3: LinearCircuitAnalysis3.0 Introduction3.1 LinearSystems3.2 Superposition3.3 Node-VoltageAnalysis3.4 Mesh-CurrentAnalysis3.5 EquivalentCircuits3.6 ThevininEquivalents3.7 NortonEquivalents3.8 SourceTransformations3.9 MaximumPowerTransfer3.10 EquivalentResistanceRevisited3.11 Examples3.12 Summary3.0IntroductionWhenacarpenter works onahouse or amechanic tunes anengine, theyeachhave manytools andvarious options forcompletingtheirtasks. Mosttasksarestraightforward: thecarpenterusesahammertopoundinnails, andthemechanicusesaspeciallydesignedwrenchtoremovetheheadgasket. Othertasksareslightlymoreambiguousandrequiremoreinformation.Considerthecarpenterworkingonadoorframe. Whetherhechoosestouseahandsaw,apowersaw,orapocketknifedependsontheprecisionrequired,thenumberofdoorframestobebuilt,andquitepossiblythetoolshehasimmediatelyavailable. Themechanicmayhaveevenmoretroubleindeterminingwhethertouseahalf-inchor13mmwrenchfortighteninganassembly.The analysis methods presented in this chapter will provide you with a toolbox capable of tackling almost any linear DC circuit.The various methods will invariably terminate at the same answer, but some are clearly more ecient in solving particular circuits.Our goalisto construct boththe toolboxand an understanding for what problems eachtooliseective. More importantly,thesetoolswillprovideyouwiththenecessarypreperationtoanalyzemorecomplexcircuits: thosecontainingdiodes,transistors,anddigitallogic.Beginningwithsimpleapplicationsof OhmsandKirchoslaws, wewill useequivalentimpedancesanddividerstolaythefoundationforourremainingcircuitanalyses. Largercircuitsusingbothcurrentandvoltagedividerswillcreatetheframework.Theproperties of basiclinear systems will allowus toperformanalyses bysuperpositionandNode/Meshequations, furthereshingoutouroptions. Thelastanalysisstepwillbetocreatesimplecircuitequivalentsforanylinearcircuit,thuscompletingourtoolbox. Finally, wewilladdressthevariousmethodsinageneralsettinganddemonstrateafewtricksthatcanbeusedtoviewtheproblemsmoreeasily.Wherepossible, wewill demonstratemultiplemethodsinanalyzingthecircuits: youneedonlypicktheonewithwhichyoufeelmostcomfortable. Thebenetofsolvingacircuitmultiplewaysliesintheabilitytocheckyouranswersyoucanbeprettywellcertainthatyouarecorrectwhenyousolvebycompletelydierentmethodsandobtainthesamesolution.WewillbeginwithasimpleexampletodetermineanoutputvoltagegivenaDCinputvoltageasshowninFigure3.1.Example: DeterminethecurrentI50travelingdownwardinthe50resistor.Figure1Weknowtwomethodssofartosolvethisproblem:121. Wemaydeterminetheequivalentresistancelookingtotherightfromthevoltagesourcetothenndthetotal currententeringthecollectionofresistors. Applicationofacurrentdividerthengivesthecurrentthroughthe50 resistor.Method1: Req= 25 + 50(30 + 20) = 50 IT=2V50= 40 mAI50=(30 + 20)(30 + 20) + 50IT= 20 mAFigure22. Wemaycombinethe20, 30, and50Ohmresistorsasanequivalentandthenndthevoltageacrosstheequivalent(sameasthevoltageacrossthe50 resistor). SimpleapplicationofOhmslawthengivesusthecurrent.Method2: Rcomb= 50(30 + 20) = 25 Vcomb=RcombRcomb + 25(2V ) = 1 VI50=Vcomb50= 20 mAFigure3In many ways, this is a very typical problem to solve: a circuit containing one source and perhaps four or ve resistors. Lets tryanother,moredicult,examplethatveriestheconservationofenergyprinciple: allenergydissipatedbyaresistorisgeneratedbythesource.Example: VerifytheconservationofenergyinthecircuitshowninFigure3.4.3Figure4WeuseacombinationofdividersandOhmslawapplicationstocalculatethecurrentsandthepowerdissipated.Currentcalculations:I10=1V10= 100 mA I15=1V(15 + 630) =1V20= 50 mAI6=3030 + 6 I15= 41.67 mA I30=630 + 6 I15= 8.33 mA I1V = I10 +I15= 150 mAPowercalculations:P1V = (1V )(150mA) = 150 mW P6= I26(6) = 10.41 mW P10= I210(10) = 100 mWP15= I215(15) = 37.5 mW P30= I230(30) = 2.08 mWAfullylabeledcircuitisshowninFigure3.5.Figure5Toverifythattheconservationof energyholds, comparethepowergeneratedbythevoltagesourcetothesumof powersdissipatedintheresisters.150 mW.= (10.41 + 100 + 37.5 + 2.08) mWNotice that there is an implicit assumption in this last step: we have shown that the powers are equal,but not the energy! Understeady-stateconditions,therateofenergyintoacircuit(thepower)isconstantandthereforetheconservationofinstantaneouspowerimpliesconservationofenergyoveranytime-scale.3.1: LinearSystemsOneofthemostcommonmodelsstudiedinengineeringisthelinearsystem. Thegenericlinearsystemischaracterizedbythepropertyof linearity, whichrequiresthattheoutputisaweightedsumof theinputs. Figure3.6showstheblockdiagramof abasiclinearsystem: forinputsx1, x2, ..., xntherearecorrespondingoutputsy1, y2, ..., yn.4Figure6Whilethisdenitionsoundstrivial,answerthefollowingquestion: isf(x) = x +1alinearsystem?Actually,no! PlottingthefunctiononasetofaxesasshowninFigure3.7,youwillseetheanswerjumpoutatyou.Figure7Ifwescaleaninput,sayx1,thentheoutputisthesamemultipletimesthecorrespondingoutput, y1. Likewise,ifweaddtwoinputsandprocessthemsimultaneously,sayx1 +x2,thentheoutputisthecorrespondingsumy1 +y2.Now, letsreturntothequestionof whetherf(x)=x + 1qualiesasalinearsystem: considerthethreevaluesf(0)=1,f(1)=2, andf(2)=3. If f(x)waslinear, thenf(0 + 1)=f(0) + f(1), but2 =1 + 2=3. Further, if f(x)werelinearthenf(2 1) = 2 f(1)whichisagaincontradicted. Theonlyconclusionisthatf(x) =x + 1isnonlinear,despitebeingaline;alineneednotbemathematicallylinear! Ifthelinehappenstopassthroughtheorigin,thenitwillbelinear.Combiningthepropertiesdiscussed,weobtainthesystemresponseyk= ck xkk Nsuccinctlystatedas: theoutputisaweightedsumof theinputs. Twoimportantcharacteristicsof thelinearsystemtonoticearethatazeroinputalwayscorrespondstoazerooutputandthattheweightingterms, ck,donotdependonanyoftheotherinputs.3.2: SuperpositionThe bridge between mathematical denitions of linear systems and electrical circuits is a circuit technique called superposition.Wewill considermanycircuitsinthischapter, all of whicharelinearsystems: theinputsof thecircuitasalinearsystemarevoltageandcurrentsources, andtheoutputsareanyof thenodevoltagesorcurrentsinthecircuitelements. Aprocesscalledsuperpositionentailsactivatingonesourceatatimeandzeroingall others, allowingustodeterminetheresponseofthelinearsystem (nd eachck). Adding the contributions from each of these voltage and current sources, we will then be able to obtain theoverall responseofourlinearcircuits. Thelimitationwill bethatwecanonlyusesuperpositiontodeterminelinearquantities:power calculations (quadratic), diode calculations (exponential), and transistor analyses (mixture) will require additional methods.Example: Usesuperpositiontodeterminethecurrentinthe5resistorshowninFigure3.8.5Figure8Uptonow, wewouldhaveattackedthis problembysomecombinationof Ohms lawcalculations or voltageandcurrentdividers. Withtwopowersources, wehavetotrysuperposition. Westatedabovethatif wezerotheothersources,wemaycalculatetheresponseof eachsourceindependently. Statedmoresimply, wecangureoutthecontributionsof eachsourcetotheoverallresponseI5.So,howdowezeroasource? Firstconsideranidealvoltagesource: ithasaxedvoltageandwillallowanycurrenttoowthrough it. The only circuit element we have that will allow any current to ow through it, but is guaranteed to have zero voltageacrossitsterminalsisashortcircuit. Tozeroacurrentsource, wefollowthesameprocess: acurrentsourcetransmitsaxedcurrent independentofthevoltageacrossit,so a zero current source would beidenticaltoanopencircuit,whichisguaranteedtostoptheowof current(thereforezero)independentof voltage. Wemayalsoobtainthesezeroedsourcesfromtheinternalresistancesof theideal voltageandcurrentsourcesasderivedinChapter2: avoltagesourcehaszerointernal resistanceandthuszerostoashortcircuit,whileacurrentsourcehasinniteinternalresistance,corresponidngtoanopencircuitwhenzeroed.Figure3.9showsagraphicalrepresentationofzeroedsources.Figure9ReturningtotheexampleinFigure3.8, wendthatbyzeroingthesourcesoneatatime, weobtaintwoseparatecircuits,showninFigure3.10,correspondingtotwoseparate(andlinear)contributionstocurrentinthe5resistor.Figure10Thecontributionsfromeachsource,I53AandI59V ,maythenbeaddeduptoobtaintheactualcurrent,I5.I5= I59V+I53A=(9V )(10 + 5)+1010 + 5 (3A) = (0.6A) + (2A) = 1.4 AWhataboutpowercalculations?Powerisrelativetothesquareofeithervoltageorcurrent(whichweknowtobothbelinearquantities). Provetoyourself thatsuperpositionmaynotbeusedtodirectlycalculatethepowerinacircuit; thatis, compareI25(5) and (I253A +I259V )(5). The correct calculation of power always depends on the net current,I5, not the contributions from6eachsource.Example: UsesuperpositiontodeterminethevoltageacrosstheresistorR1inFigure3.11.Figure11Repeatingthesameprocessasbefore,wezeroallbutonesourcetocalculatethecontributionfromeachvoltageandcurrent;Figure3.12showsthecircuitwithVxzeroed.Figure12VR1, Ix= Ix Req1Req2,where Req1= R3(R4 +R5 +R6) and Req2= R1R2RepeatingtheanalysisforthecontributionofVx, wezerothecurrentsourceandreplaceitwithanopencircuitasshowninFigure3.13.Figure13VR1, Vx= (Vx) Req2Req1+Req2where Req1= R3(R4 +R5 +R6) and Req2= R1R2Thetotalvoltage,VR1,canbewrittenasthesumofcontributionsfromeachsource.7VR1= VR1, Ix+VR1, Vx= Ix Req1Req2+ (Vx) Req2Req1+Req2= Ix R1R2R3(R4 +R5 +R6) Vx R1R2R1R2 +R3(R4 +R5 +R6)Example: Solveforthevoltageacrossthe50resistorinFigure3.14.Figure14Firstconsiderthe20Vsource.Figure15V5020V= (+20 V ) 70(20 + 50)70(20 + 50) + 155020 + 50= 10 VNextconsiderthe2Asource.Figure16V502A= (+2 A) [50(20 + 7015)] = 39.29 Vandnallyconsiderthe10Vsource.8Figure17V5010V= (+10 V ) 5050 + 1570 + 20= 6.07 VAddinguptheindividualcontributions,weobtainthetotalvoltage,V50.V50= V5020V+V502A+V5010V= 55.36 VWhatif wethrowinatwisttothelastproblem? Wendoutthatthecircuitwestartedwithwasdrawnincorrectly: theengineerspecieda25Vsourceinplaceofthe10Vsource. Doweneedtore-dotheentireprocess? Notachance! Insteadofrecalculatinganycircuits, wecandeterminetheratioofchange(25V10V=2.5)inthevoltagesourceinputand, bylinearity, carrythatthroughtotheoutput. Therefore, thenewresponseV5025Vis2.5 V5010V=15.18 V . Thiseasycorrectionisoneof thebenetsofsuperpositionoverotheranalysismethods.Thenalexampleofsuperpositionwillhavesourcesthatcanbecombined(seriesvoltagesourceandparallelcurrentsources)onceanothersourceinthecircuitiszeroed. Theanalysiswill workjustaswell consideringeachsourceindividually, butwillproceedquickerbycombiningtheminasmartfashion.Example: UsesuperpositiontosolveforthevoltageVxacrossR1inFigure3.18.Figure18Ifwezerobothofthevoltagesources, weendupwithtwoparallelcurrentsources, whoseequivalentissimplythedierenceofI1andI2.9Figure19Likewise,whenwezerobothcurrentsources,thereisasingleseriesloopcontainingtwovoltagesources.Figure20BysolvingthecircuitsinFigures3.19and3.20, wereducetheworkloadbynearlyhalf (twocircuitsinsteadof four). Theoverallsolutionis:Vx= VxI1I2+VxV2V1= (I1 I2) R1(R2 +R3) + (V2 V1) R1R1 +R2 +R3Toconcludethesection,westatethegeneralstepsnecessarytosolveanycircuitusingsuperposition.1. Establishtheunknownsvoltagepolarityorcurrentdirection(andkeepconsistentthroughouttheanalysis).2. Redrawacircuitcontainingkvoltageorcurrentsourcesintokseperatecircuitswhereonlyonesourceisactiveandallothersarezeroed. Remember: zerovoltagesourcesareshortcircuitswhilezerocurrentsourcesareopencircuits.3. SolveforeachsourcescontributiontotheoverallvoltageorcurrentusingOhmslawanddividersasappropriate.4. Addtheindividualcontributionstoobtainthedesiredoverallquantity.A nal comment on superposition: nearly every mainstream textbook on linear circuit analysis has a disclaimer that you cannotperformsuperpositiononcircuitscontainingcontrolledsources. Thisnotionisfalse!1Wewill routinelyusesuperpositionwhenconsideringoperational amplierandtransistorcircuits, treatingall controlledsourcesasindependent(non-controlledsources)andrectifyingthealgebralater.OtherLineartechniquesAnother consequence of circuits as linear systems is that we may write systems of equations and employ linear algebra to solve.TheequationswillultimatelytintoOhmslaw: wewillre-writethetraditionalmatrixequation[A]x =

