电路分析基础(英文版)课后答案第二章.pdf

2 Some Circuit Simpli cation Techniques Drill Exercises DE 2 1 16k64 12 8 12 8 7 2 20 20k30 12 a v 5 12 60 V b p5A del 5 60 300 W c iA 60 30 2 AiC 3 64 80 2 4 A iB 5 2 3 Ap10 2 4 210 57 6 W DE 2 2 a vo no load 200 75 100 150 V b 75k150 50 k thereforevo 200 50 75 133 3 V c i 200 25 000 8 mA p25k 8 10 3 2 25 000 1 6 W d Maximum dissipation at no load since vois maximum p v2 o 75 000 0 3 W 27 28CHAPTER 2 Some Circuit Simpli cation Techniques DE 2 3 v30 6 4 0 825 9 3 V i30 v30 30 0 31 A i6 i30 0 825 1 135 A i10 0 825 0 31 1 135 A v30 6ib v20 10i10 0 v20 9 3 16 1 135 27 46 V i20 27 46 20 1 373 A i5 i6 i20 2 508 A i30 0 31 A i6 1 135 A i10 1 135 A i20 1 373 A andi5 2 508 A DE 2 4 Problems29 i 72 12 6 A a v 72 12 8 48 V i120V 120 57 6 20 3 12 A b va 6 9 6 57 6 V p120V del 120ia 374 40 W DE 2 5 a 110 V source acting alone Re 10 14 24 35 6 i0 110 5 35 6 132 13 A v0 o 35 6 132 13 770 13 V 4 A source acting alone 5 k10 50 15 10 3 10 3 2 16 3 30CHAPTER 2 Some Circuit Simpli cation Techniques 16 3k12 48 13 Hence our circuit reduces to It follows that v00 a 4 48 13 192 13 V and v00 o v00 a 16 3 10 3 5 8v 00 a 120 13 V vo v0 o v 00 o 770 13 120 13 50 V b p v2 o 10 250 W DE 2 6 70 V source acting alone v0 70 4i0b i0s v0 b 2 v0 10 i0a i0b 70 20i0a v0 b i0a 70 v0 b 20 Problems31 i0b v0 b 2 v0 10 70 v0 b 20 11 20v 0 b v0 10 3 5 v0 v0 b 2i 0 b v0 b v0 2i0b i0b 11 20 v 0 2i0b v0 10 3 5ori0b 13 42v 0 70 42 v0 70 4 13 42v 0 70 42 orv0 3220 94 1610 47 V 50 V source acting alone v00 4i00 b v00 v00 b 2i00 b v00 50 10i00 d i00 d v00 50 10 i00 s v00 b 2 v00 50 10 i00 b v00 b 20 i00 s v00 b 20 v00 b 2 v00 50 10 11 20v 00 b v00 50 10 v00 b v00 2i00 b i00 b 11 20 v 00 2i00 b v00 50 10 ori00 b 13 42v 00 100 42 Thus v00 4 13 42v 00 100 42 orv00 200 47 V Hence v v0 v00 1610 47 200 47 1410 47 30 V 32CHAPTER 2 Some Circuit Simpli cation Techniques Problems P 2 1 a p4 i2 s4 12 24 576 W p18 4 218 288 W p3 8 23 192 Wp6 8 26 384 W b p120V delivered 120is 120 12 1440 W c pdiss 576 288 192 384 1440 W P 2 2 a From Ex 3 1 i1 4 A i2 8 A is 12 A at node x 12 4 8 0 at node y 12 4 8 0 b v1 4is 48 Vv3 3i2 24 V v2 18i1 72 Vv4 6i2 48 V loop abda 120 48 72 0 loop bcdb 72 24 48 0 loop abcda 120 48 24 48 0 P 2 3 1 Req 1 6 1 10 1 15 10 30 1 3 Req 3 v 2 8 5 20 3 60 V i 2 8 5 60 15 4 A P5 4 2 5 80 W P 2 4 a Req 2 2 1 4 1 5 1 20 1 6 ig 120 6 20 A v4 120 2 2 20 40 V io 40 4 10 A Problems33 i 15 5 40 15 5 2 A vo 5 2 10 V b i15 2 A P15 2 2 15 60 W c P120V 120 20 2 4 kW P 2 5 a Req RkR R2 2R R 2 b Req RkRkRk kR n R s Rk R n 1 R2 n 1 R R n 1 R2 nR R n c One solution 700 200 500 1000 5 1000 2 1 k k1 k k1 k k1 k k1 k 1 k k1 k d One solution 5 5 k 5 k 0 5 k 2 k 2 k 1 k 0 5 k 2 k 