing Example 8.4, which examines a circuit
whose response is underdamped. In summary, note that the overall
processfor finding the underdamped responseis the same as that tbr the
overdampedresponse,althoughthe responseequalionsand the simultaneous equalions used to find the constants are slighdy different.
Response
ofa ParattetRlr
Cncuit 297
8.2 TheFomsoftheNatunL
NaturalResponse
of a Parattet
flc Circuit
Findingthe Underdamped
In the circuit shown in Fig. 8.8, Vo:o,
Io:
12.25mA.
and
a) Calculate the ioots of the characteristic equation.
b) CalculateD and do/dt at t : 0'.
c) Calculate the voltage responsefor t > 0.
d) Plot o(t) versus I for the time interval
0<l<11ms.
explicitly. However, this example emphasizes
$b! r' a0d\? arekno$.nascomple\frequencies.
b) Becauseo is the voltageaqossthe terminalsof a
capaclror,we nave
0(0)=u(0-)=Yo:0'
Because?(0+) = 0, the current in the rcsistive
branch is zero at t = 0+. Hence the current in
the capacitor at I = 0- is the negative of the
hductor current:
tc(0') =
0.125pF
( 12.25)= 12.2sr]l,4.
Thereforethe initial valueof the derivativer
d u { o + ) { 1 2 . 2 5 ) (r1) 0 - - - - - , ,
d/
Flgure8.8 a Theci(uitfo' E{anrpte
8.4.
:
-:9Xli(n)V/s.
(0.r2s)(10
)
c) From Eqs.8.30and 8.31,Br = 0 and
B, =
oR
non
-'''''
c 100V.
Solution
Substitutingthe numerical valuesof a, od, 81,
and 82 into the exprcssion for ?r(t) gives
a) Because
1
106
2RC 2(2q1d9.12s\
',77
: 200rad/s,
- "f-.'ff
= 103md/s,
(8X0.125)
\
.3 > "'.
0(t) = looa'misin 97980r v,
t > 0.
d) Figure8.9 showsthe plot oI t (t) ve$us I for the
fkst 11 ms after the stored eneryy is released.It
clearly indicatesthe damped oscillatorynature
of the underdampedresponse.The voltage 2{t)
approachesits final value,alternating between
values that are greater than and less than the
fiml value.Fu hemore, theseswingsabout the
final value deqease exponentialy wittr time.
Therefore, the responseis underdamped. Now,
a6: ^,/S ]:
r,,iFJxld:
roor,66
: 979.80radls,
st:
a + jod:
s2 = -a - jad =
-200 + j97980rad/s,
200
j979.80rad/s.
For t}le underdamped case,we do not ordinadly
solvefor J1 and 12becausewe do not usethem
, (v)
80
60
40
20
0
-20
40
7 8 9
tesponse
for Exampte
8.4.
Figure8.9 A ThevoLtage
,(*)
298
Naturatand
StepResponses
ofntaCircuih
Characteristics
of the Underdamped
Response
The underdampedresponsehasseveralimporranl characterisrics.
Firsr.as
the dissipatiyelossesin the circuil decrease.
rhe pcrsistenceof thc oscillations incrcaser and the frcquencv of the oscillalionsapproachcso0. I'r
other words.asR+,6. thcdissipationin thecircuir in Fig.8.ltapproaches
zero bccausep : i)7R. As R-- co. a+ 0. which tells us lhat o,j + rr0.
When a : 0. the maximum amplirudeof rhc voltage rentainsconslant:
thus the oscillationat .r0is sustained.In Example8.4,if /t wcrc increascd
to infinily, the solutionfor o(r) \\'ouldbccorne
1Jt) : 98 sin 1000rV.
r > 0.
'lhus,
iD this casethe oscillarior is sustaincd,the maximum amplitudeof
thc voltageis 98 V and the frequencyof oscillationis 1000rad/s.
Wc Draynow describequalitarivelythc differencebetwccnan undcrdiDped and an ovcrdampedresponsc.In an underdampcdsystem.thc
lcspoNe oscillates,or "bounces,"aboul ils fiDal value.This oscilLalionis
also rcferred to as rinqing. In an ovcrdamped system,the responsc
approachesits final value without ringing or in what is sometjmcs
dcscibed asa "sluggish"manner.Whcn specifyingthe dcsircdresponscof
a sccondordersystcmryou mav want to reachthe final value in the sho(csl tiDe possiblc.and you may not be concernedwith smaLloscillations
about that final value.If so,you would designthe systcmcomponentsto
achievean underdampedresponse.On the other hand.you nlay bc concornedthat thc responsenot exc€edits final value.pcrhapsto ensurcthat
oomponentsarc not damaged.Insucha case,youwould dcsignthe systcm
componentsto achievean overdampedresponse,and you would havc to
:rc(€pri relalivcl)\lu\ ri.e ro rhclinJlvalu(.
obj€ctivel-Be ableto detemine the naturaland the st€p responseof paraltelilc circuits
8.4
A 10 trI}I inductor,a 1pF capacitor,and a variable resistorare connectedin parallelin the
circuit shown.The resistoris adjustedso thal
the roots ofthe characteristicequationarc
-8000 ,r
i6000 rad/s.The initial volrageon the
capacitoris 10Y and the initial cuffent in thc
inductor is 80 mA. Find
a) Ri
b) dr(o+)ldt;
c) Br and 82 in the solutionfor ?);and
a);r(r).
NOTE: Aho tty ChdpterProblems8.3 antl8.20.
Answ€r: (a) 62.5O;
(b) -240,000v/s;
(c) -Br: 10v, B, = -80/3 V;
(d) rr.(r)= 10e3mo,[8cos6000t
+ (82/3)sin600011
mA whenI > 0.
Response
of a PanLtel
RrcCncLrit 299
8.2 TheForms
ofthe NaturaL
TheCriticattyDamped
VoltageResponse
The second-ordercircuit in Fig.8.8is critically dampedwhen ofr : o', or
o0 : a. When a circuit is critically damped,the responseis on the verge of
oscillating.In addition, the two roots of the characteristicequation are
real and equalt that is,
1
2RC
(8.32)
When this occun, the solution for the voltage no longer takesthe form
of Eq. 8.18.This equation breaksdown becauseif rr = 12 = a, it predictsthat
D = \ A 1+ 4 ) e d :
AGd,
(8.33)
where ,40 is an arbitrary constant. Equation 8.33 cannot satisfy two inde'
pendent initial conditions (y0, 10) with only one arbitrary constant, A0.
Recallthat the circuitparametenR and Cfix d.
We can trace this dilemmaback to the assumptionthat the solution
takesthe form of Eq.8.18.wllen the roots of the characteristicequation
are equal, the solution for the differential equation takes a different
folm, namely
/8 14) < Vottage
naturalresponse-criticall,
parattelf,lCcircuit
damped
Thus in the caseof a repeated root, the solution involves a simple exponenlial telm plus the product of a linear and an exponential term. The justification of Eq. 8.34 is left for an intrcductory course in differential
equations.Finding the solution involves obtaining ,1 and D2 by follovring
the same pattem set in tle overdamped and underdamped cases:We use
the initial values of ttre voltage and the derivative of the voltage with
respect to time to write two equations containing Dr andlot D2.
From Eq. 8.34,the two simultaneousequationsneededto detemine
D, ard D, are
a(.r)=uo:D2,
do(o) _ ic(o*)_
Dr
(8.35)
aD2.
(8.36)
As we can see,in the case of a critically damped response,both the
equation for o(l) and the simultaneous equations for the constants,r and
D2 diJfer ftom tlos€ lor over- and underdamped responses,but the general
approach is the same.You will rarcly encounter critically damped systems
in practicg largely becauseoo must equal a €xactly.Both of these quartities depend on circuit parameten, and in a real circuit it is very difficult to
choosecomponent values that satisfy an exact equality relationship.
Example8.5illustratesthe approachfor finding the criticallydamped
resoonseof a oarallelRIC circuit,
300 NatuDtandStepResponses
ofiltOrcurts
Findingthe CriticattyDamped
NaturaIResponse
of a Paratlel
ntc Circuit
a) For thc circuit in Example8.4 (Fig.8.8),find the
valucoIn thairesultsin a criticallydampedvolt
Eqs.8.35and 8.36,r, : 0 and ,i = 98,000V/s.
Substjtutinglhesevaluesfor a. ,1, and ,2 inlo
Eq.8.34gives
b) Calculate!(r) for r > 0.
c) Irlot 1r(l)vcrsusrfor 0 < t = 7 ms.
? ) ( i )= 9 8 , 0 0 0 t i ? r o m ' rv>, 0 .
c) Figure 8.10 showsa plot of 2,0) versus1in thc
intervall)= I < 7ms.
Sotution
- r f f o m I \ J n r n , c8 . 4 .$ < k n o w r h " r u ;
Thereforefor crilical damping.
l0'.
1
)RC.
"(v)
-12
21
106
: 4000().
(2000)(0.12s)
8
0
b) Fron the solutionof Example8.4.we know that
u(0*) - 0 and dr(0+\ldt = 98,000v/s. From
1
, (mt
Figure8.10,i llre vottage
rcsponse
for Exanrph
8.5.
obiedive 1-86 abteto determin€the nai|jraland the st€p r€sponse
of paralletf,lCcircuits
8,5
The resistorin ihe circuit in Assessment
PfoblemR.4i. idtu.led tof crilicaldamping
The inductanccand capacitance
valuesare
0.4H and 10 /..F,respectively.
Ihe inirial energy
storedin thc circuit is 25 mJ ard is distributed
equallybetweenthe ind ctor and capacitor.
Find (a) R:(b) yor(c) 10;(d) Dr and ,, in the
soiution Ior t';and (e) in. I > 0*.
Answer: (a) 100O:
(b)s0v;
(c) 2s0mA;
(d) -50.000v/s,50 V;
(e) tn(r) = (-500re500.+0.50rs00)A,
t=t)'
NOTE: AIso *j ChapterProblems8.4dnrl8.21.
A Summary
of the Results
We concludeour discussionofthe parallcl R/,Ccircuir\ naturalresponse
with a bricf sunmary of the results.The first slep iD finding the natural
responscis to calculatethe roots of the characteristicequation.You thcn
knowimmediatelywhetherthe rcsponseis overdamped.underdampcd,or
criticallydamped.
If the roots are real and dislincl (o3 < a2), the responseis over
dampcdand the voltageis
n(.t): A'd."'+ Aze''.
8.1 TheStepRespo6e
ofa Pa,altetRtC
Circuit
\/7-',
1
2RC'
"
-
l
"u
Lc
The values of -41and ,42 are determined by solving the following simultaneousequatlons:
D(0*)= A1 +A2,
dr'(0+r l.{0+l
d
t
L
If the ioots are complex ..rA> d' t}le response is underdamped ard
the voltageis
D(t) - B1e "'cosadt + B2e-"'sinadt,
-,
r-a
"
The values of Br and B, are found by solving t}le following simultaneous
1)(0-):uo:Br,
dDt\+\ - ' i.tq+
|
(
=-aB -u"B
dr
li the roots of the characteristicequation are real and equal (dA : a'),
the volfage responseis
a(t) = D1rc-a'+ Dp d',
where a is as in the other solution forms, To determine values for the constants,1 and D2,solvethe folowing simultaneousequations:
a(O-):Uo=D2,
/u(01
ic(u+)
8.3 TheStepResponse
of a Paraltel
RICCircuit
Finding the step responseof a parallel RaC circuit involvesfinding rhe
voltage acrossthe parallel branchesor the current in the individual
branchesas a result of the sudden application of a dc current source.Therc
may or may not be energy stored in the circuit when the current source is
applied.The task is representedby the circuit shown in Fig. 8.11.To Figlre8.111A circuitLrsed
to de(fibethestep
of a paGlletfr circuit.
develop a general approach to finding the step responseof a second-order rcsponse
301
102
Natunt
andSiepResponses
of,?rrCircujts
circuit, we focus on finding the current in the inductive branch (ir). This
current is of particular interest becauseit does not approachzero as I
increases.Rather, after the switch has been open for a long time, the inductoi current equals Uredc sourcecurrent I Becausewe want to focus on tlle
lechdque for finding rhe step response,we assumethat tie initial energy
stored in tlle circuit is zero.This assumptionsimplifies t]le calculations and
doesn't alter the basic processinvolved. In Example 8.10 we will see how
the presenceof idtially stored energy enters into the general procedure.
To find the inductor current ir, we must solve a second-order differ,
ential equation equated to the forcing lunction 1, which we derive as fol
lows.From Kirchhoff's curent law. we have
iL+iR+ic:1,
i,r!*c!:r.
(8.37)
Because
- cllL
(8.38)
we get
(Lu
dt
- r12i,
dt2'
(8.3e)
SubstitutingEqs.8.38and 8.39into Eq.8.37gives
L (li,
iL+=-:1
i
LC.:-L
dr
(8.40)
For convenienc€,we divide througl by -LC and rearrange terms:
d2i,
dt!
I
dit
RC d!
i,
t
LC
LC
(8.41)
Compadng Eq. 8.41 with Eq. 8.3 reveals that the presence of a nonzero
term on the dght-hand side of the equation alters the task. Before showing
how to solve Eq. 8.41 directly, we obtain the solution indirectly. When we
know the solution of Eq.8.41, explai ng the direct approach will be easier.
TheIndirectApproach
We can solve foi ir indhectly by first finding the voltage o. We do this with
the techniquesintroducedin Section8.2,because
the differentialequation
that ?Jmust satisfyis identicalto Eq. 8.3.To seetlis, we simply return to
Eq. 8.37and expressi, as a functioDof r; thus
l[',0,*i*rolt='
18.42)
8.3 lheSteD
ResDonse
of a Panttet
flrCircuit 303
Differentiating Eq. 8.42 once with respecl to t reduces the righfhand side
to zero because1 is a constant.Thus
o
L
d2u
ldu
t< dt
^dz\
ar
1 cla
RC dt
^
u
LC
(8.43)
As discussed
in Section8.2,the solutionfor t, dependson the roots ofthe
characteristic equation. Thus the three Dossiblesolutions are
b=Aretn+A2es|t,
(8.44)
b = Bpn'
(8.45)
a=
Dle-"t
cos tJdt + B2e-"'sin adt,
+ Dze-"t.
\8.46)
A word of caution: Because there is a souce in the circuit for l > 0, you
must tate into account the value of the sourcecurrent at t = 0* when you
evaluate the coefficients h Eqs.8.4ll-8.46.
To fhd the three possiblesolutions for ir, we substitute Eqs.8.44-8.46
into Eq. 8.37.You should be able to vedry, when this has been done, that
the three solutions for i, vrill be
iL:I+Aid't+Aie"1t,
iL:
I + B'Ie-'cosaat + Bie "Isind,1t.
iL=I+Dita"t+Die"t,
\8.47)
(8.18)
(8.4e)
where Ai, Ai, Bi, Bi, Di, and Di, arc arbitrary constants.
Ir each case,the primed constantscan be found indiiectly in tems of
the arbitrary constantsassociatedwith the voltage solution. However, this
approach is cumbe$ome.
TheDired Approach
It is much easier to find tle primed constants directly in tems of the initial valuesof the responsefunction.Foi the circuit being discussed,we
would find the pimed constants from iz(o) and dir(0)/dt.
The solution for a second-order differential equation witl a constant
forcing function equals the foiced responseplus a responsefunction identical in form to the natural response.Thuswe can alwayswrite the solution
for the steo resoonsein the form
I lunctionof lhe samefolm I
I = r/ +
asthenat"*i *"p.*.J'
|
(8.50)
304 Naturat
andStepResponses
of Rtf Circujts
ol thesameformI
' = -.
vr + lfunctron
I astheDot-"r*"p.'*J'
(8.51)
where 1t and yi represent tbe final value of the response function. The
fillal value may be zero, as was,for example,the casewith the voltage x' in
tlle c cuit in Fig.8.8.
Examples 8.G8.10 illustrate ttre technique of finding the step
responseofa parallelRaCcircuit usingthe direct approach.
Findingthe overdamped
StepResponse
of a ParalteI
flc Circuit
The initial energystoredin the circuit in Fig.8.12is
zero. At t : 0, a dc current source of 24 mA is
applied to the circuit. The value oI the resistor is
400 ().
a) Wlat is the initial value of ir?
b) What is the initial value of diLldt?
c) wllat are the roots of the characteristicequation?
d) W}lat is t}le numerical expression for iz(r) when
t>0?
c) From tlle circuirelements.weobtain
.a-==
o
r h. r 0 " ,
a' =25r103.
Becauseofr < l, fie roots oI the chamcteristic
equationarercal anddistinct.Thus
,1 : 5 x 104+3 x 104= -20,000
rad/s,
.r2:
Figure8.12.AThecircuiitorExampte
8.6.
^:=
\z))\z))
1
loe
'RC
')(4uu,(25,
t
5x10*
3 x 10'= -80,000
rad/s.
d) Because
the rootsof the characteristic
equation
arerealanddistinct,theinductorcurent response
will be overdamped.
Thusir(l) takest}le form of
Eq.8.47,namely,
iL= If + A\es)t+ Aietxt
Solution
a) No erergy is stored in the circuit prior to the
applicationofthe dc curent source,so the initial
current in the inductor is zero. The inductor prohibits ar instantaneouschangein inductor current; therefore il(o) = 0 immediately after the
switchhasbeenopened.
b) The initial voltage on the capacitor is zero
before the switchhas been opened;ther€foreil
will be zero immediately after. Now, because
u: LdiL/dt,
!:tot : o.
> Indudorcurrentin overdamped
parattet
RICcircuitstepresponse
Hence,ftom thissolution,the two simultaneous
equations
thatdetermine
A\ andAi are
i,(0)=./,+A'.+A\=0.
;(0)
= sL,ai+ 'rAi:0
Solvingfor ,4i and Ai gives
, 4 j: 3 2 m A a n d A i = 8 m A .
Thenumerical
solutionfor ilo) is
iLo : Qa , 32e20,m&
+ 8e-s0,1\J0'
) mA, I > 0.
8.1 lhesteoResoonse
ola Para.teL
flt LtrLut
305
Findingthe Underdamped
StepResponse
of a Parattet
RICCircuit
The resistorin the circuit in Example8.6 (Fig.8.12)
is increasedto 625O. Find ,z(1)for I > 0.
Here, d is 32,000
rad/s, rrr is 24,000
rad/s, and
is
24
mA.
4
As in Example8.6,Bi andBi aredetermined
from the initial conditioN Thusthe two simultare-
Solution
rr(0)=1f+Bi=0,
BecauseI, and C remain fixed, oA has the same
value as in Example 8.6; that is, o; = 16 x 103.
Increasing n to 625O decreases a to
3.2 x 10amd/s. With ofr > a', the roots of the
charactedsticequationare complex.Hence
*Q)
3.2 x 10"+ j2.4 x 10"rad/s,
= dBt - "B't: o.
B't : -21m4
3.2 x 10- j2.4 x 10"rad/s.
The cu{ent responseis now underdamped
and
givenby 8q.8.,18:
Bi:
-32 mA.
Thenume cal solutionfor ir(t) is
itul = It + Btc"'cosd.t. Be't:inuut.
paratler
> Inductorcurrentin underdamped
ruCcircuitstepr€sponse
iLO = (24
24e 32ac4t
cos24,0}0t
- 32e'2ooo'sin
24.ooor)
mA. r > o.
Findingthe CriticatlyDamped
StepResponse
of a ParallelRlc circuit
The resistorin the circuitin Example8.6 (Fig.8.12)
is setat 500O. Find ir for I > 0.
Again, Di and Di are computed from initial
i L ( O )= I r + D ! 2 = 0 ,
Sotution
We know tlat oAremainsat 16 x l0s.WithRsetat
500O, d becomes4 x 104s 1, which corresponds
to c tical damping. Therefore tlre solution for i_(1)
takesthe form of Eq.8.49:
!!,or=o',-oo,,:0.
d
'
Thus
,i =
itltl:
l
mA/s and Di:
960,000
24m1..
It + D\te "' + Diad.
Thenumericalexpression
for iz(t) is
> Inductorcunentin criticalLydampedpanlleL
nlc circuit step response
iLO : Q1 - 960,000rerq0m'24t10m0')
mA, r>0.
306
NatuEtand
SiepResponses
0fRlfCncuits
Comparing
the Three-Step
Response
Forms
a) Plot on a single graph, over a range lrom 0 to
220!,s, the overdamped, underdamped, and
cdtically damped responses derived in
Examples8.6-8.8.
b) Use the plots of (a) to find the time requiredfor
tr to reach90o/.ofits final value.
c) On the basisolthe rcsultsobtainedin (b), which
responsewould you specityh a designt}lat puts
a premiumon reaching90% ofthe final valueof
tlte output in the shortest time?
d) W}lich responsewould you specify in a design
that must ensure that the final value of the current is neverexceeded?
Sotution
a) SeeFig.8.13.
b) The final value of ir. is 24 mA, so we can read the
timesoffthe plots correspondiq to ir = 21.6mA.
Thusrd,l= 130/.rs,r.r = 91 ps,andt"d : 74 ps.
c) The underdampedresponseieaches90o/oof the
final value in the fastesttime,so it is the desired
responsetype when speedis the most importart
designspecification.
(/l = 625O)
Underdanped
26
22
18
l4
10
1n=+oooy
fi- licalltlo'!'a-'p.a
=
(1l
daDpcd
500O)
tC
2
i)
140
180
I(it
Figure8.13.rr Thecun€niptotsfor Example
8.9.
d)From the plot, you can see that the underdamped responseove$hoots the final value of
current, whereasneither the critically damped
nor the overdampedrespoNe producescurents
in excessof 24 mA. Although specifying either of
the latter two responseswould meet the design
specificatior, it is best to use the overdamped
fesponse.
Ir $ould be impraclicalto requirca
design to achievethe exact component values
tlat ensurea criticallydampedresponse.
FindingStepResponse
of a ParallelRLCCircuitwith Initial StoredEnergy
Energy is stored in the circuit in Example 8.8
(Fig.8.12,withR = 500 O) at the instantthe dc cur,
rent source is applied.The initial current in the
inductoris 29 mA, and the initial voltageacrossthe
capacitor is s0 v Find (a) iL(o); (b) diL(o)/dt:
(c) tr(r) for I > 0; (d) u(D for t > 0.
b) The capacitor holds the initial voltage acrossthe
inductor to 50V Therefore
r;(o-)
: 50,
lo1l:fl
" 103:2oooA/s.
So[ution
a) Therecannotbe an instartaneouschangeof cu.
rent in an inductor, so the initial value of ir in the
first instant after the dc curent source has been
appliedmust be 29 mA.
") From the solution of Example 8.8,we kno\q that
the currentresponseis citically damped.Thus
iL(t) = Ir + D\te "' + Die ""
8.3 TheStepResponse
ofa Panttetilrcncujt 307
Thus the nume cal expressionfor ir(l) is
d
^:^
4 0 . 0 0r0a d / s a n d / , - ' 4 m A .
Notice that the effect of the nonzero initial
stoied energy is on the calculations for the constants Di and ,i, which we obtain tom the nntial conditions.First we use the initial value of
the inductorcurent:
iLC) = (.24+ 2.2 x l06reroooor
+ se-aoooo)
mA,
t > 0.
d.)We can get the expressionfor ?J(t),t > 0 by
using the relationshipbetweenthe voltage and
cment in an inductor:
tr(0) = 1r + Di - 29mA.,
T(, : L;
from which we get
Di:29
24
' 1o r) 2.2
Thesolutionfor ,i is
106)(-4o.ooo)reroooo'
+ 2,2 X I06e-4aIUt
=(0-)
+ (5)( 40,000)?ro!m/l x 10 3
:
+ 50? {,000,V, I > 0_
2.2 x 106tua1,auJt
D'r=2000+aDtz
: 2000+ (40,000)(5
x 10 )
: 2200Als : 2.2 x 106nA/s.
To check this result, let's verify that the initial
voltage acrossthe inductor is 50 V:
D()t)-
2.2x 106(ox1)
+ 500):50v.
StepResponses
of RtCCircuits
308 NatuGtand
8.4 TheNaturalandStepResponse
of a SeriesRICCircuit
+
Figure
8.14A A circuiiused
to itLustrate
thenaturat
response
of a series
Rlrcircuit.
The procedures for finding the natural or step respoNes of a seriesRZC
circuit are the same asthose used to find the natural or step responsesof a
parallel RIC circuit, becauseboth circuits are describedby differertial
equationsthat have t}le sameform. We begin by sumnine the voltages
aroundthe closedpath in the circuit shownin Fig.8.14.Thus
,ii
Ri
i
L",'
fl
^lidttva-j.
dt
(8.52)
cJo
We now differentiateEq.8.52oncewitl respectto I to get
_ d' i - i - o
.tr
c
_di
ar
(8.53)
which we can reanange as
dzi
Rdi
i
,t,'-iA-i
"'
18.54J
ComparingEq. 8.54with Eq. 8.3 revealsthat tiey have the sameform.
Therefore,lofind the solutionof Eq.8.54,we follow the sameprocessthat
led us to the solutionofEq.8.3.
From Eo.8.54. the characteristic eouation for the sedesRIC circuit is
Characteristic
equation-series
f,aCcircuit >
(8.55)
R
2t.
/RY
\i)
(8.56)
(8.57)
The neperfrequency(d) for the sedesRIC circuit is
Neperfrequenq'-series
Rlf circuit>
. = ,| ,"0u.
(8.58)
andtheexpression
for theresonant
radianfrequency
is
Resonant
radianfrequency-series
nlc circuit>
(8.5e)
Note tlat tle neper frequency of the seriesRIC circuit differs ftom that of
the parallelRaC circuit,but the resonantmdian frequenciesare the same.
8.4 TheNatuaLand
StepResponse
of a5e esffrCjrcujt 309
The currentresponsewill be overdamped,underdamped,or critically
damped accordjng to whether rofr < c', rofr > c', or tafr = a', respectively.
Thusthe threepossiblesolutionsfor the curent are asfollows:
i(t\ = nt""'
+ "\rer,t (overdamped),
i(t, : Bf-"' cos.Dat+ B2e "'sinadt (underdamped),
i(t) = 1;rt" "r + Dr€-"! (criticallydamped).
(8.60)
f8.ol) < CurrentnaturaL
response
formsin series
,- _- l(ta orcults
When you have obtainedthe natural current response,you can find the
natural voltage responseacrossany circuit element.
To verifl that the procedure for finding the step responseof a series
RaCcircuit isthe sameasthatfor a parallelRIC circuit,we showthat ihe
differentialequationthat descibes the capacitorvoltagein Fig. 8.15has
the sameform as the differential equation l]rat describesthe inductor current in Fig.8.11.For convenience,
we assumethaazero energyis storedin
the circuii at the instantthe switchis closed.
Applying Kirchhoff'svoltagelawto the circuitshowr in Fig.8.15gives Figure8.15A A circuitusedto ittushate
thesiep
rcsponse
of a se.ies
fircncuit.
V - R i + L ! + D . '.
(e.63)
/11
Th€ current(i) is relatedto the capacitorvoltage(oc) by the expression
(8.64)
from which
(8.65)
Substitute Eqs.8.64 and 8.65 into Eq.8.63 and write the resulting
d?t4
R dua
* - i a -
:!_
LC
V
I.C
(8.66)
Equation8.66hasthe sameform as Eq. 8.41;thereforethe procedurefor
finding ,c parallelsthat for finding ir. The three possiblesolutionsfor z'c
are asfollows:
D . - V + A\e"lt+ A?\r (oveidamped),
(8.67)
r c = V f + Bie "' cosodt + Bie-"' sln @,1 (undeidamped), (8.68) { Capacitor
vottagestepresponse
formsin
series
nlC
circuits
"t
D c : u f + Dtte + Diad (criticallydamped),
(s.6e)
whereyl is the final valueof rc. Hence,from the circuil shownin Fig.8.15,
the final value of z'c is t}le dc source voltage y.
Example8.11and 8.12illustratethe mechanicsof finding the natural
and steDresDonses
ofa seriesRIC circuit.
310
NaturaLand
StepResponses
ofnlf Circuits
Findingthe Underdamped
NaturalResponse
of a SeriesRLCCircuit
The 0.1pF capacitorin the circuit shown rn
Fig.8.16is chargedto 100V At l : 0 the capacitor
is dischargedthrougha seriescombinationof a
100mH inductoranda 560O resistor.
a) Findj(r) for r > 0.
b) Findoc(r)for r > 0.
the initial conditions.The inductor current is
zero before the switch has been closed, and
henceit is zeroimmediatelyafter.Therefore
(0)=0=81.
Io find 82.we evalualeJir0 l Jr. Fromlhe cil
cuit, we note that,becausei(0) : 0 immediately
alter the switch has been closed,there vall be no
voltagedrop acrcssthe resistor.Thus the initial
voltageon the capacitorappearsacrossthe ter
minals of the inductor, which leads to the
di(0-]'
Figur€8.16 A lhe circuiifor E)Gnpte
8.11.
di(o-)
Solution
a) The filst step to finding i0) is to calculatethe
root\of rhecharacleri(lic
equadon,
For rbegi\en
@ 6 =_
(ld)(r01
(100x0.1)
= 1000A/s.
Because
B1 : 0,
co,so00r- Tsinab0o/j.
: - a00?,cuu'(24
Ll1
Thus
d(0)
= 960082,
R
" 2 L
8, :
= -!!L ^ |n'
2(100)
: 2800rad/s.
Next,wecornparearfrto a2andnotethat oA > 02,
l=7.84x106
= 0.0784x 103.
At thispoinl.$e know lbal lbe re\poDse
is
underdamped
andthat the solutionfor i(t) is of
the folm
i(t) : B( "t cosa,tt+ B2a"t sinridt,
where o = 2800rad/s and od: 9600md/s.
Thenumericalvaluesof B1 and ,, comefrom
=T=rr*'t"
t000
s 0.1M2 A.
9600
The solution foi i(t) is
t(/)
go00/A. r j 0.
0.104)e-2m'\in
b) To find oc(t), we can use either of the followitrg
relationships:
"= -tl''o'*'ooo'
,1,
.- = t R + L 1 .
llt
Wichever expressionis used (the secondis reconmended), the result is
oc(r) = (100cos9600r+ 29.17sin 9600t)e'm'V,
r > 0.
8.4 TheNatunL
andStepResponse
of a sedesSLrCjrcuii 311
Findingthe Underdamped
Step Response
of a Seriesffc Circuit
No energyis storedin thc 100mI-I inductor or the
0.4r.F capacitor when the switch in the circuit
shownin Fig.B.l7 is closed.Find z,c(t)for t = 0.
The roots are complex.so the voltagc respoise is
underdamped.'nrus
0c0):
'18 + Bid lroo'cos4800r
+ rle rao(r'sin4800r,r > 0.
No energyis stored in thc circuil initially. so both
,c(0) and d?rc(o-)/lt arc zcro.Ther,
0c(0):0=,18+Bi,
F i g u r e8 . 1 7h T h e. i r . u i tf o rE a n r p l8e. i 2 .
dt).(0'\
__:t)
Soh'ingfor ai and Bi yields
5olution
ai : -18v,
The roots ofthe characteisticequationare
:
=4'rn0Ri 1400R
280
//zso\. -
+ 1 / l
|
0.2 v\0.2/
Bi:
106
tn.r)(rl4)
- 14V.
Therefore,the solutionfor ,(.(r) is
: ( 1400+ i1800)rad/s.
Dc(t): (48 - 48eraoo'cos4s00r
r, = (-1100 - j1800)rad/s.
- 14r r{orsin 4800r)V.
objective2-Be ableto determine
the naturalresponse
andthe stepresponse
of s€riesftf circuits
8.7
The switchin the circuitshownhasbeenin
positiona tor a long 1ime.At t = 0- it movesto
positionb. Find (a) l(0+);(b) "c(0+);
(c) dl(0')/dr; (d) .rl, rr:and (e) t(D for I > 0.
Answer: (a) 0;
(b)s0Y
(c) 10,000
A/s;
(d) ( 8000+ 16000)radls,
( 8000 j6000)rad/s;
(e) (1.67€Mo'sin60001)
A for r > 0.
NOTE: ALto try Chteter Problems8.15-8.17
Findoc(l) for I > 0 for thecircnitin
Assessment
Problem8.7.
Answer: [100 e 3mi(50cos
6000t
+ 66.67sin60m01V for r > 0.
8,8
r > 0.
312
Natunt
andStepR€sponses
offtr Circuits
8.5 A Circuitwith TwoIntegrating
Amplifiers
A chcuit containirg two inregraring amplifiers coDnectedin cascadel is
also a second-order circuit; that is, the output voltage of the second integrator is related to t}le input voltage of the first by a second-order ditrerential equation.We begin our analysisof a circuit containing two cascaded
amplifierswith tle circuit shownin Fig.8.18.
We assumethat the op amps are ideal. The task is to derive tlle differential equation that eslablishes the relationship between ?, and Dp.We
beeh tle derivation by summirg the currents at the inverting input terminal of the first inte$ator. Becausethe op amp is ideal,
0-u"
n
=K tj + c , d; tt o
u . , ,:)0 .
(8.70)
FromEq.8.70,
dDot
dt
I
(8.71)
&Cl"g'
Now we sum the cullents away ftom the inverting input teminal of t}le
second integrating amplifierl
0-z)-,
R,
dbo
rJt
Differcntiating
Eq.8.73gives
Euo
dt'
c,frtoo"t: o.
1
l.8.t2)
(8.73)
&C)""
7 doo1
R{2 dL
(8.74)
We find the differetrtial equation that govems the relationship betweetr o,
and ?rsby substitutingEq.8.71into Eq.8.741
d2u" :
1
1
d,'
&(\-R C:*
(8.75)
CI
+
Figure8.18A Twointegratjng
ampLifieu
connected
in Gscade.
' In a cascademnnection, the output siglal oI the ii6r mplifrd
signal tor the secondanplifier.
(ror in Fig. 8.18) is the inpul
8.5 A CiftuitwiihTwoInt€grating
AmpLjfiels 313
Example 8.13 illustrates tlle step responseof a circuit containing two cascaded integrating amplifierc.
Analyzing
TwoCascaded
IntegratingAmplifiers
No energyis storedin the circuit shownin Fig.8.19
when the input voltage ?s jumps instantareously
from 0 to 25 mV
0 . 1F F
1/,F
9V
a) Derive the expressionfor r'"0) for 0 < t< r""t.
b) How long is it before the circuit satuiates?
Sotution
Figure
8.r9A Thecjrcuit
forExampLe
8.13.
a) Figure 8.19irdicates that the amplifier scaling
1
1000
n1c1 (2s0x0.1)
1
RrC,
1000
(500)0)
-
No\ becauset's : 25mV for I > 0, Eq. 8.75
-
= (40X2)(25. 10 )) : 2.
To solvefor z',,we let
da^
dt'
then,
dp(r\
::2,
a n d d R 1 )= 2 d t .
Hence
becausethe energy stored in the circuit ini
tially is zero, and the op amps are ideal. (See
Problem 8.53.)Then,
dr^
i=2t,
and Do: ( + 1).(0).
But ,"(0) - 0, so the experssionfor rn becomes
p.:t2, o<t<ts^.
b) The second integrating amplifier saturates whetr
o, reaches9 V or t = 3 s. But it is possiblethat
the first integrating amplifier saturatesbefore
1 = 3 s.To explorethis possibility,useEq. 8.71to
tind d1).ldt,
du^, -40(25\
' l0 ':
=
,1,
1.
Solvingfor 0oryields
tc\4
|
ldx,
I d.y-2
.rE(o)
Jo
frcm which
80)-8(0):2r.
However,
Be)=4#:o,
0ot:
t
Thus, at r = 3 s, Dor= 3 V, and, becausethe
power supply voltage on the first integrating
ampli{ieris +5 V, the circuit teachessaturation
when the secondamplifier saturates.
When one
of the op ampssaturate$we no longer can use
the linear model to predict the behavior of the
NOTE: AsselsJout understandingof this materiul b), tying Chaptet Prcblem 8.58.
314
Natuct
andSteD
R€sDonses
ofRlrCncuits
TwoIntegratingAmptifierswith Feedback
Resistors
Figue 8.20depicts a variation of the circuit shown in Fig.8.18. Recall from
Section 7.7 that the reason the op amp in the integrating amplifier saturates is the feedback capacitor's accumulation of charge.Here, a resistor is
placed in parallel wittr each feedback capacitor (C1 and C2) to overcome
this problem. We rederive the equation for the output voltage,oo,and
detemine the impact of these feedback resistors on the integrating amplifiels ftom Example 8.13.
We begin the derivation of the second-orderdifferential equation that
relates o,1 to Dsby summing ttre currents at the inverting input node of the
firct irtepmtor:
u
us 0 uo\ ^d.^
=R:+c;(0R"
u,r=
) 0.
(8.76)
WesimplifyEq.8.76to read
th:or
dr
1.
R cr"''
rc
RJ r'
18.77)
For convenience,we let zt = R1C1and write Eq. 8.77 as
rlu"r
dt
u.\
rr
-rs
R,C,
(8.78)
The next stepis to sumthe currentsat the invertinginput terminalof the
secondintegrator:
?f.?
+c,*@_1,.)=0.
Figure8.20l Cascaded
inhgratinganpLifien
withfeedback
resistorr.
(8.7e)
8.5 A Circuit
MthTwolntegnting
Amplifier 315
Werew te Eq.8.79as
-Do1
dD.
uo
i
,r= nc,
(8.80)
where 12 = R2C2.DifJerentiating Eq.8.80 yields
&uo,lduo
1. d o . 1
: -RC
-;
dt
dt'
dP
(8.81)
From 8q.8.78,
door
-por
ilt
rr
_
oe
RuCt'
{8.82)
andftom Eq. 8.80,
o"r=-Rcr*-ryr..
(8.83)
We use Eqs 8.82 aDd 8.83 to elimiuate du.l/dt from Eq.8.81 and obtain
the desired relstionship:
*
(:,
',)o'"(i),. - a.,',1r,,, (8.81)
From Eq. 8.84,the chamctedstic equation is
.
/1
1\
\ar
r2,/
r'+i---ls
|
r -0.
7172
(8.85)
The roots of the characteridtic equation ate real, mmely,
-1
(8.86)
-1
(8.87)
Example8.14illustratesthe analysisof the stepresponseof two cascaded
iDtegating amplifien when tlle feedbackcapacitorsare shuntedwith
feedbackrcsistors.
316 Naturat
andStepR€sponses
0f RarCircuitj
Resistors
Analyzing
TwoCascaded
IntegratingAnplifierswith Feedback
The parametersfor the circuit shown in Fig. 8.20
are R. - 100kO, Xr = 500kO, Cr = 0.1pF,
'Ihe
Rb : 25 kO, R2 : 100kO, and C, = 1 r!F.
power supply voltage for each op amp is t6 V. The
signal voltage (os) for the cascadedintegrating
amplifiers jumps from 0 to 250 mV at t : 0. No
energyis stored in the feedbackcapacitomat the
instantthe signalis applied.
a) Find tlle numerical expressionof the differential
b) find r'.0) for I > 0.
c) Fhd the numedcal expression of the differential
equation for t,1.
d) Find o,(4 for t > 0.
Solution
a) Fromthe numeical valuesof the circuit paramete$, we have z1 : R1C1- 0.05s;r: - RuCz
: 0.10s, and or/RuCrRrC,: 1000v/s'. Substituthg tlesevaluesinto Eq.8.84gives
d'u^
du^
---+
+ 301: + 200r- = 1000.
d(
The solutionfor uotbus takestbe forml
u"=5+A\erat+Ate-t't.
With tr,(0) : 0 ard dr.,"(o)/dr= 0, the numedcal valuesof ,4i and Ai arc A\ : 10V and
,4i - 5 V. Therefore,the solulionfor z', is
.,,()=(5-lOe
o
5 ? - 2 0v?. r I - - .
The solution assumesthat neither op amp
saturates.We have already noted tlat the final
value of ?,,is 5 V which is lessthan 6 Y hencethe
secondop amp does not satuiate,The final value
of r,,1 is (250 x 1{r)(-500/100), or -1.25v.
Therefore, the first op amp doesnot saturate,and
our assumptionand solution are co ect,
c) Substituting the numerical values of the parameters hto Eq. 8.78 generatestle desired differcntial equation:
++2oD^.= 25.
dt
b)The roots of the charactedstic equation are
sr = -20md/s and s2 = -1orad/s. The final
value of o, is the input voltage times the gain of
ea€h stage,becausethe capacito$ behave as
open circuits as t + oo. Thus,
d) we have aheady noted tie hitial and final values of o,r, along wit]I the lime constant rt. Thus
we wdte the solution in accordancewith the
technique developed in Section 7.4:
aar= -1.2s+lo-
(
' I 0 )^r 500) 100)= 5 v .
, , ( m )= 1 2 5 0
t00 15
(-1.25)Ienat
: -1.25 + 1.25e1vV, t > 0.
NOTE: Assessyow andetstandinq of this matetial by tjing Chapter Prcblem 8.59.
P,acticalPerspective
317
PracticaIPerspective
AnlgnitionCircuit
NowLetus returnto the conventionat
ignitionsystem
introduced
at the
beginning
of the chapter.
A cjrcujtdiagram
of the systemis shownjr
Fig.8.21.Consider
the circujtcharacterjstics
that provide
the energyto
jgnitethefuet-airmixture
in thecytinder.
First,the maxjmum
vottage
avai.L
ableat thesparkptug,tr"p,mustbehjqhenough
to ignitethefuet.Second,
the voltageacross
the capacitor
mustbe limitedto prevent
arcingacross
jn theprimary
points.Thjrd,thecurrent
theswitchor distributor
winding
of
theautotransformer
mustcause
sufficient
energy
to bestoredin thesystem
to ignjtethe fuet-airmixturein the cyfinderRemember
that the energy
storedjn the circuitat the instantof swjtching
is proportionaL
to the primarycurrent
squared,
thatjs, 4,0= laf(o).
I
EXAMPLE
a) Findthe maximum
vottageat the sparkptug,assuming
the fottowingval
uesin the circuitof Fig.8.21: Ud.= 12Ir/,R= 4A, L=3mH,
C:0.4pF,anda:100.
o ) Whatdistancemustseparatethe swjtchcontactsto preventarcingat
the time the vottageat the sparkplugis maximum?
Solution
a) Weanatyze
the circuitin Fig.8.21to find an expression
for the spafk
pLug
voLtage
r'"r.Wetimitouranatysis
to a studyof thevoLtages
in the
circuitpriorto the firingof the sparkptug.Weassume
that the current
in the pdnarywindingat thetjmeof switching
possihasits maximum
blevalueydJR,whereR is the total resistance
primary
in the
cjrcuit.
Weatsoassume
that the ratjoof the secondary
voltage(oJ to the pri(r, is the sameastheturnsratjoN2/N1.Wecanjustjry
maryvottage
thisassumption
asfotlows.
Withthesecondary
circuitopen,thevottaqe
induced
in thesecondary
windjng
is
_- d i
1v,.
a- .
llt
(8.88)
ard ll-evoltage
indJced
in tl'e pr'mary
w noingis
,di
xt - L--,:.
.lt
(8.8e)
It fottows
fronrEqs.8.88and8.89that
(8.e0)
1
L
'
Figure8,21A ThecircLrit
djagnmofthe conveF
tionatauiomobite
ignitionsystem.
318
Naturatand
St€pRespons€s
offLr Circuits
It is reasonable
is thesamefortheftuxes
to assume
thatthepermeance
in
iron-core
hence
autotansforner; Eq.8.90reduces
to
d11anddz1 the
u2
NlN2g
N2
n = - n T =N ' = ^
(8.e1)
Wearenowreadyto analyze
the voLtages
in the ignitioncircuit.
Thevatues
ofR, a, andC aresuchthat whentheswitchis opened,
the
primarycoiLcurrentresponse
is underdamped.
lJsingthe technjques
devetoped
in Section
8.4andassuming
f = 0 at theinstanttheswitch
pdmary
is opened,
theexpression
forthe
coiLcurrent
is foundto be
, =*" -[-,-n . (;).'"*,],
(8.e2)
(seeProbtem
in the primary
8.62(a).)
Thevottage
induced
winding
is
of theautotransformer
Va, ^, .
..li
(8.e3)
(SeeProbtem
8.62(b).)It foLtows
from Eq.8.91that
bz:
-av* -, .
St[@dl.
M?
(8.e4)
ThevoLtage
across
thecapacitor
canbederived
ejtherby usingthe
relationship
".:lf''a,,,"101
(8.e5)
or by summing
the vottages
aroundthe meshcontaining
the pfimary
winding:
o':U'r'-iR-Li'
/11
(8.e6)
In eithercase,wefind
D": ud"[I
a"'cosadt + Ke-"'sinadtl,
(8.e7)
PracticatP€rspectjve
319
rl
"- .= A \
r
"\
).
RC
(SeeProblem8.62(c).)As canbe seenfrcm Fjg.8.21,the vottage
acrossthe sparkptugis
t'sp=Yd.+02
= v* - *"-'
""n'ot
=r".lt- -l=n* "^,",1,.
(8.e8)
Tofindthe maxirnum
positive
vatueof 0sp,wefindthesmattest
vatueof
timewheredosp/dtis zeroandthenevaluate
Dspat this instant.The
expression
for tnaxis
r . * = I , u n' ( " 1
a,t
\d
(8.ee)
/
(SeeProbLen
8.63.)Forthecomponentvatues
in theprobLem
statement,
-'
R
2L
4 x 1 d:
666.67ftd/s.
and
$
= 28.8sa.81rad/s.
rooo.arr'
Substit0tjng
thesevalues
into Eq.8.99gives
/",{ - 53.63
ss.
NowuseEq.8.98to findthe maximum
sparkptugvottage,
o,p(r--):
tse(r-*) = -25,975.69
V.
D)
Thevottage
across
thecapacitor
at lnaxis obtained
fromEq.8.97as
ac?n,J : 262.15Y.
ThedieLectric
strengthof air is approximatety
3 X 106V/m,so this
resuLttelts us that the switch contactsmust be separatedby
x 106,ot 87.38,
262.1513
arcingat thepointsat tffi.
/,Lmto prevent
320
offr Cncujts
Naturatand
stepResponses
must
consideration
andtestjngof ignitionsystems,
In the design
ofthespa* pluq
thewidening
fuet-airmixtures;
begivento nonuniform
theretatjonship
of theptugetectrodes;
gapovertimedueto theerosion
speed;
thetimeit takes
andengine
sparkptugvottage
between
avaitabte
the prinarycurrentto buitdup to its initiaLvalueafterthe switch
to ensureretjabie
required
of maintenance
is ctosed;
andthe amoLlnt
oPeration.
jgnitionsysten
of a conventional
anatysis
wecanusethepreceding
ngin
mechanical
switchi
hasreptaced
switching
whyetectronic
to exptain
and
on fuel econorny
First,the currentemphasis
today'sautomobites,
a sparkptlg witha widergap.'[his,jn turn,
requires
exhaust
emissions
(up
highervottages
sparkptugvoltageThese
requires
a higheravajtabLe
Etectronic
mechanjcaL
switching.
with
be
achieved
40
kV)
cannot
to
jn the primary
windjngof
atsopermjtshigherinitjalcurrents
switching
is
in thesystem
energy
theinitialstored
Thismeans
theautotransformer.
condiand
running
mixtures
range
of
fuelair
a
wider
and
hence
larger,
crrcuite[mswitching
Finalty,
theeLectronic
tionscanbeaccommodated.
effects
thedeteterious
Thismeans
inatestheneedforthepointcontacts.
from
the
system.
be
removed
point
arcjng
can
contact
of
bytryingchapter
Perspective
yourunde5tanding
of theProctical
N?TE:Assess
8.64
and
8,65.
Problens
Summary
The characf€risticequation for both the parallel and
seriesRLC circuitshasthe form
s2+20s+o3:0'
where d : 1/2RC for the parallel circuit, d : R/24 for
the seriescircuit, and (,6 = 1/aC for bot}l the paralel
and seriescircuits.(Seepages287and 308.)
The roots of the characteristic equation are
"'z =
o + \/"'-
(Seepage288.)
ofseriesard
The form of the naturalard stepresponses
parallel RtC circuitsdependson the valuesof I and
rofi;suchresponsescan be oterdamp€d,underdamped,
or critically rlamped.These terms describe the impact of
the dissipative element (R) on the respons€ The nePer
ftequency,a, reflectsthe effectofR. (Seepage289.)
The responseof a second-ordercircuit is overdamped,
underdamped, or critically damped as shown in
Table8.2.
In determinhg the nafuml respoNe of a second-order
clrcuit, we fust detemine whether it is over-, uncler', or
critically damped,and then we solve the appropiate
equationsas shownin Tirble8.3.
In determining the st€p respons€of a second-order cir'
cuit, we apply the apFopriate equations depending on
the damphg, as shown in Table 8.4
For each of the three forms of response,the unknown
coefficients(i.e.,the,4s, B s, and ,s) are obtained by
evaluating the circuit to find the initial value of the
response,r(0), and the initial value of the first derivadi(0)/lr.
tive of the response,
When two integrating amplifien v,/ith ideal op amps are
connected in cascade,the output voltage of the second
integrator is related to the input voltage of the fiISt by an
ordimry second-order differential equation Therefore'
the techniquesdeveloped in this chapter may be used to
analyze the behavior of a cascadedintegrator' (See
page312.)
We can overcome the limitation of a simple integrating
amplifier-the saturation of the op amp due to charge
accumulatingin the feedbackcapacitor-by placing a
resistor in parallel with the capacitorin the Ieedback
path. (Seepage314.)
Prcbtenrs321
TABLE
8.2
TheResponse
ofa S€lond-Order
Clrcultis overdanped.
Und€rdanped,
or Critic.ttyDamped
TXe Circuit is
When
Qualitrlive Nrture of the ResDorse
"'>
Ttre voltageorcurrent approaches
its final valuewithout oscillation
,3
Underdamped
d <.4
The voltaSe or cunent oscillales about its final value
Criticallydanped
i =;t,
The voltage or current is on the verge of oscillaring about its firal value
TABLE
8,3 In Determining
the NattllatRespons€.of
a Second-Order
Circuit,WeFirstDetermine
Wherher
it is Over-,Und€r,
or Criticatty
oamped,
andlhenWeSotvethe Appropriate
Equations
Danping
Natural R€sponseEquations
CoefiicientEqurlions
x(t)=ACt'+Aze,t
r(0):Ar+A,:
dr/tlt(o) - 4\+
x(r) : (Br cos@dr+ B, sino,rre-"r
Azsz
rlo) = Bi
dx/.tt(ot:-aBt+@a8,.
where
o, = \,ttCrilicaly darnped
TABLE
8.4
x(r) : (Dn + D)e "t
xlo) : Dr,
.1r/dt(0)= Dt
dD2
In D€termining
the StepResponse
of a Second-order
Circult,WeApplythe Appropriat€
tquationsDepending
on the Danping
Danping
Step Respome Equationsr
Coefricient EquatioIs
r(t)=xr+A\e\t+Aferl
x(0)=xt+A\+Ail
dx/ddo): A\ sr + Ai s2
x(t) = xt + (B'rcos@dt+ Bi sitioat)e-"t
x(0)-Xr+B\:
dx/.11(D): -dB\ + oraBi
Giticallydamped
x(t) - Xt + D\te-"' + Die "'
r(0\=xt+Dil
dx/dt(o)=D\-dDi
'wheie xl is thefinalralue ofr(r).
Frob[ervls
Sections8.1-8.2
8.1 The resistance,inductance,and capacitancein a
parallelnlc c cuit are 5000O, 1.25H, and 8 nF,
respeclively.
a) Calculate the roots of the characteristic equation
tllat describethe voltageresponseofthe circuit.
b) Will the responsebe over-,under-,or critically
damped?
.) What value of R will yield a dampedfrequency
of6 krad/s?
What are the roots of the characteristicequation
for tlle valueoIR found in (c)?
e) WtratvalueofR will resuitin a criticallydamped
response?
322
NatuEtand
StepResponses
ofRICCiruits
8.2 Suppose the capacitor in the circuit shown in
Fig. 8.1 has a value of 0.05 pF aDd an initial voltage
of 15 V. The initial curent in the inductor is zero.
The rcsulting voltage responsefor I > 0 is
'noor
+ 20P 2oolro'V
,(l, 5e
8.7 The resistance in Problem 8.6 is increased to
m'j.r 2.5kO. Find the expressionfor ?,(/)for t > 0.
8.8 The resistance in Pioblem 8.6 is increased to
EFlc 12500/3 O. Find the expressionfor o(t) for r > 0.
a) Determine the numerical values of R, a, a,
b) CalculatetRo), ir(t), and ic(t) for t > 0*.
8.3 The natural voltage response of the circuit in
Fig.8.1is
o(t) : 125ar0mr(cos
30001 2sin3000Dv, r>0,
when the capacitor is 50nF. Fird (a) t; (b) n;
(c) Yo;(d) 10;and (e) ir0).
8.4 The voltage responsefor the circuit in Fig. 8.1 is
known to be
D(t) = Dfe4aoot+ D2e-aatu, t > o.
The initial current in the inductor (It is 5 mA, and
the initial voltage on the capacitor (yd is 25 V The
inductor has an hductance of 5 H.
a) Fitrd the value of R, C, D\, and.D2.
b) Find ic(t) for / > 0+.
8.5 The idtial value of the voltage ?Jin the circuil in
Fig. 8.1 is zero, and the initial value of the capacilor
cunent, ic(O-),is 15 mA. The expressionfot the
capacitor currert is known to be
i"(t) = AF16t+
A2eat' t>o+,
when R is 200 O. Find
a) the valueofa, (,o,a, C,Aband A)
( ,,,,.0''!o)=
.liL(0)
r?ip(o) u(0)
d t :
1 rc(o)\
L - R
c
)
b) the expressionfor r(t), t > 0,
c) the expressionfor in(t) > 0,
d) the expressionfor ir(l) > 0.
8.6 The ciicuit elementsin the circuit in Fig.8.1 aie
3 ' c R : 2 k O , C = 1 0 n F ,a n d a : 2 5 0 m H . T h e i n i tial inductor curert is 30mA, and the initial
capacitorvoltageis 90V
a) Calculatethe initial current in each branch of
the circuit.
b) Find 0(t) for t > 0.
c) Find iz(r) for r > 0.
8.9 The natuml responsefor the circuitshownin Fig.8.1
is known to be
-o'-.'300)v.
/>-.
u ( / )- - 1 2 ( r
If C : 18 pF, find i1(0') in milliamperes.
8,10 In the circuil shown in Fig. 8.1, a 5 H inductor is
6tr4 shunted by a 8 nF capacitor, the resistor R is
adjusled for c tical damping, Vo25V, and
1o = -1 mA.
a) Calculate tle numedcal value of R.
b) Calculate r'(t) Ior t > 0.
c) Find 0(t) when tc(t) = 0.
d) What percertage of the initially stored energy
remains stored in the circuit at the instant
ic(r) is 0?
8.11 In the circuit in Fig. 8.1, R = 2r}, a = 0.4H,
Bdo c:0.25F,U.J:0V,and10 = -3A.
a) Find r,(t) for t > 0.
b) Find the filst three values of I for which do/dt
is zero. Let these values of I be denoted t1, t2,
and 13.
c) Showthat 13- 11= Zt.
d) Show thaft2 - 11: Zd/2.
e) Carcuhte?,(11),
?,(tr),and 0(t3).
f) Sketch0(l) versustforO < t < t2.
8,12 a) Find 2(l) for I > 0 in the circuit in koblem 8.11
if the 2 O resistor is removed from the circuit.
b) Calculate the ftequency of t (t) in hertz.
c) Calculatethe ma,\imum amplitude of r,(t) in volts
8.13 Assume tlle underdamped voltage response of the
circuitin Fig.8.1is *ritten as
r(t) = (Ar + A)a"'cos a,i + j(At - Az\e-tsinadt
The initial value of the inductor current is 10, and
t}le initial value of the capacitor voltage is y0. Show
tlat ,4, is the conjugate of,41. (lllrt Use the same
processas oudinedin the text to find,41 and ,42.)
Prob.€ms
323
8.14 Show that the results obtained from Problem 8.13rharis.rhe expression\
tor,4r atrd.4r-are consistent with Eqs.8.30and 8.31in the text
820 The resistor in the circuit of Fig. P8.19is inffeased
tstrc from 1.6 kO to 2 kO and t]le inductor is deffeased
from 1 H to 640 mH. Find zJ.(t) for t > 0.
8.15 The resistor in the circuit in Example 8.4 is changed
strr! to 4000/\2 o.
8.21 The resistorir the c cuit of Fig.P8.19is deqeased
6dc from 1.6kO to 800O, andtle inductoris deqeased
from 1 H to 160 mH. Find oo(t) for I > 0.
a) Find the numerical expressionfor o(1) when
t>0.
b) Plot rr(1) ve$us t for the time intenal
0 < t < 7 ms. Compare this response with
the one in Example 8.4 (R=20kO) and
Example8.5 (R - 4 kO). In particular,compare
peak values oI o(t) and the times when these
peak values occur.
8,16 The switch in the circuit of Fig. P8.16 has beetr in
6plft position a for a long time. At t = 0 the switch
moves instantaneously to position b. Find ?r,(t) for
t>0.
Figffe P8.16
t=0
Section 8.3
8.22 For the circuit in Example 8.6, find, for 1> 0,
6*E (a) ,0); (b) iR(t); and (c) tc(t).
8.23 For the circuit in Example 8.7, find, for , > 0,
Em (a) ,,0) and (b) ic(1).
8.24 For the circuitin Example8.8,find o(t) for t > 0.
8.25 Assume that at fhe instant the 15 mA dc current
Er!4 souce is applied to the circuit in Fig. P8.25,the initial curent in t}le 20 H inductor is -30 mA, and the
initial voltage otr the capacitor is 60 V (positive at
the upper terminal). Find the expressionfor ir(r)
for t > 0 ifR equals800O.
Figure
P8.25
8.17 The capacitor in the circuit of Fig. P8.16 is
decreasedto 1 rF and the itrductoris mueasedto
10H. Find o,(t) for 1 > 0.
E.18 The capacitor in the circuit of Fig. P8.16 is
decreasedto 800pF and t]le inductor is increasedto
12.5H. Find t"(t) for r > 0.
8.19 The two switchesin the circuit seen in Fig. P8.19
6flc operate synchronously.When switch 1 is in position
a, switch 2 is in position d. When switch 1 moves to
position b, switch 2 moves to position c. Svritch t has
beenin positiona lor a long rime.Ar r 0. rhe
switchesmove to their altenate positions.Find
!,aG)forr>0.
l) '.,ullr" *,t
826 The rcsistancein the circuit in Fig. P8.25is changed
Atrc ro 1250O. Fird ir(, for r > 0.
827 The resistancein the circuit in Fig. P8.25is changed
tsdi! to 1000O Find ir(r) Ior, > 0.
8.28 The switch in the circuit in Fig. P8.28has been open
E .r for a long time before closing at t = 0. Find o"(t)
for r > 0.
figureP8.28
Figure
P8.19
1ko
60v
62.5nF
1.6kO
6.25pF
324
Natuntand
SieoResDonses
of{r Cncuits
8.29 a) For the circuit in Fig.P8.28,find i, for I > 0.
""" b) Showthat your solutionfor i, is consistentwith
the solutionfor r,. in Pioblem8.28.
Figure
P8.33
1kO
830 Thereis no eneryystoredin the circuit in Fig.P8.30
6pr.rwhenthe switchis closedat r = 0. Find r"(t) for
t>0.
,-r- 10/,r'
1.6H
FigoreP8.30
r:0'
|
834 Use the circuit in Fig. P8.33
'strcr
a) Find the total energy deliveied to the incluctoi.
b) Find tlle total energy deiivered to the equivalent
+
6.25pF,1c) Find the total energy delivered to the capacitor.
d) Find the total energydelivered by the equivalenl cturenl source,
e) Check the results of parts (a) through (d)
against the conseraation of energy pdnciple.
8.31 a) For tle circuit in Fig. P8.30,find i, for I > 0.
"""
b) Show that your solution for i. is consistent with
the solution for 0, in Problem 8.30.
8.32 The switch in the circuit in Fig. P8.32 has been
6trc open a long time before closing at t = 0. At the
time the switch closes,the capacitorhas no stored
energy.Find tto for t > 0.
8.35 The switch in the circuir in Fig. P8.35 has been
EPI.[ open a long time before closingat t - 0. Find iL(l)
fort > 0.
tigureP8.35
Figure
P8.32
3000tr+ 31.25
rnH=500
72V
1 25rlF
8.33 The switch in the circuit in Fig. P8.33has been open
^"" a long time before closing at t : 0. Fhd
a) ?o(t)for, > 0+,
b) iro) foi t > 0.
8.36 Switches 1 and 2 in the circuit in Fig. P8.36are syn6trt! chronized. Wlen switch 1 is opened, switch 2 closes
and vice versa. Switch t has been open a long tirne
beforeclosirg at. : 0. Find ir0) for r > 0.
rigureP8.35
5kO
\
)240v
J w r c o1r v
Switch
-1-
\,=0
Switch 2
-5 sF
trl 8 0 H
1ko
(
60mA
PlobLems325
Section 8,4
8.37 The initial energy stoied in the 50 nF capacitor in
the circuit in Fig. P8.37 is 90 /iJ. The initial energy
stored in the inductoi is zero. fhe roots of the characteristic equation that describesthe natuml behavior of the current i are -1000 ai and 4000 s-1.
a) Find the numerical values of R and a.
b) Find the numerical values of i(0) and di(0)/d1
lrnmediately after the switch has been closed.
c) Find t(t) for t > 0.
d) How many miqoseconds after the switch closes
does the curent reach its maximum value?
e) What is the maximum value of i h milliamperes?
f) Fbd 0r(t) for r > 0.
FlgureP8,37
FigueP8.40
{rvJ
72Ov
g.41 In the circuit in Fig. P8.41,the resistoris adjusted
Mc for critical damping. The initial capacitor voltage is
90 V and the initial inductor current is 24 mA.
a) Find the numerical value of R.
b) Find the numerical values of i and dt/d/ immediately after the switch is closed.
c) Find 0c0) for t > 0.
Flgure
P8,4r
8,38 The current in the circnit in Fig. 8.3 is knom to be
= Bre-3m!cos600t + B2?{e sin 600t, I > 0.
The capacitor has a value of 500 /rF; ttre initial value
of the curent is zero; and the initial voltage on the
capacitor is 12 V Fhd the values of R, a, 81, ard ,2.
8,39 Fhd the voltage aqoss the 500 ,.F capacitor for the
circuit describedin Prcblem 8.38.Assume tle refererce polariry for the capacitor voltage is positive at
t]le upper termiftl.
8.42 fhe switch in the circuit in Fig. P8.42has been In
position a for a long time. At I = 0, the swrtch
moves iNtartaneously to position b.
a) What is the initial valueof?)d?
b) what is the initial valu.eof dDJ dt?
c) What is the nume cal expressionfor 2,,(t)for
t>0?
FlgoreP8.42
8.40 The switch in the circuit shown in Fig. P8.40 has
6rft been closed for a long time. The switch opells at
I : 0. Find
a) t"(l) for t > 0,
b) 0"(r) for t > 0.
326
Natu6tand
StepResponses
ofRlCCircuiis
8.43 The make-before-break switch in the circuit shown
6PIcr in Fig. P8.43has been in position a for a long time.
At , = 0, the switch is moved instantaneouslyto
position b. Find i(t) for t > 0.
FigureP8.43
100mH
8.48 The switchin the circuit shownin Fig.P8.48has
beenclosedfor a long time beforeit is openedat
t - 0. Assumethat the circuit parametersare such
that the rcsponseis underdamped.
a) Derive tle expressionfor o,(t) as a function of
us, ct,od, C, andR fot t > O.
b) Derive the expression
for the valueof I when
the magnitudeof r', is ma-\imum.
FlglreP8.48
8.44 The switch in the circuit shown in Fig. P8.44has
B{c been closed for a long time. The switch opens at
t : 0. Find ?a(t) for t > 0.
tigun P8.44
30rI
10()
8()
10f}
100v
8.45 The initial energy stored in tle circuit in Fig. P8.45
6n(r
i. ?er^
Fin,l
r /,\ +^r,
>
n
8.49 fhe cftcuitparametenin the circuitof Fig.P8.48
a r e R = 1 2 0 O , a : 5 m H , C : 5 0 0 n F F ,a n d
0s = -600 V
a) Expressr'"(t) numericallyfor t > 0.
b) How manymiqosecondsafter the switchopens
is t}le inductorvoltagemaximum?
c) Wlat is the maximumvalue of the inductor
voltage?
d) Repeat(a)-(c)withR reducedto 12O.
8.50 The circuit showtr in Fig. P8.50 has been in opera6'IG tion for a long time. At t = 0, the sourcevoltage
suddenly drops to 100V Find o,(r) for r > 0.
tigureP8.45
Figure
P8.50
4()
40 mH
8.,16the capacitor in the circuit shown in Fig. P8.45 is
changed to 100 nF. The idtial energy stored is still
zero Find od(t) for r > 0.
8.47 The capacitor in the circuit shown in Fig. P8.45 is
changedto 156.25nF. The initial energy stored is
still zerc. Find 0o(l) for t > 0.
8.s1 The swilch in the circuil of Fig. P8.51 has been in
position a for a long time. At I = 0 the switch
moves instantaneously to position b. Find
a) r'10')
b) du.(o')ldt
c) ?,'o(t)for t > 0.
tigureP8.51
the switch has been closed.At the time when the
switch is closed, tlere is no energy stored in either
the capacitoror the inductor. Find the rume cal
values of -Rand L. (H,nL Work Problem 8.53firsr.)
