电路分析基础(英文版)课后答案第五章

5TheNaturalandStepResponseofRLandRCCircuitsDrillExercisesDE5.1[a]ig=8e¡300t¡8e¡1200tAv=Ldigdt=¡9:6e¡300t+38:4e¡1200tV;t0+v(0+)=¡9:6+38:4=28:8V[b]v=0when38:4e¡1200t=9:6e¡300tort=(ln4)=900=1:54ms[c]p=vi=384e¡1500t¡76:8e¡600t¡307:2e¡2400tW[d]dpdt=0whene1800t¡12:5e900t+16=0Letx=e900tandsolvethequadraticx2¡12:5x+16=0x=1:45;x=11:05;t=ln1:45900=411:05¹s;t=ln11:05900=2:67mspismaximumatt=411:05¹s[e]pmax=384e¡1:5(0:41105)¡76:8e¡0:6(0:41105)¡307:2e¡2:4(0:41105)=32:72W[f]imax=8[e¡0:3(1:54)¡e¡1:2(1:54)]=3:78Awmax=(1=2)(4£10¡3)(3:78)2=28:6mJ[g]Wismaxwheniismax,iismaxwhendi=dtiszero.Whendi=dt=0,v=0,thereforet=1:54ms.DE5.2[a]i=Cdvdt=24£10¡6ddt[e¡15;000tsin30;000t]=[0:72cos30;000t¡0:36sin30;000t]e¡15;000tA;i(0+)=0:72A151152CHAPTER5.NaturalandStepResponseofRLandRCCircuits[b]iµ¼80ms¶=¡31:66mA;vµ¼80ms¶=20:505V;p=vi=¡649:23mW[c]w=µ12¶Cv2=126:13¹JDE5.3[a]v=µ1C¶Zt0¡idx+v(0¡)=10:6£10¡6Zt0¡3cos50;000xdx=100sin50;000tV[b]p(t)=vi=[300cos50;000t]sin50;000t=150sin100;000tW;p(max)=150W[c]w(max)=µ12¶Cv2max=0:30(100)2=3000¹J=3mJDE5.4[a]Leq=60(240)300=48mH[b]i(0+)=3+¡5=¡2A[c]i=1256Zt0+(¡0:03e¡5x)dx¡2=0:125e¡5t¡2:125A[d]i1=503Zt0+(¡0:03e¡5x)dx+3=0:1e¡5t+2:9Ai2=256Zt0+(¡0:03e¡5x)dx¡5=0:025e¡5t¡5:025Ai1+i2=iDE5.5v1=0:5£106Zt0+240£10¡6e¡10xdx¡10=¡12e¡10t+2Vv2=0:125£106Zt0+240£10¡6e¡10xdx¡5=¡3e¡10t¡2Vv1(1)=2V;v2(1)=¡2VW=·12(2)(4)+12(8)(4)¸£10¡6=20¹JDE5.6[a]i=µ1203+5¶µ¡3036¶=¡12:5A[b]w=0:5(8£10¡3)(12:5)2=625mJ[c]¿=LR=8£10¡32=4msProblems153[d]i=¡12:5e¡250tA;t¸0[e]i(5ms)=¡3:58A;w(5ms)=(0:5)(8)£10¡3(3:58)2=51:3mJw(dis)=625¡51:3=573:7mJ%dissipated=µ573:7625¶100=91:8%DE5.7[a]iL(0¡)=6:4µ1016¶=4A=iL(0+);t0Req=(4)(16)20=3:2−;¿=0:323:2=0:1sTherefore1¿=10;iL=4e¡10tALeti1equalthecurrentinthe10−resistor.Letthereferencedirectionfori1beup.Theni1=µ420¶iL=0:8e¡10tA;vo=¡10i1=¡8e¡10tV;t¸0+[b]v4−=LdiLdt=0:32(¡40)e¡10t=¡12:8e¡10tV;t¸0+p4−=v24−4=40:96e¡20tW;t¸0+w4−=Z1040:96e¡20tdt=2:048Jwi=12Li2=12(0:32)(16)=2:56J%dissipated=µ2:0482:56¶100=80%DE5.8[a]v(0)=7:5(80)150#50=200V[b]¿=RC=(50£103)(0:4£10¡6)=20ms[c]v=200e¡50tV[d]w(0)=0:5(0:4£10¡6)(200)2=8mJ[e]w(t)=0:5(0:4£10¡6)(4£104)e¡100t=8e¡100tmJ8e¡100t=2;t=(ln4)=100=13:86ms154CHAPTER5.NaturalandStepResponseofRLandRCCircuitsDE5.9[a]Fort0:i=1575;000=15mA;v5(0¡)=4V;v1(0¡)=8V¿5=(20£103)(5£10¡6)=100ms;1=¿5=10¿1=(40£103)(1£10¡6)=40ms;1=¿1=25Thereforev5=4e¡10tV;t¸0;v1=8e¡25tV;t¸0;vo=v1+v5=[8e¡25t+4e¡10t]V;t¸0[b]v1(60ms)»=1:79V;v5(60ms)»=2:20Vw1(60ms)=(1=2)(1)(1:79)2»=1:59¹Jw5(60ms)=(1=2)(5)(2:20)2»=12:05¹Jw1(0)=12(10¡6)(64)+12(5£10¡6)(16)=72¹Jwdiss=72¡13:64=58