baseither

V= [Z]

Ior[Y ]

V=

I,where[Z]isthematrixofimpedances([Y ]isthematrixofadmittances)relatingvoltagestocurrents. Thesamebasicrulesfromlinearalgebraapply: thematrix[Z]or[Y ]must beinvertible(have anon-zero determinant)forthereto beasolution,1Leach: something10andyouneedasmanyindependentequationsasunknownstofullysolve.3.3: Node-VoltageAnalysisNode-voltageanalysisexpandsKirchoscurrentlawtowriteequationsateachnodeinacircuit, andthensolvewithlinearalgebra(ideallyusingacomputer). Wealsodrawonthefactthatvoltageisaconservativeeld, whosepotentialscanbetakenrelativetoanarbitraryreferencepoint; thuswewill beallowedtodeneagroundnodeas0V(typically, wechoosethenodethathasthemostconnectionstoothernodesinanattempttosimplifytheequations).Considertherstcircuitthatweanalyzedusingsuperposition.Example: Usenode-voltageanalysistosolveforthecurrentthroughthe5resistorinFigure3.21.Figure21The most basic step is to dene the nodes: we will rst choose the ground node to be the one at the bottom since it is commontothemostothernodes, nodeAasaconnectingpointbetweenthetworesistorsandthecurrentsource, andnallynodeBastheinterfacebetweenthevoltagesourceand10resistor.Figure22Beextremelycarefultonoticethatasinglenodemayhavemanypointsandconnections;oftentimesstudentswillerroneouslydenetwodierentnodesinplaceofnodeA,oneabovethe3Asourceandanotherabovethe5resistor. Thosetwopointsaredirectlyconnectedbywire,whichconductscurrentfreelyand,moreimportantly,willnotallowadierenceinpotentialacrossit.Thevoltageatthepointabovethe3Asourceisalwaysidenticaltothevoltageabovethe5resistor,sowedeneitasasinglenode. Similarly,thegroundnodehasthreeconnections,yetisstillasinglenode.Figure2311Wewill writetwoindependentequationsrelatingtheunknownvoltagesatnodesAandBtotheinputsourcesandpassivecomponents. First,writingaKCLequationatnodeA(assumeallcurrentstobeleavingthenode),weobtain:3A+VA 05+VA VB10= 0Normally,wewouldtrytowriteasecondequationatnodeB,butwendourrstproblem: wedonotknowthecurrentintoanidealvoltagesource(thepotentialisxedcompletelyindependentofthecurrentowingthroughthesource).VBVA10+I9V = 0I9V =???Thereisawaytobypasstheindeterminatenatureofthecurrentthroughthevoltagesource: simplyrealizethatKCLworksfor arbitrary black boxes (linear) as well, and thus state that the current entering the voltage source is equal to the current leavingthevoltagesource.Figure24Theactualnameforthistreatmentofthevoltagesourceisthatofasupernode. Writingthenewequation,weendupwithanotherKCLequation.VBVA10+0 VA5+3A = 0Simplifyingthisequationshowsanotherinterestingfalsestarttonode-voltageanalysis: thisequationisexactlythesameasthe one written at node A! Multiply both sides by a (1) and you get the other equation. If the two equations are the same,thentheyarealsolinearlydependent;tosolvealinearsystemconsistingoftwounknowns,youmusthavetwoindependentequations(ingeneral,thenumberofindependentequationsmustequalthenumberofunknowns). WeveexhaustedthenodesatwhichtowriteKCLequations,butwedoknowaverysimpleequationrelatingthetwonodeshavingthevoltagesourcebetweenthem.VB0 = 9VThus, taking the voltage source as a xed potential between two points, we obtain the nal equation. Writing the two equationstogether,wecanformalinearsystemofequations(i.e. amatrix)tosolve.15+1101101 0VAVB=3A9VNoticethattheelementsof thematrixaresimplythecoecientsof theunknownnodevoltages. Further, youshouldcheckthat the units of each element in the expression make sense (matrix multiplication being across a row of the matrix and down thecolumnvectoroftheunknowns).Oncetheequationsaresetintoamatrix, wehaveaslewof choiceshowtosolve: Gaussianelimination, Cramersrule, orinverting the matrix. In general, you should be well versed in solving linear equations either by hand or with the aid of computationsoftware,sowewillnotcoverthosehere.2Evenso,wearenotquitenished. Whenaspecicvariableisrequested(weaskedforthedownwardcurrentinthe5 resistor),wemustsolveforthatquantityintermsofthe,nowknown,nodevoltages.

