耦合电感(双语)课件

LinearCircuitAnalysis

MagneticallyCoupled

Circuitsand

Transformers1.IntroductionExperimentalevidenceshowsthatachangeinthecurrenti1willgenerateavoltagev2,calledtheinducedvoltage,acrosstheopencircuit.Althoughphysicallyisolated,eachpairofcoilsinfigures18.1aand18.1bissaidtobemagneticallycoupled.Howdoesonequantitativelyaccountformagneticcoupling?Theansweristointroduceanewcircuitquantity,calledmutualinductance,forcoupledcoils.

Magneticallycoupledcircuitshavemanyimportantengineeringapplication.Anextremelyimportantmagneticallycoupleddevicesisthetransformer,whichisusedtotransformvoltagesandcurrentsfromoneleveltoanother.

Inaddition,transformershavenumeroususesinelectronicsystems,including(1)steppingacvoltagesupordown,(2)isolatingpartsofacircuitfromdcvoltages,and(3)providingimpedancelevelchangestoachievemaximunpowertransferbetweendevices.2.MutualInductanceandtheDotConvention

Experimentalevidencedemonstratesthatifthetwocoilsinfigure18.1aarestationary,theinducedvoltagev2(t)isproportionaltotherateofchangeofi1(t).Confiningourattentioninitiallytothemagnitudesofvoltageandcurrentonly,wecanwritethelinearrelationshipaswhereM21isaconstantofproportionalitylinkingv2tothechangeini1.if,infigure18.1a,thevoltagesourceismovedtocoil2whilecoil1isopen-circuited,oneobservesasimilarrelationship,viz.,(18.1)(18.2)Asverifiedinalatersection,M21andM12areequalandthereforecanbedesignatedbyasinglepositiveproportionalityconstant,Calledthemutualinductanceofthecoupledinductors.Removingtheabsolute-valuesignsfromequations18.1and18.2,andusingequation18.3,impliesthat

Equation18.1alsosuggestsapossibleexperimentalprocedurefordeterminingthevalueofM:applyarampcurrenti1(t)=Kr(t)(k＞0)tocoil1,andmeasuretheconstantopen-circuitvoltagev2inducedincoil2.ThenM=∣v2∣/K.(18.3)(18.4)(18.5)

Todeterminethesign,considerfigure18.2andapplyanincreasingcurrenti1(t)tocoil1.Sincedi1/dt＞0forallt,oneoftheterminalsofcoil2mustbeatahigherpotentialrelativetotheotheratalltimes.Thisterminaliseasilyidentifiedbyconnectingavoltmetertocoil2.Placeadotontheterminalatwhichenters.Letbeincreasing,.Placeanotherdotontheterminalwiththehigherpotentialthatisontheotheropen-circuitedcoil.RuleForInducedVoltageDropDuetoMutualInductance

Thevoltagedropacrossonecoil,fromthedottedterminaltotheundottedterminal,equalsMtimesthederivativeofthecurrentthroughtheothercoil,fromthedottedterminaltotheundottedterminal.Or,equivalently,

Thevoltagedropacrossonecoil,fromtheundottedterminaltothedottedterminal,equalsMtimesthederivativeofthecurrentthroughtheothercoil,fromtheundottedterminaltothedottedterminal.(18.6a)(18.6b)Example18.1Infigure18.2,ifi1(t)=2tu(t)A,avoltagevCD=-10u(t)mVisobserved.Determinethe(new)placementofthedotsandthevalueofM.Sincei1(t)isincreasingandDisatahigherpotentialthanC,thedotsmustbeplacedat(A,D)or(B,C).

