电路分析基础(英文版)课后答案第一章.pdf

1 Circuit Variables and Circuit Elements Drill Exercises DE 1 1q Z 1 0 20e 5000tdt 4000 C DE 1 2i dq dt te t di dt 1 t e t di dt 0when t 1 Therefore imax 1 e 1 0 03679e 10 A DE 1 3 a Therefore a v 20V i 4A b v 20V i 4A c v 20V i 4A d v 20V i 4A b Using the reference system in Fig 1 3 a p vi 20 4 80W so the box is absorbing power c The box is absorbing 80 W 1 2CHAPTER 1 Circuit Variables and Circuit Elements DE 1 4p vi 20 104e 10 000tW w Z 1 0 20 104e 10 000tdt 20 J DE 1 5p 800 103 1 8 103 1440 106 1440 MW from Oregon to California DE 1 6 The interconnection is valid is 10 15 25 A p100V 100is 2500 W absorbing p10A 100 10 1000 W generating 100 vs 40 0so vs 140 V p15A 15 140 2100 W generating p40V 15 40 600 W absorbing X pdev p10A p15A 3100 W X pabs p100V p40V 3100 W X pdev X pabs 3100 W DE 1 7 a vl vc v1 vs 0 ilRl icRc i1R1 vs 0 isRl isRc isR1 vs 0 b is vs Rl Rc R1 DE 1 8 a 24 v2 v5 v1 3i5 7i5 2i5 12i5 Therefore i5 24 12 2 A b v1 2i5 4 V c v2 3i5 6 V d v5 7i5 14 V Problems3 e p24 24 2 48 W therefore 24 V source is delivering 48 W DE 1 9 i2 120 24 5 A i3 120 8 15 A i1 i2 i3 20 A 200 20R 120 0 R 80 20 4 DE 1 10 a Plotting a graph of vtversus itgives Note that when it 0 vt 25 V therefore the voltage source must be 25 V When vtis zero it 0 25 A hence the resistor must be 25 0 25 or 100 A circuit model having the same v i characteristic is a 25 V source in series with a 100 resistor 4CHAPTER 1 Circuit Variables and Circuit Elements b it 25 125 0 2 A p 0 2 2 25 1 W DE 1 11 a Since we are constructing the model from two elements we have two choices on interconnecting them series or parallel From the v i characteristic we require vt 25 V when it 0 The only way we can satisfy this requirement is with a parallel connection The constraint that vt 0 when it 0 25 A tells us the ideal current source must produce 0 25 A Therefore the parallel resistor must be 25 0 25 or 100 b 0 25 vt 100 vt 25 0 5vt 25 vt 5 V p v2 t 25 1 W Problems P 1 1i dq dt 24cos4000t Therefore dq 24cos4000tdt Problems5 Z q t q 0 dx 24 Z t 0 cos4000y dy q t q 0 24sin4000y 4000 t 0 But q 0 0 by hypothesis i e the current passes through its maximum value at t 0 so q t 6 10 3sin4000tC 6sin4000tmC P 1 2p 6 100 10 3 0 6 W w 0 6 3 60 60 6480 J P 1 3Assume we are standing at box A looking toward box B then p vi a p 120 5 600 Wfrom A to B b p 250 8 2000 Wfrom B to A c p 150 16 2400 Wfrom B to A d p 480 10 4800 Wfrom A to B P 1 4 a p vi 40 10 400 W Power is being delivered by the box b Entering c Gain P 1 5 a p vi 60 10 600 W so power is being absorbed by the box b Entering c Lose P 1 6 a Looking from A to B the current i is in the direction of the voltage rise across the 12 V battery thereforep vi 12 30 360 W Thus the power ow is from B to A and Car A has the dead battery b w Z t 0 pdx Z t 0 360dx w 360t 360 1 60 21 6 kJ 6CHAPTER 1 Circuit Variables and Circuit Elements P 1 7p vi w Z t 0 pdx Since the energy is the area under the power vs time plot let us plot p vs t p 0 6 15 10 3 90 10 3W p 216 ks 4 15 10 3 60 10 3W w 60 10 3 216 103 1 2 216 30 16 2 kJ Note 60 hr 216 000 s 216 ks P 1 8 a p vi 30e 500t 30e 1500t 40e 1000t 50e 2000t 10e 3000t p 1 ms 3 1 mW b w t Z t 0 30e 500 