工程电路分析第五单元答案

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1、thEngineeringCircuitAnalysis8EditionChapterFiveExerciseSolutions1.(a)flinear=1+x(b)xxflineare%error-6101.0000011.000001000--4101.0011.00100005--2101.011.0100501670.005%0.11.11.1051709180.5%122.71828182826%(c)Somewhatsubjectively,wenotethattherelativeerrori

2、slessthan0.5%forx<0.1sousethisasourestimateofwhatconstitutes“reasonable.”Copyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.thEngineeringCircuitAnalysis8EditionChapterFiveExerciseSolutions2.yt()4sin2t4(2)8t

3、t8tt4sin2(a)Define%error1004sin2tt8t4sin2t%error-6-6-6108×108.000×100%(to4digits)-4-4-4108×108.000×100%(to4digits)-2-2108×100.079990.01%-1-1108×100.79470.7%1.08.03.63755%(b)Thislinearapproximationholdswell(<1%relativeerror)evenuptot=0.1.Abovethatvaluean

4、dtheerrorsareappreciable.Copyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.thEngineeringCircuitAnalysis8EditionChapterFiveExerciseSolutions3188163.i6A.i2A86Aonly82Vonly38113811Copyright©2012TheMcGraw-H

5、illCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.thEngineeringCircuitAnalysis8EditionChapterFiveExerciseSolutions4.(a)Wereplacethevoltagesourcewithashortcircuitanddesignatethedownwardcurrentthroughthe4resistorasi'.Then,i'=(10)(9)/

6、(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

7、.07692.Thus,Ix=111.1mACopyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.thEngineeringCircuitAnalysis8EditionChapterFiveExerciseSolutions145.(a)Replacingthe5Asourcewithanopencircuit,i31.75A.x3Aonly14105R

8、eplacingthe3Asourcewithanopencircuit,i51.316A.x5Aonly19(b)-I(5/19)=-1.75.Thus,I=6.65A.Copyright©2012TheMcGraw-HillCompanies.Permissionrequiredforreproductionordisplay.Allrightsreserved.thEngineeringCircui