12kO
,.,,, mv
28v **+
8.52 The two switchesin ttre circuit seen in Fig. P8.52
sflft operate s]'nchronously.men switch 1 is in position a,
switch 2 is closed.wlten srvitch1 is in position b,
switch 2 is open. Svritch t has been in position a for
a Iong time. At t : 0, it movesinstantaneouslyto
position b. Find ?.G) for t > 0.
8.55 Show that, if no energy is stored in the circuit
showr in Fig. 8.19 at the instant rJsjumps in value,
then db./dt eqlralszeroatt =0.
8.56 a) find the equarion foruo(r) for 0 < r < r.urin
the circuit sho$n in Fig.8.19if o,1(0) : 5 V and
?,,(0) = 8 V.
b) How long does the circuit take to reach
satumtion?
8.5? a) Rework Example 8.14with feedbackresistors
R1 and R2removeil.
b) Rework Example 8.14with r',(0) : -2v ^nd
0,(0)= 4v.
Figule
P8.52
200r}
400O 100mH
b\
|
+
160v1600o
8.53 Assume that te capacitor voltage in tle circuit of
Fig. 8.15 is underdamped.Also assumethat no
energyis stored in the circuit elementswhen the
switch is closed.
a) Showthat doc/dr = (SolaiVe-t sn(l.at.
b) Show that doc/dt :0 when t = nr/r,,r, where
n : O,7,2,....
c) Let t"= nnld,j, and show that !'c(t,)
:v - v (-1)"e-*"1"
d) Showthat
"
1. oc1i-V
Td'- uc(t) - V
whereZd = t3 - tr,
854 The voltage acrossa 200 nF capacitor in the circuit
of Fig. 8.15 is described as follows: After the switch
has been closedfor severalseconds,the voltage is
constant at 50 V The fbst time the voltage exceeds
50 Y it reaches a peak of 63.505V This occurs
?r/12 ms after the switch has been closed.The second time the voltage exceeds50 V it reachesa peak
of 50.985V This secondpeak occursd4ms aftei
Section 8,5
E.5E The voltage signal of Fig. P8.58(a) is applied to
tsr.r the cascaded integrating amplifiers shown in
Fig. P8.58(b). There is no energy stored in the
capacitorsat the instant the signalis applied.
a) Derive the numerical expressionsfor to(t) and
r',l(f) for the time intervals 0 < t < 0.2s and
0.2s<t<t"d.
b) Compute the value of t",,.
Figure
P8.58
,s (nv)
328
Naturatand
StepResponses
off|r Cjrcuits
8,59 The circuit in Fig. P8.58(b) is modified by adding a
*fl( 250 kO resistor in parallel with the 2 pF capacitor
and a 250kO resistor in parallel with the 4pF
capaciror.
As in Problem8.58.thereis no energ!
stored in the capacitorsat the time the signal is
applied. Derive the numerical expressionsfor ,o(l)
and o,1(t) for the time intervals0 < I < 0.2 s and
r > 0.2s.
8.60 a) Derive the differential equation tlat rclates the
output voltage to the input voltage for the cir,
cuit shownin Fig.P8.60.
b) Compare the result with Eq. 8.75 when
RrCr: R2C2= RC in Fig.8-18.
c) What is the advantageof the circuit shown in
Fig.P8.60?
is availableand by successive
integrationsgenerates d.r/dt ard r. We can synthesizethe coeffi,
cients in the equations by scaling amplifien, and
w^ecan^combine the terms requircd to genemte
d:x/dt" by rlsir]'g a summing amplifiei. With
theseideasin mind, analyzethe inlerconnectron
shown in Fig. P8.61(b).In pa iculai, describe
the purpose of each shadedarea in the circuit
and describethe signalat the points labeledB,
C, D, E, and R assumingthe signalat A representsdr4df.Also discussthe parametersR;Rl,
C1; R2,C2; R3, Ra; R5,.lR6;
and R7, Rs in terms
of the coefficients in the differential equation.
Seclions8.1-8.5
8.62 a) Deive Eq.8.92.
,lli;L!{"',b) Derive Eq. 8.93.
c) DeriveEq.8.9.
FlgureP8.50
8.63 Derive Eq.8.99.
8.64 a) Using the same numedcal values used in the
pl$fi*ii€ Practical Perspective example ir the text, find
the instant of time when the voltageacrossthe
capacitor is ma-{imum,
b) Find the maximumvalue of 2c.
c) Comparethe valuesobtainedin (a) and (b) witl
t-". and 2,.0-aJ.
8.61 We now wish to illustmte how several op amp circuits can be interconnectedto solve a differential
equation.
a) Derive the differential equation for the springmass system shown in Fig. P8.61(a). (See
page 329.)Assume tlat the force exerted by ttre
springis directly proporional to the springdisplacement,that the massis constant,and that
the ftictional force is diectly prcportional to the
velocity of the moving mass.
b) Rewrite the differential equation derived in (a)
so that the highest order derivative is expressed
as a ftrnction of all the other terms in the equa
tion. Now assumethat a \olt^ge equalto C xI dt2
8.65 The values of the parameters in the circuit in
pP#fl*i;lFig.8.21 are R = 3o; Z, = 5mH; C = 0.2spF;
Vd"= 72Y; ard a = 50. Assume the switch opens
whell the primary windingcurrentis 4A.
a) How much energy is stored in the circuit at
1=0-?
b) Assumethe sparkplug doesnot fire.Wlat is the
ma-ximumvoltage available at the spark plug?
c) What is the voltage acrossthe capacitorwhen
the voltageacrossthe spark plug is at its maximum value?
tigureP8.61
SinusoidaI
Steady-State
AnaLysis
Thusfar, we havefocusedon
9.1 Th€Sinusoidat
Sour.ep. 332
p. 335
9.2 Th€Sinusoidat
Response
p- 337
9.3 ThePhasor
9.4 ThePassive
CncuitEtements
in the
Frequencf
Donainp. 34?
9.5 Kirchhofi's
Lawsin th€ FreguencJ
Doln'ainp. 346
9.6 Series,
Parattet,
andDetta-to-Wye
p. 348
SimpLifications
9.7 Source
Transfonnations
and
Th6venin-Norton
Equivat€nt
Circuitsp. 155
9.8 TheNode-VoLtage
l,lethodp. 359
9.9 Theliesh-Cuffent
l4ethodp. 360
p. 361
9.10TheTransformer
9.11ThetdealTransform€r
p. 365
9.12 PhasorDiagramsp. 372
phasorconcepts
1 Understand
andbe ableto
perfofma phasortransformard an inveEe
2 8e abteto transform
a circuitwith a sinusoidat
sourcejnto the frcquency
domainusjngphasor
3 Knowhowto usethe fottowjnqcircuitanatysjs
techniques
to sotvea circuitjn the ftequency
.
.
.
Knchhoffstaws;
Series,parattet,
anddelta-to-wye
.
Vottageandcunentdivision;
Th6venin
and Nortonequivatents;
.
Nod€-voitage
method,and
.
iqesh-cuffent
method.
4 8e abLeto anatyze
circuitscoitainingtjfear
transformers
usingphasormethods.
5 UndeEtand
the ideattraisformerconstraints
andbe abteto anatyze
circujtscontainingideat
tnnsformeuusingphasormethods.
330
with constantsources;in
this chapterwe are now ready to considercircuits energizedby
time-varyingvoltageorcurent sources.lnparticular,weare interestedin sourcesin whichthe valueofthe voltageor currentvades
sinusoidally.
Sinusoidalsourcesand their eflecton circuit behavior
fbrm an impo ant areaof studyfor severalreasons.
First,the generation, transmission,distribution.and consumptionof electic
energyoccur under essentiallysinusoidalsteady-state
conditions.
Second,an understandingofsinusoidalbehaviormakesit possible
to predict the behavior of circuits with nonsinusoidalsources.
Third, steady-state
sinusoidalbehavioroften simplifiesthe design
of electricalsystems.Thus
a designercanspellout specifications
in
termsof a desiredsteadystatesinusoidalresponseand designthe
circuit or systemto meet thosecharacteristics.
If the dcvicesatisfies the specifications,
the designerknows that the circuit will
respondsatisfactoly to nonsinusoidalinputs.
The subsequentchaptersol this book are largely basedon a
thoroughunderstandingof the techniquesneededto analyzecircuitsddven by sinusoidalsources.
Fortunately,the circuit anatysis
and simplificationtechniquesfirst introduced in Chapters 1-4
work for circuitswith sinusoidaias well as dc sources,so someof
the materialin this chapterwill be vety familiar to you.The challengesin first approachingsinusoidalanalysisincludedevelopilg
the appropdatemodeling equationsand working in the mathematicalrealm of complexnumbers.
PracticaI
Perspective
A Househotd
Distribution
Circuit
Power
systems
thatgenerate,
transnrit,
anddjstribute
eLectricatpoweraredesjgr€d
to operate
in the sjnusoidaL
steady
state.Thestandard
househotd
distribution
circuitusedin the
ljnitedStatesis the lhrce-wlte,
240/72A
V circuitshownin
theaccompanying
figure.
Thetransformer
is usedto reduce
the utititydjstrjbution
voltage
frofi 13.2kVto 240V.Th€center
tap onthesecondarywjndingprovides
the 120V service.
Theoperating
frequency
of powersystems
in the ljnitedStates
is 60 Hz.Soth
50 and60 Hzsystems
arefoundoutside
the ljnitedStates.
Thevottage
ratingsatluded
to abovearermsvatu€s.
Thefeason for definingan rmsvalueof a time-varying
sjgnaljs
in Chapter
10.
explained
331
332
sjnusoidaL
Steady
State
Anatysis
9.1 TheSinusoidat
Source
A sinwoidal voltag€ source (independent or dependent) produces a volt
age that varies sinusoidally with time. A sinusoidal curr€nt source (independent or dependent) produces a curent that varies sinusoidally with
time. In rcviewing t]te sinusoidal function, we use a voltage source,but our
obseFations also apply to current sotuces,
We can express a sinusoidally varying function with either the sine
function or the cosine function. Although either works equally well, we
cannot use both functional forms simultaneously.We will use the cosine
functionthroughoutour discussion.
Hence,we write a sinusoidallyvarying
voltageas
0=Y-cos(ot+d,).
(e.1)
To aid discussionof the parametefi in Eq.9.1, we show the voltage
venus time plot in Fig.9.1.
Note that the sinusoidalfunctior repeatsat regularinte als.Sucha
functionis calledperiodic.One panmeter ofinterest is the lengthof time
requiredfor the shusoidalfunctionto passthroughall its possiblevalues.
This time is rcfered to as the pedod of the function and is denoted Z. It is
measued in secondsThe reciprocalof 7 givesthe number of cyclesper
second,or the ftequency,of the sinefunction and is denotedi or
figur€9,1A A sinusoidaL
vottage.
r=|
\9.2)
A cycle per second is referred to as a hertz, abbreviated Hz. (The term
ctcla per se@nd ftrely is used in contemporary technical literature.) The
coefficient of t in Eq. 9.1 contains the numerical value of f or /. Omega (id)
representsthe angular frequency of the sinusoidal function, or
a = 2Tf :21tlT (tadians/seco ).
(e.3)
Equation 9-3 is based on the fact that t}le cosine (or sine) function passes
through a complete set of values each time its argument,o,, passes
though 2' rad (360'). From Eq. 9.3,note that, whenevert is ar integral
multiple of f, the aryumentot increasesby an integmlmultiple of 2r md.
The coefficient y, gives the maximum amplitude of the sinusoidal
voltage.Because+1 boundsthe coshe function,ty- boundsthe amplitude. Figure 9.1 shows these chancte stics.
The angle4 in Eq.9.1 is known as the phaseangle of the sinusoidal
voltage.It detemines the value of the sinusoidalfunction at I : 0; thereforc, it fixes the point on the periodicwave at which we start measudng
time. Changing the phase angle d shifis the sinusoidal function along the
9.1 TheSinusoidatSourc€
333
time a-xisbut has no effect on either the amplitude (y.) or the angular fre'
quency ((r). Note, for example,that reducing 4 to zero shifts the sinusoidal
functionshoM il Fig.9.14/@ time units to the riglt, asshoxn in Fig.9.2.
Note also that if d is positive, tle siousoidal function shifts to the left,
whereasif d is negative,the function shifts to the right. (See Problem 9.4.)
A conrment vrilh iegard to the phase angle is in ordei o, and d must
carry the same units, becauset}Iey are added together in the argument of
tlle sinusoidal furction. With (dt expressedin iadians, you would expect d
to be also. However, d nomally is given in degrees,and .rt is converted
frcmFis.9.1
vottase
from radians to degreesbefore t]le two quantities are added,We continue tigure9.2A Thesinusoidat
to thenshtwhen
this bias toward degees by expiessing the phase angle in degrees.Recall shjfted
d : 0.
from your studies of tdgonometry tlat the convenion from radians to
degreesis givenby
''//N,/A
i[",
'"qf".v
\.v
(numberof degrees)= 1q{lnumber of radians).
(e.4)
Arother important characteristic of the sinusoidal voltage (or currenr) is its rms value.The rms value of a periodic function is defined as the
square root of the mean value of the squared funclion, Hence, if
D = V^cos(ot + d), tlle rms valueof lJis
=rEl'*\,;
,,^"
*",,.*
*,
(e.5)
Note ftom Eq. 9.5 that we obtain ttre mean value of the squared voltage by
integrating I over one period (tlat is,ftom t0 to t0 + I) and then dividing
by the range of integration, r. Note furtller that the starting point for the
inregratjon
roi. arbirrar'.
The quanril! underlhe fadicai.igDiD Eq.4.5reduces
to vt 2. (See
Problem 9.6.) Hence the rms value of o is
(9.6) < rns vatueof a sinusoidal
voltagesolrce
The rms value of the sinusoidal voltage depelds only on t]re maximum
amplitude of ?, namely, y-. The rms value is not a function of either the
frequency or the phaseangle.We stressthe importance of the Ims value as
it relatesto power calculationsin Chapter10 (seeSection10.3).
Thus,we can completely describea specificsinusoidalsignaliJ we know
its frequency,phaseargle, and amplitude (either the ma-'rimumor tlte rrns
value).Examples9.1,9.2,and 9.3 ilustrate thesebasicpropediesof the
sinusoidalfunction. In Example 9.4,we calculate the rms value of a periodic
function, and in so doing we clariry the meaning oI root mean squarc.
334
SinusoidatSteady-StateAnatysis
Findingthe Characteristics
of a SinusoidaI
Current
A sinusoidal current has a ma-\imum amplitude of
20 A. The current passesthrough one complete cycle
ir 1 ms.The magnitude of the cu ent at zerc time
is 10A.
a) What is the ftequency of the current in hertz?
b) What is the frequencyitrradiansper second?
c) Wrire lhe expressioD
lor itr' u(ing lhe cosine
function. Express d in degrees.
d) Wlat is the rms valueof the current?
Solution
a) From the statementof ttre problem,Z : 1 ms;
hence/:1/Z:1000H2.
b) ti : 2nf = 2mut radls.
c) Wehavet(r) : I,cos (ot + d) : 20cos(20002r
+ d), bur (0) : 10A. Therefore10 - 20cosd
andd = 60".'Ihus the expiessionfor i(l) becomes
t(r) = 20cos(2000't+ 60').
d) Fromtle derivationof Eq.9.6,thermsvalueof a
sinusoidalcurrent is 1-/\4. lherefore the rms
valueis 20/\4, or 14.14A.
Findingthe Characteristics
of a Sinusoidat
Vottage
A sinusoidal voltage is given by the expresston
1' : 300cos(1202, + 30').
a) What is the period of the voltage in milliseconds?
b) What is the frequencyin hertz?
c) What is the magnitude of a at t = 2.778ms?
d) Whatis the rms valueof ?J?
Sotution
a) From tllle erFession for t, a : 120r :f2'dls.
Bec?l$ea=2n/T,T -2nla =e s,
or 16.667ms.
b) The frequency is 1/2, or 60 Hz.
c) Frcm (a), o = 2n1L6.667:th]uf.,nt t = 2.'7'78
m\
ol is nearly 1.047 rad, or 60'. Therefore,
0(2.778ms) = 300cos(60" + 30") = 0v.
:21213v.
d)u^, - 300l.va
Transtating
a SineExpression
to a CosineExpression
We can translatethe sine function to the cosine
tunclionby subtracting90' (r/2 rad) from the aryumentoflhe sinefunction.
Sotution
a) Verification involves direct application of the
trigonometric identity
c o s ( a- B ) : c o s a c o s B + s i n d s i n B .
Weleta: (dt + d andB : 90".As cos90' = 0
and sin90" : 1, we have
a) Verify this tmnslationby showingthat
sin(.,,t+ d) : cos(rrt+ d - 90').
co(a
b) Usethercsultin (a) to expresssin((or+ 30")as
a cosinefunction.
B ) = s i n a - s i n ( ' r t + d ) = c o s ( o l+ P
b) From (a) we have
sin(.rt + 30") - cos(or + 30' - 90') - cos((,r
90").
60").
9.2
lhe Sjnusoidat
Response 335
Catculating
the rmsVatueof a Triangutar
Waveform
Calculate tle Ims value of the peiiodic tdangllar
cureDrsho\ n in Fig.c.3.Express)our a0s$efiD
terms of the peak cunent 1/.
rP. r/4 0
Figure9.4 A r' versurr.
The anal)'ticalexpressionfor i in the interval 0 to
T/4is
4T
Fiqurc9.3 l Pefiodic
tnangular
current.
i=1r.0<t<T14.
I
Solution
FromEq.9.5,thermsvalueofi is
The area under the squaredfunction for one
periodis
+1,"^"
Interpretingthe integralunderthe radicalsignas
the areaunderthesquaredfunctionfor aninterval
of one period is helpful in finding the rms value.
The squaredfunction with the areabetween0 and
f shadedis sho\r.nin fg. 9.4,which also indicates
that for this particular function,the areaunderthe
squarcdcurrent for an inteflal of one period is
equalto four timesthe areaunderthesquared
currentfor theintenal0 to f/4 seconds;that
is,
p+r ^
t2,T
tr416t',^
i'dt:4 |
I
-fit='. )
tJh
Ja
Themean,or avemge,valueof the functionis simply
the areafor oneperioddividedby the period.Thus
" t2T
The Ims value of the crment is the square root of
this mean value. Hence
I.
NOTE: Assesslow undelstanding of thit mate/ial b! tying Chapter Prcblems 9.1,9.5,9.8.
Response
9.2 TheSinusoidaI
Before focusing on the steady-state response to sinusoidal sourceq let's
consider the prcblem in broader termq that is, in terms of the total
rcsponse.Such an ove iew will help you ke€p the steady-statesolution in
perspective.fhe circuit shown in Fig. 9.5 describes the general naturc of
the prcblem.There,o" is a sinusoidalvoltage,or
0,:
Y-cos(@l+ d).
(e.7)
bya sinusojdat
Figure9,5 r AnRl cncuitexcited
voltase
source.
336
Sjnusoidatst€ady-StateAnatysjs
For convenience,we assumethe initial current in the circuit to be zero and
measurctime from the moment the switchis closed.The task is to derive
the expressionfoi i(l) when t > 0. It is similar to finding the step response
of an Rl, circuit, as in Chapter 7. The only difference is that the voltage
sourceis now a time-varyingsinusoidalvoltagerather than a constant,or
dc, voltage. Direct application of Kircbhoff's voltage law to the circuit
shownin Fig.9.5leadsto the ordinarydifferentialequation
tfi , ai = y- cos(.,r+ d),
(e.8)
t]le formal solution of which is discussedin an introductory course in differential equations.We ask those of you who have not yet studied differentialequationsto acceDtthat the solutionfor r N
w,
\,F;;t:
cos(<,- 0)e 'R tt '
, ,u- .
vR'
ul
,cos(dr
d
d).
(9.9)
where d is defined as the angle whose tangent is da/R. Thus we can easily
detemine d for a circuit driven by a sinusoidal source of knor,n frequency.
We can check the validity oI Eq. 9.9 by dererminingthat ir sarisfies
Eq. 9-8for all valuesof t > 0; tlis exerciseis left for your explorationin
Problem9.10.
The fiIst teim on the dght-hand side of Eq. 9.9 is refered to as the
tmnsi€nt component of the current because it becomes infinitesimal as
time elapses.The second term on the dght-hand side is known as the
steady"state component of the solution. It exists as long as the switch
remainsclosedand the sourcecontinuesto supplythe sinusoidalvoltage.
In this chapter,we develop a techniquefor calculatingthe steady-state
response directly, thus avoiding the problem of solving the differential
equation. However, in using this tecbnique we fodeit obtaining either the
tnnsient component or the toral response,which is the sum of the transientand steady-state
components,
We now focuson the steadystateportion of Eq. 9.9.It is impo ant to
remember the following chamcteristics of the steady slate solution:
1. The steady-state
solutionis a sinusoidalfuncrion.
2. The fiequercy of the responsesignal is identical to the frequency of
t]le source signal. This condition is always true in a linear circuit
when the circuit parameters,R, Z, and C, are constant. (ff frcquencies in the responsesignalsare not presentin the sourcesignals,
thereis a nonlinearelementin the circuit.)
3. The maximum amplitude of the steady-stateresponse,in general,
diJfers ftom the maximum amplitude of the source.For the circuit
beingdi.scussqd,-lhe
maximumampliludeof rhe fesponse
signalis
UI V R' d L andthe maKimum
amplitudeol $e .ignalsource
is U-.
4. Th€ phase angle of the responsesignal, in general, differs from the
phase angle of the source.For the circuit being discussed,the phase
angle of the current is d
I and that of the voltage source is 4.
These characteristics are worth remembering because they help you
unde$tand tlle motivatior for the phasor method, which we intoduce in
Section9.3.In particular,note that once the decisionhas been made to
find only the steady-stateresponse,the task is reduced to finding tlle mlximun amplitude and phase angle of the response signal. The waveform
and frequencyof tlle responseare alreadyknown.
NOTE: Assessyow underctandingof this matefial by ttying Chaptel
Prcbkm9.9.
9.3 ThePhasor
The phasoris a complexnumberthat caries the amplitudeand phase
angleinformationof a sinusoidal
function.rThephasorconceptis moted
in Euler'sidentity,whichrelatesthe exponential
functionto the trigonometricfunction;
e ' r "= c o s d l i s i n d .
(e.10)
Equation9.10isimpo antherebecause
it givesusanotherwayof exprcsshg thecosheandsinefunctions.Wecanthinl( of the cosinefunctionasthe
realpart of the exponentialfunctionandthe sinefunctionasthe imaghary
partof theexponential
function;that
is,
cosd: n{d'},
(e.11)
sin0 = S{dr},
(9.12)
ancl
where $t means"tle real part of" and S means"theimaginarypafi oi"
Becausewe have aheady chosen to use the cosine function in analyzing t}le sinusoidalsteadystate (see Section9.1),we can apply Eq.9.11
directly. In particular, we write the sinusoidal voltage function given by
Eq.9.1in the form suggested
by Eq.9.11:
0=Y-cos(ol+d)
= v,,$?idG'+d)]
- vnn Ietatetaj
(9.13)
Wecanmovethe coefficienty. insidethe aryumentof the rcal part of the
function without alteringthe rcsult.we can alsoreversethe order of the
twoexponential
functionshsidetheaigumentandwriteEq.9.13as
t = n{vneEd@'}.
I II you feel a bit uneasyabour cotupler nhbe4
peruse Apperdix B
\9.14)
338
SinusordaLsteddy-SiateAnalysis
In Eq.9.14,notethat the quantity y,er't is a complexnumber that carries
the amplitude and phase angle of the given sinusoidalfunction. This
complex number is by definition the phssor reprcsentation,or phasor
lransform. of the qiver sinusoidal function. Thus
Phasor
transform>
(9.15)
where t})e notation g{yncos (a,t + 4)} is read "the phasortransfom of
y- cos (ot + d)." Thus the phasor tmnsform transfers the sinusoidal function from the time domain to the complex-number domain, which is also
called the fr€queng, domain, since the responsedepends,in general, on .r.
As in Eq. 9.15,throughoutthis book we representa phasorquantity by
using a boldface letier.
Equation9.15is the polar form of a phasor,butwe alsocan expressa
phasor in rectangular form. Thus we rewrite Eq. 9.15 as
Y:U^cosf+jU-si\6
(e.16)
BotI polar and rectangular folms are useful in circuit applications of the
phasorconcept.
One additionalconrmentregardingEq. 9.15is in order.The frequent
occu ence ot the exponential function ete has led to an abbreviation that
lendsitsell to text malerial This abbreviationis the anslenotation
11i!: = letE'
Weusethis notationextensivelyin the materialthat follows.
InversePhasorTransform
So far we have emphasizedmoving frcm ttre sinusoidal function to its phasor tmnsform,However,we may also revene the process,That is, for a
phasor we may write the expression for the sinusoidal function. Thus for
V = 1.001_4:, the expressionfor o is 100cos(or - 26") becausewe
havedecidedto usethe cosinefunction for all sinusoids.
Obsene that we
cannot deduce the value of d from the phasor. The phasor ca ies only
amplitude and phase information. The step of going from the phasor
tmnsform to the time-domain expression is rcferrcd to as frnding the
inversephasor tnnsform and is fomalized by the equation
g- rlv^ei4\ - n {v^erQ
etd},
\9.t7)
9.3 ThePhasor 339
l{y-ere} is read as"the inversephasortransformof
wherethe notation@
"
yfter{ Equation 9.1? indicates that to find tlle inve$e phasor ffansform,
we multiply the phasor by e,oiand then extract the real part of the product.
The phasor transform is useful in circuit analysis becauseit reduces
the task of finding the maximum amplitude and phaseangle of the steadystate sinusoidal responseto the algebra of complex numbe$. The following observations verify t}lis conclusion:
1. The transient component vanishes as time elapses,so the steadystate component of the solution must also satisfy the differential
equation.(SeeProblem9.10[b].)
2. ln a linear circuit diven by sinusoidalsources!the steady-state
response also is sinusoidal, and the frequency of t}le sinusoidal
responseis the sane as the ftequency of the sinusoidal source.
3. Usirg the notationintroducedinEq.9.11,we canpostulatethat the
steady-statesolution is of the form n{Aejpetd}, where A is the
maximumamplitudeof the responseand B is the phaseangleof the
response.
4. When we substitute the postulated steady-srat€solution into the
diJferential equation, the exponential term d'' cancelsout, leaving
the solution for,4 and B in the domai[ of complex numbers.
We illustrate these obse ations with the circuit shown in Fig.9.5 (see
D.33t. We know that the steadv-statesolution for the current i is of the folm
i""(t) : n{I^eiqetd},
(9.18)
where the subscdpt "ss" emphasizesthat we are dealing with t}le steady
statesolution.When we substituteEq. 9.18into Eq. 9.8,we generatethe
expressron
n{jaLl^eiqeid}
+ n{RI^etpeid} = nlune!'tejdt}.
(e.1e)
In dedvingEq.9.19we recognizedthat both differcntiationand multiplicatioflby a constantcan be taken insidethe real part of an operation.We
alsorewrotethe right-handside of Eq.9.8,usingthe notation of Eq.9.11.
From t]le algebra of complex numbers, we know that the sum of the real
parts is the same as the real part of the sum.Thereforc we may reduce the
left-handsideoI Eq.9.19to a singletelm:
n{(joL
+ R)Ineiqejd} : nlv-eiadd}.
(e.20)
Recall that our decision to use t}le cosine Junction in analyzing the
responseof a circuit in the sinusoidalsteadystateresultsir the useoI the
340
SjnusojdalSteady-StaieAnatysis
n operatorin de ving Eq. 9.20.If insteadwe had chosento usethe sine
function in our sinusoidal steady-state analysis!we would have applied
Eq.9.12directly,in placeofEq.9.11, and the resulrwould be Eq.9.21:
3 {(jaL + R)I -eipetd} : 3 {v.etdei.,}.
(e.21)
Note that the complexquantitieson either sideofEq.9.21 are identicalto
thoseon either sideofEq.9.20.Whenbolh the real and imaginarypartsof
two complex quantities aie equal, then the complex quantities are themselvesequal.Therefore,tom Eqs.9.20and 9.21,
(j@L + R)I^etu = Unda,
R+ joL
\9.22)
Note that db' has been elimimted from the detemination of the amDlilude r/.r and phaseangle{6r ot lhe respon.e.t}us. tor lhis cLcujl,ihe
taskof finding/. and p in\olve.rheatgebraic
manipulation
of rhecomy-C'and R - idl. Nole thal \ e encounrered
ple\ qudntities
bolh polaf
andreclanguiar
iormc.
An important warning is in order: The phasor transform, along with
the inversephasortransform,allowsyouto go back andforlh betweenthe
time domain and the frequency domain. Therefore, when you obtain a
solution,you are eithei in the time domain or the ftequencydomain.you
cannotbe in both domaim simultaneously.
Any solution that containsa
mlxture of time domain and phasordomainnometrclatureis nonsensical.
The phasor transfom is also useful in circuit analysisbecauseit applies
direcUy to the sum of shusoidal functions. Circuit analysis involves summmg curents and voltages,so the importanceof this observationis obvi_
ous,We can formalize this DtoDertv as follows: If
1J:IJr+
D2+
- +Dn
1s.23)
whereall the voltagesor tle right-handsideare sinusoidalvoltagesofthe
samefiequency, ther
V=V+Vr+..+Vn.
19.?4)
Thus the phasor representation is the sum of the phasors of the individual
terms.We discussthe developmentofEq.9.24 in Section9.5.
9.3 ThePhasor 341
Before applying the phasor tramfom to circuit analysis,we illusfate
its usefulnessin solving a problem witl which you are already familiar:
adding sinusoidal functions via trigonometdc identities. Example 9.5
showshow the phasor transform greatly simpiifies this type of pioblem.
AddingCosines
UsingPhasors
ff)r = 20cos((rl 30') and yr: 40cos(al + 60'),
express) = )rr + ),2as a single sinusoidal function.
a) Solveby usingtrigonometricidentities.
b) Solveby ushg the phasorconcept.
3732
in thesoLutjon
torl.
Figure9.6 r A rjghttriangte
used
Solution
a) Filst we expand both y1 and 12, using the cosine
of the sum of two angles,to get
b) we can solve the problem by using phasorsas
follows:Because
)r : 20cos,)tcos30" + 20 sin.dt sin30';
Y=YltY2.
)2 = 40cos.rtcos60' 40sin(l)t sin 60".
Adding )r and )r, we obtain
Y=Y1 +Y,
y - (20cos30+ 40cos60)cosrrt
+ (20 sin 30
= 3'7.32cos
at
/'\1 1)
\tv+ tz
cosdt
: 201_!: + 401!L
40 siII 60) sin {r,
: (17.32
U.64sir]'at.
: 37.32 + i24.64
To combhe these two terms we teat the
co-efficients of the cosine and sine assidesof a dg:ht
triangle (Fig. 9.6) and then multiply and divide the
righFbandsideby lhe hypotenuse.
Our expres'ion
y : 44.721-:
then,from Eq.9.24,
)4 64
]
\
sin ot J
j10) + QO + j34.64)
= 44;72//33.43'.
Once we know the phasor Y, we can write the
corresponding trigonometdc tunction for y by
taking the invene phasortmnsform:
: n{44;72ei33a3et"t}
y = g 1I44;tzei3343\
/
- 44.?2(cos33.43'cosdt - sin33.43"sinot).
Again, we invoke the identily involvhg the
cosine of the sum of two argles and wdte
y : 44.12cos(@t+ 33.43").
= 44;72cos(at+ 33.43").
The superiority of the phasor approach foi
addirg sinusoidal fulctions should be apparent.
Note that it requ es the ability to move back
and forth between the polar and rectargular
forms of complex numbers.
342 SinusoidaL
Steady-State
Anatysjs
objective1-Undetstandphasorconcepts
andbeableto performa phasortnnsformandanlnversephasortransform
9,1
(c)r1.r8/ 26.s7'A;
(d)33e.e0M!L mv.
Find the phasortransformof eachtigonometric function:
a) lJ - 170cos(377r 40') V.
b) t : tOsin (1000r+ 20') A.
c) t: [scos(or + 36.87')+ 10cos(a,t
- s3.13)lA.
d) u : l300cos(20,000,r
+,15')
100sin(20,000rr
+ 30')l rnv.
Answer: (a) 1701 lp: Vi
(b)10l zq A;
9.2
Find the time-domainexpressioncorespon,
ding to eachphasor:
a)v=18.61:1fv.
b) | : (201!!: - 501:lq) InA.
c) v : (20+ J80 301J!) v.
Answer: (a) 18.6cos(ol - 54') V;
(b) 18.81cos(.dr + 126.68') mA;
(c) '/2."/9cos(at + 97.08') v.
NOTE: Also tt)rChapterPrcblen 9.12
9.4 TheFassive
CircuitF[en'lents
in the Freqr:ency
Ssrmailr
Thc systematicapplicatior of thc phasor transform in circuit analysis
rcquirestwo steps.First, we must establishthe relationshipbetweenthe
phasorcurrent and the phasorvoltageat the terminalsof the passjvccir
cuil elements.Second,we must develop the phasor-domainversion of
Kirchhoff'slawq which we discussin Section9.5.In this section.wc establish the relationshipbetweenthe phasorcurrent and voltageat rhc lerminalsof the rcsislor,inductor,and capacitor.We bcgin with the resjstorand
usethe passivesignconventionin a1lthe derivations.
TheV-I Relationship
for a Resistor
R
figure9.? .:hA rcsistjve
eLement
carryjnq
a sinusoidat
From Ohm'slaw,if the currcnt in a resistorvaricssinusoidallywirh time
thatis,ifl = l cos(@t+ di) the voltageat the rerminalsofthe rcsislor,
as shorvnin Fig.9.7,is
t] = RU_ cos((,r + drl
= R/_[cos(or+ r)].
(e.25)
where1u is the maximumamplitudeof the currenrin amperes
and di is
thcphaseangleof thecurrent.
Domarn
9.4 ThePassive
Circuit
Etements
in theFrcquercy
The phasor transfom of this voltage is
V:RI^dd,:RI.ll.
|s.26)
But 1-14 is the phasor representation of the sinusoidal cunetrq so we can
write Eq.9.26as
,v-;$
\9.27)
which statestlat the phasor voltage at the terminals of a resistor is simply
the resistance times the phasor current. Figure 9.8 shows the circuit diagram for a resistor io the ftequency domain.
Equations 9.25 and 9.27 both contain another important piece of
information-namely, that at the teminals of a resistor, there is no phase
shift between the cuirent and voltage. Figure 9.9 depicts this phase relationship, wheie the phase angle of both the voltage and the currert waveforms is 60'. The signals arc said to be in phos€ becausethey both reach
corresponding values on their rcspective curves at the same time (for
example,they are at thek positive maxima at the sameinstant).
phasorvoltageand
< Relationship
between
phasorcurrentfor a resisior
H^-a
R
*Jy_
I
equivaLent
cncuitof
Figure9.8 AThe frequency-domajn
for an Inductor
TheV-I Retationship
We derive the relationship between the phasor cu ent and phasor voltage
at the lerminals of an inductor by assuminga sinusoidal cuffert and using
Ldildt to est,blish the coresponding voltage. Thus, for i = 1- cos (,)t
+ di), the expressionfor the voltage is
-aLl#n(at
+ q.
andcurrigure9.9A A ptotshowins
thatthevoLtage
arcin phase.
rentatthetemjnabofa rcsktor
(9.28)
WenowrewriteEq.9.28usingttrecoshefunction:
u = -aLl^cos(at
+ 0i
90").
The phasor representation of the voltage given by Eq. 9.29is
phasorvottageand
(9.30) < Retationship
between
Dhasor
curent for anindudor
344
Sinusojd
at Steady-State
AnaLysis
Note tlat in deriving Eq. 9.30we used the identity
e F' : cos90. _ i siD90. = _j.
Equation 9.30statesthat the phasorvoltage at the terminalsof an
inductor equalsja,Z times the phasor cuffenL Figure 9.10 shows the
frequercy-domainequivalentcircuit for the inductor. lt is important to
I
Irotethat the relationshipbetweenphasorvoltageand phasorcurrent for
figuE 9.10 Theftequency-domain
equivaLent
circuit an inductor appliesas well for the mutual hductancein one coil due to
currentflowingin anothermutuallycoupledcoil.That is,the phasorvoltageat the terminalsof one coil in a mutuallycoupledpair of coilsequals
i&rM timesthe phasorcurrentin the otler coil.
Wecanre*.riteEq.9.30as
v = (aL/p:)r ^19
= aLIn /@L + 90)'.
rigure9.11 A pLotshowing
thepharepLationrhip
betl,een
thecunentandvoLtage
at th€tenninals
of an
inductor
(tr = 60").
(e.31)
which indicatesthat the voltageand current are out of phaseby exactly
90'. In paiticular,
thevoltageleadsthecurrentby 90', or,equivalently,
the
currentlagsbehindthevoltageby 90".Figure9.11illustrates
thisconcept
of voltageleadingcurrentot cunentlaggingvohage.For example,the voltage ieachesits negativepeak exactly90" before the cufient reachesits
negativepeak.The sameobservationcan be madewith respectto the
zero-going-positive
crossingor the positivepeak.
We can alsoexprcssthe phaseshift in secondsA phaseshift of 90'
corespondsto one-founhof a period;hencethe voltageleadsthe current
by T/4, or a7second.
TheV-I Relationship
for a Capacitor
We obtain the relatiodship between the phasor current and phasor voltage
at the terminals of a capacitor ftom the derivation of Eq. 9.30. In other
words. if we note that for a caDacitor that
andassume
that
D=U-cos(at+0),
then
| : j.,,cv.
{e.32}
9.4
Donain 345
Th€Palsive
CircuitEt€mentsin
th€ Ftquency
Now if we solve Eq. 9.32 for the voltage as a function of the current, we get
(e.33) < Relationshipbetweenphasorvoltageand
phasorcurrentfor a capacitot
Equation 9.33demonstratesthat t]le equivalent circuit for the capacitor in
the phasordomainis asshownin Fig.9.12.
The voltage acrossthe terminals of a capacitor Iagsbehind t]Ie current
by exactly 90". We can easily show this relationship by reniting Eq. 9.33as
1/jac
-(.---.
iFigure
domain
equivalent
circuit
9,12a Thefreqoency
v : \
7''oo't^707
:4rp,
not.
(e.34)
lte altemative way to exprcssthe phaserelationship contained in Eq.9.34
is to say that the current leads the voltage by 90". Figure 9.13 shows the
phase relationship between the curent and voltage at the terminals of a
Impedance
andReactance
Flgule9.13l A pLotshowing
thephas€r€tationship
We conclude t}lis discussion of passivec cuit elements in the frequency between
the cunentandvottage
atth€t€rminals
ofa
domain with an impo ant observation.Wlen we compare Eqs. 9.27,9.30, capacitor
(0i = 60').
and 9.33,we note that tley a-reall of the folm
(9.35) < Deflnitionof impedance
where Z represenfsthe impedence of the circuit element. Solving for Z in
Eq.9.35,youcanseethat impedanceis the ratio of a circuitelement'svoltagephasor to its cunent phasor.Thus the impedance of a rcsistor is R, the
impedance of an inductor is io-L, the impedance of mutual inductance is
jdM, afld the impedance of a capacitor is 1/loc. In all cases,impedance
is measuredin ohrns.Note that, altlough impedance is a complex number,
it is not a phasor,Remembei, a phasor is a complex number that showsup
as the coefficient of d''. Thus, althougl all phasors are complex numbers,
not all complex numbers are phasols.
Impedance in t]Ie frequency domain is the quantity analogous to
resistance,inductance,and capacitanceir the time domain. The imaginary
part of the impedance is called r€actance.The values of impedance and
readance for each of the component values are sunrmarizedin Table 9,1.
And finally, a reminder. If the refererce direction for the curent in a
passivecircuit element is in tlle d ection of the voltage dse acrossthe ele'
ment, you must insert a minus sign illto the equation that rclates the volt'
age to t]te cullenl.
CiIcuia
Element
Impedence
Resistor
R
j( 1/"'c\
-r/@C
346 5inusojdat Steady-StateAnatysis
l
9.5 Kirchhoff'sLaws
in the Frequency
Domain
Wepointedoutin Section9.3,withreference
to Eqs.9.23
ard 9.24,rharrhe
phasortranslormis usefulin circuit analysisbecauseit appliesto the sum
of sinusoidalfunctions.We illustratedthis usefulnessin Example9.5.We
no$ lofmalizelhisobser\drion
by de!elopingKifchbotts taw;in lhe trequencydomain.
Kirchhoff'sVoltageLawin the Frequency
Domain
We begin by assumingthat 01 - o, representvoltagesarourd a closed
path in a circuit.We alsoassumethat the c cuit is oDeratinein a sinusoidal
slead)srate.TbusKirchhoftr\oltage
lawfequiresihdr
Dt+x2+
+un=0,
(e.36)
which in the sinusoidalsteadystatebecomescomplex
y,1cos(a,t + 0) +
V,1cos(at + 0, + .
+ V^.cos(d + Ah)=0.
\9.37)
We now useEuler's identity to write Eq. 9.3?as
n{V.eia,ej^} + n{vh,eta,ei't} +
. + nIv-,eta"et,t}
(e.38)
9.5 Krchhoffs
tawsin theFr€quency
Domain
which we rewrite as
Yi{v^,4,ejd + v,.4'ejd +
. + v^,ejs"etd}=t).
(e.3e)
Factoring the term d't lrom each term lelds
nltv-ce -v,,.ce--.
n{(V +v,+ .
v - c a t e t " t-l. .
r v , ) e , o ' }: 0 .
(e.40)
Bur d'i + o, so
(e.41) < KVLin the frequency
donain
which is the statement of Kirchhoff's voltage law as it applies to phasor
voltages.Inotler words,Eq.9.36appliesto a set ofsinusoidalvoltagesin
the time domain, and Eq. 9.41 is the equivalent statement in the frequency domain,
Kirchhoff'sCurrentLawin the Freguency
Domain
A similar derivation aDDliesto a set of sinusoidal currents Thus if
\9.4?)
then
(9.43) < KCLin the frequency
donain
where11,I2,-. , I, are the phasorrepresentations
of the individualcui
Equations
9.35,9.41,
and9.43form thebasisfor circuitanalysis
in t}le
frequencydomain.Note that Eq. 9.35has the samealgebmicfolm as
Ohm'slaw,andthat Eqs.9.41and9.43stateKirchhoff'slawsfor phasor
quantities.Therefore
you mayuseall the techniquesdevelopedfor analyzing resistivecircuitsto find phasorcurents and voltage$You needlearn
no new analytictechniques;the basiccircuit analysisand simplification
toolscoveredin ChapteN2+ canall be usedto analyzecircuitsin the frcquencydomain.Phasorcircuit analysisconsistsof two fundamentaltasks:
(1) You must be able to col1struclthe frequetrcy-domain
model of a circuit; and (2) you must be able to manipulatecomplexnumbeE and/or
quantitiesalgebmically.We illustiate theseaspectsof phasoranalysisin
th€ discussionthat follows,begiffing with seiies,parallel,and delta{olinr'esimDlitications.
349
SinusojdalSteady-SiaieAnalyris
9.6 Series,Para[[e[,
andDelta-to-Wye
Simplifications
The rules for combining impedances in sedes ot parallel and for making
deltato-wye tnnsformations are the same as those for resistofi, The oDly
difference is that combinirg impedancesinvolves tle algebraic manipulation of complex numbers
Combining
Impedances
in Series
andPara[[el
tiguE 9.14
Impedancesin sedes can be combined into a single impedance by simply
adding the individual impedances,The circuit shown in Fig. 9.14defines the
problem in general terms.The impedancesZr, 22, . . . , Z, arc connectedin
seriesbetween terminals a,b.Wlen impedancesare in series,they carry the
samephasor cunent I. From Eq. 9.35,t]Ie voltage drop aqoss each impedan\t is ZJ,221, .. , Z"1, and ftom Kircbloff's voltagelaw,
impedances
in sedes.
Yab:Zl+
Z2l+ .. + Z"I
=(21+22+,,. + z")1.
(e.44)
The equivalent impedance between terminals a,b is
: zr+ 22+
. + Zn-
Example 9.6 ilustrates a numerical application of Eq. 9.45.
Ie.15)
Parattet,
9.6 Senes,
andDetia
to Wye
Simptificalions
349
Impedances
in Series
Combining
A 90 O resistor,a 32 mH inductor, and a 5 pF
capacilorare connectedin se es acrossthe terminils uf r .inu.oiJcl \olrrg< suurce.r\ shoqn in
Fig.9.15.Thesteady-state
expressionfor the source
voltage?."is 750cos(5000t + 30") V.
a) Construct thc ficqucncy domain equivalenl
circuit.
b) Calculatethe sleadystalecurrenlibl the phasor
melhod.
The phasor iransform of ?,"is
v = 7s0llqv.
Figure 9.16 illustrates the ftequency-doman
equivalentcircuitof the circuii shownin Fig.9.15.
h) we computethe phasorcurrentsimplyby dividirg rherolr"geoi rhe\ olrage\oufceb) rheeqli\ alentjmpedancebetweenrhe ierminalsa,b.From
Eq.9.,15,
z^b = 9lt + j16o
90O
32nH
j40
- e0+ i120: 1sol!l!a J).
Thus
'750
/30"
Figure9.15 J. ThecircuitforExamph
9.6.
W e m a ) n o ! \ ! \ J r e t l ' e . t e a d ) - s t a r ee \ p r e . s i o n
lbr i directly:
Solution
i : scos(s0oor23.13')
A.
a) From thc cxprcssion lor 11, we have
rr = 5000 md/s. Thereforethe inpedanceoI lhe
32 mH inductor is
x 10 ) : 1160-,,
zL = jaL = j(5000:)(.32
140{}
and the impedanceof the capacitoris
"'
-t
' aC
1n6
' (sooo)(s)
Figlre 9.16 ,t Thefrequency-domain
equivatent
cncuitofthe
circuiishownin Fig.9.15.
objective3-Know howto usecircuitanatysis
techniques
to sotvea circuitin the frequency
domain
9.6
Usingthevaluesofresistance
andinductance
in
the circuil of Fig.9.15,let V" = 125l:q
V
b) the magnitudcofthc steady-state
output
curenf ,,
a n d @ - 5 0 0 0 r a Lcl F i n d
a) the value of capacjaance
that yields a
steady'stateoutput current I with a phase
angleof 105'.
NOTE: Also try Chaptet Prcbleng.2l.
Answet (a) 2.86pF;
(b) 0.982A.
350
SinusojdaLSteady-StateAnalysis
Impedancesconnectedin parallelmay be reducedto a singleequivalent impedance by the reciprccal relationship
1
1
1
1
\9.46)
Figue 9.17depicts the parallel connection of impedances.Note that when
impedancesare in parallel, they have ttre same voltage acrosstheir terminals. We de ve Eq. 9-46 directly from Fig. 9.1? by simply combidng
Kirchhoff's current law with the phasor-domainvenior of Ohm's law, that
is,8q.9.35.From Fig.9.17,
I = 1 1+ 1 2 + . . . + I , ,
figlre9.17 Impedancesin
panLLeL.
\9.47)
Canceling the common voltage term out of Eq. 9.47rcveals Eq. 9.46.
From Eq.9.46,for the specialcaseofjust two impedancesin pamllel,
Z.Z,
(e.18)
We can also expressEq. 9.46 in terms of admiltatrce, defined as the reciprocal of impedance and denoted Y. Thus
1
^
7 , -
lB (siemens).
\9.49)
Admittanceis, of couise,a complexnumber,whosercal part, G, is called
cotrdu€itarce and whose imaginary part, B, is called susceptar€e. Lite
admittance, conductanc€ and susceptanceare measured in siemens (S).
Using Eq.9.49in Eq.9.46,weget
Yoo:Yt+Yz+.
+Y".
(9.50)
The admittance of each of the ideal passive circuit elements also is
worth noting and is summarized in Table 9.2.
Example 9.7 illustrates the application of Eqs. 9.49 and 9.50 ro a spe,
cilic circuit.
Circuil El€ment
Adnfttrnce (Y)
j(-r/aL)
Capacitor
1/.,1
9.6
Sedes.
ParatLel.
andDeLta-to'Wve
SimDtifications351
Combining
Impedances
in Seriesandin Parallel
The sinusoidalcurrent souce in the circuit shown in
Fig. 9.18produces the cunent ir : 8 cos200,000tA.
a) Construct the frequency-domain equivalent
crcull,
b) Find the .leady-srare
expre<sioos
tor u. ir. tr.
ancli3,
Figure9.18l ThecircuiiforExampte
9.7.
Solution
a) The phasor transfom of the current souce is
8 A; tle resistors transform directly to tle ftequency domain as 10 a.nd 6 O; the 40 /rH
inductor has an impedance of j8 O at the given
frcquency of 200,000 rad/s; and at this ftequency the 1 pF capacitor has an impedance of
q.lq sho$srbelrequenc)
-domain
i5 O. Figure
equivalent circuit atrd symbols representing the
phasor tiatrsforms of the unknowns.
b) The circuit shownin Fig. 9.19indicatesthat we
can easily obtain the voltage across the current
source once we know the equivalent impedance
of the tlree pamlel branches. Moreover, once
we know V, we can calculatethe three phasor
currents11,12,and 13by usingEq. 9.35.To find
the equivalent impedance of the three branches,
we first find the equivalent admiltance simply
by adding the admittances of each bnnch. The
admittanceof the first branchis
1
v,:_:0.1S.
the admittance of the secondbmnch is
Y"
1
6+18
6-i8 =
0.06 - j0.08 S,
100
Ir
Figure9.19 A Thefrcquency-donain
equjvat€nt
cjrcujt.
TheVoltagev is
v: zr:40/ 36.87"V.
Hence
40/ -36.87' r4 /-36.87' - 3.2 j2.4 \.
,^
40/ -36.87'
b+/a
and
40/ -36.8'7"
= I /s3.r3" : 4.8+ j6.4A.
13=
5 ,/ 90'
Wecheckthe computationsat this point by vedfying that
and the admittance of the third bmnch is
Y . : -l/ =
,o.zs.
J
The admittarceof the tbree branchesis
Y:Yr+Y2+Y3
= 0.16+ ./0.12
- 0.2149:s.
The impedance at the current souce is
z=!=s/-36.87"a.
1 1+ 1 2 + 1 3 = I
Specificay,
3 . 2 j 2 . 4 j 4 + 4 . 8+ j 6 . 4 = 8 + i 0 .
steady-slare
Tbe corresponding
Iime-domain
- 36.87')V,
t, = 40cos(200,000t
t1 - 4 cos(200,000t 36.87')A,
- 90")A,
t2 : 4cos(200,000t
j3 = 8cos(200,000t
+ 53.13')A.
352
SinusoidatSteady'StateAnatysjs
donain
techniques
to sotvea circuitin the frequency
3-Know howto usecircuitanatysis
Objective
9,7
A 20 O resistoris connectedin parallelwith a
pardllelcombinalion
is
5 mH Inducror.Thir
connectedin serieswith a5 O resistorand a
25tF cdpacilor.
ol lhi\ Inlera) Calculate
the impedaDce
the
frequency
is 2 krad/s.
conneclion if
br Repear(a' for a frequenc)of8 krad s.
c) At what finiie frequency does tlle impedance of the interconnection become purely
resistive?
at tbe trcquency
d) Whari. the impedance
founJin (c'l
Answer:(a)9 - j12O;
. ( b ) 2 1+ i 3 O ;
(c) 4 krad/s;
(d) 1so.
9.8
fte interconnection described in Assessment
Problem9.?is connectedacrossthe teminals
of a voltagesourcethat is generating
, = 150cos4000 \. Wlal i! lhe ma\imum
'amplituale
olthe curent in the 5 mH inductor?
Answ€r: 7.07A.
\OTE: Al5u try ch!ryn ProbLcn,a 24-o 23
Transformations
Detta-to-Wye
The A to-Y transformalionlhat we discussedin Section3.7 with regard
to resistivecircuits also appliesto impedances.Figure 9.20 delines the
A connected impedances along with the Y-equivalent circuit. The
Y impedancesas functionsofthe A impedancesare
Z&.
za+zb+2.'
z2
tnnsfomatjon.
Figure9.20, Thedetta-to-wye
z)
z.z^
z^+ 26+ zc
Z,Zr
z\+zb+2.
(e.51)
(e.52)
(e.53)
The A to-Y transfonnationalsomay be reversed;thatis,we can start
with the Y structureand replaceit with an equivalentA structure.The .\
impedancesasfuncdonsoftheY impedalcesare
z t z ) + z z z 3 +z 1 z 1
zl
z1z2 + z2z3 + z1z1
z2
ztz2+ z2z1+ z3zr
z3
\9.54)
9.6 Sedes,
Pa6ltet,
andDetta-to-Wve
SimDtifications
The processusedto deriveEq$ 9.51-9.53or Eqs.9.54-9.56
is the same
as that used to derive the corresponditrg equations for pure resistive circuit$ In fact, comparing Eqs. 3.44-3.46 'r th Eqs. 9.51-9.53, and
Eq$ 3.47-3.49witl Eqs.9.5,19.56,revealsthat the symbolZ hasreplaced
t}le symbol R. You may want to review Problem 3.61 concerning the deri
vation of the Alo-Y transformation.
Example9.8illustratest]le usefulness
ofthe A-to-Y transfomation in
Dhasorcircuit analvsis.
Usinga Delta-to-Wye
Transform
in the Frequency
Domain
Use a A-to-Y impedance transformation to find Io,
Ir,12,13,Ia,Is,V, and V2in the cicuit inFig.9.21.
the Y imDedanceconnectins to teminal c is
10( j20) 30 + j40
3.2 j2.4 A,
and the Y impedance connecting to termillal d is
2.4{l
120&lv+b
j20a
_
(20+ j60x j20)
^
JU + 14|J,
Inse ing the Y-equivalent impedarces into the circuit, we get ttre circuit showr in Fig 9.22, which we
can now simplify by series-pamlel reductions. The
impedence of the abn branch is
Z^h"= 12 + j4 - j4 = DA,
Figur€9.21 l Theciicuitfor Exampte
9.8.
and the impedance of the acn branch is
z^_:
Solution
First note that the circuit is not amenable to series
or parallel simplification as it now stands,A A-to-Y
impedatrcetansformation allows us to solve for all
the branch currents without rcsofiing to eithei the
node-voltage or the mesh-curent metlod. If we
rcplace either the upper delta (abc) or the lower
delta (bcd) with its Y equivalent, we can further
simplify the resulting circuit by series-parullel combhations. In decidn€ which delta to replace, the
sum of the impedances arourd each delta is worth
checkingbecause
this quanrirllormsthe deoomi
nator for the equivalent Y inpedance$ The sum
around tle Iower delta is 30 + j40, so we chooseto
eliminate it from the circuit. The Y impedance connecting to terminal b is
(20 + j60X10)
: 12 + j4A,
30 + 140
63.2+ j2.4
j2.4
3.2 : 60 A.
j2.4A
jn Fjg.9.21,wiihthe tow*
Figure9.22 a Thecircuitshown
dethr€ptaced
byjis €qujvatent
wve.
354
sjnusoidatsteady-stateAnaLysjs
Note that the abn brarch js in palalel with the acn
bmnch,Thereforc we may replacethesetwo branches
with a singlebranch having an impedanceof
I!-
"
18()
160)r12)
-juo
Combiningthis 10O resistorwith the impedance
betweenn and d reduces
the circuitsho$nin
Fig.9.22to tle oneshownin Fig-9.23.Fromthelat
I1r =
1201!
18 - j24
= 4 /s3.r3' = 2.4+ j3.2A.
Once we know I1],we can work back through the
equivalentcircuits to find the branch currents in
the original circuit. We begin by noting that Io is
the curent in lhe branchnd of Fig.9.22.Therefore
v"d: (8
j24)t - 96
j32V.
F i g u r e9 . 2 3a A r m p l i 5 erde ' s i o ol f l \ . , i l r i h o a n 1
lig.9.22.
To find the branch cunents 13,I1, and 15,we must
fint calculate the voltages V and Vr. Refering to
Fig.9.21,we note that
1)R
v.:1)0/0'
( i 4 ) L= = +
I
Y2:12019:
(63.2+ j2.4)r2- 96
t8v.
104
jiv. _ ,
We may now calculate the voltage Vai because
Wenowcalculate
thebranchcurrents13,Ia,and15:
V=V",+Y,a
and both V and Yndare known.Thus
vai : 120
96 + j32 :21 + j32v.
We now computethe branchcu[ents I"b. and laci:
21 + i32
I
=2+iJA.
I,bn:
12
,
)4+ j32
,l
8
In termsof the branchcurrentsdefinedin Fig.9.21,
r,=!"I=J-'if
e,
r,=ffi=] ;r.oe,
v^
rs = _1
=;
)6
+ i4.rJA.
We checkthe calculationsby noting that
1 1: I " b . : 2 + j : A ,
2 2 6j1.6+ j4.8 = 2.4 + j3.2 : t{,,
14+15= +
3
t5
I,=I..,=fi*;f,e
' ',
l, I t, - 1 , i - , u - / |.o= 2 - I 8 - | .
We check the calculations of Ir and l, by noting that
4
l1+ 12: 2.4 + j3.2 = l1'.
A
l,r12 .1
,n t
t -' Ri -
R
t6
n/;-;;+/a.8-1,.
9.7
5orceTratrformation5
andTh6venin-Norton
Equivahnt
Circuits
objective3-Know howto usecircuit analysistechniquesto solvea circuit in the frequencydomain
9.9
Use a A-to Y transfoimationto find thc current I in ihe circuit showD.
14(])
AnswenI= 4/28.01'A.
NOTE: Also trf ChaptetPrcbkng34.
9.7 $oureeTrnnsfqarmatiens
ar:ri
Thdweilf
cr-hlorton
iquivatentCircuits
The souce lralsformationsintroducedin Section4.9 and the Thdvenin,
Norton equivalentcircuitsdiscussedin Scction 4.10are analyticalrechniqucs that also can be applied to frequencydomain circuits.We prove
ihe validily of these techniquesby follorving the sameprocessused in
Scctions4.9 and,l.10,exceptthat $'e substituteinpedance (Z) for resistance (R). Figure 9.24 shows a sourcctrunslormationequivalentcircuit
Figure9.24A A souretransfomation
in the
qilh the Ionr<llclrlrfeor rhe lrequcn(rdomain.
Figure 9.25 illustratesthe frequcncydomain version of a Th6venin
equivalerltcircuit.Figure9.26showsthe lrequercy-domainequivalcntof
a Nortor equivalent circuit. Thc techniquesfor finding the Th6vcnin
equivalertvoltageard impedanceare identicalto thoseusedfor rcsistive
circuits,exceptthat the frequcncydomait equivalentcircuit involvesthe
mayconrarn
J
manipulatior of complex quanlities.The same holds for finding the
Nortor equivalentcurrenl and impedance.
Example9.9dcmonsualesthe applicationof the sourcetranslomatiol
equivalentcircuit to frcquency-domain
analysis.Example9.10ilustrates Figure9.25t Thefiequenq'-domajn
version
ofa
the detailsoffinding a Thdveninequivalentcircujtin thc hequencydonain. Th6venin
equivaLent
circuit.
Frequency-domain
maycontain
->
Figrre9,26Ir ThefrcqLrenry-domain
version
of a Norton
SinusoidaL
St€ady.State
Anatysis
PerfornlngSource
Tnnsformations
in the Frequency
oonain
Use the concept ofsource transformation to find tle
phasor voltage Vo in the circuit shown in Fi& 9.27.
j3a
lo
o2o io6o
O2o i0.6o
Figure9.28 a Thefilstsiepjn reducing
th€ circujtshov/n
in Fig.9.27.
tigure9,27,t Theci(uitforExanple
9.9,
1.8O j2.4O
O.2A j0.64
Solution
Wecanreplacethe sedescombination
of the voltagesource(404) andtheimpedance
of I - i3 O
with the parallel combinationof a current source
andthe 1 + j3 O impedance.The
sourcecurrentis
zt{l
40
=
r ! _____::r + J r ;(l
lu
_ /3) = a _it2A.
Thuswe canmodily the circuit shownin Fi9.9.27to
the one shownitr Fig. 9.28.Note that the polarity
referenceof the ,10V sourcedeterminesthe referencedirectionfor I.
Next, we combirc the two parallel branches
into a sitrgleimpedalce,
( 1+ i 3 x e - j 3 ) =
10
Also note that we have reduced the circuit to a
simple series circuit. We calculate the current Io by
dividitrg the voltage of the source by the total series
1.8+ j2.4 A,
which is in parallel with the cu[ent source of
4 i12 A. Another source tmnsformation converts this parallel combination to a se es combination consisting of a voltage source in serieswith the
impedanceof 1.8 + t2.4 O.The voltageofthe voltage source is
v = (4
Figure9,29 ^ Thesecond
st€pin reducing
thecircuitshown
in Fig.9.27.
36 jr2
12 jr6
.
-
=
r2(3 jD
4(3 - j4)
-'1 0 + i ) 7
-:- = 1.56+ ,1.08A.
It
j12X1.8+ j2.4)- 36 - j12V.
Using this sourcetransfomation, we rcdraw the
circuitasFig.9.29.Notethe polarityof the voltage
source.We addedthe current Io to the circuit to
expeditethe solutionfor Yo.
We now obtainthe valueof V0by multiplying lo by
theimpedance
10 - i19:
V0= (1.56+ i1.08)(10- j19) :36.12 - jr8.84v.
9.i
Source
Tmnsfomations
andThavenin-Norton
Equivatent
Circuiis 357
Findinga Th6venin
Equivatent
in the Frequency
Donain
Find the Th6veninequivalentcircuitwith respectto
terminalsa.b for the circuit shownin Fig.9-3w.
j44{r
-j40 (l
risure9.31A A sjmptified
veBion
ofthecircuit
shown
jn Fjg.9.30.
We relateth€ controllingvoltageY to the currentI
by noting ftom Fig.9.31thal
tigure 9.30 A lhe circuitfo, ENample
9.10.
v,=100-10L
Then,
Solution
We first determinethe Thdvenin equivalenl volf
age.This voltageis the open-circuitvoltageappearingat terminaisa,b.We choosethe referencefor the
Th6veninvoltageas positiveat terminal a.We can
make two source transformations relative to the
120V 12 O, and 60 O circuit elementsto simplify
this pofiion of the circuit.At the sametime, tiese
transformationsmust prese e the identity of the
controlling voltage \ becauseof the dependert
voltagesource.
We determinethe two sourcetransformations
by first replacing t}le se es combination of the
120 V sourceand 12 O resistorwith a 10A currenl
sourceill parallel with 12 O. Next, we replacethe
parallel combinatior of the 12 and 60 O resistors
$rith a single 10 O resistor. Fina y, we replace the
10 A souce in parallel with 10 O with a 100 V
souce in serieswith 10 O. Figure 9.31 showsthe
resultingcircuit.
We addedthe currert I to Fig.9.31to aidturther
discussion,
Note that once we know the current I,
we cancomputethe Th6veninvoltage.Wefind I by
summingthe voltagesaroundthe closedpath in the
circuitshownin Fig.9.31.Hence
100 : 10I
j40l + 120I + 10v":
(130 - j10)I + 10v..
I :
30
900 :
18/ -126.8'7'A.
J40
we now calculate va:
Y,:
100
180/-126.87' :20A + jt44V.
Finally,we note from Fig.9.31that
Yrh=10V,+120I
= 2080 + 11440 + 120(18)/
- 784 j)qQ, 815).
126.87"
)0.t7'V.
To obtainthe Th6veninimpedance,wemay use
any of the techniquespreviouslyused to find the
Thdvenin resistance.We ilustrate the testsource
method in this example.Recall that in using this
mcthod, we deactivate all independent sources
Irom the circuit and then apply either a test voltage
sourceoI a test current sourceto the terminaisof
interest.The ratio of the voltage to the current at
the sourceis the Th6veninimpedance.Figure 9.32
showsthe result of applyjngthis techniqueto the
cncuit shownin Fig.9.30.Note that wc chosea lest
voltage sourceVr. Also note that we deactivated
the hdependentvoltagesourcewithan appropriate
\horr-circuir
andpresenedlhe idenrir)of V.
358
Steady
StateAnatysis
Sinusoidal
_ -v.(e + i4)
-i4o o
120(1 j4)
Ir=I.+It
9+i4\
1 2 )
v? (.
10 140\
v?(3
12(10
YZrh:7:
i4)
j40)'
0 12
/384O
equivateni
forcatcutating
thelh6venin
Figure9.32a Acjrcuit
Figure9.33depictstle Theveninequivalentcircuii.
The branch curents I, and Ib have been added 10
the circuil to simplify the calculation of l? By
straightforwardapplicationsof Kirchhoff's circuit
laws,you should be able to verify the following
relationships:
T
=
10
-138.4O
v- - 10L,
cncuitshown
equjvahntforthe
Figure9.33 a TheTh€venin
Fig.
in
9.30.
lr' =
to sotvea circuitin the frequeicydomain
techniques
objective3-Know howto useorcuit anatysis
in
expre.sionfor L'o(r)
9.10 | ind lhe steady-slale
lhe circuitshowoby u.inglhe lechniqueol
the .inusoidalvollage
sourcerranslormalions.
solucesare
wilh respecr
lo
9.11 FindrheThe!eni0equivalenl
. t6rmiuls a,bin thd circuit shown.
z,1= z0cos(4ooor+ 5:li3")v,
zb : 96 si! 40001V.
15mH
-
,!.r-l
Answer: 48
"''"
cos(4000t
*-: + 36.87')v .