x 30e 1500 x 40e 1000 x 50e 2000 x 10e 3000 x dx 21 67 60e 500t 20e 1500t 40e 1000t 25e 2000t 3 33e 3000t J w 1 ms 1 24 J c wtotal 21 67 J P 1 9 a v 20 ms 100e 1sin3 5 19 V i 20 ms 20e 1sin3 1 04 A p 20 ms vi 5 39 W Problems7 b p vi 2000e 100tsin 2150t 2000e 100t 1 2 1 2 cos300t 1000e 100t 1000e 100tcos300t w Z 1 0 1000e 100tdt Z 1 0 1000e 100tcos300tdt 1000 e 100t 100 1 0 1000 e 100t 100 2 300 2 100cos300t 300sin300t 1 0 10 1000 100 1 104 9 104 10 1 w 9 J P 1 10 a 0 t 10 ms v 1000t V i 0 6 mA p 0 6t mW 10 t 25 ms v 10 V i 0 6 mA p 6 mW 25 t 35 ms v 75 2500t V i 0 mA p 0 mW 35 t 60 ms v 50 1000t V i 0 4 mA p 20 400t mW 60 t 70 ms v 50 1000t V i 0 mA p 0 mW 70 t 80 ms v 20 V i 0 5 mA p 10 mW 80 t 90 ms v 180 2000t V i 0 mA p 0 mW 90 t 95 ms v 180 2000t V i 0 9 mA p 162 1800t mW 95 t 100 ms v 200 2000t V i 0 9 mA p 180 1800t mW 8CHAPTER 1 Circuit Variables and Circuit Elements b w 25 1 2 6 10 6 15 120 J w 60 120 1 2 15 6 1 2 10 4 145 J w 90 145 10 10 45 J w 100 45 1 2 10 9 0 J P 1 11 a p vi 2e 500t 2e 1000t W dp dt 1000e 500t 2000e 1000t 0 at t 1 4 ms pmax p 1 4 ms 0 5 W b w Z 1 0 2e 500t 2e 1000t dt 2 500e 500t 2 1000e 1000t 1 0 2 mJ P 1 12 a p vi 900sin 200 t cos 200 t 450sin 400 t W Therefore pmax 450 W b pmax extracting 450 W c pavg 200 Z 5 10 3 0 450sin 400 t dt 9 104 cos400 t 400 2 5 10 3 0 225 1 cos2 0 d pavg 180 1 cos2 5 180 57 3 W P 1 13 a q area under i vs t plot h 1 2 5 4 10 4 1 2 8 4 8 6 1 2 3 6 i 103 10 40 16 48 9 103 123 000 C Problems9 b w Z pdt Z vidt v 0 2 10 3t 90 t 15 ks 0 t 4000s i 15 1 25 10 3t p 135 8 25 10 3t 0 25 10 6t2 w1 Z 4000 0 135 8 25 10 3t 0 25 10 6t2 dt 540 66 5 3333 103 468 667 kJ 4000 t 12 000 i 12 0 5 10 3t p 108 2 1 10 3t 0 1 10 6t2 w2 Z 12 000 4000 108 2 1 10 3t 0 1 10 6t2 dt 864 134 4 55 467 103 674 133 kJ 12 000 t 15 000 i 30 2 10 3t p 270 12 10 3t 0 4 10 6t2 w3 Z 15 000 12 000 270 12 10 3t 0 4 10 6t2 dt 810 486 219 6 103 104 4 kJ wT w1 w2 w3 468 667 674 133 104 4 1247 2 kJ P 1 14 a p vi 400 103t2e 800t 700te 800t 0 25e 800t e 800t 400 000t2 700t 0 25 dp dt fe 800t 800 103t 700 800e 800t 400 000t2 700t 0 25 g 3 200 000t2 2400t 5 100e 800t Therefore dp dt 0 when 3 200 000t2 2400t 5 0 so pmaxoccurs at t 1 68 ms b pmax 400 000 00168 2 700 00168 0 25 e 800 00168 666 mW 10CHAPTER 1 Circuit Variables and Circuit Elements c w Z t 0 pdx w Z t 0 400 000 x2e 800 xdx Z t 0 700 xe 800 xdx Z t 0 0 25e 800 xdx 400 000e 800 x 512 106 64 104x2 1600 x 2 t 0 700e 800 x 64 104 800 x 1 t 0 0 25e 800 x 800 t 0 When t 1 all the upper limits evaluate to zero hence w 400 000 2 512 106 700 64 104 0 25 800 2 97 mJ P 1 15 a p 0t 3 s p vi t 3 t 6 4t 18t 18t2 4t3mW0 t 3 s dp dt 18 36t 12t2 12 t2 3t 1 5 dp dt 0when t2 3t 1 5 0 t 3 p9 6 2 3 p3 2 t1 3 2 p3 2 0 634 s t2 3 2 p3 2 2 366 s p t1 18 0 634 18 0 634 2 4 0 634 3 5 196 mW p t2 18 2 366 18 2 366 2 4 2 366 3 5 196 mW Therefore maximum power is being delivered at t 0 634 s b pmax 5 196 mW delivered c Maximum power is being extracted at t 2 366 s d pmax 5 196 mW extracted e w Z t 0 pdx Z t 0 18x 18x2 4x3 dx 9t2 6t3 t4 w 0 0 mJw 2 4 mJ w 1 4 mJw 3 0 mJ Problems11 P 1 16 a p vi 12 105t2e 1000tW dp dt 12 105 t2 1000 e 1000t e 1000t 2t 12 105te 1000t t 2 1000t dp dt 0 at t 0 t 2 ms We know