V=VAVB= [Y ]1 I=7V9VI5=VA5 = 1.4 A2AgreatreferenceonlinearalgebraforsolvinglinearsystemsofequationsbyhandisApostol: LinearAlgebra12Example: Usenode-voltageanalysistosolveforthevoltageacrossR1inFigure3.25.Figure25Westartoutbydeningthebottomleftnodeasground(sinceitiscommontothemostothernodes), andthenbywritingthevoltageequationatthesupernodecontainingthevoltagesource.VA +Vx= VBBefore writing any current equations,we will dene two equivalent resistances,and also collapse nodes C and D. The fact thatweareabletocollapsethenodesmeansthatwecanwriteadependentequationintermsofVAandVBforeachnode.Req1= R3(R4 +R5 +R6) Req2= R1R2VC=R5 +R6R4 +R5 +R6 VB VD=R6R4 +R5 +R6 VBThesimpliedcircuitisshowninFigure3.26.Figure26Usingthereducedcircuit(andasupernodeinplaceofthevoltagesource),wecanwritethelastequationbyinspection.VAReq2+VBReq1= IxPuttingeverythingtogetherinmatrixformat,wehave:

1Req21Req11 1!VAVB=IxVxTherequestedvoltageacrossR1isthesameasthevoltageacrosstheequivalentReq2,orVA.VR1= VA= Vx Req2Req1+Req2+Ix Req1Req2Example: Usenode-voltageanalysisagaintosolveforall of thenodevoltagesinFigure3.27. Specically, determinethevoltageaccrossthe50resistor.13Figure27The rst stepis to write the easyequations betweennodes having voltage sources betweenthem, andthencollect anyimpedancesintoequivalents.VA 0 = 20 VVD VC= 10 VSincewehavefourunknownvoltages,andsofartwoequations,weneedtowritetwoindependentKCLequationsasfollows.VBVA15 +VB070 +VBVC20 = 0VCVB20+VD 050 2 A = 0Wehaveappliedthesupernodeconditioninthesecondequation: thecurrentleavingnodeCintothevoltagesourceisthesameasthesumofthecurrentsleavingnodeD. Combiningthefourequationsintomatrixformat(wewill nowomittheunitssincetheycancel),wereducethelinearcircuitto[Y ]

V=

I.0BB@1 0 0 00 0 1 1115115+170+12012000 1201201501CCA0BB@VAVBVCVD1CCA=0BB@2010021CCANotonlydoessolvingthissystemvianodal analysisresultinthesamenodevoltagesthatweobtainedinthesuperpositionanalysis,buttheprocessndsallofthevoltagesatonce!

V=0BB@VAVBVCVD1CCA=0BB@20 V27.5 V45.36 V55.36 V1CCATrivially,wesolveforthevoltageacrossthe50resistor,V50= VD= 55.36 VAnycircuitparameterthatwewishtondisimmediatelyavailableasafunctionof thenodevoltages. Asanexercise, youshould verify the conservation of energy using the solution above (you will need to use the voltages across the 15 and 50 resistorstoobtainthecurrentsinthe20and10Vsources,respectively).Example: Usenode-voltageanalysistosolveforthevoltageacrossR1inFigure3.28.14Figure28There are three independent nodes (two of which form supernodes) and one dependent node (B). The resulting set of equationsare:VC0 = V1VA VCR1+VA VDR2 +R3+I2 I1= 0VD 0 = V2 VB=R3R2 +R3 (VA VD)withthematrixformat(weneedonlyconsidertheindependentequations).0@1R1+1R2+R31R11R2+R30 1 00 0 11A0@VAVCVD1A=0@I1 I2V1V21AThevoltageacrossR1isthenthedierenceofVAandVC.Summaryof thestepsforNode-VoltageanalysisThebasicstepsinsolvingacircuitvianode-voltageanalysisare:1. Establishagroundorreferencenode.2. Label remainingindependentnodesnodesaredependentif theycanbewrittenasavoltagedividertimesothernodevoltages.3. Foreachsupernode,writeanequationforthexedpotentialandcollapsethenode.4. Combineimpedanceswherepossible.5. Foreachremainingindependentnode, writeoneKCLequation. Ifyoualwaystakecurrentsasleavingthenode, thentheentriesinyourmatrixwillinpracticehaveaminimalnumberofnegativesigns.6. Combineall KCLequationsintothematrixformat[Y ]

V =

I andsolve. PossiblesolutionmethodsincludeGaussianeliminationbyhand,pluggingintoacomputer(ideal),orusingCramersrule.7. Afterdeterminingthevector

V ofunknowns,solveforthespecicvoltagesorcurrentsrequested.3.4: MeshCurrentAnalysisSimilar to node-voltage analysis, mesh-current analysis is a method used to completely solve a linear circuit, but by using meshorloopcurrentsastheunknownsinsteadofnodevoltages. EverywherethatwewroteKCLequationsbefore,wewillnowwriteKVLequationsbasedontheseunknowncurrents. Considerthefollowingexamples.15Example: Usemesh-currentanalysistosolveforthecurrentdownwardsinthe5resistor.Figure29Theloopcurrents Ia andIbrepresent imaginaryvortices of current: whiletheyarenot measurableinalaboratory, theyarequiteuseful fortheoretical analysis. Thedirectionthatthecurrentows(clockwise/counter-clockwise)isarbitrary, butasuggestionistotakeallloopcurrentstobeorientedclockwise(thematriceswillagainhaveaminimalnumberofminussigns).WewillbegintheanalysisbyseeingthatthecurrentIamustbeequalinmagnitude,butoppositeinsign,tothe3Acurrentsource.Ia= 3 AFor the other equation,we apply KVL around the second loop. The net current through the 5 resistor is the dierence of thecurrentsIaandIb;ifweareconsideringaclockwiseloopgoingupwardsacrossthe5resistor,thentherstsignofthevoltageacrosstheresistorispositivewhenthenetcurrentistakenupwards,orI5=Ib Ia. Ibistheonlycurrentinthe10resistor,andwearegoingrighttoleftforaclockwiseloop,sothenetcurrentisjustIbIb (10 ) + 9 V+ (Ib Ia) (5 ) = 0Puttingthesetwoequationstogetherinmatrixformat,weobtainamatrixequationoftheform[Z]

I=

V .1 05 10 + 5IaIb=39Whichgivesthesolution:

I=IaIb=3A1.6AWe use this solution to calculate the current in the 5 resistor. As solved for previously,the current downwards in the resistorisIa Ib.I5= Ia Ib= (3) (1.6) = 1.4 AExample: Usemesh-currentanalysistosolvefortheallofthecurrentsinthecircuitshowninFigure3.30. Also,determinethevoltageacrossthe50resistor.Figure30Weareunabletodirectlydeterminethevoltageacrossacurrent source,soareunabletowriteaKVLequationcontainingthe2 Amp source. Instead,we write a current equation relating the current source toIbandIc(notice thatIchas the same directionasthe2Asourceandthuscontributespositively,whileIbisintheoppositedirectionandcontributesnegatively).16Ic Ib= 2 AWethenwriteourKVLequationssuchthattheybypassthecurrentsource(alsocalledcreatingasuper-mesh).20 + 15 Ia + 70 (Ia Ib) = 070 (Ib Ia) + 20 Ib 10 + 50 Ic= 0Combiningthesethreeequationsintomatrixformat,weobtainanothersetoflinearequations,whosesolutionisthevectorofunknownmeshcurrents.0@0 1 115 + 70 70 070 70 + 20 501A0@IaIbIc1A=0@220101A=

I=0@0.5A0.893A1.11A1ATocomparethissolutiontoourothermethods,noticethatV50= 50 Ic= 55.36V justasbefore.Example: SolveforthevoltageacrossR1usingmesh-currentanalysis.Figure31Therststepistodeterminehowmanyequationwereallyneedoutofthefourpossiblemeshes: bycombiningtheresistorsintoappropriateequivalentsasbefore,wecanreducethenumberofmeshestotwo.Figure32Req1= R3(R4 +R5 +R6) Req2= R1R2Writingthetwoindependentequations,Ia Req2Vx +Ib Req1= 0Ib Ia= Ixwecansetuptheappropriatematrixequation,Req2Req11 1IaIb=VxIx17andnallysolve.

I=IaIb=

VxIxReq1Req1+Req2Vx+IxReq2Req1+Req2!Noticethatthissolutionyieldsthesameexpressionsaswhenwesolvedbysuperpositionandnodalanalysis.VR1= Ia R1R2=Ix Req1VxReq1+Req2 Req2= Ix Req1Req2Vx Req2Req1+Req2Example: Usemesh-currentanalysistosolveforthevoltageacrossR1inFigure3.33.Figure33Thetwocurrentequationswilleliminatetheleftloopandcombinethelargerloops.Ia= I1 Ib Ic= I2TheremainingequationisobtainedbyaclockwiseloopaboutthecurrentsourceI2.V1 +R1 (Ib Ia) + (R2 +R3) Ic +V2= 0Combiningtheseequationsandsolvingyetanotherlinearsystem,weobtainthecurrentsineachofthemeshes.0@1 0 00 1 1R1 R1 R2 +R31A0@IaIbIc1A=0@I1I2V1 V21A=

