电路分析基础(英文版)课后答案第五章.pdf
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1、5The Natural and Step Response ofRL and RC CircuitsDrill ExercisesDE 5.1a ig=8e300t 8e1200tAv=Ldigdt=9:6e300t+38:4e1200tV;t 0+v(0+)=9:6+38:4=28:8Vb v=0when38:4e1200t=9:6e300tort=(ln4)=900=1:54msc p=vi=384e1500t 76:8e600t 307:2e2400tWddpdt=0whene1800t 12:5e900t+16=0Let x=e900tand solve the quadraticx
2、2 12:5x+16=0 x=1:45;x=11:05;t=ln1:45900=411:05s;t=ln11:05900=2:67msp is maximum at t=411:05se pmax=384e1:5(0:41105)76:8e0:6(0:41105)307:2e2:4(0:41105)=32:72Wf imax=8e0:3(1:54)e1:2(1:54)=3:78Awmax=(1=2)(4 103)(3:78)2=28:6mJg W is max when i is max,i is max when di=dt is zero.When di=dt=0,v=0,therefor
3、e t=1:54ms.DE 5.2a i=Cdvdt=24 106ddte15;000tsin30;000t=0:72cos30;000t 0:36sin30;000te15;000tA;i(0+)=0:72A151152CHAPTER 5.Natural and Step Response of RL and RC Circuitsb i80ms=31:66mA;v80ms=20:505V;p=vi=649:23mWc w=12Cv2=126:13JDE 5.3a v=1CZt0idx+v(0)=10:6 106Zt03cos50;000 xdx=100sin50;000tVb p(t)=v
4、i=300cos50;000tsin50;000t=150sin100;000tW;p(max)=150Wc w(max)=12Cv2max=0:30(100)2=3000J=3mJDE 5.4a Leq=60(240)300=48mHb i(0+)=3+5=2Ac i=1256Zt0+(0:03e5x)dx 2=0:125e5t 2:125Ad i1=503Zt0+(0:03e5x)dx+3=0:1e5t+2:9Ai2=256Zt0+(0:03e5x)dx 5=0:025e5t 5:025Ai1+i2=iDE 5.5v1=0:5 106Zt0+240 106e10 xdx 10=12e10t
5、+2Vv2=0:125 106Zt0+240 106e10 xdx 5=3e10t 2Vv1(1)=2V;v2(1)=2VW=12(2)(4)+12(8)(4)106=20JDE 5.6a i=1203+53036=12:5Ab w=0:5(8 103)(12:5)2=625mJc =LR=8 1032=4msProblems153d i=12:5e250tA;t 0e i(5ms)=3:58A;w(5ms)=(0:5)(8)103(3:58)2=51:3mJw(dis)=625 51:3=573:7mJ%dissipated=573:7625100=91:8%DE 5.7a iL(0)=6:
6、41016=4A=iL(0+);t 0Req=(4)(16)20=3:2;=0:323:2=0:1sTherefore1=10;iL=4e10tALet i1equal the current in the 10 resistor.Let the reference directionfor i1be up.Theni1=420iL=0:8e10tA;vo=10i1=8e10tV;t 0+b v4=LdiLdt=0:32(40)e10t=12:8e10tV;t 0+p4=v244=40:96e20tW;t 0+w4=Z1040:96e20tdt=2:048Jwi=12Li2=12(0:32)(
7、16)=2:56J%dissipated=2:0482:56100=80%DE 5.8a v(0)=7:5(80)150#50=200Vb =RC=(50 103)(0:4 106)=20msc v=200e50tVd w(0)=0:5(0:4 106)(200)2=8mJe w(t)=0:5(0:4 106)(4 104)e100t=8e100tmJ8e100t=2;t=(ln4)=100=13:86ms154CHAPTER 5.Natural and Step Response of RL and RC CircuitsDE 5.9a Fort 0:i=1575;000=15mA;v5(0
8、)=4V;v1(0)=8V5=(20 103)(5 106)=100ms;1=5=101=(40 103)(1 106)=40ms;1=1=25Thereforev5=4e10tV;t 0;v1=8e25tV;t 0;vo=v1+v5=8e25t+4e10tV;t 0b v1(60ms)=1:79V;v5(60ms)=2:20Vw1(60ms)=(1=2)(1)(1:79)2=1:59Jw5(60ms)=(1=2)(5)(2:20)2=12:05Jw1(0)=12(106)(64)+12(5 106)(16)=72Jwdiss=72 13:64=58:36J%dissipated=(58:36