p is a minimum at t 0 since v and i are zero at t 0 b pmax 12 105 2 10 3 2e 2 649 61 mW 12CHAPTER 1 Circuit Variables and Circuit Elements c w 12 105 Z 1 0 t2e 1000tdt 12 105 e 1000t 1000 3 106t2 2 000t 2 1 0 2 4 mJ P 1 17 a From the diagram and the table we have pa vaia 46 16 6 276 96 W del pb vbib 14 16 4 72 66 8352 W abs pc vcic 32 6 4 204 80 W abs pd vdid 22 1 28 28 16 W del pe veie 33 60 1 68 56 448 W del pf vfif 66 0 4 26 40 W del pg vgig 2 56 1 28 3 2768 W abs ph vhih 0 4 0 4 0 16 W abs XP del 276 96 28 16 56 448 26 40 387 9680 W XP abs 66 8352 204 80 3 2768 0 16 275 072 W Therefore XP del6 XP absand the subordinate engineer is correct b We can also check the data using Kirchho s laws From Fig P1 17 the following equations should be satis ed ia ib id 0 ok ib ic ie 0 no if ia ic 0 ok id ig ok ig ie ih 0 no ih if ok Using Kirchho s current law it appears ieis in error From Kirchho s voltage law we have vb va vc 0 ok vd vb ve vg 0 ok ve vc vf vh 0 ok Therefore all the voltages are consistent with Kirchho s voltage law Assume ieis in error Therefore ie ib ic ig ih 4 72 6 40 1 28 0 4 1 68 A So the error is in the sign of ie ieequals minus 1 68 A Correcting ieleads to XP del XP abs 331 52 W Problems13 P 1 18pa vaia 48 12 576 W abs pb vbib 18 4 72 W del pc vcic 30 10 300 W abs pd vdid 36 16 576 W abs pe veie 36 8 288 W del pf vfif 54 14 756 W abs pg vgig 84 22 1848 W del XP del 72 288 1848 2208 W XP abs 576 300 576 756 2208 W Therefore XP del XP abs 2208 W P 1 19 a From an examination of reference polarities the following elements employ the passive convention a c e and f b pa 56 W del pb 14 W del pc 150 W abs pd 50 W del pe 18 W del pf 12 W del XP abs 150 W XP del 56 14 50 18 12 150 W P 1 20 a 9 b 7 c 4 d va Ra vb Rb vc Rc e 6 f 1 va Ra Rd Rb vb 2 Rd Rf Re 3 vb Rb Rd Rf Rc vc 4 vc Rc Rf Ra va 5 va Ra Rf Re Rb vb 6 va Ra Rd Re Rc vc 7 vb Rb Re Rc vc 14CHAPTER 1 Circuit Variables and Circuit Elements P 1 21The interconnect is valid since it does not violate Kirchho s laws 60 20 40 0 KVL 8 4 12 0 KCL X Pdev 4 60 8 60 720 W X Pabs 12 20 12 40 720 W X Pdev XP abs 720 W P 1 22 a Yes Kirchho s laws are not violated b No because the voltages across the independent and dependent current sources are indeterminate For example de ne v1 v2 and v3as shown Kirchho s voltage law requires v1 20 v3 v2 100 v3 Conservation of energy requires 8 20 8v1 16v2 1600 24v3 0 or v1 2v2 3v3 220 Problems15 Now arbitrarily select a value of v3and show the conservation of energy will be satis ed Examples If v3 200 V then v1 180 V and v2 100 V Then 180 200 600 220 CHECKS If v3 100 V then v1 120 V and v2 200 V Then 120 400 300 220 CHECKS P 1 23 a Yes independent voltage sources can carry whatever current is required by the connection independent current source can support any voltage required by the connection b 30 V source absorbing 10 V source delivering 8 A source delivering c P30V 30 8 240 W abs P10V 10 8 80 W del P8A 20 8 160 W del XP abs XP del 240 W d Yes 30 V source is delivering the 10 V source is delivering and the 8 A source is absorbing P30V 30 8 240 W del P10V 10 8 80 W del P8A 40 8 320 W abs P 1 24The interconnection is valid because it does not violate Kirchho