I=0B@I1I1R1+I2(R2+R3)+V1V2R1+R2+R3V1V2+(I1I2)R1R1+R2+R31CASummaryof thestepsforMesh-CurrentanalysisThebasicstepsinsolvingacircuitviamesh-currentanalysisare:1. Drawandlabelasmanymeshcurrentsasnecessary.2. Writeequationsforanymeshescontainingcurrentsources. Replacewithsuper-meshesasappropriate.3. Combineimpedanceswherepossible.4. Writeoneindependentequationforeachoftheremainingmeshes. Solvetheequationsbyhand,calculator,orcomputer.5. Determinethedesiredquantitiesusingtheresultingvectorofcurrentvalues.ComparisonofAnalysisTechniquesSofar, wehaveseenthreedierentanalysistechniquesforlinearcircuits. Voltageandcurrentdividersworkwell, butonlywhenthereisonepowersource. Superpositionreducestheanalysisof complicatedcircuitstosimpliedcircuits(withzero-ed18sources) where we can use dividers to calculate the contribution of each source to a single desired quantity. Node-voltage analysisuses KCL equations to determine all of the node voltages in a circuit, and mesh-current analysis uses KVL equations to determineall themeshcurrents, butneithersolvesdirectlyforspecicvoltagesorcurrents. Node/meshanalysesdohoweverprovidealltheinformationinthecircuitatonce. Withthisinformation,thequestionbecomes: howdoyouchoosewhichmethodtouseinsolvingacircuit?Wewill explorevariousfactorsthataectwhichmethodweuse: time, accesstocomputingtools, analysisofperturbations,and desired form of outputs are just a few to consider. Strictly speaking, analysis of circuits can take the form of any combinationof thesemethods. Oneremainingwaytosolvelinearcircuitsistocreatecircuitequivalents(containingbothimpedancesandpowersources),reducingthecircuitasmuchaspossiblebeforeanalysing.3.5: EquivalentCircuitsNow that we know the basic steps to analyze a circuit, what happens if we want to use the circuit in a system, especially if thatsystemchanges?Commondevicesasseeminglysimpleasanalarmclockmayconsistofhundredsofcircuitelements,soweneedamodelingtechniquethatallowsusreplacethelargecircuitsbymuchsmaller,andmucheasiertocalculate,equivalentcircuits.ThetwotypesofrepresentationsweintroducenowarecalledThevininandNortonequivalents. TheThevininequivalentofanylinearcircuitisdenedasavoltagesourceinserieswithasingleresistance(impedance)havingthesameterminalcharacteristicsastheoriginal circuit, whileaNortonequivalentisdenedbyacurrentsourceinparallel withthesameThevininresistance(impedance), still withidentical terminal characteristics. BoththeThevininandNortonequivalentscouldreplacetheoriginalcircuitwithoutusbeingabletotellthedierencefromoutsidethecircuit.Consider the proverbial blackboxthat has two output terminals andconsists of anyunknownnumber of linear circuitcomponents (sources, resistances, openor short circuits, capacitors andinductors, etc.). Fromthis box, we candetermineboththeThevininandNortonequivalentsbymakingthethreemeasurementsshowninFigure3.34.Figure34Thersttwomeasurementsarewiththeblackboxconnectedtowhateversourcesthatpowerthecircuitelements. Wemea-suretheopen-circuitvoltage, Voc, astheterminal voltagewithnothingconnectedtotheoutput(hencetheopencircuit). Theshort-circuitcurrent, Isc, isobtainedbyplacingashortcircuitacrosstheoutputterminalsandmeasuringthecurrentowingthroughit. Thenal measurementistozeroall of thesourcesinsidethebox(forareal device, unplugtheboxfromthewallsocket)andthenmeasuretheequivalentresistance, Rth(impedancedenotedZth), seenlookingintothetwoterminals. ThesevaluesmaythenbeusedtoconstructeachofourThevininandNortonequivalentcircuits.3.6: ThevininEquivalentsAsstatedpreviously, thegoal of solvingforaThevininequivalentistoreducealargecircuitintoanexternallyequivalentcircuitconsistingofonlyasinglevoltagesourceinserieswithanimpedance.19Figure35Thevalueof thevoltagesourceis equal totheopen-circuit or Thevininvoltage, Voc, andtheimpedanceis theThevininimpedance,Zth.Example: Calculate the Thevinin or open-circuit voltage and the Thevinin impedance to the left of the 50 resistor in Figure3.36. VerifythattheequivalentcircuitproducesthesamevoltageV50asbefore.Figure36Therststepistodesignatewhatisinsidetheblackboxwearetryingtomodelandwhatisoutsidethebox. Asdesignatedontheschematic,wearelookingfortheThevininequivalenttotheleftofthedashedline. Thisspecicallymeansthatthe50 resistorisnottobeconsideredpartofthecircuitformodelingpurposes,resultinginatrueopencircuitatthedashedlines. Tocalculatethisopen-circuitvoltage, wemayuseanyof thecircuitanalysistechniqueslearnedpreviously. The10 V sourcedoesnotcontributeanycurrenttotherestofthecircuit(duetoitsplacementinserieswithanopencircuit),andweareonlylookingforonespecicvoltage: superpositionwillbethebestchoice.ThethreecircuitsinFigure3.37showthedecompositionof Figure3.36intosub-circuitswithonlyonesourceactive. WecalculatethecontributionsfromeachsourcetothevoltageVocandaddthesecontributionstoobtainthetotalVoc.Figure37Thetrickiestofthethreesub-circuitsisthethirdone: sincenocurrentisabletoowthroughtheopencircuit,thereiszerocurrentinanyoftheresistors. Thisresultsinazerovoltageacrosstheresistors, whosepotential isthenaddedinseriestothepotentialdierenceofthe10V source. Forsimilarreasons,thereisnocurrentowinginthe20resistorintherstcircuit,andthusVocisthesameasthevoltageacrossthe70resistor.20Voc= Voc20V+Voc2A+Voc10V=7070 + 15 (20V ) + (2A) (20 + 1570) + (10V ) = 91.18 VTo calculate the Thevinin resistance (impedance), we zero all of the independent sources, and solve for the equivalent impedancelookingintothetwoterminals. Figure3.38showsthecircuitwithallsourceszeroed.Figure38Byinspection,Rth= (20 + 1570) = 32.35 Puttingthesetwoparameterstogether,weformaseriescombinationofVocandRthandreplaceeverythingtotheleftoftheoriginaldashedlinewithournewequivalent.Figure39Howdowecheckifthisnewequivalentcircuittrulyhasthesameeectastheoriginal?Reconnectingthe50 resistorasaloadontothenewcircuit,wecancalculatethevoltageV50,whichshouldbethesamevoltagewesolvedforinpreviousexamples.Figure40Usingasinglevoltagedivider,V50= Voc5050 +Rth= 55.36 VNowwhyintheworlddidwegothroughthatmuchworktoobtainthesameanswerasbefore? Consideracasewheretheleftsideofourcircuit(forwhichweobtainedtheThevininequivalent)isastereoamplierandthe50 resistorisacollectionof speakersthatwewanttohookup. Whathappensif wethenwanttoadd/removeaspeaker? Eachof theothernumericalanalysesperformedwouldbeuselessonceanimpedancevalueischanged,whereastheThevininequivalentmakesthecalculationeasyinanycase;simplyrecalculatethevoltagedividerattheend. Letstryafewmoreexamples,andthenwewillexplorethe21relationshipsbetweenmethodsinmoredepth.Example: UseaThevininequivalent circuit tosolvefor thecurrent throughthe5resistor inFigure3.41. Repeat thesolutionwiththe5resistorchangedto25.Figure41Althoughsmaller, thiscircuitishardertoconceptualize. Wemayredrawthecircuitwithoutthe5resistorandstretchoutthenodesforbettervisualization.Figure42SolvingforVocbysuperposition,andZthbyzeroingallofthesources,weobtainthevaluesfortheThevininequivalent.Voc= 9V+ (3A) 10 = 21V Zth= 10Tocalculatethevoltageacrossaloadresistorconnectedtotheoutputnodes,weattachanarbitraryload,Rload,asshowninFigure3.43,anduseOhmslawtogetthecurrent.Figure43IRload=21V10 +RloadForloadresistorsof5and25,wecalculate1.4Aand0.6A,respectively.Somecircuitssimplyhavenoeasysolutionwiththeprevioustechniquesdiscussed. ThevininandNortonequivalentsprovideanewmethodthatworkswhereothersfail.Example: ConsiderthebridgecircuitinFigure3.44: determinethevoltageacrossR5.22Figure44Initially, theoptions lookrather bleak; node-voltageanalysis will work, but thesolutionwill beextremelytedious. If