本文格式為Word版，下載可任意編輯——電路原理題目第一章

1.圖中所示電路中，已知a點、b點的電位分別為φa=10V,φb=5V。

則電動勢E＝____V:電壓U＝____V:由參考方向，得

E=φa?φb,U=φa?φb

由已知得E=10?5=5V,U=10?5=5V

2.蓄電池端紐a和b兩端的電壓Uab和電動勢Eab的關(guān)系是：

Eab=Uab

Eab=?UabEab=?Uab-

由參考方向知道，電壓Uab是指電壓從a點到b點的降低。電動勢Eab是指電壓從a點到b點的升高。由此得Eab＝－Uab

3.圖中所示電路中，已知a點、b點的電位分別為φa=10V,φb=5V。

則電動勢E＝____V:電壓U＝____V:

由參考方向，得

E=φb?φa,U=φa?φb

由已知得E=5?10=?5V,U=10?5=5V

4.

圖中所示電路中，已知a點、b點的電位分別為φa=10V,φb=5V。

則電動勢E＝____V:電壓U＝____V:由參考方向，得

E=φa?φb,U=φb?φa

由已知得E=10?5=5V,U=5?10=?5V

5.圖中所示電路中，已知a點、b點的電位分別為φa=10V,φb=5V。

則電動勢E＝____V:電壓U＝____V:

由參考方向，得

E=φb?φa,U=φb?φa

由已知得E=5?10=?5V,U=5?10=?5V6.二端元件的端電壓和流過的電流如下圖。

則二端元件發(fā)出的功率P＝____W:

由電流的參考方向與電壓降的方向一致,得，發(fā)出功率P＝－2×3＝－6W7.二端元件的端電壓和流過的電流如下圖。

則二端元件發(fā)出的功率P＝____W:

由電流的參考方向與電壓降的方向不一致,得，發(fā)出功率P＝2×3=6W

ThereferencedirectionofthevoltageUacrosstheresistorRisdesignatedas:rightispositiveandleftisnegative,whilethereferencedirectionofit’sthecurrentIisalongthearrowfromlefttoright,thentherelationshipofUandIis:

U=RIU=RI-U=?RI

BasedonOhm’slaw,notingthatthereferencedirectionsofUandIfortheresistorRarenon-associated.

8.Forthecircuitinthefigure,thevoltageUabis:

Uab=?14VUab=?14V-Uab=11VUab=?5VUab=1V

MarkthereferencedirectionofthevoltageUasshowninfigure.

Weget

U=3?1=3V

AccordingtoKVL,thevoltageis:

Uab=?8+3=?5V

9.Forthecircuitinthefigure,thecurrentI=____A:

MarkthereferencedirectionofthecurrentI1asshowninfigure.

BasedonOhm’slaw,

I1=12/2=6A

AccordingtoKCL,

I=4?I1=?2A

10.Forthecircuitinthefigure,thecurrentI=____Aandthe

voltageU=____V:

MarkthereferencedirectionsofthecurrentI1andthevoltageU1asshowninfigure.

BasedonOhm’slaw,

I1=12/2=6AU1=3?8=24V

AccordingtoKCLandKVL,

I=8?I1=2AU=U1+12=36V

11.Forthecircuitinthefigure,thecurrentI=____Aandthe

voltageU1=____V，U2=____V:

MarkthereferencedirectionsofthecurrentI1andthevoltageU3andU4asshowninfigure.

BasedonOhm’slaw,

I1=9/1=9A

U3=3?2=6VU4=2?2=4V

AccordingtoKCLandKVL,

I=3+2?I1=?4AU1=U3+9=15VU2=U4+9=13V

12.Forthecircuitinthefigure,thecurrentI=____AandthevoltageU1=____V，U2=____V:

MarkthereferencedirectionsofthecurrentI1andthevoltageU3andU4asshowninfigure.

BasedonOhm’slaw,

I1=9/1=9A

U3=3?2=6VU4=2?2=4V

AccordingtoKCLandKVL,

I=3+2?I1=?4AU1=U3+9=15VU2=U4+9=13V

13.Giventhecircuitasshowninthefigure,(1)whenI=4A，the

currentofcurrentsourceIS＝______A;(2)whenU=9V，thecurrentofcurrentsourceIS＝______A:

WritetheequationapplyingKVL,

U1=3U1+I×1

AccordingtoOhm’slawandKCL,

U=3×(U1/1+I)

AccordingtoOhm’slawandKCL,

IS=(U+U1)/2+U1/1+I

Combineandsolvetheequationsabovebyusingtheknownvalues.（1）當(dāng)I=4A時，Is=4A（2）當(dāng)U=9V時，Is=6A

14.Inthefigure,thecurrentofcurrentsourceIS=(A)?