Fromequation18.1,thevalueofthemutualinductanceisSOLUTIONExample18.2Figure18.3bshowstwowaveformsi1(t)andv2(t)asdisplayedonanoscilloscope.Determinetheplacementofthedotsandthevalueofthemutualinductanceforthecircuitoffigure18.3a.1-112+-Signalsource2-2(b)Figure18.3SOLUTIONForthetimeinterval0＜t＜0.5ms,i1(t)isarampfunctionandv2(t)isconstant,Theinformationissimilartothatgiveninexample18.1.Thereforewecansolvetheprobleminthesameway.Thecurrenti1isincreasingandv2ispositive.Accordingtofigure18.2,thedotsmustbeplacedat(A,C)or(B,D).Equation18.5becomesThemeasuredvaluesduring0＜t＜0.5sgivev2=2Vanddi1/dt=1/0.0005=2000A/s.ThusM=2/2000=0.001H.Example18.3Inthecircuitoffigure18.3a,ifi1(t)=2(1-e-100t)u(t)A,findv2(t).SOLUTIONInthiscase,di1/dt=200e-100tu(t).Fromexample18.2,M=0.001H.Fromequation18.5,Itisworthwhiletorememberthefollowingbasicfacts:1.Letthenumbersofturnsofthetwoinductorsinfigure18.1beN1andN2.Thentheselfandmutualinductanceshaveapproximatelytheratio2.Iftwoinductorsareplacedinanonmagneticmedium,bringingtheinductorsclosertogetherincreasesthevalueofM.3.Ifoneinductorofapairisrotated,thenalargervalueofMresultswhentheaxesoftheinductorsareparalleltoeachother.ThesmallestvalueofMoccurswhentheaxesareperpendiculartoeachother.4.ChangingthecoreonwhichtwoinductorsarewoundfromanonmagneticmaterialtoaferromagneticmaterialmayincreasethevaluesofL1,L2,andMbyseveralthousandtimes.3.DifferentialEquation,LaplaceTransform,andPhasorModelsofCoupledInductors

Figure18.4showsapairofinductorswithmutualinductanceM.Besidesthemutualinductance,eachinductorinthepairalsohasaself-inductance,denotedbyL1andL2,respectively.Thereferencedirectionsforvoltagesandcurrentsmaybechosenarbitrarily,withthoseshowninfigure18.4beingtypical.Ourobjectivehereistodevelopequationsrelatingthecurrentsandvoltagesforthedotplacementsoffigures18.4aand18.4b.(1)DifferentialEquationofcoupledinductors

Thevoltagedevelopedacrosseachinductoristhesumofthevoltageduetotheself-inductanceLk(dik/dt)andthevoltageduetothemutualinductance±M(dij/dt),withthe±signsdulyconsidered.Infigure18.4a,currentsareenteringatthedottedterminals,andvoltagedropsarefromthedottedterminalstotheundottedterminals;hence,themutualtermshaveplussigns,accordingtoequation18.6a.Applyingsuperpositionleadstothefollowingsetofequationsgoverningthecircuitoffigure18.4a,(18.7a)(18.7b)

Ontheotherhand,thevoltagev2(t)infigure18.4bisfromtheundottedterminaltothedottedterminalofcoil2.Thecurrentthroughcoil1,fromtheundottedterminaltothedottedterminal,is–i1,Therefore,accordingtoequation18.6b,(18.8b)

Infigure18.4b,v1(t)isthevoltagedropfromthedottedterminaltotheundottedterminalofcoil1.Thecurrentthroughcoil2,fromthedottedterminaltotheundottedterminal,is–i2.Therefore,accordingtoequation18.6a,(18.8a)Again,thedevelopmentbasedonfigure18.1presupposeslinearity.Ifthetwoinductorsareplacedinanonmagneticmedium,thisistrue.Iftheinductorsarecoupledthroughaferromagneticmedium,thenthelinearrelationshipsholdonlyifbothcurrentsaresufficientlysmallsothatthemagneticmediumhasnotreachedsaturation,aphenomenondiscussedinothercoursesormoreadvancedtexts.Ourinvestigationconsidersonlythelinearcase.Example18.4Apairofcoupledinductorsisconnectedintwodifferentways,asshowninfigure18.5.Ineachcase,findthedifferentialequationrelatingvandi,andthenfindtheequivalentinductance“seen”atthetwoterminalsofeachbox.SOLUTIONForfigure18.5a,wecanapplyequations18.7directlytoobtain+--+--Figure18.5Equivalentinductanceoftwoseries-connectedinductors.ComparingthisrelationshipwithWeobtainSimilarly,forfigure18.5b,weapplyequations18.8toobtainTherefore,Theprecedingexampleimpliesthat(18.9)