werememberthatpotentialsarethesameanywherealonganode, wecansimplifythecircuitbysplittingthevoltagesourceintotwoparallelsourcesofthesamevalue,showninFigure3.45.Figure45Withidenticalpotentialsatboththetopandbottomnodes,wecanseethattherewillbenocurrentowingatthetopofR1andR2orbelowR3andR4. Therefore,removingtheconnectionswillnotaectouranalysisofthecircuit.23Figure46Nowwereinbusiness: takeThevininequivalentstotherightandleftof R5, andthenreplacetosolveforVR5. BothsidessolveasinFigure3.47.Figure47TheresultingcircuitwithThevininequivalentsinsertedshowsasingleseriesloopfromwhichtocalculateVR5.Figure4824VR5= V [R3R1 +R3R4R2 +R4] R5R1R3 +R5 +R2R43.7: NortonEquivalentsANortonequivalentistheotherequivalentwherewereduceanylinearcircuittoasinglecurrentsourceinparallel withanimpedance.Figure49ThecurrentistermedeithertheNortoncurrentortheshort-circuitcurrent, andtheimpedanceisagaincalledtheThevininimpedance (for reasons we shall see). The short-circuit current,Isc,is calculated by externally applying a short circuit to the twonodesandthenmeasuringthecurrentacrosstheshort. TheThevininimpedanceiscalculatedthesamewayasbefore: zeroallindependentsourcesandndtheequivalentimpedancelookinigintothetwoterminals.Example: UseaNortonequivalentcircuittocalculatethevoltageacrossthe50resistorinFigure3.50.Figure50LetsndtheNortonequivalenttotheleftofthe50resistor. Wewillsolvefortheshortcircuitcurrent(viasuperposition)owingbetweenthetwooutputnodes.25Figure51Someofthedividertermsaretrickyatrstglance. Intherstcircuit,calculatethecurrentintothe20resistorasadividerbetweenthe20and70resistors; noticethatthetworesistorsareinparallel duetotheadditionoftheexternal shortcircuit.Whencalculatingtheequivalentimpedanceforthetotalcurrentoutofthe20V source,weconsidertheimpedancelookingrightfrom the voltage source,not left into the equivalencynodes. The secondcircuit is easy: current takes the path ofleast resistance,soallofthecurrentgoesthroughtheshortcircuit. ThethirdcircuitisasingleapplicationofOhmslaw.Isc=7020 + 7020V(15 + 2070) + (2A) +10V(1570 + 20) = 2.818 ANow,wezeroalloftheindependentsourcesandcalculatetheThevininimpedance.Zth= 20 + 1570 = 32.35Puttingthesetwotermstogether,weredrawastheNortonequivalentcircuit.Figure52Toverifyourcalculations,wendthevoltageacrossthe50resistor.V50= Isc (Rth50) = 55.35 VWeagainseeanimmediatebenetoftheequivalentcircuit: responsesmaybeeasilycalculatedforanyloadresistor!Example: CalculatetheNortonequivalentofthecircuitshowninFigure3.53,totherightofR1,andthenustheequivalenttodeterminethevoltageVR1.26Figure53Wereducethecircuitbytakingtheequivalentresistanceoftherightside,andbeginusingsuperpositiontosolvefortheshortcircuitcurrent,followedbytheThevininimpedance.Figure54SolvingfortheNortonequivalentvalues:Isc= Ix +VxReq1Zth= R2Req1TheresultingNortonequivalentwithR1reattachedisshowninFigure3.55.Figure55WewillsolveforthevoltageVR1inordertoverifyequivalency.VR1= Isc (RthR1) = (Ix +V xReq1) (R1R2Req1) = Ix Req1R1R2 + (Vx) R1R2Req1+R1R2Example: UseaNortonequivalentcircuittosolveforthecurrentthroughthe5resistorinFigure3.56.27Figure56Replacingthe5resistorbyashortcircuitinFigure3.57,wemaycalculateourtwoparameters: IscandZth.Figure57Isc= (3A) +9V10= 2.1 A Zth= 10 Replacingthe5resistorandcalculatingacurrentdivider,weobtainthecurrentI5.I5=1010 + 5 (2.1A) = 1.4AExample: UseaNortonequivalentcircuittosolveforthevoltageacrossR1inFigure3.58.Figure58Usingsuperpositionagain(withashortcircuitinplaceof R1), weobtaintheexpressionsforshortcircuitcurrent(orienteddownward)andThevininimpedance.Isc= I1 I2 +V2 V1R2 +R3Zth= R2 +R328AndthennallysolveforthevoltageVx.Vx= Isc R1(R2 +R3) = (I1 I2) R1(R2 +R3) + (V2 V1) R1R1 +R2 +R33.8: DualityofEquivalents: SourceTransformationsWehavesofar derivedtwoentirelydierent methods for modelingalargelinear circuit as asimpleequivalent;if bothmethodsproducecircuitsequivalenttotheoriginal,thentheyshouldalsobeequivalenttoeachother. ConsiderthetwocircuitsinFigure3.59,whichwewilltaketobetheThevininandNortonequivalentsofyetathirdcircuit,notpictured.Figure59Bytakingthe Thevininequivalent of the Nortonequivalent, andlikewise takingthe Nortonequivalent of the Thevininequivalent,weshallndadirectrelationbetweenthem.Rth= ZthVoc= ZthIsc Isc=VocZthThustondbothequivalentcircuits, weonlyneedtondoneoftheequivalentsandthenconverttotheother. Inpractice,thispropertyallowsustobuildinanothervericationsteptotell uswhetherouranalysisiscorrect. Theeasiest, andtherebytheleastlikelytogenerateerrors,ofthethreequantities(Voc, Isc,andZth)istheequivalentimpedance. Asaresult,thesurestmethodistocalculatebothVocandIscandthencomparetheratioVocIsctothecalculatedvalueofZth.Convertingasinglevoltagesourceinserieswithanimpedancetocurrentsourceinparallelwiththesameimpedanceiscalledasourcetransformation. Thistransformationcanoftensimplifycircuitsdramatically; letsagainconsiderthecircuitshowninFigure3.60forwhichwefoundThevininandNortonequivalents.Figure60We stillwant to nd the Thevinin/Norton equivalentto the left of the dashed line,but we are going to systematicallyperformsource transformations from the far left until we have completed the equivalent. The rst step will be a source transformation fortheseriescombinationof20 V voltagesourceand15 resistor,showninFigure3.61.29Figure61TakingtheNortonequivalentof justthe20Vsourceandthe15resistor, wecalculateashort-circuitcurrentIsc =20 V15 =1.333 A,andthesame15 impedance. Substitutingthisequivalentintothecircuit,weobtaintheoneshowninFigure3.62.Figure62Wenowhavethe15and70 resistorsinparallel,creatinganequivalentimpedanceof12.35 . Reversingthesourcetransfor-mationprocedurewiththenewimpedance(takingaThevininequivalentofthe12.35resistorinparallel with1.33Asource),weobtainthecircuitshowninFigure3.63.Figure63Absorbingthe20 resistorandtakingonemoresourcetransformation, weseeanewsimplication: twocurrentsourcesinparallel.30Figure64Thetwocurrentsourcesadddirectly, soweareineectabsorbingthecurrentsourceintoourmodel. Transformingonelasttime,weobtaintwoseriesvoltagesources(whichadddirectly)andasingleimpedance,whichistheoverallThevininequivalentandthesameequivalentcircuitweobtainedpreviously.Figure65Weshallconsideronemoreexampleofusingsourcetransformationstoreduceacircuit.Example: DeterminetheNortonequivalenttotherightofR1inthecircuitshowninFigure3.66.Figure66Usingthesameequivalentresistance, Req1=R3(R4+ R5+ R6), asbeforeandasingletransformationtotherightof thecurrentsource,wenearlyendupwiththeoverallNortonequivalent.31Figure67Combiningparallelcurrentsourcesandresistances,weobtainthesameequivalentasbefore,butinonlytwosimplesteps!Onewordofcaution: equivalentcircuitsaswithanyotherequivalentarenotthesamethingastheoriginal. Theequivalentcircuitsarewonderful mathematical toolsformodelingasystem, butthisconveniencedoescomeatacost: trackingvariablesinternal to the collapsed circuit is nearly impossible. The equivalent circuit derived at one two-port junction will be dierent thananothertwoelementsaway. BynomeansdoesthisrealizationmakeThevininorNortonequivalentsunimportantwesimplyseethatNorton/Thevininequivalentsaresuitableforsomeapplicationsandnotothers.Ingeneral,thestepsusedtondtheThevininandNortonequivalentsofacircuitare:1. Separatethoseelementsthatarepartoftheequivalentandthosethatareexternal. Clearlylabel thepolarityof Voc andthedirectionofIsc,makingsuretheylineupwiththepolarity/directionofthedesiredequivalent.2. SolveforVocandIscusinganymethoddesired: superposition,nodalormeshanalysis,etc.3. Atanypoint,youmaywishtoemployasourcetransformationtoreducethecircuittoasimplerform.4. Replacethelargercircuitbyitscalculatedequivalentandproceedwithanyexternalloadcalculations.3.9: