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Chapter4,Problem1.CalculatethecurrentiointhecircuitofFig.4.69.Whatdoesthiscurrentbecomewhentheinputvoltageisraisedto10V?Figure4.69Chapter4,Solution1.+−51411i=+=Ω=+4)35(8,===101i21ioPROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.0.1ASincetheresistanceremainsthesamewegeti=10/5=2Awhichleadstoio=(1/2)i=(1/2)2=1A.Chapter4,Problem2.FindvPROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.ointhecircuitofFig.4.70.Ifthesourcecurrentisreducedto1μA,whatisv?oFigure4.70Chapter4,Solution2.A21ii21==,3)24(6Ω=+,41i21i1o====ooi2v0.5V0.5μVIfi=1μA,thenvso=Chapter4,Problem3.(a)InthecircuitinFig.4.71,calculatevandIPROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.oowhenv=1V.s(b)Findvoandiwhenvos=10V.(c)WhatarevandIoowheneachofthe1-Ωresistorsisreplacedbya10-Ωresistorandvs=10V?Figure4.71Chapter4,Solution3.+−+−+vo(a)WetransformtheYsub-circuittotheequivalentΔ.,R43R4R3R3R2==R23R43R43=+2vvso=independentofRio=v/(R)o0.5V0.5AWhenvs=1V,vo=,io=5V5A(b)Whenv=10V,v=so,io=(c)Whenv=10VandR=10Ω,svo=5V500mA,i=10/(10)=oChapter4,Problem4.UselinearitytodetermineiinthecircuitinFig.4.72.oFigure4.72Chapter4,Solution4.IfIo=1,thevoltageacrossthe6Ωresistoris6Vsothatthecurrentthroughthe3Ωresistoris2A.+v1.A34vio1==Ω=263,v=3(4)=12V,oHenceIs=3+3=6APROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.IfIs=6AIo=1Is=9AIo=9/6=1.5AChapter4,Problem5.ForthecircuitinFig.4.73,assumevo=1V,anduselinearitytofindtheactualvalueofv.oFigure4.73Chapter4,Solution5.+−V2131V1=+⎟⎠⎞⎜⎝⎛=Ifv=1V,o310v322V1s=+⎟⎠⎞⎜⎝⎛=310Ifv=v=1so=15x103PROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.Thenvs=15vo=4.5VChapter4,Problem6.ForthelinearcircuitshowninFig.4.74,uselinearitytocompletethefollowingtable.ExperimentVVso112V4V2--16V31V--4---2VLinearCircuit+_VsVo+–Figure4.74ForProb.4.6.Chapter4,Solution6.Duetolinearity,fromthefirstexperiment,13osVV=Applyingthistootherexperiments,weobtain:VExperimentVso24816V0.333VPROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.31V4-6V-2VChapter4,Problem7.PROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividualcoursepreparation.IfyouareastudentusingthisManual,youareusingitwithoutpermission.UselinearityandtheassumptionthatVo=1VtofindtheactualvalueofVinFig.4.75.o.+_4Ω3Ω2Ω4V+_Vo1ΩFigure4.75ForProb.4.7.Chapter4,Solution7.IfVo=1V,thenthecurrentthroughthe2-Ωand4-Ωresistorsis½=0.5.Thevoltageacrossthe3-Ωresistoris½(4+2)=3V.Thetotalcurrentthroughthe1-Ωresistoris0.5+3/3=1.5A.Hencethesourcevoltage=+=11.534.5VsvxIf4.51svV=⎯⎯→1440.8889V4.5svx=⎯⎯→==888.9mVThen.Chapter4,Problem8.inthecircuitofFig.4.76.Usingsuperposition,findVo+_1Ω3Ω9V3VVo4Ω5Ω+_Figure4.76ForProb.4.8.PROPRIETARYMATERIAL.©2007TheMcGraw-HillCompanies,Inc.Allrightsreserved.NopartofthisManualmaybedisplayed,reproducedordistributedinanyformorbyanymeans,withoutthepriorwrittenpermissionofthepublisher,orusedbeyondthelimiteddistributiontoteachersandeducatorspermittedbyMcGraw-Hillfortheirindividual