ApplyingtheLaplacetransformtoequations18.7and18.8,whileinvokingtheassumptionthati1(0-)=i2(0-)=0,yieldsComparingequation18.10withequations18.7and18.8yields

onlytwodifferences:V(s)andI(s)replacev(t)andi(t).2.TheLaplacetransformvariablesreplacestheoperatord/dt.(18.10a)(18.10b)(2)LaplaceTransformofcoupledinductorsExample18.6Inthecircuitoffigure18.7,R1=4Ω,R2=2Ω,L1=8H,L2=6H,andM=4H.IfE=36V,i2(0-)=0,andSisclosedatt=0:(a)Findi1(t)andi2(t).(b)ShowthatV2(t)anddi1/dtarepositiveforallt＞0.SOLUTION(a)Writingtwoloopequationsinthes-domainproducesFigure18.7EInmatrixnotation,SolvingforI1andI2byCramer’srule,weobtainInverseLaplacetransformingI1(s)andI2(s)yields(b)Usingtheexpressionsfoundinpart(a)impliesandFromequation18.11and18.12,V2(t)＞0anddi1/dt

＞0forallt＞0,fort＞0

TheLaplacetransformmethodisveryusefulforfindingthecompletesolution.Inpractice,however,therearemanyoccasionswheretheexcitationtoastablecircuitissinusoidalandonlythesteady-statesolutionisofinterest.ThephasormethodofanalysisproveseasierthantheLaplacetransformmethodinsuchcases.PhasorequationsresembleLaplacetransformequationsundertheassumptionofzeroinitialconditions.TheonlydifferenceisthatVandIarenowphasors(complexnumbers),Withsreplacedbyjω.Inphasorform,equations18.10becameThefollowingexampleillustratestheuseofthephasormethod.(18.13a)(18.13b)(3)PhasormodelsofcoupledinductorsExample18.7Considerthecircuitoffigure18.8.Findthesteady-statecomponentsofV1(t)andV2(t)atthefrequencyof1rad/sforthefollowingtwocases:(a)The2-Fcapacitorisdisconnected.(b)The2-Fcapacitorisconnected.(a)Withthecapacitordisconnected,=0.Henceinducesnocomponentin,andtheequationforisFigure18.8SOLUTIONFirstwenotethatω=1rad/sandthatthephasorforthevoltagesourceis1∠0V.and(b)Withthe2-Fcapacitorconnected,thetwoloopequationsareFromwhichitfollowsthatAnd,since,Thesolutionsoftheseequationsare:so,inwhichcase

Clearly,

areoutofphase!