2 k 2 k 2 2 k 4 2 k 2 k 2 k k2 k 2 k k2 k k2 k k2 k 34CHAPTER 2 Some Circuit Simpli cation Techniques P 2 6 a 12 k24 8 Therefore Rab 8 2 6 16 b 1 Req 1 24 k 1 30 k 1 20 k 15 120 k 1 8 k Req 8 k Req 7 15 k 1 Rab 1 15 k 1 30 k 1 15 k 5 30 k 1 6 k Rab 6 k P 2 7 a For circuit a Rab 15k 18 48k16 10 For circuit b 1 Re 1 20 1 15 1 20 1 4 1 12 30 60 1 2 Re 2 Re 16 18 18k18 9 Rab 10 8 9 27 For circuit c 48k16 12 12 8 20 20k30 12 12 18 30 30k15 10 10 10 20 40 Rab 40k60 24 b Pa 202 10 40 W Pb 1442 27 768 W Pc 62 24 864 W Problems35 P 2 8 a 5k20 100 25 4 5k20 9k18 10 20 9k18 162 27 6 20k30 600 50 12 Rab 5 12 3 20 b 5 15 20 30k20 600 50 12 20k60 1200 80 15 3k6 18 9 2 15 10 25 3k6 30k20 2 12 14 25k75 1875 100 18 75 26k14 364 40 9 1 18 75 11 25 30 Rab 2 5 9 1 3 4 15 c 3 5 8 60k40 2400 100 24 8k12 96 20 4 8 24 6 30 4 8 5 2 10 30k10 300 40 7 5 45 15 60 Rab 1 5 7 5 1 0 10 P 2 9 a Rcond 845 0 0397 33 5465 Rtotal 2 1 2 Rcond 33 5465 Ploss 2000 2 33 5465 134 186 MW Pcalif 800 2 134 186 1465 814 MW E ciency 1465 814 1600 100 91 61 b Pcalif 2000 134 86 1865 814 MW E ciency 93 29 c Ploss 3000 2 2 1 3 845 0 0397 201 279 MW Poregon 3000 MW Pcalif 3000 201 279 2798 7 MW E ciency 2798 70 3000 100 93 29 P 2 10 i10k 18 15 40 6 75 mA v15k 6 75 15 101 25 V i3k 18 6 75 11 25 mA v12k 12 11 25 135 V vo 101 25 135 33 75 V 36CHAPTER 2 Some Circuit Simpli cation Techniques P 2 11 a v1k 1 1 5 30 5 V v15k 15 15 60 30 6 V vx v15k v1k 6 5 1 V b v1k vs 6 1 vs 6 v15k vs 75 15 vs 5 vx vs 5 vs 6 vs 30 P 2 12 60k30 20 i30 25 75 125 15 A vo 15 20 300 V vo 30i30 750 V vg 12 25 750 vg 1050 V P 2 13 5 k20 4 4 6 10 10k40 8 Therefore ig 125 8 2 12 5 A i6 40 12 5 50 10 A io 5 10 25 2 A P 2 14 a 40k10 8 i75V 75 10 7 5 A 8 7 15 i4 3 7 5 30 45 5 A 15k30 10 io 5 10 50 1 A b i10 i4 3 io 5 1 4 A P10 4 2 10 160 W Problems37 P 2 15 a v9 1 9 9 V i2 9 2 1 3 A i4 1 3 4 A v25 4 4 9 25 V i25 25 25 1 A i3 i25 i9 i2 1 1 3 5 A v40 v25 v3 25 5 3 40 V i40 40 40 1 A i5k20 i40 i25 i4 1 1 4 6 A v5k20 4 6 24 V v32 v40 v5k20 40 24 64 V i32 64 32 2 A i10 i32 i5k20 2 6 8 A vg 10 8 v32 80 64 144 V b P20 v5k20 2 20 242 20 28 8 W P 2 16 a Let isbe the current oriented down through the resistors Then is Vs R1 R2 Rk Rn and vk Rkis Rk R1 R2 Rk Rn Vs b is 200 5 15 30 10 40 2 A v1 2 5 10 V v2 2 15 30 V v3 2 30 60 V v4 2 10 20 V v5 2 40 80 V 38CHAPTER 2 Some Circuit Simpli cation Techniques P 2 17 a vo 25 25 20 20 V b vo 25 5 Re Re Re 20 12 32 7 5 k vo 25 12 5 7 5 15 V c vo 25 20 25 0 80 d vo 25 15 25 0 60 P 2 18 a No load vo R2 R1 R2 Vs Vs R2 R1 R2 Load vo Re R1 Re Vs Vs Re Re R1 Re R2RL R2 RL R2RL R1R2 RL R1 R2 ButR1 R2 R2 R1 R2 R2 R2RL R2 R2 R2 RLR2 Problems39 RL R2 1 1 RL or R2 1 1 RL RL R2 1 1 RL 1 R2 1 RL R1 1 R2 RL b R1 