电路分析基础(英文版)课后答案第五章.pdf

5 The Natural and Step Response of RL and RC Circuits Drill Exercises DE 5 1 a ig 8e 300t 8e 1200tA v Ldig dt 9 6e 300t 38 4e 1200tV t 0 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 dp dt 0whene1800t 12 5e900t 16 0 Let x e900tand solve the quadraticx2 12 5x 16 0 x 1 45 x 11 05 t ln1 45 900 411 05 s t ln11 05 900 2 67ms p is maximum at t 411 05 s 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 78A wmax 1 2 4 10 3 3 78 2 28 6mJ g W is max when i is max i is max when di dt is zero When di dt 0 v 0 therefore t 1 54ms DE 5 2 a i C dv dt 24 10 6 d dt e 15 000t sin30 000t 0 72cos30 000t 0 36sin30 000t e 15 000tA i 0 0 72A 151 152CHAPTER 5 Natural and Step Response of RL and RC Circuits b i 80 ms 31 66mA v 80 ms 20 505V p vi 649 23mW c w 1 2 Cv2 126 13 J DE 5 3 a v 1 C Z t 0 idx v 0 1 0 6 10 6 Z t 0 3cos50 000 xdx 100sin50 000tV b p t vi 300cos50 000t sin50 000t 150sin100 000tW p max 150W c w max 1 2 Cv2 max 0 30 100 2 3000 J 3mJ DE 5 4 a Leq 60 240 300 48mH b i 0 3 5 2A c i 125 6 Z t 0 0 03e 5x dx 2 0 125e 5t 2 125A d i1 50 3 Z t 0 0 03e 5x dx 3 0 1e 5t 2 9A i2 25 6 Z t 0 0 03e 5x dx 5 0 025e 5t 5 025A i1 i2 i DE 5 5v1 0 5 106 Z t 0 240 10 6e 10 xdx 10 12e 10t 2V v2 0 125 106 Z t 0 240 10 6e 10 xdx 5 3e 10t 2V v1 1 2V v2 1 2V W 1 2 2 4 1 2 8 4 10 6 20 J DE 5 6 a i 120 3 5 30 36 12 5A b w 0 5 8 10 3 12 5 2 625mJ c L R 8 10 3 2 4ms Problems153 d i 12 5e 250tA t 0 e i 5ms 3 58A w 5ms 0 5 8 10 3 3 58 2 51 3mJ w dis 625 51 3 573 7mJ dissipated 573 7 625 100 91 8 DE 5 7 a iL 0 6 4 10 16 4A iL 0 t 0 Req 4 16 20 3 2 0 32 3 2 0 1s Therefore 1 10 iL 4e 10tA Let i1equal the current in the 10 resistor Let the reference direction for i1be up Then i1 4 20 iL 0 8e 10tA vo 10i1 8e 10tV t 0 b v4 LdiL dt 0 32 40 e 10t 12 8e 10tV t 0 p4 v2 4 4 40 96e 20tW t 0 w4 Z 1 0 40 96e 20tdt 2 048J wi 1 2Li 2 1 2 0 32 16 2 56J dissipated 2 048 2 56 100 80 DE 5 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 100tmJ 8e 100t 2 t ln4 100 13 86ms 154CHAPTER 5 Natural and Step Response of RL and RC Circuits DE 5 9 a Fort 0 i 15 75 000 1 5 mA 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 25 Thereforev5 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 20V w1 60ms 1 2 1 1 79 2 1 59 J w5 60ms 1 2 5 2 20 2 12 05 J w1 0 1 2 10 6 64 1 2 5 10 6 16 72 J wdiss 72 13 64 58 36 J dissipated 58 36 72 100 81 05 DE 5 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 DE 5 11 a v R 1 L Z t 0 vdx Vs R Problems155 1 R dv dt v L 0 dv dt R L v 0 b dv dt R L v dv dt dt R L v dt dv v R L dt Z