电工学原理及应用electricalengineering经典双语详解

Chapter4TransientsElectricalEngineeringandElectronicsII4CourseHours4.Transients4.1Introduction4.2Initialstateandsteadystate4.3First-orderRCCircuits4.4First-orderRLCircuitstENewsteadystatetransientCOldsteadystateKRE+_SwitchKisclosed4.1IntroductionNewsteadystateRUs+_ConceptionofsteadystateandtransientstateWhent=0，uc(0)=0Whent=∞,uc(∞)=UsOldsteadystateWhythetransientresponsehappens?NotransientIResistivecircuitt=0ER+_IKResistorisaenergy-consumptionelement,currentisproportionaltovoltage,notransientresponsewillhappenevenifchangingsourceEnergycannotchangeinstantlybecauseofaccumulatingordecayingperiod.ChargingordischargingChangegraduallyElectricfieldenergyEKR+_CuCEtMagneticfieldenergyChangegraduallyKRE+_t=0iLtE/REnergycannotchangeinstantlybecauseofaccumulatingordecayingperiod.Thecausesoftransients1.Energystorageelements-inductorsandcapacitorschangegradually；2.Changingcircuit,suchasswitchingsource.TransientsThetime-varyingcurrentsandvoltagesresultingfromthesuddenapplicationofsources,usuallyduetoswitching.Bywritingcircuitequations,weobtainintegro-differentialequations.Theseequationscanbeconvertedtopuredifferentialequationsbydifferentiatingw.r.ttime.Thestudyoftransientsrequireustosolvedifferentialequations.4.2InitialstateandsteadystateAssumechangingcircuitwhent=0,thent=0–isendpointofoldsteadystate;t=0+isthestartpointoftransientstate.Fromt=0–tot=0+,iL、uCcannotchangesuddenly.t=0tt=0-t=0+Thelawofchangingcircuit(换路定则)Exa4.2Solu(1)BroforeswitchWeknowthatBythelawofcircuitchangingKnowns：beforecircuitchanging，C、L

havenoenergy;Find：thecurrentsandvoltagesofallelements。S(a)CU

R2R1t=0+-LHowtogetinitialvalueExample4.2,changingmoment，capacitanceshort,changingmoment，inductanceopeniC

、uL

altersuddenly(2)Bythecircuitatt=0+,FindotherunknownsSCU

R2R1t=0+-L(a)circuitiL(0+)U

iC(0+)uC

(0+)uL(0+)_u2(0+)u1(0+)i1(0+)R2R1+++__+-(b)t=0+

circiutHowtogetinitialvalueExercise1:AssumingoldcircuitisinDCsteadystatebeforeswitchKisclosed.howtogetuC(0+),iR(0+)?iRR14k12VKt=08kR22mFuCSolution:Whent=0-,capacitorisconsideredasopencircuit,wegetequivalentcircuit.R14k12VuC(0–)8kt=0-R14k12VuC(0–)8kiRR14k12VKt=08kR22mFuCsubstitutingvoltagesourceforuC(0+)iR(0+)8kR2+–

uC(0+)t=0＋HowtogetinitialvalueExercise2:GivenbyR1=4Ω,R2=6Ω,R3=3Ω,C=0.1µF,L=1mH,US=36V,switchSisclosedforalongtime.OpentheswitchSwhen

t=0,howtogettheinitialvaluesofallelements?HowtogetinitialvalueAnswer:Uc(0)=12V,iL(0)=4A,iR(0)=2AConclusions

1.Atthemomentofchangingcircuit，uC、iLcannotaltersuddenly,butic,uLcanchangesuddenly.