9、=72)(100)=81:05%DE 5.10 a i(0+)=24=2=12Ab v(0+)=10(8+12)=200Vc =L=R=(200=10)103=20msd i=8+12 (8)e50t=8+20e50tA;t 0+e v=0+200 0e50tV=200e50tV;t 0+DE 5.11 avR+1LZt0vdx=VsRProblems1551Rdvdt+vL=0dvdt+RLv=0bdvdt=RLvdvdtdt=RLv dt:dvv=RLdtZv(t)v(0+)dyy=RLZt0+dxlnyv(t)v(0+)=RLtlnv(t)v(0+)#=RLtv(t)=v(0+)e(R=
10、L)t;v(0+)=VsR IoR=Vs IoR:v(t)=(Vs IoR)e(R=L)tDE 5.12 aIsR=Ri+1CZt0+idx+Vo0=Rdidt+iC+0:didt+iRC=0bdidt=iRC;dii=dtRCZi(t)i(0+)dyy=1RCZt0+dx156CHAPTER 5.Natural and Step Response of RL and RC Circuitslni(t)i(0+)=tRCi(t)=i(0+)et=RC;i(0+)=IsR VoR=IsVoR:i(t)=IsVoRet=RCDE 5.13 avo=60+90e100tVvA vo8000+vA16
11、0;000+vA+7540;000=020vA 20vo+vA+4vA+300=025vA=20vo 300vA=0:8vo 12vA=48+72e100t 12=60+72e100tV;t 0+b t 0+DE 5.14 a vc(0+)=50Vb vc(1)=302520=24Vc Find the Th evenin equivalent with respect to the terminals of thecapacitor:vTh=24V;RTh=20k5=4;Therefore=4(25 109)=0:1sd i(0+)=50+244=18:5Ae vc=24+50 (24)et
12、=24+74e107tV;t 0f i=18:5et=18:5e107tA;t 0+DE 5.15 a vc(0+)=(9=12)(120)=90Vb vc(1)=1:5(40)=60VProblems157c Find the Th evenin equivalent with respect to the terminals of thecapacitor:vTh=60V;RTh=50k=RThC=1ms=1000sd vc=60+(90+60)e1000t=60+150e1000tV;t 0Thereforet=ln(150=60)1000=916:3sDE 5.16 a For t 0
13、,the circuit reduces toThereforei(1)=60=5=12mAc =(400=5)106=80sd i(t)=12+13+12e12;500t=12 e12;500tmA;t 0158CHAPTER 5.Natural and Step Response of RL and RC CircuitsProblemsP 5.1p=vi=40te10t 10te20t e20tW=Z10pdx=Z1040 xe10 x 10 xe20 x e20 xdx=0:2JThis is energy stored in the inductor at t=1:P 5.20 t
14、0b p=viv(200ms)=100e2(1 2)=13:53mVi(200ms)=50(0:2)e2=1:35Ap(200ms)=13:53 103(1:35)=18:32mWc deliveringd w=12Li2=12(2 103)(1:35)2=1:83mJedidt=0whent=110s=100msimax=50(0:1)e1=1:84Awmax=12(2 103)(1:84)2=3:38mJP 5.4a 0 t 1ms:i=1LZt0vsdx+i(0)=106300Zt06 103dx+0=20 xt0=20tA1ms t 2ms:i=106300Zt103(12 103 6
15、x)dx+20 103:i=40t 10;000t2 10 103A2ms t 1:i=106300Zt2103(0)dx+30 103=30mAb160CHAPTER 5.Natural and Step Response of RL and RC CircuitsP 5.5ai=0t 0i=16tA0 t 25msi=0:8 16tA25 t 50msi=050ms tb v=Ldidt=375 103(16)=6V0 t 25msv=375 103(16)=6V25 t 50msv=0t 0v=6V0 t 25msv=6V25 t 50msv=050ms tp=vip=0t 0p=(16
16、t)(6)=96tW0 t 25msp=(0:8 16t)(6)=96t 4:8W25 t 50msp=050ms tw=0t 0w=Zt0(16x)6dx=96x22t0=48t2J0 t 25msw=Zt0:025(96x 4:8)dx+0:03=Zt0:02596xdx Zt0:0254:8dx+0:03=96x22t0:0254:8xt0:025+0:03=48t2 4:8t+0:12J25 t 50msw=050ms tP 5.6a 0 t 1s:v=100ti=15Zt0100 xdx+0=20 x22t0Problems161i=10t2A1s t 3s:v=200+100ti(
17、1)=10A:i=15Zt1(100 x 200)dx 10=20Zt1xdx 40Zt1dx 10=10(t2 1)40(t 1)10=10t2 40t+20A3s t 5s:v=100i(3)=10(9)120+20=10Ai=15Zt3100dx 10=20t 60 10=20t 70A5s t 6s:v=100t+600i(5)=100 70=30i=15Zt5(100 x+600)dx+30=20Zt5xdx+120Zt5dx+30=10(t2 25)+120(t 5)+30=10t2+120t 320Ab i(6)=10(36)+120(6)320=720 680=40A;6 t