v t v 0 dy y R L Z t 0 dx lny v t v 0 R L t ln v t v 0 R L t v t v 0 e R L t v 0 V s R Io R Vs IoR v t Vs IoR e R L t DE 5 12 a IsR Ri 1 C Z t 0 idx Vo 0 R di dt i C 0 di dt i RC 0 b di dt i RC di i dt RC Z i t i 0 dy y 1 RC Z t 0 dx 156CHAPTER 5 Natural and Step Response of RL and RC Circuits ln i t i 0 t RC i t i 0 e t RC i 0 IsR Vo R Is Vo R i t Is Vo R e t RC DE 5 13 a vo 60 90e 100tV vA vo 8000 vA 160 000 vA 75 40 000 0 20vA 20vo vA 4vA 300 0 25vA 20vo 300 vA 0 8vo 12 vA 48 72e 100t 12 60 72e 100tV t 0 b t 0 DE 5 14 a vc 0 50V b vc 1 30 25 20 24V c Find the Th evenin equivalent with respect to the terminals of the capacitor vTh 24V RTh 20k5 4 Therefore 4 25 10 9 0 1 s d i 0 50 24 4 18 5A e vc 24 50 24 e t 24 74e 10 7t V t 0 f i 18 5e t 18 5e 10 7t A t 0 DE 5 15 a vc 0 9 12 120 90V b vc 1 1 5 40 60V Problems157 c Find the Th evenin equivalent with respect to the terminals of the capacitor vTh 60V RTh 50k RThC 1ms 1000 s d vc 60 90 60 e 1000t 60 150e 1000tV t 0 Thereforet ln 150 60 1000 916 3 s DE 5 16 a For t 0 the circuit reduces to Thereforei 1 60 5 12mA c 400 5 10 6 80 s d i t 12 13 12 e 12 500t 12 e 12 500tmA t 0 158CHAPTER 5 Natural and Step Response of RL and RC Circuits Problems P 5 1p vi 40t e 10t 10te 20t e 20t W Z 1 0 pdx Z 1 0 40 x e 10 x 10 xe 20 x e 20 x dx 0 2J This is energy stored in the inductor at t 1 P 5 20 t 0 b p vi v 200ms 100e 2 1 2 13 53mV i 200ms 50 0 2 e 2 1 35A p 200ms 13 53 10 3 1 35 18 32mW c delivering d w 1 2Li 2 1 2 2 10 3 1 35 2 1 83mJ e di dt 0whent 1 10 s 100ms imax 50 0 1 e 1 1 84A wmax 1 2 2 10 3 1 84 2 3 38mJ P 5 4 a 0 t 1ms i 1 L Z t 0 vsdx i 0 106 300 Z t 0 6 10 3dx 0 20 x t 0 20tA 1ms t 2ms i 106 300 Z t 10 3 12 10 3 6x dx 20 10 3 i 40t 10 000t2 10 10 3A 2ms t 1 i 106 300 Z t 2 10 3 0 dx 30 10 3 30mA b 160CHAPTER 5 Natural and Step Response of RL and RC Circuits P 5 5 a i 0t 0 i 16tA0 t 25ms i 0 8 16tA25 t 50ms i 050ms t b v Ldi dt 375 10 3 16 6V0 t 25ms v 375 10 3 16 6V25 t 50ms v 0t 0 v 6V0 t 25ms v 6V25 t 50ms v 050ms t p vi p 0t 0 p 16t 6 96tW0 t 25ms p 0 8 16t 6 96t 4 8W25 t 50ms p 050ms t w 0t 0 w Z t 0 16x 6dx 96x 2 2 t 0 48t 2J 0 t 25ms w Z t 0 025 96x 4 8 dx 0 03 Z t 0 025 96xdx Z t 0 025 4 8dx 0 03 96x 2 2 t 0 025 4 8x t 0 025 0 03 48t2 4 8t 0 12J25 t 50ms w 050ms t P 5 6 a 0 t 1s v 100t i 1 5 Z t 0 100 xdx 0 20 x 2 2 t 0 Problems161 i 10t2A 1s t 3s v 200 100t i 1 10A i 1 5 Z t 1 100 x 200 dx 10 20 Z t 1xdx 40 Z t 1dx 10 10 t2 1 40 t 1 10 10t2 40t 20A 3s t 5s v 100 i 3 10 9 120 20 10A i 1 5 Z t 3100dx 10 20t 60 10 20t 70A 5s t 6s v 