Observethatareinphase.Figure18.84.Applications:AutomobileIgnitionandRFAmplifierExample18.8Figure18.9showsanautomobileignitionsystemandasimplifiedcircuitmodelthereof.Thisignitionsystemistypicalofoldermodelcars.Theignitioncoilisapairofinductorswoundonthesameironcore.Thecoilconnectedtoapowersourceiscalledtheprimary,whereasthecoilconnectedtoaloadiscalledthesecondary.Theprimaryhasafewhundredturnsofheavywire,thesecondaryabout20000turnsofveryfinewire.Whentheignitionpoint(orcontact)opensbycamaction,avoltageexceeding20000Visinducedacrossthesecondary,causingthesparkplugtofire.Today’signitionsystemsdonottypicallyhavepointsandacondenser;switchingindoneelectronically.Theboxindashedlinesinfigure18.9aisreplacedbyanelectronicignitionmodule.Nevertheless,thebasicideaofgeneratingahighvoltagetocauseasparktooccurattheplugisaccomplishedbyabasicRLCcircuitcontainingaswitchthatrepresentsthepointoftheignitionsystem.SolutionAsmentioned,thefollowingfigureisasimplifiedcircuitmodelfortheignitionsystem.Supposethattheswitchhasbeenclosedforalongtime.Sincethesecondaryisopen-circuited,ithasnoeffectonthesolutionfortheprimarycurrent.Usingthemodelforaninitializedinductorresultsinthes-domainequivalentcircuitoffigure18.10.Accordingly,at,wehaveFigure18.9bItssimplifiedcircuitmodel+_Theprimarycurrentissimplythenetdrivingvoltagedividedbythetotalimpedanceintheseriescircuit,SubstitutingthegivencomponentvaluesintotheequationyieldsTakingtheinverseLaplacetransformofyieldsHavingobtained,wecalculatefromthebasicrelationship,usingtheplussigninthepresentcaseofdotmarkings:Fromthisexpression,thevoltagereachesamagnitudeof36000Vinabout(one-fourthofacycleoftheoscillations).Thisvoltageishighenoughtocausethesparkplugtofire.Inpractice,whentheengineisrunning,aresistancewireoranactualresistorisplacedintheprimarycircuittolimittheamountofcurrentflowthroughthebreakerpoint.Thecapacitor(condenser)servesasimilarpurpose—thatofprotectingthebreakerpointbysuppressingthearcthatresultswhenthepointopens.5.CoefficientofCouplingandEnergyCalculation

WebeginthissectionwithajustificationofourassumptionthatM12=M21=M.Thejustificationstemsfromaphysicalpropertythatapairofstationarycoupledcoilscannotgenerateaveragepower.OnecanjustifythatM12=M21=Mbytheprinciplesofmagneticcircuits,butthisisbeyondthescopeofthistext.Asaconsequence,wewillshowthatthemutualinductanceMhasupperbound,i.e,themutualinductancecanneverexceedthegeometricmeanoftheself-inductances.PASSIVITYPRINCIPLEFORINDUCTORSApairofstationarycoupledinductorsisapassivesystem;i.e.,theycannotgenerateenergyand,hence,cannotdeliveraveragepowertoanyexternalnetwork.JustificationthatM12=M21=MThispropertyisaconsequenceoftheprinciplesofelectromagneticfieldtheoryandcannotbeprovenbyapuremathematicalargument.Theprinciplesoffieldtheory,however,arenotcommonlyknownbystudentsstudyingintroductorycircuitanalysis.Tomakeourapproachmoreaccessible,wewillbuildourjustificationonthepassivityprincipleforinductors.

SupposethatM12≠M21.Then,insteadofhavingequations18.7,thedifferentialequationsforthecoupledinductorsoffigure18.4amustaccountforthisdifferenceandtaketheform

Letusapplyi1=sin(t)Aandi2=cos(t)Atotheinductors.Fromequations18.20,theterminalvoltagesare(18.20a)(18.20b)

Thetotalinstantaneouspowerdeliveredtotheinductorsis,therefore,(18.21)Tocalculate,theaveragepowerdeliveredtothecoupledinductors,weusetheidentities.Itfollowsimmediatelythatthefirstandlasttermsinequation18.21makenocontributionto,whereasthetermsinvolving

andleadto(18.22)Thisresultshowsveryclearlythatwith,theaveragepowerofcoupledinductorisalwayszeroforarbitrarysinusoidalexcitations.Hencecoupledinductorsaresaidtobelossless.CalculationofStoredenergy(18.23a)(18.23b)Considerthecoupledinductorsshowninfigure18.4.Thevoltage-currentrelationshipsattheterminalsaregivenbyHavingprovedthat,weshallnowshowthatthereisalimittothevalueofthatisattainableonceandarespecified.Thisisagaindonebytheuseofthepassivityprinciple.Whered(i1,i2)isthetotalderivativeoftheproducti1i2andisequaltoi1di2+i2di1.