18、1162CHAPTER 5.Natural and Step Response of RL and RC CircuitscP 5.7a i(0)=A1+A2=0:05didt=2500A1e2500t 7500A2e7500tv=50A1e2500t 150A2e7500tVv(0)=50A1 150A2=10:5A1 15A2=1But from the equation for i(0),5A1+5A2=0:25Solving,A1=0:175and A2=0:125Thus,i=0:175e2500t 0:125e7500tA;t 0v=8:75e2500t+18:75e7500tV;
19、t 0b p=vi=4:375e10;000t 1:53125e5000t 2:34375e15;000tWp=0when 4:375e10;000t 1:53125e5000t 2:34375e15;000t=0Letx=e5000t;then4:375x 1:53125x2 2:34375=0Solving,x=0:7143;x=2:143If x 0 must bex=2:143e5000t=2:143sot=152:43sProblems163P 5.8a From Prob.5.7 we havei=A1e2500t+A2e7500tAv=50A1e2500t 150A2e7500t
20、Vi(0)=A1+A2=0:05v(0)=50A1 150A2=100:A1+A2=0:05andA1+3A2=2:A2=0:975A;A1=0:925AThus,i=0:925e2500t+0:975e7500tAt 0v=46:25e2500t 146:25e7500tVt 0bi=0when0:975e7500t=0:925e2500t:e5000t=1:0541t=(ln1:054)=5000=10:53sv=0when46:25e2500t=146:25e7500t:t=ln3:1622=5000=230:25s164CHAPTER 5.Natural and Step Respon
21、se of RL and RC Circuits:Energy is being stored between 10:53s and 230:25s;energy is beingextracted between 0 and 10:53s and between 230:25s and innity.c p=vi=180:375e10;000t 42:78125e5000t 142:59375e15;000tW:Wstored=Zt2t1pdt+w(0)Wstored=103(18:0375e10;000tt2t1+8:55625e5000tt2t1+9:50625e15;000tt2t1)
22、+25 106=8:55625e5000t2+9:50625e15;000t2 18:0375e10;000t28:55625e5000t1 9:50625e15;000t1+18:0375e10;000t1+0:025mWwheret1=10:52s;t2=230:11sWstored=1:23mJ:Wextracted=Zt10pdt+Z1t2pdt=Zt10(180:375e104t 42:78125e5000t142:59375e15;000t)dt+Z1t2(180:375e104t 42:78125e5000t142:59375e15;000t)dt=103(18:0375e10;
23、000tt10+8:55625e5000tt10+9:50625e15;000tt1018:0375e10;000t1t2+8:55625e5000t1t2+9:50625e15;000t1t2)=f18:0375e10;000t2 8:55625e5000t2 9:50625e15;000t2+8:55625e5000t1+9:50625e15;000t1 18:0375e10;000t10:025gmJWext:=1:23mJ:Wstored=WextractedProblems165P 5.9a vL=Ldidt=125sin400te200tV:dvLdt=25;000(2cos400
24、t sin400t)e200tV=sdvLdt=0whentan400t=2:t=2:77msAlso400t=1:107+etc.Because of the decaying exponential vLwill be maximum the rst timethe derivative is zero.b vL(max)=125sin1:107e0:554=64:27VvLmax=64:27VNote:Whent=(1:107+)=400;vL=13:36VP 5.10ai=100050Zt0250sin1000 xdx 5=5000Zt0sin1000 xdx 5=5000cos100
25、0 x1000t0 5=5(1 cos1000t)5i=5cos1000tAbp=vi=(250sin1000t)(5cos1000t)=1250sin1000tcos1000tp=625sin2000tWw=12Li2=12(50 103)25cos21000t=625cos21000tmJw=312:5+312:5cos2000tmJ:166CHAPTER 5.Natural and Step Response of RL and RC CircuitsProblems167cAbsorbing power:Delivering power:0:5 t ms0 t 0:5ms1:5 t 2
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