100t 600 i 5 100 70 30 i 1 5 Z t 5 100 x 600 dx 30 20 Z t 5xdx 120 Z t 5dx 30 10 t2 25 120 t 5 30 10t2 120t 320A b i 6 10 36 120 6 320 720 680 40A 6 t 1 162CHAPTER 5 Natural and Step Response of RL and RC Circuits c P 5 7 a i 0 A1 A2 0 05 di dt 2500A1e 2500t 7500A2e 7500t v 50A1e 2500t 150A2e 7500tV v 0 50A1 150A2 10 5A1 15A2 1 But from the equation for i 0 5A1 5A2 0 25 Solving A1 0 175and A2 0 125 Thus i 0 175e 2500t 0 125e 7500tA t 0 v 8 75e 2500t 18 75e 7500tV t 0 b p vi 4 375e 10 000t 1 53125e 5000t 2 34375e 15 000tW p 0when 4 375e 10 000t 1 53125e 5000t 2 34375e 15 000t 0 Letx e5000t then 4 375x 1 53125x2 2 34375 0 Solving x 0 7143 x 2 143 If x 0 must be x 2 143 e5000t 2 143sot 152 43 s Problems163 P 5 8 a From Prob 5 7 we have i A1e 2500t A2e 7500tA v 50A1e 2500t 150A2e 7500tV i 0 A1 A2 0 05 v 0 50A1 150A2 100 A1 A2 0 05andA1 3A2 2 A2 0 975A A1 0 925A Thus i 0 925e 2500t 0 975e 7500tAt 0 v 46 25e 2500t 146 25e 7500tVt 0 b i 0when0 975e 7500t 0 925e 2500t e5000t 1 0541 t ln1 054 5000 10 53 s v 0when46 25e 2500t 146 25e 7500t t ln3 1622 5000 230 25 s 164CHAPTER 5 Natural and Step Response of RL and RC Circuits Energy is being stored between 10 53 s and 230 25 s energy is being extracted between 0 and 10 53 s and between 230 25 s and in nity c p vi 180 375e 10 000t 42 78125e 5000t 142 59375e 15 000tW Wstored Z t2 t1 pdt w 0 Wstored 10 3 18 0375e 10 000t t2 t1 8 55625e 5000t t2 t1 9 50625e 15 000t t2 t1 25 10 6 8 55625e 5000t2 9 50625e 15 000t2 18 0375e 10 000t2 8 55625e 5000t1 9 50625e 15 000t1 18 0375e 10 000t1 0 025mW wheret1 10 52 s t2 230 11 s Wstored 1 23mJ Wextracted Z t1 0 pdt Z 1 t2 pdt Z t1 0 180 375e 10 4t 42 78125e 5000t 142 59375e 15 000t dt Z 1 t2 180 375e 10 4t 42 78125e 5000t 142 59375e 15 000t dt 10 3 18 0375e 10 000t t1 0 8 55625e 5000t t1 0 9 50625e 15 000t t1 0 18 0375e 10 000t 1 t2 8 55625e 5000t 1 t2 9 50625e 15 000t 1 t2 f18 0375e 10 000t2 8 55625e 5000t2 9 50625e 15 000t2 8 55625e 5000t1 9 50625e 15 000t1 18 0375e 10 000t1 0 025gmJ Wext 1 23mJ Wstored Wextracted Problems165 P 5 9 a vL Ldi dt 125sin400t e 200tV dvL dt 25 000 2cos400t sin400t e 200tV s dvL dt 0whentan400t 2 t 2 77ms Also400t 1 107 etc Because of the decaying exponential vLwill be maximum the rst time the derivative is zero b vL max 125sin1 107 e 0 554 64 27V vLmax 64 27V Note Whent 1 107 400 vL 13 36V P 5 10 a i 1000 50 Z t 0250sin1000 xdx 5 5000 Z t 0 sin1000 xdx 5 5000 cos1000 x 1000 t 0 5 5 1 cos1000t 5 i 5cos1000tA b p vi 250sin1000t 5cos1000t 1250sin1000tcos1000t p 625sin2000tW w 1 2Li 2 1 2 