^'-*:'
)
'IOTL: Al'o try Chapw ProblPnta 40.att. ando 1-
Vir,= v"h: r0/4s' v;
9.8
TheNode-Voltage
Method 359
9.8 TheNode-Voltage
Method
In Sectio s 4.2-4.4,we intioduced ttre basicconceptsof the node-voltage
method of circuit analysis.The same concepts apply when we use the
node-voltage method to analyze frequency-domain circuits. Example 9.11
illustrates the solution of such a cicuit by the node-voltage technique.
AssessmentProblem 9.12 and many of the Chapter Problems give you an
opportunityto usethe node-voltagemethodto solvefor steady-state
sinu
soidalresponses,
Usingthe Node-Vottage
Methodin the Frequency
Domain
Use the node-voltage method to lind the branch
currents Ia, Ib, and t in the circuit shown in Fig.9.34.
1()
10.6/E
iza
t l$
{ 1r0ooo
v1(1.1+ 10.2) y2:10.6 + j21.2.
Summingthe curents awaylrom node 2 gives
t ,
i)
5f,}
Multiplying by 1 +/2 and collecting the coefficientsof Y1and V2generatesthe expression
l t b
r.,+-i5r)
20t,
v,
v2.v2
!
-
r+ tl
20I, ^
----=
f
)
/)
The controlling current I* is
Figure9.34 A Thecjrcujtfor Erampt€
9.11.
I,:
Solution
We can describe the circuit in terrns of two node
voltagesbecauseit containstbree essentialnodes.
Four branchesterminateat the essentialnode that
stretchesaqoss the bottom of Fig.9.34,so we useit
as the relerencenode.The remainingtwo essential
nodesare labeledl and 2, and the appropriatenode
voltages are designatedV and V2. Figuie 9.35
reflectsthe choiceof referencenode and the terminal labels.
1
j2A 2
J- ; + , , .
to
|
s0
20r,
Summingthe curents awayfrom node 1 yields
v
v-v,
10.6+-+-r_^:=0
ru r+ tz
1+j2'
Substituting tlis expression for I, into the node 2
equation, muitipling by 1 + j2, and collecting
coefficientsof V1and V2producesthe equatron
-sY+(4.8+J0.6)Yr:0.
Thesolutionsfor V andV2are
v = 68.40 j16.80V,
v, = 68 j26 V.
Hencetie bmnchcurrentsare
I. :
t0
= 6.84 y'.68A,
v -v
I, = ,I a
Ib =
v,
v
Figur€9.35L lhecircuit
shown
in Pig.9.34,withthenode
vottages
defined.
Y-V
I,.: =
lz
20r-=
.-.....
J]
: 3.76+ lr.bSA,
-1 44
j1192 A,
= 5 . 2+ 1 1 3 . o A .
To checkour work, we note that
Ia + I. =
=
Ir =
=
6.84 - j1.68 + 3.76 + j1.68
10.6A,
Ib + !c:
1.44 j11.92+ 5.2 + j13.6
3;76+ j1.68A.
360
Sieady
StateAnatysis
Sinusojdal
objective3-Know howto lse circuitanatysis
techniqlesto sotvea circuitin the frequency
domaih
9.12 Use lhe nodc voltagemethodto lind the sleady
fbr o0) in the circuitshown.The
slateexpression
sirusoidalsourcesare i = l0cos(,tArDd
L, - 100si d/ V. $ lrer<d - 50 kriJ
204
,t, Irr"
Answer: o(r) = 31.62cos(50,000t 71.57") V.
NOTE: ALto ttr ChapterProblens 9.51tud9.56.
StretXr*d
9.9 The$4esh-{urrcnt
Wc can also usc thc ncsh curcnt mcthod to analyzcficqucncy-domain
circuits.The proceduresused in frequency-domainapplicationsare the
sameas thoseused in anal]'zingresistivecircuits.In Sections4.5 .1.7,we
introduccdthc basiclcchniqucsof thc mcsh currcnt mcthodl we demonstratc thc cxtcnsion of this mcthod to frcqucncydomain circuits in
Example9.12.
Usingthe Mesh-Current
Method
in the Frequency
Domain
Use the mesh-currentmethod io find the voltages
V r .\ ' . " n d \ n r h e c i f c . r.i hr . $ I i r l : 9 . o . J h .
tir
t2O1
lO
/l!)
72Q
Sotution
150i0:
The circuit has two meshesand a dependentvoitage source.so we rnusl wrrle two mesh-current
equationsand a constraintequation.Thereference
direction for the mesh currentsIr and I, is clockwise.as shownin Fig.9.37.Oncewe know Ir and Ir,
q e c . n e d . l ) f n d r h - u n k n ^ \ n\ o l r a C cS. .u n m , n !
the voltagesaroundmesh I gives
rer,
t ,t,
i,'
Figure9,36...!.Thecjrcuitfor Exampte
9.12.
1s0= (1 + l2)rr+ (r2- i16)(rr r,).
j2tl
1s0 : (13 - r14)Ir - (12 - 116)I,.
1 2O : r I .
Sunming the voltagesaroundmesh2 generaleslhe
0 = (12
j16)(1,
/3{}
11)+ (1 + j3)r' + 39r..
Figure9.37revealsthat the controllingcurrentI, is
ttre differencebetweenIr and lrr that is. the conl- : Ir
12.
V
i--
/160
Figure9.37r| l4esh
cuftents
used
to solve
thecncuitshown
jn Fiq.9.36.
9.10 TheIransformer 361
Substitutingthis consrraintlnto lhe mesh 2 equa
rionandsimplihing rhefesulring
e\pre*ion grve.
- ( 2 6+ 1 1 3 ) r r .
0=(27+j16)r1
Solvingfor Ir ard I, yields
Ir = -26 j52 A.
Iz = -21
t 5 0+ v 1 + v :
12
jt04 + 150 j130
78+j234=0,
jr04v,
v,: (12 ji6)r, = 12 + j104y.
1 5 0+ 7 8 - j 1 0 1+ 1 2
+ j101: 0,
v2 + vr + 39I. =
Thcthreevoltages
are
7iJ+ i234V.
We chocklhesecalcularions
bv sumningthe voliagesaroundclosedpaths:
j5lt A,
l,=-2+j6A.
vl =(1 + j2)Ir
39I, =
1 5 0 + Y 1+ v 3 + 3 9 I a = 1 5 0 + 7 8 j 1 0 4 + 1 5 0
v3: (r + j3)r, = 1s0 J130v.
78 + i234 - 0.
jr30
objective3-Know howlo usecircuitanatysis
technques
to solvea circuitin the freguency
domain
9.13 Use the mesh-currentmethodto find the phasor curent I in the circuit shown.
1{l
i2a
0.75V,
\'+
Answer:I:29+
j2 -29.U 13!t:A.
NOTE: Also try ChapterPrcblens9.58and9.61.
9.10 TheTransforrmer
A transformeris a devicethat is basedon nragneticcoupling.Tiansformers
areusedin both communicatioD
and powercjrcuils.Incommunicaiioncir
cuits.the transformeris usedto matchimpedances
and eliminatedc signals
tiol1] portions of the system.In power circuits,UaNformers are used to
establishac voltagelevelsthat facilitatethe transnission,distribution,and
consumptionof electricalpolver.A knowlcdgeof the sinusoidalsteady,
statebehaviorof lhe lransformeris requiredi]l the anaiysisof both communication and power systems.In this section.we will discussihe
sinusoidalstcadystatebehaviorof the linear transform€r,which is found
primarily in comnuication circuits.In Section9 11,we will deal with the
id€al transformer,which is usedto modcl the ferromagnetictransformer
Iound in powersystems.
l5o
362
Sinusoidat
Steady
StateAnalysis
Before shning we make a usetul observation.When anal'zing circuits
cortaining mutual inductanceusethe meshorloop-currert method for writing circuit equations The node-voltagemetlod is cumbersometo usewhen
mutual inductarce in involved. This is becausethe currents in the vadous
coils c-annotbe *ritten by inspection as lurctions of tlle node voltages.
TheAnatysis
of a LinearTransformer
Circuit
A simple iraNform€r is formed when two coils are wound on a single core
to ensure magnetic couplirg. Figure 9.38 showstle ftequency-domain circuit model of a system that uses a transformer to connect a load to a
source.In discussingthis circuit, we refer to the transformer wirding connected to the source as the primary winding and the winding connected to
the load as the secordary winding. Based on this terminology, the transfomer circuit parameten are
Fisure
9.38l lhefr€quency
domain
cncuiinrodetfora
trandormer
used
to connect
a toad
to a source.
R1 = the resistanceof the p mary winding,
R2 : the resistanc€of the secondary\ dnding,
a1 = the self-inductance of the primary witrding,
12 = the self-inductance of the secondarywinding,
M = the mutual inductance.
The internal voltage of the sinusoidal source is v', ard the intemal
impedance of the source is 2,. The impedance ZL representsthe load connected to the secondary winding of t}re transfolmei, The phasor currents
I1 and 12represent the piimary and secondarycurents of the transfonner,
iespectively.
Analysis of the circuit in Fig. 9.38 consistsof finding 11and 12as functions of the circuit paramergrcY", 2,, Rb Lr, L2, R2,M, ZL, ald a.We arc
also interested in finding the impedance seenlookirg into the transformer
ftom the terminals a,b.To find 11and 12,we fiIst write t}le two mesh-crlrrent
equations that descdbe the circuit:
- jaMr2,
(e.57)
0 = -joMr1 + (R2 + jal2 + Z)r2.
(e.58)
vs= (zs+ Rr+ jaL)\
To facilitatethe algebraicmanipulationof Eqs.9.57and 9.58,we let
Zr=Z"+Rr+j'tLl,
(e.5e)
Zn= Rz+ jtiL2+ ZL,
(e.60)
where Z1r is the total self-impedance
of the meshcontainingthe pdmary
winding of the transfomer, and Zn is the total self-impedance of tle
mesh containing the secondarywinding. Based on tlle notation intoduced
inEqs.9.59and 9.60,thesolutionsfor lr and 12lrom Eq$ 9.57and 9.58are
(e.61)
z\zn +
ZnZ22 !
v
=',
lL
(e.62)
9.10 IheTGnsformer353
To the intemal source voltage \, tle impedanco appea$ as V"/I1, or
v" =
zi"'
i
znz22 + ozMz
(e.63)
The impedanceat the teminals olthe sourceis Zinl
2 " " = 2 . .+ + -
z " : R .+ i u L , +
-Z.,so
0jM'
(R2+ jaL2+ Z)'
(e.61)
Note tllat the impedance Z.b is independent of tle magnetic polarity
of the tiansformer. The reason is that the mutual hductance aoDearsin
fq. q.64a. a \quaredquanriry.
This impedance
i( ot paruicuta;
inreresr
becauseit showshow the tmrsformer affects the impedance of tle load as
seen hom the source. Without the transformet, the load would be connected directly to the source, and the source would seea load impedance
of ZL; with tle transformer, the load is connected to the source through
the transfoner, and the sourceseesa load impedancethat is a modilied
versionofZL, asseenin the third term ofEq.9.64.
Reflected
Impedance
The third term ir Eq. 9.64is caled the r€flected impedance (2.), because
it is t]le equivalent impedance of the secondary coil and Ioad impedance
tlansmitted, or reflected, to the pdmary side of the transformer. Note that
the reflected impedance is due solely to the existence of mutual induc,
tance;that iq if t]te two coils ate decoupled, M becomeszero, Z, becomes
zero, and Zubreduces to the self-impedanceof the primary coil.
To consider reflected impedance in more detail, we first exprcss the
load imDedancein rcctansular form:
ZL:
RL + jXL,
(e.65)
wherc the load reactance-YL cardes its own algebmic sign. In other words,
-YL is a positive number if the load is inductive and a negative number if
tlle load is capacitive.We now use Eq.9.65 to *rite the reflectedimpedancein rectansularform:
R2+ RL+ j@L2 + XL)
a'M2l(R, + RL) - j(oL2+ xL)l
(R2 + RL)? + (alz + XL)'
=
ffin*
- j(aL,+ xL)1
+ RL)
(e.66)
The derivation of Eq. 9.66 takes advantage of the lact that, when ZL is
written iD rectangular form, the self impedance of the mesh contai ng the
secondarywinding is
Zn=
R2+ RL+ j@L2+ XL\.
(e.67)
SinusoidaL
Steady
StateAnaLysis
Now observefrom Eq.9-66that the self-impedance
of the secondary
circuit is reflected into the primary circuit by a scaling factor of
(oMl Zrr\', atfi, that the sign ol the reactive component (ol2 + XL) is
revelsed.Thus the linear transformerreflectsthe conjugateof the selfimpedanceof the secondarycircuit (ZiJ into the primary winding by a
scalarm tiplier. Example9.13illustratesmeshcurrent analysisfor a cir
c t containinga linear transfbrmer.
Analyzing
a LinearTransformer
in the Frequency
Domain
The parametersof a certainlinear transformerare
R1 : 200 O, R2 : 100O,1-r = 9 H, L2 = 4H, and
,t = 0.5. The transformer couples an impedanc€
consistingof an 800{} resistorin seies with a 1 /rF
capacitorto a sinusoidalvoltagesource.The 300V
(rms) sou(ce has an internal jmpedance of
500 + j100 O and a frequencyo1400rad/s.
a) Construct a ftequency-domain equivalent circuit
of the system.
b) Caiculate the self impedance of the primary
transformer. In constructing the circuit in
Fig.9.39,we
madethefollowingcalculations:
jo.1 = i(400xe) = j3600o,
jo., = j(a00)(a): j1600o,
M=0.5{e)(,0:3H.
j{,M = i(400x3) : 11200
o,
1
d) Calculatetlle impedarce reflectedinto the primary winding.
e) Calculate the scaling factor for the reflected
impedance.
f) Calculale the impedance seen looking into t}le
pdmary terminals of the transformer,
g) Calculatethe Th6veninequivalert witl respect
to the terminalsc,d.
a) Figure9.39shows
thefrequency-domain
equiva
lentcircuit.Notethattheintemalvoltageof the
phasor.andthat
sourcese es asthe reference
the terminalvoltages
of thc
V andU represent
^=
ldL
l4uu
zr = 500+ j100+ 200+ j3600: 700+ i3700o.
c) The self-irnpedance
of the secondarycircuit is
ZL = 100 + j1600+ 800 - j2500 - 900
vL i3600o
figure9.39& Ihefrcquency-domajn
equivatent
circuittor
Exampte
9.13.
j900 O.
d\ The impedance re0ecr(Ll iDro rhe primary
winding is
=
/
"t "r n
"n
\ l9oo j9o0
;(900
\2
l/on^!;onn'
+ i900) = 800+ i800o.
e) Thescalingfactorby whichZi, is reflectedis 8/9.
Ir
300!t v
l2s00o.
h) t}e ,elf-impedance
of rhepflmdrycircuilis
7 =l
"'
Sotution
t06
_=
c) Calculatethe self-impedanceof the secondary
j2s00o
s.11 TheIdeatTnnsformer
365
f) The irnpedanceseen looking inlo the primary
terminalsofthe transformeris the impedaDceof
l h e n r i m a r yu i n L l n gp l u . r n e r e l l c ( l e di m p e d -
TheTh6veninimpcdancewil be equalto thc impedance of the secondarywindilg plus the impedance
reflectedfrom the prinary when ihe voltagesource
is replacedby a shot circuit.flus
zdh :200 + j3640 + 800 + i800 : 1000+ j4400 ().
Zrh: 100 F 11600
+ (
g) The Thdvcninvoltagewill equalthc open circuit
valuc oI Vcd.The open circuil v.1lueof V.d will
equal j1200 limes the open circuil value of lr.
The opcn circuit valueof lr is
Ir:
^#$-)','nn ,.'0,
= 1'71.09
+ j1224.26
A.
'IheTh6vcnin
equivalent
is shownin Fig.9.40.
3001!:
700+ j3700
1 7 1 . 0{9)
O
/t22.1.26
19.67/ 79.29'mA.
Therefore
= i120aQ9.6'7
/ 19.29')t 10 3
: 95.60
/10.71'V.
Figure9.40;i TheThevenin
equivalent
circuittortxampLe
s.13.
objective4-Be abteto anatyze
circuitscontaining
lineartransform€rs
usingphasornethods
9.14 A linear transformercouplcsa load consisting
ofa 360 O resistorin serieswirh a 0.25H
induclor to a sinusoidalvollagesource,as
shown.The voltagesourcehasan internal
impedanceof 184 + JOO and a mlximum volt,
ageof245.20Y and jt is operatingat 800rad/s.
The transformerparamelersare Xr : 100O,
l,1 = 0.5 H, R' : ,10o, lz = 0.125 H, arrd
k : 0.4.Calculale(a) t}le reflectedjmpedarce;
(b) .he primary currerq ard (c) thc secondary
Answer: (a) 10.24 j7.68Q,
(b) 0.5cos(800t- 53.13")A;
(c) 0.08cos8001
A.
NOTE: Also ttf Cha4er Problems 9.72and 9.73.
9.11 TheXdeal
Tramsfornrer
An ideal tmnsformerconsistsoftwo magneticallycoupledcoilshavingN1
and N2 turns.rcspcctively,
and exhibitingthcsethreeproperries:
1. The coefficientof couplingis unity (ft = 1).
2. The self-inductance
ofcach coil is infinite (L1 = l,, : ,ro).
3. The coil losscs,due !o parasiticresistance,
are negligible.
366
Anatysis
Sinusoidat
Steady
State
Understanding the behavior of ideal transformen begins with Eq. 9.64
which describesthe impedance at the terminals ofa sourcecolnected to a
Iinear transformer. We reDeatthis eouation below and examine it further.
ExploringLimitingValues
A useful relationship between the input impedance and load impedance,
as given by Z.h in Eq. 9.68, emergesas -L1and a2 each become infinitely
large afld, at the sametime, the coefficient of couplhg approachesunity:
zzb:
-
zt1 a --
=\+jaLj+
zs
(e.68)
(R2 + jaL2+ ZL)
Tiansformers wound on ferromagnetic cores can approach tl s condition.
Even though such transformers are nonlinear, we can obtain some useful
informationby constructingan idealmodelthatignoresthe nonlinearitiet
To show how Zub changeswhen & : 1 and lt and a? approach infinity, we first irtroduce the notation
zn= R2+ RL+j@L2+ x)
= R2+ jxL
andtlen rearrange
Eq.9.68:
7,"
" - R,
d2M2R,,
t
---------+
t-iloL
'\
Rt
xz,
-2M2x,, \
- --_=
|
Riz xiz'
= R,b + jx,b.
(e.6e)
At this point, we must be careful with the coefficient of j in Eq. 9.69
because,as -L1and -L2approachinfinity, this coetficientis the dillerence
betweentwo large quantities.Thus,before letting Z1 and -L2increase,we
write the coefficientas
Xdb = 'tLI
(d'L)(dL)
X22
RL + x,?2
: d L , \/ t
aL.X-
4;;rL
\
(,.70)
wherewe iecognizetha1,when k = 1, M' = tr1l2. Puttingthe telm multiplying oar over a common denominator gives
X^6= rLt(
R4.+ -L,x, + xi \
# l
Riz + xi
,/
(e.71)
9.11 TheldealTEnsformer
Factoring(dl, out ofthe numeratorand denominatorofEq.9.71yields
xL + (4,, + x?)lak
Lr
Lz (Rzl.L)2 + [1 + (xJdh)]2'
\9.72)
As & approaches1.0,the ratio lr/lz approachesthe constantvalue of
(&/Nr)', which followsfrom Eqs.6.54and 6.55.Thereasonis rhar,as the
couplingbecom€sextremelytight,the two permeances
91 and g2 become
equal.Equation9.72then reducesto
(**)'
XL,
(9.73)
asal + oo,l4 +,ro, andt + 1.0.
The samereasoning leads to simpliJication of the reflected resistance
in Eq.9.69:
a2M2R22
L.
R,+ x4,
L2
--
(t)'*,
\9.14)
Applying the resultsgivenby Eqs.9.73and 9.74ro Eq.9.69yields
Z*=
R t . (**)'
Rz+
(+)'
(Rr + jxr.
le.75)
Comparethis resultwith t]le result in Eq.9.68.Herewe seethat when the
coefficient of coupling approaches unity and the selJ-inductancesof the
coupled coils approachinfinity, the tmnsformer reflects the secondary
windingresistanceand the load impedanceto the primary sideby a scalhg factor equal to the tums ratio (N/Nr) squared. Hence we may
describe the terminal behavior of the ideal translormer in terms of two
characteristics.
First, the magnitudeof the volts per tuln is the samefor
eachcoil.or
l"l=i-'l
-i
(e.76)
Second,themagnitudeofthe ampere-turNis the samefor eachcoil,or
llrNr : llzNz.
(e.11)
!r
We are forced to use magnitudesignsin Eqs.9.76and 9.77,becausewe
(b)
havenot yet establishedieferencepolaritiesfor the currentsand voltages;
we discussthe rcmoval of the magnitude signsshortly
Figure9.41A Theci(ujis us€dto veriirth€ vottsFigure 9.41showstwo lossless(Rl = R2 : 0) magneticallycoupled pertum andampere-turn
rcLabonships
for anjdeal
coils.We useFig.9.41to validateEqs 9.76and 9.77.InFig.9.41(a),coil2 is
368
SinusojdalSteady"StateAnaL!^h
open; in Fig. 9.41(b), coil 2 is shorted. Althougi we carry out the following
analysis in tems of shusoidal steady-state operation, the results also
apply to instantaneousvalues of ?,and i,
Determining
the VottageandCurrentRatios
Note in Fig. 9.41(a) that tle voltage at t}Ie terminals of the open-circuit
coil is etrtirely the result of the curent in coil 1;therefore
\, = jaMlr
(9.78)
The current in coil 1 is
\ :
19.19)
i_L,
From Eqs.9.78and 9.79,
,.-to,.
For unity coupling,the mutual inductanceequals lEf,
(e.80)
so Eq. 9.80
E;
(e.81)
For unity coupling, the flux linking coil 1 is the same as the flux linking
coil2, so we need only one permeatrce to describe the self-inductance of
eachcoil.ThusEq. 9.81becomes
n:ffin=ffn
R',- i,
for anideal
Vottage
retatlonship
>
transfonner
r
lv
(e.82)
(981)
Summing
tie voltages
aroundtheshortedcoilof Fie.9.a1(b)lelds
0=-jaM\+juL2r1,
(9.81)
fiom which,for & = 1,
\ =h = L,
lz M
\/Lrh
(e.85)
s.11 TheIdeatTmnstornrer
369
Equation9.85is equivalentto
I1N1 : I2N2.
(9.86) { Currentretationship
for anideal
iransformer
Figure 9.42 shows the graphic symbol for an ideal transfomer. The
verticallinesin the symbolrepresentthe layersof magneticmaterialfrom
which terromagneticcoresare often made-Thus,the symbol remirds us
that coilswound on a ferromagneticcore behavevery much like an ideal
-;.lt.
^,1I l',
2 r l
I IdealI
symbotfor
anjdeat
There are seveml reasonsfor tiis. The ferromagnetic material creates Figure9.42A Thegraphic
a spacewith high permeance.Thus most of th€ magnetic flux is trapped
inside the core material, establishingtjght magneticcoupling between
coils that share the same core. High permeancealso meaff high selfinductance,becausea : N'S. Finally, feromagnetically coupled coils
efficiently transfer power from one coil to the other. Efficiencies in excess
of 9570are common,so neglectinglossesis not a cripplingapproximation
lbr manyapplications.
Determining
the Potarityof the Vottage
andCurrent
Ratios
We now tuln to the removal oI the magnitude signs from Eqs. 9.76 and
9.77.Note that magnitudesignsdid not show up in the derivationsof
Eqs-9.83and 9.86.Wedid not needthem therebecausewe had established
reference polarities for voltages and relerence directions for currents. In
addition,we knew the magneticpolarity dotsof the two coupledcoils.
The rulesfor assignirgthe proper algebraicsigr ro Eqs.9.76and 9.77
arc asfollows:
If Olecoil voltagesVland V2are bothpositiveor negativeat the dot
markedtermiMl, use a plus sign in Eq. 9.76.Otherwise,usea negatrvesrgn,
Ifthe coil currents11and I, are both dircctedinto or out of the do!
marked teminal, use a minus sign in Eq.9.77. Otherwise,use a
plus sign.
"{ Dotconvention
for idealtransformers
The four circuitsshownin Fig-9.43illustmte theserules.
.-ll"i I
u,,,ll I
i ldeal
vt=vt.
Nrll =
(a)
NrI,
[:E'
&r,-&t:
(b)
vr -Yu,
NrIr = NuIu
G)
l/'= &'
Nrlr = NrI,
(d)
figure9.43A Circujts
thatshowthepropsatgebran
signsfor retating
thetermjnat
voLtag€s
andcunent5
ofan ideattranstormer.
370
Sjnusojdalsieady-StaieAnatysis
Nr = 500
.;l
N, = 2500
F. +
+;l 1r5[. +
" lll v: ",llf v:
lIdealL
(a)
The ratio ofthe turns on lhe two windingsis an important parameter
of the ideal tnnsformer. The turns ratio is defined as either Nr/& or
N/Nt; both mtios appearin various writings.In this text, we use d to
denotethe mtio Ny'N1,or
lldeal
(b)
N,
(e.87)
+ ; l 1 A : 1 F .+
" lli v.
) t (
lra*rL
k)
Figule9,44A Thrcewaysto showthatthetumsratio
of anideattransfornrer
is 5.
Figure 9.44showsthree waysto representth€ turns ratio of an ideal
transformer.Figure9.44(a)showsthe numberof turns in eachcoil explicitly. Figure 9.44(b)showsthat the rario NxlNr is 5 ro 1, and Fig. 9.44(c)
showsthat the ratio Nx/Nr is 1to i.
Example 9.14illustmtestle analysisof a circuit containingan ideal
transformer,
Anatyzing
an ldealTransformer
Circuitin the Frequehcy
Domain
The load impedanceconnectedto the secondary
windingofthe ideal transformerin Fi9.9.45consists
of a 237.5mO resjstor in series with a 125pH
inductor.
If the sinusoidalvoltagesource(r,s)is generat
ing the voltage 2500cos400t V, find the steady
stateexpressions
fof:(a) t1;O) t1; (c) tr;and (d) or.
0.25O 5 mH
237.5m{}
tigurc9.45A The(ircuitforExanipte
9.14.
Solution
a) We begin by constructing the phasor domain
equivalentcircuit.The voltage sourcebecomes
25004 V; the 5 mH inductor convertsto an
impedanceof t2 O; and the 125/rH inductor
conveitsto an impedanceof j0.05 O. The phasor
domainequivalentcircuit is showninFig.9.46.
It followsdirectlytom Fig.9.46that
0.25(}
_rl
l2soou
) v
./2|l
0..2375
o
10r1
_---Il
v.
i0.05o
Figure
9.46A Phasor
domain
circuit
forExampte
9.14.
2s001y= (0.2s+ j2)Ir + v1,
and
+ jo.0s)rrl.
v : 10v,: 10(0.2375
Because
12: 10Ir
v1 = 10(0.237s
+ 10.05.)1011
: (23.7s+ js)rr
Therefore
p:
2s00
\ = 100/ -16.26'A.
Thusthesteady-state
er?ression
for i1is
tr : 100cos(400t 16.26')A.
b) V = 2s00lp: $00 // 16.26)(0.2s+ j2)
= 2500- 80 - jl85
= 2120 j185: 427.06/-4.3'7' y.
Hence
(2a+ 17)ry
xr : 21n.06cos(400t 4.37')Y.
Iransformer 371
9.11 TheldeaL
c) I? = 10Ir : 1Un /-16.26' A.
Therefore
t, = i000cos(400r- 16.26)A.
d) V, = 0.lV :242.11/-4.37'V,
glvlDg
Dz= 242;7Icos(400t- 4.37')V.
TheUseof an IdealTransformer
for Impedance
Matching
Ideal transformerccan alsobe usedto raiseor lower the impedancelevel
of a load.The circuitshownin Fig.9.47illustratesthis.The impedanceseen
by the practicalvoltagesource(Vsin sedeswith Zr) is Vr/Ir. The voltage
and currentat the terminalsof the load impedance(V2and Ir) are related
to H and Ir by the transforrncrlurns ratio;thus
v'=
(e.8e)
Thereforethe impedanceseenby the praclicalsourceis
Yr
L
-
u ' [ )
Y
/
._,
-
J
l :
+-
.r
+
.
i
q
.
+
l
\ L ] | ' l z ,
)
T
-I
.
(9 88) fitut" 9.47g Ur;rganideat
tnnsform€r
to coupte
a
toadt0 a source.
lt = alz'
-.
l
L+.---_lldeall
v"I
and
ztN=it-
- I r
.), Z,t
1V
aj t,
(e.s0)
but the ratio Vz/I, is the load impedanceZL, so Eq.9.90becomes
z, =iz,
Thus,the ideal transfo(mer'ssecondarycoil reflectsthe load impedancc
back to the primary coil, with the scalingfactor 1/ar.
Note that the idealtransformerchang€sthe magnitudeof ZL but does
not alfect its phaseangle.wlether ZrN is grealeror lessthan ZL depends
on the turns ratio d,
The ideal transformer or ils practical counterpart, the ferromagnetic
core transformer can be usedto matchthe magnitudeof Zr, to the magnitude of 2.. We will discusswhy this may be desirablein Chaptcr10.
obtectlire5-Be ableto anatyzealrcuitsreithideal transformers
9.15 The sourcevoltagein lhe phasordomaincircuit
in the accoEparyingfigureis 25 A kV. Fird
the amplitudeand phaseangleof Vzalld 12.
Answer V2= 1868.15
//142,39'Yi
h = 125 '/216.87' A.
NOTE: Also try ChapterProblem9.n.
372
SinuroidatSteady-StaieAnatysis
As we shall see,ideal transformersarc used to increaseor deoease
voltagesftom a sourceto a load.Thus,ideal transformersare usedwidely
in t}le elect c utility industry,where it is desirableto decrease.or step
down,the voltagelevel at the powerline to saferresidentialvoltagelevels.
9.12 Phasor
Diagrams
When we are Ning the phasormethod to analyzethe steady-state
sinusoidaloperationofa circuit,a djagramofthe phasorcwents and voltages
may give further insight into the behavior of the circuit. A phasor diagram
shows the magnitude and phase angle of each phasoi quantity in the
complex number plane,Phaseanglesare measuredcounterclockwiseftom
the positive real a\is, and magnitudes are measuredfrom t}re origin of the
axes.For example,Fig. 9.48showsthe phasor quafiines n 1q, D 1lE:,
Figure9.48A A guphicrcpGsentation
of phasols.
Figlre9.49A Thecomplex
numb€r
-7 - j3 :' 7.62 /-156.80'.
51_15:,an.dB
/ 170".
Constructingphasordiagramsof circuit quantitiesgenerallyinvolves
both currents and voltages,As a rcsult, two diflerent magnitude scalesar:e
necessary,one for currents and one for voltages,The ability to visualize a
phasorquantityon the complex-numberplare canbe usefulwhenyou are
checkhg pocket calculator calculations. The twical pocket calculator
doesn'toffer aprinloul ofthe dataentered.Butwhenthe calculatedangle
is displayed,you can compare it to your mertal image as a check or
whether you keyed in the appropriate values.For example, suppose that
you are to computethe polar form of -7 - i3. Witlout making any calculations,you should anticipatea magnitudegreaterthan 7 and an angle
in the third quadrant that is more negativeihan 135' or lesspositive
than 225",asillustratedin Fig.9.49.
Examplesf.i5 and 9.16ilustrate the constructionand use ol phasor
diagrams.We use such diagramsin subsequentchapten wheneverthey
giveadditionalinsightinto the steady-state
sinusoidaloperationof the circuit under irvestigation.Problem 9.76showshow a phasordiagramcan
help explainthe operationof a phase-shifting
circuit.
usingPhasorDiagrams
to Analyzea Circuit
For the circuit in Fig. 9.50,use a phasordiagmm to
find tlre value of R that will cause the curent
rbroughthrr fesislor.
t/?.ro lagrhesourcecunenl.l,.
by 45" when o = skrad/s.
.ll,.
t)
,.,]o.z-n
f *nopr
Figure9.50 A Theci,cuiiforEranple
9.15.
R
Solution
By Kirchhoff's curent laq the sum of the currents
IR, lr, and Ic must equal the source current Ir. If we
assumethat the phaseangie of the voltage V- is
zero,we can draw the current phason for each of
tlle components.The cur.ent phasor for the induc
tor is givenby
lL=
v- 1!:
i(s0n0x0.2r0 )
- v^ / -90' ,
9.12 Phasor
Diagmms373
the current phasor for the capacitor is
given by
u^
1!:
L = j7(s000x800 106) : 4U,,1y:,
see.summingthe phaso$ makesan isoscel€striangle..orhelengrhof lhe curfenrphasorfor there(i\
tor must equal 3y-. Therefore,the value of the
resistoris i o.
and the currentphasorfor the resistoris givenby
rn=
v. 1l:
t)4,
= :!!!
/^.
R
Thesephasorsare shownin Fig. 9.51.The pha,
sor diagramalso showsthe sourcecurrent phasor,
sketchedasa dotted line,whichmust be t}le sum of
the curent phason of the threecircuit components
and must be at an angle that is 45' more posjtive
thar the current phasorfor the resistor.As you can
r
I
l
I a = V,'lR
Fiqure
9.514 Thephasordiag€m
forthecurr€nts
in tig.9.50.
UsingPhasorDiagrams
to AnalyzeCapacitive
LoadingEffects
The circuit in Fig. 9.52has a load consistingof the
parallel combhation of the rcsistor and inductor.
Use phasor diagramsto explore the effect of adding
a capacitor acrosstbe terminals of the load on the
amplitude of Y if w€ adjust Vr so that the amplitude
of Vr remains constant. Utility companies use this
teclmique to control the voltage drop on their lines.
I
v, n' J r,,1i,r,
L
Figure9.53 d, Thefrequ€ncy-domain
equivatent
ofthecncuit
in Fig.9.52.
Ll
.
I ),"
]
-
+
".
R.J
L,
Figure9.52 A Th€circuitfor Examph
9.16.
Solution
We begin by assumingzero capacitanceacrossthe
load.AJter constructinglie phasordiagramfor the
zero-capacitance
case,wecanadd the capacitorand
study its eflect on the amplitude of Y, holding the
amplitudeof VL constant.Figure9.53showsthe fre,
quency-dom"in
equi\alenro[ rhe circul shownin
Fig.9.52.Weaddedthe phasorbranchcurrentsI,Ia,
and Ib to Fig.9.53to aid discussion.
Figure9.54showsthe stepwiseevolutionof the
phasordiagram.Keepin mind that we are not interestedin specificphasorvaluesand positionsin this
example,but ratherin the generaleffectof addinga
capaciloracrossthe terminaisof the load.Thus,we
want to developthe relative positionsof the phasorsbefore and after the capacitorhasbeenadded.
Relaling the phasor diagram to the circuit
shownin Fi9.9.53revealsthe following poinls:
a) Becausewe arc holding the amplitude of the
load voltageconstant,wechooseV! asour referencc-For convenience,
we place this phasor on
the posilivereal a-\is.
b) We know that I" is in phasewith Vr and that ils
magnitudeis Vrlnz. (On the phasor diagam,
374
sinLrsoidatSt€ady-Stat€Aratysjs
R,I\
\I
(5)
phasor diagnm shown ir Fig. 9.57 depicts tlese
obse ations.Thedotted phasomrepresentthe pertinent currents and vollages before tlle addition of
the capacitor.
Thus,compadngthe dotted phason of I, RrI,
j.darl, and V" witl tlren sold counteryarts clearly
showsthe effect of adding C to the circuit. In particular, note that this reduces t]te amplitude of the
source volrage and still maintains tle amplitude of
the load voltage. Practically, this rcsuli means that,
as the Ioad inffeases(i.e.,as Ia and Ib increase),we
can addcapacitorsto the system(i.e.,inqeaseL) so
tlat under heavy load conditions we cao maintain
VL withoul ircreasing the amplitude of th€ source
voltage.
Fiqure9.54A Thestep-by-step
evoLuiion
ofthephasor
jn Fiq.9.53.
djagram
torthecircuit
the magnitude scale foi t]le cunent phason is
independent of the magnitude scalefor the voltagephasois.)
c) We know that Ib lags behind VL by 90" and that
its magdtude is IVL/o42.
d) The line curent I is equal to the sum of Ia and Ib.
e) The voltage drcp across -R1is in phase with the
line curent, and the voltage drop across ioal
leadsthe line currentby 90".
f) The source vollage is the sum of tte load voltage
and the drop along the Line; that jsr Vs : Vt
Flgure9.55,| Theaddition
0f a capacitorto
the circujtshown
in Fi9.9.53.
+ {& + jdL)r.
Nore rhatrhecompleledphasordiagfamshownin
step 6 of Fig. 9.54 clearly shows the amplitude and
phase angle relationships among all the currerts
and voltages in Fig. 9.53.
Now add tbe capacitor branch sho\rn in
Fig. 9.55.We are holding VL constant,so we conslrucl the pbasordiagfamtof lhe circuil in fig. r).55
followingthe samestepsastlose in Fig.9.54,except
that, in step 4, we add the capacitor currert I to the
diagram. In so doing, I. leads VL by 90', witl its
magnitude being lVL(,Cl. Figure 9.56 shows tlle
elfect of Ic on the line clmenr Both the magnitude
and phase angle of the litre cunent I change with
changesir the magnitude of lc. As I changes,so do
the magnitude and phase angle of the voltage drop
along the line. As the drop along the line changes,
the magnitude and phase angle of Vs change.The
Figure9.55 A The€ff€ctof the capacitor
cunentId onthe tjne
I
Figure9.57 Theeffectofaddinq
a load-shuntjng
capacitorto
jn Pig.9.53ifV, is hetdconsiant.
thecircuitshown
NOTE: Assessyow ndentanding of this material by ttying Chaptet Ptoblens 9.81 and 9.82.
sunrmary375
Practical
Perspective
A Househotd
Distribution
Circuit
Letusreturnto the househotd
djstribution
circuitinhoduced
at the beginningof thechapter.
Wewittmodjfu
the circuitsLightty
byaddingresistance
to eachconductor
onthesecondary
sideofthehansformerto
simutate
more
accuratety
the residentiat
wiringconductors.
Themodified
cicujt is shown
in Fig.9,58.In Probtem
9.85youwitl catcuLate
thesix bEnchcurrents
on
the secondary
sideof the distributjon
transforner
andthenshowhowto
, dlcuLate
in theprimary
thecurrcnt
winding.
yourunderstanding
NoTE:Assess
ofthis PrddicolPerspective
byttyingChdpter
Prablems
9.85and9.86.
1()
-r1
12O/X
_ -v, - 2{ , r- }
1201!Y
-v
1r)
|
| {20 r)
L
roo
{11
l:{o
t6
+T.
Figure9.58,{ Distribution
cjrcuit.
5ummary
. The general equation for a sinusoidsl source ls
x = U^cos(at + d) (voltag€source),
i = I -cos(dt + 4) (currentsource),
where y- (or 1-) is the maximum amplitude,o is the
tequency,and 4 is tie phaseangle.(Seepage332.)
. The frequency,(o,of a shusoidal responseis the sameas
the frequency of the sinusoidal soorcedrivirg the circuit.
The amplitude and phaseangle of the respome are usually differentfrom thoseof the source.(Seepage335.)
. The best way to find the steady-statevoltages and currents ir a circuit d{iven by sinusoidal sourc€sis to perform the analysisin the frequency domain.The following
mathematical transfoms allow us to move between tie
time and frequency domains.
. The phasor transform (from the time domain to the
frequency domain):
Y = V-eto = g{Yu cos(d, + d)}.
. lhe inve$e phasor transform (ftom the frequency
domainto the time domain):
s) {u^eto= ft {v^daet""\.
(Seepages
337-338.)
Wlcn working with sinusoidally varying signals,
rememberthat voitageleadscurrent by 90' at the terminalsof an inducloi, and curent leadsvoltageby 90'
at the terminalsof a capacitor.(Seepages342-345.)
376
SinusoidatSteady-StateAnatysis
hnpedance(Z) plays the same role in the frequency
domair as resistance,
induclance,and capacitanceplay
in the time domain. Specifically,the relationship
betweenphasor curent and phasor voltagc for resistors,inductorEand capacitorsis
v:zt,
where the referencedirectior for I obeys the passive
sign convention. The reciprocal of impedance is
admittance(y), so anotier rvayto er?ressthe cullentvoltagerelationshipfor resistors,inductors,and capacitors in the frequencydomainis
V = UY,
(Seepages
345and350.)
All of the circuit analysis techniquesdeveloped in
Chapters2-4 for resistivecircuits also apply to sinusoidal steady-statecircuits in the ftequency domain.
ThesetechniquesincludeKVL, KCL, se.ies.and parallel combinationsof impedances,voltage and current
division, node voltage and mesh curent methods,
source transformations and Th6venin and Norton
The two-winding linear traNfonner is a coupling device
made up of two coilswound on the samenonmagnetic
core.R€ll€ctedimpedanceis the impedanceof the sec,
ondarycircuit asseenfrom the terminalsofthe prjmary
circuit or viceversa.Thereflectedimpedanceof a linear
tansformer scenlrom the primary sideis the conjugate
ofthe self impedanceof the secordarycircuit scal€dby
thelactor ('rMl Z22l)2.(.Seepages361and363.)
Thc two winding id€al transforn€r is a linear transformer with the fo owing special properties:perfect
coupling (k : 1), infirite self-inductancein each coil
(Lt = L2 = .r.), and losslesscoils (R1 : R, = 0). The
circuit behavior is govemedby the tuns ratio d = Nr,/Nl.
In particular,the volts per turn is the same fot each
winding,or
and the anpere tums are the samelor eacbwinding,or
NrIl = + NrIr.
(Seepase36s and 366.)
TABLE
9,3
Impedan(eand ReLated
Valu€s
Inpedanre (Z)
j(
t/@c)
Adnittance (Y)
Rer(tance
tl@c
of
j(
I/aL)
-1/aL
Froblems
Section9.1
9.1 A sinusoidalvollageis givenby the exprcssion
lj = 100cos(240rr + 45") mV.
Find (a) / in hertz; (b) r in mitliseconds;(c) y-:
(d) o(0);(e) din degreesandradiansr(f) the smaltcst
positivevalueof/ at whicht) : 0: and (g) the small
est positive value of r at which .lz,/dl = 09.2 Ina singlegraph,sketchc : 60cos(or + d) versus
ot for d = -60', 30",0",30',and60'.
a) Statewhetherthe voltagefunction is shifting !o
the right or left asd becomesmore positive.
b) What is the direclior of shift if d changesfrom
0ro 30'l
9.3 Att :
250/6 ps. a sinusoidalvoltageis known to
be zero and goingpositive.The voltageis next zero
aLt = 1250/6/.s. Il is also known that lhe voltage
is75Vatl = 0.
a) What is the frequcncyot Din hefiz?
b) What is the expressionfor ??
9-4 A sinusoidalcu ert is zero at l:150!,s
and
rncrea.ing
dt x rrte or 2
l0az { q. fhc ma\imum
amplitudeof the voltageis 10A.
a) w}Iatis the ftequencyof ? in radianspersecond?
b) What is the expressionfor u?
probtems377
Considerthe shusoidalvoltage
o(r) = :170cos(1202t
60') V.
a) mat is the maximum amplitude of the voltage?
b) What is the frequency i]) heitz?
c) Wlat is the tiequency in mdians per second?
d) mat is t}le phase angle in mdians?
e) What is the phaseanglein degees?
f) What is the period h miliseconds?
g) What is the fi^t time after l : 0 that zr : 170V?
h) The sinusoidal tunction is shifted 125/18 ms to
the right along the time axis.What is the expression for 0(t)?
i) What is the minimum number of milliseconds
that the functioD must be shifted to the dghl if
the expressionfor t'(t) is 170sin 1202t V?
j) What is the mhimum number o{ milliseconds
tlat the functiotr must be shifted to the left if the
expressionfor o(l) is 170cosl2htt y?
9.6 Showthat
th+T
/
Jr'
c) Find the nume cal value of i after the switch has
been closed for 1.875rns.
d) Wlat arc the maximum amplitude, frequercy
(in radians per second), and phase angle of the
steady-state
current?
By how many degreesare the voltage and the
steady-state
cment out of phase?
o
9.10 a) Ve Jy that Eq.9.9 is the solutionofEq.g.s.This
canbe done by substitutingEq.9.9 into the lefthard side of Eq. 9.8 and tlen noting that it
equalsthe dght-hand side for all values of I > 0.
At l - 0, Eq. 9.9 should reduce to the initial
value of the current,
b) Because the tiansient component vanishes as
time elapsesand becauseour solution must satisly the differential equation for all values of t,
the steady-statecomponent, by itself, must also
satisfy the differential equation. Verify this
observation by showing that the steady-state
componentof Eq.9.9 satisfiesEq.9.8.
v2 T
v'. cosz(dr+ 6S,tt= :ir
L
The Ims value of the sinusoidal voltage supplied to
the convenience outlet of a US. home is 120 V
What is the maximum value of the voltage at the
outlet?
9.8 Find the rms value of the half-wave rectified sinu
soidalvoltageshown.
Figu|e
P9.8
r=v^sin+t,0<t<rP.
Sectiotr9.2
9.9 The voltageappliedto the circuit shownin Fig.9.5
atl = 0is 100cos(400t + 60') V. The circuitrcsistance is 40 O and the initial current in the 75 mH
irductor is zero.
a) Find t(r) for I > 0.
b) Write the expressions for the transient and
steady'statecomponentsof i(t).
Sections9.3-9.4
9.11 Use the conceptof the phasorto combinetlle following sinusoidalfunctionsinto a singl€ trigono
metricexpression:
a) -y: 100cos(3001
+ 45') + 500cos(300r 60'),
:
b) y 250cos(3771
+ 30") 150sin(377,+ 140"),
y
c)
60cos(100t+ 60') - 120sin(100r- 125")
+ 100cos(100t+ 90"),and
d) ) : 100cos((,,t+ 40') + 100cos(ot+ 160")
+ 100cos(o,- 80").
9.12 A 50 Hz sinusoidal voltage with a maximurh amplitude ot 340 V at t : 0 is applied acrcss the teminals of an inductor. The maximum amplitude of the
steady-statecurrent in the inductor is 8.5 A.
a) wlat is the frequency of the inductor culrent?
b) If the phaseangleof the voltageis zero,what is
the phaseangleofthe current?
c) w}lat is t}le inductive reactanceof the inductor?
d) W}Iat is the inductance of the inductor in
millihenrys?
e) What is the impedance of the inductor?
378
SjnusojdalSteady'StateAnatysis
9.13 A zl0kHz sinusoidalvoltagehas zero phaseangle
and a maximum amplitude of 2.5 mV When this
voltageis applied acrossthe teminals of a capacitor, the resultingsteadystate cunent has a maxrmum amplitudeof 125.67/..A.
a) What is the hequency of the currcnt in radiafls
per second?
b) What is the phaseangleof the current?
c) What is the capacitivercactanceof the capacitor?
d) What is the capacitanceof the capacitor in
microfarads?
e) W}lat is tie impedance of the capacitor?
9.14 The expressionsfor the steady-statevoltage and
9.16 A 400 O resistor, a 87.5 mH inductor, and a
adc 312.5nF capacitor are conrected in series.The
series-connectedelements are energized by a sinusoidal voltage sourcewhosevoltage is 500cos (8000t
+ 60")v.
a) Draw the frequency domain equivalent crcurt.
b) Referencethe curent in the direction of the
voltageiise adoss the source,and find the pha
sor current,
c.) Find the steady-state
er?ressionfor i(r).
9.17 a) Show thar at a given frequency o, the circuits in
Fig.P9.17(a)and (b) will have the sameimpedancebetweenthe terminalsa,b if
current at the temhals of the circuit seen rn
Fig.P9.14are
p-=
-''
R2
1 + azR34'
z's : 150cos(800011+ 20") V,
1 + l/lR74
ts = 30sin (8000rr + 38") A
a) Wlat is the impedanceseenby the source?
b) By how many micmsecondsis the currert out of
phasewith the voltage?
.'4c,
b) Find fte values of resistance and capacitance
thal whenconnectedirseries will havethe same
impedance
at 80 Lrad/. as rharot a 5n0O re(i.
tor connectedin parallelwith a 25 nF capacitor.
tigureP9.14
Figure
P9.l 7
^ ,1"
1
-l
"l
G)
Sections9.5 and 9.6
9.15 A 20 O rcsistorand a 1 pF capacitorare connected
En( in parallel.This parallelcombinationis also in par
allel with the sedescombinationof a 1 O resistor
and a 40],.Hinductoi.Thesetlueeparallelbranches
are diven by a sinusoidalcurrent source whose
- 20') A.
currentis 20 cos(50,0001
a) Draw t}le frequency-domain equivalent circuit.
b) Reference the voltage acrossthe currenr source
as a rise in the direction of the sourcecurrent,
and find the phasorvoltage.
c) Find the steadystateexpressiorfor ir(r).
9.18 a) Show that at a given frequency rr, the circuits in
Fig 9.17(a)and (b) will have the sameimped,
ancebetweer the terminalsa,b if
p^:
"
-
'
f + d'R?C?
.rn,cl
cl
r + d,Rici
(Hirt: The two circuits will have the same
impedanceif they havethe sameadmittance.)
Problens379
Find the valuesof resistanceand capacitancethat
when connected in paralel will give the same
impedanceat 20 krad/s as that of a 2 kO rcsistor
connecred
in qerierrilh a capdcirance
ot50 r
9:19 a) Showthat, at a givenftequency{r, the circuitsln
Fig.P9.19(a)and (b) will have the sameimped
ancebetweenthe terminalsa,b if
9.22 a) For the circuit shown in Fig. P9.22,find the frequency (in radians per second) at which tlle
impedanceZ.b is purely resistive.
b) Find the valueof Z.b at the frequencyof (a).
FigureP9.22
.,LiR2
ff + &rl'
R) + -' L!'
Find the valuesof resistanceandinductancethat
when connecled in series will have the sane
impedance
al 20 krad/sas rharol a 50 kO resistor connectedirl parallelwith a 2-5H inductor.
^+"
"?.
9.23 Find the impedance Z,b in the circuit seen in
FigureP9.19
Fig.P9.23.ExpressZ"b in both polar and rectangular form.
' \
figureP9.23
-110o
(a)
10f,)
j10o
9.20 a) Show that at a given frequency d, the circuits in
Fig.P9.19(a)and (b) will have the sameimpedancebetweenthe terminalsa,b if
R1+ a'Ll
_
j20a
R +.?L1
(alirt The two circuits will have the same
impedance
if lheyha!e rhesameadmilrance.)
b) Find the valuesof resistanceand inductancetnar
when connectedin parallel will have the same
impedanceat 10 krad/s as a 5 kO resistorconnectedin serieswith a 500mH inductor.
9.21 Three brancheshaving impedancesof 4
j3 O,
rc+ jnA,
and -i100 O, rcspectivelyj are
connected in parallel. What are the equivalent
(a) admittance,(b) corductance,and (c) susceptance of the pamllel connection in millisiemens?
(d) lf theparallelbranche.are e\citedtroma sinusoidalcurent sourcewhere i = 50 cos.rt A, what
is the maximum amplitude of the current in the
purely capacitivebrarch?
9.24 Find the admittance yab in the circuit seen in
Fig.P9.24.Er?ress yd, in both polar and rectangular form. Give the value of yabin millisiemens.
Pigure
P9.24
y'2.8O
/10O
13.60
380
SinusoidatSteady-StateAnatysjs
9.25 Find the steady-stateexpiession for io(t in the cirBla cuit in Fig. P9.25if o" : 750 cos50001mV.
FiglreP9.25
0.4/,F
9.29 The phasor current L in tlle circuit shown in
sPrc Fig. P9.29is 40 A rllA.
a) Find Ib, \, and Vs.
b) Ifra = 800 rad/s, wdte the expressions
for ib(l),
t"(r), and ?,s0).
Figur€
P9.29
25'|"
trzoo
r{l
r
l40 + t80mA
9.26 The circuit sholin in Fig. P9.26 is operating m the
sinusoidalsteadystate.Find the valueof., if
L = 100sin(ot + 81.87")mA,
t's = 50 cos(ot - 45") V
FiglreP9.26
4000
930 a) For the circuit shownin Flg.P9.30,find the steadystate exprcssionfor oo iJ is : 5 cos (8 x 105t)A.
b) By how many nanosecondsdoes 2l. lag is?
tjqureP9.30
40mH
12()
9.n Find the steady-stateexpressionfor o, in the circuit
ofFi9 P9.27it iR : 200cos5000tmA.
9.31 Thecircuitir Fig.P9.31is opelatingin tie sinusoidal
6*c steadystate.Fird rr,0) iJ i"(, : 15cos8000rn .
Figore
P9.31
rbrreP9.27
5ko
+
240A7u,
r ) I 2.5pF
800
48rnH
9.32 Find Ib and Z in the circuit shown in Fig. P9.32 iJ
928 The circuit h Fig. P9.28is openting ir tle sinu-
vs : 60av andr, : 5l:2q:A.
Eilc .oidalsleady(late.Findlbeslead)-srare
e\pression tigureP9.32
for od(t) if ?Js: 64 cos 8000t V.
Figure
P9,28
31.25nF
j5o+
r,J
-18o
Probtems
381
9.33 End the value of Z in the circuit seenin Fig. P9.33if
vs = 100- i50v,Is:20 + i30A,andvr = 40
+ i30v.
Flgure
P9.35
figureP9.33
10ko
j5o
F -i10o
9,37 The frequency of the sinusoidal current souce in
E,IG the circuit in Fie. P9.37 is adjusted until o. is in
phasewith ls.
a) Wlat is the value of o in radians per second?
b) If i! = 2.5 cosl.)t rnA (where o is tle frequency
found in [a]). whal is lhe slead]-stareer?ression
lor 1).'l
9.34 Find Z,b for tlle circuit shown in Fig P9.34.
Flgule
P9.34
Figure
P9.37
9,35 The frequency of the sinusoidal voltage source in
6plft the circuit in Fig. P9.35is adjusted until ttre current
i, in phasewith I's.
a) Find the frequency ir hertz.
b) Find tne steady-state expression for i. (at the
frequency found in [a]) if rs - 10 cos(,t V.
FlglreP9.35
1500
10()
4mF
2H
9.38 The circuit shom in Fig. P9.38 is operating in the
sirusoidal steady state. The capacitor is adjusted
until tlte current is is in phase $rith the sinusoidal
voltage I's.
a) Specify the capacitarce in miffofarads if
?Js: 250 cos 1000t V.
b) Give t}le steady-state expression for is when C
hasthe valuefound in (a).
Figore
P9.38
9.36 a) The ftequency of the sourcevoltage in the circuit
in Fig. P9.36 is adjusted until is is in phase with
r,s.W}Iat is the value of o in radians per second?
b) If o, = 45"o".t V (where (l) is the frequency
foundin [al).whatis thestead]-srate
expression
382
Sinusoida
l Stea
dy-state
Anatysis
9.39 a) The sourcevoltagein the circuit in Fig.P9.39is
os : 96 cos10,00OtV. Find thevaluesofl, such
that is is in phase with os when the circuit is
opemting in the steady state.
b) For the valuesof a found in (a), find rhe sreadystate er?ressions foi is.
Figure
P9.39
62.5nF
9.43 Find the Th6venin equivalent circuit with respect to
the terminalsa,b for the circuit shownin Fig.P9.43.
figureP9,43
j12 tL
12f,l
Section 9,7
9.40 Use souce transfomations to lind ttre Thdvenin
equivalentcircuit with respectto the terminalsa,b
for.the circuitshownin Fig.P9.40.
DA
Figure
P9.40
/40 ()
9.44 The device h Fig. P9.44 is represented it the fre-
9.41 Use souice tmnsformationsto find the Norton
equivalent circuit with respect 10 the terminals a,b
lbr the circuitshownin Fig.P9.41.
quency domain by a Norton equivalent. When an
inductor having an impedance of j100 O is
connectedacross the device, the value of Vo is
1007_l![ mV. W]en a capacitor havhg an impedance of j100 O is cornected acrossthe device,the
value of I0 is -31?lq
mA. Find the Norton current IN and the Norton impedance ZN.
rigureP9.44
Figure
P9.41
J30O
16[A
9.42 The sinusoidal voltage source in the c cuit
in Fig. P9.42 is developirg a voltage equal to
22.36cos(5000r+ 26.565')v.
a) Find the Th6veninvoltage with respecrto rhe
terminalsa,b.
b) Fhd the Th6venin impedancewith respectro
the terminalsa,b.
c) Draw the Th6veninequivalent.
9.45 Find Zabin the circuit shownin Fig.P9.45whenthe
circuitis operating
at a frequency
of 1.6Mracvs.
FiguleP9.45
15ra
Probtems
383
946 Find the Th6venin impedance seen looking into the
telmhals a,b of the circuit in Fig. P9.46 if the frcquency of operation is 25 krad/s.
Figrrc P9.46
5
t 0 _
".'l
1ko
9.50 The circuit shown in Fig. P9.50is operating at a frcquency of 10 krad/s. Assume a is rcal and lies
between 50 and + 50,that is, 50 < a < 50.
a) Find the value of a so that the Th6venin impedance looking into the terminals a,b is purely
_
19i,\ a nr
What is tlle value oftheTh6ve n impedance for
the a found in (a)?
c) Can a be ddjusredso lhal lhe fierenjn
impedanceequals5 + /5 O? If so,what is the
d) For what values of a vrill the Th6venin imped-
ancebe inductive?
9.47 Find the Norton equivalentwith respectto terminalsa,bin the circuit of Fig.P9.47.
Figure
P9.50
FigueP9.47
Section 9.8
9.51 Use the node-voltage method to find Va in the circuit in Fig.P9.51.
9,48 Find the Thdvenin equivalent circuit with respect to
t}le terminals a,b of the circuit shown in Fig. P9.48.
Figure
P9.5r
j40a
rigureP9.48
600O j150O
1004 v
+
60{)
40r)
v,
j20a
9.52 Use the node-voltagemethod to find the steadyAflft stateexpression
for o,(t) in the circuit in Fig.P9.52if
9.49 Find the Norton equivalentcircuit with respectto
the terminalsa,bfor the cLcuit shownin Fig.P9.49
whenVr : 25 A V.
Figure
P9.49
?Jsr= 10cos(5000t+ 53.13')V,
0s2= 8 sin50001V.
figuru
P9.52
./250O
384
Sinusoidat
Steady-state
Anatysjs
9.53 Use the node voltage method to fhd t}Ie steady-state
expressionsfor t}le branch curents ia and ib in tle
circuit seen in Fig. P9.53if za : 100sin10,0001V
axd ?b : 500cos10,000t
V.
Sectior 9.9
9.57 Use the mesh-curent method to find the st€ady
slateexpressionfori.(r) in the cncuit in Fig.P9.57if
t)x = 60 cos40,0001V,
Figure
P9.53
tb = 90sin(40,000r
+ 180')v.
5 r.F
Figure
P9,57
1.25tlF
20tl
9.54 Use the node-voltagemethod to find the phasor
voltageVs in the circuit shownin Fig.P9.54.
9,58 Use the mesh-currentmethod to find the sleadystateer?ressionfor oo(r)in the circuit in Fig.P9.52.
tigureP9,54
9,59 Use the mesh-currentmethod to find the phasor
curent Is in the circuit in Fig.P9.54.
/3()
9.60 Use the mesh-currenrmethod to find the bmnch
currenlsla. Ib. lc. and ld in lhe circuil sho$n in
Fig.P9.60.
510"A
Figure
P9,60
5-9!I v
2CA
9.55 Use the rode-voltagemethod to find V, and I, in
the circuitseenin Fig.P9.55.
r.,i50
FiguEP9.55
5o
i:o
100r1ry
v
500 v
j50o
9.56 Use the node-voltage method to find tle phasor
voltage V. in the circuit showrl in Fig. P9.56.Express
the voltagein both polar and rectangularform_
figureP9,55
lr
9.61 Use the mesh-currentmethod to find the steadysrft state expressionfbr o, in the cfucuit seen in
Fig.P9.61.ifrr equals72cos5000rV.
FigrrcP9.61
500nF
15-CA
Secrions9.5-9.9
9.62 Use the concept of voltage division to find the
6nc steady-state expression for ?.(1) in the circuit in
Fie.P9.62it ts = 75cos50001V.
tigureP9.62
300O 400mH
9.63 Use the €oncept of curent divisior to find the
ErI( steady-stateer?ression for ,, in tlre circuit in
Fig.P9.63if is - 125cos500t
9.67 The op amp in the circuit seenir Fig. P9.67is ideal.
tsdc Find the steadystate expressior for o,(t) when
?Jr: 20 cos10bl V
Figure
P9.57
Figure
P9.63
40 ko
250c)
50r}
20 pF
1H
r0k0 |
r-\?6 v
9.64 The sinusoidal voltage source in the circuit shown
tsPld in Fig. P9.64 is generaring tie
volrage
os = 1.2cos1001V.If the op amp is ideal,what is
the steady-state
expressionfor o.(l)?
Fig(reP9.54
200ko
9.65 The 1 /,F capacitor in the c cuit seenin Fig. P9.64is
6PIG replaced witl a variabl€ capacitor. The capacitor is
adjusteduntil the output voltage leads the input
voltageby 120".
a) Find the values of C in microfarads.
b) Write the steady-state
er?ressionfor uo(r)when
Chas the valuefound in (a)9.66 The op amp in the circuit in Fig.P9.66is ideal.
"""
a) Find the steady-state
expressionfor o,trl.
b) How largecanthe amplitudeofos be beforethe
amplifiersaturates?
9.68 The opeiational amplfier in the circuit shown in
sPI.r Fig. P9.68is ideal. The voltage of the ideal sinu\oidalsuurce
is ,s = I0co.2
l0sr \ .
a) How small can C, be before the steadystate
output voltage no longer has a purc sinusoidal
wavelbrm?
b) For the value of C, found in (a), write the
steady-state
expressionfor r),.
Figurc
P9.58
12.5nF
386
SinusoidatSteady-stat€Anatysi,
9.69 a) Fhd tle hput impedance Zab for the circuit in
Fig. P9.69.Express Zab as a function of Z and I<
wheref : (R,/nr.
b) If Z is a pure capacitive elemenr, what is the
capacitanceseen looking into the teminals a,b?
tigureP9.69
Seclion 9,10
9.72 a) Flnd the steady-state expressions for the currents is and iL in the circuit in Fig. P9.72 when
?s : 200 cos 10,000tV.
b) Find the coefficient of coupling.
c) Find t}le energy srored ir the magnetically coupled coils at, = 50zps and t : 1002,.s.
F,gweP9.72
5()
0.5mH
1nH
9.70 For the circuit in Fig.P9.57,suppose
D, = 5 cos80,000rV
,,, = -2.5 cos320,000t
V.
15r}
lmH
9.73 The sinusoidal voltage source in the circuit seen in
6prft Fig. P9.?3 is operating at a frequency of 50 kmd/s.
The coefficient of couplhg is adjusted until the
peak amplitude of ir is maximum.
a) What is the valueof ,t?
b) Wlat is the peak amplitude of i1 if
.'s : 369cos(5 x 104r)v ?
Figur€
P9.73
a) What circuit a.nalysis
techniquemustbe usedto
fhd the steady-state
expressionfoi i.(t)?
b) Findthe steady'state
expression
for t,(r)?
20o"
75[)
1000
3000
5 rnH
9.fl For the circuit in Fig.P9.71suppose
rr1= 20cos(2000r 36.87")V
+ 16.26)V
r', = 10cos(5000r
a) Wlat circuit analysistechniquemustbe usedto
find thesteady-state
expression
for 2,(1)?
b) Findthesteady-state
expression
for o.(t).
Figure
P9.71
100pF
9.74 For t}le circuit in Fig. P9.74, find the Th6venin
equivalent with respect to t}le terminals cd.
figuleP9.74
15r}
225U +
v Gns)
jsoo
300
j20o
9.75 The value of I itr the c cuit in Fig. P9.75is adjusted
so that Z,b is purely resistive when zo- 25 krad/s.
F]]trdZ^b.
PobLers 387
40()
b) Show that if the polatity terminal of either one
of the coils is reveNed that
-
Z
L
/
N \,
t1--:t
9.?6 A series combhation of a 150 O resistor and a
20 nF capacitor is comected to a sinusoidal voltage
sourceby a Uneartransformer.Tbesourceis oper
ating at a iiequercy of 500 krad/s.At this frequency,
the intemal impedanceof the sourceis 5 + i16 O.
The rms voltage at the terminals of the souce is
125V when it is not loaded. The parameters of the
Lhear transformer are n1 : 12 O, 11 : 80 pH,
R2 : 50 O, L2 : 500 pH, and M : 100 PH
a) Wlat is the value of the impedance rellected
into the primary?
b) What is the value of the impedance seen ftom
the terminals of the pmctical source?
Section9.11
9.7? Find ttre impedance Zabin the circuit in Fig. P9.77if
zL=2
tigureP9.79
9.80 a) Show that the impedance seen lookhg into t]le
termimls a,b in th€ ci(cuit in Fig. P9.80is given
by the expression
+ i1'50
A
'^, = ('
FigureP9.77
.t)'"
Show that iI the polarity terminals of either one
of the coilsis revened,
9.78 At first glance,it may appear trom Eq. 9.69 that an
inductive load could make the reactanceseenlooking irto the pdmary terminals (i.e., Xab) look
capacitive. Intuitively, we know tlis is impossible.
Show that -l.ab can never be negative if ,YL is an
inductive rcactance.