:36¹J%dissipated=(58:36=72)(100)=81:05%DE5.10[a]i(0+)=24=2=12A[b]v(0+)=¡10(8+12)=¡200V[c]¿=L=R=(200=10)£10¡3=20ms[d]i=¡8+[12¡(¡8)]e¡50t=[¡8+20e¡50t]A;t¸0+[e]v=0+[¡200¡0]e¡50tV=¡200e¡50tV;t¸0+DE5.11[a]vR+1LZt0vdx=VsRProblems1551Rdvdt+vL=0dvdt+RLv=0[b]dvdt=¡RLvdvdtdt=¡RLvdt¢::dvv=¡RLdtZv(t)v(0+)dyy=¡RLZt0+dxlny¯¯¯¯v(t)v(0+)=¡µRL¶tlnv(t)v(0+)#=¡µRL¶tv(t)=v(0+)e¡(R=L)t;v(0+)=µVsR¡Io¶R=Vs¡IoR¢::v(t)=(Vs¡IoR)e¡(R=L)tDE5.12[a]IsR=Ri+1CZt0+idx+Vo0=Rdidt+iC+0¢::didt+iRC=0[b]didt=¡iRC;dii=¡dtRCZi(t)i(0+)dyy=¡1RCZt0+dx156CHAPTER5.NaturalandStepResponseofRLandRCCircuitslni(t)i(0+)=¡tRCi(t)=i(0+)e¡t=RC;i(0+)=IsR¡VoR=µIs¡VoR¶¢::i(t)=µIs¡VoR¶e¡t=RCDE5.13[a]vo=¡60+90e¡100tVvA¡vo8000+vA160;000+vA+7540;000=020vA¡20vo+vA+4vA+300=025vA=20vo¡300vA=0:8vo¡12vA=¡48+72e¡100t¡12=¡60+72e¡100tV;t¸0+[b]t¸0+DE5.14[a]vc(0+)=50V[b]vc(1)=µ¡3025¶20=¡24V[c]FindtheTh¶eveninequivalentwithrespecttotheterminalsofthecapacitor:vTh=¡24V;RTh=20k5=4−;Therefore¿=4(25£10¡9)=0:1¹s[d]i(0+)=¡50+244=¡18:5A[e]vc=¡24+[50¡(¡24)]e¡t=¿=¡24+74e¡107tV;t¸0[f]i=¡18:5e¡t=¿=¡18:5e¡107tA;t¸0+DE5.15[a]vc(0+)=(9=12)(120)=90V[b]vc(1)=¡1:5(40)=¡60VProblems157[c]FindtheTh¶eveninequivalentwithrespecttotheterminalsofthecapacitor:vTh=¡60V;RTh=50k−¿=RThC=1ms=1000¹s[d]vc=¡60+(90+60)e¡1000t=¡60+150e¡1000tV;t¸0Thereforet=ln(150=60)1000=916:3¹sDE5.16[a]Fort0,calculatetheTh¶eveninequivalentforthecircuittotheleftandrightofthe400-mHinductor.Wegeti(0¡)=¡260=20=¡13mAi(0¡)=i(0+)=¡13mA[b]Fort0,thecircuitreducestoThereforei(1)=¡60=5=¡12mA[c]¿=(400=5)£10¡6=80¹s[d]i(t)=¡12+[¡13+12]e¡12;500t=¡12¡e¡12;500tmA;t¸0158CHAPTER5.NaturalandStepResponseofRLandRCCircuitsProblemsP5.1p=vi=40t[e¡10t¡10te¡20t¡e¡20t]W=Z10pdx=Z1040x[e¡10x¡10xe¡20x¡e¡20x]dx=0:2JThisisenergystoredintheinductoratt=1:P5.20·t1iL=1034Zt030£10¡3e¡3xdx+2:5=7:5e¡3x¡3¯¯¯¯t0+2:5=5¡2:5e¡3tA;0·t·1P5.3[a]v=Ldidtdidt=50[t(¡10e¡10t)+e¡10t]=50e¡10t(1¡10t)Problems159v=(2£10¡3)(50)e¡10t(1¡10t)=100e¡10t(1¡10t)mV;t0[b]p=viv(200ms)=100e¡2(1¡2)=¡13:53mVi(200ms)=50(0:2)e¡2=1:35Ap(200ms)=¡13:53£10¡3(1:35)=¡18:32mW[c]delivering[d]w=12Li2=12(2£10¡3)(1:35)2=1:83mJ[e]didt=0whent=110s=100msimax=50(0:1)e¡1=1:84Awmax=12(2£10¡3)(1:84)2=3:38mJP5.4[a]0·t·1ms:i=1LZt0vsdx+i(0)=106300Zt06£10¡3dx+0=20x¯¯¯¯t0=20tA1ms·t·2ms:i=106300Zt10¡3(12£10¡3¡6x)dx+20£10¡3¢::i=40t¡10;000t2¡10£10¡3A2ms·t·1:i=106300Zt2£10¡3(0)dx+30£10¡3=30mA[b]160CHAPTER5.NaturalandStepResponseofRLandRCCircuitsP5.5[a]i=0t0i=16tA0·t·25msi=0:8¡16tA25·t·50msi=050mst[b]v=Ldidt=375£10¡3(16)=6V0·t·25msv=375£10¡3(¡16)=¡6V25·t·50msv=0t0v=6V0t25msv=¡6V25t