50 10 3 25cos21000t 625cos21000tmJ w 312 5 312 5cos2000t mJ 166CHAPTER 5 Natural and Step Response of RL and RC Circuits Problems167 c Absorbing power Delivering power 0 5 t ms0 t 0 5 ms 1 5 t 2 ms t 1 5 ms P 5 11i B1cos5t B2sin5t e t i 0 B1 25A di dt B1cos5t B2sin5t e t e t 5B1sin5t 5B2cos5t 5B2 B1 cos5t 5B1 B2 sin5t e t v 2di dt 10B2 2B1 cos5t 10B1 2B2 sin5t e t v 0 100 10B2 2B1 10B2 50 B2 150 10 15A Thus i 25cos5t 15sin5t e tA t 0 v 100cos5t 280sin5t e tV t 0 i 0 5 6 70A v 0 5 150 23V p 0 5 6 70 150 23 1007 00W absorbing P 5 12 a v 20 s 12 5 109 20 10 6 2 5V end of rst interval v 20 s 106 20 10 6 12 5 400 10 3 10 5V start of second interval v 40 s 106 40 10 6 12 5 1600 10 3 10 10V end of second interval b p 10 s 62 5 1012 10 5 3 62 5mW v 10 s 1 25V i 10 s 50mA p 10 s vi 62 5mW p 30 s 437 50mW v 30 s 8 75V i 30 s 0 05A 168CHAPTER 5 Natural and Step Response of RL and RC Circuits c w 10 s 15 625 1012 10 10 6 4 0 15625 J w 0 5Cv2 0 5 0 2 10 6 1 25 2 0 15625 J w 30 s 7 65625 J w 30 s 0 5 0 2 10 6 8 75 2 7 65625 J P 5 13 a 0 t 50 s C 0 5 F 1 C 2 106 v 2 106 Z t 0 20 10 3dx 20 v 40 103t 20V0 t 50 s v 50 s 2 20 22V b 50 s t 200 s v 2 106 Z t 50 10 6 40 10 3dx 22 8 104t 4 22 v 8 104t 26V50 t 200 s v 200 s 8 104 200 10 6 26 10V c 200 t 1 v 2 106 Z t 200 10 6 0dx 10 10V200 s t 1 d P 5 14iC C dv dt 0 t 1 vc 20t3V iC 0 8 10 6 60 t2 48t2 A 1 t 3 Problems169 vc 2 5 3 t 3V iC 0 8 10 6 7 5 3 t 2 1 6 3 t 2 A P 5 15 a v 5 106 Z 250 10 6 0 100 10 3e 1000tdt 60 6 500 103 e 1000t 1000 250 10 6 0 60 6 500 1 e 0 25 60 6 50V w 1 2Cv 2 1 2 0 2 10 6 50 2 250 J b v 500 60 6 439 40V w 1 2 0 2 10 6 439 40 2 19 31mJ 19 307 24 J P 5 16 a w 0 1 2C v 0 2 1 2 0 40 10 6 25 2 125 J b v A1t A2 e 1500tv 0 A2 25V dv dt 1500e 1500t A1t A2 e 1500t A1 1500A1t 1500A2 A1 e 1500t dv dt 0 A1 1500A2 i C dv dt i 0 C dv 0 dt dv 0 dt i 0 C 90 10 3 0 40 10 6 225 103 225 103 A1 1500 25 Thus A1 2 25 105 3 75 104 262 500 V s 170CHAPTER 5 Natural and Step Response of RL and RC Circuits c v 262 500t 25 e 1500t i C dv dt 0 40 10 6 d dt 262 500t 25 e 1500t i d dt 0 105t 10 10 6 e 1500t 0 105t 10 10 6 1500 e 1500t e 1500t 0 105 157 5t 15 10 3 0 105 e 1500t 0 09 157 5t e 1500tA t 0 90 157 500t e 1500tmA t 0 P 5 17 a i 50 10 3 10 10 6 t 5 103t0 t 10 s i 50 10 310 t 30 s q Z 10 10 6 0 5 103tdt Z 30 10 6 10 10 650 10 3dt 5 103 t2 2 10 10 6 0 50 10 3 20 10 6 5 103 1 2 100 10 12 1000 10 3 10 6 1 25 C b i 200 10 3 5 10 3t30 s t 50 s q 1 25 10 6 Z 50 10 6 30 10 6 200 10 3 5 103t dt 1 25 10 6 200 10 3 20 10 6 5 103 t2 2 50 10 6 30 10 6 1 25 10 6 4000 