9.79 a) Show tiat the impedance seen looking into the
terminals a,b in the circuit in Fig. P9.79is given
by the expression
(.,
r^":
('-ft)'"'.
388
sinusoidatSteady-StareAnatysjs
Section9.12
Sections9.1-9.12
9.81 The pammeters in the circuit showl in Fig. 9.53 are
Rr = 0.1 O,arl1 = 0.8 A. R2 : 24 A,oL2 : 32 A.
andvl=240+J0V.
a) Calculatethe phasorvoltagey.
b) Connect a capacitor in parallel with rhe inducror,
hold VL constant, and adjust the capacitor until
the magnitude of I is a minimum. What is the
capacitive reactance?What is the value of V,?
c) Fird the value of the capacitivereactancethat
keeps the magnitudeof I as small as possible
and thal at the sametime makes
lYl : Yr =240v
9.82 Show by using a phasordiagramwhat happensro
lhe magnirude
andphasedngleo[ the vollage.,, in
the circuit in Fig.P9.82as R, is vaded from ze{o to
infinity. The amplitude and phase angle of the
sourcevoltageare held constantasRr varies,
Figure
P9.82
9,83 a) For the circuit shown in Fig. p9.83,compurey
and V.
b) Constructa phasor diagram showingthe relationshipbetweenV", V, and the load voltageof
4401!: v.
c) Repeat parts (a) and (b), given rhar rhe load
voltageremainsconstantat 440 A
V. when a
capaqtrve reactance of -22 O is connected
acrossthe load terminals,
FiguEP9.83
+ 0.2rl
-
_
l
j1.6O +
14011!:
V
j22n j22A-;:
9.84 You may have the opportunity as an engineering
gracluateto sene as an expe witnessin lawsuits
involving either personalinjuy or property damage.As an example of the type of problem on
which,ou may be askedlo gj!e an opinion.consider the following event.At the end of a day of
fieldwork,a famer returnsto his famstead,checks
his hog confinemert building,and finds to his dismay that the hogsare dead.Theproblem is rraced
to a blown fuse that causeda 240V lan motor to
stop. The loss of vertilation led to the suffocation
of the livestock.The interupted fuse is locatedin
the main switchthat connectsthe farmsteadto the
electrical service. Before the insurance company
settlesthe claim,it wantsto know if the electriccircuit suppling the farmsread functioned properly.
The lawyersfoi the insurancecompanyatepuzzled
becausethe farmer'swife,who wasin the houseon
the day of the accidentconvalescingfiom minor
surgery,was able to watch TV dudng the aftei
noon. Furthermore,when she went to the kitchen
to start ptepaing the evening meal, the electric
clock indicatedthe conect time. The lawyershave
hired you to explain (1) why rhe etectricctock in
the kitchen and the televisionsetin the living room
continued to operate after the fuse in the main
switch blew and (2) why rhe second fuse in rhe
main switchdidn't blow after the fan motor stalled.
After ascertainingthe loads on the tlnee-wire distribution circuit prior to the interruption of fuseA,
you ate able to constructthe circuit model shown
m Fig. P9.84on page 389.The impedancesof rhe
line conductors and the neutral conductot are
assumednegligible.
a) Calculatethe branchcurents 11,I2j13!14,15,
and
16priorto rbe inlerruplionot fuseA.
b) Calculate the bmnch currents after |nc ulerruption of fuse A. Assume the stalled fan motor
behavesas a shoit circuit.
c) Explain why the clock and televisionset were
not affectedby the momentaryshort circuit that
intefupted fuseA.
d) Assume the fan motor is equipped with a rhermal cutout designedto interrupt the motor circuil if lhe molor currenrbecomeserce.si\e.
Would you er?ect the thermal cutout to operate?Explain.
e) Explain why fuseB is not interrupredwhen the
fan motor stalls.
ProbL€ms
389
FisunP9.84
FuseA (100A)
lro-irta.y
1201U'
circuil
nrernPrs
fusc A
| 1,-
X
I t2
; :-
24l)
t.l
15A
1 2{ }
r20[t
brunchcarryingI, is modelingthe neutral conductor. Our purposein analyzingthe circuit is to show
the importance of the neutral conductor in the satisfactory operation of the circuit. You are to choose
the method lor analyzing the circuit.
a) Showthat I, is zero if -R1:.R2.
15A
b) Showthat V : Vz if R' = Pr.
c) Open the neutral branch and calculate Y and V2
if R1 : 60 o, R2 = 600o, and n.r = 10 o.
d) Closethe neutralbranchand repeat(c).
e) On the basis of your calculations, explain why
the neutral conductoris never fused in such a
manner that it could open while the hot conducrors are energveo.
o
16.r0
F;;;"t..
FuseB (100A)
935 a) Calculatethe branchcurrentsll 16inthecircuit
J,liil'jfr.. in Fig.P9.s8.
b) Fhd tie primary curent Ip.
tigureP9.87
_IL
9.86 Supposethe 40 O resistancein the distributioncirRndL cuit in Fig.P9.58is replacedby a 20 O resistAnce
. +
a) Recalculate the branch curent in the 2 O
resistor,12.
14tffkv
b) Recalculate
lhe pfima4 cufrenr.I!
c) On the basis of your answers,is it desirable
to have the rcsistanceof the two 120 V loads
be equal?
9.87 A residentialwidng circuitis shownin Fig.P9.87.In
,lH;lLliEthis model, the resistor R3 is used to model a 240V
appliance (such as an electric range), and the resistors Rr and R2 are usedto model 120V appliances
(such as a lamp, toaster,and iron). The branches
carrying 11 and I, are modeling what electricians
refer to as the hot conductors in the circuit, and the
0.02O
0.020
j0.02o"
t0.02tt
rzsffv
v,Jn,
o
10.03
' + o.o3
o
:f;'
l2stffv
_
O
0.02
-1
Rr
v.€R,
jDUA
+tz
9.88 a) Find the pdmary current Ip foi (c) and (d) in
Problem9.87.
b) Do your answersmake sensein termsofknown
circuitbehavior?
SinusoidaL
Steady-State
Power
CaLculations
Powerengineeringhas evolvedinto oneof the importantsub10.1 Instantaneous
Powerp. J92
10.2 Averageand R€activePowerp. J94
10.3 Therms Vatueand Pow€r
p. 398
CaLculations
1a.4 Compter ?owet p. 401
10.5 Power(aLculationsp. 40J
10.6 l4arimumPowerTransf€rp. 410
Understand
the fotLowinq
ac powerconcepts,
thejr rctationships
lo oneanother,aid howto
catcuLate
themin a circuit:
. Instant€feous
powet
. Averag-a
(reat)power;
.
power;ard
Comptex
Und€rtardthe conditioifor maximum
reaL
powerd€tjvered
to a toadjn an ac circuitandbe
abteto calcutate
the toadimpedance
reqLired
to
d€Ljver
maximum
reatpowerto the load.
Beabtelo catcutate
att foms ofac powerin
accircuitswirh tireartransfornrers
andii
accircuitswith ideattnnsformers.
390
disciplineswithin electricalengineering.The range of problcms
dealingwith the delivery of energyto do work is considerable,
from determining the power rating within which an appliance
operatessafclyand efficiently.to designingthe vast array of gcn
crators,tlansfbrme$, and wires that provide electric energy to
householdand industrialconsumers.
Nearly all electlicenergyis suppliedin the tbrm of sinusoidal
voltagesand curents. Thus, atter oul Chapter g discussionof
sinusoidalcircuits,this is thc logical place to considetsinusoidal
steadystatc power calculations.We are pimarily intcrcsled in
the averagepower deliveredto or suppliedfron a pair of terminals as a result of sinusoidalvoltagesand currents.Other meas
ures. such as reactive power, complcx power, and apparent
power,will also be presentcd.The conceptof the rnrs value of a
sinusoid,briefly introducedin Chapter9, is particularlypcrtincnt
to power calculations.
We begin and cnd this chapterwith two coDceptsthat should
be very familiar to you hom previouschapters:thc basicequation for power (Section 10.1) and maximum power transler
(Section10.6).In bctween,we discussthe generalprocessesfor
analyzingpo\\,er,which will be familiar from your studies in
Chapters1 and 4, althoughsome additional malhematicaltech
niqucs are required here to deal with sillusoida],rather than dc,
signals.
Perspective
PracticaI
App[iances
Heating
voLtages
andcur
the steady-state
In Chapter9 we caLcutated
In thjs
sources.
rentsin eLectrjc
circuitsdfivenby sinusoidal
we
power
The
techiiques
in
such
circuits.
chapterweconsjder
rnanyof the etect caLde\rces
developareusefuIfor anatyzing
sourcesare the prewe encounterdaity,becausesinusoidat
doninant meansof providingeLectricpowerin our homes,
schoots,
andbusinesses.
is heaters,which
one cornmonclassof electricaldevices
jnctude
energy.
ExampLes
transforfir
etectricenergyinto thermaL
electric atovesand ovens,toasters,irons, etedric water
dryers,andhairdryers
eLectric
cLothes
heaters,
spaceheaters,
in " hedle is poie-'oI
onFol rherrilicatde.iqrroncerrs
js
Themorepower
sumption,Power impoffantfor two reasonsl
andthe moreheat
a heateruses,the moreit coststo operate,
it canproduce.
I4anyelectricheatershavedifferentpowersettingscorre
spondingto the amountof heatthe devjcesupptiesYoumay
wonderjusthowthesesettjngsresuttin differentanrountsof
exampleat the end of
heatoutput.ThePracticatPerspechve
hair dryer
the designof a handheLd
this chapt€rexanrines
figure).
with threeoperatiigsettingsGeethe accompanying
provides
for
three
different
YouwjLtseehow the design
powerlevets,which correspordto three djfierentL€vetsof
heatoutPut.
391
392
SinusoidaL
Steady
State
Power
catcutations
10.1 Instantaneous
Fower
We begin our investigationof sinusoidalpower calculationswith the
familiar circuitin Fig.10.1.Heie, o and i arc steady-state
sinusoidalsignals.
Using the passivesign convention, the power at any instant of time is
p - ui.
figur€10.1A Theblackboxreprcsentatjon
of a circujt
usedforcalcutating
power.
(10.1)
This is inslatrfaneouspower. Remember that iJ the reference direction of
tlle current is in the direclion of the voltage rise, Eq. 10.1must be w tter
with a minus sign.Instantaneouspower is measuredin watts \qhen the
voltage is ir volts and the current is in amperes.First, we write expressions
lbr ?)and i:
r,: Vhcos(at + 0.),
i:
(10.2)
I -cos (at + 0),
(10.3)
wherc d. is the voltage phaseangle,and 0; is the currelt phaseangle.
We are operating in the sinusojdal steady state,so we may choose any
convenient relerence for zero time. Engineers designing systems that
transfer large blocks of power have found it convenient to use a zero time
coirespondingto the instantthe curent is passingthrougha positivemaximum. This reference system requires a shift of both the voltage and cur
ient by pr.ThusEqs.10.2and 10.3become
r-U-cos(at+0,-0),
(10.4)
(10.5)
Whenwesubstitute
Eqs.10.4
and10.5intoEq.10.1,rheexpression
Ior rhe
instantaneous
powerbecomes
p = V^I ^cos (at + d0
0i) cosor.
(10.6)
We could useEq. 10.6directlyto find the averagepower;however,by simply applyirg a coupleof t gonometricidentities,we canpur Eq. 10.6irto
a much more inlormative form.
We beginwith the t gonometricidentityr
co.ocosB
]..,,"
to e4and Eq. 10.6;lettinga : at + p!
o :T*"r,
L Secenlry 8 in Appendir F,
ei)+ W
u'
]*.,"
- B,
,i and B = al gives
cos(2at
+ o, - oj).
(10.7)
Powet 393
10.1 Instantaneous
Now use the trigonomeidc identity
c o s ( d+ B ) :
c o s a c o s P- s i n d s i n B
to expandthe secondterm on the right-handsideofEq.10-7'which gives
e =).ns(0,-
+Lf
o1)
cos
@,
cos2at
01)
!t!o"in1o,- e,1"a2,,
(10.8)
Figure 10.2 depicts a representative relationship among t', ', and p,
basedtn t}le assumptionsd, = 60' and d = 0" You can seethat the freouencv of the instantaneouspower is twice the frequency of the voltage or
c'urrent.Ttris observation ah; follows directly from the second two terms
on the right-hand side of Eq. 10 8. Therefore, the iNtantaneous power
goes through two complete cyclesfor every cycle of either the voltage or
ihe curt.ot. Also ttote that the instantan€ouspower may be negative for a
po(ion of each cycle,even if the network between the terminals is passive
in a completely passive network, negative power implies that energy
storealin the inductors or capacitors is now being extracted.The facf that
the instantaneous power varies with time in the sinusoidal steady_state
oDeration of a circuit explains why some motor-driven appliances(such as
refrigerators) experienci vibration and require resilient motor mountings
vibration.
to Dreventexcessive
\ e a r en o wr e a d \t o u s eE q l U . 8 l o l i n d l h e a \ e r a g e p o w e r a l l h e l e r minals of ttre circuit ;epresented by Fig. 10 1 and, at the same time, introduce the concept of reactive power'
3V-r^
v^I2
0
4,
ilM
Gadiant
power,
andcun€mvdsusor for
vottage,
figure10.2l. Instantaneous
opention
sinusoidal
steady-state
394
Sinusoidat
Steady
StatePower
CatcLrLations
10.2 Average
andReactive
Forser
We beginby noting that Eq.10.8hasthree terms,which we can rewrite as
follows:
p:P+Pcos2ot-Qsin2at,
(reat)powerF
Average
Reactivepowerts
(10.e)
r =h!cos(e,- e,:t,
(10.10)
Q:2;Lslr (o"-a).
(10.11)
P is called the averagepower, and O is called the r€active pow€r. Average
power is sometimescalledreal power,becauseit descdbesthe power in a
circuit thal is transformedfrom electricto nonelectricenergy.Altho gh
the two terms are inierchangeable, we prinarily use the term dvenge
It is easyto seewhy P is calledthe averagepower.The averagepower
associatedwith sinusoidalsignalsis the averageof the instantaneous
power over one period,or,in cquationlorm,
+1,.^"
(10.12)
where 7 is the period of the sinusoidalfunction.The limits on Eq. 10.12
imply that we caninitiate the integmtionprocessat any convenienttime ro
but that we must terminatethe integrationexactlyone period later. (We
could integrateovcr nZperiods,wherer is ar integer,providcdwe multi
ply the integralby 1/rI.)
We couldfind the avemgepolverby substitutingEq.10-9directlyinto
Eq. 10.12and then performingthe integration.But note that the average
value ofp is given by the fi6t term on the right-handside oI Eq. 10.9,
becausethc integral of both cos2ol and sin2ot ovcr one pe od is zero.
Thusthe avemgepower is givenin Eq. 10.10.
We candevelopa better understandingofall the temsinEq.10.9 and
the relationshipsamongthem by examiningthc power in circuitsthat are
purely resistive.purely hductive, or purely capacitive.
Powerfor PurelyResistive
Circuits
If the circuit betweenthe terminalsis purely resislive,thevoltageand current are in phase,whichmeansthat d! : dr.Equalion 10.9then reducesto
0
0.005 0.01 0.015 0.02 0.025
rine G)
Figlre10.3A Instantaneous
reatpowerandaverage
powerfora purelyresistilecjrcujt.
p = P + Pcos2ot.
(10.13)
power cxpressedin Eq. 10.13is referred to as the
The instantaneous
instatrlatr€ous
real power. Figure 10.3sholvsa graph of Eq. 10.13for a
Power 395
andReactive
10.? Averaqe
purely resistivecircuit,assumingo = 377 rud/s.By defini
representative
tion, the averagepower,P, is the averageof p over one period Thus it is
easyto seejust by looking at the graph that P = I for this circuit Note
from Eq. 10.13that the instantaneousreal power can never be negative,
whichis alsoshownin Fig.10.3.Inother words,power cannotbeextracted
from a purely resistivenetwork. Rather, all the electiic energyis dissi_
patedin the form ofthermal energy.
Powerfor PurelyInductivecircuits
If the circuit betweenthe terminalsis purely inductive,the voltage and
currentare out ofphaseby precisely90'.In particular,thecurrentlagsthe
voltage by 90" (that is, dr: d! 90'); therefore 0u - 0t: 199' 11.
power then reduccsto
expressionfor the instantaneous
p : -Qsir,2dt.
(10.14)
In a purely inductive circuit, the average power is zero. Therelbre no
transfomation of energy ftom electdc to nonelectric form takes place
The inslantaneouspower at the terminalsil a purely inductivecircuit is
continually exchangedbetween the circuit and the source driving the circuit, at a ftequencyof 20. In other words,when 2 is positive,cnergy is
with the inductiveelements,
beingstoredh the magneticfieldsassociated
and when p is negative,energy is being extracled from the magnetic fields
A measureof the power associatedwith purely inductivecircuitsis
the reactive power O. The narne reactivepower comesfrom the characterization of an inductor as a reactiveelement;itsimpedanceis purely reactive. Note that averagepower P and reactivepower O cary the same
dimension.To distjnguishbetweenaverageand reactivepower,we usethe
units na, (W) for avemge power and var (vrlt-amp ledctive, ot yAR) tor
power for a represenreactivepower.Figure 10.4plots the instantaneous
:
377 rad/s and O : 1 VAR
lative purely inductive circuit, assumingo
1.1)
c (\ AR)
-0.5
1.0
0
0.005 0.01 0.0i5 0.02 0.025
rine G)
rcatpower,average
Fiqlre10.4A Instantaneous
inductive
circuit
powerfor a purcLy
power,
andreactiv€
Circuits
for PurelyCapacitive
Power
If the circuit betweenthe terminalsis purely capacitive,the voltageard
current are precisely90' out of phase.In this case,the current leadsthe
voltageby 90' (that is,0' = do + 90'); thus,dN 0i : -90'. Thc er?respower then becomes
sionfor tie instantaneous
p =
Q sin2ar.
(10.15)
1.{J
6
E
0.5
0
-0.5
Again, the averagepower is zero,so there is no transformationof energy
from electric to roflelectnc form. ln a purely capacitive circuit, the power
-1.0
is conthually exchanged betweer the source driving the circuit and the
a
with the capacitiveelements.Figure10.5plots the
electic field associated
1.5
power
for a represenaativepurcly capacitive circuit, assuminstafltaneous
0
0.005 0-01 0.015 0.02 0.025
=
=
1VAR.
it.rgu 377 radlsandQ
rine G)
Note that the decisionto use the current as the referenceleadsto Q
reatpowerandaveraqe
beingpositivelor inductors(that is,0? di = 90" ard negativefor capac tigrre 10.5s Instantan€ous
pur€ly
capacitive
cjrcuit.
powerfor
a
in
recognize
tris
difference
(that
Power
engineets
is,d,, 9'=
90'.
itors
396
Srnusoidal
Steady
StatePower
taLcutatjons
the algebraic sign of O by saying that inductors demand (or absorb) magnetizingvars,and capacitoNfumish (or deliver) magnetizingvars.We say
more about tlfs convention later.
ThePowerFactor
The angle d,
P, plays a role in the computation of both avemge and
power
reactive
and is referred to as the power factor angle.The cosine of
this angle is called the pow€r faclor, abbreviated pf, and the sine of this
angleis calledthe reactivefactor,abbreviatedrl Thus
Powerfactor>
(10.16)
rr : sin(d,
0,).
(10.17)
Knowing the value of tlle power factor does not tell you the value of t}le
power factor angle,because €os(9!
d,) : cos (di - d!). To completety
describethis angle,we use the descriptive phraseslagging power factor and
leading power f|clor. Lagging power factor implies that current lags vol!
age-hence an inductive load. Leading power factor implies that current
leadsvoltage hencea capacitiveload. Both t}te power factor and the ieactive factor are convenient quantities to use in descdbingelectrical loads.
Examplel0.l illustratesthe interptetationofP and O on the basisof
a numedcalcalcurauol
Calculating
Averdge
andReadivePower
a) Calculatethe averagepower and the rcactive
po\reral the tefminalsol lhe ner$orksho$n in
Fig.10.6if
100cos(or + 15') V,
I : 4sin ((,)t- 15') A.
b) State whether the network inside the box is
absorbingor delivedngaveragepower.
c) Slale whelher rhe net$ork inside lhe bor i.
absorbingor supplyingmagnetizingvars-
Figure10.6 ^ A pairo'teTinaL used_0rca.cutaring
poqe..
Solution
a) Because i is expressedin terms of the sine function, the filst step ir the calculation for P and O
is to rewrite i as a cosine function:
t - 4 cos(ot
105') A.
We now calcnlate P and C directly from
Eqs.10.10and 10.11.Thus
P-
I
(
= -100w.
;(100)ra)cosIls Iosij
= 173.21
vAR.
1rs- ( 105)l
Jrool+;.in
b) Note from Fig. 10.6the use of the passivesign
convention. Because of this, the negative value
of -100W meansthat the network inside the
box is delivering avemge power to t]te terminals.
c) The passivesignconventionmeansthat,because
O is positivg t]le network inside the box is
absorbingmagnetizingvaISat its terminals.
Power 397
10.2 Avenqe
andReactjve
to oneanother,
andhowto catruateth€min a circuit
their retationships
Obiective
1-Unde6tandac powerconcept!,
10.1 for (dctol lhelollo$ing.etsolvolr"gernJ
currenl,calculatethe real and rcactivepower in
the line betweennetworksA and B in the clrcuit shown.In eachcase,statewhetherthe
powerflow is from A to B or vice versa.Also
statewhethermagnetizingvars arc beingtrallsferred from A to B or vice versa.
a) 1)= 100cos(or 45")V;
+ 15")A.
i = 2ocos(a,r
Answer: (a) P = 500 W (A 10B),
B66.03VAR (B to A);
a=
( b ) P : -866.03w (B toA),
a =500vAR
( . ) P = 500w (A to B),
866.03vAR (A to B);
(d)P:
500 W (B to A),
o - -866.03VAR (B ro A).
b) 0:100cos(dt 4s')
t = 20cos(@t+ 165')
t:
(A to B)r
100cos(.,t - 45") V ;
20 cos(or - 105')
100cosol V;
20 cos((,r + 120")
10.2 Computothe power Iactor and the reactivefac_
tor lor the network insidethc box in Fig.10.6,
whosevoitageand curent are describedin
Example10.1.
Hirt:Use -i to calculatethe power and reac_
Answer pf - 0.5leading:f - -0.866.
NOTE: Also tJ Chaptet Prcblen 14.2.
Ratings
Appliance
Avcragepower is usedto quantity lhe powcr needsof householdappliances.Theaverageporvcrrating and eslimatcdannualkilowatt hour consumptionof some comn1onappliancesarc presentedin Tablc l01.lhe
cnergy consumplionvalues are obtained by estimatingthe number of
hoursannuallythat the appliancesare in use.For example,a coilccmaker
hasan cstimatedannualconsumptionof 140kwh and an averagcpower
consumptionduring operation of 1.2 kW. Thcrcfore a cofleenaker is
assumed10be in operation1,10/l-2,or 116.67.hourspcr year.or approxi
narely l9 ninutes per da),.
Exampie 10.2usesTablc 10.1 to determinc whether tbur common
appliancescan all be in opcrationwithoul excccdingthe currenl carrying
capacjtyof the household.
398
Sinusojda
t Stea
dy'State
Power
catcuLaiions
MakingPowerCalculations
InvolvingHousehotd
Appliances
The branch circuit supplyhg the outlets in a typical
home kitchen is wired with #12 conducror and is
protectedby either a 20 A fuse or a 20 A circuit
breaker.Assume that the following 120V appliances
are in operation at the same time: a coffeemaker,
egg cooker, hying pan, and toaster. Will the crcurt
be interruptedby the protectivedevice?
Avenge
Wattag€
Est kwh
Consumed
AnnurlJl
Solution
From Table10.1,thetotal averagepower demanded
by 0re four appliancesis
P : 1200+ 516 + 1196+ 1146 = 4058W.
The total cunent in the protectivedeviceis
40s8
-]]:1".
_ : 12{l e 33.82A.
Yes,the protectivedevicewill interrupt the circuit.
trsr.kwh
Appliance
Average
Consumed
Wrttrge
Annurryr
Health and beauly
1,200
Dishwashel
Egg cooker
Fryins pan
Mixer
oven. microsave (only)
1.201
516
1,196
127
1,450
12,20t)
1,146
140
165
14
100
2
190
596
39
LrDdry
Clothes dryer
Washingmachi.e. automatic
103
4,219
4,811
Fan (circulaiins)
Heater (portable)
860
25'7
88
7,322
860b
37'1
43
1'76
600
15
279
25
0.5
Home entertrinmenl
Radio
Tclevision,color.tube type
Sotid-staterype
993
Comforl conditioning
Dehumidifier
Sunlamp
Clock
4.856
572
Quick recovery t}?e
Air onditioner (roon)
Han dryer
'71
86
240
145
528
320
2
630
1
7
46
a) Basedon nomal usage.When
usingtheseligureslorprojectionn
such factors as thc size ol the specilic .ppliane, the geographiml
area ot use,and nrdividual usageshould be laken into consideration, NoIe Lhat the wartagesarc not additive, since all units are
.ormauy noi in oPeralionai the samctime
b) B.sed oD 1000hous of operatiod per year.This figure will vary
Nidelt depending on the area and the speciiic sire of the unit. See
EEI-lub *76 2, Air Condidoring Usage Srudy." for an estinare
JdL/.z: Edisor Elecrric Insiituie.
NOTE: Assessyur uru)erytandingof this materiljl4i trying Chapter Ptoblen n3.
catcutations399
10.3 ThennsVatue
andPowe,
10.3 ThermsValueandPower
Catculations
In introducing the rms value of a shusoidal voltage (or €urrent) in
votiase
apptjed
to ihe
Section9.1,we mentionedtlat it would play an important role ir power Figure10.7A A sinusojdat
calculations.We can now discussttris role.
Assume that a sinusoidal voltage is applied to the terminals of a resistor, as shown in Fig. 10.7,and that we want to delermine the average
power deiiveredto the resistor.From Eq.10.12,
'^= i .1l
['r+rv)-cos'(dt + O) ,
"'
R
,,
:*l j/'v;-.l.t+6,1
^ L t J n
I
1.
(10.18)
ComparingEq. 10.18with Eq. 9.5 revealsthat the avemgepower delivered to R is simply the rms value of the voltage squared divided by R, or
(10.19)
R
If the resistoris carrying a sinusoidalcurent, say.I-cos (i'))t+ dr),
tlle averagepower delivered to the iesistor is
P : rl-"R.
(10.20)
The rms value is also referred to as the effective value of the sinusoidal voltage (or cullent). The rms value has an interestingproperty:
Given an equivalent resistive load, R, and an equivalent time period, Z,
the rms value of a sinusoidal source delivers the sameenergy to R as does
a dc sourceof the samevalue.For example,a dc souce of 100V delivers
the sameenergyin ? secondsthat a sinusoidalsourceof 100V!,." delivers,
assumingequivalentload resistarces(see Problem 10.11).Figure 10.8
demonstratesthis equivalence.Energywise,the effect of the two sources
is idertical. This has led to the term efrectire value being used interchangeablywith rmr ral,ra
The average power given by Eq. 10.10 and the reactive power given
bv Eo.10.11canbe written in tems of effectivevalues:
r:)cos@,
e1)
:!,$"""t'"-'t
: %ffI.ncos(d, - 0J;
+
?, = 100V (rnt R
+
Y. : 100v (dc) R
vatueof!" (100V ms) detiversthe
Figure10,8A lhe effective
v" (100V dc).
samepowertoR asthedcvottage
(10.21)
400
SinuroidatSteadjr
5tatePor,/er
Caru aiions
and,by sirrilar maripulaLion,
O = I/.i}I.irsin(d"
d).
170.22)
The effectivevalue ofthe sirusoidalsignalin powcr calculationsis so
widcly used that voltage and current ralings oI circuils and equipment
involvcd in porer utilizationare given ir terms o[ rms valucs.For examplc. thc voltagerating of residentialelectdcwiring is oltcn 240V/120 V
voltagcs
seNice.Thesc|'oliagclcvcls are the rms valuesof the siDrrsoidal
supplicdby thc utilit)' companfwhich prorides power allwo voltagclevcls !o accommodatelow-voliage appliances(such as lclcvisions) and
highcr voltagc appiiances(such as electric raDges).Appliancessuch as
eleclriclamps.irons.and toastersall carry rms ratingson their namcplates.
For examplc,a 120V 100w lamp hasa resistanoeof 120'/100,or 144O.
ard drawsan rnis currcnt ot 120/144.of 0.833A. The peak valuc of thc
lamp currentis 0.833\4, or 1.18A.
The phasorlransfbrmof a sinusoidalfunction may also be expressed
in lerms of the nns valuc.Thcmagnitudeof the rns phasoris equalto thc
rms valLreol lhc sinusoidalfunction.If a phasoris baled on thc rms valuc.
we indicalethisbycithcr an explicitstatement,aparenlhelicai"rms'adja
cenl1|)lhe phasorquantit]'.or the subscript"efl" asin Eq.10.21.
ln Exanplc l0.3.wcillustratetheuseofrms values{or calculalingpowcr.
bysinusoidaI
vottage
Determining
Average
PowerDelivered
to a Resistor
a) A sinusoidalvoltagehaving a n dmun anplitude of 625 V is applied to the terninals of a
50 O resistor.Find the averagepower delivered
Eq- 10.19.the averagepower delivered1()$c
50 f) rcsistoris
r44lrJ4rl
|::] -=lqnb.2sw.
b) Rcpcat (a) by first finding the c rrent in the
SoIution
a) Thc rms value of the sinLrsoidaivoltage is
625iy2. ot approxi ately 4,11.94V From
1J
b) Thc maximum amplitude of the culrenl in thc
resislor is 625/50.or 12.5A. The rms value of
lhe currcnt is 12.5/rD. or approximaleiy
8.84 A. Hcnce the averagepower delivered lo
P = (8.84)'50: 3906.25w.
obiective1-ljnderstand ac powerconcepts,their relationshipsto one another,and howto calculatethem in a orcuit
10.3 The periodictrlangularcurrentin Example9.4,
repcatedhere.hasa peak valueof 180mA.
Find thc avcragcpo\r'crthat this currentdelive$ to a 5 k() rcsistor.
NO'[E: Aln nj ChapterPrcblen 14.13.
Power 401
10.4 CompLex
Power
10.4 Complex
Before proceedingto the vadousmethodsof calculatingreal and reactive
power ir circuits operating in t]le sinusoidal steady state,we need to introduce and define complexpower.Complex power is the complexsum of
realpower and reactivepower,or
('10.23) .{ ComptexPower
As you will see,we can compute the complex power directly from the voltageand curent phasorsfor a circuit.Equation 10.23car then be usedto
computethe averagepower and the reactivepower.becauseP : !t{S}
andO = s{s}.
Dimensionally, complex power is the same as average ol ieactlve
power. However, to distinguish complex power ftom either average or
reactive power, we use the units volt-amps (VA).Thus we usevolt-amps for
complex power, watts for average power, and vars for reactive power, as
summarizedin Table 10.2.
Another advantageof usingcomplexpoweris the geometricinterpretation it provides. When working with Eq. 10.23,think of P, Q, and S as
the sidesofa right tdangle,asshownin Fig.10.9.Itis easyto showthat the
angle I in the power tdangle is the power factor angle d. 0i. For the
right triangleshownin Fig.10.9,
_o
Unils
(r0.24)
of P andI (bqs'll0.l0l and[10.I l]. recpecri\ely).
But ftomlhe delinirioD\
P : ateragc poler
Figure10,9 a A powettiangte.
a
(ubl^/2)sin(er 0)
(v^I ^12) cos(0. - 0)
= tan(d, - 0J.
(10.25)
Therefore, d : 0,
Pi.The geometric rclations for a right triangle mean
power
triangle dimensions(tie three sidesand the
also that the foul
power factor angle) can be determined if any two of the foui are kno{'n.
The magnitude of complex power is referred to as apparent pow€r.
Soecificallv.
power
(i0.26) d Apparent
Apparent power, like complex power, is measuied jn volt-amps The
apparentpower,or vollamp, requhementofa devicedesignedto convert
electric energy to a nonelectric form is more important than the average
power requirement.Although the averagepower representsthe useful
output of the energy-converting device,the apparenl power representsure
vollamp capacityrequired to supplylhe averagepower.As you can see
402 SinusoidaL
Steady-state
Power
Catcuhtjons
from the power triargle in Fig. 10.9,unlessthe poiver factor angleis 0'
(that is, the deviceis purely resistive,pf: 1, and O = 0), the volt amp
capacity required by the device is larger than the average powet used by
the device.As we will see in Example 10-6,it Dakes senseto operale
devicesat a power lactor closeto 1.
Many useful appliances(suchas refrigerators,fans,air conditioners,
fluorescentlighting fixtures,and washingmachines)and most industrial
loadsoperateat a laggingpower factor.The power factor of theseloads
sometimesis correctedeither by addinga capacitorto the deviceitself or
by connectingcapacitorsacross the line leeding the load; the latter
method is often used for large industdal loads.Many of the Chapter
Problemsgive you a chanceto make somecalculalionsthat conect a laggingpower factor load and improve the operationof a circuit.
Example 10.,1usesa power triangle to calculateseveralquaniities
associated
with an electricalload.
Catcutating
Power
Complex
An electricalload operatesat 240V rms.The toad
absorbsan dveragenower of 8 kw ar d lagging
power factor of 0.8.
a) Calculatethe complexpower ofthe load.
b) Calculatethe impedanceofthe load.
Solvingfor /cn,
I cn = 416'7A'
We alreadyknow the angle of the load impedance,because
it is the power factor angle:
0:
Solution
a) The power lactor is describedas lagging,so we
know that the load is inductive and that the
algebraicsign of the reactivepower is positive.
From the power triargle shownin Fig.10.10,
P:
l s l c o sd ,
Q : l s l s i0n.
Now,because
cosd = 0.8, sind - 0.6.Therefore
P
cos '(0.8): 36.87".
We also know that , is positive becausethe
power lactor is lagging,indicatirg an inductive
load. We compute the magnitude of the load
impedarcefrom its definition as the ratio of the
magnitudeof the voltage to the magnitudeof
_,
v,n)
24n
IL\l/
= 10kvA.
sl' = r]] =:t:_j:
(l-ll
cosft
O = iosin0:6kVAR,
z : 5.'76/36.8'7"A = 4.608+ j3.456().
s:8 + i6kvA.
b) From the computation of the complex power of
thc load,weseethat P = 8 kW. UsingEq.10.2l,
P :
feflI,ffcos (0,
: (240)1.r(0.8)
: 8000w.
A)
Figure10.10A A powe,irianqle.
10.5 Power
Catcuiations
403
10.5 PowerCatculations
We are now ready to develop additional equations that can be used to calculate rcal, reactive,and complex power.We begin by combining Eqs.10.10,
10.11.and 10.23to set
s =bfcos(e,
\/l
0,t+ ja;sinrc,
- dj) + jsin (r, =k!r*"re.
:\t!r,lu,
) -
0,1
dr)l
u,t: !vJ-/ lo, - a,t.
(to.?7)
If we use the elfective values oI the sinusoidal voltage and curent,
Eq.10.27becomes
s : v"ttr"11_!3"_4.
(10.28)
Equations10.2?and 10.28are importantrelationshipsin power calculations becausethey show that if the phasor current and voltage are
knowr at a pair of terminals, the complex power associatedwith that pair
of terminalsis either one half the productof the voltageand the conjugate
ofthe current,or the productofthe rms phasorvoltageand the conjugate
of the rms phasor culrent. We can show this lor the rms phasor voltage
Flgure10.11AThe phasorvottage
andcunentassocjand curent inFig.10.11asfollows:
atedwjrha panofhnninals.
s:V
.t11!ui)
: V"rl "rreilo' o)
: Ve[eio,Iexe ja,
: v"d:r.
power
(10.29) <l Complex
Note that {fi : Iefier', follows from Euler's identity and the trigonomef
ric identities cos( d) : cos(d)and sin(-0) = -sin(9)
lene-ra = /effcos( 0,) + jlensin(
- 1effcos(9r)
= I3r'
jlcffsin (d,
di)
404
SinusoidatSteady-StatePowerCatcuLations
[he \amedefivalionlechnrque
couldbe dppliedro Fq. 10.27lo yield
s: fvl
(10.30)
Both Eqs.10.29and 10.30are basedon the passivesignconvention.lf the
current refeience is ilt the direction of the voltage dse aqoss the terminals,we inse a minus sign on the right-hand side of each equation.
To illustratethe use of Eq. 10.30in a power calculation,let's use the
samecircuit that we usedin Example10.1.Expressedin termsofthe phasor representationofthe teminal voltageand current,
v : 100lll
v,
|= 41 !!lA.
Therefore
I
s = (100Z!114 ,, 105'
200Llq
= -100 + j173.21VA.
Oncewe calculatethe complexpower,we canread off both the real and
rcaclivepowen,because
S : P + jO. Thus
P:
100w,
Q : 113.21vAR
Theinterpretations
signsonP andO areidenticalto tlose
ofthe algebraic
givenin thesolutionofExample10.1.
AlternateFormsfor Complex
Power
figure10.12 Thegenentcjrcujtof
Fjg.10.11
reptaced
withanequivatent
impedance.
Equations10.29and 10.30haveseveralusefulvadations.Here,we usethe
Ims valueform of the equations,becauserms valuesare the most commoD
type of represertation for voltages ard curents in power computations.
The first variarion of Eq. 10.29is to replace t}le voltage with the product of the currenttimesthe impedance.
In other words, e can alwaysrepresentthe circuit insidethe box ofFig.10.11by an equivalentimpedance,
asshownin Fie.10.12.Then.
Ys = Zlat
(10.31)
10.5 Power
GLculrtions405
Substituting
Eq.10.31into Eq.10.29ields
S=ztd*t
: r"nl'Z
: LdlR + jx)
= l.nl'R + jll.n2x = P + jQ,
(10.32)
frcm which
P-lr,tr2R:r2^R,
=:P.x.
o = lr"$l,x
(10.33)
(10.34)
In Eq. 10.34,-Y is the reactatrce of either the equivalent inductance or
equivalent capacitanceof the circuit. Recal from our earlier discussionof
reactancetlat it is positive for hductive circuits atrd negative for capacitive cfucuits.
A second useful variation of Eq. 10.29comes ftom replacing the current with t]le voltage divided by the impedance:
/ v-"\,
s-Y,rl - l=
\ z
/
lv-",= P - i O .
:;
(10.35)
Notethat if Z is a Dureresistiveelement.
-
lv"nl'
R '
(10.36)
andif Z is a pule reactiveelement,
^
V"al2
(10.37)
In Eq. 10.37,X is positive for aDinductor and negative for a capacitor.
The folo$.ing examples demoDstrate various power calculations in
circuits opeiating in the sinusoidal steady state.
406 Sinusoidal
Steady-state
PowoCatcuLaiiom
Calculating
Average
andReactive
Power
In the circuit shownin Fig. 10.13,a load having an
impedanceof 39 + j26 O is fed ftom a voltage
source through a Line having ar impedance of
1 + j4 (). The effective,or rms,value of the source
voltageis 250V
a) Calculatethe load cunent IL and voltageVL.
b) Calculatethe aveiageand reaclivepower delivered to the load.
c) Calculatethe averageand reactivepower deliveied to the line.
d) Calculaiethe averageand reactivepower sup
plied by the source.
lQ
i4a
'
)Ji9*,
Source -F
39f,l
1,1
P6p"
Line
*l+Load
Figure10.13,AThecircuitfor
Examph
10.5.
c) The average and reactive power delivered to the
line are most easily calculated from Eqs. 10.33
and 10.34becausethe line currentis known.Thus
P = (51(1)= 2sW
Solution
vAR.
o = 6),e) : 1oo
a) The Lineard loadimpedancesare in seriesacross
the voltagesource,sothe ioad currentequalsthe
voltdgedi! idedb) lhe loralimpedance.
or
lr
250'/0'
,c u^ + . , ^
a
/ru
1 3- 5
- 3 o8 - ' A 0 m ,
Becausethe voltage is given in terms oI its
rms value,the curent also is rms.The load volt
ageis the product of the load current and load
impedance:
vL = (39 + /26)lL = 234
= T4.36/
j13
Note ihat the reactivepower associatedwiththe
line is po.iri\e becduserhe line reaclancei(
inductive.
d) One way to calculatethe averageand reactive
power deliveredby the souce is to add the complex powerdeliveredto the line 1()that delivered
to the load,or
S:25+i100+975+./650
: 1000 + j750 vA.
The complexpowerat the sourcecanalsobe calculatedfrom Eq.10.29:
s" = -250ri.
3.r8'v (nnt.
b) The averageand reactivepower delivered10the
load can be computedusingEq. l0.29.Theiefore
Theminussignis insertedin Eq.10.29whenever
thecurrentreference
is in thedirectionofa volt
agerise.Thus
s": -250(4+ j3) :
S : vlli = (234-./13)(4+ 13)
- 975+ j650vA.
Thus the load is absorbingan averagepower of
975W and a reactivepower of650VAR.
(1000+ j7s0)vA.
power
Th€minussignimpliesthal both average
andmagnetizing
reactivepowerarebeingdeliveredby the source.
Note that this resullagrees
wilh the previouscalculationof S, as it must,
because
the sourcemustfunish all the average
andreactivepowerabsorbedby the line andload.
10.5 Pow€r
Catcutations
407
Calcutating
Powerin ParallelLoads
The two loadsin the circuit showniD Fig.10.14can
be desoibed as follows: l-oad 1 absorbs an average
powerof BkW at a leadingpowerIactorof 0.8.Load
2 absorbs20 kVA at a laggingpo\eerfactor of0.6.
0 05O
s kw t36 87'
ervln *
,'ffi
{)
/O.50
16KVAR
12kW
(b)
(a)
10KVAR
20kw
G)
Figure10.14A Thecjrcuit
forExampte
10.6.
a) Determine the power factor of the two loadsin
parallel.
b) DetelmiIle the apparent power required to supply t})e loads,tie magnitude of tlre current,I", and
the aveiage power loss in tlre transmissionline.
c) Given that the frequencyof the sourceis 60 Hz,
compute the value of the capacitorthat would
correct the power factor to 1 if placed in parallel
with the two loads.Recomputethe valuesin (b)
for the load with the correctedpower factor.
tisur€ 10.15 a (a)ThepoweftdangL€
for toad1. (b)The
powertrianste
for Load
2. (c)Thesun ofthe powertrianstes.
It follows that
s = 20.000+ j10.000vA,
and
j10.0n0
+
r ; =20,n00
250
= 80 + 140A.
Therefore
Solution
I":
a) All voltageand curent phasorsin this problem
are assumedto represent effective values, Note
ftom the circuit diagram in Fig. 10.14 that
I, : 11+ 12.The total complexpower absorbed
by the two loadsis
s = (250)rl
= (250)01
+ rr)'
= (2s0)ri+ (2s0)ri
= .t1+ s2.
We cansumthe complexpowersgeometricallyl
usingthe powertrianglesfor eachload,asshown
in Fig.10.15.
By hlpothesis,
roru '1499
(.li)
8000- j6000vA,
+ j20,000(.8)
20,000(.6)
12,000+ i16,000vA.
80
J40: 89.44/
26.57'A.
Thus the power factor of the combined load is
pf - cosro
26.57') - 0.8q44iaggints.
The power factor of the two loadsjn parallel is
laggingbecausethe net reactivepowerispositiv...
b) The apparentpower which must be suppliedto
theseloadsis
kvA.
ls : 20 + i10l = 22.36
The magnitudeof the cunent that suppliesthis
apparentpower is
rJ :
80
j40l : 89.44A.
The averagepower lost in the line resultsirom
the currentflowing throughthe line resistance:
: 4oor.v
4r^": l\l'n : (B9.44)'(o.os)
Note thatthe powersuppliedtotals20,000+ 400
- 20.400w.even rhougbrhe load. requne a
total oI only 20,000W.
408
SinusoidatSt€ady-Stat€
PowerCatcutatjons
c) As we can see from tlle power triangle in
Fig.10.15(c).we
cancorrectthe powerlactor to 1
if we place a capacitor in parallel with the existing
loads such that the capacitor supplies 10 kVAR
of magnetizingreactivepower.The value of the
capacitoris calculatedas follows.First,find the
capacitivereactancefrom Eq. 10.37:
apparentpower once the power factor hasbeen
l s l: P : 2 0 k v A .
The magnitudeof the current that suppliesthis
apparentpower is
lr,l:
lv->
o
The average power lost in the line is tlus
(2s0),
: 320w.
PI"" : r"l,R: (80),(0.0s)
10.000
:
Now, the power supplied totals 20,000 + 320
= 20,320W. Note that t}le addition of tie capacrtor hasreduced the line Iossfrom 400W to 320W
6.25A.
Recallthatthereactiveimpedance
o{ a capacitor
is l/oc, and @= h(60) : 376.99rad/s, if
the sourceftequencyis 60IIz. Thus,
-1
ax
-l
20.00080 A.
250
= 44.4u.F.
(3'76.99)(.
6.2s)
The addition of the capacitor as the third load is
representedin geometricform as the sum of the
two power tiangles sho$n in Fig. 10.16.When
the power factor is 1, the apparentpower and
the averagepowerare the same,asseenftom the
power triangle in Fig. 10.16(c).Theiefore, the
1OKVAR +
20kw
(")
10KVAR
(b)
20kw
(c)
figure 10.16d (a)Thesumofthe powertiangtes
for Loads
1
and2. (b) Thepower
inangtefot a 421.1ttFcapacitot
at 60 Hz.
( c ) I h es u mo ft h ep o w e r i n a n si hn sG ) a n d( b ) .
Batancing
PowerDetivered
with PowerAbsorbed
in an acCircuit
a) Calculatethe total averageand reactivepower
deliveredto eachimpedancein the circuitshown
in Fig.10.17.
jO
l2O+l
v-
tO
v, lI,
/lO
3qr"
b) Calculatethe averageand reactivepowen associatedwith €achsourcein the circuit.
\=1s0l{rv
c) Veril, that th€ average power deliveied equals
tlte averagepower absorbed,and that the magnet
izing reactive power delivercd equalsthe magnet
izing reactive power absorbed.
V
vl = (78 1104)
=
v2 (72+i104)v
v
vi = (150 1130)
Ir = ( 26 ./s2)A
11: ( -2 +./6) A
Ir=( 24 r58)A
Figure10.17A Thecircuilwithsotutjon,
for Example
10.7.
10.5 Power
Catculations
409
Solution
b) The complex power associated with the independentvoltagesourceis
a) The complexpower deliveredto tle (1 +J2)O
impedanceis
l
s' " = - : v " r = P " + r o "
2 "
51 =
=
l
;vJi
= PI+ jQ1
l
: -:(150X 26 + i52)
:1950
l
26+ ls2)
t(78 t104X
I
= :(3380 + 16760)
j3900vA.
vollagesourceis
Nore lhdr tbe independent
absorbingan averagepower of 1950W and
deliverhg 3900VAR. The complexpower associated with the current-controlled voltage
= 1690+ j3380VA.
Thus this impedanceis absorbhg an average
power of 1690 W and a ieactive power of
3380VAR. The complexpowerdeliveredto the
(12 j16) O impedance
is
I
s. = -(]gr.)(li): P. + io.
2
-
=:(
-
1
s2:
;v2r,
:
;(72
: P2 + jQ2
+ tI04\(-2 - j6\
Thereforethe impedance
in the veiticalbmnch
is absorbing240W anddelivedng320VAR. The
complexpower deliveredto the (1 + j3) O
impedance
is
l
sr=ivrr,=Pr+lol
I
s8s0 i5070vA.
Both av€ragepow€r and magnetizirgreactive
power arc being deliveredby the dependent
c) fhe total powerabsorbedby the passiveimped!ollagesourceis
aDces
aDdthe iDdependeDl
: 240 - j320VA.
:;fi50
78 + /234)(-24+ 158)
i13u)( 24 + i58)
: 1970+ j5910VA.
This impedanceis absorbing1970 W and
5910VAR.
Por,wra : Pr + P2 + P3 + Ps : 5850W.
The dependentvoltagesourceis the only circuit
elementdeliverhg averagepower,Thus
Pd"ri,.,.d: 5850W.
Magnetizingreactive power is being absorbed
by the two horizontalbranches.Thus
Qou,o,r"a: Or + 03 : 9290 VAR.
Magnetizing reactive power is being delivered
by the independentvoltagesource,tlrecapacitor
ir the vertical impedance branch, and the
dependentvoltagesource.Therefore
Qaai,-.a : 990 VAR
410
Power
Catcutations
SinsoidatSteady-state
acpowerconcepts,
theirretationships
to oneanother.
andhowto catcutate
themin a circuit
objective1-Understand
(c) 23.52W. 94.09vAR:
(d) 1152.62w'-376.36VARi
10.4 The load impedancein the circuit shownrs
shuntedbt' a capacirorhavirg a capacitivereactanceof 52 O. Calculaie:
a) the rms phasorsVL and IL,
b) the averagepower aDdnagnetizingreactive
power absorbedby rhe (39 + j26) o load
c) the averagepower and magnetizingreactive
power absorbedby rhe (1 + 14) o line
d) lhe averagepower and nagnetizingreactive
power deliveredby thc source,and
e) the magnetiziDg
reacdvepower deliveredby
the shuntingcapacilor.
10
jl tl
\
+ \ 25010.
t)ii,:;
sourcc -
Lnro
-t-
Answer: (a) 2s2.20/ 4.s4'V(ms),
s.38/ 38.23'A(rms);
(b) 1129.09w,752.73
VAR;
''
39{)
(e) 1223.18VAR.
10.5 The rms voltageat the lermirlalsof a load is
250VThe load is absorbjngan avcragcpower
of40 kW and deliveringa magrclizingrcactive
power of30 kVAR. Derive hvo equivalcnl
impedancemodelsof the load.
Answer: l (t rn\e e. uilh tr.-SQ ol .nnocrii\e
{) ii parallclwith 2.083O
reactance:1.5625
of capacitivereactance.
10,6 Find tho phasorvoltageV, (rms) in the circuit
shownif loadsLr and l, are absorbing15 kVA
1 l 0 . hf l l r g g r r r g a n d 6 k V A0dR p l e a d : n g .
respectively.
ExpressY in polar form.
tlo
t26lI
l
load
Answer: 251.61
/15.91'Y.
and10.22.
NOTE: ALtotry ChaptetPlohlent 10.17,10.21,
10.6 $'iaxin*r$Psw*r'nr*n$r*er
power
Figlte10.18,1 A circuitde$rjbingmaxjmum
Recallfrom Chapter4 that certaiDsysrems for exanplc.th osethat transmit information via electricsignals deperd on bcjng ablc to transfera
maximumamountofpower from the sourceto lhe load.Wc now rccxamine maximum powef transferin the contert oI a sinusoidalsteadystatc
network. beginning$'ith fig. 10.18.We musr determinethe load impcd
anccZr that rcsultsin thc dcliveryof maximum averagepower 1lrlermi
nals a and b. An]' lincar nctwork n1aybe viewedfrom the terninals oI lie
load in tcrms of a Thivcnin equivalentcircuit.Thus the task reduces1()
finding the valueof Zr_that resultsin maximun averageporverdelivcrcd
io Zr. in thc circuit sho\\,nin Fig.10.19.
10.6 Maximum
Power
I€nsfer 411
For maximum averagepower tmnsfer, ZL must equal tie conjugate of
the Thdvenin imoedance:that is.
(10.38) < Condition
for rnaximum
avengepower
transfer
WededveEq. 10.38by a straightforwardapplicationof elementarycalculus.Webeginby expressingZrh and ZL in rectangularform:
Rrh+ jxrn,
(10.3e)
RL + jXL.
(10.40)
Zrh:
ZL:
In botl Eqs. 10.39and 10.40,the reactance telm caries its own algebraic
sign positive for inductance and negative for capacitance.Because we
are making an average-power calculation, we assumethat tie amplitude
of the fh6venin voltage is er?ressed in terms of its ims value.We also use
the Th€venin voltage as tlle reference phasor. Then, from Fig. 10.19,the
Ims value of the Ioad current I is
-t =
v..
"
(nr, + R, + j(xrh + x)'
110.4t)
The averagepower delivered to the load is
P : lrl'Rl.
\rc.42)
yields
Substituting
Eq.10.41
inroEq.10.42
lvrr'l'Rr
(Rrh+Rrf+lxrn+lf;2
(10.43)
When workiq with Eq. 10.43,always remember that Vrh, Rrx, and -Yrh
are fixed quantities, whereas R1 and -YL aie independent variables.
Therefore, to maximize P, we must lind the values of RL and -YL where
aPlARL andAPIAX L arc both zero.From Eq. 10.43,
aP _
dxr
-lV11'2RL().r +
[(Rr + Rr,)'+ (xL + xrif]''
(10.44)
ap _ lvl.hl'l(Rl+ nrh)'+ (xL+ x]x), - 2RL(RL+ Rft)]
'
dfir
[(Rr + Rr + 6L+ Xr,.),],
(10.45)
412
SjnusoidatSteady-StatePowerkLcutations
FrcmEq.10.44,aP
laXL is zerowhen
Xt:
-Xn.
(10.46)
From Eq. 10.45,aP/aRLis zero when
R': \/6 + 6';t#
(10.41)
Note that when we combineEq.10.46with Eq.10.47,bothdeivatives are
zerc when ZL - Zit.
PowerAbsorbed
TheMaximum
Average
The maximum average power tiat can be delivered to Zr when it is set
equal to the conjugate of Zrh is calculated directly ftom the circuit in
Fi.9.1O.19.When ZL = Zih, the Ims load cu(ent is V11,/2R1,and the maximum averaseDower delivered to the load is
vrr,l'Rr 1 v*l'
4ElL
4 Rr-
(10.48)
If the Thdvenin voltage is er?ressed in terms of its maximumamplitude
ratherthan its rms amDlitude.Eq. 10.48becones
-
1V;
(10.49)
MaximumPowerTransfer
WhenZ is Restricted
Maximum averagepower can be delivered to ZL only if ZL can be set
equal to the conjugateof Zrh. Th€re are situationsin which this is not
possible.First, RL and -l.L may be restricted to a limited range of values.In this sitlration,the optimum condition for RL and -YLis to adjust
xt-:q
!9C! !9 4I0 as possible and then adjust RL as close to
VRih + (xL + xr 'as possible(seeExample10 9).
A second type of restdction occuN when the magnitude of ZL can be
vaded but its phase angle cannot. Under this restriction, the greatest
amount of powei is lransferred to the load when the magdtude of ZL is
setequalto the magnitudeof Zrh;that is,when
lz'l = zrnl
The pioof ol Eq.10.50is left to you asProblem10.40.
(10.50)
10.6 l4aximum
Powerlmnsfer 413
For purely resistivenetworks,maximum power transferoccurswhcn
the load resistanceequals the Th6venin resistarce.Note that we first
dedved this result in the introduction to maximum power transfer in
Chapter4.
Examples10.8-i0.11 illustrate the problem of obtaidng maimum
power traNfer in the situationsjustdiscussed.
Determining
Maximum
Power
Transfer
withoutLoadRestrictions
a) For lhe circuit shownin Fig.10.20,determinetl1e
impedanceZL that resultsin maximum average
Powertransferredto ZL.
b) What is the maxirnumaveragepowertranslered
to the load impedancedelerminedin (a)?
Solution
5{}
20t111+
jlo
20f)3 ioo
ZL
Figure10.20A ThecircuitfofExampte
10.8.
a) We beginb! dererminints
tbeThd!eninequi\alent with respectto t1leload teminals a,b.After
two source lransformations involving the 20 V
source,the 5l) resistor,andthe 20 O resistor,we
simplifythe circuitshownin Fig.10-20to the one
sbownin Fig.10.21.Then,
16 /0'
v * = ;q + l F. r,-(l o
i30
lo)
53.13' - 11.52- 115.36V.
= 19.2 /
4r)
figure 10.21n A sinptification
of Fjg.10.?0bysource
we lind the Th6veninimpedanceby deacrivar
ing the independentsourceand calculatingthe
impedanceseen looking into the terminals a
and b. Thus.
i1.68o
Zn
{- i6Jr4+ r3l
+i/r
/o
For maximum averagepower transfer,the toacl
impedancemust be thc conjugateofZrh, so
+j1.68o
zL = 5.76 + j\.68 0.
b) We calc ate the maximun averagepower deliveredto Zr from the circuil shownin Fig.10.22,in
which we replacedthe orighal network with ils
Th6veninequivalent.From Fig. 10.22.the rms
magdtudeofthe load currertI is
: 1.1785
A.
2(s;76)
F i E n e7 0 . 2 2t l l F r ' L | . \ o ^ i l t ' 9 . 1 0 .0 . r i r ' l l .
originat
network
reptaced
byjts Thdvenin
equivaLent.
The averagcpoNer deliveredto the load is
P = 1:ft(5.76)- 8W.
414
sinusoidatSt€ady-StatePowercaLcutatjons
Maximum
Power
Transfer
with LoadImDedance
Restriction
Determining
a) For t]Ie circuil shown in Fig. 10.23,what value of
ZL results in maximum averagepower fiansfer [o
ZL? What is t]Ie maximum power in milliwatts?
b) Assume tlat the Ioad resistance can be varied
between 0 and 40000 and that the capacitive
rcactance of the load can be varied between
0 and 2000 O. Wlat settings of -RL and XL
hansfer the most average power to the load?
What is the maximum averagepower that can be
hansferredundortheserestrictions?
Solution
a) If therc are no restrictionson RL and XL, the
load impedanceis set equal to t})e conjugateof
the outputor theTh6veninimpedance.Therefore
RL : 3000O and -Yr -
4000O,
ZL : 3000 J4000o.
Becausethe sour€evoltagejs givenin terns of its
rmsvalue,the averagepowerdeliveredto ZL is
I
Inl
'\
: inw
P: ii:
4 Jt't J{)
.t
= 8 . 3m
3 w.
b) BecauseRL and -trL are restdcted,we fi^t set
-trr as close to -4000 O as possible: thus
X, - :?@Q!. !qt. \ e set R, as clo.e to
Ril. + (xL + xft)'as possible.Thus
o.
Rt = \r6000t+ F2000+ 4000t= 360s.55
30000 l4O00Oa
101!l
b
figure 10.23 a Tle c'(Lit fo tvarple5 l0.a dld t0.t0.
Now,becauseRL can be varied ftom 0 ro 4000 O,
we can set RL to 3605.55O. Therefore,the Ioad
impedanceis adjustedto a valueof
Zt = 3605.55 - j2000 A.
With ZL set at this value, the value of the load
cufent is
lar =
101!
: 1.M89 /-16.85' 'lj{.
+ j2000
6605.55
The averagepower delivered to the load is
p : (1.4489x 10-3f(3605.s5): 7.57mW.
This quantity is the maidmum power that we can
deliver to a load, given the restictions on RL
and XL. Note that tlis is less than the power
that canbe deliveredif there are no restrictions;
in (a) ne toundrhar$e candetiver8.31mW
Power
Transfer
AngteRestridions
FindingMaximum
with Impedance
A load impedance having a constart phaseangle of
36.87"is connectedacrossthe load terminals a and
b in the circuit shown in Fig. 10.23.The mag tude of
ZL is varied until the averagepower delivercd is the
most possibleunder the given restriction.
a) Specify ZL in rectangular form.
b) Calculate the averagepower deliveied to ZL.
Solution
a) From Eq. 10.50,we kl)ow that the magnitude of
ZL must equal the magnitude of Z.rh.Therefore
lz"l = lztl:
= s000
o.
3000+ J40001
10.6 Maxinun
Power
Transfer415
No\ as wc know that the phaseanglc of Zr is
36.87',wc have
zL = 5000/ -36.8'7' :4000
j3000l).
b) With ZL sel equal to ,1000 13000O, the load
lar =
Iu
: 1.4U2/-8.13'mA,
7000+ j1000
and thc averagepower deliveredto the load is
P : (1.4142x 10 )' (4000) :8mw.
This quantilyis the maximumpower that can be
delivered by this circuit to a load impedance
whoseangleis constantat 36.87'-Again, this
quantilyis lessthanthe maximumpowcrthal can
bc deliveredif thereareno restrictionson ZL.
0bjediv€ 2-Understand the conditionfor maximunreal powerdetiveredto a Loadin an ac circuit
10.7 The sourcecurrentin the circuit shownis
3 cos5000r A.
3.6mH
a) Wlat impedanceshouldbe connecred
acfussrenninals
d.blof narimum rrcrage
power lransfer?
What is the averagepower transferredto
the impedancein (a)?
c) Assumethat the load is resrrictedto pure
resistance,
What sizeresistorconnccted
acrossa.b will result in the maximum aver,
agepower transferred?
What is the averagepower transferredto
the resistorin (c)?
j10 o;
(b) 18W:
(c) 2n6 a;
(d) r7.00w.
Answen (a) 20
NOTE: Also tt! Chopkt Ptoblens 10.11,10.19,and
10.50.
FindingMaximum
Power
Transfer
in a Circuitwith an IdealTransformer
The variable rcsistor in lhe circuit in Fig. 10.24is
adjusteduntilmaximum averagepoweris delivcrcd
ro RL.
8.10
/0'
o) Whatis rhcvclu( oi Rr jn ohmsl
b) What is the maximum averagepower (n wattsl
deliveredto Rz?
v t..)
b
Figure10,24.t Thecircuitfortxamph10.11.
416
Sjnusoid
aLsteady-state
Power
Calcutations
Sotution
a) We first find the Th6venin equivalent wlth
respectto the terminals of Rr. The circuit for
delermrning
lhe opencrrcuir!ollagein sho\rnin
Fig. 10.25.The variablesVr, Y2, 11,and 12have
beenaddedto expediatethe discussionFirst we note the ideal transformer imposes
the following constraints on the vadables H, V2,
11,and 12:
Ir+
840!ry
v (nnt
F i g u r e1 0 . 2 5d T 5 pL i r r i . u . " d. o f i i d l f e l h e v e - i 1 v o t r a 9 e .
v,- = lv.
4
t, :
lr,
4 -
The open circuit value of 12is zero, hence 11is
zero. It follows that
lr+
qV= 8a0Av,
\z = 2101!Y.
From Fig. 10.25we note that Vrh is the negative
of V2,hence
84otll
vrh= -210av.
The circuit shown in Fig. 10.26is usedto determine the short circuit current. Viewing 11and I,
asme.hcurrenrs.lhe
lwo mesbequalions
afe
840A:80Ir
0 :2012
20I, + V,
35()
2011+!2.
Wlen thesetwo meshcurent equationsare
combined
with theconstraint
equations
we get
840l!|=
Figure10.26 A Thecircujtusedto catcuLat€
theshodcircuii
210d1
35{)
40Ir+ L
v
b
Loaded
tot nraxjmum
tigure 10.27 A TheThcvenin
equjvatent
0 = 2 5-L + 4- .
Solving for the short circuit value of I, yields
r, = 6A.
Thereforethe Theveninresistanceis
)10
Rrh=-:=35O
Maximum power will be deliveredto Rr when
nr equals35 O.
b) The mlximum power delivered to -Rr is most
easily determined using the Th6venin equivalent.
From the circuitshownin Fig.10.27we have
/
",^\2
/--*=l-:^
l(r)r=rr)w.
\ / u . /
P,adkaL
Pe,spedr@ 417
0bjective3-ge abLeto catcutate
all formsof ac powerin accircuitswith tineartransformers
andideal
transformers
10.8 Find thc averagepower deliveredto the
100O resistorin the circuit shownif
rs : 660cos5000tV.
340
4O-trH -
::
^ 100 LnH
375{}
l^u-qg.
,.,1-\,-"
--j
100ll
Answer:612.5w'
10.9 a) Find the averagepower deliveredto thc
400O resislorin the circuitshownif
0s = 248cos10,000rV.
b) Fird the averagepower deiiveredto the
375 O resistor.
c) Find the powerdeveiopedby the idealvol|
agesource.Checkyoul resultby showjngthe
powerabsorbedequalsthe powerdeveloped
400f)
Answer: (a) 50W;
(b)49.2w;
(c) 99.2W 50 + 49.2= 99.2W.
10.10SolveExample10.11ifthe poladtydol on the
coilconnected
to terminala is at thelop.
Answer: (a) 15 ();
(b) 73sw.
1 0 . 1 15 o l \ cF \ a m p l e
l 0 . l l i f I l < \ u l r a g e . o L ri c. e
reduced10146lq V rmsandtheturnsratiois
reversed
to 1:4.
Answer: (a) 1460O;
(b) s8.4w.
NOTE: Also trt ChaptetPrcbkns 10.41,10.45,10.58,
and 10.59.
PracticaI
Perspective
Heating
Apptiances
A handheldhair dryercontainsa heatjngelement,whichis just a resjstor
heatedby the sinusoidalcurrentpassingthroughit, and a fan that btows
the warmair surrounding
the resistorout the front of the unit. This js
shownschematicaLV
in Fig.10.28.Theheatertub€in this figureis a resis
tor madeof coiLednichromewire.Nichrome
is an attoyof iron, chromium,
andnjckel.Twopropetiesmakeit idealfor usein heaters.Fj6t, it is more
resistivethan rnostotherrnetals,so lessiraterialjs required
to achjevethe
neededresistance.
