电气专业毕业设计外文翻译

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1、附录AAfewexampleswillrefreshyourmemoryaboutthecontentofChapter8andthegeneralapproachtoanodal-analysissolution.EXAMPLE17.12DeterminethevoltageacrosstheinductorforthenetworkofFig.Solution:Steps1and2areasndicatedinFig.17.22.Step3:NoteFig.17.23fortheapplicationofKirchhoff’scurre

2、ntlawtonodeV1:Fig.17.22Fig.17.23∑Ii=∑I00=I1+I2+I3V1-E/Z1+(V1/Z2)+(V1-V2)/Z3=0Rearrangingterms:V1[1/Z1+1/Z2+1/Z3]-V2[1/Z3]=E1/Z1(17.1)NoteFig.17.24fortheapplicationofKirchhoff’scurrentlawtonodeV2:0=I3+I4+IV2-V1/Z3+V2/Z4+I=0Rearrangingterms:V2[1/Z3+1/Z4]-V1[1/Z3]=-I(17.2)Fig

3、.17.24Groupingequations:V1[1/Z1+1/Z2+1/Z3]-V2[1/Z3]=E1/Z1V1[1/Z3]-V2[1/Z3+1/Z4]=I1/Z1+1/Z2+1/Z3=1/0.5kΩ+1/10kΩ+1/2kΩ=2.5mS∠-2.29°1/Z3+1/Z4=1/2kΩ+1/-5kΩ=0.539mS∠21.80°andV1[2.5ms∠-2.29°]-V2[0.5mS∠0°]=24mΑ∠0°V1[0.5mS∠0°]-V2[0.539mS∠21.80°]=4mΑ∠0°with24mΑ∠0°-0.5mS∠0°4mΑ∠0°-0.

4、539mS∠21.80°V1=2.5ms∠-2.29°-0.5mS∠0°0.5mS∠0°-0.539mS∠21.80°=(24mΑ∠0°)(-0.539mS∠21.80°)+(0.5mS∠0°)(4mΑ∠0°)/[(2.5ms∠-2.29°)(-0.539mS∠21.80°)+(0.5mS∠0°)(0.5mS∠0°)]=-10.01ν-j4.81ν/-1.021-j0.45=11.106ν∠-154.33°/1.116∠-156.21°V1=9.95∠1.88°MathCadThelengthandcomplexityoftheabovem

5、athematicaldevelopmentstronglysuggesttheuseofanalternativeapproachsuchasMathCad.NoteinMathCad17.2thattheequationsareenteredinthesameformatasEqs.(17.1)and(17.2).BothV1andV2weregenerated,butbecauseonlyV1wasaskedfor,itwastheonlysolutionconvertedtothepolarform.Inthelowersoluti

6、onthecomplexitywassignificantlyreducedbysimplyrecognizingthatthecurrentisinmilliamperesandtheimpedancesinkilohms.Theresultwillthenbeinvolts.K:=10³m:=0.01rad:=1V1:=1+jV2:=1+jdeg:=π/180GivenV1·[1/5·k+1/10j·k+1/2·k]-V2·1/2·k≈24·mV1·[1/2·k]-V2[1/2·k+1/-5j·k]≈4·mFind(V1,V2)=9.9

7、44+0.319jVolts1.786-0.396jVoltsV1:=9.944+0.319jV1=9.949arg(V1)=1.837·degRecognizingthatcurrentinmAresultsehenZisinkilohmns,analternativeformatfollows:GivenV1·[1/5+1/10j+1/2]-V2·1/2≈24V1·1/2-V2[1/2+1/-5j]≈4Find(V1,V2)=9.944+0.319jVolts1.786-0.396jVoltsV1:=9.944+0.319jV1=9.9

8、49arg(V1)=1.837·degMATHCAD17.2DependentCurrentSourcesFordependentcurrentsources,theproced