MaximumPowerTransferWe have stated that both Thevinin and Norton equivalents can be used to model a linear circuit for use in calculating responsesduetomanyloads. Themostcommongoalinelectronicsistotransferinformationwiththeleastlossinthesignal: incircuits,thisgoal translatestodeliveringthemaximumamountof powerpossibletoaload. Howdoweachievesuchmaximumpowertransfer?ConsideraloadedThevininequivalent,whereboththeThevininimpedanceandtheloadarepurelyreal(resistors),asshowninFigure3.68.Figure68Thepowertransferredtotheloadisaproductof itscurrentandvoltage(otherformsgivethesameresult). Wemaywritethispoweras:Pload=VthRth +RloadVth RloadRth +Rload32Tomaximizethis power relation, wedierentiatewithrespect toRload, andsolvefor theconditiononRloadtomakethederivativezero.ddRloadPload= 0 = Rload= RthWendthatthemaximumpowertransferoccurswhenRload=Rth; or, equivalently, maximumpowertransferoccurswhentheloadresistancetoanydeviceisequal totheThevininresistanceof thedrivingcircuit(forgeneral impedances, maximumpowertransferoccurswhenZload= Zth).To verify this condition onRloadqualitatively,consider again the power delivered toRloadas the product of the voltage acrossitandthecurrentthroughit. IfRloadistoosmall,thenthevoltageacrossitisalsosmall,whileifRloadistoolarge,thecurrentthroughbothRthandRloadwill bemadesmall. Somewhereinbetweenweobtainthemaximumof thisproduct, asshowninFigure3.69.Figure69ThevalueofthemaximumrealpowerthatacircuitcandelivermaybewrittenusingtheThevininresistanceandeithertheshort-circuit current or the open-circuit voltage. (Notice that for Rload= Rth, the value of both the voltage divider for the voltageacrosstheloadandthecurrentdividerforthecurrentthroughtheloadareequalto12.)Pmax=V2oc4Rth=I2scRth4Wewillreconsiderthislimitonpowertransferforgeneralimpedanceslateron.Example: Solvefor theloadresistancethat maximizes thepower transfer fromthecircuit inFigure3.70. What is themaximumpowerthatcanbetransferred?Figure70Solution: Tosolvefortheappropriateloadresistance, weonlyneedtondtheThevininresistanceofthecircuit, andthenspecifytheloadresistancetohavethesamevalue.Rload= Rth= 15 + 630 = 20 To determine the amount of power delivered to the load, we need to calculate either the open-circuit voltage or the short-circuitcurrent. Theexpressionsforbothareshown.33Voc= (12 V ) 3030 + 6+ (1 A) (630) = 15 VIsc=(12 V )6 + 15303015 + 30+ (1 A) 630630 + 15= 0.75 AWiththesevaluescalculated,wedeterminethepowerdeliveredtoa20 loadresistor.P20 =(15 V )24 20 =(0.75 A)2 20 4= 2.81 WExample: What is the value of R in Figure 3.71, assuming that the 10 load resistor was chosen for maximum power transfer?Howmuchpowerisdelivered?Figure71TodetermineR,ndtheThevininresistanceandsetitequaltothe10 toensuremaximumpowertransfer.Rth= R + 2 + 0 = 10 = R = 8 Noticethatzeroingthevoltagesourcesalsoshortsoutthe15 resistor, resultinginthezerotermabove. Next, tondthepowerdelivered,ndeithertheopen-circuitvoltageortheshortcircuitcurrent,Voc= (2 V ) + (1 A) (2 ) = 4 V Isc=(2 V )2 + 8+ (1 A) 22 + 8= 0.4 Aandcalculatethepower.P10 =(4 V )24 10 =(0.4 A)2 10 4= 0.4 WExercise: Explainwhythe15 resistor intheprevious circuit playednopart intheanalysis or solution. Consider thepropertiesofanidealvoltagesource.Youwill oftenhearelectrical engineersspeakof impedancematching: oneofthemostcommonerrorsinsystemdesignisfor subsystemstohaveanimpedancemismatch, therebywastingelectrical power. Theideaof maximumpower transferandimpedancematchingdrives goodcircuit design. Wewill comebacktothis ideaindiscussions of complexpower andsystemintegration.3.10: EquivalentResistanceRevisitedWehaveseenthatsomecongurationsof resistances(impedances), forexamplethebridgeconguration, havenoseriesorparallelcomponents. Toobtaintheequivalentresistancesofthesecircuits,wemusttryanewmethodbesidessuccessiveparallelandseries equivalents. If we use our knowledge of ThevininandNortonequivalent circuits, we mayexpress the equivalentresistance(impedance)asaquasi-ratioofaopen-circuitvoltageandshort-circuitcurrent. Moreexactly,weattacha1Vvoltagesourceora1Acurrentsourceandtestforthecurrentorvoltage,respectively.34Figure72Thenumericalvalueoftheinducedvoltage,VT ,orthenumericalvalueoftheinverseofthecurrentITwillbethesameastheequivalentresistance(impedance),Req.VT= (1A) Req = Req=VT1A= VT [] IT=(1V )Req= Req=1VIT=1IT[]Thebracketnotationattachedtothenal answeristoremindyouof theappropriateunitsassociatedwiththequantities.Letsbeginbyconsideringasimpleexamplethatcouldbesolvedusingtheoriginalmethod.Example: UsethemoregeneralizedsolutionmethodtodeterminetheequivalentresistanceofthecircuitinFigure3.73.Figure73Solution: Choosinga1Acurrentsourceasaninputtothecircuit,wewilltestforthecorrespondingvoltageacrossthetwoterminals. With1Athroughthe10resistor, weseea10Vvoltageinserieswiththevoltageacrosstherestof theelements,6(25 + 5) 1 A = 5 V . Thistotalvoltage,15V,isthennumericallyidenticaltotheequivalentresistanceof15 .Clearly, the last example used a more round-about approach to solving for the equivalent resistance than introduced previously,butletusconsideragaintheresistanceoftheWheatstonebridgecircuitshowninFigure3.74.Figure74First, weattacha1Vtestsourcetotheterminals; tothendeterminetheequivalentresistance, wemustmeasurethecurrentowingoutofthesourcewhichisidenticaltothesumofthecurrentsowingintoR1andR3. Theproblemisthatwecannotdothisdirectly: insteadwewill repeatourprevioussolutionforthevoltagesatnodesCandD(usingThevininequivalentstotheleftandrightofR5).35VC= (1 V ) R4R3 +R4+ (1 V ) R2R1 +R2R4R3 +R4R5 +R3R4R1R2 +R3R4 +R5VD= (1 V ) R4R3 +R4+ (1 V ) R2R1 +R2R4R3 +R4R3R4R1R2 +R3R4 +R5Now,returningtotheoriginalcircuit,wemaycalculatethecurrentsdownthroughR1andR3,IR1=VA VCR1IR3=VA VDR3andthenaddthemusingaKirchosCurrentLawequationatnodeAtoobtainthecurrentthroughthevoltagesourceandourequivalentresistance.IT= IR1+IR3Req=1IT[] = .... =R5 (R1R3) +R1R3R1 +R3 +R5+R2R4Wewillnotwastespacewithmassagingthealgebraicformulationintotheclosedform;rather,wewillpresentonenumericalexample,andshowthatbothrepresentationswork.Example: DeterminetheequivalentresistanceofthebridgecircuitinFigure3.75.Figure75Attachinga1Vtestsource,wecalculatethevoltagesatnodesCandD.VC= (52 + 5V ) + (41 + 452 + 5) V3 + 2514 + 3 + 25= 0.787 VVD= (52 + 5V ) + (41 + 452 + 5) V2514 + 3 + 25= 0.738 VReturningtotheoriginalcircuit,wecalculatethecurrentintothe1 and2 resistors.I1 =(1 V ) (0.787 V )1 = 0.213A I2 =(1 V ) (0.738 V )2 = 0.131AWethencalculateatotal currentfromthevoltagesourceof IT =0.344 A. Finally, theoverall equivalentresistancemaybecalculated.Req=(1 V )(0.344 A)= 2.90 Wecomparethistotheclosed-formsolutiontoverify.Req=3 (12) + 1 21 + 2 + 3+ 45 = 2.89 The two solutions are identical, but you should notice that roundo errors due to truncating values CAN aect the nal answer,asinanyengineeringcalculation...36Exercise: Inmanypowersystems, bridgecircuitareusedonaregularbasis; explainhowyouwouldmodifythepreviousanalysisfortwobridgecircuitsinparallel(hint: thinkaboutthestepwhereKCLisapplied)?Delta-WyeTransformsToconclude our discussiononequivalent resistances (impedances), we state twowell-known, but rarelyused, results forconvertingtoandfromtheandYcongurations of genericimpedances showninFigure3.76. Wehavederivedall of theanalysistocomeupwiththeseresultsinpreviousexamples,buthaveomittedthealgebraicmess.Figure76ZA=Z1Z3Z1 +Z2 +Z3Z1=ZAZB+ZAZC+ZBZCZBZB=Z2Z3Z1 +Z2 +Z3 Z2=ZAZB+ZAZC+ZBZCZAZC=Z1Z2Z1 +Z2 +Z3Z3=ZAZB+ZAZC+ZBZCZC3.11: LinearCircuitExamplesWithoutadoubt, theprerequisitetotrulyunderstandinglinearcircuitsispractice. Theindividual