WritetheequationapplyingOhm’slawandKVL,

U=?100U1+40×2=?100U1+80

AccordingtoOhm’slaw,

U1=0.2IS

AccordingtoKCLandOhm’slaw,

U=5(IS?2)

Combiningtheaboveequations,wehave

IS=3.6A

15.Refertothecircuitofthefigure,thepowerdeliveredbythe

independentsourceisP=_____W.

WriteKCLontheuppernode,

U/2+U/3+0.5U=12

Thus,

U=9V

Thepowerdeliveredbytheindependentsourceis

Pgen=12×9=108W

16.Inthefollowingcircuit,findthenodevoltageV3=__8.75____V.

(hint:usesimulationtools.)

其次章

1.AsFig2-1shown,theinputresistanceofthetwocircuits

are______Ωand______Ω

Simplifythecircuit,weget

Thebridgeisbalancedandthe5ohmresistorcanberemoved,weget

Fromtheseries-parallelformula,thesolvedresistanceis

R=10Ω

Solutionfor(b)：

Simplifythecircuit,weget

Fromtheseries-parallelformula,thesolvedresistanceis

R=10Ω

2.ThecircuitinFig2-2canbesimplifiedas：Fig（a）,Fig（b）,Fig

（c）,Fig（d）.Choose（）or（）.

解析：IntheFig.2-2,the3ohmresistorconnectedinserieswiththe5Acurrentsourcedoesn’tcontributetotheportvoltage,itonlyaffectsthevoltageofcurrentsource.Soitcanberemovedinthesimplification.

UsingsourcetransformationtothecircuitinFigabove,wegetthecircuitshowninFigbelow

3.ThecircuitinFig2-3canbesimplifiedas：Fig（a）,Fig（b）,Fig

（c）,Fig（d）.Choose（）.

&：Removethe10ohmresistorconnectedinparallelwiththe6Vvoltagesource,becauseitdoesn’tcontributestotheportvoltage,thenweget

Transformthe2Acurrentsourceinparallelwiththe10ohmresistorintoavoltagesourceinserieswitharesistor,thenweget

It’sfinallysimplifiedas

4.

UsingsourcetransformationtosolvethecurrentI(/A)ofthecircuitinthefigure:

Usingsourcetransformationtothe8Vvoltagesourceinserieswiththe4ohmresistor,weget

It’ssimplifiedas

Accordingtothecharacterof“virtualshortcircuit〞,wehave

uc=uo

Accordingtocharactersof“virtualopencircuit〞and“virtualshortcircuit〞,writeKCLequationonthenoninvertinginputterminal,weget

Combiningtheaboveequations,wehave

Reorganizeequationsandget

u=0.5(ui+uo)

ApplyingOhm’slawtotheoutputresistorRo,andbasedonthecharacterof“virtualopencircuit〞fortheidealop-amp,

Substituteknownvaluesintoit，thus

13.IntheFigure,theu-irelationshipofelementXisi=Aeu/B,whereinAandBareallconstants.Analyzetherelationshipbetweentheoutputvoltageuoandtheinputvoltageui,andselecttheoperationfunctionofthecircuit.

Accordingtothecharacterof“virtualshortcircuit〞,wehave

u+uo=0

thatis

u=?uo

Accordingtocharactersof“virtualopencircuit〞and“virtualshortcircuit〞,writeKCLequationontheinvertinginputterminal,weget

Reorganizeequationsandget

Hence,thecircuitrealizedthelogarithmoperation.

Step1Selectareferencenode,anddesignatethevoltageofindependentnodeasUa,seeFigure(a).WriteKVLequationfortheindependentnode(1/13.7+1/17.2+1/11.9)Ua=5/13.7+15/17.2+10/11.9+1Step2WecangetthevoltageofindependentnodeisUa=14.31VStep3ThevoltageacrosstheindependentcurrentsourceisU=Ua+1×5.5=19.8VStep4BranchcurrentsareI1=Ua?5/13.7=0.680AI2=Ua?15/17.2=?0.0401AI3=Ua?1011.9=0.362A5.SolvethecurrentIflowsthrough8VvoltagesourceandthevoltageUacross1.5Acurrentsourcebyusingnodevoltageanalysisforthecircuitasshowninfigure5-5.I=_____A,U=_____V.解析：Inoriginalcircuit,designatethereferencenodeandindependentnodes,asshowninfigure(a).WriteKVLequationsfornodesaandb

Rewritetheseequationsas

Step2

Solveequations,weget

Ua=10.75VUb=4.031V

Step3Thecurrentis

Step4

Thevoltagecrossthecurrentsourceis

U=Ua+4×1.5=10.75+6=16.8V

6.Writethenodevoltage