This atLowsthe heaterto be very cornpact.Second,
untik-"manyotherrnetats,
nichrofire
doesnot oxidiz€whenheatedredhot in
ajr Thusthe heateretementtastsa longtime.
js shownin Fiq.10.29.This
A circuitdiag|am
for the hairdryercontroLs
is the only partof the hairdryercjrcuitthat givesyou controloverthe heat
setting.Therestofthe cjrcuitprovides
powerto the fan motorandis not of
inter€sthere.Thecoitedwirethat comprises
the heatertube hasa connec,
tion partwayatongthe coit,dividingthe coitinto two pieces.Wemodelthis Figur€10.28!. Schenratic
Eprerntationofa
jn Fig.10.29with two seriesresjstors,
Rr and R2.Thecontrolsto turn the
418
SinusojdatSteady-StatePowerCatcutations
switchin whjchtwo
dryeronandsetect
theheatsettingusea four-position
pairsoftermina15
in thecircuitwiLtbeshorted
together
by a pairof diding
whichpairsof terminaLs
metalbars.Thepositionof the switchdetermines
Themetalbarsareconnected
areshorted
together.
by aninsutator,
sothere
is noconduction
Dathbetween
the Dairs
of shotedtermina15.
Thecjrcujtjn Fig.10.29contains
fuse,Thisis a protective
a thermaL
1 : 1 1
Butif the temperature
neaf
device
that
normalty
acts
like
a
short
cjrcuit.
OFF L M H
the heaterbecomes
dangerousty
high,the thermalfusebecomes
an open
Figure
forthehairdryer circujt,djscontinuing
10.29.4A circuiidiagnm
the flow of curfentandreducjng
the rjskof fire or
protection
injury.Thethermalfuseprovides
in casethe motorfajtsor the
is not
airftowbecornes
btocked.
Whjtethe designof the protection
system
Thermalfuse
partof thG exampte,
it is important
to pointoutthat safetyanai.ysjs
is an
pat of anetectrjcaI
essentjaI
enginee/s
work.
Nowthat we havemodeted
the controts
for the hajrdryer,lets deter
mjnethecl'rcuitcomponents
that arepresent
forthethreeswjtchsetbngs,
Tobegin,the circujtin Fig.10.29js redrawn
in Fjg.10.30(a)
for the Lolf
switch
setting.
The
open-circujted
wires
have
been
removed
for
ctariry.A
1 1 1 1
OFF L M H
simptified
equivatent
circuitis shownin Fiq.10.30(b).
A sjmilarpairof fig(a)
uresis shownfor the ilEorur'4
setting(Fig.10.31)and the rrcr'rsethng
(Fig.10.32).Notefromthesefiguresthat at the rowsettjng,the vottage
source
seestheresistors
Rl andR2in series;
at ther,4E0rui1
settinq,thevottagesource
seesonLythe R2 resistor;
andat the HrGH
settinq,the vottage
jn paralLel.
source
seestheresistors
Figure10.30a (a)ThecircujtjnFis.10.29red€wn
fortheL switchsetiing.(b) A simptjfied
equjva-
Thernalfuse
i l 1 1
OFF L
M H
1
1
f
OFF L M H
i
G)
rigurc10.31,a(a)IhecncLrit
in tiq.10.29
edrawn
(b)AsimpLjfied
forthe,roruswitch
settinq.
equjv-
Figure10,32a (a)Th€cjrcujtjnFjs.10.29
redrawn
equivatorih€ HI6H
switch5ettins.(b)A simptjfied
lentcircuiifor (a).
your undestonding
NoTE:Assess
af this PrddicalPerspective
by ttyingChapter
ProbLems
10.66-10.68.
Summary419
Summary
. tNtantaneous power is the producl of the instartaneousteminal voltage and current,or ? = trt. The
positive sign is used when the referenced ection for
the current is from the positive to the negativereference polarity of the voltage.The frequency of the
instantaneouspower is twice the freqoency of the voltage(or current).(Seepage392.)
. The power faclor is the cosire of the phase angle
betweenthe voltageand the currenl:
. Av€rage power is the averagevalue of the instantaneouspower over one period.It is the power converted
Irom electric to nonelectric form and vice versa. This
conversionis the reason that averagepowet is also
refeired to as real power.Averagepower.witl the passivesignconventior,isexpressedas
The reactiye lactoi is the sine of the phase angle
betweenthe voltageand the current:
- e,:t
r = lv; ^"o"1e,
pf=cos(d,,9r.
The terms laSginSand leadtnSadded to tle description
of the power factor indicate whether t}le current is lag'
gingor leadingthe voltageand thus whetherthe load is
inductiveor capacitive.(Seepage396.)
rf : sin(or
d,).
(Seepage396.)
Complex pow€r is the complex sum of the real and reac-
S-P+jQ
- 4n/encos(o, 0,).
:
(Seepag€394.)
ftt'= *t'u
v?n
z'
Reactiv€ power is the electdc power exchanged
between the magnetic field of an inductor and the
sourcethat drives it or betweenthe electriclield of a
capacitorand the sourcethat drivesit. Reactivepower
is never converted to noneiectric power. Reactive
power,with the passivesignconvention,is expressedas
(Seepage401.)
' Apparentpoweris the magnitudeof the complexpower:
sl=\/P\e'
(Seepage401.)
1
Q = ;u^I -sin(.o,
0)
= %dd sin(d, - d').
Both average power and reactive power can be
eeressed in terms of ejther peak (y-, 1-) or effective
(%fi, 1ed current and voltage. Effective values are
widely used in both householdand industrial applica
tior].s.Effectire ,alue and /ms value are interchangeable
termsfor the samevalue.(Seepage394.)
The watt is usedas the unit for both instantaneous
and
real power.The mr (volt amp reactive,or VAR) is used
as the unit for reactive power. The volt-arnp (VA) is
usedas the unit for complexand apparentpower.(See
page401.)
Maximum pow€r transfer occurs in circuits operating in
the shusoidal steady state when the load impedance is
the conjugate of the Th6venin impedanceasviewed from
the terminalsof t})eload inpedance.(Seepage410.)
420
SinusoidatSt€ady'StatePowerCatcutatjons
Problems
Sections10.1-10,2
figureP10.5
80nF
10.1 Showthat the maximumvalueof the instantan€ous
po\rergivenbyEq.Io.qi. P VP) ' STandthal
theminimumvalueis P - t/ e' 9 .
102 The following setsof valuesfor o and i pe ain to
the circuit seenin Hg. 10.1.For eachset of valueE
calculateP and I and state whether the circuif
insidethe box is absorbingor delivering(1) average
powerand (2) magnetizingvals.
a) o = 340cos(ot+ 60')V,
j = 20cos(.dr+ 15') A.
b) ?,:75cos(ot- 15') V,
j = 16cos(.dt+ 60') A.
c) ?J- 625cos(ot+ 40")V,
i - sln(at + 240")A.
d) 2 = 180sin(ol+ 220')V,
t = 10cos(ar+ 20")A.
10,6 Find the averagepower dissipatedin the 20 O
Btrc resistor in the chcuit seel1 in Fig. P10.6 if
is = 15cos10,0001A
FiglreP10.5
10.3 a) A universitystudentis makhg coffeewhile listeningto a Buck'sgameon themdio At the same
time,her roommateis watchingthe Packenol1a
solid-stateTV All theseappliancesaie supplied
ftom a 120V branchcircuit protectedby a 20A
circuit breaker, If the roommate tuns on a
portableheater,is the circuitbreakertdpped?
b) Will the roommatebe able to use the heaterif
t}le coffemakeris unplugged?
10.4 Find tlle averagepower delivered by the ideal
curent source in the circuit in Eg. P10.4 if
mA.
is = 30cos25,0001
Figure
P10,4
10ta
lmH
1) r"Jfz.si,r
20o'
10.7 The load impedarce in Fig. P10.7 absorbs 40 kw
and 30 magnetizing kvars. The sinusoidal voltage
source develops 50 kW.
a) Find the values of capacitive line reactance that
will satisfy these constraitrls.
b) For each value of line reactarce found in (a),
show that the magretizing vars developed
equalsthe magnetizirgvars absorbed.
2A
t)
f +od
figue P10.7
40r.H
10.5 The op amp in the circuit shown in Fig. P10.5 is
ideal. Calculate t}le average power deiivered to the
1 kO resistor when t's = 4 cos50001V.
n" ^
2500rc|
v(nnt
-/Xo
Problems
421
10.8 A load consistingof a 1350 O resistorin parallel
with a 405 mH inductor is connected aooss the terminals of a sinusoidalvoltage source og, where
oc = 90 cos25001V
a) What is the peak value of the instantaneous
pow€r deliveredby the source?
b) Wlat is the peak value of the instantaneous
power absorbedby the source?
c) What is t]le averagepower delivered to the load?
d) What is the rcactivepowerdeliveredtotheload?
e) Does the load absorb or generate magnetiz
ing vars?
f) Whar is the power factor of the load?
g) What is the reactivefactor of the load?
10.9 a) C-alculaiethe real ard reactive power associatecl
with each circuit element in the cifcuil in
Fig.P9.56.
b) Verify that the avemge power generated equals
the avemgepower absorbed.
c) Verify that the magnetizing vars generated
equalthe magnetizingvarsabsorbed.
10.10 Repeat Problem 10.9 for the circuil shown in
Fig.P9.57.
Figure
P10.13
10,14 Fhd the rms value of the periodic current shor,n in
Fig.P10.14.
figureP10.14
i (A)
Section103
10.11 A dc voltage equal to fdcV is applied to a resistor of
R O. A sinusoidal voltage equal to z)sV is also
appliedto a resistoroI R O. Show that the dc voltagewill deliver th€ sameamount of energy in 7 seconds (where r is the period of the sinusoidal
voltage) as the sinusoidal voltage provided yd"
equals the rms value of o,. (1tir?t Equate the two
er?ressionslbr the energy delivered to tle resistor.)
10.12 a) A personalcomputer with a monitor and keyboard requires 60 w at 110V (rms). Calculate
the rms value of the cuirent carried by its
power cord.
b) A laser printer for the peisonal computer in
(a) is ratedatB0W at 110V (ms).If this printer
is pluggedinto the samewall outlet as the computer, what is the rms value of the current drawn
from the outlet?
10.13 a) Find tne rms valueof the peiodic voltageshown
in Fig.P10.13.
b) If this voltage is applied to the terminalsof a
12 O resistor,what is the averagepower dissipatedin the resistor?
40
s0
.(mt
10.15 fhe pedodiccurrentshownin Fig.P10.14dissipates
an averagepower of 24 kW in a resistor.wlrat is the
valueof the resistor?
Sections10.4-10.5
10.16 Find the averagepower,the reactivepower,ard the
6'c apparentpower absorbedby the load in t}le circuit
in Fig.P10.16if
isequals30cos
100tmA.
Figure
Pr0.16
422
Power
5inusoid
aLSteady-State
Calcutations
10.17 a) Find the average power, the reactive power, and
ttre apparent powei supplied by t]le vollage
source in the circuit in Fig. P10.17 if
0g = 50 cos10sl V.
b) Check your answer in (a) by showins
c) Check your answer in
0a- = )8"t"
(a)
by
1()
i8r)
2a0.llV (rmt
showing
10.20Two 660V (rms) loads are connectedin parallel. The
tigu|ePl0.u
two loads draw a total averagepower of 52,800W at
a power factor of 0.8 leading. One of t]Ie loads dmws
40 kVA at a power factor of 0.96lagging.Wlat is the
power factor of the other load?
7.50
10.18 The voltage Vs in the frequency-domain circuit
shownio Fig.P10.18is 340 A v (rms).
a) Find the average and reactive power for the
voltage source,
b) Is the voltage source absorbingor deliverirg
averagepower?
c) Is the voltage source absorbing or delivedng
magnetizingvars?
d) Find the average and reactive powers associated
with each impedance branch in the circuil.
e) Check fie balan€e between delivered and
absorbed averagepower.
f) Check the balance between delivered and
absorbedmagletiang vars.
tigureP10.18
50()
- )
rigureP10.19
80()
1
-lloo o
j60()
10.19 a) Find VL (rms) atrd d for the circuit in Fig. P10.19
iJ the load absorbs 250 VA at a lagging power
factor of 0.6.
b) Consrucl a phasordiagramof each solulion
obtainedin (a).
ln.2I The thrce loads in lhe circuit in Fig. P10.21can be
described as follows: Inad 1 is a 12 o resistor in
series with a 15 mH inductor; load 2 is a 16 rrF
capacitor in serieswith a 80 O resistor; and Ioad 3 is
a 400 Q resistor in sedeswith tle pamlel combination of a 20 H inductor and a 5 pF capacitor. The
ftequency of the voltage source is 60 Hz.
a) Give the power factor and reactive factor ot
each load.
b) Give the power factor and reactive factor of the
composite load seenby the voltage source.
Figure
P10.21
1022 Three loads are connectedin parallel acrossa 2400V
(ms) line, as sho$n in Eg. P10.22.Load 1 abso$s
18 kw and 24 kvAR. Load 2 absorbs60 kvA at 0.6pf
lead.I-oad 3 absorbs18 kW at uDrrypower tactor.
a) Find the impedance that is equivalent to the
tbreeparallelloads.
b) Find t}le power factor of the equivalent load as
seenfrom tbe line'sioput terminals,
Flgure
P10.22
Probtems
423
1023 The tlree loads in Problem 10.22are fed frcm a line
having a sedesimpedance0.2 + j1.6 O, as shown
inFig.P10.23.
a) Calculate the ms value of the voltage (Y) at the
sendhg end of the line.
Calculate the average and reactive poweN associated with the line impedance.
") Calculate the averageand reactive powen at the
sending end of the line.
d) Calculate the efficiency (T) of tie line if the efficiency is defhed as
4 = (P6"6/P.."ai"*"") X 100.
tigureP10.23
0.2o j1.6o
Figure
P10.25
0 . 0 1 O1 0 . 1 O
+
0.01oj0.1o
ll.2i
The three loads in the circuit shown in Fig. P10.26
are 51 : 5 + j2 kVA, S, : 3.75 + j1.5 kVA, and
s3=8+j0kvA.
a) Calculate the complex power associated wit]I
each voltage source,\1 and Vs2.
b) Vedfy that the total real and reactive power
delivered by the sources equals the total rcal
and reactive power abso$ed by the network.
figuruP10.26
2400i0"
0.050
12s48V (mt
t0.u
The three parallel loads in the circuit shown in
Fig. P10.2 can be described as follows: Load 1 is
absorbingan avenge power of 24 kW and 18 kVAR
of magnetizing varsi load 2 is absorbing an average
power of 48 kW and generating 30 kVAR of magnetiziiig reactive powert load 3 coNists of a 60 O
rcsistor in paiallel with an inductive reactance of
480 O. Fhd t})e rms magnitude and tlre phase angle
of Vs iJ V. = 2400 lq V(Ims).
tigurcP10.24
j10o
+
10.25 The two loads showr in Fig. P10.25can be described
as follows: Load 1 abso$s an average power ol
24.96 kW and 47.04 kVAR magnetizing reactive
power; load 2 has an impedance of 5 - j5 O.
fhe voltage at the teminals of the loads is
48O^vAcos12ont Y .
a) Find the Ims value of the souce voltage.
b) By how many microseconds is the load voltage
out of phasewith the source voltage?
c) Does t]Ie load voltage lead or lag the source
voltage?
12s{r V (rmo
to27 The three loads in the circuit seen in Fig. ?10.27 are
desffibed as follows: Load 1 is absorbing1.8 kW
and 600VAR; load 2 is 1.5kVA at a 0.8pf lead;load
3 is a 12 O rcsistorin parallelwith an inductor that
hasa reactanceof 48,(!.
a) C-alculatethe average power and t])e magnetizing reactive powei delivered by each source if
Y"t = Ysz:120 lq V (rms).
b) Check your calculations by showing your results
are consistentwith the requirements
SP.
= Sp.
Sn.
: Sn.
Figure
P10,27
424
sinusoidatsieady-StatePow€rCalculatjons
10.28 Supposethe circuit shown in Fig. P10.27represents
a residential distribution circuit in which the impedances of the service conducton are lregligible and
Vrl = Vgz= 120 7!|| V(rms). The three loads in
the circuit are L1 (a coffeemaker,a frying pan and
an egg cooker); L, (a hairdryer, a sunlamp, a fan,
and an automatic washing machine);and q (a
quick recovery water heater and a range with an
oven).Assumethat all tlese appliancesare in operation at the sametime.The senice conductorsare
protected with 100A circuit breakerr Will the serv
ice to this residencebe interrupted?Explain.
10.29 The steady-statevoltage drop between the load and
the sendingend of the iine seenin Fig. P10.29is
excessive.A capacitor is placed in paralel with the
250 kVA load and is adjusted until the steady-state
voltageat t}le sendingend of the line hasthe same
magnitude as the voltage at the load end, that is.
2500V (ms). The 250 kVA load is operatingat a
power factor of 0.96 lag. Calculatethe size of the
capacitoi in microfarads if the circuit is operating at
60 Hz. In selectingthe capacitor,keep in mind the
need to keep the power lossin the line at a reasonablelevel.
tigureP10.29
1f}
figurcP10.30
20
j20o I
1380
7200[f
v G-t
+
Souce !'--Line
i460o
'F
Load
10.31 A group of small appliances on a 60 IIz system
r€quires25 kVA at 0.96pf laggingwhenoperatedat
125V (rms).The impedanceof the feedersupplying
the appliancesis 0.006+ j0.048O. The voltage at
the load end ofthe feederis 125v (rms).
a) What is the rms magnitude of the voltage at the
source end of the feeder?
b) What is the averagepower Iossin the feeder?
c) What sizecapacitor(jn microfarads)at the load
end of the feederis neededto improve the load
power tactor to unity?
d) Aftef the capacitor is installed, what is t}le Ims
magnitude of the voltage at the source end of
the feeder if t}le Ioad voltage is maintained at
125V (rms)?
e) What is the averagepower loss in the feeder
for (d)?
isI)
25ixl& V (rms)
10.30 a) Find the averagepower dissipatedin the lire m
Fig.P10.30.
b) Find the capacitive reactance that when connected in parallel with the load will make the
load look purcly rcsistive.
What
c)
is the equivalent impedanceof the load
in (b)?
d) Find lhe averageponer dissipaled
i0 lhe line
when the capacitive reactance is connected
acrossthe load.
e) Expressthe power lossin (d) as a percentageof
the power lossfound in (a).
10.32 A factory has an electrical load of 1800kW at a lagging power factor of 0.6. An additional variable
powei factor load is to be added 10 the factory. The
new load will add 600 kW to the real power load of
the factory. The power faclor of the added load is to
be adjustedso ttrat the overall power factor of the
factory is 0.96lagging.
a) Specifythe reactivepower associated$rith the
addedload.
b) Does the addedload absorbor deliver magnetizing vars?
c) w}lat is the power factor of t]le additional load?
d) Assumethat the voltageat the input to the factory is 4800V (rms).What is the rms magntude
of the currert into the factory before the variablepower factor load is added?
e) Wlat is the Ims magnitude of the current into
the factory after the variable power factor load
hasbeeDadded?
Probtems
425
10.33 Assume the factory describedin Problem 10.32is fed
from a line having an impedance of 0.02 + j0.16 O.
The voltage at the factory is maintained at
4800V (mN).
a) Fird the avemge power loss io the Lhe before
and after the load is added.
b) Find the magnitude of the voltage at the sending
end of the line before and after t]le Ioad is added.
tlgurcP10.36
o
1oo a .is!-q
3qc
v (r'nt
+
;-
i30
^t29-9.
j70.o
I
L$a
j100o
b
10.34 Considerthe circuit describedin Problem9.76.
a) What is the Ims magnitude of the voltage adoss
the load impedance?
b) What percentage of the average power devel
oped by the practical source is delivered to the
load impedance?
10.37 a) Find the average power delivered by the smusoidalcurrentsourcein the circuitof Fig.P10.37.
b) Find the average power delivered to the 1 kO
rcsistor,
Flgure
P10.37
10,35 a) Find t}le six branch currents la Ir in tle circuit
in Fig.P10.35.
b) Find the complex power ir each branch of the
crcult.
c) Check your calculations by veriflng that the
average power developed equals the average
power dissipated.
d) Check your calculations by vedfing that the
magnetizing vars genemted equal the magnetizing vaIS absorbed.
101UmA
Flgure
P10.35
10,38 a) Find the average power dissipated in each resistor in the ctucuit ir Fig. P10.38.
b) Check your answer by showing that the total
power developed equals the total power
absorbed.
1()
j2 o /jlo\
" tl+
)lt"
ff'"
20E | . ,
It'
I
jl o
Id
iio
rr
10
Pig(eP10.38
40
10.36 a) Find the averagepower deliveredto ttre 40 O
resistorin thecircuitin Eig.P10.36.
b) Find the averagepower developedby the ideal
sinusoidalvoltagesource.
o Fjrrd Z ab.
d ) Show tlat the average power developed equals
the averagepowei dissipated.
-j8 0
426
SinusojdatSteady
StatePowerCatculations
10.39 The sinusoidal voltage source in the circuit in
Fig. P10.39is developingan rms voltage of 680V
The 80 O load in the circuit is absorbing 16 times as
much averagepower as the 320 O load. The two
loadsare matchedto the sinusoidalsourcethat has
an interml impedanceof 136 A kO.
a) Speci$'the numedcalvaluesof dr and dr.
b) Calculatethe power deliveredto the 80 O load.
c) Calculatethe rms valueof the voltageaqossthe
320 O resistor.
10.42 The phasor voltage Vrb in the cjrcuit shown m
Fig. P10.42is a80 A VtmO when no external
load is connectedto the terminalsa,b.When a load
having an impedanceoI 100 + j0 O is connected
acrossa,b.thevalueof\b is 240 j80 V (rms).
a) Find the impedancethat should be connected
acrossa,b for ma-\imum avemge power transfer.
b) Find the maximumaveragepowertransferredto
the load of (a).
Figu|e
P10.39
680U
10.43 The load impedanceZL for the circuit shown in
Fig. P10.43 is adjusted until ma-\imum average
poweris deliveredto ZL.
a) Find the maximum average power delivered
to ZL.
b) what percentageof the total power developed
in the circuit is deliv€redto ZL?
Seclion 10.6
10.40 Prove that if only the magnitude of the load
impedance
cdn be vafied.mosl a\erdgepouer i\
transferred to the load when ZL - Z\. (Hint:
In deriving the expressionfor the averageload
power,write the load impedance(ZL) in the fom
ZL: ZLlcos9 + j ZL s;\0, and note that only
lZLl is variable.)
10.41 a) Determine the load impedancelor the circuit
shownir Fig.P10.41that will resultin maximum
averagepower being transferredto the load if
o = 10 krad/s.
b) Determine the ma-\imum average power
delivered to the load ftom part (a) if r)s:
90 cos10,000tV.
Figure
P10.41
figue P10.43
l10o
2500'V lrms)
lD.U For the frequency-domaincircuit in Fig. P10.4,1,
calculate:
a) the Ims magflitude of V,.
b) the averagepower dissipatedin the 70 O resistorc) the percentageol the averagepower gen€rated
by the ideal voltage sourcethat is deliveredto
the 70 O toadresistor.
Figure
P10.44
2.5nF
j40 {)
10o
'
36011I
v (rno
l
l
+
r 3 o n - ] ; + o ov , 70o"
PrcbLenis
427
10.45 The 70 O rosistor in the circuit in Fig. P10.44 is
replaced with a variable impedance-Z,. Assume
Z, is adjustedfor maximum avemgepower transfet to za.
a) What is the maximum average power that can
be delivered to Z,?
b) What is the average power developed by the
ideal voltage source when maximum average
poweris deliveredto Zo?
10,46 Suppose an impedance equal to the conjugate of
the Th6venin impedance is connected to the terminals c,d of the circuit sho*lr in Fig. P9.74.
a) Find the average power developed by the sinusoidalvoltagesource.
b) What percentageofthe power developedby the
source is lost in the linear transformer?
10.47 The variable resistor in the circuit shown in
Fig. P10.47is adjusteduntil the averagepower it
absorbsis maximlrm.
a) Find R.
b) Find the maximum averagepower.
tigureP10.47
18()
\ 630xr
i6r}
8o
llso
F;
Figure
P10.48
100
j.29a
600Cv Gmt
10.49 The peak amplitude of the sinusoidal voltage
souce in the circuit shown in Fig. P10.49 is
15014 V, and its period is 2002],.s.The load resistor can be vaded ftom 0 to 20O, and the load
inductoi canbe variedftom 1 to 8 mH.
a) Calculate the average power delivered to the
load when Ro : 10 O and L, : 6 mH.
b) Detemine the settingsof R, and Z, that will
rcsult in the most avemgepower being translelred to Ro.
c) w}lat is the most avenge power in (b)? Is it
gieater than tlle power in (a)?
d) If tlere are no constaints ofl & and L,, what is
the maximum average power that can be delivered to a load?
e) wllat are the values of R, and -L. for the condition of (d)?
f) Is the avemgepower calculatedin (d) larger
than thal calculatedin (c)?
figureP10.49
10.44 The variable resistor Ro in the circuit shown m
Fig. P10.48 is adjusted until maximum avetage
poweris deliveredto Ro.
a) What is the value of Ro in ohms?
b) Calculatethe aveiagepower deliveredto R..
c) If R" is replaced with a variable impedance .Z,,
what is the mlximum avemge power that can be
deliveredto Zo?
d) In (c), what percentageof the circuit's developedpower is deliveredto the load Z.?
10.50 a) Assume tlat R, in Fig. P10.49 can be varied
between 0 and 50O. Repeat (b) and (c) of
Problem10.49.
Is the new avenge power calculatedin (a)
geater than that found in Problem10.49(a)?
c) Is the new avemgepower calculatedin (a) less
than ttrat found in 10.49(d)?
Power
Catcutations
425 SjnusoidaL
Steady-State
10s1 The
sending-end voltage in the c cuit seen in
Fig. P10.51is adjusted so that the rms value of the
load vokage i. dLra)s 4800V The variablecapacitor is adjusted until the averagepower dissipatedin
the lire resistanceis minimum.
a) If the frequency of tle sinusoidal source is
i0
o0 Hz. whal is rhe \alue ol lhe capacilance
miuofarads?
b) If the capacitor is removed ftom the circuit,
what percentageinqease in the magnitude of V"
is necessaryto maintain 4800V at the load?
c) If ttre capacitor is removed from t]le circuit,
what is the percentage inuease in line loss?
10,54 The polarity dot on a1 itr the circuit in Fig. P10.52is
a) Find the value of k tlat makes t , equal to zero
b) Fird the power developed by the souce when k
has the value found in (a).
10.55 The impedance ZL il the circuit in Fig. P10.55 is
adjusted for maximum average power transfer to
ZL. The intemal impedance of the sinusoidal voltagesourceis 8 + J56O.
a) What is the maximum average power deiivered
to ZL?
b) What percentageof tle averagepower delivercd
to the lirlear transformer is delivered to ZL?
Figure
P10.51
1()
Figure
P10.55
i8fl
- - l
I
a800
10:v (rmo
7600a
j40o
v (rmt
10.52 The values of the paiameten in the circuit shown
i n F i g . P 1 0 . 5 2a r e L r 4 0 m H : / ) - l 0 m H :
k-0;75|. Rs= 20O; and Rr-140o.
ff
:
find
?,s 240\4 cos 4000t V,
a) the rms magnitude of ?),
b) the averagepower delivered to RL
c) the percentage of the avenge power generated
by the ideal voltage sourcethat is delivered to RL.
P10.52
Figure
Rs
+\
L1
k
,l
L,1
-'
source+L
+
in the circuit in
Fig. P10.52is adjustable.
a) wlat value ot RL will resull iD the ma"\imum
averagepo\{er being transfe ed to RL?
b) What is the value of the maximum power
transferred?
j*Load*
10.s6 a) Find the steady-stateexpressionfor the currents
is and iL in the circuit in Fig. P10.56 when
t's = 70 cos50001V.
b) Find the coefficient of coupling.
c) Find the energy stored in the magnetically coupled coils a : 100,7 /rs and I : 200r' ps.
e)
1)
g)
h)
10,53 Assume the load resistor (R,
_ _ _Tralslo!!9r
Find the power deliveredto the 30I) resistor.
If tle 30 O resistor is replaced by a variable
rcsistor RL, what value of RL will yield maxl_
mum averagepower tmnsfer to RL?
What is the maximum averagepower in (e)?
Assumerhe 30 O resisloris replacedb' a variable impedance ZL. What value of ZL will result
in maximum averagepower transler to ZL?
What is t}le ma-{imum average power in (g)?
tigue P10.56
10O
zr,Il
30[}
Probtems
429
10.57 Find the impedanceseenby the ideal voltage source
in the circuit in Fig. P10.57when Z, is adjusted for
maximum averagepower transfei to Zo,
tigureP10.57
i{Qr
I0o
801!8
10.58 Findthe averagepowerdeliveredto the 4 kO resistor in lhe circuitofFlg.P10.58.
10.61 a) ff & equals2520turnq how many tums should be
plac€d on the N2 winding of the ideal transfom)er
in the circuit in Fig P10.61so that maximum averagepower is delivered to the 50 O load.
b) Find the average power delivered to the
50O load.
Find the voltage V.
d) What percentageof the power developed by the
ideal current source is delivered to the 50O
resistor?
tigureP10,61
Figure
P10.58
10(}
5ko
-e;l1,2.sft
1:4f.lfl
loo1fvA I llt ll I l
{nas,Y//
|
llt \
lt\
|Ideal[1 ]Ideall
160
lsko '-ll
t
l
10.59 The ideal transformer connected to the 4kO load
in Problem 10.58 is replaced with ar ideal transfomer t])at has a tums mtio of 1:l?.
a) W}lat value of d results in maximum average
power being delivered to the 4 kf,l resistor?
b) Wlat is the maximum averagepower?
10.60 a) If Nl equals 1500 tums, how many tums should
be placod on the N2 winding of the ideal tmnsfomer in the circuit seen in Fig. P10.60 so that
maximum average power is delivered to the
3600O load?
b) Find t}te average power delivered to the 3600O
rcsistor.
c) wltat percentage of the average power delivered by the ideal voltage source is dissipated in
the linear tra.nsformer?
10.62 The variable load resistor RL in the circuit shown ttr
Fie. P10.62is adjusted for maximum average power
transfer to RL.
a) Fhd the maximum averagepower.
b) What percentageof the averagepower developed
by the ideal voltage source is delivered to RL
when RL is absorbingmaximum avenge powei?
c) Test your solution by showing that the power
developed by the ideal voltage source€quals the
power dissipated in the circuit.
Figure
P10.62
Figurc
Pl0.60
t-200
JO;--\
it#,9 ,"1
I
JzO
t24 A
4f}
'l [ra.,r
&1 |
3ooo
o
2sE
v G.t
0.25f}
430
SinusoidatSteady-StatePowercaLcutatjons
10.63 Repeat Problem 10.62 foi the circuit shown in
4trc Fig.P10.63.
tigureP10.65
30ko j10 kO
Figure
PI0.63
20ko
lko
10ko
1001ry
10.64 The sinusoidal voltage source h the circuit in
Fig. P10.64is operating at a frequency of 50 kmd/s.
The variable capacitive reactance in the circuit is
adjusted until the average power delivered to the
160 O resistor is as large as possible.
a) Find the value of C in micofarads
b) When C has the value found in (a), what is r}le
averagepower delivered to tlle 160O resistor?
c) Replace the 160O resistor with a va able resistor R,. Specify t}le value of R. so that mlximum
averagepower is delivered to Rd.
d) What is the maximum average power that can
be deliveredto R,?
Sections10.1-10.6
10,66 The hair dryer in the Practical Perspective uses a
tl$ll&rr 60 Hz sinusoidal voltage of 120V (rms). The heatei
element must dissipale 250 W at the Low setting,
500 W at the MEDTUMsetting, and 1000 W at tie
HrcHsetting.
a) Fhd the value for resistor R2 using the specification for the MEDrI,.Msetting, using Fig. 10.31.
b) Find the value for tesistor using t]te specification for the Low setting, using the results frcm
pal1 (a) and Fig.10.30.
c) Is the specification for the HrGHsetting satisfied?
Figure
P10.54
25r}
2s0Lqf
20o./50O
160r)
ill''
i
(
*,
IdealI
10,65 The load impedance ZL in the circuit in Fig. P10.65
is adjusted until maximum average power is tansIened to ZL.
a) Specifythe valueof Zr if Nr = 15,000turnsand
N2:5000turns.
b) Specily the values of IL and VL when ZL is
absorbing ma-\imum avemge power,
10.67 As seenin Problem 10.66,only two independent
plllflI#hpower speciflcationscan be madewhen two rcsis;;iA-__tors makeup t}le heatingelementof the hair diyer.
a) Show tiat the er:prcssionfor the rxcH power
iating(PIr)is
-"
P4.
Pv-Pr'
where Pff = the MEDIUMpower rating and
Pt = the Low power mthg.
b) ff Pr - 250 W and PM = 750W, what must the
r{rGHpower rating be?
431
10.68 Speci$'the valuesof Rr and R2in the hair dryer cirplgljil - cuit in Fig. 10.29if the low power rating is 240W
6e and ttre high power tating is 1000W Assumethe
supply voltage is 120 V (rms). (flird: Work
Problem10.67fint.)
tigureP10.59
R3
Rr
R3
It"l"3"
10,69 If a third resistoris addedto the hair dryer circuit in
Fig.10.29,it is possibleto designto ttrreeindependp?Jiflfii€
LOW
MEDIUM
HIGH
-;;if- ent power
specificationsIf the resistorR3is added
in serieswith the Thermalfuse,then the correspondingro\\. MEDIUM
. and FncHpowercircuitdja- 10.70 You havebeengiventhe job of redesignhgthe hair
grairs are as shown in Fig. P10.69.ff the three
dryer described in Problem 10.66 for use itr
J€l|;:,#;E
power settirgs are 600 W, 900 W and 1200W,
The\randardsupplyvoltaeejn Eoglandis
o6c, England.
\,/r'balresisrorlalueswill you usein
respectively,
(rms).
whenconnectedto a 120V (rms) sup220
V
;i
ply,whatresistorvaluesshouldbeused?
your designto meetthe samepowerspecifications?
Ba[anced
Three-Phase
C'ircuits
Thre€-Phase
1 1 . 1BaLanced
Vottagesp. 434
17.2 Three-Phas€
VottageSourcesp. 4J5
1 1 . 3Analysisofthe Wye-WyeCircuit p. 4i6
r\.4 Anatysisofthe Wye-DeltaCircuit p. 44?
in Batan(edThree-Phase
1 1 . 5Po$erCaLcuLations
11.6 i\,leasuring
AveragePowerinThfee-Phas€
(now howto anaLyze
a baLanced,
thrce,phase
wY€wY€cornectedcircujt.
Knowhowto aratyzea batanced,
thlee-phase
wye-detta
coinectedcircuit.
power(averag€,
B€abteto catcuLat€
reactive,
jn anyihrc€ phasecircuit.
andcomptex)
432
Generating,transmitting, distributing, and using largeblocks
of electricpower is accomplishedwith three-phasecircuits.The
complehensivcanalysisof such sysfemsis a field of study in its
own right; \ve cannot hope to covcr it in a singLechapter.
Fortunatel),,an uldcrstanding of only the steady-statesinusoidal
behavior of balancedthree-phasecircuits is sufficientfor enginccrswho do not specializein powcr systems.
We deline what .rve
mean by a balancedcircuit later in the discussion.
Thc samecircuit analysistechniquesdiscussedin carlier chapter-scan be
applied to either ulbalanced or balancedthree-phasecircuits.
FIeLewe use thesc familiar techniquesto developsevcralshortcutsto the analysisof balancedthrcc phasecircuits.
Fol economic rcasons, three-phase systems are usually
designedto operatein the balalced statc.Thus.in this irllroductory treatrnent,we canjustify consideing only balancedcircuits.
The analysisol unbalancedthree-phasecircuits,which you will
encounterifvou studyelectricpowe. in later coulses,reliesheavi l ) o n r n u n d ( | . t a n J i nogl b J l c n c ecdi r c u i r . .
ffle basicstructureof a three-phasesystemconsistsof voltage sourcesconnectedto loads by meansof translonnersand
tlansmissionlil1es.
To analyzesucba circuit,we can reduceit to a
voltagesou.ceconnectedto a load via a linc.The on1ission
oi the
transformersimplitiesthe discussionwithout jeopardizinga basic
understalding of the calculations involved. Figure i].1 on
pagc434 showsa basiccircuit.A dcfining cbalacteristicof a bal
anced three-phasccircuit is that it contains a set of balanced
three-phasevoltagesat ils source.We begin by consideringthese
voltagcs.and then we move to the voltage and currenl relationshipsfor the Y-Y and Y A circuits.After consideringvoltageand
current in suchcircuits,we concludewith sectionson power and
power rneasutement,
PracticaI
Perspective
Transnission
andDistribution
of tlectricPower
In this chapterwe introducecircuitsthat are designedto
handtelargebtocksof etectrjcpower.Theseare the cjrcujts
that areusedto transportelectricpowerfromthe generating
ptantsto bothindustrialand residential
customers.
Weintroducedthe typicalresidentiat
customercircuitas usedjn the
linitedStatesasthe designperspective
in Chapter9. Nowwe
introducethe type of circuitusedto deiiveretectdcpowerto
an entireresidentiai
subdivision.
imposedon the desjgnandoper
0ne of the constraints
is
ation of an etect c utitjty the requirement
that the utitjty
maintainthe rms voitage[eve[at the custome/sprcmjses.
Whethertighttytoaded,as at 3:00 am, or heavityloaded,as
at midafternoon
on a hot, humidday,the utitity is obtjgated
to supptythe samermsvottage.RecatL
frofi Chapter10 that
a capacjtorcan be thought of as a sourceof magnetizing
vottageLeveLs
vars.Therefore,
onetechniquefor maintaining
on a utitity systemis to placecapacitorsat strategic[ocationsin the distributionnetwork.Theideabehindthis techniqueis to use the capacitorsto supptymagfetizjngvars
ctoseto the loadsrequirjng
then, as opposed
to sending
themoverthe linesfrornthe generator.
WeshalLitLustrate
this conceptafterwe haveintroduced
the anatysis
of bal:n.a/ rhra.-nh:(o.ir."it(
434
Balanced
Thrce-Phde
Circuitr
figure11.1a A basicthrce-phase
c'rcuit.
Voltages
11.1 Balanced
Three-Phase
A set of balancedthiee-phasevoltagesconsistsof three sinusoidalvolt
agesthat have idenlical amplitudes and frequencies but are out of phase
with eachotler by exacdy120'. Standardpmcticeis to refer to the three
phasesas a, b, and c, and to use the a-phaseas the referencephase.The
tlnee voltages are refened to as the a-phasevoltage, the b-phas€ voltag€,
and the c-phasevoltage.
Only two possiblephaserelationshipscan exist betweenthe a-phase
voltage and the b- and c-phasevoltages.One possibility is for the b-phase
voltageto lag the a-phasevoltageby 120', in which casethe c-phasevoltage must lead the a-phasevoltage by 120'. This phaserelationshipis
known as the abc (or posifiye) phase s€quence.The only other possibility
is lor the b-phasevoltageto lead the a-phasevoltageby 120",in which
casethe c-phasevoltagemust lag the a-phasevoltageby 120'.This phase
relationship is knowr as the acb (or negative) phase s€quenc€.In phasor
nolation,thetwo possiblesetsof balancedphasevoltagesare
Y"- v^1g:,
vb = v- / -r20"'
vc = u^ / +'t20',
v^= v- 1!1.
Yh = u^ / +720",
v" = u^// -120".
FiguE1r.2 A, Phasor
diagBms
ofa batanced
s€tof
thee-phase
vottases.
(a)Iheabc(positivdfquence.
(b)rheacb(nesatjve)
seqLrence.
\11.?)
EquatioDs 11.1 arc for the abc, or positive, sequence.Equations 11.2
Figure 11.2showsthe phasordiaare for the acb,or negative,sequence.
gramsof the voltagesetsin Eqs.11.1and 11.2.Thephasesequenceis the
clockwise order of the subscripts around the diagram ftom V.. The fact
tlat a thee-phase cftcuit can have one of two phase sequencesmust be
taken into account whenever two such circuits operate in panllel. The circuitscan operatein parallelonly il they havethe samephasesequence.
Voliag€
Sou(es
11.2 Three-Phase
Another important characteristic of a set of balanced three-phase
voltages is that the sum of the voltages is zero. Thus, from either Eqs. 11.1
or Eqs.11.2,
\+vb+Y=0.
Becausethe sum of the phasorvoltagesis zero,the sum of the instanta
neousvoltagesalsois zero;that is,
(11.4)
wi!ding
Now that we know t}le natue of a balancedset of three-phasevolf
ageq we can state the filsl of the analyticalshortcutsalluded to in the
iniroductionto this chapter:Ifwe know the phasesequenceand one voltagein the set,we know the entire set.Thusfor a balancedthree-phasesystem,we can focuson determiningthe voltage (or curent) in one phase,
becauseonce we know one phase quantity, we know the others
NOTE: Assessyow understandingof three-phasewltages by trying
ChapterProbkms 11.1and 11.2.
Vottage5ources
11.2 Three-Fhase
A three-phasevoltage source is a genemtor with tbree separate windings winding
distributed around th€ periphery of the stator. Each winding comprises
voLtage
soLrrce.
onephaseof the generator.The rotor oI the generatoris an electomagnet Figure11.3A A skekhofa thrce-phase
diven at synchronousspeedby a prime mover,suchasasteamor gasturbine. Rotation of the electrcmagnet induces a sinusoidal voltage in each
winding.The phasewindingsare designedso that the sinusoidalvoltages
induced in them are equal in amplitude and out of phase witl each other
by 120'. The phasewindingsare stationarywith respectto the rctating
so the frequencyof the voltageinducedin eachwindingis
electromagnet,
the same.Figure11.3showsa sketchof a two_polethree-phasesource.
Theie arc two ways of interconnecting the separatephasewindings to
form a three'phasesource:in either a wl,e (Y) or a delta (A) conliguratior. Figure11.4showsboth,with ideal voltagesourcesusedto model the
phasewindingsof the three-phasegenerator.The commonterminalin the
/l in Fig.11.4(a),iscalledthe neuhal terminal
Y-conrectedsource,labeled
of the source.The neutral teminal may or may not be available for exterSometimes,tle impedanceof each phasevrindingis so small (compared with other impedancesin the circuit) that we need not account tbr it
in modeling the generator;the model consistssolely of ideal voltage
sources,asin Fig. 11.4-However, if the impedance of eachphasewinding is
not negligible, we place the whding impedance iII series with an ideal
sinusoidal voltage source. All windings on the machhe arc of the same
(b)
construction, so we assumethe whding impedances to be identical. The
ofanideat
wjnding impedanceof a tluee-phasegeneratoris hductive Figure 115 rigure11.4A Thetwo basicconn€ctions
(a)A Y connected
source
and thrce-phas€
soutce.
showsa model of sucha machine,in which is the windingresistance,
(b) A A-connected
sourc€.
of thewinding
) , i. rheinductivereaclance
436
Batan.ed
ThrcePhase
Ci(uits
Flglre11.5A A modeLofa
thrce-phase
soLrfte
wiih winding
inrpedance:
source;
and
G)a y-connected
(b) a A-connectedsource.
Becausethree-phasesourcesand loadscanbe eitherY-connected
or A-connected,the basic circuit in Fig. 11.1represents
four different
configurations:
Load
Y
A
A
A
We begin by analyzing the Y-Y circuit. The remaining tluee arrangements
can be rcduced to a Y-Y equivalent circuit, so anal]sis of the Y Y chcuit is
the key to solving all balanced three-phase arangements. We then illustrate tie reduction of tle Y-A arrangement atrd leave the analysis of the
A-Y and A-A arrangements
to you in the Problems.
11.3 Anatysisof the Wye-Wye
Circuit
Figure 11.6illustrates a geneml Y-Y circuit, in which we included a fourth
conductorthat connectsthe sourceneutml to the load neutial,A fourth
conductor is possible only in the Y-Y arangement. (More about this
later.) For convenience,we tmnsformed the Y connections into ..tipDed
o!ertees. ln f-,g.ILo.7sa.7sa.afidZc. represenl
lhe inlerDalimped;nce
associatedwith each phasewinding of the voltag e genento\ Zh, Z h, nnd
Zr" represent the impedance of the Linesconnecting a phase of the source
to a phaseofthe load;Zo is the impedanceol the neutral conductorconnectingthe sourceneutral to the load neutral;and.Za, ZB, ard Zc rcpreseDlthe impedance
of eachphaseo[ the load.
Circuit 437
11.3 Anatysis
oftheWye-Wye
YYsystem
figure11.5A Aihte+phase
We can describ€this circuit with a single node-voltageequation.
Ushg the source neutral as the refercnce node and letting VN denote the
noalevoltagebetweenthe nodesN and n, we fird that the node-voltage
equationis
vN
za
\
za+ zb+ ze
zR+z1b+Zgr
Zcl
v",
Zb* Zs"
This is the general equation for any circuit of the YY configuntion
depictedin Fig. 11.6.But we can simplify Eq. 11.5significantlyif we now
coNider the formal definition of a balancedthree-phasecircuit. Such a
circuit satisfiesthe following criteria:
1. The voltage sourcesform a set of balanced three-phase voltages.
In Fig. 11.6,tlis meansthat V, ., Vb., and Vcn are a set of balanced
tbree phasevoltages.
2. The impedanceof each phaseof the voltage sourceis the same-In
Fig.11.6,this meansthatZsa: Zcb = Zcc.
3. The impedance oI each line (or phase) conductor is the same.In
Fig.11.6,
thismeansthatZ1u: Zyo: 26
4. The impedanceof eachphaseof the load is the sameln Fig 116,
this meansthat ZA = Zs : Zc.
There is no restriction on the impedance of a neutal conductor; its
value has no effect on whethei the systemis balanced.
If the circuit in Fig.11.6is balanced,wemay rewrite Eq.115 as
"^(;.;)
va,i+vb.+v."
Z a = Z ^ + Z r ^ + Z E a =Z B + 2 6 l
zb
216: Zc+ Zk+ Zc.'
three-phase
for a balanced
4 conditions
circuit
438 Batanced
Ihrce-Pha5€
Circuitr
The dghFhandsideofEq.11.6 is zero,becauseby hypothesisthe numerator is a set of balanced three-phase voltages and Zd is not zero. The only
valueof VN that satisfiesEq.11.6 is zero.Therefore,for a balancedthreephasecircuit,
Y.r = 0.
(11.j)
Equation 11.7is exhemely importart. If VN is zero, there is no differencein potential betweenthe sourceneuhal, n, and the load neutral,N;
consequendy,tlre curent in the neutal conductor is zerc. Hence we may
either remove the neutral conductor from a balarced Y-Y configuration
(lo = 0l or replaceit with a perfedsbortcircujtberreenrhenodisn and
N (\ - 0). Both equivalentsare convenientro usewhen modelingbalanced three-phasecircuits.
We now turn to the effect that balanced conditions have on the three
line currents.With reference to Fig. 11.6,when the system is balanced, the
threeline currentsare
V"', - VN
z a + z r a + z c ^ z4 ,
vuo Yrq
\s:
Zsl
I"c =
Figurc11.7A A singte-phase
equjvaLent
cjrcuit.
Zlot
zt'
Zs6
Y'" - VN
v"'"
z . + 2 . " + 2 , . " zb
(11.8)
(11.9)
(11.10)
We seethat the three line curents form a balanced set of three-phasecurrents;that is, the current in each line is equal in amplitude and frequency
ard is 120' out of phasevrilh the other two line currents.Thus. if we calculate the curent I,A and we know the phase sequence,we have a shortcut
for Iinding IbB and lcc. This procedure paraltels the shorcur used to find
the b- and c-phasesource voltages from the a-phasesource voltage.
We can useEq.11.8 to constructan equivalentcircuil for the a-phase
of the balancedY-Y circuit. From this equation, the curent in the a-phase
conductor line is simply tl)e voltage generated in the a-phasewinding of
the generator divided by t]le total impedance in the a-phaseof the circuit.
Thus Eq. 11.8describesthe simplecircuit shownin Fig. 11.7,in which rhe
neutral conductor has been replaced by a pedect short circuit. The circuit
in Fig. 11.7 is referred to as the singl€-phase equivalent circuit of a balanced three-phasecircuit. Because of the establishedrelationships
between phases,otrce we solve this circuit, we can easily write down the
voltages and cu(ents in the other two phases.Thus, drawing a singlephase equivalent circuit is an important first step itr analyzing a t]lieephasecircuit.
A word of caution here. The current in the neutral conductor in
Fig.11.7is IaA,which is not the sameasthe curent in the neutml conduclor o[ tbe bdlanced
rbree-pha(e
ciJcuit.
whichi.
Io: I.a + IbB + lcc.
(11.11)
Thus the circuit shown in Fig- 11.7gives tlle correct value of the line current but only the a-phasecomponent of the neutral current. Whenever this
oftheWye-Wye
Circuii 439
11.3 Anaiysis
single-phaseequivalent circuit is applicable, the line currents form a balancedtluee-phas€set,and the ight-hand sideofEq.11.11 sumsto zero.
Oncewe know the line currentin Fig.11.7,calculatioganyvoltagesof
interest is relatively simple. Of particular interest is the relatlonship
between the line{oljne voltages and the line{o-neutral voltages. We
establish this relationship at the load termhals, but our obsenations also
apply at the source terminals. The Line{o-line voltages at fie load termi_
nalscan be seenin Fig.11.8.They are VaB,VBc,and VcA,where the double subscript notation indicates a voltage drcp from the first-named node
to the second.(Becausewe are interesledin the balancedstate,we have
C
omitted the neutralconductorfromFig.11.8.)
vottages
The line{o-neutral voltages are VAN, VBN, and VcN. We can now Figure11.8A Linetoljne andtjne-to-neutmt
descibe the line+oline voltages in terms of the line-to-neutral voltages,
using Kirchhoff's voltage law:
VAB = VAN
VBN,
(lt.r?)
VBc = VBN
vcN,
(11.13)
VcA:VcN-VAN.
(11.14)
To show the relationship between the iine+oline voltages and t]le
Ushg the
line{o-neutralvoltages,we assumea positive,or abc,sequence.
reference,
line{o-neutralvoltageofthe a-phaseasthe
var : voA,
(11.15)
YsN: Va/ -720' ,
(11.16)
YcN = u6 / +120',
111.17)
where /d, representsthe magnitude of the line{o-neutral voltage.
yields
SubstitutingEqs11.15 11.17into Eqs.11.1211.14,respectively,
\;/ts- vo 191- vb/ 120'= \/av6/30",
(11.18)
Ysc= va/ 120'- u,t/120" = '\a3va
/ -9o',
(11.19)
Yce=Vq,
/1s0".
/120" V$1!:= \a3vb
\\1.24)
revealthat
Equations11.18-11.20
1. The magnitude of the Line+o-line voltage is a5 dmes the magnitude of the line-to-neutralvoltage.
2. The line-toline voltagesform a balancedthree-phaseset of voltages.
volt3. The set oI line-to-linevoltagesleadsthe set of Line-to-neutral
agesby 30".
We leave to you the d€monstration that for a negative sequence,the only
changeis that the set ollinetoline voltageslagsthe set of line to-neutral
440
BaLanced
lhrce'Phase
Cncuitr
voltagesby 30".The phasordiagramsshownin Fig. 11.9summarizetlese
observations.
Here, again,is a shortcutin the analysisof a balancedsystem: If you know the lirle-to-neurralvoltage at somepoint in the circuit,
you can easily determinethe line-toline voltage at the samepoint and
rigurc11.94 Phasordjaqnms
showjng
the r€taijonshipbetwe€n
tineio-tineandtine-to{eutrat
voltages
in
a batanced
(a)Theabcsequence.
system.
(b)Theacb
We now pauseto elaborate on teiminology. Line voltage refers to the
voltageacrcssany pair oflines;phssevoltag€refersto the voltageacross
a single phase.Line current refers to the curent in a sillgle line; phase
cullent refers 10 current in a singlephase.Obse e that in a A connec
tion,line voltage and phasevoltage are identical,and in aY connection.
line cunent and phasecurrent are identical.
Becausethree-phasesystemsare designedto handlelarge blocks of
electric power, all voltage and current specilications are given as rms vai
ues.Wlen voltage ratings are given, they refer specifically to the rating of
the line voltage.Thus when a three-phasetransmissionline is rated at
345kV, the nominal value of the rms line-to-lhe volraseis 345.000v' In
fiis chapterwe er?re..all vokagesandcurrenlsar r*t "ufr...
Finally,the Greek letter phi (d) is widely used in the literature ro
denotea per-phasequantity.Thus Vp,I!r, Zd, Pd, and Od are interpreted
as voltage/phase,current/phase,impedance/phase,power/phase,and
reactivepower/phase,
respectively.
Example11.1showshow to usethe observationsmadeso far to solve
a balancedthree-phase
Y-Y circuit.
Anatyzing
a Wye-Wye
Circuit
A balanced three-phaseY-connected generator
wit}l positive sequence has an impedance of
0.2 + j0.5 O/d and an internal voltageof 120y/d.
The genemtoi feeds a balanced tbiree-phase
Y-connected load having an impedance of
39 + J28O/d. The impedanceof lhe lire connecring the generatorto the load is 0.8 + i1.5 O/d.The
a-pha.eintemal\ollageol the generalori, specified asthe referencephasor.
Sotution
a) Figure 11.10showstle single-phaseequivalent
circuit.
b) The a-phaseline cunent is
12019:
(0.2+ 0.8+ 39)+ j(0.5+ 1.s+ 28)
=
a) Construct the a-phase equivalent circuit of
Ine system,
1201y
40+ i30
= 2.4/
36.8'7'
A.
b) Calculatethe threeline curents IaA,IbB,and Icc.
c) Calculatethe three phasevoltagesat rhe load,
VaN, Vs1, and VcN.
d) Calculate the line voltages VAB,VBc, and VcA at
the termhals of the load.
a, 0.2o /0Jo
a 0.8o j1.5o A
39f)
-
)tzoruvv",
e) Calculatethe phasevoltagesat the terminalsof
the generator, V-, Vb,, and Vm.
l) Calculate the line voltages Vab,Vbc,and Vcaat the
terminalsof the generator.
g) Repeat(a) (0 for a negativephasesequence.
i28O
N
rigure11.10A ThesjngLe-phas€
€qujvatenr
cjrcujr
for
Example
11.1.
Circuit 441
11.3 AnaLysis
of theWye-Wye
For a positivephasesequence,
lbB = 2.4/ -'t56.87"A,
r.c = 2.4/83.13"A.
c) Thephasevoltageat the A terminalof the loadis
v^tr = G9 + j28)(2.4/ -36.87' )
= 115./2/ -1.19" V.
For a positivephasesequence,
vBN= 115.22/ -121.19'v ,
Vcr = 11522 ,/11881' V
tl1eline voltages
d) For a positivephasesequence,
leadthe phasevoltagosby 30"; ttrus
vAB: (\6 /30')vA}r
= 199.58/28.81" V,
g) Changing the phase sequence has no ellect on
the single-phaseequivalent c cuit.Th€ three lme
l"A:2.4 /-36.87' A,
r6s= 2.4191jl:A,
rcc= 2.4l:E5g:4.
Thephasevoltagesat the load are
v^N : 1L5.22
/ 1.19"v,
vBN: 115.22
/118.81"V,
YcN = 115.22/
the line voltages
For a negativephasesequence,
by 30':
lagthephasevoltages
vAB : (\61:lq)vAN
Vnc = 199.58
/ 9119'V,
: 199.58
/-31.19'V,
Vcr = 19958,/14881' V'
vollage
lhesource
is
al thea lermjnalo[
el Thepha(e
van= 120- (0.2+ j0.5)(2.4
/ -36.87")
= 120 1.29/31.33"
: 118.90- j0.67
= 118.90
/ 0.32"V.
Fora positivephasesequence,
/ 12032'Y,
\." = 118.90
V.
v* : 118.90,/119.68"
0 The line voltagesat the sourceterminalsaie
v"b = (1,6 /30')v-
121.19"Y
vBc = 199.58
/88.81"V,
vcA - 199.58
/-151.19'V.
at theteminalsof thegenerThephasevoltages
atof are
V,. = 118.90
/ 0.32"V,
Vb. = 118.90
/119'68"V,
!6 - 118.90/
12032' V.
The line voltagesat the terminalsof the genera_
v"b: (\4/
3o')v-
- 20s.94
V,
/ 30.32'
= 205.94
/29.68' V,
ybc: 2n5.94/ 90.32'V,
\.
vq - 205.94/ 149.68'V.
va:
= 2fr5.94
/89 68"Y '
205.94/ 150.32'v.
442 BatanedThree'Phas€
Circuits
objective1-Know how to analyzea balanced,thrce-phasewye-wyecircuit
11.1
'l'he
voltagefromA to N in a balancedtbreephasccircui!is 2,10/ 30' V. 11lhephase
sequcnceis positive,what is the valueof VRcl
Answ€r:
4,56el
120."
11,2 The c-phasevoltageof a balancedthree-phase
Y conlectedsystemis 450/ 25'V.If the
phasesequelceis negatjve.whatis the va]uc
ofv^ts?
Answet '779.421j! V.
11,3 The phasevoltageal the terminalsot a balancedthree'plras()Y-connectedload is 2400V
The load hasan impedanceof 16 + j12 O/.,
and is fed from a line havingan impcdanceof
0.10 + i0.13{)
O/.r. The Y-connectedsourceat
the sendingcnd oI lhe line hasa phase
sequenceofacb and an internal impedanceoI
0.02 + /0.16 O/.r. Use rhe a-phasevolrageat
the load as thc relerenceand calculatc(a) the
[:,""Tlill."t*"1'+;Jt].:;!r],:i:iTi"
internal phasclo'neutral voltagesa!
the source.
Va'n.x',,. and Vcn.
Answer: (a) I"A = 120/-3687' A,
IDB= 120/83.13" A. and
rcc : 120/ 156.87' A;
(b) V,b = 4275.02/ 21J.38"V,
Ir''".= 1n5.02 /9162' y, ar'd
Vca= 1275.02/ 14838' Vi
(c) Y,,. - 2182.05L):g
V,
Vb.,,- 2482.05/121.93' V, and
V.,. - 2182.05/ 118.07' V_
NOT E: ALU)tn ChaptetPnhLens Il.8 11.10.
11.4 A*eNy*ris
clf:igt*';tliy*-ileLi:a
Cireui.i
U the load in a thrcc phasecircuit is conncctedit a detra,ir can bc rrans
iormed into a wyc by usingthe delta-tow]'e translormationdiscusscdin
Seclion9.6.Whcn the load is balanced.ihc impedanceof eachlcg ol lhe
wl'e is one third thc iilpedance of eachlcg ol the delta.of
Retationship
between
three-phase
detta-connected
andwye-connected
r,'
imPedance
rigurc11.11., A single?hdeequivatent
cncuit.
-
z!
, : ]
(11.21)
which follows direclly fiom Eqs 9-51 9.53.After the I ioad has been
replacedb,r'its Y equivalent,the a-plrasccan be modeled bl' rhe single,
phaseequivalcntcilcuit shoiJnin Fig-11.11.
We usc this circuit to calculatcthe line currents,and wc then usethe
tine c flcnts lo find the currcntsin eac]rle.qof the orignralI load.The
relationshipbelweenthe line currcntsaDdthe currentsin cachleg of the
delta cai be derivedusingthc circuit shownin Fig.11.12.
Whcn a load (orsourcc)is conrectedin a delta.thc currerrineach leg
of ihc dclla is tire phasecurrcnl, and rhe voltage acrosseach leg is rhe
phasevoltage.Figure 11.12showsthal, in rhe I configuralion.fie phase
voltagcis identicalto the linc vollage.
Cncuit 443
11.4 Anatysis
ofih€Wye-Detta
To demonstrate the relationship between t]le phase currents and Line
curents, we assume a positive phase sequence and let 1d represent ttre
masnitudeof the Dhasecurrent.Then
les : I o19:,
111.22)
rec= Io 1-lT,
(11.23)
rcl'= Ia1lT.
\11.24)
In wriling these equations, we arbitrarily selected IAB as the reference
pllasor,
we can write the line currents in terms of the phase currents by direct
applicationof Khchhoff'scunent law:
used
to estabtish
the
Figure
11.t2a A circuit
Ljne
cunents
andphase
curr€ntsin
rctationship
between
Iua-Ias-Io\
: Io 1!l - Io1)4)
- ',Bto 1 !,
(11.25)
IbB:IBc-IAB
: lo / 120'- 1,r19:
= \nI6 / -15o",
tc:
111.26)
Ica Inc
= I+ 1&l
lao/ 12o"
= .\trto12Y.
\11.21)
revealsthat the magniCornparingEqs.11.2511.27with Eqs.11.22-11.24
the
magnitude
of t]le phase currents
tude of the line curents is \4 times
and that the set of line currentslagsthe set ofphasecurrentsby 30".
We leaveto you to verify that,for a negativephasesequence,theline
curents are 1/3 times larger than the phase currents and lead t}le phase
(b)
cunenls by 30'. Thus,we have a shoitcut for calculating line curents from
(or
A-connected
phasecuirents
vice versa) for a balancedthree-phase
diagrams
shov/ing
the
Figure11.13A Phasor
load. Figure 11.13summarizesthis shotcut graphicaly.Examp]e 11.2 relationship
tinecurrents
andphase
curcntsin
b€tween
(b)The
illustiates the calculations involved in analyzing a balanced three phase a A-connected
sequence.
toad.G)Thepositjve
load.
circuithavhg a Y-connectedsourceand a A-connected
444
Batanced
Thr€€-Phase
Circuits
Anatyzing
a Wye-Detta
Circuit
The Y'connectedsourcein Example 11.1 feeds a
A-connected load through a distribution line having an impedance of 0.3 + i0.9 O/d. The load
impedanceis 118.5+ i85.8 O/d. Use the a-phase
htemal voltage of the generator as the reference.
39.50
a) Consfuct a single-phaseequivalenr circuit of the
three-phase system.
b) Calculalethe line currentsIaA,IbB,and Icc.
c) Calculatethe phasevoltagesat the load terminals.
d) Calculate the phase currents of the load.
e) Calculatetle line voltagesat the soruceterminals.
Solution
a) Figurc 11.14showsthe sitrgle-phase
equivalent
circuit.Theloadimpedanceof theY equivalentis
118.5+ 185.8
= 39.s+ j28.60/0.
b) The a-phaseLinecurent is
1201l:
r,_r: (0.2+
0.3+ 39.s)+ i(0.5+ 0.9+ 28.6)
120
= -; .:/0' = 2.4/
4u + lJu
36.87"A
Hence
lB = 2.4/-156.87' A,
t c:2.4 /83.13"A.
c) Becausethe load is A connected,the phasevoltages are the same as the line voltages To calcu,
late the line voltages,we first calculate VAN.
vAr,{= (39.5+ p8.6)(2.4 / 36.87')
= 117.04/ 0.96" y.
Becausethe phasesequenceis positivg the line
voltage VABis
P8i A
N
N
Figsr€11.14l. Thesingle-phase
equivalent
circujtfor
Example
11.2.
d) The phasecurrents of the load may be calculated
directly from the line currents:
/ 1
\
r ^ a = l - , -//3 0 ' l t .,' "
\vJ
= 1.39/ 6.87'A.
Oncewe know IAB,we alsoknow tlle other load
pnaseculrenls:
lec = 1.39/ 126.8'7'
A,
lc^ - 1.39/113.13'A.
Note that we cancheckthe calculationof IaB by
using the previously calculated VAB and the
impedanceof the A-comectedload;that is,
202.72/29.04"
\aa
_
26
118.5+ /35.8
= 1.39/ 6.n' A.
e) To calculatethe litre voltageat tte terminalsof
the souce, we first calculateVm. Figure 11.14
showstlat Vr is the voltagedrop acrossthe line
impedanceplustlle load impedance,so
Yn : G9.8+ j2e.s)(2.4
/ -36.87')
: 118.90
/-032'y.
Theline voltageVabis
v"b: (15 /30")v"",
vAB : (\6 /30') vAN
= 202;72
/29.04"v.
v,b:205.94
V.
/29.68"
\Bc:202.72 / 90.96"y,
\tcA : 202;72/149.04"V.
Therefore
\r" = 205.94
/-90.32' Y,
v.": 205.94/149.68"V.
Thereforc
l1.r
. B.,, "d \ "e +.,e t' . i
P o t r e( .r. . t . c .o
445
wye-delta
conne<ted
circuit
a batanced,
thrce-phase
obiedive2-Know howto analyze
11.4 The currenl IcA in a balancedtluee-phase
d-comecled load is 8 / 15' A. I[ the phase
sequenceis posjtjve,x'hatis the valueofI"c?
11.6 The line voltagcVAr at the terminalsoI a balancedthree-phaseA-connectedload is
4160A V. Thc line current I"A is
6e.28
1 )!: A.
Answer: 13.86/
a) Calculatethe per-phaseimpedanceof the
load if ihe phasesequenceis positive.
b) Rep€at(a) for a negativephasescquence.
45- A-
11.5 A balancedthree-phase,\-connecledload js
fed from a balancedthree-phasecircoil.The
referenceibr thc b'phaseline cullenl is tot'arcl
the load.Thevaluc of the currentir thE
b-phas<
i. ' '_lq A. Tfthe pltu'..<uuet'c.i.
regative,what is the valueoflAts?
Answer: 6.c)3/
Answer: (a) L041 lLttl
(b) 104lllq
o.
11.7 The line voltageal lhc tcrmjnalsofa balanced
A-connected
load is ll0V Eachphaseof the
load consistsofa3.667 O resistorin pamllel\rilh
is the l1]ag
a 2.75() inductiveimpcdance.What
nit de of the ounenl in thc lioe feedingthe load?
85' A.