conceptsareinpracticeextraordinarilysimple: OhmslawisjustV= I R,powerisP=I V ,andKirchoslawsaresimplesums. Puttingtheconceptstogethertosolveanyparticularcircuitisamorediculttask. Wecompletethisratherlongdiscourseonsolvinglinearcircuitsbygivingtwomorebasiccircuitsandsolvingwitheachofthemethodspresented.Therstsetofexamplesusesalarge,butrelativelysimple,linearcircuitshowninFigure3.77. Wewillseethatfundamentalconceptslikeequivalentresistancesaswellasvoltageandcurrentdividersareusedoverandover.37Figure77Example: Tostartout,solvefortheequivalentvoltageVxacrossthe6resistor.Wesolvebyzeroingall butonesourceatatime, calculatingthecontributionof eachsourcetothevoltageVxi, andnallyaddingthecontributionstodeterminethetotal voltageVx. Figure3.78showseachof thefourcircuitswithonlyonesourceactivated.Figure78Eachoftheindividualcircuitsreducestoavoltageorcurrentdivider,addinguptoobtainthevoltageVx.Vx= Vx0.5 A+Vx15 V+Vx1 A+Vx22 V= (0.5 A) 1212 + (8 + 650) (650) + (15 V ) 6(8 + 12)6(8 + 12) + 50+ (1 A) 8 + 128 + 12 + 650 (650) + (22 V ) 650650 + 8 + 12= (1.268 V ) + (1.268 V ) + (4.225 V ) + (4.648 V ) = 2.958 VExample: Next,setupandsolvethenecessarynode-voltageequationstoobtainVx.Theredrawncircuit,withallnodeslabeled,isshowninFigure3.79.Figure7938Wewritethefourequationsnecessarytosolve,VCVB= 22VVA12+VA VB8= 0.5A VD= 15VVBVA8+ (1 A) +VC6+VCVD50= 0and then compute the four node voltages. You should readily notice that the desired unknown,VX, is the same as node voltageVC.VA= 9.028 V VB= 19.046 V VC= Vx= 2.954 V VD= 15 VWeseethatthenodevoltageVCis2.954V,ascomparedtothe2.958Vfoundpreviously(identicalwithinroundo).Example: Now,setupandsolvethenecessarymesh-currentequationstosolveforVx.Theredrawncircuit,withallmesheslabeled,isshowninFigure3.80.Figure80Writingtheequationsforthefourmeshes,I1= 0.5 A (I2 I1) (12 ) +I2 (8 ) 22 V+ (I3 I4) (6 ) = 0I2 I3= 1 A (I4 I3) (6 ) +I4 (50 ) + (15 V ) = 0andsolvingtoobtainVxI1= 0.5 A I2= 1.2521 A I3= 0.2521 A I4= 0.2408 AVx= (I3 I4) (6 ) = 2.597 VExample: Next,letsndVxusingaThevininequivalent. Todoso,weremovethe6resistorfromthecircuitandproceedtondtheopencircuitvoltage(viasuperposition)andtheThevininequivalentresistanceasshowninFigure3.81.Figure81Theopencircuitvoltageisobtainedbysuperposition,Voc= (0.5 A) [12(8 + 50)] 508 + 50+ (1 A) [(8 + 12)50] + (22 V ) 5050 + (12 + 8)+ (15 V ) (8 + 12)50 + (8 + 12)= (4.286 V ) + (14.286 V ) + (15.714 V ) + (4.286 V )= 10 Vandtheequivalentresistanceiscalculatedbyrstzeroingallthesourcesandthenndingtheequivalent.39Zth= 50(8 + 12) = 14.286 ThenalcalculationistouseavoltagedividerforthevoltageVxacrossthereplaced6resistor.Vx= Voc 6Zth + 6= (10 V ) 614.286 + 6= 2.958 VExample: One more method available is to use the Norton equivalent circuit. Find the Norton equiavlent and then use OhmslawtodeterminethevoltageVx.Onceagain, weremovethe6resistor, butthistimeattachashortcircuitinitsplaceasshowninFigure3.82. CarehasbeentakentomakesurethatthepositivedirectionofIsccorrespondstotheproperpolarityforVx.Figure82Usingsuperpositiononelasttime,wesolvefortheshortcircuitcurrent,Isc= (0.5 A) 128 + 12+ (1 A) +(22 V )(8 + 12) +(15 V )50 = (0.3 A) + (1 A) + (1.1 A) + (0.3 A)= 0.7 AandthensolveforthevoltageacrosstheparallelequivalentofThevininresistance(sameasbefore)andthe6resistor.Vx= Isc (Zth6) = (0.7 A) (14.2866) = 2.958 VExample: Thenal methodtosolvethiscircuitistousesourcetransformations. Takingtheoriginal circuit, weiterativelyreduce the circuit to either a Thevinin or Norton equivalent. Performing one source transformation on both the left and the rightsideofthecircuit,wereduceittotheoneshowninFigure3.83.Figure83ThreemoreiterationsshowninFigures3.84,3.85,and3.86givethesameNortonequivalentassolveforpreviously.40Figure84Figure85Figure86Withthenalstep,wecalculatethevoltageVx.Vx= (0.7 A) (14.2866 ) = 2.958 VExercise: Whatarethebenets/limitationsforeachofthemethodspresentedinthissection?Wewillrepeatthefullanalysisforasecondcircuitthatcontainsanumberoftransparenttricks. Theanalysisforthevariousmethodswillactuallybelessinvolved,butrequireyoutoseefundamentalsimplication.Example: SolveforthevoltageVxacrossthe15resistorinFigure3.87usingeachofthecircuitmethods.Figure8741Beforewebegintheanalyses,weshouldtakeamomenttoguesstimatewhattheoverallcharacteristicswillbe:1. Thecircuithasfourindependentsources,sosuperpositionwillrequiresolvingfourseparatecircuits.2. Thecircuithas6independentsources,butthreevoltagesourcesthatwillmakeforsimplenodeequations.3. Thecircuithas5independentmeshes,yetonlyonecurrentsourceforasimpleequation.4. The3V, 12V, and1Awill contributetocurrentowingupwardsinthe15resistor(creatinganegativecontributiontoVx),whilethe22VsourcewillcontributetoadownwardcurrentandapositiveVx.5. The4resistorinserieswiththecurrentsourceandthe8resistorinparallel withthe3Vvoltagesourcewill notaectanyoftheothercircuitvariables;thevoltageacrossacurrentsourceandthecurrentthroughavoltagesourceareindeterminate.Anyvalueofresistancecanbeusedforthesetwowithoutchangingothercircuitvariables.6. Althoughnon-trivialtodetermine,byfartheeasiestmethodtosolvethiscircuitwouldbesourcetransformationsincom-binationwithsuperposition.Wewillcommentontheseobservationsinmoredetailsasweproceed.Superposition: Werepeatthesameprocessasbeforetosolvethecircuitbysuperposition, butsimplifybyremovingthe4resistor(shortingto0)andthe8resistor(openingto)whenredrawingtheequivalentcircuits. ThefoursimpliedcircuitswithasinglesourceactiveareshowninFigure3.88.Figure88TocalculatethevoltageVx,wetakeeachcontributionandthenaddthemup.42Vx1 A= (1 A) 66 + 102 + 1530 (1530 ) = 3.396 VVx3 V= (3 V ) 15301530 + 102 + 6= 1.698 VVx22 V= (22 V ) 15(6 + 210)30 + 15(6 + 210)= 3.182 VVx12 V= (12 V ) 2(6 + 1530)10 + 2(6 + 1530)15306 + 1530= 1.132 VVx= Vx1 A+Vx3 V+Vx22 V+Vx12 V= 3.04 VNode-VoltageAnalysis: WelabelthenodesasshowninFigure3.89,choosingthebottomnodeasground.Figure89Wehaveleftthe4and8resistorsinthecircuit,butcanquicklyshowthattheydonotaecttheremainderofthecircuit.Thecurrentthroughthe4resistoristhesameasthe1AsourceVA VB4 + (1 A) = 0whilethepotentialdierencebetweennodeEandgroundisjustthevoltagesource.VE0 = (3 V )Changingthevaluesofthesetworesistorswillchangeeitherthecurrentthrough(the8)orthevoltageacross(the4)theresistors. Writingtheequationstosolveforthenodevoltages(takingallcurrentsleavingthenodes),VE0 = (3 V ) VCVD= (12 V ) VF0 = (22 V )(1 A) +VBVE6+VBVD2+VBVC10= 0VD VB2+VCVB10+VD 015+VD VF30= 0Solving5equationssimultaneouslyisapain,buttheresultsareidentical.VA= 6.77 V VB= 2.77 V VC= 8.96 V VD= 3.04 V VE= 3 VVx= VD= 3.04 VMeshCurrentAnalysis: First,welabelthemeshesasshowninFigure3.90,43Figure90Disregardingthe8resistor,weendupwithfourmeshes,resultinginthefourequations:I1= (1 A) (I2 I3) 2 +I2 10 + (12 V ) = 0(3 V ) + (I3 I1) 6 + (I3 I2) 2 + (I3 I4) 15 = 0 (I4 I3) 15 +I4 30 + (22 V ) = 0Solvingtheseequations,weobtainthefourmeshcurrents,I1= 1 A I2= 1.173 A I3= 1.038 A I4= 0.835 AandtheoverallsolutionforVx.Vx= (I3 I4) (15 ) = 3.044 VThevininandNortonEquivalentsNowwewillrepeatthesolutionforthevoltageVxusingbothThevininandNortonequivalentcircuits;sincethetwomethodsaresoclosetooneanother,wewillsolvethemtogether. Werstremovethe15resistorfromthecircuitandsolvefortheopencircuitvoltage,Voc,acrosstheterminalsusingsuperposition.Figure91ThesolutionofVocinthiscircuitisidenticaltotheanalysisoftheoriginalcircuitwiththe15resistorincreasedto,oranopencircuit. Toshowthismodicationinalittlemoredetail, rsttakethepreviousexpressionforVxusingsuperpositionandincludingthe15resistor(whichwecallV