s laws i 25 A 6i 150 V 200 50 150 0 But the power developed in the circuit cannot be determined as the currents in the 200 V 50 V and 6i sources are unspeci ed P 1 25The interconnection is not valid because it violates Kirchho s current law 3 A 5 A 6 8 A 16CHAPTER 1 Circuit Variables and Circuit Elements P 1 26 i 4 A so ig 12 A vo 100 V 60 v1 100 so v1 160 V v2 80 100 so v2 180 V X Pdev 180 4 100 8 60 12 2240 W CHECK XP diss 160 12 80 4 1920 320 2240 W CHECKS P 1 27The interconnection is valid because it does not violate Kirchho s laws pV sources 100 60 5 200 W P 1 28First there is no violation of Kirchho s laws hence the interconnection is valid Kirchho s voltage law requires v1 v2 150 50 100 V The conservation of energy law requires 20v1 10v1 10v2 500 1500 0 or v1 v2 100 Hence any combination of v1and v2that adds to 100 is a valid solution For example if v1 80 V and v2 20 V Pabs 80 20 10 20 50 10 2300 W Problems17 Pdev 1500 80 10 2300 W If v1 60 V and v2 40 V Pabs 60 20 10 40 500 2100 W Pdev 60 10 1500 2100 W If v1 100 V and v2 200 V Pabs 10 100 10 200 10 50 3500 W Pdev 20 100 10 150 3500 W P 1 29 a 1 6 ig ia 80ia 1 6 30 90 192therefore ia 2 4 A ig ia 1 6 2 4 1 6 4 A b vg 90 1 6 144 V c XP dis 2 42 80 1 62 120 768 W XP dev 4 192 768 W Therefore XP dis XP dev 768 W P 1 30 a vo 8ia 14ia 18ia 40 20 800 V 800 10io io 800 10 80 A b ig ia io 20 80 100 A c pg delivered 100 800 80 000 W 80 kW P 1 31 a 20ia 80ibig ia ib 5ib ia 4ib 50 4ig 80ib 20ib 80ib 100ib ib 0 5 A therefore ia 2 Aandig 2 5 A 18CHAPTER 1 Circuit Variables and Circuit Elements b ib 0 5 A c vo 80ib 40 V d p4 i2 g 4 6 25 4 25 W p20 i2 a 20 4 20 80 W p80 i2 b 80 0 25 80 20 W e p5V delivered 5ig 125 W Check X Pdis 25 80 20 125 W X Pdel 125 W P 1 32 a vo 20 8 16 15 400 V io 400 80 5 A ia 25 A P230 supplied 230 25 5750 W ib 5 15 20 A P260 supplied 260 20 5200 W b XP dis 25 2 2 20 2 8 5 2 4 15 216 20 22 5 2 80 1250 3200 100 3600 800 2000 10 950 W XP sup 5750 5200 10 950 W Therefore XP dis XP sup 10 950 W Problems19 P 1 33 a v2 80 4 12 128 V v1 128 24 2 80 V i1 v1 16 80 16 5 A i3 i1 2 5 2 3 A vg v1 24i3 80 72 152 V vg 4i4 v2 4i4 vg v2 152 128 24 V i4 24 4 6 A ig i3 i4 3 6 9 A b p8 2 2 8 32 Wp4 6 2 4 144 W p12 2 2 12 48 Wp6 5 2 6 150 W p4 2 2 4 16 Wp10 5 2 10 250 W p24 3 2 24 216 Wp12 4 2 12 192 W c vg 152 V d XP dis 32 48 16 216 144 150 250 192 80 4 1368 W XP del 152 9 1368 W Therefore XP dis XP del 20CHAPTER 1 Circuit Variables and Circuit Elements P 1 34 a v2 180 100 80 V i2 v2 8 10 A i3 4 i2 i3 10 4 6 A v1 v2 v3 80 6 10 140 V i1 v1 70 140 70 2 A b p5 82 5 320 W p25 4 2 25 400 W p70 22 70 280 W p10 62 10 360 W p8 102 8 800 W c X Pdis 320 400 280 360 800 2160 W Pdev 180ig 180 12 2160 W P 1 35 a Problems21 b v 20 V i 10 mA R v i 2 k c 2i1 3is i1 1 5is 40 i1 is 2 5is is 16 mA d vs open circuit 40 10 3 2 103 80 V e vs open circuit 55 V f Linear model cannot predict the nonlinear behavior of the practical current source P 1 36 a Plot the v i characteristic From the plot R v i 125 50 15 0 5 When it 0 vt 50 V therefore the ideal current source has a current of 10 A 22CHAPTER 1 Circuit Variables and Circuit Elements b 10 it i1and5i1 20it Therefore 10 it 4itso it 2 A P 1 37 a b R 24 18 24 0 6 24 0