M35_KNIG9404_ISM_C35.pdf

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1、35ACCIRCUITSConceptualQuestions35.1.(a)a:–100Vb:+60Vc:+80V.Theemfisthex-componentofthecounterclockwiserotatingvectors.(b)a:Decreasingb:Decreasingc:Increasing35.2.(a)1.0A.UseIVRRR=/.(b)4.0A.UseIVRRRR==//ε0.(c)2.0A.IRdoesnotdependonfrequency.35.3.(a)4.0A.UseICc=ωε0

2、forallpartsofthisquestion.(b)4.0A(c)4.0A135.4.(a)fc=100Hz.Useωπcc==2.fRC1(b)100Hz.Useωπcc==2.fRC(c)200Hz.Thecrossoverfrequencydoesnotdependonthepeakemf.35.5.(a)1.0IL=A.UseIVXLLL==//εω0(L)forallpartsofthisquestion.(b)4.0IL=A.(c)1.0IL=A.135.6.(a)1000Hz.Useω0=.Resis

3、tancedoesnotmatter.LC1(b)1000Hz=707.1Hz21(c)1000Hz=707.1Hz2(d)1000Hz.Peakemfdoesnotmatter.35.7.Lessthan.Herethecurrentleadstheemf,soweknowthatφ<0(seeEquation35.22).FromEquation35.27,wefind?1??XXLC?XLφ=

4、ightsreserved.Thismaterialisprotectedunderallcopyrightlawsastheycurrentlyexist.Noportionofthismaterialmaybereproduced,inanyformorbyanymeans,withoutpermissioninwritingfromthepublisher.35-135-2Chapter35ThereactancesaregivenbyXCc=1/(ω)andXLL=ω,andtheresonancefrequen

5、cyisω0=1/LC.CombiningtheserelationshipsgivesXL21=<ωωLC1?<=ω0XCLC35.8.Wearegiventhatωω0<.FromthelastrelationshipoftheanalysisinQ35.7,weseethatthisimpliesthatXXLC>,soφ>0andthecurrentlagstheemf.35.9.ThepowerwillincreasewhenthepeakcurrentIincreases.Thiswillincreasewh

6、enyou(1)decreaseR,(2)setXXLC=.35.10.(a)8.0W.UsePI=12R.RR2(b)16W.Doublingthepeakemfdoublesthecurrent.(c)8.0W.DoublingtheemfwoulddoublethecurrentexceptthatwhentheresistorisdoubledIRremainsthesame.PRincreasesduetodoublingR,asinpart(a).ExercisesandProblemsSection35.1

7、ACSourcesandPhasors35.1.Model:Aphasorisavectorthatrotatescounterclockwisearoundtheoriginatangularfrequencyω.Solve:(a)ReferringtothephasorinFigureEX35.1,thephaseangleisπrad262rad..262rad2ωωt=°180?°30=°150×=.262rad?===2210rad/s×180°t1510s×?3(b)Theinstantaneousvalue

8、oftheemfisEE==0cosωt(12V)cos(262rad).=?10VAssess:Becarefultochangeyourcalculatortotheradianmodetoworkwiththetrigonometricfunctions.35.2.Model:Aphasorisavectort