10 9 5 103 2500 900 2 10 12 1 25 C Since q vC v 1 25 0 25 5V c i 300 10 3 5 10 3t50 s t 60 s Problems171 q 1 25 10 6 Z 60 10 6 50 10 6 300 10 3 5 103t dt 1 25 10 6 300 10 3 10 10 6 5 103 3600 2500 2 10 12 1 C v 1 10 6 0 25 10 6 4V w C 2 v2 1 2 0 25 10 6 16 2 J P 5 18v 60V t 0 C 0 4 F v 15 15e 500t 5cos2000t sin2000t V t 0 a i 0 t 0 c no d yes from 0 to 3 mA e v 1 15V w 1 2Cv 2 1 2 0 4 225 10 6 45 J P 5 1910k 15 25 8H 8k12 4 8H 44k 1 2 4 8 5 28H 21k4 3 36H 5 28 3 36 8 64H 172CHAPTER 5 Natural and Step Response of RL and RC Circuits P 5 206k14 4 2H 15 8 4 2 20H 20k60 15H 15 5 20H 20k80 16H 16 24 40H 40k10 8H Lab 12 8 20H P 5 21From Figure 5 17 a we have v 1 C1 Z t 0 i v1 0 1 C2 Z t 0 idx v2 0 v 1 C1 1 C2 Z t 0 idx v1 0 v2 0 Therefore 1 Ceq 1 C1 1 C2 veq 0 v1 0 v2 0 P 5 22From Fig 5 18 a i C1 dv dt C2 dv dt C1 C2 dv dt Therefore Ceq C1 C2 Because the capacitors are in parallel the initial voltage on every capacitor must be the same This initial voltage would appear on Ceq P 5 23 1 4 1 6 5 12 Ceq 2 4 F 1 4 1 12 4 12 Ceq 3 F Problems173 1 24 1 8 4 24 Ceq 6 F P 5 24 1 C1 1 8 1 32 5 32 C1 6 4nF C2 5 6 6 4 12nF 1 C3 1 18 1 12 10 72 C3 7 2nF C4 12 8 7 2 20nF 174CHAPTER 5 Natural and Step Response of RL and RC Circuits 1 C5 1 8 1 20 1 40 1 5 C5 5nF P 5 25 a io 0 i1 0 i2 0 5A b io 1 10 Z t 01250e 25x dx 5 125 e 25x 25 t 0 5 5 e 25t 1 5 5e 25tA t 0 c va 3 6 d dt 5e 25t 450e 25tV vc va vb 450e 25t 1250e 25t 800e 25tV i1 1 8 Z t 0800e 25xdx 10 4e 25t 4 10 i1 4e 25t 6At 0 d i2 1 32 Z t 0800e 25xdx 5 e 25t 1 5 i2 e 25t 6A t 0 Problems175 e w 0 1 2 8 100 1 2 32 25 1 2 3 6 25 845J f wdel 1 2 10 25 125J g wtrapped 845 125 720J P 5 26vb 1250e 25tV io 5e 25tA p 6250e 50tW w Z t 0 6250e 50 xdx 6250e 50 x 50 t 0 125 1 e 50t W wtotal 125J 80 wtotal 100J Thus 125 125e 50t 100 e50t 5 t 32 19ms P 5 27 a i t 1 7 5 Z t 0 1800e 20 xdx 12 240e 20 x 20 t 0 12 12 e 20t 1 12 i t 12e 20tA b i1 t 1 10 Z t 0 1800e 20 xdx 4 180e 20 x 20 t 0 4 9 e 20t 1 4 i1 t 9e 20t 13A 176CHAPTER 5 Natural and Step Response of RL and RC Circuits c i2 t 1 30 Z t 0 1800e 20 xdx 16 60e 20 x 20 t 0 16 3 e 20t 1 16 i2 t 3e 20t 13A d p vi 1800e 20t 12e 20t 21 600e 40tW w Z 1 0 pdt Z 1 0 21 600e 40tdt 21 600e 40t 40 1 0 540J e w 1 2 10 16 1 2 30 256 3920J f wtrapped 1 2 10 13 2 1 2 30 13 2 3380J wtrapped 3920 540 3380Jchecks g Yes they agree P 5 28 a vo 10 9 12 Z t 0900 10 6e 2500 xdx 30 75 000e 2500 x 2500 t 0 30 30e 2500tV t 0 b v1 10 9 20 900 10 6 e 2500 x 2500 t 0 45 18e 2500t 27V t 0 c v2 10 9 30 900 10 6 e 2500 x 2500 t 0 15 12e 2500t 27V t 