NOTE: AIso tty ChapterProblems11.11-l1.16.
in S*la*e*d
11.5 F*rvsr{t*i*uiatisnr
{'ireuits
Thrse-Phase
So lar. we have limited our analvsisof balancedthree phasccitcuits to
dclcrmining curfents aDd voltages.we no\\, discussthree phasc po$'er
We begin by considcringthe averagepower delivcrcd to a
calculations.
balancedY connectedload-
AveragePowerin a BatancedWyeLoad
Figurc11.15showsaY conncctcdload,alongwith its perlinentcurrentsand
with aDyone phascbi'
vollagcs.Wc calculatethe averagcpoNcr associated
ulirg thc tcchniquesintroducedin Chapter10.WithEq. 10.21as a starung
\lith the a phaseas
poinl,wc cxpressthe averagepowcr associated
PA :
v,\\ llaAlcos('?,A
diJ,
r,:
A
(11.28)
whcrc 0rA and PiAdenole the phascanglesof VANaDdI,,^, rcspcctively
Using llrc notation introducedin Eq. 11.28,we can find the powcr associatedwilh thc b- and c-phases:
(d"B ,iB);
Pn = XrNllbulcos
P. = LVCN
ll.c cos(d"c
,ic).
(tt.?9)
Y toadusedto introduce
Fig!re11.15,i A batanced
power
in
thre+plrase
circuits.
avsage
caLcutatioB
(11.30)
are \rrillenin icrms
currcntsandvoltages
In Eqs.11.2811.30,a1lphasor
oI lhe Ims vahe of the siDusoidalfunctionthey represent.
446 Eatanced
Three-Phase
Circuits
In a balancedth ree-phasesystem,the magnitude of eachline-to-neutral
voltage is the same,as is the magnitude of each phase current. The argument of the cosine functions is also the same fot all three phases-We
emphasizetheseobsenationsbyintroducing the following notation:
: VrNl = YcNl,
Yp= V,111l
(11.31)
1{= IoAl- IbBl:
\11.32)
Iccl,
and
0a: AA
qiA = 0B
eiB = 0,c - qic
(11.33)
Mor€over,for a balancedsystem,the power deliveredto eachphaseof th€
load is the same,so
PA: PB = Pc = Pi - Uil6cosei,
(17.34)
whereP4 representsthe aveiagepower per phase.
The total averagepower deliveredto the balancedY-connectedload
is simplythreetimesthe power per phase,or
ga.
Pr : 3P,b: 3U6I,bcos
(11.35)
Expressingthe total power in termsof the rms magnitudesof the line volt,
age and cment is also desirable.If we let yL and 1L representthe rms
magnitudesof the lire voltage and current,iespectively,we can modify
Eq. 11.35asfollows:
Tota[reatpowerin a baLanc€d
three-phase
loadF
r, =z(ft)t*"e,
(11.36)
In deriving Eq. 11.36,we recognizedthat, for a balancedY-connected
load,the magnitudeof the phasevoltageis the magnitudeofthe line voltage divided by \6, and that the magnitude of the line current is equal ro
the magnitudeofthe phasecurrenl.WhenusingEq.11.36to calculatethe
total power deliveredto the load, rememberthat 9d is the phaseangle
betweenthe phasevoltageand current.
ComplexPowerin a BalancedWyeLoad
We can also calculatethe reactivepower and complexpower associated
with any one phaseof a Y-connectedload by usingthe techniquesintroducedin Chapter10.For a balancedload,the expressions
for the reactive
TotaLreactivepowerin a batanced
three-phaseload ts
Qt: Uil$lr'sd,
\11.37)
Qr=3Qo=\f3VLILsin,i.
(11.38)
11.5 Power
CatcuLatiom
in Batanced
Three-Phase
Cncuits 447
Equation 10.29is the basisfor expressingthe complexpower associated
with anyphase.For a balancedload,
Sd = VaN{A : VBN4B = Vo,fic : VdU,
(11.39)
where Yd and I4 repr€sent a phase voltage and curent taken from the
samephase.Thus, in general,
\11.40)
(11.41) < Totatcomptex
powerin a batanced
threephaseload
PowerCatculations
in a Balanced
DeltaLoad
If the load is A -connected,the calculation of power-reactive or complex
is basicallythe sameas that for aY-connectedload.Figure 11.16showsa
A-connected load, along with its pertinent currents and voltages"The
Dower associatedwith each Dhaseis
Pa = lvABlIABlcos(d,AB- daB),
111.12)
PB = lvBcllBc cos(d,s6 - d;6d,
\r1.43)
Pc : lvcAlllcAlcos(dvcA oicr.
(1144) rigur"rl.ro r I a-connected
toadused
to discuss
power
catcuLatjons.
Foi a balancedload,
lVag:lYsc:lVcr:V6
(11.45)
lle-e = llecl = llcc =10,
(11.16)
0,as-oies:d,sc-diBc
= 0&^
9ic^ - 00,
PA = PB = Pc = P6=Uol6cos0o
B r,,^1
(17.47)
(11.48)
Note that Eq. 11.48is the same as Eq. 11.34.Thus,itr a balalced load,
regardlessof whether it is Y- oi A -connected,the aveiage power per phase
is equalto the productof the rms magdtude of the phasevoltagg the rms
magnitudeoI the phasecwent, and the cosineof the anglebetweent]le
Dhasevoltaseand current,
448
Balanced
Ihree-Phde
Ci,cuits
The lotalpoqerdelivered
ro a balanced
A-connecred
loadrs
Pr = 3Pq = 3UoIacos06
=-(*)*"*
: \6yLlLcosd+.
(11.49)
Nore that Eq. 11.49is the sameas Eq.11.36.Theexpressions
for reactive
power and complex power also have the samefolm asthose developed for
the Y load:
sitt0q;
Q4,= V,pI,p
(11.50)
Qr : 3Qo: 3U,bI$ine6.
(11.51)
Sd=Pa+jO{=Va4;
\11.52)
sr=3sc=\a3vLILA4.
(11.53)
Instantaneous
Powerin Three-Phase
Circuits
Although we are primarily interestedin average,rcactive,and complex
power calculationq the computation of the total instantaneous power is
alsoimportant.In a balancedthree-phasecircuit,this power hasar interesting propefiy: It is invariant vrith time!Thus the torque developed at t]le
shaft of a three-phasemotor is constant,which in turn meanslessvibration in machinerypoweredby three phasemotors.
Let the instantaneousline to-neutral voltage ?rANbe the reference,
and,as before,dd is the phaseangle0,A - 9jA.Then,for a positivephase
sequence,the instantaneouspower in each phaseis
?A = oANiuA= U-I ^cos ar cos@t
ed),
pB = oBNibB: U^I-cos(ot - 120') cos(ot
d{
pc = ocNi"c= U^I^cos(at + 120')cos(@t
dd + 120"),
120"),
where y- and 1- representthe maximum amplitudeof the phasevoltage
and line currert, respectively.The total instantaneous powei is the sum of
phasepowerqwhich rcducesto 1.5y-1-cosdd; that is,
the instantaneous
pr:
pA+ pB+ pc:7.5VhI-cos06.
11.5 Power
CaLcutations
in Batanced
Three-Phase
Cncuits 449
Note this result is consistent with Eq. 11.35 s:f;lce V^= \/=2Vo and
1- = \41d (seeProblem11.29).
Examples11.3 11.5illustmte power calculationsin balancedthreephasecircuits.
Calculating
Power
in a Three-Phase
Wye-Wye
Circuit
a) Calculatethe averagepower pei phasedelivered
to the Y'connectedload of Example11.1.
b) Calculate the total average power delivered to
the load.
c) Calculatetle totalaveragepowerlostin the line.
d) Calculate the total averagepower lost in the
generalor,
e) Calculate the total number of magnetizing vars
abso$ed by the load.
f) Calculatethe total complexpower deliveredby
c) Thetotalpowerlostin thelineis
= 13.824
Prne- 3(2.4F(0.8)
w.
d) Thetotalinternalpowerlostin thegenerator
is
= 3.4s6w.
Ps- = 3(2.4)r(0.2)
varsabsorbed
e) Thetotal numberof magnetizing
by the load is
sin3s.68"
o? = \6(199.s8X2.4)
= 483.84
VAR.
50lution
f) The total complex power associatedwith the
a) FromExample11.1,y6 = 115.22y, 16 = 2.4 A,
andgd = 1.19 (-36.87): 35.68'.Therefoie
Pa = (11s.22)(2.4)
cos35.68"
:224.64W.
The powerper phasemay alsobe calculated
from I'aRd,or
Po: Q.q' Ge):224.64w.
b) The total aveiagepowerdeliveredto the load is
Pr = 3Pa = 673.92W. We calculatedttre line
voltagein Example11.1,so we may alsouse
Eq.i1.36:
Pr : v3(199.58)(2.4)
cos35.68'
:673.92W.
sr = 3s{ = -3(120)(21)/3687'
= -691.20- j518.40VA.
Theminussignindicates
thattheintemalpower
andmagnetizingreactivepowerare beingdeliveredto t}le circuit.Wecheckthis resultby calculatingthe total ard reactivepowerabsorbed
by
the circuit:
P : 673.92+ 13.824+ 3.456
: 691.20w(check),
I : 483.84+ 3(2.4f(1.s)+ 3(2.4f(0.5)
= 483.84+ 25.92+ 8.64
: 518.40
VAR(check).
450
Three-Phase
Circuits
Baknced
Wye-Delta
Circuit
Power
in a Three-Phase
Calculating
a) Calculate the total complex power delivered to
the A-connectedload of Example11.2.
b) Wlat percentage of the avemge power at the
sending end of tle Lineis delivered to the load?
Solution
b) The total power at the sendhg end of t]le distribution Iine equals the total power deliveied to
the load plus the total power lost in the line;
therefore
p",,*:
a) Using the a-phase values from the solution of
Example11.2,weobtain
682.56+ 3(L4\,(0.3)
= 687.74W.
Yo = \as:20272
/29O4'Y'
-6.8'7"
:
:
A.
t4 IAB 1.39/
Using Eqs.11.52and 11.53,wehave
sr : 3(N2;72
/2e.M")(1..3e
159:)
= 682.56+ j494.21Y4.
The percentageof the avemgepower reaching
or 99.25ol".Nearly
the load is 682.561687.74,
power
the
average
at the input is
100'/. of
impedanceof
to
the
Ioad
because
the
delivered
quite
to the load
line
is
small
compared
the
impedance.
Load
Three-Phase
Power
with an Unspecified
Catculating
A balanced tlree-phase load requfues480 kW at a
lagging power factoi of 0.8. The load is fed fmm a
lhe having an impedance of 0.005 + j0.025 A/0.
The line voltage at the tem)inals of the load is 600V
a) Consrruct a single-phase equivalent chcuit of
the system.
b) Calculate tle magnitude of the line current.
c) Calculate the magnitude of the Line voltage at
t]le sending end of the line.
d) Calculate the power factor at the sending end of
the lille.
Solution
/6on\
(=lt;=(160+j120)1f.
t:^ = 5Tt.35/36.87' A.
Therefore,I.{= 577.35,z 36.87"A. Themagnitudeof the Linecurent is the magnitudeof IaA:
A.
IL- 577.35
We obtain an altemative solution for 1L from
the expression
Pr = \f3vLlLcos 0p
a) Figure 11.17 shows tle single-phase equivalent
circuit. We arbitmrily selectedthe line-fo'neutral
voltage at the load as the reference,
0.005
o
b) Theline currentIh is givenby
p.m5(l
: \€(600)lL(0.8)
: 480,000
w;
480,000
160kW at 0.8lag
16(600x0.8)
1000
.J3
circuiitor
Hgure11.17A Thesingt€-phase
equivatent
Exampte
11.5.
: 5'17
.35A.
11.5 Power
Cahutations
in Batanced
IhreePhase
Cncuits 451
c) Ib calculatcthc magniludeof the line voltagcat
the sending cnd. we firsl calculate Vxf Fron
Fig.11.17,
s : (160 + jl20)10r + (577.35t(0.005+ j0.025)
V""=Va1;i21I.,1
600
10.rrn5 ,0 02<){s-7r\ /
An alternatjvemethodlor calculatingthe power
factor is to first calculatethe complexpower at
the sendingend of the line:
16r- )
= 16r.67+ j128.33kVA
= 206.41
/38.41'kVA.
= 35'7.st
1lA:v.
Thus
The po$'crfactor'is
pf = cos38.44'
tJ_: v3I,"
= 619.23
V.
d) Thepowcrlaclorat lhe sendingendof the line
is ihc cosineoI the phaseanglebetr,cenVrn
andI.^:
pf: cos[1.s7' ( 36.87')l
= 0.783lagghg.
Fina y, if we calculatethe total complexpower
al the sending end, after first calculatingthe
magniludeof the lire current,we may use this
valueLocalculateYL.'fhatis.
\a3YL1L= 3(206.4r)x 10r,
= cos311.44'
3(206.4r
) x 10r
\6(577.35)
= 0.783lagging.
V.
619.23
0bjective3-Be abteto catculate
power(average.
reactive,
andcomptex)
in .ny three-phas€
circuit
11.8 The lhree-phase
averagepowerratingof the
central processirg udt (CPU) on a mainframe
digiialcomputeris 22,659WIhe three-phase
line supplying the con1puterhasa line voltage
rating of 20BV (ms). The line curent is 73.BA
(rms). The computer absorbsmagnetizirg VARS.
a) Calculatethe total magnetizingreaclive
power absorbedby the CPU
b) Calculatethe powerfactor.
11,9 The complexpower associated
with eachphase
ofa balancedload is 384 + l28B kVA. The line
voltageat the terminalsofthe load is 4160V
a) w}Iat is the magnitudeof the line currenf
feedingthe load?
b) The load is delta connected,and the impedance of each phase consistsof a resistancein
pa.allel with a reactance.Calculate R and -L.
c) The load is wye connected.and the irl1pedanceof eachphaseconsistsofa resistancein
serieswith a reactance.CalculateR and-Y.
Answerr (a) 199.95A:
Answer: (a) 13,909.50
VAR:
(b) 0.852lagging.
NOTE: Also tryCh.tptetPrcbtems
I1.24and 11.25.
(b)R - 4s.07o,x = 60.09o;
( c )R : 9 . 6 O
1 ,x : 7 . 2 1 O .
Ihrce-Phase
circuits
452 Batanced
Power
11.6 Measuring
Average
in Three-Phase
Circuits
rerminals
The basjcinstrumentusedto measurepower in three-phasecircuitsis the
el€ctrod].namometer
wattmeter.It containstwo coils-One coil, calledthe
currentcoil,is stationaryand is designedto carry a cunent proportionalto
the load current.The secondcoil, calledthe potentialcoil, is movableand
carriesa currentprcportional to the load voltage.The impo ant features
ofthe wattmeterare sho\r'I)in Fig.11.18.
The average deflection of the pointer attached to the movable coil is
proportional to tle product of the effective value of the current in the currcnt coil, the effective value of the voltage impressedon the potentral coil,
and the coshe of the phase angle between the voltage and current. The
direction ilr which t}re pohter deflects dependson the instantaneouspolarity of the currentcoil current and the potential-coil voltage.Therefore each
coil has one terminal with a polarity mark usually a plus sign but sometimes t]le double polarily maik t is used.The wattneter deflects upscale
when (1) the polaity-marked termhal of the current coil is toward the
source,and (2) the poladty-marked teminal of tle potential coil is con
nected to the sameline in which the current coil has been inserted,
Pointer
figure11.18a. Thekeyfeatures
ofthe
ehctrcdynamometer
wattmeter.
TheTwo-Wattmeter
Method
poweris
Figure
11.19A A genentcircuii
whose
supptjed
by, conductor.
I""
z , t =z 8
\\r
ia
Fisure
r1.20A A circuit
used
to anatyze
the
power
method
two-wattmeter
of measuring
average
d€tivercd
to a batanced
Load.
Considei a generalnetwork hside a box to which power is suppliedby
r conductinglines.Sucha systemis shownin Fig.11.19.
If we wish to measure the total power at the terminals of the box, we
need to know n - 1 currents and voltages. This follows because if we
choose one teminal as a reference, ttrere are only n - 1 independent
voltages.Lilewise, only n - 1 independent curents can exist in the n conducto$ entering the box-Thus the total power is the sum of n - 1 product
terms;thatis,p : 1)i1 + azi2+ .. + a" i" 1.
Applying this general observation,
that for a threeconductorcircuit,whetherbalancedor not, we need only two wattmete$
to measurethe total power.For a four-conductorcircuit, we need three
wattmeters if the three-phase circuit is unbalanced, but only two
wattmetersifit is balanced,becausein the latter casethereis no currentin
the neutml lhe. Thus, only two wattmeters are needed to measure the
total averagepower ill any balanced thrce phase system.
The two-wattmeter method reduces to detemining the magnitude
and algebraic sign of the average powor indicated by each wattmeter. we
can describethe basicproblem in terms of the circuit shownin Fig.11.20,
wh€re the two wattmetersare indicatedby the shadedboxesand labeled
Iry1and W2.The coil notarions cc and pc stand for curent coil and potential coil, rcspectively. We have elected to insert the curent coils of the
wattmeters in lines aA and cC Thus, line bB is the rcference line for the
two potential coils.The load is connected as a wye, and the per-phaseload
impedanceis designated^s26= Zl4!.Tlis s ^ generalrcpresentation,
as any A-connected load can be repiesented by its Y equivalenl fu hermoie, for the balanced case,the impedance angle 0 is unafrected by the
A-to-Y transformation.
11.6 l4easudng
Average
Powerin
Three-Phase
Circuits
We now develop general equations for the rcadings of the two
wattmetef. We assumethat the curent drawn by the potential coil of the
wattmeter is neglgible compaied with the line current measuredby the current coil. We further assumethat tlle loads canbe modeled by passivecircuit
elementsso that the phaseangle of the load impedance(d in Fig. 11.20)lies
between 90" (pure capacitance)
and +90' (pure inductance).
Finally,we
assumea Positive phasesequence.
From our introductory discussion ol the avetage deflection of the
wattmeter, we can see that wattmeter 1 will resDond to the Droduct of
lvasl. I a. andrhecosineot theanglebetweenVABaDdl,A. ll we denole
this wattneter reading as Wl, we can wite
W1 :
VABIIa\ cos 01
- I/111cosd1,
(11.54)
It followsthat
IV, : VcBltcl cosd,
: VLILcosq2.
(11.55)
In Eq. 11.54,91is tlre phaseanglebetweenVABard l"A, ard in Eq- 11.55,
02is tlte phaseangle between VcB and lcc.
To calculate Wr and Wr, we express 0t and d2in terms oI tle impedance angle 0, which is also tlle same as the phase angle between the phase
voltageand cuffent.For a positivephasesequerce,
e r - 0 + 3 0 "= 0 0 + 3 0 ' ,
(11.56)
02=0 30":00-30'.
(11.57)
The derivationof Eqs. 11.56and 11.57is left as an exercise(see
hoblem 11.32).
wllen we substitute
Eqs.11.56and 11.57into Eqs.11.54
and11.55,
respectively,
we get
w\=uLILcos(0,b+30'),
(11.58)
try2= VLILcos(0,, - 30').
(11.59)
To find the total power, we add W1and W2;thus
Pr : Wr + W2 = 2YLILcosd{ cos30"
: \a3YL1Lcosd!r,
(11.60)
which is the expressionfoi the total power in a three-phasecircuit.
Therefore we have confimed that the sum of the two wattmeter readings
yieldsthe lotal averagepowei.
454
Batanced
lh€e'Phase
Cncuits
A closerlook at Eqs.11.58and 11.59revealsthe following about the
readingsof the two wattmeteN:
1. ff the powerfactoris greaterthan0.5,both wattmetersreadpositive.
2. Ifthe powerfactor equals0.5,onewattmeterreadszero.
3. If the power factor is lessthan 0.5,one wattmeterreadsnegative.
1. Revening the phasesequencewill interchangethe readingson the
two wattmeters.
These obse ations are illustrated in the following example and in
Problems11.,13F11.51.
Wattmeter
in Three-Phase
Computing
Readings
tircuits
Calculatethe readingof eachwattmeterin the circuit in Fig. 11.20if the phasevoltageat the load is
120v and (a) zd) = B + j6 o; (b) zd : 8 j6 O;
(c) Zd : 5 + j5V3 o; ar,d (.d)Z$: 10 1_ll:4.
(c) VerjJy for (a) (d) thar th.. sum of the wattmeter
rcadingsequalsthe total power delivercd to t}le load.
c) zo = s(I + j\5) - 101!!
o, vL : 120\6v,
and IL : 12 A.
wr = (i20vr(12) cos(60' + 30') : 0,
w, = (120\aO(12)
cos(60' 30')
: 2160W.
d) zb = 101 lt:a'vL
IL=124.
Solution
a) z4 : 10/36.8'7'A,VL = 120\6V, and
IL - 120/10- 12A.
tv1: (120\4)(12)cos(36.87'+ 30")
:979;75W,
r, : (120\6)(12)cos(36.87'- 30')
- 2416.25
W.
b t l , ! - l 0 - 3 0 . 8 70 . l r - l 2 0 v l v . x n d
IL : 120/10: 12A.
u/1: (120\6)(12)cos(-36.87"+ 30')
= 24'76.25
W,
= 120\4v, and
lll
= (120\6X12)cos( 75'
w2 = (120\6X12)cos( 75"
w,
30'): 1763.63
30'):
W.
645.53
e) Pr(a) = 3(12)'(8)= 34s6w.
w + t4r2= 979.75+ 24'76.25
= 3456W,
Pr(b) = Pr(a) : 3456W
w1 + w2 = 24'76.25+ 979.75
= 3456W
Pr(c) : 3(12),(s)= 2160w,
wr+w2-0+2160
: 2160W,
: 1118.10w,
Pr(d): 3(12f(2.5882)
w, = (120\4)(12)cos(36.B7' 30')
= 979.75
W.
wt + w2 = 1163.63 645.53
= 1118.10W.
your anderstading of the tuo-trattmetermethodby ttying ChapterPrcblems11.41and11.42.
NOTE: Assess
PracticatPe6pectiv€
455
PracticaI
Perspective
Transmission
andDistribution
of Etectric
Power
At thestartof this chapter
we pointedout the obLigation
utiLities
haveto
maintajn
premises.
the rmsvoLtage
levelat theircustome/s
Atthough
the
acceptabte
deviation
froma nominal
[eve[mayvaryamong
different
utititjes
we witl assume
for purposes
of discussion
that an attowabte
toterance
is
15.8%.Thusa nornlnal
rmsvottage
of 120V couLd
range
frcm113to
pointed
127V. WeaLso
outthatcapacjtors
strategjcalLy
located
onthesystemcoutdbeusedto support
vottage
levels.
Thecjrcuitshown
in Fiq.11.21represents
a substation
ona Midwestern
municipal
systen.Wewitl assume
the system
is batanced,
the [ine-to{ine
vottage
at thesubstation
is 13.8kV, the phase
impedance
ofthe djstribution tineis 0.6 + J4.8O,andthe toadat the substation
at 3 PMon a hot,
humiddayin Jutyis 3.6MW and3.6maqnetizing
MVAR.
lJsjng
the Line-to-neutral
vottage
at the substation
asa reference,
the
sjngte-phase
equivalent
circuitfor the system
in Fiq.11.21js shownin figurc11.21.{ A substation
connecied
to a power
Fig.11.22.TheUnecurr€ntcanbe calcutated
fromthe expression
for the ptantviaa thr€€-phase
tine.
powerat thesubstation.
compLex
Thus,
(r.2+ j1.2)106.
#$rtIt fotiows
that
{-,, = 1s0.61+ j150.61A
IaA= 150.61- 1150.61
A.
n
rigure11.22A A singte
phase
equivaLent
circuit
for
thesyst€m
in Fig.11.21.
ptantis
The[ine-to-neutraL
voLtage
at thegenerating
V"-
t -l R
. -O: n
VJ
-1 0
r 0 . 6- 1 4 . 8 x l 5 0 . o L
il50.ol)
= 8780.74
+ j632.58
= 8803.50
lllq
v.
Therefore
plantis
the magnitude
ofthe linevoLtage
at thegenerating
v,bl = \6(8803.50)= 15,248.11v.
W€areassuming
theutiLityis required
to keepthevoLtage
Levet
within
+ 5.8o/o
of thenominal
vaLue.
Thismeans
themagnitude
ofthe Line-toiine
voltageat the powerptantshoutdnot exceed
14.6kV nor be lessthan
13kV. Therefore,
themagnitude
ptant
ofthe linevoLtage
at thegenerating
probLems
coutdcause
for customers.
When
the magnetizing
varsaresuppfied
bya capacitor
bankconnected
to thesubstation
bus,thelinecurrent
IiA becomes
I"a = 150.61+ J0A.
N
456
Eatanced
Ihree-Phase
cncLrits
ptantnecessary
Therefore
the vottageat the generating
to maintaina
[ine-to-tine
vottageof 13,800
V at the substatioris
1J-800
v, 1 4 . 8 ) f l 5 0 -. /60t )
\3 '/0"-(0.6= 805'7
.8O+ j'722.94
: 80eo.r'7
1:13:v.
Hence
v.
lv"bl : \6(8090.17) : 14,012.58
lhis voLtage
tevethLtswithin the aLLowabLe
langeof 13 kv to 14.6kV.
your undetstonding
Njft: Assess
of this PtocticalPerspective
by tryingAoptel
Prcblems
11.52(a)-(b) dnd11.53-11.55.
Sunrmary
Wlen analyzing balarced thee-phase circuits, the fust
stepis to transform any A connectionshto Y connections,
so that the overall cicuit is of t}te Y-Y configuration. (See
page436.)
A single-phaseequivalent circuit is used to calculate the
line curent and the phasevoltagein one phaseof the
Y Y structure. The a-phase is normally chosen for this
purpose.(Seepage438.)
Once we know the line curent and phasevoltage in the
a-phaseequivalentcircuit,we cantake anallaicalshorf
cutsto find any current or volragein a balancedthreephasecircuit,basedon the followingfacts:
. The b- and c-phasecurents and voltagesare identical to tle a-phasecurrent and voltage exceptfor a
120'shift h phase.Ina positive-sequence
circuit.the
b-phasequartity lags tlle a-phasequantity by 120",
and the c-phasequantity leadsthe a-phasequantity
by 120".For a negativesequencecircuit,phasesb and
c are interchangedwith respectto phasea.
. The set of line voltagesis out of phasc with the set of
Fha\e\ollagesh) LJ0".Theplucor mtnu.,igncorre
spondsto positive and negativesequencgrespectively.
. In a Y-Y circuit the magnitudeof a line vollage is
\6 timesthe magnitudeof a phasevoltage.
. The set of line currentsis out of phasewith the set of
phasecurents in A-connectedsourcesand loadsby
+30".The minusor plus sign correspondsto positive
and negative sequence,respectively.
. The magnitudeof a line currentis \4 timesthe mag
nitude of a phasecurent in a A connectedsource
or load,
(Seepages439 and 443.)
The techniques for calculating per-phase average
power, rcactive power, and complex power are identical
to thoseintroducedin Chapter10.(Seepage445.)
The total real, reactivg and complex powei can be determined either by multiplying the conesponding per phase
quantity by 3 or by using tlle expressionsbased on line
currentandline voltage,asgivenby Eqs 11.36,11.38,
and
11.41.(Seepages146and447.)
powerin a balancedthree-phase
The total instantaneous
circuit is constant and equals 1.5 times the average
powerper phase.(Seepage448.)
A wattmetermeasuresthe avemgepower deliveredtoa
load by using a current coil connectedin serieswith
the load and a potential coil connectedin parallel
wirh rhe load. (see page,l52.)
The total averagepower in a balancedthree-phasecircuit can be measuredby surnmingthe readingsof two
wattmetersconnectedin two different phasesoI the
circuit.(Seepage452.)
Probhms457
Problems
All phasor voltagesin th€ folowi|rg Probl€ms ar€ stated in
t€rmsofthe rms value,
e) ?,, : 339cos(dt + 30") V,
?)b: 339cos(4,1- 90') V,
Section11.1
0c = 393cos(ot + 240') V.
11.1 What is the phasesequenceof eachof the following
setsof voltages?
a) 1]a= 120cos(o, + 54') V,
t)b : 120cos(or
66') V,
I) r'" = 3394sin((,t + 70') V,
t,b : 3394cos((rr - 140")V,
lJ" : 3394cos(.rt + 180")V.
?]": 120cos(.,,1+ 174")V.
b) 1]i : 3240cos(ot
26') V,
ll,3 Vedfy that Eq. 11.3is true for eitler Eq. 11.1 or
Eq.11.2.
,b : 3240cos(ol + 94") V,
tr. = 3240cos(rt - 146') V.
Section11.2
11.2 For eachset of voltageEstatewhetler or not t]te voltagesform a balancedthrce-phaseset.ff the setis balanced,statewhether the phasesequenc€is positive or
negative.If the set is not balanc€d,erylain why.
11.4 Reler to the circuitin Fig.11.5(b).Assumethat there
are no erlemal comections to the terminals a,b,c.
Assume furtler that t}le three windings are from a
balancedthree,phase generator. How much currenr
will circulate in the A-connected genemtor?
a) ?a : 339cos37?l
V,
?)b= 339cos(37?r
120') V,
t'. = 339cos(377r
+ 120")V.
b) ," : 622sin377rV,
Db: 622sn(3'7'7t- 240')v,
D.: Ozsln(311t + 240')Y.
c) ?]a= 933sin377rV,
0b = 933sin(377r + 240') V,
0" = 933cos(377t + 30') V.
d) ?,! : 170sin(,))t+ 60') V,
0b = 170sin (or + 180') V,
t'c = 170cos(.rr - 150")V.
Section11.3
11.5 a) Is the circuit in Fig. P11.5a balancedor unbalancedthree-phasesystem?Explain.
b) Find I,.
Figure
P11.5
458
Eatanced
Three-Phase
Circuits
11.6 a) Find I, in the circuit in Fig.P11.6.
6a€ b) Find vAN.
c) Find VAB.
d) Is the circuit a balancedor unbalancedthreephasesystem?
11.7 Find the Ims value of L in the unbalanced three
6dc phasecircuit seenin Fig.P11.7.
1L8 The time-domain er?ressiois for three line{o-neutral
voltagesat the teminals of a Y-connectedload are
oAN: 7967cosol V,
t'BN: 7967cos((,t + 120') V,
ocN:7967cos(ot - 120")V.
What are the time-domain expressionsfor the three
lhe-to'line voltages?.,A8,
,Bc, and ?cA?
11.9 The magnitude of the line voltage at the terminals
of a balancedY-connected load is 12,800V The load
impedanceis216 + j63 O/d. The load isfed from a
line tiat hasanimpedanceof 0.25 + j2 Ab.
a) What is tlre magnitude of the line cu[ent?
b) What is the magdrude of tlre line voltage at
the source?
11.10 The magnitude of t]le phase voltage of an ideal balanced tlree-phase Y-coinected source is 4800 V
The souce is contrected to a balancedY-connected
load by a distribution line that has an impedance of
2 +j16 O/d. The load impedanceis 190+ /40 O/d.
The phase sequence of ttre source is acb. Use the
a-phase voltage of the source as the referenc€.
Specify the maglitude and phase angle of the following quantitiesr (a) the three line currents, (b) the
threeline voltagesat the source,(c) the tbreephase
voltagesat the load, and (d) tle thee line voltages
at the load.
11.11 A balanced tlree-phase circuit has t}le folowing
chamctenstics:
. Y-Y connected;
. The line voltage at tlte source, Vab, is
120^v51yv;
.
.
.
FigurcP11.6
0.2O ll.6O
a
0.8O jJ.4O A
03a
b
0.7o
j5.6o
B
u0/-t20" v
0.4O .i2.8O c
L6A
j1.24
C 78O j50O
0.8o
j4o
B loo
i4O
C 1590 j114.4O
t2.4
7aO
/ 5 to
5O
Ta
j52a
Figure
P11.7
0 . 2o
jt.6o
b
480t128V
480!2q" V
o2A
jL6A
i 2 4 .o
4
2OA
c
0.8O
The phase sequenceis positive;
is2 . i3ab:
The lineimpedance
fhe load impedanceis28 + j3'7 AlO.
a) Draw the single phase equivalent cfucuit for the
a-phase,
b) Calculatedthe line currentin the a-phase.
c) Calculated the line voltage at the load in the
a-phase.
probtems459
Section11,4
11,1j, A balancedA-connectedload hasan impedanceof
360 + i105 O/d. The load is fed througha line havhg an impedanceof0.1 + j1 O/d. The phasevottage at the teminals of the load is 33 kV The phase
sequenceis positive. Use VABas the reference.
a) Calculate the three phase currents of tle load.
b) Calcutate the three line currents.
c) Calculate the thiee line voltages at the sending
end of the line.
11,13 A balalced Y-connected Ioad having an impedance
of 96 - j28 O/4 is coinected in parallelwith a balanced A-connectedload having an impedanceof
144 + j42 A/ 6.me paralleled loads are fed frcm a
line having an inpedance of i1.5 O/4. The magni,
tude of the Line-to-neutral
voltage of the Yload is
7500
v
a) Calculate tie magnitude of the current ir t}le
line feedingthe loads.
b) Calculate the magnitude of the phasecurent it
the A-connectedload.
c) Calculate the magnitude of the phase current in
tie Y-connected load.
d) Calculate the magnitude of rhe line volrage ar
t}le sendhg end of the line.
11.14 A balanced, three-phase circuit is characterized as
follows:
. Y-A contrected;
. Souice voltagein the b-phaseis 201:9q V;
. Sourcephasesequenceis acb;
. Lhe impedanceis 1 + j3 O/4;
. Load impedanceis 117 - j99A/0.
a) Draw the singlephaseequivalent for the a-phase.
b) ( alculared
lhe a-phase
linecurreDl.
c) Calculated the a-phase line voltage for the
three-phaseload.
11,15 In a balanced ttree,phase system, the source $ a
balanced Y with art abc phase sequenceand a line
voltage Vab= 208 q V. The load is a balanced1'
in pamllel with a balancedA. The phaseimpedance
of the Y is 4 + i3 O/4 and the phaseimpedance of
the A is J
iq O d. The tine impedance is
1.4 + j0.8A/0. Draw the single phaseequivalenr
circuit and useit to calculated tle line voltage at t}Ie
load in the a-phase.
11.16 An abc sequencebalancedthree-phaseY-connected
sourcesuppliespower to a balanced,three-phase
A-connected load with an impedance of
12 + j9 O/d. The sourcevolragein the a-phaseis
1204qV. The line impedance is 1 + j1Ab.
Draw the single phase equivalent circuit for the
a-phaseand useitto find the currentin the a-phase
of the load.
11.f
A balanced three-phase A-connected source is
showninFig.P11.17.
a) Find the Y-connected equivalent circuit.
b) Show that the Y-connected equivalent circuit
delivers the same open-circuit voltage as the
original A-connectedsource.
c) Apply an external short cftcuit to the terminals
A, B, and C. Use the A,connected source to find
the thee line curents I"A, IbB, and lcc.
d) Repeat (c) but use the Y-equivalent source ro
tind ttre thrce line currents.
rigurcP11.17
'72U,!tLl2w
460
Three-Phase
cncLrits
Salanc€d
1L18 The A-connectedsourceof Problem 11.17is con'
nected to a Y-connectedload by meansof a balanced three-phase distribution line. The load
impedance is 957 + j259 O/d, and t}le line impedanceis1.2 + j12 A/6.
a) Construct a single-phaseequivalent c cuit of
me sysrem,
b) Determinetlre nagnitude of the line voltageat
the teminals of the load.
c) Determine the magnitude of the phase current
in the A-source.
d) Detemine the magnitudeof the line voltageat
the terminals of th€ source.
11.19 A three-phaseA connectedgeneratorhasan internal impedanceoI0.6 + j4.B O/d. Wlen the load is
removed from the generator, the magdtude of the
terminal voltageis 34,500V The generatorfeedsa
A-connectedload through a transmissionline with
an impedanceof 0.8 + j6.4o"/6. The per-phase
impedanceof the load is287'7 j864A.
a) Construct a single-phaseequivalent circuit.
b) Calculatethe magnitudeofthe line current.
c) Calculate the magnitude of the line voltage at
the terminalsof the load.
d) Calculate the magnitude oI the line voltage at
the terminals of the souce.
e) Calculatethe magnitudeof the phasecurent in
the load.
f) Calculafe the magnitude of the phase current in
lne soutce.
11,20 The impedance Z in the balanced three-phase circuir in Fig.P11.20is 600 + j450 o. Find
a) tAB.IBc, and IcA,
b) laA,IbB,and lcc,
c) Ib,, tb, and Iac.
tigureP11.20
a
69[40: kV
I
"
{
A
11,21 For the circuit shown in Fig. P11.21,find
""t' a) the phasecurrentslAB,IBc, and IcA
b) the line currentsIaA,lbB,and \c
when Zl : 4 8 + j1.4A. 22: 1.6 jDA,
2 3 : 2 5 + j 2 5[ 1 .
arl'd
tigureP11,21
72OD2y
S€ct'ion11,5
1112 A balanced three-phase sourceis supplying 90 kVA
at 0.8 lagging to two balanced Y-connected paiallel
loads The distribution line connecting the source to
the load has negligible impedance.Load 1 is purely
resistive and absorbs 60 kW Find the per-phase
impedanceof Load 2 if the line voltageis 415.69V
arein sefies.
andlhe impedance
componenls
1L23 The total apparentpower suppliedin a balanced,
three-phaseY-A systemis 3600VA. The line voltage is 208 V If the line impedance is negligible and
the power factor angleof the load is 25".determine
the impedanceof the load.
11.24 In a balanced tbiee phase system, the source
has an abc sequence, is Y connected, and
The sourcefeedstwo loadq both
van = 1204qv
are
Y-connected.
The impedanceof load 1
o{ which
j6
power for the a-phase
+
O/{.
The
complex
is 8
2
is
60013!iy4.
Find
the total complex
of load
power suppliedby the sourc€.
11.25 A three-phase positive sequ€nce Y-conrected
source supplies 14 kVA with a power factor of 0.75
lagging to a paralel combination of a Y-connected
load and a A-connectedload.The Y-connectedload
uses9 kVA at a power factor of 0.6laggingand has
an a-phasecurrent of 101:lq A.
a) Find t}le complex power per phase of the
A-connectedload.
b) Find the magnitudeof the line voltage.
Prcbtenrs
461
11,26 Calculate the complex power in each phase of the
unbalancedload in Problem11-21.
1127 Threebalancedthrce-phaseloadsare connectedin
parallel.Load 1 is Y-conrectedwith an impedance
ol 300 + j100 O/d; load 2 is d-connectedwitl an
impedanceof 5a0[ - P700glf; and load 3 is
112.32+ /95.04kVA. The loadsare fed from a distribution line with an impedanceof 1 + i10 O/d.
The magnitude of the line{o-neutral voltage at the
load end ofthe line is 7.2 kV.
a) Calculatethe total complexpower at tlte send
ing end of the line.
b) What percentageof the averagepowei at the
sendingend of the Lineis deliveredto the loads?
11.28 a) Find tlle rms magnitude and the phase angle of
IcA in the circuit shownin Fig.P11.28.
b) What percent of the averagepower delivered by
the three-phasesourceis dissipatedinthe threephaseload?
Figure
Pl1,28
5031I)
j1380O
t6o
b 3()
5031r)
j24a
B
5031()
O
11380
1,1,000r12E
v
j6{I
j1380O
c 3O
j 2 4A
0.80pf lag,and L = 168kW and 36 kVAR (magnetizing). The magnitude of the Linevoltage at the
teminals of the loads is 2500\6 V.
a) What is the magnitudeof the line voltageat the
sendingend of the line?
b) Wlat is ttre percert efficiency of the distribution
line with iespectto averagepower?
11.31The three pieces of computer equipment described
below are installed as part of a computation center.
Each pieceof equipmentis a balancedthree phase
load rated at 208V Calculate(a) the magnitudeof
the line cuirent supplyingthesethree devicesand
(b) the power factor ofthe combinedload.
. Hard Ddve:5.742kW at 0.82pflag
. CD/DVD drive: 18.566kVA at 0.93pf lag
. CPU:line current 81.6A, 11.623kVAR
tL32 A
three-phase line has an impedance of
0.5 + j4 oft. Tl:'e Line feeds two balanced tl eephaseloads connectedin parallel.The first load is
absorbinga lotal ol 6q1.2 kW and delirering
201.6 kVAR magnetizingvars. The second load
is A-connected and has an impedance of
622.08+ j181.44A/6. The line to,neutral voltage
at the load end of the line is 7200V What N ue
magnitude of the line voltage at the source end of
the line?
1133 At full load, a commercially available 200 hp, threephase hduction motor opemtes at an efficiency of
96% anda powerfactoro{ 0.9lag. The motor is supplied from a three-phase oudet with a line-voltage
rating of 208V
a) What is the magnitude of the line current dmwn
fiom the 208V outlet?(1 hp = 74614';
b) Calculare the reactive power supplied to
the motor.
11,34 The line-to-neutral voltage at the teminals of the
11:9 Show that the total instantaneouspower in a balancedtbree-phasecircuit is constantand equal to
\NhereU- and 12 represent the max1.5U^I- cos0a2,
imum amplitudesof the phasevoltage and phase
curent, respectively.
11.30 A balancedtbree-phasedistribution line has an
impedanceof 1 + j5 O/d. This line is usedto sup
ply three balancedthree-phaseloadsthat are con,
nected in parallel. The tiree loads are
Lr = 75 kVA at 0.96 pf lead, lz = 150kVA at
balancedlhfee-phase
load in rhe circuilsho$n in
Fig-P1l.34 (page462) is 480V At this voltage,the
load is absorbing60 kVA ar 0.8pflag.
a) Use VAN as the reference and express I.o in
polar form.
b) Calculate the complex power associatedwith
the ideal thrce'phasesource.
c) Check that the total averagepower delivered
equals the total averagepower absorbed.
d) Check that the total magnetizing reactive power
deliveied equalsthe total magnetizingreactive
Powerabsorbed.
462
Batanced
Thr€€-Phase
Circuits
1L38 The total power delivered to a balanced three-
FigurcP11.34
02o
v.n
/0.4o
-js o
11.35 A
balanced three-phase source is supplying
1800kVA at 0.96pf lead to two balancedY-connected
panllel loads The distribution line connecting t}te
souice to the load has negligible impedance.The
powei associatedwith load 1 is 192 + i1464 kVA.
a) Detenine the impedanceper phaseof load 2 iJ
the line voltageis 6400V3 V and the impedance
componentsale m senes,
b) Repeat (a) with t}le impedance components in
parallel.
11.36 A balancedthree phaseload absorbs190.44kVA at
a leading power factor of 0.8 when the line voltage at
rhererminakor rheloadis lJ.800vrind fourequivalent circuits that can be used to model this load.
tL37 The output of tlle balarced posilive'sequence
three-phasesourcein Fig. P11.37is 78 kVA at a
leadingpower factor of 0.8.Theline voltageat the
sourceis 208\5 v.
a) Find the magnitude of the line voltage at the load.
b) Find the total complex power at the terminals of
the load.
phase load when operating at a line voltage of
6600V3-V is 1188 kW at a lagging power {actor ol
0.6.The impedanceof the distibution line supplying the load is0.5 + j4 o/d. Under theseoperating
conditionsr tle drop in the magnitude of the line
voltage between the sending end and the load end
of the line is excessive.To compensate, a bant of
A-connectedcapacitorsis placed in parallel with
the load. The capacitor bank is designed to furnish
i920 kVAR of magnetizing reactive power when
operated at a line voltage of 6600\,€ v.
a) w1lat is ttre mag tude of the voltage at tie
sendingend of the lhe when the load is opemt
ing at a line voltage of 6600\4 V and the capacitor bank is discoDnected?
b) Repeat(a) witl the capacitorbank conn€cted.
c) What is the averagepower efficiency of the line
in (a)?
d) wllat is tlle averagepower efficiercy in (b)?
e) If the systemis operating at a ftequency of 60 Hz,
what is the size of eachcapacitor in microfarads?
11.39 A balanced bank of delta-connected capacitom is
connectedin paralel witl t]le load describedin
AssessmentProblem 11.9.The effect is to place a
capacitor in parallel with the load in each phase.
The line voltage at the terminals of the load thus
remains at 4160V The circuit is operating at a frequency of 60 Hz.The capacitors are adjusted so that
the magnitude of the line current feeding t})e paiallel combinationofth€ load and capacitorbank is at
its minimum.
a) What is the sizeof eachcapacitor in miqofarads?
capacitois
b) Repeat(a) for wye-connected
c) Wiat is t}le magnitude of the line current?
F l g u rP
e1 1 . 3 7
Section11.6
11.40 Dedve Eq$ 11.56and 11.57.
11.41 The two-wattmeter method is used to measure the
power at the load end of the line in Example 11.1.
Calculatethe readingof eachwattmeter.
Probtems
463
11.42 The two wattmete$ in Fig. 11.20 can be used ro
computethe total reactivepower ofthe load.
a) Pro\e rhi. sralement by .houiog rhal
^''3(W- W\: ^/3v'1'"i"0,r.
b) Compute the total reactive power from the
wattmeter readirgs for each of the loads in
Example 11.6.Checkyour computationsby calculating the total reactive power directly from
the given voltage and impedance.
11,43 In the balanced three-phase circuit shown in
Fig.P11.43,thecurrentcoil ofthe wattmeteris con,
nected in line aA, and the potential coil of the
wattmeter is connected across lines b and c_Show
that the wattmeter rcading multiplied by 1,6 equals
the total reactivepowei associatedwilh the load.
The phasesequenceis positive.
11.46 The balarc€dtbree-phaseload showninFig.P11.46
is fed from a balanced,positive-sequcnce,
rnreephase Y-connected source. The impedance of the
line connecting the souce to the load is negligible.
The line-to-neutral voltage of the source is 4800V
a) Find the readingofthe wattmeterin warts.
b) Explain how you would connect a second
wattmeter in t}le circuit so t]tat the two
wattmeterswould measurethe total power.
c) Calculatethe readingof the secondwacmeter.
d) Verify that the sum of the two wattmeter readings equals the total averagepower delivered to
the load.
figureP11.46
it ----
Figure
P1r,43
- I
- 1 -
600kvA
<{B
0-96pf lag
w-
---{c
11,47 a) Calculatethe readingof each wattmeterin the
circuit shol*tr in Fig. P11.47.The value of Zo is
601j!:4.
b) Verify that the sum of the wattmeter readings
equalsthe total averagepower deliveredto the
A-connectedload.
1L44 The line,to-neutral voltage in the circuit in
Fig.P11.43is 720V the phasesequenceis posirive,
and the load impedance1s96+ j72 A/0.
a) Calculatethe wattmeterreading.
figurc Pt7-47
b) Calculate the total reactive power associate{:l
with the load.
4801qv
11.45 a) Calculatetlrc complexpower associatedwith each
phaseof the balatced load in Problem 11.20.
b) Ifthe two-wattmetermethodis usedto measure
ure averagepower deliveredto the load,specify
the readingof eachmeter.
oi . n
480/-120"V
i w ,
464
circuits
Thre€-Phase
Batanced
11,48 a) Find the reading of each wattmeter in the circuit
sho\rn in Fig. P11.48 tt 2^=60/-30"{l'
ZR- 24lyq
andZc = 80A O.
b) Show that the sum of the wattmeter readings
equals the total average power delivered to the
load
unbalancedthree-Phase
figureP11.46
trtolrzo"v!---
11.49 The wattmeten in the circuit in Fig. 1120 read as
w'
folows: u1 = 114,29164 w, and w, = 618,486.24
V.
is
7600v3
voltage
of
the
line
The magnitude
2,.
posiri!e.
Fnd
is
Thephasesequence
fl50
a) Calculate the reading of each watfmeter in the
circuit sho\an h Fig P11.50 when z =
216 - j2M A.
b) Check that the sum of the two wattmeter readings equals the total power delivered to t]Ie load
c) Check that \4(W1 - W2) equals the total magnetizing vars delivered to the load.
figureP11.50
c) Show that the sum of the two wattmeter readings equals the total power delivered to the
unbalancedload.
Sections11.1-11.6
1L52 Refer to the Pmctical Perspective example:
,lli|l#h a) Constru"t u power triangle for the substation
Ioad before the capaciton aie connected to
the bus.
b) Repeat (a) aftel the capaciton aie connected to
the bus
c) Using ttre line-to-neutral voltage ar the substation as a reference, construct a phasor diagram
that depicts the relationship between VAN and
Ve before the capacitors are added.
d) Assume a positive phasesequenceand construct
a phasor diagram that depicts fhe relationship
between Vas and Vu6.
1L53 Refer to the Practical Perspectiveexample.Assume
mamrrthe fteouenc! of the utilitYis 60 H2.
a) What is tle /-!F mting of each capacitoi if the
capacitors are delta-connected?
b) What is the pF rating of each capacitor if the
capacito$ are wye-connected?
11.54 In the Pmctical Perspective example,what happens
volldge level al lhe generalingplant if lhe
-rg11a;
"_"'"" to the
ar Il8 kV the subslation
subslalion
is maintained
and the added capacitor
to
zero,
load is reduced
connected?
bank remains
I
I
I
I
6900
L 12Ev
6900
&t2c v
11.51 The two-wattmeter method is used to measure
the power delivered to the unbalanced load in
Problem 11.21.The current coil of wattmeler I rs
placed in line aA and that of wattmeter 2 is
placed in line bB
a) Calculate the reading of wattmeter 1
b) Calculatethe readingof wattmeter2.
11.55 In the Practical Perspective example, calculate the
kw beforeand after tbe c?pacitors
;fffl#l toral Uneloss_in
are connecteclto tle suDstallonDUs
1L56 Assume the load on tle substation bus in t]le
mamdtPracticalPersDecli\eexampledrops lo 240kW and
""""*'
o00tugn.Liting kvAR Al;o assu;e lbe capacilofs
remain connected to the substation
a) What is the magnitudeof the line-to-line voltage at the generating plant that is required to
maintain a line+o-line voltage of 13 8 kV at the
substation?
b) Will this power plant voltage level cause problems for other customem?
Prublems
465
11.57 Assumein Problem11.56ttrat when the load drops
,ll|ffSt! to 240 kW and 600 magnetizilg kVAR the capacitor bank at tlle substation is discomected. Also
assume tlat t]le line-toline voltage at the substation is maintained at 13.8kV
a) What is the magnitude of the line+oline voltage
at the generating plart?
b) Is the voltage Ievel found in (a) withh the
acceptablerange of variation?
c) What is the total line loss in kW when the capacitors stay on line after the load drops to
240 + j600kVA?
What is tle total iine loss in kW when the capacitofi are removed after the Ioad drops to
240+ i600kvA?
e) Based on your calculatiom, would you recommend disconnectingthe capacitors after the load
drops to 240 + j600 kVA? Explain.
Introduction
to the
Transform
LapLace
12.1Definitionofth€ L.plaeeTranstotilp.467
12.2TheSteprunctionp. 468
p. 420
12.3TheImputseFunctjon
p. 474
12.4Fundionat
Transforms
p. 475
12.5opehtionaL
Transforms
p- 481
12.6Apptying
th€ L.placeTransfolr't
p. 482
12.7Inverse
Transforms
12.8?olesandzercsof rlst p.494
p. 495
12.9InitiaL-andFinal-Vatue
Theorems
transformofa
1 Beableto catcutate
th€ Laptace
functjofLrsing
the defifjtioi of Laptace
tmnsfom,the Laptace
tlanslomtabL€,and/ora
tabteof opentionat tB nsforms2 B€abteto calcuLate
the inve6e LapLace
transformusiig pattialfiactionexpansioiafd
the Laptace
transform
tabLe.
3 Underctard
andknowhowto usetheinitiaL
vatuetheorcm
aid thefinalvaLue
theorcm-
We now introduce a powerful analytical technique that is
widely used to study the behavior of lineal, lumped-paramctcr
circuits.The methodis basedon the Laplacetransform,wlrichwe
define mathcmaticallyin Section12.1.Betbre doing so,we need
to expiainwhy another analyticaltechniqueis needed.Fimt. we
wish to considcrthc transientbehaviorof circuitswhoseclescribing equationsconsistofmore than a singlcnode voltaEleor meshcurrentdillerelltial equation. In other words,we want io considcr
multiplc node and multiple-meshcircuits thal are describedby
selsoI linear differentialequations.
Second.rve \vish to determinethe transientresponseof cir
cuits whose signalsourccsvary in ways more complicatedtha11
the sinple dc level jumps consideredin Chaptcrs7 and 8. Third,
wc canusc thc Laplacetranslbrmto introducethe colcept ol the
transfer function as a tool for analyzingthe steady-statesinusoidalresponseoI a circult when the frequencyof the sinusoidal
sourccis varicd.Wc discussthe transferfunclion in Chapter 13.
Finally,we wish to relate,in a svstematicfashion,thctime domain
behaviorof a circuit lo its ftequency-domainbehavior-.
Using the
Laplacetransformwjllprovide a broadcrunderstandingolcircuit
functions.
In this chapter,we introduce the Laplace tlanslorrn.discuss
ils pertinent characteristics,
and presenta systcmaticmethod lbr
transformingfrom the liequencydomainto the tirne domain.
12.1 D€finjtion
TGnsform 467
ofthe Laplace
12.1 Detinitionofthe Laptaee
Transform
The Laplace lmnslbrm of a lUnction is given by the expression
s\f(t)]
-I
f(t)e-"'dt,
(12.1) { Laptacetransform
wherethesymbolg{f(/)} is read"th€ Laplacetransformof/(r)."
TheLaplacetransfbrmof/0) is alsodenotedF(r): thatis,
F(0: s{/o}.
\12.2)
This notationemphasizes
that when the integralin Eq. 12.1hasbeeneval
uated,the resultingexpressionis a functionofs.In our applications,
t rep
resents the time domain, and, becausethe exponent of e in the integral of
Eq. 12.1must be dimensionless,
s must have the dimensionof reciprocal
lime, or ftequen+ The Inplace transform transforms tlre problem from
lhe time domain to the frequency domain. After obtaining the frequencydomain expressioflfor the unkrrowl we inverse-transform
it back to the
time domair,
II the idea behind the Laplace transform seemsforeign, consider
anotherfamiliar mathematicaltransfom. Logaritbmsare usedto change
a multiplication or division problem, such as A : BC, into a simpler
addition or subtraction problem: log A = log BC = logB + log C.
Antilogs are usedto caffy out the inverseprocess.Thephasoris another
transform;aswe knowftom Chapter9, it convertsa sinusoidalsignalinto
a complex number for easier,algebraiccomputation of circuit values.
After determinirg the phasorvalueof a signal,wetransformitback to its
time-domainexpression.Both of theseexamplespoint out the essential
feature of mathematicaltmnsforms:They are designedto create a new
domain!o make the mathematicalmanipulationseasier.After finding the
unknownin the new domain,weinversetransformit back to the original
domain.In circuit analysis,we use the Laplacetmnsfoim to transforma
setol integrodifferenlialequationsfrom the time domain 10a setof alge
braic equationsin the ftequencydomain.W()thereforesimpiify the solution for an unknom quantity to the manipulationof a set of algcbraic
equations.
Beforewe illustratesomeof the impo ant propertiesof the Laplacc
transform,somegenemlcommentsare in order,First,note thal the irte
gral in Eq.12.1isimproperbecausethe upper limit is iniinite.Thuswe are
confronled inrmediately with the question of whether the integral converges. In other words, does a given /0) have a Laplace transform?
Obviously,the lunclions of primary interestin engineednganalysishave
468
Transfom
to th€Laptace
Intoduction
Laplace transforms; otlena'rse we would not be intelested in the tfansform. In Linear circuit analysis,we excite circuits with sources that have
Laplace tiansforms. Excitation functions such as I or ?', which do not
have Laplace transforms, are of no interest here.
Sec;nd,becausethe lower limit on the integral is zero' the Laplace
lransform ignores /(r) for negative values of t Put another way, F(s) is
aleterminedby the behavior of /(t only fdr positive values of t To emphasizethat the iower limit is zero,Eq 121 is frequentlyreferred to as the
otr€-sialed,or udlatenl, Laplace transform.In the two-sided' or bilateral'
Laplace transform, tle lower limit is co We do not use the bilateral
folm herel hence F(.r) is understood to be the one-sided transform
Another point regarding the lower limit concerns the situation when
origin as,
/(t) has a discontiouity at the odgin lf /(D is continuous at the
ioi example,in Fig.12.1(a) /(0) is not ambiguous.However'iff(r) hasa
finite discontinuity at the origin-as, for example, in Fig 12.1(b)-the
question adses as to whether the Laplace transform integral should
i;clude or exclude the discontiouity ln other words, should we make the
lower limit 0- and include ttre discontinuitv, or should we exclude the discontinuity by making the lower limit 0+? (We use th€ notation 0- and 0- to
denote values of I just to the left and right of the origin, respectively.)
to
For reasoDs
we may chooseeither aslong aswe are consistent,
Actually,
(b)
(a)
be eiDlainedlater.we choose0.' asthe lower limil.
function
anddisconlinuous
Figuru12.1A A continuous
Becausewe are using 0- as the lower limit, we note immediately that
atthe orisin.
attheoriqin.(a)/(r) k continuous
the integation from 0- to 0+ is zero The only exception is when the dis_
(b)/(r) k discontinuous
aithe oriqin.
continuity at the origin is an impulse function, a situation we discussin
Section 12.3.The important point now is that tlte two Iunctions shown in
Fig. 12.1have the sam€unilateralLaplacetransformbecausethere is no
impulse function ai the origin.
The one-sided Laplace transform ignor€s /(t) foi t < 0 what hap_
pens pdoi to 0- is accounted for by the initial conditions Thus we use the
Laplace tiansform to predict the rcsponseto a disturbancethat occurs
after initial conditions have been established.
In the aliscussionthat follows, we divide fhe Laplace transfoms into
two tlpes: functional transforms and operational transforms.A funcdonal
trsnsform is the Laplace transfom of a specific function, such as sin ol. I'
t", and so on.An op€rationaltransformdefin€sa genelalmathematical
property of the Laplace ffansform, such as finding the tiansfom of the
derivative of /(t). Before considering functional and operational tiansfoms, however, we need to inftoduce the step and impulse functions
72.2
'lhe StepFunction
we mayencounter
lunclionslhal havea discontinuil)or iump ar lhe origin. For example,we know froD earlier discussionsof transient behavior
that switching operations create abrupl changesin currents and voltages
We accommodatethese discontinuities mathematically by introducing the
steDand imDulse functions.
12.2 Th€StepFunction 469
Figuie 12.2illustrutes the step function. It is zero for . The symbol for
the step function is fr(4. Thus, the mathematical definition of the step
funotion is
ru0)=0, t<0,
K u ( t ) =K , t > 0 .
(123)
rigure
rz.z.l trrestep
runctton.
If1(is 1,the functiondelinedby Eq.12.3is the unil step.
The stepfunctionis rot defined at I = 0. In situations wherc we need to
define the rransition between0 and 0+.we assumethat it is linear and that
ru(0) - 0.s1(.
(r?.4)
As befoie, 0 and 0r represent symmetric points arbitrarily close to the
left and dght of tle origin. Figure 12.3illustrates t}le linear traNition lrom
0- to 0'.
A discontinuity may occLrrat some time other than I : 0; for example, in sequentialswitching.A step that occuis at 1 = a is expiessedas
(r.,(t d).Thus
Ku(t-a):0,
t<a,
Figure12.34, Thetjnearapprcximaiion
to thesiep
Ku(t-a)-K,
t>a.
\12.5)
If 4 > 0, the step occurs to the ght of the origin, and if a < 0, the step
occun to the left of the origin.Figure 12.4illustratesEq. 12.5.Note that
the step function is 0 when the argumentI - /I is negative,and it is K
when the argumentis positive.
A step function equal to Kfor t < d is written as Ku(d - r). Thus
Ku(a
t):K.
t<a,
Figure12.4a A sreo'un iol o(L.dngat I - d
Ku(a-t)-0,
t>a.
The discontinuity is to t})e left of the origin when a < 0. Equarion 12.6is
shownin Fi9.12.5.
OIIe application of the step function is to use it to wdte t]le mathe
matical expressionfor a function that is nonzero for a finite duration but is
defined for all positive time. One example useful in circuit analysis is a
finite-width pulse, which we can create by addirg two step furctions_ Th€
tunction -K[,i(l 1) ,,0 - 3)] has rhe value r for 1< I < 3 and rbe
value 0 everfwherc else,so it is a finite-width pulse of height fa itritiated at
I = 1 and teminated at 1 = 3. In defining this pulse using step functions,
it is helpful to think of the step tunction !.(l - 1) as ,'rurning on,'the constafltvalueK at I = 1, and the stepfunction -&(t
3) as,,tuning off,the
constant value l. at t : 3. We use step functions to tum olr and turn off
linearfunctionsat des ed timesin Example12.1.
Figue12,5A A stepfunctronKu\d - t) fot a > 0.
470
lranstorm
Introduction
to theLaplace
to Represent
a Fundionof FiniteDuration
UsingStepFunctions
Use step functions to write an expression for the
functionillustratedh Fig.12.6.
+ 2 r 8 , o n a t t = 3 , o f f a t t - 4 - Thesestraight
line segments and their equations
Fig.12.7.Theexpressionfor /(l) is
f(t) :2tlu(t)
&(
- u(t - 1)l + ( 2t + 4)lu(t- L)
3)l + (2t - 8)["(r - 3) u(t 4\1.
forExampte
12.1.
tigure12.5A Thetunctjon
Solution
The function shown in Fig. 12.6is made up of Linear
segmentswith brcak points at 0, 1, 3, and 4 s.To con
stiuct this function, we must add and subtract lhear
functions of the proper slope.We use the step function to initiate and terminate these linear segments
at ttre proper times.In other words,we usethe step
function to tum on and turn off a straight line with
the following equations:+2t, on at t = 0, off at
t = 1; -Zt + 4, on at t - 1, off at t =3t and
tuned
Figure12,7 a Definiiion
ofthethreeljnesegmentr
formthetunctjonshou/n
0nandoff withstepfunctionsto
in Fig,12.6.
NOTE: Assessyour unde/standing of stepfrnctians by trJinq Chapter Prcbkns 12.1 and 12.2.
12.3 TheImpulseFunction
When we have a finite discontinuity in a function, such as that illustrated
in Fig.12.1(b),the derivativeof the funclion is not defined at the point of
the discontinuity. The concept of an impulse ftrnctionl enablesus to define
the dedvative at a discontinuity, and thus to define the Laplace tiansform
of tlat derivative. An imp se is a signal of infinite amplitude and zero
duiation. Such signals don't exist in nature, but some circuit signals come
very close to approximating tlis definifion, so we find a mathematical
model of an impulse useful. Impulsive voltages and currents occul in cir_
cuit analysiseither becauseof a switchitrg operation or becauset]le circuit
is excited by an impulsive source. We will analyze these situations in
Chapter 13,but here we focus on defining the impulse function generaly
i The inpulse function is aho known as the Dirac delta iunctio..
12.3 TheImpulse
Function 471
To define the derivative of a function at a discontinuity, we first
assumethat the funcfior varies linearly aqoss ttre discortinuity, as shown
in Fig. 12.8,where we observe that as €+0, an abrupt discontituity
occurs at the origin. Wher we dilferentiate the function, the derivative
between € and +€ is constant at a value of 1/2€. For I > €, the derivative is aa'(' (). Figure 12.9 shows these observations graphically. As €
approaches
zero,the valueof /'(t) betweenf€ approaches
infhity. At the
same time, the dumtion of this large value is apprcaching zero.
Furthermore,the areaunder/'(t) between+€ remainsconstantas € +0.
In this example, the area is unity. As € approacheszero, we say that the
funcrion betweenre approachesa unit impulsefunclion. denored5().
Thus the derivative of/(/) at the origin approachesa unit impulse function as € approacheszerc, o{
/'(0) -6G)
:r + 0.5
Figure12,84 A magnified
vjewofthe disco'rtjnujty
in
-€
Fis.12.1(b),
assuftring
a lineartransjtjon
between
f'(! )
as€ +0.
If the area under the impulse function curve is other thar unity, the
impulse tunction is denoted K6(t), where ( is the area.1<is often referred
lo as lhe shengthof rbe impulsefunction.
To sunrmarize,an impulse function is createdfrom a variable-parameter
function whose parameter approacheszero The vadable-parameter function must exlibit the following three charactedslics as the parameter
1. The amplitude approacheshfinity.
2. The dumtion of the function approacheszerc.
3. The area under the variable-parameter function is comtant as the
parameter changes,
Figure
12.9t' Th€dedvativ€
ofth€tunctjon
shown
jn Fig.12.8.
Many differelt variable-parameter functions have the aforementioned
characteristics.
In Fig. 12.8,we useda linear tunction /(t) = 0.5r/€ + 0.5.
Another example of a variable pammeter function is the exponential
function:
/(,)
: Ln'r,
2F-
\12.7)
As € apprcacheszero, the function becomesinJinite at the origin and at the
same time decays to zero in an inJiniresimal le gth of time. Figue 12.10
illushates the character of /(,) as € + 0. To show that an impulse function
is created as € + 0, we must also show thal the area under the function is
indeDendentof €. Thus
K
t'ea= | !a''ar+ I !"-''a,
z.
Ja.e
.tn
=' ' G l ' ' e-t/' l-* 1"1,
K
-
e,/.lo
K
K
2+
Z=
K
K.
tigure12,10A A vadabLe-panmeter
tunction
used
to
generat€
animputse
tunctjon.
(12.8)
which tells us that t]Ie area under the cuve is constant and equal to tr units.
Therefore,as € + 0, /(1) -16(r).