Nextweapplydrivingsourcestotheinductorstobringthecurrentsuptoi1=I1

andi2=I2att=T.Theenergydeliveredtotheinductorsduringthetimeinterval(0,T)is(18.24)Forthisreason,theenergygivenbyequation18.24iscalledthestoredenergy.Thephysicsofthesituationshowsthattheenergyisstoredinthemagneticfieldproducedbythecurrentsintheinductors.Acoupleofthingsaboutthisresultareworthnoticing:Thefinalintegralinequation18.24dependsonlyonthefinalvaluesI1

andI2.Theexactwaveformsofi1(t)andi2(t)during0≤t≤Tareimmaterial.2.TheenergyW(T)deliveredbythesourcesduring0≤t≤T

isnotlost,butmerelystoredinthesystem.18.24UpperBoundforMandtheCoefficientofCouplingTheenergyW(T)mustbenonnegativeforarbitraryvaluesofI1

andI2.Otherwisetheinductorswillbegeneratingenergyduringthetimeinterval0≤t≤T

andthusviolatethepassivityprinciple.ToensureanonnegativeW(T)forallI1andI2,thevaluesofL1,L2,andMmustsatisfytheinequalityorToseethis,werewriteequation18.24intheform(18.26a)(18.26b)Equation18.27showsthatisnegativewheneverisnegative.Nowisasecond-degreepolynomialinwithapositivecoefficientfortheterm.Fromanalyticgeometry,dependingonthesignofthediscriminant,thecurvemayormaynotintersecttheaxis,asillustratedinfigure18.12.18.27Figure18.12PlotofstoredenergyversuscurrentratioxFromthefigureitisobviousthatif,therewillbesomecurrentratiothatyieldsanegativeandhenceanegative,againviolatingthepassivityprinciple.Therefore,,whichyieldsequaton18.26Fromequations18.27and18.28,(18.28)(18.29)ThedegreetowhichMapproachesitsupperboundisexpressedbyapositivenumber,calledthecoefficientofcoupling,definedasWhen,isalsozero,andtheinductorsareuncoupled.When,theinductorshaveunitycoupling,anidealizedsituationimpossibletorealizeinpractice.Example18.10Inbothcircuitsoffigure18.4,supposethatL1=5H,L2=20H,M=8H,I1=2A,andI2=4A.Find:(a)Thecouplingcoefficientk.(b)Thestoredenergy.SOLUTION(a)Forbothcircuits,(b)Forfigure18.4a,Forfigure18.4b,Example18.11Inthecircuitoffigure18.13,I1=6A.FindtheminimumvalueofthestoredenergyandthecorrespondingvalueofI2.SOLUTIONFromequation18.24,Figure18.13Coupledinductorsforcalculatingstoredenergyinexample18.11

ThisyieldsI2=2A,andthecorrespondingminimumstoredenergyWmin=18-36+72=54J.Followingthestandardmethodincalculusforfindingthemaximumandminimum,wesettozeroandsolvefor:6.IdealTransformerasACircuitElementAndApplicationsIdealization1.Thecoupledinductorshaveunitycoupling.i.e.,M2=L1L2,orthecouplingcoefficientk=1.EffectofIdealization1.Underthisidealizedconditionofunitycoupling,thepairofcoilshasthevoltagetransformationpropertyTwocoupleddifferentialequationscontainingthreeparametersL1,L2,andMcharacterizethecoupledcoilsoffigure18.4.Byimposingtwoidealizedconditionsontheseparameters,someveryusefulcircuitpropertiesresult.Whereaisaconstantandbothv1andv2arethevoltagedropsfromdottedtoundottedterminalsofthecoils,asthe

figure.