0 Problems177 d p vi 30e 2500t 900 10 6 e 2500t 27 10 3e 5000t w Z 1 0 27 10 3e 5000tdt 27 10 3 e 5000t 5000 1 0 5 4 10 6 0 1 5 4 J e w 1 2 20 10 9 45 2 1 2 30 10 9 15 2 20 25 10 6 3 375 10 6 23 625 J f wtrapped 1 2 20 10 9 27 2 1 2 30 10 9 27 2 10 15 27 2 10 9 18 225 J CHECK 18 225 5 4 23 625 J g Yes they agree P 5 29C1 1 1 5 2 5nF 1 C2 1 2 5 1 12 5 1 50 1 2 C2 2nF vd 0 va 0 vc 0 40 15 45 100V a vb 10 9 2 Z t 050 10 6e 250 xdx 100 25 000e 250 x 250 t 0 100 100 e 250t 1 100 100e 250tV t 0 178CHAPTER 5 Natural and Step Response of RL and RC Circuits b va 109 12 5 Z t 050 10 6e 250 xdx 15 4000e 250 x 250 t 0 15 16 e 250t 1 15 16e 250t 1V c vc 109 50 Z t 050 10 6e 250 xdx 45 1000e 250 x 250 t 0 45 4 e 250t 1 45 4e 250t 41V t 0 d vd 10 9 2 5 Z t 050 10 6e 250 xdx 40 20 000e 250 x 250 t 0 40 80 e 250t 1 40 80e 250t 40V t 0 CHECK vb vd va vc 80e 250t 40 16e 250t 1 4e 250t 41 100e 250tV checks e i1 10 9 d dt 80e 250t 40 10 9 20 000e 250t 20e 250t A t 0 f i2 1 5 10 9 d dt 80e 250t 40 1 5 10 9 20 000e 250t 30e 250t A t 0 CHECK i1 i2 50e 250t A ib Problems179 P 5 30 a w 0 1 2 2 5 40 2 1 2 12 5 15 2 1 2 50 45 2 10 9 54 031 25nJ b va 1 1V vc 1 41V vd 1 40V w 1 1 2 2 5 40 2 1 2 12 5 1 2 1 2 50 41 2 10 9 44 031 25nJ c w Z 1 0 100e 250t 50e 250t 10 6dt 10 000nJ CHECK 54 031 25 44 031 25 10 000 d delivered 10 000 54 031 25 100 18 51 e w 5 10 3 Z t 0e 500 xdx 104 1 e 500t nJ 104 1 e 500t 5000 e 500t 0 5 Thus t ln2 500 1 39ms P 5 31 a v i R 100e 80t 4e 80t 25 b 1 80 12 5ms c L R 12 5 10 3 L 12 5 25 10 3 312 5mH d w 0 1 2L i 0 2 1 2 0 3125 16 2 5J e wdiss Z t 0 400e 160 xdx 2 5 2 5e 160t 0 8w 0 0 8 2 5 2J 2 5 2 5e 160t 2 e160t 5 Solving t 10 06 ms 180CHAPTER 5 Natural and Step Response of RL and RC Circuits P 5 32 a t 0 i iT 200 300 2 3iT vT 50i iT 100 200 300 50iT 2 3 200 3 iT vT iT RTh 100 3 200 3 100 L R 200 100 10 3 1 500 iL 6e 500tA t 0 b vL 200 10 3 3000e 500t 600e 500tV t 0 c vL 50i 100i 150i Problems181 i vL 150 4e 500tAt 0 P 5 33w 0 1 2 200 10 3 36 3 6J p50i 50i iL 50 4e 500t 6e 500t 1200e 1000tW w50i Z 1 0 1200e 1000tdt 1200e 1000t 1000 1 0 1 2J dissipated 1 2 3 6 100 33 33 P 5 34 a i 0 125 25 5A b L R 4 100 40ms c i 5e 25tA t 0 v1 Ldi1 dt 4 125e 25t 500e 25tVt 0 v2 80i 400e 25tVt 0 d pdiss i2 20 25e 50t 20 500e 50tW wdiss Z t 0 500e 50 xdx 500e 50 x 50 t 0 10 10e 50t J wdiss 12ms 10 10e 0 6 4 51J w 0 1 2 4 25 50J dissipated 4 51 50 100 9 02 P 5 35 a t 0 15k k15k 7 5k ig 0 9 15 7 5 103 0 4mA i1 0 i2 0 0 4 10 3 15 30 0 2mA 182CHAPTER 5 Natural and Step Response of RL and RC Circuits b i1 0 i1 0 0 