472
Iniroduction
to theLaptace
lransform
Malhematically, the impulse futrctiotr is defined
K6(t)dt : K;
11,2.e)
60) : 0, r + 0.
f(,
K6(/
a)
(12.10)
Equation 12.9states tlat the area under the impulse function is constant.
This area represents the strength of the impulse. Equation 12.10 states
that the impulse is zerc everywhere except at t : 0. An impulse that
occnrs at r = l' is denoted K6(t - a).
The graphic slmbol for the impulse function is an alrow The strength
of the impulse is given parenttreticaly next to t]le head of the arrow
Figure12.11showsthe impulsesf6(t) and (6(t - d).
An important prcperty of the impulse function is the sifting propefty,
which is expressedas
f(t)6(t-a)dt-f(a),
tigure12.11i A qraphic
representation
oftheimpube
16(,) andr5(t
a).
112.11)
where the function /(t) is assumedto be continuous at t : d; that is, at the
location of the impulse.Equation 12.11showsthat the impulsefunction
sifts out everything except the value of /(t) at I - a. The validity of
Eq- 12.11follows from noting that 60
") is zero everywherc except at
r = a, and hence the integral can be written
I:
/I f(t)8(t a)dt:
Ja
I
.l "-,
f(t)6(t
- a)dt.
(12.1?)
But becausef(t) is continuousat d, it takeson the value/(a) as t *.r, so
I :
td+.
ta+€
f(a)6(t a)dt= f(a).1.
.1".
,6(t
: f(a).
f (.t)
atdt
We use the sifting prope y oI the impulsefunction to lind its Laplace
transfornl
= / u1,;"a,= [" a|)a,=t.
r1u1,;1
(b)
Figurc12.12l Thefint dedvative
of theimpulse
function.
functionusedto
G) Theimpuke-genenting
(b)Thefilst
definethefirstdenvative
of ihe impulse.
dedvatjve
0f theimpulse-generatjng
function
that
apprcaches
6'(t) as€+0.
(12.13)
112.14)
whichis animportantLaplacetmnsformpair that we makegooduseof in
circuit analysis.
We can also define the derivativesof the impulsefutrction and the
Laplace transform of these derivatives.we discussthe first derivative,
along with its transform and then state the resulr for the higher-order
derivatives,
Thefunctionillustratedin Fig.12.12(a)generatesan impulsefunction
as € + 0. Figue 12.12(b)showsthe dedvativeof tlis impulse-generating
function,whichis defitredasthe dedvativeof the impulse[6'0)] as€ - 0.
12.3 TheImputse
Function
The derivative of the impulse function sometimes is referred to as a
momenttunction,or unit doublet.
To find the Laplace traNform of 6'(r), we simply appty rhe defining
integral to the function shown in Fig. 12.12(b)and, after inregmting,ter
€ + 0. Then
: s[f. .],,", [(-i)*,"]
.r,,r,rr
1im
s-?'. + ,t-e '_
2s
(12.15)
In derivingEq. 12.15,we had to usel'H6pital's rule twice to evaluatethe
indeterminateform 0/0.
Higher-orderderivativesmay be generatedin a manner similar to
that used to generatethe first derivative (see Problem 12.11),and the
defining integral may then be used to find its I-aplace transfom. For the
,th derivativeof the impulsefunction,we find that its Laplacetransform
simplyis s'; that is,
rr{6('\t} : s'.
112.16)
Finally,an impulsefunction canbe thoughtof as a derivativeof a sfep
function;that is,
ut,t= o"!t)
0
\12.!7)
f'lt)
Figure12.13presentsthe graphicinterpretarionofEq.12.17.Thefuncrion
shownin Fig.12.13(a)approachesa unit stepfunctionas € +0. The furction shown in Fig. 12.13(b)-the dedvarive of the Iunction ir
Fig.12.13(a)-approachesa unit impulseas € +0.
The impulse function is an extremely useful concept in circuit analysis,and we say more about it in tlle following chapte$. We introduced t]te
concepthereso that we car includediscontinuitiesat the odgin in our definition of the Laplace transform.
NOTE: Asse.Js)Joul undentanding of the impulse function by nying
ChapterPrcblems12.5 12.7.
ftgsr€12.13A Thejmpulse
functjon
asthedenvabve
(a)/O J r(r) as€-0, and
0rth€stepfunction:
(b)/'(r)+ 5(r)as€+0
474
Intmduction
totheLaptace
T€nstornr
12.4 FunctionaI
Transforms
A furctional transform is simply the Laplace lransfom of a speciJiedfunction of t. Becausewe are limiting our intoduction 10 the unilateral, or onesided,Laplace tansfom, we defire all functions to be zero for t < 0 .
We derived one lunctional transform pair in Section 12.3,wheie
we showed that the Laplace transform of the udt impulse function
equals1;seeEq. 12.14.A secondillustration is the unit step function of
Fig. 12.13(a),where
t 1 , q 4 1= [ y 6 n . o , = [ - , " " 0 ,
:;".:;
(12.18)
1
Equation 12.18shows that tle Laplace transform of the unit step function
is 1/s.
The Laplace tmnsform of the decaying exponential function shown in
Fig.12.14is
tl""'l -
Pigure12.14^ A decaying
e{ponentiat
/
-
ede4dt=
Jo
Joe''"'ar-,
l
o.
o?.re)
In derivingEqs.12.18and 12.19,we usedthe fact that integrationacross
the discontinuity at the origitr is zero.
A tlird illustratiotr of findhg a functional transfom is the sinusoidal
function shownin Fig. 12.15.The expressionfor l(l) for t > 0- is sin dl;
hen€ethe I-aplace transform is
e{sind1}:
l"
ut.un"' o,
=
[("'';"'^)-'.
Fiquer2.r5 A A sinEoidat
tunction
fo' I > 0.
I
t l
= - t -
dt
2j
t
-
r \
" + lr)
112.20)
Table 12.1 gives an abbreviatedlist of Laplace transform pairs. It
includes the functions of most interest in an introductory coune on circuit
aDDlications,
12.5 0peBtionat
Tnnsfoms 475
Tlre
,f(, c> o-)
6(r)
(imp'ilse)
u(t)
rG)
1
1
1
;'
(ramp)
I
(exponenlial)
(cosine)
1
(danped ranp)
(damped sine)
G+;Fia
12.5 0perationalTransforms
Operational transforms indicate how mathematical operations oerformed
oDeirher/(ri or F(,) areconvenedinto lheopposiridomajn.iteoperations of pdmary interest are (1) multiplication by a constantt (2) addirion
(subhaction); (3) differentiatio{ (4) integmrion; (5) rranslarion in the time
domain; (6) translation in the hequency domain;and (?) scalechanging.
Muttiptication
by a Constant
Fiom the defining integral, if
e{f(t)J = F(s),
s tKf (t)j : KF(r.
(12.21)
476
Introduction
tothe Laplace
T6nsfom
Thus,multiplicationof /0) by a constantcorespondsto multiplying F(s)
by thesameconstant,
Addition(Subtraction)
Addition (subtraction)in the time domair rranslatesinto addition (subtraction)in the ftequencydomain.Thusif
s{l(r} : r(r),
s{/r(r)} = rr(r),
s{/3(r)}= r3(t,
then
g{ft(t)
+ f2O - /:(4}
= r(")
+ Fls) - F3(s),
(12.2?)
which is dedved by simply substituting the algebraic sum of time-domain
functions into the defining integral.
Differentiation
Differentiation in the lime domain conesponds to multiplying F(r) by r
and ttren subtracting t}le initial value of f(t-that
is, /(0 )-fton
this
prcduct:
' { + } = r F ( s ) - r { o) .
112.23)
which is obtahed directly from the definition of the Laplace transfom, or
'{+\fl+)"",,
lrz.24)
We evaluate the integral h. Eq. 12.24 by integrating by parts. Letting
u = e ", ajnddt = ld f (t)ldtldt yletds
.\+\ : n'',rl*I*',u,*''0,
\12.25)
Because
we areassuming
that /(t) is Laplacetransformable,
the evalua
tion of e-Y0) at I : oois zero.Thereforethe right-handsideof Eq. 12.25
f((D + sr
rr(r) /(0-).
ftr)e'-.crt:
12.5 operationat
Transtorms477
This observation completes the derivation of Eq. 12.23.It is an imporant
result becauseit states that differentiation in the time domain teduces to
an algehaic operation in the s domain.
We determine the Laplace tmnsform of higher-order dedvatives by
using Eq. 12.23 as the starting point. For example, to find rhe Laplace
transtbm of the second dedvative of /(t), we first let
. .
df(t)
112.26)
dt
NowweuseEq.12.23to write
G(r) = rF(s)
/(0 ).
\12.27)
But because
ds(t) _ d2f(t\
'
dt
dtz
"{ot:.")
- "{t'!!)
l d t )
l d f
)
-scrr) - B(0).
(1?.28)
Combhing Eqs.12.26,12.27,and12.28gi'tes
I atrr,tl
.t O I
ri=rj'i - "'rr',- ,tro-t "'),'.
1tz.zs:
We find the Laplace ffansfom of the rth derivative by successively
applying the preceding process,which leads to the general result
* l o ' t ' ! \ = s , F r , ) r , , / { o1 - , ^ z l l t o t
I d," I
d!
J-,-3af9-)
,--
d" lfg )
'---:-,
Lrz,ru)
Integration
Integration in the time domain correspondsto dividing by r ir the s domain.
As before,we establishthe relationship by the defining integral:
'\lro^| =Ill[to,,,1,"""
(12.31)
478
Intrcductjon
t0 theLaplace
Transform
We evaluatethe integralon the right-handsideofEq.12.31 by integrating
by partq first letting
Then
: f(t)dt,
The htegratior-by-padsfomulayields
-;f'r,r,,l;.
.{[:o,*1=
f 1,,r0,,,,,
The first telm on the right'hand side of Eq. 12.32is zero at bot}l the upper
and lower limits. The evaluation at t]le lower limit obviously is zero,
whereasthe evaluation at the upper limit is zero becausewe are assuming
that /(l) has a Laplace tiansfolm. The secondierm on the dght-hand side
of Eq. 12.32is F(s)/s; t}leiefore
'\lrta*j:?,
(12.33)
which rcveals that the operation of inte$ation in the time domain is transfomed to the algebraic operation of multiplyiDg by 1/r in the r domain.
Equation 12.33and Eq. 12.30form the basis of the earlier statement that
ttre Laplace transform translates a set of integrodiffeiential equations into
a set of algebraic equations.
Translation
in the TimeDomain
IJ we start witll any function /(t),l(t), we can represetrt the samefunction,
tra0slated
in time by rhecon<lanla. as J\t - a)u\! - /l).- Tran.larionin
the time domain coresponds to multiplication by an exporertial in the
ftequencvdomain.Thus
:t{f(t
o)u(t a)}:e'"F(s), a>0.
(12.34)
For example,knowingthat
s{ta(t)}
1
, Notethat throushoulwemultiply anyarbirra.yfnDcrion by the unii steplmctjon ,G)
l(r)
to ensurethat the resultinsfunctionis definedio. all Dositivetine.
12.5 0peratjonat
lransfolms 479
Eq. 12.34 permits writing the Laplace tmnsform of (t
directly:
a)u(t - a)
9{(t - a)u(t a)}=t".
Tbeproofof Eq.12.34
follows
fromthedetining
integrai:
].*
I u ( t- a ) f ( t a ) e - " t d t
sIG-a\u(t-a)\:
.ln'
=
, o)n"a,.
l-
(12.35)
In writing Eq. 12.35,we took advantage of r(t
d) = 1 for r > /i. Now
we change t]le va able of integration. Specifically,we let r = r - lr. Then
r : 0 when r = a, .r : co when I = co, and !r.r - dt. Thus we write the
integralin Eq.12.35as
a)u(t- 4j =
sIf(t
|
ftOe {*o a,
="-"'J*rav*a'
: e_"F(s)'
which is what we set out to prove.
TEnslationin the Frequency
Domain
Tlanslatiotr in tlle frequency domain coresponds to multiplication by an
exponential in the time domain:
9[e-"'f(t)]
= F(s + a),
(1236)
which follows from the defining integral. The derivation of Eq. 12.36is left
to Problem12.16.
We may use the relationship in Eq. 12.36 to dedve new transfom
pai$. Thus,knowing that
i/ I cosorl
we useEq.12.36to deducethat
j/t?
-
co(@l}
\ ' a
(t + a)2 + e|'
Scate
Changing
The scale-changeproperty gives the relationship between /(t) and F(r)
when the time variable is multiplied by a positive constant:
sua,*:in(i),,, o,
(12.37)
480
Ttnsform
to theLaplace
Introduction
property
the derivationof whichis left to Problem12.22.Thescale-change
where
time-scale
work,
especially
in
experimental
particularly
useful
is
are madeto facilitatebuilding a modelof a system.
chaDges
use
We Eq.12.37to formulatenewtransformpails Thus,knowingthat
g{coit r} :
l+1'
we deducefrom Eq. 12.37that
sla
a(sl(,)z+L f +&]'
1
E{cosot} :
Table12.2givesan abbreviatedlist of operationaltransforms
NOTE: Assessyour unilentanding of theseoperutionaltransformsby
trying ChaptetPrcbkns 12.24and 12.25.
Operarior
J\r'l
r(s)
Multiplication by a constant
KT(I)
(r(r)
Additioi,/subtraction
l(, + l,(r) - ,fd4+ ...
F(r)+F,(s)-F(s)+'
first derivative (tine)
df(t)
dt
srG) /(o-)
a'fG)
df(r)
lrG) - ,,f(o ) -
second de vative (time)
nth derivative (tilne)
d'f(.t,
"'FG)
s"flf\
"
Tme integal
lf(rtdx
- df,tO )
tu2
F(r)
f(t - c)a(t
F(r + a)
Translation in hequency
Scile €hanging
Ft r derivative (r)
tf(.t,
"th deiivative (s)
{f(t\
r integral
ro
aF6)
, ,,.{!!)
t,-,,,..
^df(t\
{ |i
d' \ftD-l
d{-l
1 2 . 6 A p py i n qt h eL a pa c eT , a t r f o , m 481
objective1-ge abl.eto catcutate
the Laptace
transform
of a functionusingthe Laptace
transform
tableor a tableof
oPerational
transforms
12.2 Use Lheappropriateoperationaltransfbrm
fiom Table 12.2to find the Laplacetransfbm
of eachfunction:
2
Answer: (u) --;
+
ts aJ(b)
a:) Pe-'t
b) :(r
dl
Bs
(s + d)'
'' sinh6rJ.
(c)
Bz'
(r' + o')'
NOTE: Al\o trJ,ChaptetPrcbtems1L 17 and 11.18.
12.6 Ag:p[yimg
tl"l* l-eptaeefraslsforn']
We now illustraleholv to usethe Laplacetransformto solvethe ordinary
integrodiffercnlial equations thal descr.ibethe bchaviol of tumped
parametcrcircuits.Considerthe circuit shorJnin Fig. 12.16.We assunrc
that no initial eneryyis storedin the cfcuit at the instanl\r,henthe switch,
which is shor'lirgthe dc current soulce.is opencd.Theproblem is ro find
tiguE12.16r A paratletfrao
ort.
the timc domainexpressionfor ,(/) when I > 0.
We bcgin by writing thc integr-odifferenrial
cqualior1that ,(/) rnusl
satisf,v.
We need only a singlc node-volrageequationto describethc cir.
cuit- Summingthe currentsawaylrom the top node iD the circuit gcner
+.! l,'^,to'
+(
- = I ^ . u- -( t ) .
dr
(12.38)
Nole that in writing Eq. 12.38.we indicaredthe openingof the switctrin
the stepjump ofthe sourcecurrenrfrom zero to 1d..
After dcriving the integrodiffcrenlialequations(in this example.jusr
one),\\'e transfbrmthe equationsto the r domain.Wc lvill Dot go through
the stepsofthc transformationjn dclail,becausein Chapter13we will dis
coverhow to bvpassthem and gcneratethe s-domainequationsdirecdy
Briefly though.we use three opcralioDaltransformsand one functional
tralrsformon Eq. 12.38to obtain
?.;+
-,rr1: r.(i),
+c['r(r)
(12.39)
an algebraicequation in wlich y(s) is thc unknown variabtc.We are
assunring
lhat the circujtpararnererR.l-.and C,aswell asrhe sourcecurrent 1dc,are known; the inilial volrage on thc capacitor (0 ) is zero
becausethc iDitial enere)' sturcd in the cjrcuit is zero-Thus we ha\,e
reducedthe problemto solvingan algebraicequation.
482
Tmnsform
Introduction
to the Laptace
Next we solvethe algebraicequations(again,justone jn this case)for
the unknowns.SolvingEq.12.39for y(s) gives
ur,r(|.-!.,.) =*,
Ia./C
C+(llRc)s+(llLC)
\12.44)
To find r(t) we must inverseftansform the expressionfor v(s). We
denotethis invene operation
., (t): s1|v(s)\.
\12.41)
The next steD in the analvsisis to find the inverse tmnsform oI th€
r-alomainexpiession;thisis the subjectof Section12 7.In that sectionwe
also present a final, critical step: checkiDg the validity of the resulting
The need for suchcheckingis not unique to the
time-domainexpression.
Laplace transfom; conscientious and prudent engineers ahvays test any
dedved solution to be sure it makes sensein terms of known system
behavior.
Simplifying the notation now is advantageous.We do so by droppirg
the parentheticalt in time-domainexpressionsand the parenfheticalr in
frequency-domain expressions.We use lowercase letters for all timedomainvariables,and we representthe correspondings-domainvariables
wiih uDDercaseletlers. Thus
Y{u}:Y or u=rr{v!,
E{i}:1
or i::FU},
9{f) - F
ot f:
e'{F},
,Assessyour understanding of this matetial by trying Chapter
NO?t
Ploblem 12.26.
12.7 InverseTransforrns
The expressionfor I/(s) in Eq. 12.40is a ntional function ofr;that is,one
that can be expressedin the form of a ratio of two polynomials in s such
that no nonintegral powers of s appear in the polynomials ln fact, for linconslanllhe
circuitsqhosecomponenl\aluesare
ear.lumped-pdr0meler
voltaget
curren(t
are al$a)\
and
for lhe un-knoPn
r'domarnexpfession(
workmg
(You
by
verify
this
observation
may
rational functions of s.
of s,
mtiorlal
lunctions
Problems12.2712.30.)If we caninverse-tmnsform
12.7 Inverse
Innsforms 483
we can solve for the time-domain expressionsfor the voltages and currents.The purpose of this section is to present a straightforward and systematic technique for finding tlle inverse tansform of a rational {unction.
In general, we need to find the inverse transfom of a function that
has the folm
rc)=#
a n s n + a nf n r + . + a 1 s + s o
b ^ s ^ + b ^ - 4 f d n1 + . + 4 s + b o -
\12.4?)
The coefficients a and b are rcal constants,and the exponents m and n are
positive integers.fhe ratio N(s)/r(s) is called a proper rationd furctior
lf m > n, and an improp€r rational fir€tion if m < ,?. Only a proper
iational function can be exparded as a sum of partial ftactions. This
restriction posesro problem, as we show at the end ol this section.
PartiaIFractionExpansion:
ProperRationa[Functions
A prcper rational function is expanded hto a sum of partial fraclions by
writing a telm or a sedes of tems for each rco1 of D(s). Thus D(s) must
be in factored folm before we can make a partial fraction er?ansion. For
each disainctroot of ,(r), a shgle term appearsin the sum of partial fraction$ For each multiple root of D(r) of multiplicity r, the expansion contains r tems. For examole.in t}re rational function
s + 6
r(r+3)(s+1)' '
the denominatorhasfour rcots.Tho of theseroots are distinct-namely,
atr = 0ands - 3. A multiplercot of multiplicity2 occursat s = -1.
Thusthe pa ial fraction expansionof this functiotrtakesthe form
Kz
_
= Kr tl ,r . . l
s(r+3)(s+1),
Kr
K4
r '
l s r t l - '
1 1 2 1 3 )
The key to the partial ftaction technique for finding inveBe transforms
lies in recognizing the /(t) coresponding to each telm in the sum of partial fractions. Frcm Thble 12.1you should be able to ve fy that
s+6
.^ |
+
3Xr + 1)'
[s(s 3)(r
= 1Y, + Kre-3'+ K{e | + K4et)u(t).
(12.44)
All that remains is to establish a technique for detemhing tlle coefficients ((1, 1<2,-K3,. . .) genemted by making a pa ial fraction expallsiorlThere are four general forms this problem can take. Specifically,the roots
of D(s) arc either (1) real and distinct; (2) complex and distinct; (3) real
and repeated;or(4) complexand repeated.Beforewe considereachsituation in turn. a few seneml comments are in oidei,
484
Inhoduction
to ihe Laptace
Transform
We used the identity sign = in Eq. 12.43to emphasizethat expanding
a mtional function into a sum of partial ftactions establishesan identical
equation. Thus both sides of the equation must be tle same for all values
of the va able r. A1so,the identity relationship must hold when both sides
are subjected to t]le same mathematical operation. These chamcteristics
are pertinent to determining the coefficients, as we will see.
Be sure to vedfy that the rational function is proper. This check is
imporranr
because
nolhingi0 rbepfocedure
lor findi;g ihe \ driousKs will
ale you to nonsenseresults iJ the mtional function is imprcper. We presetrt a procedure for checking the fs, but you can avoid wasted effort by
forming the habit of asking youisell "Is F(s) a pioper rational function?,'
PartialFractionExpansion:
DistinctRea[Rootsof ,(s)
We first consider determining the coefficients in a partial fraction er?ansion when all the roots of ,(s) are real and distinct. To find a I< associated
with a term that arisesbecauseof a distinct root of D(s), we multiply both
sides of the identity by a factor equal to the denominator beneatl the
desired l(. Then when we evaluate botl sidesof the identity at the ioot corresponding to tne multiplying factor, the right-hand side is always rhe
desiredl(, and the left-hand side is alwaysits numerical value. For example,
-
a6(' - s)(s t 12)
r ( , - 8 X rl o ,
Kt
r
K)
cr8
K1
r-b
To find the valueof 1K1,
we multiply both sidesbys and then evaluateborh
sidesats=0:
a 6 ( r 5 ) ( r+ r 2 ) l
(,-8)ir or
I.o
'
,
K," I
K.sI
8 1 . , ' ^ 1 .o '
-96(5)412)
:=
K 1: 1 2 o .
112.46)
To find the valueof &, we mulriply borh sidesby s + 8 ard thenevaluare
borhsidesat s = -8:
96(s+sxr+12)
"(" + 6l
i"= s
:-
-(rts+ 8)l
+ s)l
- ^,, ' -. ft:ds
r "* 6 )l , = - " '
l , =.
"
96(-3)14)
(-8X-2)
= K-:
1)
\12.47)
Therl ri is
96(r+5)ts+i2)l
I
r(( + r<l
l,="
K-
48.
(12.48)
FromEq. 12.45andthe,(valuesobtained.
q 6 G+ s ) G+ t 2 )
120
48
s(s+il)(s+6)
\12.49)
r + 8
At this point, testingthe resulttoprotect againstcomputationalerrorsis a
good idea.As we aheadymentioned.a partial fraction expansioncrcates
an identily;thus both sidesof Eq. 12.49musl be the samefor ali r values.
Thc choiceoftest valuesis completclyopen;hencewe choosevaluesthal
are easyto verify.For example,in Eq. 12.49,testingat either 5 or 12 is
attractive becausein boih casesthe lelt hand side reduces to zero.
Choosing 5 yields
120
48
'72
=
2 4+ 4 8- 2 4 : 0 .
whereastesting 12 givcs
120
48
'72
,--10
il
tq
0.
Now confidentthat thc numericalvaluesofthe various(s are correct,we
proceedto find the invene lransform:
. f g -b.l-r + 5 "r -r + 1-2' l)|
- {l),r
/ '{
.
I r l r + E ) ( r + ( r )J
,lbp'
-2. "')ult).
|?..0)
0bjedive2-Be ableto catcutate
the inverseLaptace
transformusingpartialfractionexpansion
andthe Laplace
transform
tabte
12.3 Find/(r)if
12.a Find/0) if
6s2+2or+26
(r+1l(r+2Xs+3)
Answer: /(r) = (3e1 + 2e'''1+ e-3)u1).
NOTE: Aho ty ChltptetPrcblems12.10(a)
and(b).
7s' +63r+134
(s+lt(r+4)(s+5r'
Answer: /(r) : (4e 3t + 6e t - 3e st)u(t).
486
Introduction
totheLaptace
Tlansfom
PartiaIFractionExpansion:
DistinctComplex
Rootsof D(s)
The only difference between finding the coefficients associatedwith distirct complex roots and finding those associatedwith distinct real roots is
that the algebra in tle former involves complex numbers We illustrate by
expanding tie rational function:
_
"'"' -
t00(s | 3)
," * o1i ' o' + :1
( 1 25 1 )
We begin by noting that F(s) is a proper rational function. Next we must
find the roots of the quadratic telm s2 + 6r + 25:
f + 6r + 25 = (s + 3 - j4)(r +3 + ja).
Oz.5zl
With tlle denominator in factorod fom, we proceed as before:
100(s+ 3)
(r+6)(f+6s+25)
K
I
K
.
K
.
r-3
/4 r+3-/4
b
To find i:1, &, and i(3, we use the same processas beforel
'K- _
100{i| 3) |
- _
t'
fr + b)(,
J
''
t 1
2s
o s ' 2 s 1 ,..
l00rr- Jr
,,'
too{ 3r
t _ -
too1in)
I
t4)l. -r. ,.
{3
/4( 18.,
: 6 _ i8 = 10s 153.13,
02.55)
r00rr- Jr
|
( r + 6 X r 3 t 4 r l . -, r "
'-'
:6
t \ . . t \
100{i4)
13 j4\-j8l
+ j8 = 10erj:t3.13.
02.56.)
Then
-12
100(s.f 3)
(r+6Xl+6.t+25) s+6
r0/ 53.1s"
s+3- j4
10/s3.73.
12.7 InvereTnnstornrs487
Again, we need to make someobseNations.
First,in physicallyreaLizable circuits, complex roots always appear in corjugate pairs. Second,the
coefficients associatedwith these conjugate pairs are themselves coniugates.Note, for example,that i(3 (Eq. 12.56)is the conjugareof f2
(Eq. 12.55).Thusfor complexconjugateroots,you actuallyneedto calculate only half the coefficients.
Before inverse{mnsformingEq. 12.5?,we checkthe partial fraction
expansionnumerically.Testingat -3 is attractjvebecausethe leffhand
sidereducesto zero at this value:
f lr)--i
t 0/ - 5 J . 1 J .
t2
t0
- - - / 5J.l].
i,
= -4 + 2.5/36.a'7"+ 2.5/
36.87.
: -4 + 2.0 + j1.5 + 2.0 , j1.5= 0.
Wenowproceedto inversetransformEq.12.57:
J
.' i[
l o o r s +- ' 3 r
:-:-.'
I
| = r- t2r 6
l { r + 6 ) ( s ' +6 r + 2 s ) l
l'c 't r4rl
lge r1
+ 10?,,53.13.€_(3+14),
)r,(l).
(12.58)
ln geneial, having t]te function it t]le time domain contain imaginary components is urdesirable. Fo unately,becausethe terms involving imaginary
components always come in conjugate pairs, we can eliminate the imaginary componentssimplyby addingthe pairs:
10e t53.13
e (3 i4' + l\e js3.r1
a(1+i4)l
:10a1t(eit4t 5313')+ e-t({ '.rr"))
= 20€ircos(4r 53.13"),
\12.59)
which eMbles us to simplifyEq. 12.581
1 0 0 (+r 3 )
," ,1
I
+
6r
6)(sz
+
+
2s)J
[(r
: I 72eat+ 2].]tcos(ar
53.13')lu0).
112.6A)
Becausedistinct complex roots appearftequently in lumped-parameter
linear circuit analysis, we need to sulnmarize these results with a new
488
lntroduclion
to the Laphce
T€nsforn
transformpair. Whenever D(s) containsdistinct complexroots-that js.
jB)(r + d + ip)-a pair ot rermsof rhe form
factorsof lhe form (s + a
I<
s+a-jB
K"
s + d +j B
(12.61)
appearsin the partial Iractior expansion.where
the partial fractjoncoeffi
cientis,in gencral,a complexnumber.ln polar form.
K - lKleto= lI<11A,
(2.62)
where lK I dcnotesthe qlagnitudeof the complexcoefficient-Then
K':
Klett= KLA.
(12.63)
thc conplex conjugatepair in Eq.12.61atwaysinvcrsc Lransformsas
t.l
K
[r+a-jp
,
K'
\
s+d+jpj
=2Ke"tcos(fu+a).
\r2.64)
In applyingEq. i2.64it is importantto nole thai (is definedasrhe coefficiert associatcd
with the denominarorterms + a - i6. and K' is delined
.'slhe coerhcrenr
a..oiiaredsirh rhedcnomi||xlor,
o
iB.
obiective2-Be abteto catculate
the inveBeLaptace
transform
usingpartiatfractionexpansion
andth€ Laplace
transform
tabte
Ansrver:/(r) = (10e5/- 8.33e{r
sin12r"t/.
12.5 Find/0) it
10(r'+ 119)
(r+5)(r2+lur+1oo)
NOTE: AIto tly ChaptetPrcblems12.40(c)and (d).
PartiatFraction
Expansion:
Repeated
Rea[Rootsof ,(s)
To find the coefficientsassociated
with the terms generatedby a mukiplc
root of multiplicity /,wc mulliply both sidesof the identity by the multiplc
rool raisedto its /th power.Wefind the L appearingover the factor raised
to lhe fth power by evaluatingboth sidcso{ the identit}' ar rhe mulriple
roo1.To find the rcmaining(J' 1) coefiicients,we differentiateboth sides
of the identity (,' 1) times.Ai the end of cachdiffereDtiation.we evalu
ate bolh sidesof thc identity at the multiplc roo1.The right-handsidc is
12.7 In!€rselransfoms 489
always,thedesired,<i and the lef!:hao.dside,is:
alwayqirs ludreTipqtr
value.
For exam.ple,
100(s2
. t5 )
K,
K, ___L.
=Kt_
.
( s+ 5 ) r ( s+ 5 ) 2 r - 5 '
s{r-5)3
s
Welind Kt aspreviouslydescribed;that ii
=^
"'-ti#|"=;='.Y?"
(1?.66)
To find K2, we multiply both sidesby (s + 5)3 and then evaluateboth
sidesat -5i
loo(s+ 2s)|
_
+ -K,+ K3(s+ tl"= s
l
' t . '
+ Xi(s + 5),1
,
F2.67)
tPfl' = ) - - \
o, o x2 K3 o+ K4xo
-)t
{
(12.68)
, r:.:
.:., i,a :l'
I r . - r . J I . : : i r i i ; .. . t 1 1t . ' , 1 . ) l j , - .
To find tr3 we first mustiiriUliipli both sidesdfiEql 1255by (si+ 5)lr'Next
' !,o-,
-
,,., : , i :r ',*.#tri(j+,5Fi"=_,,
(12.69)
'*[-{i2],=-
= K3= -100.
,r . (12J0)
490 Intmduction
to the Laptace
Tnnsfonn
To find fa we fint multiply both sidesof Eq. 12.65by (r + 5)r. Next
we differertiate both sides twice with respect to:r and then evaluate both
sides at r = -5. After simplifying t]le first derivative, the second deriva-
,4[-3]",:..,*[!+ s),(2s
s2
+ o+
'1",
+
+ s)]"=,,
*lr3l"= s r4[zro{"
_40 = 2K+
(12.77)
SolvingEq.12.71for Ka gives
K4 = -20.
112.t2)
100(r+ 25) _ 20 _ 400
100 _ 20
(s+ 5)3 (r+5), r+5
s(s+s)3 s
112.13)
Then
At this point we can check our expansion by testing both sides of
Eq.12.73 at r = -25. Noting both sidesof Eq. 12.73equal zero when
r:
25 gives us confidencein the corectness of the partial fiaction
expansion.
The inversehansform of Eq. 12.73yields
+ 25)\
-! e, 1 1 0 0 ( s
l , 1 s + s 1Jr
-
"
120- 200r'e
'rlllvl.
'
luor? - 20r
\12.74)
12.7 InvereTnnsfoms491
PartialFractionExpansion:
Repeated
Complex
Rootsof ,(s)
We handle repeated complex roots in the sameway that we did repeated
real roots: the only difference is that t]le algebra involves complex numbels Recall tlat complex roots always appear ia conjugate pairs and that
the coeff;icients associatedvrith a conjugate pair are also conjugates,so
t]tat only haiJ tle trs need to be evaluated. For example,
' "'
768
112.15)
G t +* r z 5 7
After factoring the denominator pollnomial, we [aite
Kr
K2
(s+314)' -t+3-j4
(r+3l/4)' r+3-14
.
t12 76\
Now we need to evaluate only K1 and (2, because/<i and (i ate conjugate values.The value of 1<1is
..
,768
'
|
j4fl._",,4
tr+3
768
\12.77)
( i8f
The value of K2 is
R2
-*L
168
[ . s + 3 + j4)'
_ G2Q68)
r3+j4
l=.,*,.
rl,
2(768)
11?.78)
492
Introduction
to the Laptace
TGnsfor'n
FromEqs.12.77
and12.78,
Ki-
12,
(12.79)
Ki=i3=31y.
(12.80)
We now group the partial ftaction expansion by conjugate tems to obtain
-t
"^
-12
rG): [;
LTJ+3
- j4), + (s+ 3 + j 4 \ ' )
| 3 / -90'
\ { + 1
3 /c0'
i 4
\
s + l + a /
(12.81)
We now write the inverse transform of F(r):
3t
3'cos(4r- 90')]r(r).
f(t) = [-24te cos4t + 6?
(12.8?)
Note that if F(r) has a real root a of multiplicity r in its denominator,
the term in a partial fraction expansion is of the form
(r + d)'
The inve$e tmnsform of this term is
=ffi",
,,{.fo}
(12.83)
If F(r) has a complex root of a + iB of multiplicity r in its denominator,
the telm in pafiial ftaction expansion is the conjugate pair
(s+d- jBY (s+a + jB)'
The iDversetraNform of this pail is
:P1{
lG+'
"
K'
iB)' (r+d+jB)'
+
I
: f 2Kn,-," "''o'rn'* alla'1'
rx
[;
(12.84)
Equations 12.83and 12.84are the key to being able to inverse-tmnsform
any partial fraction expansiotr by inspection. One fu trer note regarding
t-hesetwo equations: In most circuit analysisproblems, r is seldom greater
tlan 2.Therefore. the inverse transform of a mtional function can be handled wit}l four transform pairs. Table 12.3lists these pairs.
12.7 Inverse
Transtorms493
TABLE12.3
Fo|JrUsefulTransfornPairs
Pair
Number
Natue of
Rools
T('
K
K
Kte rtu(t)
K
K
"
s+a jP s+d+ jB
K
K
'
(\+d
jpr (r+d+ jB).
2 (Ld ''cos(pt + t ),(r)
h KIe-'t.os(.tu+ a)u|)
N , 2 r I n p a i r s t a r d 2 , , < n a r c a l q u n t i t y , q h e r a s i n p a n s 3 a n d 4 . , < i s l n e c o n p l ,c<x Iq u a n t i t r
Obiective
2-Be abteto catcutate
the inverseLaptace
transform
usingpartia[fractionexpansion
andthe Laplace
transform
table
Answer:f(r) : (-20te 2tcost+ 20e"sintr0).
12;7 Findf0) it
-F 'r"c' r :
"
lr2+ 4t + 5r2'
NOTE: Also tty Chdpter Problem 12.41(e).
PartiaIFraction
Expansion:
Improper
RationaI
Functions
We concludethe discussionof partial fraclion expansionsby relurning ro
an observationmade at the beginning of this section, namely, that
improper rational functions pose no sedousproblem in finding inverse
tmnsforms.An improper rational function can alwaysbc expandedinto a
polyromial plus a proper rational function. The polynomiat is then
inversctransformedinto impulse functions and derivativesof impulse
functions.Theproper rational functionis invene-transformedby the tech
niques ouuined in this section.To iltustrate the procedure.we use thc
lunction
ta + 13r' + b6s2+ 20ur+ 300
s' +9s+20
(12.85)
Dividing the denominatorinto the numerator until the remainderis
proper rationalfunclion gives
F(r):r' +4s+10+
sherc lhe refmrJos
IUU'r,
30s+ 100
s ' + 9 s+ 2 0 '
- r)s | '0) r. lhe renrrindef.
(12.86)
494
Intbduction
to ihe Laptace
Tnnsfom
Next we er?and the proper rational function into a sum of partial
ftacriorls:
30s+ 100
f+9s+20
-20
30r + 100
=
(s+4Xr+5) -+-
50
-.
l\287\
Subsrituting
Eq.12.87
inroEq.12.86
yields
.
n
I r r )- s z 4 r - 1 0
5
r"'?
0
rr-"-
(12.88)
Nowwecaninverse-tmnstorm
Eq.12.88byinspection.
Hence
".. e5()
.d;t!)
t(tl--t
| 4-,
I106(r)
(20e 4t
st)est)u(t).
(12.89)
12.8 PolesandZerosof F(s)
The mtional function of Eq. 12.42also may be expressedas the ratio oI
two factored polynomials.In other wordsJwe may write F(s) as
F t s )-
K f r + . r ) ( r+ . 2 ) . l s + . , )
(s+pr)(r+pr.(s+p-)'
(12.90)
where 1{ is the constart a/6_. For example,we may also write rhe function
8s2+120r+400
2s4+ 20s3+ 70s,+ 1O0r+ 48
r("):
8(s' +15s+50)
2(r4+ 10s3+ 351 + 50r + 24)
4(s+s)(i+10)
(r + 1Xr + 2)(r + 3)(s+ 4)'
(12.91)
12.9 Initial-andFinalvalue
Theorems495
The roots of the denomimtor polynomial,thatis,-pr, -p2. h,...,
pu , are calledthe polesof F(s) rhey are rhe valuesof s ar which F(s)
becomesinfinitely large.In the function describedby Eq.12.91,thepoles
of F(r) are -1, 2, -3, and -4.
fhe roots of the numerator polynomial, that is, - zr, - 22, 21,. . . ,
zn, are calledthe zeros of F(s); they are the valuesof s at which _F(r)
becomeszero.In the function descdbedby Eq. 12.91,the zerosof tq(r)
aJe-5 and 10.
In what folows, you may find that behg able ro visualize rhe poles
and zeros of F(r) as points on a complex r plane is helpful. A complex
planeis reeded becausethe roots of the pol)'nomialsmay be complex.In
t]le complex r plane, we use the hodzontal axis to plot tlle real values of s
potes
Figure
12.17A Ptotting
andzeros
onthesptane.
and the vertical axis to plot the imaginary values of s.
As an exampleof plotting the poles and zerosof F(r), considerthe
function
'-'
10(r+ 5)(r + 3 l4i{r + 3 + i4}
(r+10.)(s+6 l8r(s+6+j8)
\12.9?)
The poles of F(r) are at 0, 10, -6 + j8, and -6 - j8. The zeros are ar -5,
-3 +
J4,and -3 - i4. Figuie 12.17showsthe polesand zerosplottedon
the s plane,whereX's rcpiesentpolesand O's representzeros,
Note that the polesand zerosfor Eq.12.90are locatedin the finite r
plane. F(s) can also have either an /th-order pole or an J'th-order zero at
infinity. For example, the tunction descdbed by Eq. 12.91 has a secondorder zero at infinity, becausefor large values of r tle function reduces to
4/r', and F(s) : 0 when r = oo.In this text,we are interestedin rhe poles
and zeros located in the finite r plane. Therefore, when we refer to the
poles and zeros of a mtioml function of s, we are refering to the finite
polesand zeros.
12.9 Initiat- andFinat-Value
Theorems
The idtial- and final-value theorems are useful becausethey enable us to
determin€ from F(s) the behavioi of f(1) at 0 and c<).Hence we can check
the initial and final values of /(t ro seeif they conform with known circuit
behavior, before actually finding the inverse trarNform of F(r).
The initial-value theorem statesthat
(1293) < Initial valuetheorem
and the final-value theorem statesthat
(12.94) < Finalva(u€theorem
The initial-value theorem is basedon ttre assumptionthat /(r) contains no
impulsefunctions.ItrEq.12.94,\4'e
must add the restriclionthat the theorem is valid only if the poles of F(s), exceptfor a fint-order pole at the
odgin,lie in the left half of t}le s plane.
496
Intrcductjon
Tnnsform
t0 theLaptac€
To prove Eq. 12.93,we stafi with the operational transfom of the first
derivative:
( sr\
_ r { L f = , r r s r l r oI
lat )
t6,tr
l - i dl
e"tlt.
Jo-
{r?.e5J
oo:
Now we take the limit as r-
lim t,r{\) - t(0 )t = rim [*!"
s.aln
'0,.
dt
rr2.e6)
Observe
thattheright-handsideof Eq.12.96may
bewrittenas
I fo dt ^
ljm
:e'tu
I I
, .d\Jo
l*dt
| :e"'dIl.
dr
.lo. dl
"- 0;
As s- &, (df/dt)e
hence tle secondintegal vanishesir the
limit.The first integralreducesto /(0')
/(0-), which is indeperdentofr.
Thus the iglt-hand sideof Eq.12.96becomes
in
r! "-',h - f(0 )
I dt
s_6 ln
/(0 ).
lLz.st)
Because/(01is irdependentofr, the left-handside of Eq.12.96may be
[mfsFrs)- /r0 )l - um[rFrr)]-/(0 ).
(12.e8)
From Eq$ 12.97and 12.98,
lirnsFG): /(0') = lja-f(r),
whichcompletesttreprcof of the initial-valuetheorem.
The prcof of the fiml-value theoremalsostartswith Eq. 12.95.Here
we takethelimit asr + 0:
lirqtrFG)-/(r)l=t\^"(fX-'")
(12.ee)
The integrationis witll respectto t and the limit operationis with respect
to s,sotheright-hand
reduces
to
sideof Eq.12.99
!L"'a,\ = If'd
r^( I
r .o\..i. .,
l
t
i,d'
.tn
(l2 roo)
Becausethe upperlimit on tle integial is iniidte, this integal may alsobe
wiiiten asa limit process:
fff"-' |^4.,,.
(r2.r01)
12.9 Initiat-andFinal-Vatue
Theorems 497
where we use ]|, as the symbol of integration to avoid confusion with the
upperlimit on the integral.Canying out the integrationprccessyields
,\*t/(/)
.f(o)l -,'1.*t/0)l- f(o-).
(72.102)
Suhstituling
Eq. 12.102
into Eq. 12.99gives
riqlrrq(s)l - /(0-) =,tij*f(r)l
"f(o ).
(12.103)
Because
Eq. 12.103reducesto the final-valuetheorem,
/(0 ) cancels,
namely,
lim sF(r)
" : lim f(r).
r-0
The fiml-value tieorem is useful orly if /( oo) exists.This condirion is true
only if all the polesof F(r), exceprfor a simplepole ar the origin,tie in the
left half of the s plane.
TheApplicationof Initial- andFinal-Va[ue
Theorems
To illustmte the applicationof the initial- and finalvalue theorems,we
apply them to a function we used to illustrate partial fraction expan
sions.Considerthe transform pair given by Eq. 12.60.The initial-value
rneoremgrves
lim sF(s) = lim
100"[ + (3/r)]
,:;s'[1 + t6lr)]U+ (6/r.)+ (2sls1l
,4,f
: -12 + 12= 0.
(r : l-12 + 20cos(-53.13')l(1)
The fural-value theorem gives
l00r(s + 3)
.
l u n \ / t , l t j m - , : - 0 ,
{-lr
J-n (j + 6)(s. + 6r + 25)
,Iij*/(l)
= limi-12€ 6! + 2Oer,cos(4r
5:.r:")lU(r) - o.
In applying the theorems to Eq. 12.60,we already had rhe timedomain expressionand were merely testingour understanding.But the
real valueof the initial and final-valuetheoremslies in beingable to test
the r-domain expressionsbefore working out the inve$e transform. For
example,considerthe expressionfor V(s) given by Eq. 12.40.Alrtough
we cannotcalculater'0) until the circuit parametersare specified,wecan
checkto see if y(s) predictsthe corect valuesof r.r(0+)and z,(oo).We
know from the statementofthe problemthat generatedy(s) that ?r(0+)is
zero.We alsoknow that ?)(6) must be zero becausethe ideal inductot is
a perfectshort circuit acrossthe dc current source.Finally,we know that
the polesof y(s) must lie in the lefr hau oI the r plane becauseR. a. and
C are posi!i!econqtanrs.
Hencethe polesot rl (r) al.o lie in the teft hall
of the s plane.
498
IntrodLr.t
on to the Lapta(e
Tmn+orm
Applying lhe inilial valuelhcor'cnlyiclds
lim jY(:!)
=
5(/dc/c)
"r:;r2[1 + 1/(^ct + 1/(...cr11
Applying the fhal-value theoremgives
l'r,"Yt'l= ilii
s(1d./c)
,2 + (rl RC) + (rrlC)
'llre
derjvedexpressionfor v(s) correctl]'predictsthe initial and final valuesof r(l).
objective3-Understandandknowhowto usethe initial vatuetheorenandthe finat vatuetheorem
12.10 Use the initial and linal valuelheorems1()fird
the inilial and Iinal valuesofl0) in Assessment
Problems12.4,12.6,
and 12.7.
Answer:7,0;4.1;and0,0.
NOTE: AIso ttj Chaptet Prcblen 12.17.
in } r5
$U{_ni
The Laplace transform is a tool for convertingtimedomarnequ"rion. ilro lrequency-donJ
n equdrion..
accordingto the following generaldefinition:
:tll!\:
I
.lo
I()e"'tt= F(s),
where l(r) is the tine-donain erpression,and F(s) is
(Seepaee467.)
the frequeDcy-domain
expression.
("(r)
The sl€p frnction
describesa function that experiences a discontiruit]' from one constant level to
anotherat somepoint intime. 1(is the magnitudeofthe
iunp; if Il = 1, d"(t) is the l|nit step function. (See
page469.)
The inpnlse funclion (5(r) is defined
I rtrt*: r
.J 60) : 0, r + 0.
r is the strengthof the nnpulse;if r : 1, K6O is the
u t inpulse function.(Seepage171.)
A functional transform is the Laplace transform of a
specificfunction. Imporiant functional transformpairs
arc summarizedin Table12.1.(Seepage471.)
Opcrutionaltr:rnslbrmsdefinethc gcncralmathematical
propelies of the Laplace lransfomr.lmportant opera'
tiondl trdDsformpairs are sur1lnarizedin Tabtc 12.2.
(Seepage47s.)
ln linear lumped-parametercircuits,I(s) is a ratronal
functionof .r.(Seepage182.)
II F(r) is a proper rational function.the inversetransfbrm is found b], a partial fraction expansion.(See
pagc,l[33.)
IfF(r) is nn inproper ftlional iunction,il canbe inverselranslomred by lirsl exparding il into a sum ol a poly'
nomialard a properralionalfunclion.(Secpagc493.)
FG) can be expressedas the ratio oftwo factoredpoly'
nonjals.The rooh of the denominatorare calledpoles
and are plotted as Xs on the complex.!plane.The roots
of the numeratorare calledzerosand are plotted as 0s
on the complexs plane.(Seepase49s.)
499
The initial-value theorem states that
The theorem is vaiid only if the poles of F(s), exceprfor
a first-order polo at the origin, Iie in the left half of fhe s
plane.(Seepage495.)
lirn- /(.) = tin sF(r)
The theorem assumes that /(t) conrains no impulse
lunclions.(Seepage495.)
The final-valuetheoremstatestlat
iim /(r) = lim sF(s).
. The inifial- and final-value theorems allow us to predict
the initial and final values of /(t) from an s-domain
(Seepage497.)
expression.
Probtems
S€clion 12.2
12.1 Use step functions to write the expressionfor each
function shom.
12.2 Use step functions to \r,Titethe expression for each
of the functionsshowr inFig.P12.2.
Figure
P12.1
Figue?12,2
/o
J('
-20
-10
(a)
fo
rG)
12.3 Make a sketchof /0) for 25s < r < 25s when
/(l) is givenby thefollowingexpressior:
f(t):
(20t+ 400)u(t+ 20) + (40r+ 400)"(r+ 10)
+ (400- 40r)r(r 10)
+ (20t - 400)u(t 20)
500
Inhoduction
to the kptaceTransfom
12,4 StepIunctionscanbe usedto definea pirdop function. Thus (t 1, u(t 4) defines a vrindow
1 unit high and3 unitswide locatedon the time axis
between1 and4.
A function/(t) is definedasfollows
f0) : 0,
tL8 Er?lain why the folowing lunction generates an
impulse function as € - 0:
t<o
:30t,
0< t = 2s
: 60,
2 s= t < 4 s
/\
= oocos[;' - rJ.
4s <, < 8s:
: 301- 300
8s = l = 10s
12,9 In Section 12.3,we used the sifting property of the
impulse tunction to show that lt{6(t)} = 1. Show
that we can obtain the same result by finding the
Laplace hansform of ttre rectangular pulse that
exists between +€ in Fig. 12.9 and then finding
the limit of tlis transform as € + 0.
Lr.10 a) Showthat
'ta,:
l* rr,n'r,
a) Sketchl(t) over the interval 2s < t < 12 s.
b) Use the concept of the window function to write
an expressionfor /(1).
t',,t.
(Hizt: Integiate by parts")
b) Use the formula in (a) to showthat
E{6' (,)} = r.
Section 123
12.5 a) Fhd the area under the function shown in
Fig.12.12(a).
b) What is the duiation of t]Ie function when € = 0?
c) What is the magnitudeof/(o) when € - 0?
x2.11The triangular pulses shoM in Fig. P12.11 are
equivalentto the iectangularpulsesin Fig.12.12(b),
becausethey both enclose the same area (1/€) and
they both approach infirlity proportional to 1/€2 as
€ + 0. Use this triangular-pulse representation for
5'(.) to find the Laplacetransformof6"(t).
12.6 Evaluate the following integmls:
t 4 "
a) / - / (r' a)[5n) a5G 2)]/r.
Flgure
P12.11
6'(r)
t4-
b) 1: / lt6(r)+ 6G+ 2.5)+ 60 s)ldr.
12.7 Findl(r) if
fro: ! [ rpten'a-,
and
12.1i, Show tlat
3+ io
F(d)=,.
.
n'(d).
E{6'r1(t} = sr.
Prcblems501
Sections12.+12.5
l r l
Lr.13 Find t]Ie I-aplace transform of each of t]le following
functions:
a) f(t) = tu-d'
b) /(t) = sinot'
c) Find iP {:=l
I
}.
vr)
d) Checkthe resultsofparts (a), (b), and (c) by first
differentiating and then transforming.
12.19 ^) Find tle Laplace tmnsform of the futrction illus-
c) "fG) = sin (or + 0)'
d) ,f0) : r;
e) /(r) : cosh(r+ 9).
(I{,rl: SeeAssessment
Problem12.1.)
XLl4 ^) Find the Laplacetmnsformof re-'r.
tated in Fig.P12.19.
b) Find the Laplace tansfolm of the first derivative of the function illustrated in Fig. P12.19.
c) Find the Laplace tmnsform of the second derrvaliveoI Lhefuncrionilluslraledin Fig.P 12.t9.
Figur€
P12.19
f (!,
b) Usethe opentional transformgivenby Eq. 12.23
to find the Laplacetransformof:
i
LIT
(/?-'' ).
c) Check your result in part (b) by fhst differenti,
ating and then transforming the resulting
expression.
Find t]te Laplace transform (when €-0)
of tle
derivative of the exponential function illustrated in
Fig. 12.8,using each of the following two merhods:
a) First differettiate the functiotr and tlen find the
tansform of the resulting function.
b) Use the opemtional tmnsform given by Eq.12.23.
I
by first integmtingand tlen transforming.
Showthat
sle-"' f(t)j = F(s + a).
t".r,^t,t ot{l u^0,}.
Fkdsl
Lr.20 a) Find the Laplacetransformof
( r
I
I ydy I.
Checkthe resultsof (a) and (b) by firsr integrar
ing andthen transfomhg.
tr.tro .rur{f,.r.,}.
o,"ro r {4.or.,}.
[d.
)
b) Check the result obtained in (a) by using rhe
operationaltransformgivenby Eg. 12.33
1221 Find the Laplacetransformof eachof the following
functions:
a) f(r):
2oe5t' 2)u( 2).
b) /(r) : (8r - 8)lu(t 1) u(t - 2)l
+ (24 qt)Iu(t - 2) u(t - 4)l
+ (8r- a0x,(t 4) l,(r- s)1.
12.22 Showthat
eIf(at)j :
:'(il
502
Intmductjon
to theLaplace
Tmnstorm
lL23 Fhd the Laplacetransform{or (a) and (b).
a, f u t:
12.27 The switch in the circuit in Fig. P12.27 has been
open for a long time. At t : 0, the switch closes.
a) Derive the integrodifferential equation tlat
govems the behavior of the voltage o, for 1 > 0.
b) Showthat
+ l e a ts i n u t l .
b) ,f0) = / .a'cnsax dx.
c) Verify the rcsults obtained in (a) and (b) by first
carrying out the indicated mathematical opemtion and then finding t]le Laplace transform.
uJll :
udc/RC
s2+(1lRC)s+(llLC)
c) Show that
L2.24 a) civen that F(s) : g{/(t)}, sho\i'that
11"):
dF(")
=.rr.r,,,.
ud.Q/RLC)
sls' +(1/RC)s
+ (l/ZCr'
ligwe Pr2.27
Showthat
t-I\';
c)
= Yt( ftttl.
Usetheresultof (b)to findg{rs}, g{r sinBr},
andg{te ' cosht}.
D2A There is no energy stored in the ckcuit shown in
12.8 a) show that if FG) = s{f0)},
Laplace-[ansformable,t]ren
and {/(t)/t} is
1,-'r"":-{*}
(H/r?rrUse the definingintegral to write
: l"-(l) ro*^")*
1"",r,0"
andthenrevercetheorderofintegration.)
b) Startwith theresultobtainedin Prcblem12.2(c)
for g {tsinBt} and usethe operationaltransform given in (a) of this pmblem to find
:r I sin Bt].
Fig.P12.28al the time the switchis opened.
a) Dedve the integrodifferential equation that
govems the behavior of the voltage o,.
b) Showthat
Ia"/c
%G): t 2 + ( 1 l R c ) s + ( I I L C )
c) Showthat
1ls) =
s' +(URC)S+(IILC)
P12.28
Fjgur€
SectionLl,6
12.26 In the circuit shown in Fig. 12.16,the dc current
sourceis ieplacedwitl a sinusoidalsouicethat
deliversa curent of 5 cos10tA, Thecircuit componentsarc -R: 1 (), C : 25mF, and a : 625mH.
Findthenume calexpression
for y(r).
12.29 There is no energy stored in the ciicuit shown ir
Fig. P12.29at the time t}le switch is opened.
a) Derive the integrodifferential equations that
govem the behavior of the node voltages u1
b) Show that
Figore
PI2.31
11"(r)
VG)= c [ s 2 + ( R l L ) s + ( 1 l L c ) l
ligtrrc?12.29
12.32 a) Wdte tle two simultaneous differential equatrons that de,scdbetle circuit showl in Eg. p12.32
in rermsol lbe meshcurren15
4 dnd/r.
b) Laplace-transformthe equarionsdedved in (a).
Assume that the initial eneryy stored in the cilcutt rs zero.
c) Solvethe equationsin (b) for 1r(r) and /r(s).
1230 The switch in tlle circuit in Fig. P12.30has been in
position a for a long time. At , - 0, the switch
movesmstantaneously
to positionb.
a) De ve the integrodifferential equation tlat govems t]le behavior of tlle current ia for t > 0+.
tigureP12.32
60()
25H
l .1 1-0 0-" -0 )lvI) \"l slH
Tl
t -"'/
|
)40o
it-"t
b) Show that
.fdclr+ (1/RC)l
Is' + (llRc)s + (llLc\l
Seclion 12.7
12,33 Find 0(t) in Problem12.26.
Figure
P12.30
1),jolo
R
= C L
12.31 The switch in tlle circuit in Fig. P12.31has been in
position a for a long time. At , = 0, the switch
moves instantaneously to position b.
a) Derive the integrodifferential equation rhat
governs the behavior of the voltage r,, for
1>0-.
1i1.34 The ctucuit parameters in the circuir in Fig. P12.27
6nr! are tR:10ko; a:800mH; and c = 100nF.If
rdcis 70 V find
a) o,(1)for I > 0
b) t,(r) for t > 0
t1,35 The c cuit parameters in the circurr seen rn
E d Fig P12.28have the followirg values:n : 4 kO,
L - 2.5H,C = 25 nF, and 1d. = 3 mA.
a) Find 0,(t) for 1 > 0.
b) Find to0) for I > 0.
c) Does your solution for i,(r) make sensewhen
1 = 0? Er?lain.
b) Showthat
Ydcls+ (R/1.)
[s' + (RlL)s + (llLC)]'
11.36 The circuit pammeters in the circuit in Fig. P12.29
*pt E are R : 2500 A. L : 500 InH; and C : 0.5 pF. If
ts(r): 15r,G)mA,find,'lt.
504
Intrcdudion
to theLapLace
lransfbln
12,37 The circuit pammetersin the circuit in Fig. P12.30
arc n:50O,
L : 37.25Ix'H, and C = 2 pF. ll
=
1d" 100mA, find i,(t) for 1 > 0.
12.42 Fhd /(t) for each of the following tunction$
10r' +85s+95
s2+6r+5
a)
The circuit parameten ir the circuit in Fig. P12.31
a r e R = 5 0 0 0O , l , : 1 H ,
and C:0.25t1F. If
Yd. = 15 v, find 0,0) foi t > 0.
Use tlle results ftom Problem 12.32 and the circuit
shownin Fig P12.32to
a) Find 4(1) and i,(/).
b) Find t1(6) and tr(co).
c) Do the solutions for i1 and i make sense?
Explain.
5(r' +8s+5)
F(r) :
s' +4s+5
0
s' +25r+150
F(,) =
s+20
12.43 Find /(1) for each of the following functions.
100(r+ 1)
r,(s, + 2s + 5)'
a)
12.40 Find/(t) for eachof the followingtunctions:
a)
18s' +66r+54
(r+1)(s+2)(r+3)
8r3+8912+311r+3oo
b)
r("): ( r20c
+1f
0
s(s+2)(r' +8r+1s)
c.)
o,,
17s2+172t+700
(r+2xr' +12r+100)
56s'+112r+5000
s(s' +r4r+625)
12.41 Find /(r) for each of the following functions.
4
s(r' -ss+s0)
rG)=
d)
e.)
sG+42
s4(r + 1)
12.,14 Derive ttre transfom pair given by Eq. 12.64.
12.45 a) Derive the transformpair givenby Eq.12.83.
b) Derive the translormpair givenby Eq.12.84.
s'(s+ 10)
10(3s' +4s+4)
r(r + 2)'
")
.7)
40G+ 2)
r(s + 1)3
Sections12.8-0.9
12.46 a) Use the initial-value theorem to find the initial
valueof ?,in Problem12.26.
13 6s' + 15s+ 50
r' (f+4s+5)
b) Can tlre final-value theorem be used to find th€
steady-state
valueof r'?W}ly?
f+os+s
(s + 2)l
t6s3 + 12tz + 216s
(s' +2s+5)'
728
Apply t}le hitial- and final-value theorems to each
transform pair in Problem 12.40.
.
12.8 Apply the inifial- and finalvalue theoremsto each
transfom pair in Problem12.41.
12.49 Apply the initial- and finalvalue theorems to each
tmnsformpai in Problem12.42.
probLems
505
12.51 Use the initial- and final-valuetheoremsto check
the initial and final values of the current and vol!
agein Problem12.28.
!2.52 Use the initial- and final-value theoiems to check
the initial and final values of the current in
Problem12.30.
12.50 Use the initial- and fhal-value theoremsto check
the initial and final values of the curent and vol!
agein ?roblem 12.27.
Apply t}le initial- ard final-value theorems to each
transformpair in Problem12.43.
Transform
TheLaplace
in C'ircuit
Analysis
13.1Circuitttenentsin thes Donainp. 508
in the 5 Domainp. 511
13.2CircuitAnatysis
p. 512
13.3AppLications
p. 526
13.4TheTransfer
Function
Funciion
in PartiatFnction
13.5TheTransfer
p. 528
Expansions
Funchon
andthe Convotution
13.6TheTransfer
Integratp. 5J1
tunctionandthe
13.7lhe Transfer
p. 5i7
SinusoidalResponse
Steady-State
in Circuit
13.8TheImputseFuncbon
1 Beabteto tGnsforma circuitinto the 5 donain
usingLaptace
t€nsforms;be sureyou
howto rcpresent
the initiat
undeGtand
in the
conditjons
on energy-stoEge
€tements
2 Knowhowto anatyze
a circujtin the s'domain
ard be abLeto tlanfom an s'domainsotution
backto the tine domain.
the definitionandsiqnifica.ceof
3 Unde6tand
ih€ transferfunctionandbe abt€to catcutate
the hansferfurctior for a circujtusjng
technjques.
s-domain
4 Knowhowt0 usea cncuifstansferfunctionto
jis
calculate
th€ cjrcuj{surjt inpulseresponse,
andits steady-state
unit steprcsponse,
input.
rcsponse
to a sjnusoidal
506
The Laplace transform has two characteristicsthat make it an
attractivetool in circuit analysis.Fi$t,it transformsa setof linear
constantcoefficient differential equations into a set of linear
polynomialequations,which are easierto manipulate.Second,it
automaticallyintroducesinto the polynomialequationsthe initial
valuesof the current and voltage variables.Thus,initial conditions are an inherent part of the transform process.(This contrasts with the classicalapproachto the solution of diflerential
equations,in which initial conditions are consideredwhen the
unknowncoefficientsare evaluated.)
We beginthis chapterby showinghow we car skip the stepof
w ting time-domainintegrodifferentialequationsand transfbrming them into the r domain. In Sectio[ 13.1,we'll develop the
r-domaincircuit modelsfor resistors,inductors,and capacitorsso
that we can write .r-domainequationsfor all circuits directly.
Section13.2reviewsOhm's and Kirchhoff'slawsin the contextof
the,t domain.After establishilgthesefundamentals,weapply the
Laplace transform method to a variety of circuit problems in
Section13.3.
Analytical and simplificatior techniquesfirst introducedwith
resistivecircuits-such as mesh-curlentand node-voltagemeth
ods and sourcetransformations-can be usedin the s domain as
well. After solving for the circuit responsein the s domain,we
inversetransformback to the time domain,usingpartial fraction
expansion(asdemonstratedin the precedingchapter).Asbefore,
checkingthe final time-domainequationsin terms of the initial
conditionsand final valuesis an impo ant step in the solution
process.
The s-domaindescriptionsof circuit input and output lead us,
in Section13.4,to the colcept of the transferfunction.Thetransfer function lor a particular circuit is the ratio of the Laplace
transformof its output to the Laplacetransform of its input. In
Chapters14 and 15,we'll examinethe designusesof the transfet
function, bu1 here we focus on its use as an analyticaltool. We
continuethis chapter with a look at the role of partial fraction
PracticaI
Perspective
Surge
Suppressors
personal
Withthe adventof home-based
computers,
modems,
fax machines,
andothersensitiveeLectfonic
equipment,
it is
necessary
to provideprotectjonfromvoLtage
surgesthat can
occurjn a househotd
circujtdueto switchinq.
A comnefcialtyavaiLabte
surgesuppressor
is showf jn the accompanyjng figure.
Howcanfljppinga switchto turn on a ijght or turn off a
hair dryercausea voltaqesurge?At the end of thjs chapter,
we witl answerthat questjonusjng LapLace
transformtech
niquesto analyzea circuit.WewjlLitlustratehow a voltage
surgecanbe createdby swjtchingoff a resistjveloadin a circuit operatingin the sinusojdaI
st€adystate.
507
508
lhe tapa.eTEnsfo,m
in CkcLrit
Anatysis
er?ansion(Section13.5) and the convolution integral (Section13.6)in
employingthe transfer function in circuit analysis.We concludewith a
look at the impulsefuflctionin circuit analysis.
in the s Domain
13.1 CircuitElements
The procedurelor developingan.r-domainequivalert circuit foi eachcir'
cuit element is simple.First, we write the time-domain equation that
relatesthe teminal voltage to the terminal culrent. Next, w.3lake the
Laplace transform of the time-domainequation.This step generatesan
algebraicrelationship between the s-domain current and voltage.Note that
the dimension of a tmnsfomed voltage is volt-seconds,and the dimension
of a transformed curent is ampere-seconds.A voltage-to-curenl ratio in
the r domain caries the dimensionof volts per ampere.An impedance
in the s domair is measuredin ohms.and an admittanceis measuredin
siemens.Finally,we constructa circuit model that satisfiesthe relation
ship betweenthe s-domaincurrent and voltage.We use the passivesign
convention in all tie deivations.
A Resistor
in the s Domain
We begin$riththe resistanceelement.From Obm's law,
1) :
Ri.
(13.1)
Because R is a constant, ttre Laplace transfom of Eq. 13.1 is
V=RI.
?
?
v:e.{D}
(13.2)
and I=gIi}.
"j n , ' t ! n j t
I
Equation 13.2statesthat the r domain equivalentcircuit of a resistoris
simply a resistanceof R obms thal caries a current of I ampere-seconds
and hasa terminal voltageof yvolt seconds.
(a)
(b)
Figure 13.1 shows the time- and frequency-domaincircuits of the
resistor.