Toderivetheconditionofequation18.30,notethattheconstraintM2=L1L2impliesthatL1/M=M/L2.DenotingtheratioL1/MbyaleadstoL1=aMandM=aL2.Substitutingtheserelationshipsintoequations18.7anddividingequation18.7abyequation18.7byieldsAunitycouplingcoefficientisanidealizationthatisnotachievableinpractice.However,couplingcoefficientsnearunityareachievablebywindingtheturnsoftwoinductorsverycloselytogethersothatnearlyallthefluxthatlinksonecoilalsolinkstheothercoil.Withunitycoupledcoils,equation18.30indicatesthatthevoltagesv1(t)andv2(t),bothfromdottedtoundottedterminals,alwayshavethesamepolarity.Withcouplinglessthanunity,itispossibleforv1andv2tohaveoppositepolaritiesatsometimeinstants,asshowninexample18.7Theconstanta=L1/M=M/L2=inequation18.30isthensimplytheratioofthenumbersofturns,denotedbyN1andN2,ofthetwocoilsandisusuallyreferredtoastheturnsratio,i.e.,a=N1/N2.ItispossibletoshowthatIdealization2.Inadditiontounitycoupling,thecoupledcoilshaveinfinitemutualandself-inductances.EffectofIdealization1and2.Underthesetwoidealizedconditions,thepairofcoilshasthecurrenttransformationproperty.Wherebothi1andi2arethecurrentsenteringthedottedterminalsofthecoils,asperfigure18.4a.LettingL1→∝,L2→∝,andM→∝,L1/M=M/L2=a

IdealTransformerTwocoupledcoilssatisfyingtherelationshipsaresaidtobeanidealtransformer,showninfigure18.14a.

Inthefigure,thetwoverticalbarsserveasareminderofthepresenceofaferromagneticcoreinthephysicaldevice.Theterm“ideal”mayormaynotappearintheschematicdiagram.Again,themathematicalmodelofanidealtransformerdependsonlyontheturnsratioa:1andtherelativedotpositions.Toavoidthenegativesigninthecurrentrelationship,thealternativelabelingofvoltagesandcurrentsshowniffigure18.14bmaybeused.

Thesubscriptpstandsfortheprimarycoils,connectedtoapowersource,andsforthesecondarycoil,connectedtoaload.Notethatipisenteringatthedottedterminalandisleavingthedottedterminal.Thenotationoffigure18.14bismorecommonlyusedinthestudyofelectricpowerflow.Forourpresentpurposesincircuitanalysis,wehavedefinedanidealtransformerstrictlyfromitsterminalvoltage-currentrelationships.Insection8wewillshowhowapracticaltransformercanbeconstructedtoachieveapproximatelytheidealizedconditionsk=1and(L1,L2,M)→∝.Oneimportantsimplificationresultingfromtheidealizationsisthatanidealtransformerischaracterizedbytwoalgebraicequationscontainingasingleparametera,theturnsratio.Thisistobecontrastedwithapairofcoupledcoils,characterizedbytwodifferentialequationscontainingthethreeparametersL1,L2,andM.

Theinput-outputpowerpropertiesofanidealtransformerareveryinteresting.Fromequations18.30and18.31,appliedtofigure18.14a,theinstantaneouspowerdeliveredtoanidealtransformerisConsideredasasingleunit,anidealtransformerneithergeneratesnorconsumesinstantaneouspower-whateverinstantaneouspowerisreceivedatonesidemusttransfertotheotherside.Furthermore,sincep(t)isidenticallyzero,soisitsintegralwithrespecttot.Thusanidealtransformercannotstoreanyenergy.Also,unlikeLorC,thepresenceofidealtransformersinacircuitdoesnotincreasethenumberofpolesofanynetworkfunction.Thisfollowsbecauseequations18.30and18.31suggestrepresentinganidealtransformerbytwocontrolledsources,asshowninfigure18.15.Sucharepresentationisusefulwhenacircuitsimulationprogramdoesnotincludetheidealtransformerasapermissibleelement,butdoesadmitallfourtypesofcontrolledsources.IfthesecondaryofanidealtransformeristerminatedinaloadimpedanceZ(s),thentheimpedancelookingintotheprimaryisZin(s)=a2Z(s),whereaistheturnsratiotakeninthedirectionfromsourcetoload.(seefigure18.16,thedotsarenotmarkedonthefigurebecausetheirpositionareimmaterialforthisapplication.)ImpedanceTransformationPropertyDerivingthispropertyisstraightforward.Wemerelyapplyequations18.30and18.31tofigure18.16toobtainthedesiredresult:(18.34)Coupledinductorsareoftenusedforoneofthethreetransformationproperties:voltage

