電阻電路例題

1、Question1aPleaseusesourcetransformationtofindthecurrentIinthefollowingcircuit.8Marks5A03,I7升Q2Ao4iSolution:Usesourcetransformationtoredrawthecircuit,wewillget:5MarksSo:0.50.5A3MarksQuestion2:Pleaseusesourcetransformationtofindthecurrentiinthefollowingcircuit.82MarksR4R4Solution:Redrawtheoriginalcirc

2、uitasthefollowing:R3Is5MarksThenthecurrentiis:2.RIi3s-2RR233MarksR3Is5MarksThenthecurrentiis:2.RIi3s-2RR233MarksQuestion3:15MarksUsethenode-voltagemethodtofindthebranchcurrentandthepowerdissipatedinthe18resistorinthecircuitinFigureP3.6FigureP312I0Solution:6Selectthelowernodeasthereferencenodeanddefi

3、nethenodevoltagesofnode112I0andnode2asVandV.ApplyKCLfornode1andnode2,andweget3Marks3MarksH.51t0122Marks2Marks2Marks2Marks2Marks2Marks2Marks2Marks2Marks2MarksFig.1VVVV12I21622003Marks6618Thecontrollingcurrentcanbeexpressedas:VIt3Marks06SolvingforVandVgives12V18V2V18Then,thebranchcurrentIt3A066Thepowe

4、rdissipatedinthe18resistoris:P喈2印2I01L8卡8印2S.818W1818|Question4:12MarksUsemeshcurrentmethodtofindthecurrentIinthecircuitinFig.1.Solution:AssumethereferencedirectionsofthreemeshcurrentsI1,I2andI3areclockwise,then,wewillget:262A262AFromthecircuit,weknowI3=2A,onlyI1,I2areunknown.BuildKVLequationsforthe

5、appointedsupermesh,wewillget:623(I2)6121301124MarksAstothecurrentsource4A,wecangetanotherequation:II4623(I2)6121301124MarksAstothecurrentsource4A,wecangetanotherequation:II4124MarksSowecanget:I1A才A21MarksSo:1MarksQuestion5:15MarksUsethemesh-currentmethodtofindthepowerdissipatedinthe6resistorinthecir

6、cuitinFigureP3.8260VFigureP38260VFigureP3Solution:8260VDefinemeshcurrentsforthetwomeshesasIandI,andbuildtwomesh-currentequationsas:816II，200.5V8260VDefinemeshcurrentsforthetwomeshesasIandI,andbuildtwomesh-currentequationsas:816II，200.5V01124Marks214160H206II.022214MarksThecontrollingvoltagecanbeexpr

7、essedas:V4I23MarksSolvingforI1andIgives2I8A,and1I1A22MarksThepowerdissipatedinthe6resistoris:2MarksPI6睢116486W612MarksQuestion6:13MarksFindthecurrentIinthecircuitinFig.3byusingtheTheveninequivalentoftheremainingcircuitexceptfortheresistorof14.Solution:Thecircuitwithout14branchisasfollowing:Inorderto

8、findI,wewillfindtheTheveninequivalentcircuitbetweenterminalaandb.2MarksStep1:FindTheveninequivalentvoltage.Obviously,theTheveninequivalentvoltageis:2.520VThV。（不E）5.RW4MarksThab2.510205Step2:Findequivalentresistor.Turnoffallindependentsources,wewillget:Sotheequivalentresistoris:Sotheequivalentresisto

9、ris:2.5/102056B4MarksSowecangettheequivalentcircuit:SothecurrentIis:7.5V1MarkV皿5ISothecurrentIis:7.5V1MarkV皿5IThRH146H14Th0.375A2MarksQuestion7:12MarksTheoperationalamplifierinthecircuitshowninFigure6isideal.Findthecurrentio3Marks3Marks83Marks83MarksSolution:Thevoltagesattheinputnodesofanidealoperat

10、ionalamplifierareequalduetovirtualshort,so:v2Va3MarksApplyKCLatnodea:v聯(lián)12聯(lián)一I03MarksSo,weget:vII30V03MarksApplyOhmsLawtothe8kIresistor,wecangetthecurrenti0:2vH睢0o-mAmA=3.5mAQuestion8:12MarksTheoperationalamplifierinthecircuitshowninFigure6isideal.Findthecurrenti0.Solution:6V6Vi0i0Solution:6V6Vi0i0+v0Figure6Definenodevoltagesv0,v1andv2asshowninthecircuitabove.Duetovirtualshortfortheoperationalamplifier,weget:v6V13MarksDuetovirtualshortfortheoperationalamplifier,weget:v6v=36V213MarksThen,thevoltagev0canbeobtained:VVV63642Vo123MarksApplyOhmsLawtothe10k-resis