0 9 0 7 0 63 126 k 40 k R2 0 9 0 7 0 7 0 1 126 k 360 k P 2 19 a Let vobe the voltage across the parallel branches positive at the upper terminal then ig voG1 voG2 voGN vo G1 G2 GN It follows thatvo ig G1 G2 GN The current in the kthbranch isik voGk Thus ik igGk G1 G2 GN b i6 25 1142 0 16 4 0 4 1 0 16 0 1 0 05 32 mA P 2 20 Re 4 8 103 500 XG 1 500 2 mS i1 2i2 2 10i3 20i4 i2 10i3 10i4 i3 i4 40CHAPTER 2 Some Circuit Simpli cation Techniques 8 20i4 10i4 i4 i4 32i4 i4 8 32 0 25 mA R4 vg i4 4 0 25 10 3 16 k i3 i4 0 25 mA R3 16 k i2 10i4 2 5 mA R2 vg i2 4 2 5 10 3 1 6 k i1 20i4 5 mA R1 vg i1 4 5 10 3 800 P 2 21 a Problems41 io 120 40 k 3 mA b va 3 20 60 V ia va 100 0 6 mA ib 4 3 6 0 4 mA vb 60 0 4 15 54 V ig 0 4 54 30 1 4 mA p75V developed 75 1 4 105 mW Check p4mA developed 60 4 240 mW XP dev 105 240 345 mW XP dis 1 4 2 15 1 8 2 30 0 4 2 15 0 6 2 100 3 2 20 345 mW 42CHAPTER 2 Some Circuit Simpli cation Techniques P 2 22 Apply source transformations to both current sources to get io 6 6 1 mA P 2 23 a vo 1 2 240 120 V io 120 24 5 A b p300V 12 5 300 3750 W Problems43 Therefore the 300 V source is developing 3 75 kW c 10 i6 7 5 12 5 0 i6 15 A v10A 4 10 6 15 0 v10A 130 V p10A 10v10A 1300 W Therefore the 10 A source is developing 1300 W d Xp dev 3750 1300 5050 W p4 100 4 400 W p40 7 5 2 40 2250 W p6 15 2 6 1350 W p42 5 2 42 1050 W Xp diss 400 1350 2250 1050 5050 W CHECKS P 2 24 Applying a source transformation to each current source yields Now combine the 20 V and 10 V sources into a single voltage source and the 5 4 and 1 resistors into a single resistor to get Now use a source transformation on each voltage source thus 44CHAPTER 2 Some Circuit Simpli cation Techniques which can be reduced to io 1 25 8 10 1 A P 2 25 First nd the Th evenin equivalent with respect to Ro RoiovoRoiovo 012020480 2102030390 67 545402 496 10660502100 154 872701 5105 P 2 26 100 k25 20 i 400 60 20 5 A Problems45 v0 o 20i 100 V 100 k60 37 5 i 500 25 37 5 8 A v00 o 37 5i 300 V vo v0 o v 00 o 100 300 400 V P 2 27 46CHAPTER 2 Some Circuit Simpli cation Techniques i0o 100 25 4 A 15 k30 10 i00 o 50 25 2 A io i0o i00 o 4 2 2 A P 2 28 15 2i0 50i1 3i0 Problems47 15 2i0 12i02 i0 i01 i02 i01 27 26 A i0 51 26 A i02 12 13 A v0 o 96 13 V 2i00 5i00 1 3i 00 i00 i00 1 i00 2 i00 i 00 1 2i00 4i00 2 8 i 00 2 8 2i 00 i00 2 64 13 A i00 1 32 13 A i00 32 13 A 8 i00 2 40 13 A v00 o 8 40 13 320 13 V vo v0 o v 00 o 96 13 320 13 32 V 48CHAPTER 2 Some Circuit Simpli cation Techniques P 2 29 a The evolution of the circuit shown in Fig P2 29 is illustrated in the following steps b Starting at the left end of the circuit and working toward the right end a series of source transformations yields Problems49 VR