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1、3TechniquesofCircuitAnalysisDrillExercisesDE3.1[a]11,8resistors,2independentsources,1dependentsource[b]9[c]9,R4¡R5formsanessentialbranchasdoesR8¡10V.Theremainingsevenbranchescontainasingleelement.[d]7[e]6[f]4[g]6DE3.2Solutiongivenintext.DE3.3Solutiongivenintext.DE3.4Solutiongivenintext.DE
2、3.5[a]Thetwonodevoltageequationsarev1v1v1¡v2¡15+++=060155v2v2¡v15++=025Solving,v1=60Vandv2=10V;Therefore,i1=(v1¡v2)=5=10A[b]p15A(del)=(15)(60)=900W[c]p5A=¡5(10)=¡50WDE3.6Usethelowernodeasthereferencenode.Letv1=nodevoltageacross1−resistorandv2=nodevoltageacross12−resistor.Thenv1v1¡v2+=4:51
3、85354CHAPTER3.TechniquesofCircuitAnalysisv2v2¡v1v2¡30++=01284Solving,v1=6Vv2=18VThus,i=(v1¡v2)=8=¡1:5Av=v2+2i=15VDE3.7Usethelowernodeasthereferencenode.Letv1=nodevoltageacrossthe8−resistor,letv2=nodevoltageacrossthe4−resistor.Thenv1¡50v1v1¡v2++¡3i1=0682v2v2¡v1¡5+++3i1=04250¡v1i1=6Solving,
4、v1=32V;v2=16V;i1=3Ap50V=¡50i1=¡150W(delivering)p5A=¡5(v2)=¡80W(delivering)p3i1=3i1(v2¡v1)=¡144W(delivering)DE3.8Usethelowernodeasthereferencenode.Letv1=nodevoltageacrossthe7.5−resistorandv2=nodevoltageacrossthe2.5−resistor.Placethedependentvoltagesourceinsideasupernodebetweenthenodevoltag
5、esvandv2.Thenodevoltageequationsarev1v1¡vnode1:+=4:87:52:5v¡v1vv2v2¡12supernode:+++=02:5102:51Wealsohave:v+ix=v2andix=v1=7:5.Solvingthissetofequationsforvgivesv=8Vv1¡60v1v1¡(60+6iÁ)60+6iÁ¡v1DE3.9++=0;iÁ=22433Thereforev1=48VDE3.10vovo¡10vo+20i¢10¡vo10+20i¢++=0;i¢=+4010201030Thereforevo=24V
6、Problems55DE3.11De¯nethreeclockwisemeshcurrentsi1,i2,andi3inthelowerleft,upper,andlowerrightwindows.Thethreemesh-currentequationsare80=31i1¡5i2¡26i30=¡5i1+125i2¡90i30=¡26i1¡90i2+124i3[a]Solving,i1=5A;thereforethe80Vsourceisdelivering400Wtothecircuit.[b]Solving,i3=2:5A;thereforep8−=(6:25)(
7、8)=50WDE3.12[a]b=8,n=6,b¡n+1=3[b]De¯nethreeclockwisemeshcurrentsi1,i2,andi3intheupper,lowerleft,andlowerrightwindows.Thethreemesh-currentequationsare¡(¡3vÁ)+19i1¡2i2¡3i3=025¡10=¡2i1+7i2¡5i310=¡3i1¡5i2+9i3WealsohavevÁ=3(i3¡i1)Solvingfori1andi3givesi1=¡1A,i3=3ATherefo