oc).44V

oc1 A= (1 A) 66 + 102 + 1530 (1530 ) = 3.396 VV

oc3 V= (3 V ) 15301530 + 102 + 6= 1.698 VV

oc22 V= (22 V ) 15(6 + 210)30 + 15(6 + 210)= 3.182 VV

oc12 V= (12 V ) 2(6 + 1530)10 + 2(6 + 1530)15306 + 1530= 1.132 VNowwereplaceeach15termwithtoobtainVocfromV

oc. Theequivalentresistanceofanopencircuit()inserieswithanyniteresistoristheopencircuit,inparallelwithanyniteresistorisjusttheresistor.Voc1 A= (1 A) 66 + 102 +30 (30 ) = (1 A) 66 + 102 + 30 (30 ) = 4.779 VVoc3 V= (3 V ) 3030 + 102 + 6= (3 V ) 3030 + 102 + 6= 2.389 VVoc22 V= (22 V ) (6 + 210)30 +(6 + 210)= (22 V ) (6 + 210)30 + (6 + 210)= 4.479 VVoc12 V= (12 V ) 2(6 +30)10 + 2(6 +30)306 +30= (12 V ) 2(6 + 30)10 + 2(6 + 30)306 + 30= 1.593 VVoc= Voc1 A+Voc3 V+Voc22 V+Voc12 V= 4.282 VBeforehavingtheanswerforVx,weneedtodeterminetheThevininresistanceasseenfromthetwoterminalswithallsourceszeroed: looking right, we see a 30 resistor and to the left, we see 6+102 (the 4 is swamped by the open circuit of the zeroedcurrentsourceandthe8resistoriskilledbythezeroed3Vvoltagesource).Rth= 30(6 + 102) = 6.106 Finally,wedeterminethevoltageVxasavoltagedividerwiththeThevininequivalentcircuit.Vx= Voc 15Rth + 15= (4.282 V ) 156.106 + 15= 3.043 VWithoutadoubt, thiscircuitwasmucheasiertosolvewiththeoriginal applicationof superposition! Forcompleteness, wewillrepeattondtheshortcircuitcurrentofthecircuitshowninFigure3.92.Figure92Again,themoststraightforwardmethodistosimplyusesuperpositiontoadduptheindividualsourcescontributions,45Isc= (1 A) 66 + 102+12 V10 + 2622 + 6+3 V210 + 6+22 V30= (0.783 A) + (0.261 A) + (0.391 A) + (0.733 A)= 0.702 AandthenapplyOhmslawtodeterminethevoltageVxacrosstheparallelcombinationofsameRthcalculatedbeforeandthereplaced15resistor.Vx= Isc (Rth15) = (0.702 A) (6.10615) = 3.046 VQuestion: WhycanyounotmodifythesuperpositionsolutionforVxtosolveforIsc?3Onehintistoconsiderthevoltageacrossashortcircuit.We found that the Norton equivalent method requires roughly the same amount of work as the original superposition solution.Afterworkingmany,many,ofthesecircuits,youwillbegintoseehowthemethodscanbeintermixedforquickersolutions. Ourlastexamplewillbeahybridmethodofsourcetransformations(single-stageThevininorNortonequivalent)andsuperposition.HybridSourceTransformationsandSuperpositionFirstofall,takeanotherlookattheoriginalcircuit.Figure93Aspreviouslydiscussed,the4and8resistorsdonotaecttherestofthecircuit,sowethrowthemaway(shortingthe4andopeningthe8). Wethennoticethatwehavethreeinstancesofsinglevoltagesourcesinserieswithsingularresistances;weperformasourcetransformationoneachofthesetermssimultaneously,obtainingthecircuitinFigure3.94.Figure943Althoughpracticallyuseless, youmaydividetheentireexpressionfor Vxasderivedusingsuperpositionbythe15resistorandthentakethelimitoftheentireexpressionasthe15 0, butwill beforcedtouseLHopitalsruleandmany, manyextrasteps.46WebeginaddingparallelcurrentsourcesandparallelresistorstoobtainthesimpliedcircuitshowninFigure3.95.Figure95Fromthispointforwards,superpositionandsourcetransformationswillrequireroughlythesameamountofeort;wechoosesuperpositionsincetheexpressioncanbewrittenbyinspectionandwithoutredrawingthecircuitanyfurther(youcaneasilyvisualizethezeroedcurrentsourcesbycompletelydetachingthem).Vx= (10 ) (1.5 A) 66 + (10 + 1.667)+ (1.2 A) 1.6671.667 + (10 + 6)+ (0.733 A) (6 + 1.667)(6 + 1.667) + 10= 3.045 VExercise: Whatcharacteristicsofthecircuitcauseeachofthemethodstobeeasierorharder?3.12: SummaryInthischapter, wehaveseenall of thebasictoolsusedtosolvelinearcircuits: voltageandcurrentdividers; superposition;node-voltageandmesh-currentanalyses;ThevininandNortonequivalents;andsourcetransformations.Voltageandcurrentdividersgivethefractionofatotalvoltageorcurrentfromasinglevoltageorcurrentsource.Superposition allows you to decompose a circuit containing multiple sources into multiple circuits (having only a single source)thatmakelinearcontributionstotheoverallcircuitquantities.Node-voltageandmesh-currentanalysesmakepossiblesimultaneoussolutionof all linearcircuitvariablesbytransformingmultipleapplicationsof Kirchoslawsintolinearsystemsof equations. Ohmslawisreinterpretedinamatrixform: eitherV =[R] I orI=[G] V . Notethatthegroundnodeissimplyaorpotential energyreferencedenedtobezero; all nodevoltagesaremeasuredrelativetothisground. Anal commentisthatmeshcurrentsarenotphysicallymeasurablequantities;nodevoltagescanbeeasilymeasuredinalabandsolvedbyacomputer.ThevininandNortonequivalentsareacircuitmodelingtechniqueprovidingasimplecircuit,eitherasinglevoltagesourceinserieswitharesistororasinglecurrentsourceinparallel witharesistor, thatcanbeusedtocalculateacircuitseectonanynumberofloads.Sourcetransformationsareaspecial caseof ThevininandNortonequivalents, whereasinglevoltagesourceinserieswitharesistanceistransformedintoacurrentsourceinparallelwitharesistor,orviceversa.Moreoftenthannot, thebestmethodof solvingalinearcircuitwill consistof ahodgepodgeof methods. Superpositionisperhaps the most anaytically straightforward, yet requires (redrawing and) solving multiple smaller circles. Node-voltage analysisis the most computationally ecient way to solve an entire circuit,while superposition is often better for a single circuit variable.Finally, thetwocircuitequivalentshelpdeterminehowanentirecircuitwill contributewhenattachedtoanexternal load; byusingtheseequivalents,weeaseexternalcalculations,butwelosetheabilitytocalculateinternalcircuitquantities.Thebottomlineinunderstandinghowtosolvealinearcircuitispractice. Nomanual, textbookorprofessorcantell youanythingbeyondV= IR,P= IV ,orsumsofcurrentsandvoltagesbeingzeroatanodeorinaclosedloop,respectively. Everyoneofthetime-domain,AC,analog,anddigitalcircuitsdiscussedthroughouttheremainderofthetextreliesonthesefourbasicconcepts.47Problem3.1WhatistheThevininequivalentofavoltagesourceinparallelwitharesistor?WhycanyounotndaNortonequivalent?Problem3.2WhatistheNortonequivalentofacurrentsourceinserieswitharesistor? WhycanyounotndaThevininequivalent?Problem3.3: ForthelinearcircuitshowninFigure3.96,determinethevoltageacrossthetwooutputterminals.Figure96Problem3.4: ForthelinearcircuitshowninFigure3.97,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.Figure97Problem3.5: ForthelinearcircuitshowninFigure3.98,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe84kresistor).(e) SolveforVxusingtheNortonequivalent(rstremovethe84kresistor).(f) SolveforVxusingsourcetransformations.Figure98Problem3.6: ForthelinearcircuitshowninFigure3.99,48(a) UsesuperpositiontodeterminethecurrentIx.(b) SolveforIxusingnode-voltageanalysis.(c) SolveforIxusingmesh-currentanalysis.(d) SolveforIxusingtheThevininequivalent(rstremovethe11kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe11kresistor).(f) SolveforIxusingsourcetransformations.Figure99Problem3.7: ForthelinearcircuitshowninFigure3.100,(a) UsesuperpositiontodeterminethecurrentIx.(b) SolveforIxusingnode-voltageanalysis.(c) SolveforIxusingmesh-currentanalysis.(d) SolveforIxusingtheThevininequivalent(rstremovethe93kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe93kresistor).(f) SolveforIxusingsourcetransformations.Figure100Problem3.8: ForthelinearcircuitshowninFigure3.101,(a) UsesuperpositiontodeterminethecurrentVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe55kresistor).(e) SolveforVxusingtheNortonequivalent(rstremovethe55kresistor).Figure10149Problem3.9: ForthelinearcircuitshowninFigure3.102,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe62kresistor).(e) SolveforVxusingtheNortonequivalent(rstremovethe62kresistor).(f) SolveforVxusingsourcetransformations.Figure102Problem3.10: ForthelinearcircuitshowninFigure3.103,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe43kresistor).(e) SolveforVxusingtheNortonequivalent(rstremovethe43kresistor).Figure103Problem3.11: ForthelinearcircuitshowninFigure3.104,(a) SolveforIxusingnode-voltageanalysis.(b) SolveforIxusingmesh-currentanalysis.(c) SolveforIxusingtheThevininequivalents.(d) SolveforIxusingtheNortonequivalents.Figure10450Problem3.12: ForthelinearcircuitshowninFigure3.105,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent.(e) SolveforVxusingtheNortonequivalent.(f) SolveforVxusingsourcetransformations.Figure105Problem3.13: ForthelinearcircuitshowninFigure3.106,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe20kresistor).(e) SolveforVxusingtheNortonequivalent(rstremovethe20kresistor).(f) SolveforVxusingsourcetransformations.Figure106Problem3.14: ForthelinearcircuitshowninFigure3.107,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent.(e) SolveforVxusingtheNortonequivalent.(f) SolveforVxusingsourcetransformations.51Figure107Problem3.15: ForthelinearcircuitshowninFigure3.108,(a) UsesuperpositiontodeterminethevoltageVx.(b) SolveforVxusingnode-voltageanalysis.(c) SolveforVxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent.(e) SolveforVxusingtheNortonequivalent.(f) SolveforVxusingsourcetransformations.Figure108Problem3.16: For thecircuit showninFigure3.109, theunknowncircuit element inthemiddleis nonlinear andthatVx =2 I2whereI is thecurrent owingthroughthedevice. Showthat superpositionis nolonger avalidmethodfor thenonlinearelement.Figure109Problem3.17: ForthelinearcircuitshowninFigure3.110,(a) UsesuperpositiontodeterminethecurrentIx.(b) SolveforIxusingnode-voltageanalysis.(c) SolveforIxusingmesh-currentanalysis.(d) SolveforIxusingtheThevininequivalent(rstremovethe78kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe78kresistor).(f) SolveforIxusingsourcetransformations.52Figure110Problem3.18: ForthelinearcircuitshowninFigure3.111,(a) UsesuperpositiontodeterminethecurrentIx.(b) SolveforIxusingnode-voltageanalysis.(c) SolveforIxusingmesh-currentanalysis.(d) SolveforIxusingtheThevininequivalent(rstremovethe58kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe58kresistor).(f) SolveforIxusingsourcetransformations.Figure111Problem3.19: ForthelinearcircuitshowninFigure3.112,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe98kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe31Vvoltagesource).Figure112Problem3.20: ForthelinearcircuitshowninFigure3.113,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe25kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe88kresistor).53Figure113Problem3.21: ForthelinearcircuitshowninFigure3.114,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe87kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe38kresistor).Figure114Problem3.22: ForthelinearcircuitshowninFigure3.115,(a) UsesuperpositiontodeterminethecurrentIx.(b) SolveforIxusingnode-voltageanalysis.(c) SolveforIxusingmesh-currentanalysis.(d) SolveforIxusingtheThevininequivalent(rstremovethe8kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe8kresistor).Figure115Problem3.23: ForthelinearcircuitshowninFigure3.116,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.54(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethecurrentsource).(e) SolveforIxusingtheNortonequivalent(rstremovethe29kresistor).Figure116Problem3.24: ForthelinearcircuitshowninFigure3.117,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe25kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe16kresistor).Figure117Problem3.25: ForthelinearcircuitshowninFigure3.118,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe12kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe6kresistor).(f) SolveforVxandIxusingsourcetransformations.Figure11855Problem3.26: ForthelinearcircuitshowninFigure3.119,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheThevininequivalent(rstremovethe23kresistor).(e) SolveforIxusingtheNortonequivalent(rstremovethe19kresistor).Figure119Problem3.27: ForthelinearcircuitshowninFigure3.120,(a) UsesuperpositiontodeterminethevoltageVxandthecurrentIx.(b) SolveforVxandIxusingnode-voltageanalysis.(c) SolveforVxandIxusingmesh-currentanalysis.(d) SolveforVxusingtheNortonequivalent(rstremovethecurrentsource).(e) SolveforIxusingtheThevininequivalent(rstremovethevoltagesource).Figure120