3.Beforecircuitschanged,ifuC(0-)0,the

capacitance

canbereplacedbyanidealvoltagesourcewithuc(0+)

att=0+;

ifiL(0-)0,theinductancecanbereplacedbyanidealcurrentsourcewithiL(0+)att=0+.2.Beforecircuitschanged,ifenergy-storageelementshavenoenergy,justafter

circuitschanged(t=0+)

thecapacitanceandinductanceareviewedasshortcircuitsandopencircuit,respectively.DCSteadyStateResponseThetransienttermsforcurrents/voltagesdecaytozerowithtime.UnderSteadyState:Forcapacitorswithdcsource,capacitancebehavesasopencircuits.Forinductorswithdcsource,inductancebehavesasshortcircuits.DCSteadyStateResponseThestepsindeterminingthesteadystateresponseforRLCcircuitswithdcsourcesare:1.Replacecapacitanceswithopencircuits.2.Replaceinductanceswithshortcircuits.3.Solvetheremainingcircuitusingmethodsinchapter2.Example4.1Findsteady-statevaluesofvxandixinthiscircuitfort>>0.Answer:vx=5V,ix=1At>>0Exercise4.3Findsteady-statevaluesoflabeledcurrentsandvoltagesfort>>0.Answer:va=50V,ia=2Ai1=2A,i2=1A,i3=1AHomeworkP4.2P4.6P4.8EquivalentcircuitofFirst-ordercircuitTwoparts:one(equivalent)capacitororinductor;atwoterminalnetworkwithresistanceandsources.NLNCorFirst-ordercircuitOnlyone(equivalent)capacitororinductorisincludedinalinearcircuit.4.3First-orderRCCircuitsAccordingtoTheveninLawNLNCorRULuLiL+-RUCuCiC+-4.3First-orderRCCircuitsDifferentialequationoffirst-orderRCcircuitRULuLiL+-RUCuCiC+-Solution:First-orderRCCircuitsExample:tofindthetransientresponseafterchangingcircuitwhent=0.First-orderRCCircuits——homogeneoussolution——particularsolutionFirst-orderRCCircuitshomogeneoussolutionFirst-orderRCCircuitsThereforeThen,thefinalsolutionisParticularsolutionFirst-orderRCCircuitsThesolutionofdifferentialequationSubstitutingtheinitialcondition：First-orderRCCircuits

Timeconstantreflectsthelengthoftransientperiod.t234567e-t/36.8%13.5%5%1.8%0.3%0.25%0.09%Afteraboutfivetimeconstants,thetransientresponseisover.Afteronetimeconstants,thetransientresponseisequalto36.8percentofitsinitialvalue.ThecurvesversustimeTheinitialslopintersectsthefinalvalueatonetimeconstant.MountingcurveDecayingcurve

Timeconstantreflectsthelengthoftransientperiod.Thesolutionofdifferentialequation——Timeconstant——Steadystatevalue——InitialvalueFirst-orderRCCircuitsSolutionofotherparametersThreeelementsmethodThreeelements:1.steadystatevaluef(∞);2.timeconstantτ;3.initialvaluef(0+).FormulaofThreeelementmethod：f(∞)——steadystatevalueτ——timeconstantf(0+)——initialvalueτ=RC——timeconstantofRCcircuitτ=？？——timeconstantofRLcircuitThreeelementsmethod解：用三要素法求解电路如图，t=0时合上开关S，合S前电路已处于稳态。试求电容电压

和电流

。(1)拟定初始值由t=0-电路可求得由换路定则ApplicationExamplet=0-等效电路9mA+-6k

RS9mA6k2F3kt=0+-C

R(2)拟定稳态值由换路后电路求稳态值(3)由换路后电路求时间常数t∞电路9mA+-6k

R

3kt=0-等效电路9mA+-6k

RR0旳计算类似于戴维宁等效电阻。即从储能元件两端看进去旳等效电阻。三要素uC旳变化曲线如图18V54VuC变化曲线tO4.4First-orderRLCircuitsTimeconstantτ=RCτ=L/RRULuLiL+-RUCuCiC+-4.4First-orderRLCircuitsThreeelementmethodInitialvalue:t=0-→t=0+

f(0+)Steadystatevalue:t=∞f(∞)Timeconstant:τ=RCτ=L/RSubstitutingthreeelementsDrawthecurveversustimeStepsLimitedCondition:1)first-ordercircuit2)DCsourceExample4.2Findvoltageofv(t)andcurrenti(t)inthiscircuitfort>0.Answer:i(0)=0,i(∞)=2AV(0)=100,V(∞)=0Exampl