工程电路分析第五单元答案

EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.1.(a)flinear=1+x(b)xflinearex%error10-61.0000011.000001000-10-41.0011.00100005-10-21.011.0100501670.005%0.11.11.1051709180.5%122.71828182826%(c)Somewhatsubjectively,wenotethattherelativeerrorislessthan0.5%forx0.1sousethisasourestimateofwhatconstitutes“reasonable.”EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.2.()4sin24(2)8ytttt(a)84sin2%1004sin2ttDefineerrortt8t4sin2t%error10-68×10-68.000×10-60%(to4digits)10-48×10-48.000×10-40%(to4digits)10-28×10-20.079990.01%10-18×10-10.79470.7%1.08.03.63755%(b)Thislinearapproximationholdswell(1%relativeerror)evenuptot=0.1.Abovethatvalueandtheerrorsareappreciable.EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.3.86Aonly3186A3811i.82Vonly8162A3811iEngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.4.(a)Wereplacethevoltagesourcewithashortcircuitanddesignatethedownwardcurrentthroughthe4resistorasi'.Then,i'=(10)(9)/(13)=6.923ANext,wereplacethecurrentsourceintheoriginalcircuitwithanopencircuitanddesignatethedownwardcurrentthroughthe4resistorasi.Then,i=1/13=0.07692AAdding,i=i'+i=7.000A(b)The1Vsourcecontributes(100)(0.07692)/7.000=1.1%ofthetotalcurrent.(c)Ix(9)/13=0.07692.Thus,Ix=111.1mAEngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.5.(a)Replacingthe5Asourcewithanopencircuit,3Aonly1431.75A1410xi.Replacingthe3Asourcewithanopencircuit,5Aonly551.316A19xi.(b)-I(5/19)=-1.75.Thus,I=6.65A.EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.6.(a)Opencircuitingthe4Asourceleaves5+5+2=12inparallelwiththe1resistor.Thus,17Av=(7)(1||12)=(7)(0.9231)=6.462VOpencircuitingthe7Asourceleaves1+5=6inparallelwith5+2=7.Assistedbycurrentdivision,14A7(1)42.154V77vThus,v1=6.462–2.154=4.308V(b)Superpositiondoesnotapplytopower–that’sanonlinearquantity.EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.7.(a)27A7(5)712.89V19v22A14(5)27.368V19v(b)Weseefromthesimulationoutputthatthe7Asourcealonecontributes12.89V.Theoutputwithbothsourcesonis5.526V,whichagreeswithinroundingerrortoourhandcalculations(5.522V).EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.8.(a)4V→8V;10V→20V(b)4V→–4V;10V→–10V(c)notpossible;superpositiondoesnotapplytopower.EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.9.3||21||312(15)2.454V(3||2)1(1||3)2xv3||21||36(10)0.5454V(3||2)1(1||3)2xv3||21||36(5)1.909V(3||2)1(1||3)2xv2.455Vxxvv(agreeswithinroundingerror)EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.10.(a)Withtheright-handvoltagesourceshort-circuitedandthecurrentsourceopen-circuited,wehave2||5=10/7Byvoltagedivision,lefthand4V1(4)0.7368V3110/7xvWiththeothervoltagesourceshort-circuitedandthecurrentsourceopen-circuited,wehave(3+1)||5=2.222.52.22242.105V2.2222v.Then,righthand4V12.1050.5263V4xvFinally,withbothvoltagesourcesshort-circuited,wefindthat2A3(1)21.105V3110/7xvAddingthesethreetermstogether,vx=1.316V(b)(0.9)(1.316)=0.7368+1.105k–0.5263Solving,k=0.8814.Hence,weshouldreducethe2Asourceto2k=1.763A(c)Ourthreeseparatesimulations:EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.Ourreducedvoltagealternative:EngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.11.Weselectthebottomnodeasthereference,thenidentifyv1withthelefthandterminalofthedependentsourceandv2withtherighthandterminal.Viasuperposition,wefirstconsiderthecontributionofthe1Vsource:11210500070002000vvvand120.2107000vvSolving,v1’=0.237VNext,weconsiderthecontributionofthe2Asource:1122500070002000vvvand120.2107000vvSolving,v1”=–2373V.Addingourtwocomponents,v1=–2373V.Thus,ix=v1/7000=339mAEngineeringCircuitAnalysis8thEditionChapterFiveExerciseSolutionsCopyr