transformation,currenttransformation,andimpedancetransformation.Whenusedinsuchmanner,theyarecalledtransformers.Example18.12Thetransformerinthesystemoffugure18.17isassumedtobeideal.Theloadsrepresentedbyresistorsconsistof10incandescentlampsinparallel,eachdrawing0.5A.

Findthevoltage(magnitude)acrosseachlamp,(b)

Findthecurrent(magnitude)deliveredbythesource.10incandescentlampsRepresentedbyresistors

Sinceonlymagnitudesareinvolvedinthisproblem,dotpositionsonthetransformerandreferencedirectionsforvoltagesandcurrentsareimmaterial.Theturnsratioisa=1760/88=20.Fromequations18.30and18.31,thevoltageacrosseachlampisSOLUTIONAndthecurrentmagnitudethroughthepowersourceisExample18.13Figure18.18showsasimplifiedmodelofanaudioamplifiercontaininganidealtransformer.Theinputvoltageisat2kHzwithamagnitudeof1Vrms.Theloadisaloudspeaker,representedbya4-Ωresistance.

Findtheaveragepowerdeliveredtothe4-Ωloadifitisconnecteddirectlytotheamplifier.

Withthetransformerconnected,andwithaturnsratioa=5,findtheaveragepowerdeliveredtotheload.(c)Iftheturnsratioaisadjustable,whatshoulditbeinordertohavemaximumpowerdeliveredtotheload?Whatisthevalueofthemaximumpower?Modelofanamplifier(a)The(magnitudeofthe)currentthroughtheloadis(b)Lookingintotheprimaryofthetransformer,theimpedanceisSOLUTIONTherefore,theaveragepoweris

ModelofanamplifierFromthevoltagedividerformula,thevoltageacrosstheprimaryis(c)Fromthemaximumpowertransfertheorem,theturnsratioashouldbesuchthatthe4-Ωresistance

reflectsbacktotheprimaryas900Ωtomatchtheinternalresistanceoftheamplifier;i.e.,a2×4=900,ora=15.Witha=15,arepeatofthecalculationsofpart(b)yieldsPmax=11.11W.ThevoltageacrossthesecondaryisTherefore,7.CoupledInductorsModeledWithanIdealTransformerGivenacircuitcontainingcoupledinductors,anaturalwaytoanalyzethecircuitistowriteloopornodeequationsandthensolvethesimultaneousequationsbyanyofthetechniquesstudiedearlier.Althoughverygeneralandsystematic,suchmethodshavethedrawbackofgrosslyinvolvedmathematicaloperations.Thisobscurestheessentialpropertiesofthecircuit.Inthecurrentsectionweshallpresentsomeusefulmodelsforapairofcoupledinductors.Thesemodelsmakeuseofanidealtransformer.Sincethethreebasicpropertiesofanidealtransformerareextremelyeasytocomprehend,substitutingoneofthemodelsforapairofcoupledinductorshelpsustounderstandthecircuitoperationsmoreeasily,withoutcomplicatedmathematics.Asafirstcase,considerapairofinductorswithunitycoupling,asshowninfigure18.20a.Figures18.20band18.20cshowtwoequivalentcircuits,eachconsistingofoneinductoran