Note that going froD the time domain to the frequencydomaiD
Fig(re13.1Alhercsjstance
ehnrent.
G)rimedomain.
doesnot changethe resistanceelemcnt.
t
i
. Ii
t tlllo ri
J
figure13.2A AnjnductoroflhenrXs
canying
an
initialcunent
of 10ampercs.
An Inductorin thes Domain
Figure 13.2showsan inductor carrying an initial current of /o amperes.
The time-domainequationthat relatesthe terminal voltageto the termi-
"=,#.
(r3.3)
13.1 Circuit
Etenents
in ther Domain 509
The LaplacetransformofEq.13.3 gives
Y = rlsl
t(u )l = sLI
Lto.
(13.4)
TWo different circuit configurations satisfy Eq. 13.4.The first consists
of an impedance of .rl, ohms in serieswith an independent voltage source
ofalo vollseconds,asshownin Fig. 13.3.Note that the polarity marks on
the loltage sourcealo agreewith the minus sign in Eq. 13.4.Note also
that 110 carriesits own algebraicsign;that is, if the initial value of j is
opposite to the reference direction for i, then 10has a negative value.
The secondr-domainequivalert circuit that satisfiesEq.13.4consists Figure
13.3A Iheseries
€quivatent
cjrcujttor
an
oI an impedanceof r-L ohms in parallel with an independentcurent inductor
of L henrys
canying
aniniijatcunent
of 10
sourceof 1o/r ampere-seconds,
as shownin Fig. 13.4.We can dedve the
alternativeequivalentcircuit shownin Fig. 13.4in severalways Ore way
is simplyto solveEq.13.4for the currentl aIld then constructthe circuitto
satisfythe resultingequation.Thus
', :
v+LIo
"z
= JV ' ; I o
!!
Two otier waysare:(1) {ind the Norton equivalentof the circuit shownin
Fig.13.3and (2) startwith the inductor currert as a function ofrhe inductor voltage and then fiird the Laplace tmnsform of the resu]ting integral
b
equation.Weleavethesetwo approachesto Problems13.1and13.2.
figure13.4
paraltelequivalent
circuittor
an
Ifthe initial energystoredin the inductor is zero,that is,if10 - 0, rhe jnductorofA Th€
a henrys
canying
aninjtiatcurcntof10
s-domain€quivalentcircuit ofthe inductorreducesto an inductorwith an
impedanceofsa ohms.Figure 13.5showsthis circuil.
*1
)
A Capacitor
in thes Domain
An initially charged capacitor also has two r-domain equivalent circuits.
Figure 13.6showsa capacitorinitially chargedto y0 volts.The terminal
current is
vlsLtl
I
tigure13.5A TheJ-domain
circuitforaninductor
rvhen
theinitiatcurentisz€ro.
,
'Iiallsforming
Eq. 13.6yields
? ,
'-t"
1 : C[sY - ?,0 )]
I
Figure13.5A A capaciior
ofC farads
injtiattycharged
to ro votts.
I=sCV-CVa,
(13.7)
which indicatesthat the r-domain current1is the sum of two bmnch currents (Jnebranchconsistsof an admittanceofrC siemens.
and the second
branchconsistsof an independentcurrentsourceofCyo arDpereseconds.
Figure13.7showsthis parallelequivalentcircuit.
510
IheLaptace
Iransfom
in circuitAnatysis
we derive the seriesequivalentckcuit for the charyedcapacitorby
solvingEq.13.7for y:
(-t)'.7
thc
b
rigure13.7A ThepaEltetequivaL€nt
cjrcujt
fora
capacitor
iniiialty
charu€d
to vovotts.
.r.
ri
hc
"T'
A
Y
(13.8)
Figure 13.8showsthe circuit that sarisfiesEq. 13.8.
In the equivalent circuits shown in Figs. 13.7 and 13.8, yo carries its
own algebraic sign. In other wordi if the polarity of % is opposite to tlre
reference polarity for o, yo is a negative quantity. If the initial voltage on
the capacitor is zero, both equivalent c cuits reduce to an impedance of
1/rC ohms,asshowr in Fig.13.9.
In this chapter, an important first problem-solving step will be to
choose between the parallel or series equivalents when inductors and
capacitors arc present. With a little forethought and some experience,the
correctchoicewill often be quite evident.Theequivalentcircuitsare summarizedin Table13.1.
va/'
figure13.8a Theseries
equivateni
circuiifora
jnjtiatly
capaciior
chaqed
io v0votts.
*1,
, *, t,,
FREQUENCYDOMAIN
TIME DOMAIN
.1"
rl
,:R
.1"
tl r5R
t,
,ln
V=RI
?=Ri
I
I
b
Figure
13.9a Thei-domain
cjrcujttor
a capacitor
js zerc.
when
theinitiatvothge
.1"
il ,lzlro
rl
I,
._r
i =!f" ,a' * 4
V=sLI-LIt'
* *
T"
,, . +cyo
.l
" =lIo ia' + vo
lsc
thc
A
I
Y
'
r
1
C
_. vr ZI as
V
r
"
13.2 Cncuit
Anatysis
in thesDomain511
13.2 CircuitAnalysis
in the s Domain
Befbreillustratinghow to usethe r-domainequivalentcircuitsin analysis,
we needto lay somegroundwork,
F st, we know that if no energy is stored in the inductor or capacitor,
the relationshipbetweenthe teminal voltageandcurrentfor eachpassive
element takes the form:
V=ZI,
(13.9) < ohm'stawin theJ-domain
whereZ refersto the.r-domainimpedanceof the element.Thusa resistor
hasan impedanceof R ohms,an inductor has an impedanceofra ohms,
and a capacitorhas an impedanceof 1/sC ohms.The relationshipcon
tained in Eq. 13.9 is also contained in Figs. 13.1(b), 13.5, and 13.9.
Equation 13.9is sometimesreferredto as Ohm'slaw for thes domain.
The reciprocal of the impedanceis admittance.Therefore, the s domain
admittance of a resistor is 1/R siemens,an inductor has an admittance of
l/sa siemens,
and a capacitorhasan admittanceol sC siemens
The rulesfor combiningimpedancesand admittancesin the.r domain
are the sameas thosefor frequency-domaircircuits.Thus series-parallel
simplificationsand A-to-Y convetsionsalso are applicableto s-domain
In addition,Kirchhoff'slawsapplyto s-domaincurrentsand voltages.
Their applicability stems from the operational transfom stating that the
Laplacetransformof a sum of time-domainfunctionsis the sum oI the
transformsof the individualfunctions(seeTabte12.2).Becausethe algebraic sum of the curr€ntsat a node is zero in the time domain.the alsehrdic.um or rbe rran.formedcuffentcis dlso,/ero.A simrtarsrateminr
holds for the algebraic sum of the transfomed voltages around a closed
palh.Thes-domain
verrionor Kirchhoifsla$ri.
alg>1 = 0,
(13.10)
alg>Y : 0.
(13.11)
Becausethe voltageand currentat the terminalsofapassiveelement
are related by an algebmicequation and becauseK chhoff's laws still
hold, all the techniquesof circuit analysisdcvelopedfor pure resistive
networks may be used in r domah analysis.Thus rode voltages,mesh
curents! sourcetransformations,and Th6venin,Nortonequivalentsare
all valid techniques,even when erergy is storedinitially in the inductors
andcapacitors.
Initially storedeneryyrequiresthat we rnodifyEq.13.9by
simply adding jndependentsourceseither in seriesor parallel with the
element impedances.The addition ol these sources is governed by
Kirchhoff'slaws.
512
jn CjrcujtAnatysjs
Th€Laptace
Tnnsform
transforms
obiective1-Be abteto transforma circlit into the 5 domainusinqLaDtace
13.1 A 500 O resistor.a 16 mH inductor,and a 25 nF
capacitorare connectedin parallela) Expressthe adniltance oI lhis parallelcom'
biMtion ofelementsas a ratioral function
b) Computethe n mericalvaluesof tlle zeros
and poles.
Answer: (a) 25 x 10 e(s2+ 80,000r+ 25 x 103)/sr
(b) z1 :
10,000 j30,000;
:
40,000+ j30.000rpr = Lr.
z2
13.2 The parallelcircuit in Asses$nenlProblem13.1
is placedin serieswith a 2000f) rcsistoI.
a) Expressthc impedanceofthis seriescombinalion as a lational function ofj.
b) Computethe Dumedcalvalueso[ thc zcros
and poles.
Answer: (a.)2000(.r+ 50,000)'/(." + 80.000s
+ 25 x 103);
(b) .zr = -?, = -50,000;
-40'000 - j30,000,
Pr =
pr=
40,000+ /30,000.
NOTE: Also try ChapterProblems13.1and 13.5.
13.3 App!ie*t'i*iis
we Dolvilluslralehow to usethc Laplacctransformto determineihe transient behavior oI severallinear lumped paramctcr circuits.We start b]'
analyzingfamiiiar circuilsliom Chaptcrs7 and 13bccanscthey representa
simple slarling piaceand bccauscthcy show that the Laplacetransform
the easeof manipuapproachyieldslhe sameresulls.In all thc cxan1ples.
lating algebraicequalions inslead of difibrcntial equations should be
apParent.
TheNaturalResponse
of an RCCircuit
Figurc13.10.,tThecapacitor
djscharge
circuit.
vo
we first revisit the natural responseof an -RC circuit (Fig. 13.10)via
Laplace transform techn'ques.(You may want to review the classical
analysjsofthis samecircuit in Section7.2).
Thc capacitoris initially chargedto vovolts.and we are interestediD
the lime domain expressions
ibr i and r.wc start by finding L ln transferring the circuitin Fig.13.10to thc r domain,wchavea choiceofrwo equivaienl circuilslor the chargedcapacitor.Bccrusewe are interestedin ihe
culletrt.thesedesequivalenlcircuiIis more attractivc:it rcsultsin a singlenesh circuil in thc ircqucnc!,domain.Thuswe constructthe s-domaincircuil shownin Fig.13.11.
Summingthc voltagcsaroundthe meshgeneratesthe expression
+:+,
+ RI.
fiqure13.11.ir AnJ domajn
equivatent
circuitfortlre
circuitshownin Fig.13.10.
( 1 3r .2 )
Sol\ingEq.13.12for / yields
I =
CV|I
uulR
R C s + 1 .' + (fRC)
(13.13)
1 3 . 3 A p p l i c a t i o n s5 1 3
Note that the expressionfor 1is a proper rarionatfunctionoI s and canbe
inverse-transformed
by inspcctioni
i =
'iR':uO,
;e
(13.14)
which is equivalentto thc expressionfor the currentderivedby the classi
cal mcthodsdiscussed
in Chapter7.ln rhal chapter,rhc currentis givenby
Eq.7.26,where7 is uscdin placeofRC.
Aller we havc fouid i, the easiestwav to dctcrmine t) is simpllr to
apply Ohm'slawrthat is,Irom the circuil,
Rt = V,f 'R'ult).
u:
(13.15)
We Dowillustratca way to find o from rhecircuitwirhoutfirstfindingi.
In this alternativeapproach,we retum to the original circuit of Fig. 13.10
cYo
and lransferjt to the s domainusingthe parallelequjvalentcircuit for the
chargedcapacitor.Using the parallel equivalentcircuit is atrractivenow
becausewecandesc be the resultingcircuitin rcrmsof a singlenode volf
Figure11.12s Anr-domain
equivalent
circuitforihe
age.Figure13.12showsthe new.r domain equivalcntcircuit.
circuitshownin Fiq.13.10.
The node-voltageequationthat describesthe ncw circuit is
t
lo, "c, = cu".
$ n
(13.16)
SolvingEq. 13.16tor Y gives
,,
uo
s + (l/nc)
113.71)
Inverse-transformiDg
Eq.13.17lcadsto the samecxpressionfor 1)givenby
Eq.13.15,
namciy.
o = Ve 'tRc = Vae'tu(,t).
(13.18)
Our purposein deriving by direct use of the transformmethodis to
showthat thc choiceof which.! domain equivalentcircuit io useis influerced by rvhichresponsesignalis ofinterest.
objective2-l(now howto analyze
a aircuitin ther domainandbeableto transform
anJ domainsotutionto the
trmedomain
13.3 The swilchin thc circuitshownhasbeenin
positior a for a long lime.At I : 0, the smtch
is thrown to positionb.
a) Find 1,I4, and y, as rationalfunctionsof r.
(b) t = 20? l'5oruG)mA,
o1 = 80e r'5o'a(r)V,
u2 = 20e 125o'u(t)
Y.
b) Findlh<I ime-domain
e\pr(ssion\tor i. , .
and 02.
Answer:(a)1- 0.02/(r+ 1250),
Yr:80/(s+1250),
V=20/(s+1250]\;
NOTE: Ako trr ChaptetProblems13.9nnd 13.10.
r00v
5ko
514
Thetaptace
Transform
in Cncuit
Anatysis
TheStepResponse
of a ParallelCircuit
Fig(re13.13A Thenep response
of a paratteL
RLrcircuit.
ldc
1
sC
R
Y sLII!
Next we analyzethe parallelRIC circuit.shownin Fig.13.13.thatwe first
analyzedin Exanple B.7.Theproblemis to tind the expressionfor iz after
the constantcurrent sourceis switchedacrossthe parallel elements.The
initial energy stored in thc circuit is zero.
As before,we begin by constructingthe r-domain equivalentcircuit
shownin Fig.13.14.Note how easilyan independentsourcecan be [ansformed from the time domain to the frequency domain. We transform the
sourceto the r domair simplyby determiningthe Laplacetnnsform ofits
time-domainfunction.Here,openingthe switchresultsin a stepchangein
the currentapplicdto the circuit.Therelorethe s-domaincurent sourceis
g{ 1d"u(1.)},
or /dJr. To lind 12,we filst solvefor y and then use
v
Figurc13.14A Thei-domajn
equivatent
circuitforthe
circuitshownin Fig.13.13.
sL
(13.19)
to establishthe s-domainexpressionfor 1r Summing the currents away
ftom the top node generatesthe expression
SCV
'N'i:+
\13.?0)
SolvingEq.13.20for Y gives
toJC
i+(llRc)s+(llLC)
(13.21)
SubstitulingEq.13.21intoEq. 13.19gives
I,JJ LC
rlrr+tr/Rc)r+(r/.c)l
(\3.2?)
Substitutingthe numericalvaluesoflR, a, C, and 1d"into Eq. 13.22yields
384x r05
s(s' +64,i)oor+16x10)
(13.23)
Before expandingEq. 13.23irto a sum of partial fractions,we factor the
quadraticterm in the denominator:
IL
384 x 10.
(13.24)
r(r + 32,000 j2a,000)(.r
+ 32,000+ j24,000)'
Now, we can test the r-domain er?ression for 1r by checking to see
whether the final-value theorem predicts the correct value for tr ai
1 = cir.All the polesof 1r, exceptfor the firslorder pole at the origin,iie
in the left half oI the s plane,so the theoremis applicable.We krow from
the bchavior of the cir€uit that after the switch has been open for a long
time, the inductor will shorFcircuit the current source.Therefore, the final
valueoI i, must be 24 mA.The limit of 11, ass+Urs
lim 11' :
384x 1d
16 x 108
(13.25)
(Currents in the s domain cary the dimension of ampere seconds,so
the dimension of s1/_will be amperes.)Thus our r-domajn expression
13.3 AppUcatjons
515
We now proceedwith the partial fraction expansionofEq.13.24:
K2
K1
s + 32,000 i24,000
K;
r+32,000+j24,000
(13.26)
The partial fraction coefficients are
384x 105: 2 4
16 x 103
10r,
(13.27)
384 \ ld
( 32,000+ j24,000)(j48,000)
: 20x 10"1J48
(13.28)
Substituthgthe numedcalvaluesof 1<1and ii2 into Eq.13.26andinversetnnsfoming the resulting expressionyields
iL : IA + 4Oe324ffit
cos(24,000r+ 126.87')lr(r)mA.
(13.2e)
The answer given by Eq. 13.29 is equivalent to the answer given tor
Example 8.7 because
40cos(24,000t+ 126.87'):
24cos24,0001 32 sin 24,000,.
If we weren't using a previous solution as a check,we would tesr
Eq. 13.29to make surethat i.(0) satisfiedthe giveninitial conditionsand
ir(oo) satisfied the known behavior of the circuit.
516
lhe taptace
Tmnsfom
in CjrcujtAnatysjs
TheTransientResDonse
of a Parattet
RLCCircuit
Anoiher exampleof ushg the Laplace transfom to find the transient
behaviorof a circuit arisesftom rcplacirg the dc currentsouicein the circuitshownin Fig.13.13
with a shusoidalcunentsource.
TheDewcurrent
sourceis
(13.30)
where 1u = 24 mA and {, = 40,000 rad/s. As before, we assumethat the
hitial elergy stored in the cncuit is zero.
The r-domain expressionfor the source curent is
s1-
(13.31)
The voltage aooss the paralel elements is
(I!:/C)s
s' +(i/RC)s+(r/rc.)
(13.32)
Sub(lilulingL.q.lJ.3l inlo Eq. 11.32rerullrin
QJC\5,
+ (11LC)l'
G'+ '1t"' + (JIRC)S
from which
IL
(.t^/LC)s
sr- 1s' + u)[r' + (l/nc)s + (r/LC)]
(13.34)
Substituting
thenumerical
valuesof 1-, (1),
R,a,and C intoEq.13.34gives
384X 10ss
(s'+ 16x 1o3XC+ 64,ooos
+ 16x 103)
(13.35)
We noq wrile the denominalorrn facloredform:
184 r 105s
IL
G - joxr + j.,,)G+ d - jB)G+ d + jB)'
(13.36)
whereo : 40,000,a = 32,000,andB : 24,000.
Wecan'ttestthe final valueof iz with the final-valuetheorembecause
1r hasa pair of poleson the imaginaryaxis;that is,polesat +i4 x 104.
Thuswe must first find i, and then checkthe validity of the expression
from knowncircuit behavior.
Whenwe expandEq.13.36
intoa sumof partialfractions,
wegenerate
theequation
'
K,
KI
K2
+
j40.000
s-/40.000 s
s+32,000-j24,000
+
r+32,n00+j24.n00
(13.37)
13.3 Appticaiions517
The numerical values of the coefficients Kt and 1(2are
Kr
384x 105040,000)
+ j16,000)(32,000
+ 164,000)
080,000)(32,000
=7.5x1o3/ 9(,,
K2
(13.38)
384x 101-32,ooo
+ j24,ooo)
(-32,000 j16,000)(-32,000
+ j64,000xj48,000)
= 12.5x 10 .12[!.
(13.3e)
Substituting
the numericalvaluesfrom Eqs.13.38ard 13.39inro
8q.13.37andinvene-transforming
theresultingexpression
yields
tr = [15cos(40,000, 90")
+ 25e-3xm.cos
(24,0001
+ 90')lrnA,
: (15sin40,tD0, 25erxom,sin
24,000i)aomA.
(13.10)
Wenow testEq. 13.40to seewhetherit makessensein termsof the
giveninitial conditionsandthe knowncircuit behaviorafter the suritchhas
beenopenlor a long time.For I : 0, Eq. 13.40predictszeroinitial current,
whichagrceswith the initial energyof zeroin the circuit. Equation13.40
alsopredictsa steady-state
curent of
rz,. : 15sin40,000tnA,
(13.41)
whichcanbeve fiedby thephasormethod(Chapter9).
TheStepResponse
of a MuttipteMeshCircuit
Until noq we avoided circuits that required two or more node-voltage or
mesh-currentequations,becausethe techniquesfor solvingsimultaneous
differentialequatioN are beyond the scopeof this text. Howevet,using
Laplacetechniques,we can solve a problem like the one posedby the
multiple-meshcircuit in Fig. 13.15.
Here we want to find the branch currents 4 and i2 that arise when t}le Figue13.15A A muttiph-mesh
Rlcircuit.
336 V dc voltage source is applied suddenly to r})e circuit. The inirial
energystoredin the circuitis zero.Figure13.16showsthe r-domainequiv8.4r
10r
alert circuitof Fig. t3.15.Thetwo mesh-currentequationsare
336 -
0:
336
(42 + 8.4s)11 4212,
42IL+ (90+ 10r)1r.
\13.42)
l)f') l"/i\
48r}
tigure13.16.r'Therdomain
equjvatentcircuittorthe
jn Fis.13.15.
(13.43) circuit
shown
Theil,aptacelTran3bh
in CircuitAnatysis
UsiDgCramer's
ftethodltdsolvefotrtririild'/t:rneobta
i'.
: .'":-a .1
. lrlaihrb..il
m"iiirrl'' "
iirl;,.1r,ifaF:.laif,iz)l i
: 84(f + 14s+ 24)
..
"
12)i.,.'
E+fu+,?)(,f',1
_y1'1y1ir:
r ,xraj.:.i jr1:/r.i:,
I
: , j i i r . " l . i) 1 i r i ] i ] . i : , i r
''
1336/s - 42 |
90+l0rl
l0
..
l+z- e.a"r:ol'l
i 1lt'
/v2- |
.t
- 1 ,2.
'.ii, : .:..1..,..\!'4i
ra.!4!1i'
.
,)
:it '
"
u
I
I
,,ji i.r frr'Jll
: ....
\"
,.
it,
J
/,.r,irri
r.i d.1!rj:,:,;
(13,441
,: t,t
(13.46)
"
Bi{iad.,6tiEdc.15.hjB:4*,|r;jr, i.: Ji.:r i.: i,.
,jl'!.:1i].,:l]:],;li1):..1.''.;|i].J.i,.rj]|':j.ii
-.'
1
Nr
40G+e)
A s G- 2 X s + l 2 ) -
\73.47)
t , . i , 1 t ;r ,. . . r r , _ , :
N2
168
'., : - ar (J- tG { 12)'
(13.48)
Expanding
11aDgl?,i$o$surtr9!Pll-rg9!,gq9tio,ns_.9Le,q.,.
.. . ,
rrr.ii
I l, ii rl]rt:i
i, : ltsi - t'it-z - trblultS4,
(13.5r)
ir= Q - &4e-a + 1.4:et})u\t) L,
(13.52)
13.3 Apptications519
Next we test the solutionsto seewhetherthey mate sellseiD termsofthe
circuit.Becauseno energyis slorcd in the circuit at the instantthe switch
is closed,both t1(0 ) and i(0 ) must be zero.The solutionsagreewith
theseinitial values.After the switch hasbeen closedfor a ]ong time, the
two inductorsappcaras short circuits.Therefore,the final valuesof i1 and
i 1 ( c o ) :M q
t:(6):
= isA,
1544-2)
=7A.
^
113.54)
One final testinvolvesthe numericalvaluesof lhe cxponentsand cal
culatingthe voltagedrop aooss the 42 O resistorby threedifferent methods.From the circuit,the vollageacrossthe 42 J) resistor(positiveat the
top) is
D- 42tt, i,t
JJh
dl 8.4
4bt,
,
l
-
d
Inl'.
{13.55,
l
You shouldverify that regardlesso{ which form of Eq. 13 55 is used,the
voltageis
It : (3s6
23s.2ea - 100.80""')u0) v.
We are thus confidentthat the solutionsfor i1 and t, are correct.
ans domainsotutionto the
anatyze
a circuitin the s domainandbeableto trensform
.(;) { - IsG + 3)l/t(s + 0.5)G+ 2)1,
+ 6)l/b(r + o.s)(r+ 2)l;
v2: 12.5(sz
=
(1s
(b) ;,
]e{5' + ler')u1r.tv,
13.5
u2=(15- + "4:' + + en')"(t)v:
a) Oeflu. tt'. "-aornainexpressionsfor yL
- 2.5 v;
(c) 1)r(0*). 0, ?.'2(0')
:
=
(d) t'r ?,, 1s V
andv2.
NOTE: Also try ChapterPtublems]3'26 and13'27.
.
520
TheLaptace
Tlansfonn
in CircuitAnatysis
TheUseof Th6venin's
Equivalent
b
Figur€13.17a A circuittobeanalyzed
using
T h 6 v e n ienqsu i v a h n ttihnes d o m a i n .
20{)
60f)
a
In this sectionwe showhow to uscThdvenin'sequivalentin the.r domain.
Figure 13.17showsthe circuit to be analyzed.The problem is to find the
capacitorcurrentthat resultstom closingthe switch.The energystoredin
the circuitprior to closingis zcro.
To find ic, we first constructthe s domainequivalentcircuit and thcn
Iind theTh6veninequivalentof this circuitwith respectto the terminalsof
the capacitor.Figurc 13.18showsthe r-domaincirc tTheTh6veninvoltageis the open-circuitvoltageacrossterminalsa.b.
Under open-circuitconditions,lhereis Do voltage acrossihe 60 O rcsis
1or.Hence
(a80/sx0.002s) 480
za.t0'
b
20 + 0.002s
s+ 104
(13.56)
The Th6veninimpedanceseenfrom terminalsa and b equalsthe 60 O
resistorin sedeswith the parallelcombination of the 20 O resistorandthe
2 mH induclor.fhus
figure13.18a Ther-domajn
modetofthe
circuit
s h o winn F i g1. 3 . 1 7 .
^ . 0.002s(20) 80(s+ 7s00)
-
.! + lOa
(13.57)
UsingtheTh6veninequivalent,
we reducethe circuitshoNr in Fig.13.18
to the one shownin Fig. 13.19.It indicatesthat the capacitorcurenl 1.
equalsthe Th6venin vollage divided by the total seriesimpedance.Thus,
80 (r + 7500)
s+104
a80/(r+ r0a)
[ 8 0 ( \+ 7 . 5 u 0G) + 1 0 4 )+] [ ( : . t o s l s ]
b
fiqurc11.19A A simptified
versjon
ofihe cjrcuit
shown
in Fig.13.18,usinga IhCvenjn
equivalent.
(13.58)
Wcsimplify
Eq.13.58
to
-'
+ 25 x 106 (s + 5ooo)r'
r, + lo,ooos
(13.59)
A partial fraction expansionofEq.13.59 generates
r.
-"
3o,ooo +
6
(s + 5000)' s+5000'
(1r.60)
the inverse transform of which is
ic : ( 30,0001e-sm0,
+ 6e sm9u0) A.
(13.61)
We now tesl Eq. 13.61to see whether it makes sensein terms of
known circuit behavior.From Eq.13.61,
ic(D : 6 A.
(I].62)
This result agreeswith the initial current in the capacitor,ascalculatedlrom
the circuit in Fig. 13.17.The initial inductor current is zero and the iDitial
capacitorvoltageis zero,so the initial capacitorcurrentis 480/80,or 6 A.
The final valueof the currentis zero,which alsoagreeswith Eq.13.61.Note
also from this equation that the current rcvenes sigD when r exceeds
6/30.000,or 200ps. The lact that ic reversessign makes sensebecause,
when the switch first closes,the capacitor begins to charge.Eventually this
charge is reduced to zero becausethe hductor is a short circuil at r = ,ro.
The sign reversal of ic reflects the chargingand dischargingof the capacitor.
13.3 Apptjcations 521
Let's assumethat the voltage drop acrossthe capacitoroc is also of
interesl-Once we krow lc, wc find ,. by integmrionin the time domain;
that is,
0 c = 2 x 105/ (6
'm'rx.
10.000.r)€
(13.63)
Allhough the integrationcailed for in Eq. 13.63is not difficult, we may
avoidit altogetherbylirst finding the r domainexpressionfor % and rhen
finding t,c by an inversetransform.Thus
,v, . : i r cI :.
2x105
6s
s
(r + 5000),
12x tos
(s + 5000)''
(13.64)
lrom which
0c : 12 x 105te5mora(r)
(13.65)
You shouldverify that Eq. 13.65is consistentwirh Eq. 13.63and that ir
alsosupportsthe observations
madewith regardto the behaviorofic (sec
Problen 13.33).
obiective2-Know howto anatyze
a circuitin thes domainandbeabLeto transform
ans-domain
sotutionto the
time domaln
13.6 The initial chargcon lhe capacitorin ahecircuit
\ho\rn i\ 7ero.
(h)/,b: [20(s+ 2.4rl[(({+ 3)(r+ 6r.
ar find rher-domdinfte\enin<qui!alenrcJr.
cuil with respect to terminals a and b.
b) Findrher-domajne\pre\siontor rhecuire r
that the circuil delivers to a load consistingof
al H inductorin serieswith a 2 O resistor.
Answen (a) yrh = y,b:
Zr. - 5f\
[20(:r+ 2.4)]/[s(s + 2)],
2 . 8 )t \
2):
'IOTE: Al'a trt (hartcr Prcblrn 13.3).
A Circuitwith MutuaIlnductance
The next exampleillustrateshow to usethe Laplacetransformto analyze
the transient tesponseof a circuit that contains mutual inductance.
Figure 13.20showsthe circuit.Thc make-before-break
swirchhasbeenin
positiona for a long time.At , = 0, the switchmovesinstantaneously
to
positionb.Tho problemis to derivethe time-domainexprcssionfor i.
0.5F
522
TheLaptace
Innsfomin circujtAnatysj,
Figure13.20A A cjrcuitcontaining
nagneticatly
coupted
coils.
(\
30
M\
0H
(Lz M)
6H
2()
IM)J2H
r0O
We begin by redrawingthe circuit in Fig. 13.20,with the switch in
position b and the magnetically coupled coils replaced with a T-equivalent
circuit.l Figure 13.21showsthe new circuit.
We now transform this circuit to the s domain.In so doins. we note that
rr(o)-g=sA,
(13.66)
i(r) = 0.
\13.67)
jn Fjg.13.20,
figur€13.21 Thecircujtshown
wiih
thenagnetjcatty
coupted
coitsrcplaced
bya T-equivatent
30
2A
10r}
10
Figure
13.22l Ther-domain
equivatent
circujt
forih€
circuit
shown
in Fjg.13.21.
Becausewe plan to use meshanalysisin the s domain,we use the se es
equivalent circuit for an inductor carrying an initial cwent. Figuie 13.22
shows the r-domain circuit. Note tlat tlere is only one independent vol!
age source.This source appearsin the vefiical leg of the tee to account for
the initial valueof t}Iecurent in the 2 H inductorof 4(0-) + ir(o ), or 5A.
The branch carrying i1 has no voltage source becausea1 M : 0.
The two r-domain mesh equations that describe ttre circuit in
Fiq.13.22 arc
(3+2s)Ir+2s12=10
(13.68)
2 s I r+ ( 1 2 + 8 s ) 1 2 = 1 0 .
(13.69)
Solvingfoi 12yields
-'
(r + 1)(r + 3)
(13.70)
ExpandingEq. 13.70into a sumof partial ftactionsgenemtes
I2
1.25
s + 1
1.25
s + 3
(13.71)
Then,
i, = (i.25€ l
r.25e-t')ult)A..
113.72)
1 3 , 3 Applications 523
Equalion 13.72reveals thal i, increasesfion zero to a peak value of
481.13n1A iD 519.31ms atlcr the slvitchis movedto posirionb. Ttrereafter,
l, decrcases
exponentiallytorvardzero.Figurc 13.23sho$'sa plot oIi, versus1.'lliis responsemakcssensein terms of the known physicalbehavior
of thc nagneticallycoupledcoils.A currcnl caDerist in thc l,2 inductor
I (nis)
only if there is a timc varyirg currenr in rhc a1 inducror.As ir decreases
trol11i$ initial valuc oi 5 A, L increascstrom zero and thcn approaches Figur€13.23.i Theptotoft: versus
tforthe circuit
zclo as il approachcsZcro.
shownin Fig.13.20.
objective2-Know howto anatyzea circuit in the s domainand be abteto transforman s-domainsotutionto the
hme domain
13.7 a) Verify from Eq.13.72that t2reachesa peak
valueof48l.l3 mA at, = 549.31nrs.
b) Fin.l4. toi I
0. tor rhecircuir.hu$n in
Fig.13.20.
c) Compuledir/dr when t2is at its peak value.
d) Expressl, as a funclion ofdir/dr whcn 12is
al its peak valuc.
e) Use the resultsobtainedin (c) and (d) to
calculatethe pcak valueof i2.
A n s w e r :( a t J r . / r / /= 0 w h e n t : l l n 3 l t i
( h l t l : 2 . 5 ( c ' + t t t ) u Q )A l
(c) -2.89 A/s;
(d) i2 =
@di/dt)l121
(e) 481.13
nA.
NOTE: Also ty ChapterProblens 13.36and 13.37.
TheUseof Superposition
Becausex'e arc analyzinglinear lunped-parametcrcircuits,we can use
superpositionto divide the responseinto componentsthat can bc idenlified with parlicular sourccsand iDitial condiriorls.Disringuishingthese
componcnlsis criticalto bcing ableto usethc lransferfunction,lvhichwe
introduceil1the Dextsection.
Figure 13.24shows our illustrative circuil. We assumcthal at the
instantwhen the two sourccsare appliedto thc circuir,rhe induclor is carrying an niidal current of p amperesand that thc capaciroris catr!,ingan
initial voltageof ,y volts-Ihc dcsiredresponseof the circuit is the voltage
acrossthc rcsislorR2.labeled1r2.
Figurc 13.25showsrhe i domain eqnivalcnrcircuit.We optcd lor the
parallcl cquivalentsfor /- and C becauser,c anlicipatedsolving lor %
usingthe node-voltagemclhod.
To lind y, by supcrposition,we calculalethe componcnloI % resulting holn eachsourccacling alone,and rhen we sum thc conlponents_
We
bcgin with y! acting alote. Opening cach oI the threc current sources
deacti!alesthem. Figurc 13.26sho\rsthe rcsulling circuit.Wc addedrhe
nodc vollageyj to aid the aDah,sis.The
primesoD la and y2indioaierhar
figure13.24,i A circuitshowing
the useofsuper,
position
in r donainanatysis.
rC
Figure13.25:l Thes-domain
equivahntfor
thecircuit
af lig.73.24.
TheLaptace
Transtorm
in (ircuitAnaLysis
they are the components of 14 and % attributable to 4 acring alone.The
lqo equatioDsthat describelhe circuitin Fis. 13.2bare
(+r+.,,),i,,,;:!,
(13.73)
Figur€
13.26 lhecircuiishown
in Fig.13.25
withy,
"cu1* (] * "c )u1= o.
\^:
/
\13.74)
For convenience,we int.oduce the notation
1
1
4r-^+-+sC;
(13.75)
YD: -scl
(13.76)
vr,=! + "c.
\13.77)
Substituting
Eqs.13.75-13.7
into Eqs.13.73
and13.74gives
YIlV't+ \2Vi:
VclRb
Yr2v' r+Yr2Vi=O.
(13.78)
\13.79)
SolvingEqs.13.78
and13.79fot Vigives
-Yi/R,
-
YrY,-Yl
'
(r3.80)
Witl the current source 1s acting alone, the circuit shown in Fig. 13.25
ieduces to the one shown h Fig. 13.27.Here, yi and yi are the components of I,i and y2 resulting from Is. If we use t]te notation irtroduced in
Eqs.13.75-13.77,
the two node-voltageequationsthat descibe the circuit
lr]'Fig.13.27 are
Figure13.27l Thecircujtshown
in Fjg.13.25,
with
Is actingalone.
YrVi+YnVi=0
(13.81)
Yr2Vi + YtVi = 1,.
(13.82)
and
SolvingEqs.13.81
and13.82for yi yields
" =
E1
Y.Yr- yit *
(13.83)
13.3 Apptications 525
To find the component of f2 resulting from the initial energy stored in
the inductor (yi), we must solvethe circuit shownin Fig.13.28,where
YiVi' + Yr2Vi : -pls,
(13.84)
Yr2vi'+Y22v!=0.
in Fig.13.25,with
(13.85) Figore13.28A Ihe circuitshown
the energized
indlctoractingatone.
Thus
v \- ' :
\o
Y.tYr yiz
(13.86)
FIom the circuit shown in Fig. 13.29, we find the component of
y, (yi\ resulting from the initial eneryy stoied in the capacitor.The nodevoltageequationsdescribingthis circuit are
\rvi'
+ Y12vi"=yC,
(13.87)
Ylzvi'+Y22vi': -'tC.
(13.88)
figurc13.29A Thecircuitshownin Fig.13.25,rvith
theeneqjzed
capacitor
actingatone.
Solvingfor Yi' yields
v!'
(&r + Ylr)c
Y,tYzz Yiz
v.
(13.89)
The expfessionfor Y, is
vr:vi+ vi+vi' +vi"
-
(\tl R) ,,
- y2n'c
YIIYD
Y-2/s
Yh2
YD
Ytr
Y11Yn yrD'e
-ctYt-Y)
|
^
YtY)-Yi)
v.
(:1.90)
We can find y2 without usingsuperpositionby solvingthe two nodevoltage equations that describe tie circuit shown in Fig- 13.25.Thus
v
Y n v t+ Y t 2 V=2+ +
,
"
yC-!.
Y.U+Y.U:I--\C.
(13.91)
/.13.e2)
You should verify in koblem 13.43that the solution of Eqs. 13.91and
13.92for y2givesthe sameresultas Eq.13.90.
526
ThetapLace
Tnnsfomin CircuitAnaLysis
objective2-Know howto anaLyz€
a circaitin the 5 domainandbeableto transform
ans-dornain
sotutionto the
timedomain
13.8 The energysroredin the circuit shownis zcrc
at the instantthe two sourcesare turned on.
Ahswer:(a) (r00/3)c'r (100/3)e+'lu0)V;
(b)L(soJr, " (50J/. 3lr/(, \
(c) [soe'' - 5o€4']r(t v.
a) Find the comporentof o for I > 0 owingto
the voltagesource.
2A
b) Fhd the compoDentof t, for t > 0 owingto
the currentsource.
c) Find the expressionfor ? when I > 0.
NOTE: Also try Cha et Prablem 13.42.
13.4 TheTransfsrFuele{t**
The transf€rfuncrionis definedasthe s-domainratio of the Laplacetransform of the output (response)to the Laplace lransform of the inpul
(source).In computingthe transferfunction,we restrict our attentionto
circuitswhere all initial conditionsare zero.lf a circuii hasmultiple independentsources,
we canfind the transferfunction for eachsourceand use
superyositionto find the responseto all sources.
The transferfunctionis
"r"r:.H.
Definitionof a transferfunction ,r
(13.93)
wherc v(r) is the Laplacelral1slormof the oulput signal,aDd,Y(s) is ihe
Laplace lranslorm o[ the input signal.Note that the transfer function
dependsoD what is defined as the output signal.Consider.for example.
the seriescircuit sholvn in Fig. 13.30.If the current is defined as the
responsesignalof the circuit.
Figure
13.30/r Asenes,?lf
cncuit.
5C
H(s)
R+sL+]/sC
t z L C+ R C t + 1
\r3.94)
In de ving Eq. 13.9,1,
we recognizedthat l correspondsto the output y(s)
and vs correspondsto the input,Y(r).
Ifthe voltageacrossthe capacitorisdefinedasthe outputsignalofthc
circuitshownin Fig.13.30,the transferfunctionrs
rlG)
v
1/'C
R+sl,+l/sc
rrac+RCs+l
(13.95J
Thus,becausecircuits may have multiple sourcesand becausethe definition
of the output signal of interest can vary a single circuit can genente many
transfer functions, Remember that when multiple souces are hvolved, no
single transfer function can reptesent tlle total output transfer functions
associatedwith each sourcemust be combined using supeiposition to yield
the total response.Example 13.1 illustrates the computation of a transfer
function for knoM numerical values of 4 L, and C
Derivingthe Transfer
Fundionof a Circuit
The voltage source ?s drives the circuit shown m
Fig.13.31.The responsesignalis the voltageacross
the capacitor, rl,.
1000o
a) Calculatethe numericalexptessionfor thetransfer furction.
b) Calculatethe numedcalvaluesfor the polesand
zerosof the transferfunction,
1000r)
Figur€13.32 A Ther-domain
equivalent
circuitforthecjrcuit
shownjn Fig.13.31.
Solving for % yields
''
+ s000)Ys
1000(s
s2+ 6000s+ 25 x 106
Hencethe transferfunctionis
Figure13.31A Thecircuittor
txamph
13.1.
v"
uE
1000(s
+ s000)
Solution
s' +6000r+25x106
a) The first step itr finding the transfer tunction is to
construct the r-domair equivalent circuit, as
shown in Fig. 13.32.By definition, the transfer
function is the mtio of y,/ys, which can be computed from a single node-voltage equation.
Summing the curents away from th€ upper node
genemles
u"s
1000
250+ 0.05r 106
b) Thepolesol H(s) aretherootsof thedenominator polynomial.Therefore
pr =
3000 J4000,
ft:
3000+ j4000.
Thezerosof l/(.r) arethe rootsof the numerator pol],nomial;
thusF1(s)hasazeroat
zr :
5000.
528
TheLaptace
lransfornin CircuitAnatysis
Objective3-Understandthe definition and significanceof the transferfunction;be abLeto derivea trensferfunction
13.9 a) Dcrive the numericalcxpressionfor the
transferfunction %/1s for the circuit shown.
Answer: (a)
(b)
a(s) = to(s + 2)/(s/+ 2r + 1o);
p 1: - 1
+ j3, pz=
1- j3,
b) Give the nurncricalvalueof eachpole and
zeroora(s).
NOTE: Also try ChaptetProblem 13.19
TheLocationof Potesandzerosof lr(s)
For lincar lumped-paramctercircuits,H(r) is alwaysa rational funclion
oI r. Complex poles and Tcros alwatrsappcar in conjugale pairs lhe
poles of H(.f) must lie in thc left half of the r plane if the responseto a
boundedsource(one whosevalueslie wirhin sornefinite bounds) is to
be boundcd.ThezerosoI A(r) may fie h eilher the right half or thc lett
half of the .! planc.
in mind, lYe ncxt discussthe rolc
With lhesc generalcharacl€r'istics
plays
response
funclion We begin with thc
in determirirg thc
that 11(s)
for
finding
partial traclion expansiontechnique
l(/).
Fc!net'isn
13.5 {h* Transfer
in F*rtialFraeti**Fxpn*si*ns
From Eq. 13.93we car witc the circuit oulput asthe producl of the transfer lunclion and the driving iunction:
YG) : H(s)x(s).
(13.96)
We have alreadynoted thai -47(s)is a rational function ofr. Reference1()
Table13.1showsthat X(s) alsois a rationalfunclion ofJ for the excilalion
fuDclionsofmost interestin circuit analysis.
Expandingthe right-lund sjde of Eq.13.96into a sum of Pdr'lialtactions produccsa term for each polc of H(s) and X(r). Rememberfrom
Chapter12 that polesare lhe roots of the deDominatorpolynomial;zeros
are the rools of thc numerator polynomial.'fhe lerl11sgeneratedby thc
polcs of H(s) give rise to the transientconponent o[ the total response,
whcreasthe terms gencratedby the polesof X(s) give rise to the steadyresponse.wc mean the
slate componeDtoI lhc rcsponse.By steady-state
responsethat exislsatter the traDsientcomponertshave bccomenegligi
ble. Example13.2illustratesthesegeneralobservations.
13.5 TheTBnsfer
Function
in Partiat
FGctjon
Apansions 529
Analyzing
the Transfer
Function
of a Cir.uit
The circuitin Example13.1(Fig.13.31)is driven by
a voltage source whose voltage increases linearly
$rith time, namely,?rs: 50t (l).
a) Use the tmnsfer funclion to find ?.,,.
b) Identily the transient component of the
rcsponse,
c) Identily the steady-state componert oI the
response,
d) Sketch0oversusrfor 0 < t < 1.5ms.
(4000r+ 79.70")
r,, = [10\6 x 10 ae3moicos
+ 10r 4 x 10 al"(r)V.
b) The ffansientcomponentof 0, is
10\6 x 10-aerm'cos(40001+ 79.70').
Note that this term is generated
by the poles
( 3000+ j4000)and (-3000 j4000) of the
tansfer function,
c) Thosteadystatecomponent
ofthe rcsponse
is
Sotution
a) From Example13.1,
H(") =
Thetime-domainexpressionfor r, is
+ s000)
1000(s
s2+6ooos+25^tu6
The transfom of the ddving voltage is 50/s2;
therefore, t}le s-domain expression for the output voltage is
1000(s
+ 5000) s0
_.
%:G'+6ooor+25x1017'
(10r 4x104),0).
Thesetwo tems are generatedby the secondorderpole(K/s') ofthe drivingvolrage.
d) Figure 13.33showsa sketchof o. versust. Note
that the deviationftom the steady-state
solution
10,000t- 0.4mV is imperceptiblealter approximately1 ms.
Thepartialtaction expansion
of % is
v
Kr
16
s + 3000 j4000
l4
K 0 0 i0 + j K4 0 0z0 ' 7 K
-, s + 3
-;
z
We evaluate the coefficients (1, 1{2, and K3 by
usingthe techniquesdescribedin Section12.7:
1<r= 5\6 x 10-4
/79.'70';
(1(i: s\5 x 1001 12-J!,
1,2
10
8
6
4
2
K2 = 10,
K3-
4 x 10 4.
Figure13.33A Th€g€phofr, ve6us
r fortuampte
13.2.
ThetapLace
Transfom
in CircuitAnatysjs
objective4-(now howto usea circuit'stransferfunctionto ca(culate
the circsits imputseresponse,
unit step
response.
andsteady-state
response
to sinusoidat
input
13.10Find(a) theunit stepand(b) theunjt impulse
aesponse
oI lhe circuitshownin Assessmenl
Problem13.9.
Answer: (a) [2 + (1013)ercos
(31+ 126.87')]z(4
v;
'cos
O) r0.s4€ (31 18.a3'),Ov.
NOTE: Also try ChapterPrcblens 13.76(a)and (b).
13.11 The unit impulscrcsponseofa circuitis
-''cos(240r
P, \
r"ti
10.000P
wheretan d *.
a) Find the transterlunclion of the circuit.
b) Find the unil stepresponseof the circuit.
Answer: (a) 9600s/(r2+ 140s+ 62,500);
(b) 40e ?o/sin240r V.
observations
onthe Useof H(s)in CircuitAnalysis
Example 13.2clearlyshowshow the transferfunction H(r.) reiatcsto the
responseof a circuil through a partial fraction expansion.Holvever,the
exampleraiscsquestionsabout the practicalityofdriving a circuil with an
increasingramp voltage that generatesan incrcasingramp r'esponse.
Eventualiythe circuil componentswill fail under thc stressoI exoessive
voltage,and when that happensour linear model is no iotger valid.The
ramp responseis ofinterest in practicalapplicationswherethe ramp function increasesto a maximumvalue over a linitc lime inleNal. If the rime
takcn to reach this maximumvalue is long conrparedwith the time constanlso[ lhe circuit,the solutionassumingan unboundedramp is valid for
this tinile time interval.
We make two additionalobservationsrcgardingEq. 13.96.First, let's
look al the responseof ihe circuit duc to a delayedinpur. If the inpur is
delayedby a seconds.
Y.\:((t
a)u(.!
,)] = " ^xG).
and,from Eq.13.96,the responsebecomes
y(J) : H(r)x(r)e ,"
If l(r) : I
\73.91)
'{11G)x(r)}, rhen,fron Eq. 13.e7.
y(t
a)u(t - d) - e rIH(s)x(.s)e"'j.
(13.98)
Therefore,dela]'ingthe inpul by., secoDdssimply delavsthe responsc
functionby d seconds.Acircuil thal exhibilsthis characteristic
is saidto bc
time invariant.
Second.if a unit impulsesourcedrivesthe circuit.the responseof the
circuitequalsthc inverselransformofthe transferfunction.Thusii
.rG)= 60), then-r.(s): I
Y(r)= ,ry(t.
(13.99)
13.6 Tle TclsterFuldiooanothe(o,votLtio.Int"gr'
531
Hence,fromEq.13.99,
y(t) - h(t),
(13.100)
wherc the inverse transform of the tansfer function equals the unit
impulse rcsponse of the circuit. Note that this is also the natural response
of the circuit becausethe application of an impulsive source is equivalent
to hsfantaneouslystorh)genergyin the circuit(seeSection13.8).The subsequent release of this stoied energy gives Iise to the natural iesponse
(seePioblem 13.86).
Actually,the unit impulseresponseof a circuit,ft(t), contains€nough
inJormation to compute the responseto any souce that drives the circuit.
The convolution integFl is used to exhact the responseof a circuit to an
arbitrarysourceas demonstratedin the next section,
13.6 TheTransfer
Function
andthe
Integral
Convolution
The convolutionhtegral relatestle output y(t) of a lhear time invariant
circuit to the input r(t) of the circuit and the circuit's impulseresponse
ft(l). The integralrelationshipcanbe expressed
in two ways:
.y(r):
l'or^t,t,^ t a ^ : l *
n,, nl,l^ta^. {13.101)
We are interested in t}le convolution integral for several reasons.First,
it allowsus to work entirely in the time domain.Doing so may be beneficial in situations where r(t) and ,(t) are known only thrcugh experimental data, ln such cases,the transformmethod may be awkward or even
impossible,as it would requte us to computethe Laplacetransfom of
expedmentaldata. Second,the convolutioninte$al introducesthe conc€pts of memory and the weighting function into analysis.we will show
how the concept of memory enablesus to look at the impulse response(or
rhe weighting function) ,(t) and predict, to some degree,how closely the
output waveform replicates the input waveform, Finally, the convolution
integral provides a formal pioc€dure for finding the inverse transform of
products of Laplace transfoms.
We basedt]Ie de vation of Eq. 13.101on the assumptiontlat ttre circuit is linear and time invariant. Becausethe cir:cuit is linear, the principle
of superyositionis valid, and becauseit is time hvariant, the amount of
the responsedelay is exacdy the sameas that of the input delay,Now consider Fig. 13.34,in which the block containing fi(t) rcprcsents any linear Figuer3.34 a A blockdhgramofa generat
ci'cLrit.
time-invadant circuit whose impulse rcsponse is known, r(t repiesents
tlle excitation signal and J(t) iepiesents the desired output sigllal.
'We
assume that r(t is the general excitation signal shown in
Fig. 13.35(a).For conveniencewe also assumethat i(t = 0 for t < 0 .
Once you seethe derivation of t}le convolution integral assumingx(t) = 0
for t < 0', the exteNion of tlle integral to include excitation functions
tllat exist over all time becomesapparert. Note also that we pemit a discortinuity ir r(r) at the odgin, that is, a jump between 0 and 0-.
532
TheLaptaceTransfom
in CjrcuitAnatysjs
Now we approximate i(1) by a sedesof rectangular pulses of unifom
width Ai, as shownin Fig.13.35(b).Thus
.!(r)- ro(t +.!r(,) +.
+ i,(r) +
.,
(13.102)
where ri(l) is a rectangular pulse that equals r(,\i) berween ,\, and ,\,+r and
is zeroelsewhere.
Note thal the ith pulsecanbe expressed
in termsof step
lunctions; that is,
r(t - a(^,X,G- ^,) rlr
(I + ^,\)l].
The next step in the approximation of r(t) is to make A,\ small
enoughthat the ith componentcan be approximatedby an impulsefunction of strengthi(i')Al. Figure 13.35(c)showstle impulse represenration, with the strength of each impulse shown in bracketsbesideeach
arrow. The impulse representation of i(t) is
-r(t) = x(,\o)d,\6(t - ,\) + '(,\1)A,\60 - ,\1) +
5
s
tcl
Flguer3.35 Theexcitation
signatofx(r).
(a)A
genelaL
€xcitation
signat.
(b)Approxjmatjnq
r(r) witha
senes
(c)Appronmatjng
ofputses.
r(r) witha series
of
+ r(DAI6(t - ^) + ..
(r3.103)
Now when r(t) is represented by a series of impulse functions (which
occur at equally spaced intervals of time, that is, at ,\, ,\1,lr, . . .), tlre
responsefunctior )(1) consistsof the sum of a seriesof unifomly delayed
impulseiesponses.
The strengthof eachresponsedependson tle strength
of the impulse driving the circuit. For example,let's assumethat the unit
impulse response of the circuit contained it t]le box in Fig. 13.34 is the
exponentialdecayfunction shown ir Fig. 13.36(a).Then the approximarion ofy(l) is the sum of the impulsercsponsesshownin Fig.13.36(b).
Analytically, tle expressionfor l(1) is
)(r) - r(,\o)A.Ir(r-,\o) + r(Ir^^ti(
+ r(,\t^^rt
,\1)
,\t +
+ r(,l'j)A,in(t D +..
(13.104)
As AI+ 0, the summation
in Eq. 13.104approaches
a continuous
integlatio4 or
it^,,r,, - ^,)^^- /-.r(^)ft(r ^)d^.
(13.105)
Therefore,
t)-
(b)
Figure
13.36A Theapprcximation
ofy(r).(a)Ihe
jmpulse
response
oftheboxshown
in Fjg.13.34.
(b)summjng
theimpube
rcsponses.
|
x(^)h(t - ^)d^.
(13.106)
If r(l) existsover all time, then the lower limit on Eq. 13.106becomes
,x: thus,in geneml,
ttt: l',{\nt, t)at
(13.107)
11.6 TheTEnsfur
Function
andtheConvotution
Inteqrat 533
which is the secondform of the convolutionintegral givenin Eq. 13.101.
We dedve the first folm of the htegral ftom Eq. 13.107 by making a
change in the vadable of integmtion. We let ,, : I ,l, and then we note
Ihat du = -di,ll : -oo when ,\ : ca, and !r: +co when .d = -co.
Now we can write Eq.13.107as
4h(u)((1u),
tfO: l-,ft
:
y1t1
l*,(t
,7n@Xa,1.
(13.108)
t(r)
But becausea is just a symbol of integration,Eq. 13.108is equivalentto
the first form ofthe convolutionintegral,Eq.13.101.
The integral relationshipbetweeny(0, ,0), and r(r), expressedin
8a.13.101.often is written in a shorthandnotation:
y(t) : h(t) * x(t) : r(t) * ft(t),
(13.109)
where the asterisk signifies the integral relationsbjp between ft(t) and r(t).
Thusfi(t) * r(t) is readas"&(t) is convolvedvrith r(t)" andimpliesthat
h(1)' Y(t)'
/I h(^\Y(t ^)d^.
'/--
(a)
,(r)
M
0
(b)
whereasr(t) + ft(t) is readas"r(t) is convolvedwith h(t)" and impliesthat
(D. rv,
t'
I rl^)hQ ^)d^
'/a
11 0
The integals in Eq. 13.101give t}le most general rclationship for t]le
convolution of two functions However, in our applications of the convolution integral, we can changethe lower limit to zero and the upper limit to t.
Thenwe car write Eq.13.101as
t
r
Jn
,lo
y ( t )= l t ( t ) t ( r - ^ ) d ^ =
lx(^)h(t- ^)d^.
G)
.r(r - I)
(!.rrol
(d)
we change the limits for two reason$ Fftst, for physically realizable circuits,t(t) is zero for t < 0. In other words, there can be no impulse
responsebefore an impulseis applied.Second,westart measuringtime at
the instantthe excitationi(l) is turned on;thereforeiro) : 0 for t < 0-.
A gmphic interpretation of the convolution inte$als contained in
Eq.13.110is importantin the useof the integralas a computationaltool.
We begh with the fi^t integral. For purposes of discussion,we assume
that the impulse responseof our ckcuit is the exponential decay function
shown in Fig. 13.37(a) and that the excitation function has the wavefom
shownin Fig. 13.37(b).In each of theseplots,we replac€dt with ,[, the
symbol of integation. Replacing ,trwith -,\ simply folds the excitation
function over the vertical a-{is, and rcplacing i with t
i slides the
folded function to the dght. See Figures 13.37(c) and (d). This folding
l'(IF(r
I)
(e)
inteQetation
ofthe
tigure13.37A A grap.hic
jntesmt
convolutjon
/;n(^)-v( .\)d^. (a)rhe
(c)Ihe
(b)Theexcitation
function.
impulse
rcspons€.
folded€xcitation
function.(d)Ihe fotdedexcitatjon
tunctjon
disptaced
t unjtr.(e)Theprcduct
t(I)x( - ^).
534
Th€Laptace
I'ansform
in ti(uitAnaLysis
operation gives dse to the term conrolution. At any specified value of r,
the response function )(r) is tlle area under the product function
ft(,\)r0 - ,t), as showl] in Fig.13.37(e).It should be apparenrfrcm tlfs
plot why the lower limit on the convolution itrlegral is zero and t]le upper
limit is t. For ,\ < 0, tle product ,(i)"r(r - I) is zero because,(,\) is zero.
For,\ > r, the prcductl'(,l')r(t
l) is zero becausex(r I) is zero.
Figure 13.38shows the second form of the convolution htegral. Note
that tlle product furction h Fig. 13.38(e) confirms the use of zero for the
Iower limit and t for the upper limit.
Example 13.3 ilustrates how to use the convolution integral, in conjunction with the unit impulse rcsponse,to find the rcsponseof a circuit.
r'(I)
.x(^)
t( r)
l?(
r)
ll(r rF(r)
tigurc13.38 A gmphic
interyeiation
oftheconvotu(a)
tjonjnteqmL
^)'(,\)d,\.
The
impuke
/;n(r
(b)Theexcitatjon
r€sponr€.
function.
k) Therotded
impulse
response.
(d)Thefolded
imputse
response
displrc€d,
units.(e)Theproductr(r i)x(I).
13.6 TheTnnsfer
Function
andtheConvotution
Inteqrat 535
Usingthe Convolution
Integralto Findan outputSignal
,(r)
The excitation voltage ri for the circuit shown in
Fig.13.39(a)is sho*n in Fic.13.39(b).
a) Use the convolution integml to find o,.
b) Plot u, over the rangeofo < r < 15 s.
0
1H
6 ,n:
Y
G)
figure13,39L Thecncuit
andexcitation
vottage
tor
Exampte
13.3.(a)The
(b)Th€excitaijon
cjrcuit.
vottaqe.
Figure13.40 A lhe impukercspons€
andthetotd€dercjtation
function
for ExampLe
13.3.
Sotution
a) The fust step in using the convolution integral is
to find the unit impulse response of the circuit.
We obtain the erpression for % from ttre
r-domainequivalertof the circuitin Fig.13.39(a):
i(\)
u
%: -10).
w})en or is a unit impulse function 6(t),
r , 0- ^ )
%=h(t)=e'u(t).
fiom which
h(^) = e ^u(^).
Using the first form of the convolution integral
in Eq.13.110,weconstructthe impulser€sponse
and folded excirarion funcrion shoun in
Fig. 13.40,which are helpful in selecting the limits on the convolution integral. Sliding the
folded excitation function to the right requires
brcaking the integration into three intervals:
0 < I < 5 ; 5 < t < 1 0 ;a n d 1 0 < t < o o .T h e
breaksin t]le excitatior function at 0,5, and 10 s
dictate ttrese break points. Egure 13.41 sho\qs
ttre positioning of the folded excitation for each
of these intenals. The analytical exprcssion foi
0; in the time intenal0 < t < 5 is
t,.=4r,0<t<5s.
?]r(r
t)
20
Figlre 13.414 Thedjtptac€rnent
of oi(t
^) tor thrce
536
]he Laplace
in (ircuitAnaLysk
Transfom
Hence, the analytical er?rcssion for the folded
excitation function in the interval r - 5 <,1
<lis
Di(
i\=4(t-,1),
r
r, (v)
20
18
T6
14
1,2
10
8
5<I<1.
We can now set up the three integral
exprcssions
for oo,For 0 < 1< 5s:
": I'o'1' ;'1"^1^
2
= 4(e'+ r - 1)V.
Fors< t < 10s,
1).=
20e^d^ +
Jo
=4(s+e-t
e\l
l' ,a{,
t)" ^at
Figur€13.42 A ThevoLtage
rcsponse
ve6ustjmetol
Exampte
13.3.
') v.
Andforl0=r<oos,
trs
I
D": I 20e^d^+ I 4(t
Jr 1n
: 4(e-t
^\.^di
Jt-s
I
1,.47
2
4.54
8.20
12.0'7
16.03
3
e-r 5) + 5d (' 10))V
b) We have computed ?r. foi 1 s intervals of time,
using the apprcpriafeequalion.The resulfsare
tabulatedin Thble 13.2and showngraphicallyin
Fi9.13.42.
5
6
,7
18.54
19.56
19.80
19.93
10
ll
t2
1l
l4
15
19.9'7
7.35
2.70
0.99
0.37
0_13
NOTE: Asse.ss,our understandingof conolution by trying Chapter Problems 13.57and 13.58.
TheConcepts
of Memoryandthe WeightingFunction
,/(r - I)
rurure
luirrrrppenr| $
i---\
I "e
|
t
\ p c c r{ h r s h a o D ( n < J )
\
t-it,r:fiI-,-^
Figur€
13.43A Thepast,prcsenl
andfutur€
vatues
of
We mentioned at the beginning of this section thar the convolution integral introduces the concepts of memory and the weighting function into
circuit analysis.The graphic interyretation of the convolution integnl is
the easiestway to begin to grasp these concepts.We can view the folding
and slidingof the excitationfunction on a timescalecharacterizedaspast,
present, and Iuture. The vertical axis, over which the excitation function
'I(t) is folded,representsthe presentvalue;past valuesof i(1) lie to the
dght oI tne vertical axis, and future values li€ to the left. Figure 13.43
showsthis descriptionof r(t). For ilustrative purposes,we usedthe excitation functionftom Example13.3.
When we combinet]le past,present,and future viewsof.r(r
r) with
the impulse response of the circuit, we see that the impulse response
weightsr(l) accordingto prcsentand pasl values.For example,Fig.13.41
showsthat the impulseresponsein Example13.3giveslessweight to past
valuesol r(t) than to the prcsentvalue of r(t). In other words,the circuit
retainsless and lessabout past input values.Therefore,in Fig. 13.42,o.
quickly appmacheszero when the present value of the input is zero (that
13.7 TheTransfer
Function
andtheSteady-state
Sinusoidat
Response
is,when / > 10 s). In other words,becausethe presentvalue of the input
receivesmore weight than the past values,the output quickly approaches
the present value of the input.
The multiplication of r0 - i) by t(l) gives rise to the practice of
relerring to the impulse response as the circuit weighting ftrctioD. The
weighting function, ir turn, determires how much memory tlle circuit has.
Memory is the extent to which the circuit's responsematchesits input. For
example' if the impulse respome, or weighting function, is flat, as shov,n in
Fig.13.44(a),it givesequal weight to all valuesof r(t), past and present.
Sucha cbcuit has a pefect memory-However,if the impulseresponseis
an impulse function, as shown in Fig. 13.44(b),it gives no weight to past
valuesofi(t). Sucha c cuit hasno memory.Thus the more memory a circuit has,the more distortion tiere is between tie waveform of tlle excita,
tion function and the waveform of the responsefunction. We can show this
relationship by assuming that the circuit has no memory, that is,
,(t : ,a6(1),and ther noting from the convolutioninte$al that
/, =
Jah(^\x(t
Figurc
13.44A Weishtins
functions.
(a)turf€ct
nremory.(b)Nomemory.
^)d^
- n)d^
t",46(^)r(r
= Att\t).
(13.111)
20
18
t6
\/
1,2
Equation 13.111showsthat, if the circuit has no memory,the output is a
scaledrcplica of the input.
The c cuit shownin Example 13.3illustratesthe distofiion between
input and output for a circuit that has some memory. This distortion is
clear when we plot the input and output waveforms on the samegraph, as
inFip.13.45.
13.7 TheTransfer
Function
andthe
Steady-State
SinusoidaI
Response
Once we have computed a circuit's traNfer function, we no longer need to
peform a separatephasoranalysisof the circuit to detemhe ils steadystateresponse.Instead,we use the transferfunction to relate the steadystateresponseto the excitationsouice.Fi$t we assumethat
r(r) : ,4cos((,t + d),
(13.112)
and then we use Eq. 13.96to find the steady-statesolution of ),(r). To find
the Laplacetmnsformofr(t), we first write x(t) as
x(t) : ,4 cos(,t cosd
Asin.,,tsird,
(13.113)
8
2
810
12 14
, (s)
Figure13.45 A Theinputandouhut waveforms
for
Exampte
13.3.
538
lhe taptace
Transfom
in CncuitAnalysis
from which
x(") =
(,4cosd)s _ (,asind)o
s2+ ,&
s2+.2
,4(scosd o sind)
f+0.?
(13.114)
Subsrituting
Eq. 13.114
into Eq. 13.96gves the.r-domain
exprcssion
for
Y(s) = //G)
,4(,'cosd
&,sin d)
(13.115)
The number
We now visualizethe paitial ftaction expansionof Eq. 13.115.
of terms in the expa$ion depends on the number of poles oI H(r).
necause a(s) is not specified beyond being the transfer function of a
physicallyrealizablecircuit,the expansionofEq.13.115is
Ki
sila
KL
s-l,)
...
+ )
terms generatedby the polesoflt(s).
(13.116)
In Eq.13.116,thefiISt two termsresultfrom tle complexconjugatepoles
of tle ddving source; that is, ,r2 + r,,2= G - jr)G + jo). However, the
terms generated by tle poles of ll(s) do not contdbute to t}le steady-stat€
responseof y(t), becauseall tlese poleslie in the left half of the s plane;
consequendy,the corresponding time-domain tems apprcach zero as t
incleases.Thus the first two terms on the right-hatrd side of Eq. 13.116
determine the steady-state response.The problem is ieduced to finding
the partial fraction coefficient .Kr:
".-' :
H(r),a(rcosd- usin,r)|
,+i.
l,=,,
- aslr'O)
, H(ja)A(j@cos6
2ja
H ( J d ) ,( c o s d + j s i n 0 )
- 1--. ._,,
,
tHuu\Ac".
(lJ.l17l
In general, H(Jd) is a complex quantity, which we recognize by x'dting it
in polar form;thus
HAa') = H(io) eiat!').
(13.118)
Note ftom Eq. 13