2mA i2 0 i1 0 0 2mA when switch is open c L R 30 10 3 30 103 10 6 1 106 i1 t i1 0 e t i1 t 0 2e 10 6t mA t 0 d i2 t i1 t whent 0 i2 t 0 2e 10 6t mA t 0 e The current in a resistor can change instantaneously The switching operation forces i2 0 to equal 0 2mA and i2 0 0 2mA P 5 36 a Fort 0 iL t iL 0 e t A t 0 L R 0 20 5 15 1 100 0 01s iL 0 1A Problems183 iL t e 100tA t 0 vo t 15iL t vo t 15e 100tV t 0 P 5 37P20 v2 o 20 11 25e 200tW wdiss Z 0 01 0 11 25e 200tdt 11 25 200e 200t 0 01 0 56 25 10 3 1 e 2 48 64mJ wstored 1 2 0 2 1 2 100mJ diss 48 64 100 100 48 64 P 5 38t 0 Find Th evenin resistance seen by inductor iT 5vT vT iT RTh 1 5 0 2 184CHAPTER 5 Natural and Step Response of RL and RC Circuits L R 50 10 3 0 2 250ms 1 4 io 25e 4tA t 0 vo Ldio dt 50 10 3 100e 4t 5e 4tV t 0 P 5 39 a t 0 t 0 220 iab 50 3 10 3 iab 200A t 0 Problems185 b Att 1 iab 220 1 220A t 1 c i1 0 50 3 1 2 12 10 3 0 167ms i2 0 10 3 2 15 60 10 3 0 25ms i1 t 50 3 e 6000tA t 0 i2 t 10 3 e 4000tA t 0 iab 220 50 3 e 6000t 10 3 e 4000tA t 0 220 50 3 e 6000t 10 3 e 4000t 210 30 50e 6000t 10e 4000t 3 5e 6000t e 4000t By trial and error t 123 1 s P 5 40 a io 0 0since the switch is open for t 0 186CHAPTER 5 Natural and Step Response of RL and RC Circuits b For t 0 the circuit is 120 60 40 ig 12 10 40 0 24A 240mA iL 0 120 180 ig 160mA c For t 0 the circuit is 120 40 30 ig 12 10 30 0 30A 300mA ia 120 160 300 225mA io 0 225 160 65mA d iL 0 iL 0 160mA e io 1 ia 225mA f iL 1 0 since the switch short circuits the branch containing the 20 resistor and the 100 mH inductor g L R 100 10 3 20 5ms 1 200 iL 0 160 0 e 200t 160e 200tmA t 0 h vL 0 0since for t 0 the current in the inductor is constant Problems187 i Refer to the circuit at t 0 and note 20 0 16 vL 0 0 vL 0 3 2V j vL 1 0 since the current in the inductor is a constant at t 1 k vL t 0 3 2 0 e 200t 3 2e 200tV t 0 l io ia iL 225 160e 200tmA t 0 P 5 41 a t 0 iR 5et mA L R 20 10 6 iR 5e 50 000tmA 188CHAPTER 5 Natural and Step Response of RL and RC Circuits vR 2 5 103 5 10 3 e 50 000t 12 5e 50 000tV v1 20 10 3 5 10 3 50 000 e 50 000t 5e 50 000tV vo v1 vR 7 5e 50 000tV b io 103 48 Z t 0 7 5e 50 000 xdx 0 3 125e 50 000t 3 125mA P 5 42 a From the solution to Problem 5 41 iR 5 10 3e 50 000tA pR 25 10 6e 100 000t 2 5 103 62 5 10 3e 100 000tW wdiss Z 1 0 62 5 10 3e 100 000tdt 62 5 10 3 e 100 000t 105 1 0 625nJ b wtrapped 1 2Leqi 2 R 0 1 2 50 10 3 5 10 3 2 625nJ CHECK w 0 1 2 20 25 10 6 10 3 1 2 80 25 10 6 10 3 1250nJ w 0 wdiss wtrapped P 5 43 a iL 0 80 40 2A io 0 80 20 2 4 2 2A io 1 80 20 4A Proble