Tài liệu Bài giải phần giải mạch P3 ppt

Chapter 3, Solution 1. 8 Ω v 2 2 Ω 40 Ω v 1 6 A 10 A At node 1, 6 = v 1 /(8) + (v 1 - v 2 )/4 48 = 3v 1 - 2v 2 (1) At node 2, v 1 - v 2 /4 = v 2 /2 + 10 40 = v 1 - 3v 2 (2) Solving (1) and (2), v 1 = 9.143V, v 2 = -10.286 V P 8Ω = () == 8 143.9 8 v 2 2 1 10.45 W P 4Ω = () = − 4 vv 2 21 94.37 W P 2Ω = () = = = 2 286.10 2 v 2 1 2 52.9 W Chapter 3, Solution 2 At node 1, 2 vv 6 5 v 10 v 2111 − +=− − 60 = - 8v 1 + 5v 2 (1) At node 2, 2 vv 63 4 v 212 − ++= 36 = - 2v 1 + 3v 2 (2) Solving (1) and (2), v 1 = 0 V, v 2 = 12 V Chapter 3, Solution 3 Applying KCL to the upper node, 10 = 60 v 2 30 v 20 v 10 v 0oo +++ 0 + v 0 = 40 V i 1 = = 10 0 v 4 A , i 2 = = 20 v 0 2 A, i 3 = = 30 v 0 1.33 A, i 4 = = 60 v 0 67 mA Chapter 3, Solution 4 At node 1, 4 + 2 = v 1 /(5) + v 1 /(10) v 1 = 20 At node 2, 5 - 2 = v 2 /(10) + v 2 /(5) v 2 = 10 i 1 = v 1 /(5) = 4 A, i 2 = v 1 /(10) = 2 A, i 3 = v 2 /(10) = 1 A, i 4 = v 2 /(5) = 2 A Chapter 3, Solution 5 Apply KCL to the top node. k 4 v k 6 v20 k 2 v30 000 = − + − v 0 = 20 V i 1 10 Ω 10 Ω 2A i 4 i 3 v 1 v 2 i 2 4 A 5 Ω 5 Ω 5 A Chapter 3, Solution 6 i 1 + i 2 + i 3 = 0 0 2 10v 6 v 4 12v 00 2 = − ++ − or v 0 = 8.727 V Chapter 3, Solution 7 At node a, ba baaa VV VVVV 3610 101530 10 −=→ − += − (1) At node b, ba bbba VV VVVV 72240 5 9 20 12 10 −=→= −− + − + − (2) Solving (1) and (2) leads to V a = -0.556 V, V b = -3.444V Chapter 3, Solution 8 i 2 5 Ω 2 Ω i 1 3 Ω v 1 i 3 1 Ω + – 4V 0 3V – + + V 0 – i 1 + i 2 + i 3 = 0 0 5 v4v 1 3v 5 v 01 11 = − + − + But 10 v 5 2 v = so that v 1 + 5v 1 - 15 + v 1 - 0v 5 8 1 = or v 1 = 15x5/(27) = 2.778 V, therefore v o = 2v 1 /5 = 1.1111 V Chapter 3, Solution 9 + v 1 – i 2 6 Ω i 1 3 Ω v 1 i 3 8 Ω + – + v 0 – 2v 0 12V – + At the non-reference node, 6 v2v 8 v 3 v 01 11 12 − += − (1) But -12 + v 0 + v 1 = 0 v 0 = 12 - v 1 (2) Substituting (2) into (1), 6 24v3 8 v 3 v 111 − += −12 v 0 = 3.652 V Chapter 3, Solution 10 At node 1, 8 v 4 1 vv 112 += − 32 = -v 1 + 8v 2 - 8v 0 (1) 1 Ω 4A 2i 0 i 0 v 1 v 0 8 Ω 2 Ω v 2 4 Ω At node 0, 0 0 I2 2 v +=4 and 8 v I 1 0 = 16 = 2v 0 + v 1 (2) At node 2, 2I 0 = 4 v 1 v 212 + − v and 8 v 1 0 = I v 2 = v 1 (3) From (1), (2) and (3), v 0 = 24 V, but from (2) we get i o = 62 4 24 2 2 2 v 4 o −=−= − = - 4 A Chapter 3, Solution 11 i 3 6 Ω v i 1 i 2 4 Ω 3 Ω – + 10 V 5 A Note that i 2 = -5A. At the non-reference node 6 v 5 4 v10 =+ − v = 18 i 1 = = − 4 v10 -2 A, i 2 = -5 A Chapter 3, Solution 12 i 3 40 Ω v 1 v 2 10 Ω 20 Ω – + 5 A 50 Ω 24 V At node 1, 40 0v 20 vv 10 v24 1211 − + − = − 96 = 7v 1 - 2v 2 (1) At node 2, 50 v 20 vv 221 = − + 5 500 = -5v 1 + 7v 2 (2) Solving (1) and (2) gives, v 1 = 42.87 V, v 2 = 102.05 V == 40 v i 1 1 1.072 A, v 2 = = 50 v 2 2.041 A Chapter 3, Solution 13 At node number 2, [(v 2 + 2) – 0]/10 + v 2 /4 = 3 or v 2 = 8 volts But, I = [(v 2 + 2) – 0]/10 = (8 + 2)/10 = 1 amp and v 1 = 8x1 = 8volts Chapter 3, Solution 14 1 Ω 2 Ω 4 Ω v 0 v 1 5 A 8 Ω 20 V – + 40 V – + At node 1, 1 v40 5 2 vv 001 − =+ − v 1 + v 0 = 70 (1) At node 0, 8 20v 4 v 5 2 vv 0001 + +=+ − 4v 1 - 7v 0 = -20 (2) Solving (1) and (2), v 0 = 20 V Chapter 3, Solution 15 1 Ω 2 Ω 4 Ω v 0 v 1 5 A 8 Ω 20 V – + 40 V – + Nodes 1 and 2 form a supernode so that v 1 = v 2 + 10 (1) At the supernode, 2 + 6v 1 + 5v 2 = 3 (v 3 - v 2 ) 2 + 6v 1 + 8v 2 = 3v 3 (2) At node 3, 2 + 4 = 3 (v 3 - v 2 ) v 3 = v 2 + 2 (3) Substituting (1) and (3) into (2), 2 + 6v 2 + 60 + 8v 2 = 3v 2 + 6 v 2 = 11 56 − v 1 = v 2 + 10 = 11 54 i 0 = 6v i = 29.45 A P 65 = =       == 6 11 54 Gv R 2 2 1 2 1 v 144.6 W P 55 = =       − = 5 11 56 Gv 2 2 2 129.6 W P 35 = () ==− 3)2(Gvv 2 2 3L 12 W Chapter 3, Solution 16 2 S + v 0 – 13 V – + i 0 1 S 4 S 8 S v 1 v 2 v 3 2 A At the supernode, 2 = v 1 + 2 (v 1 - v 3 ) + 8(v 2 – v 3 ) + 4v 2 , which leads to 2 = 3v 1 + 12v 2 - 10v 3 (1) But v 1 = v 2 + 2v 0 and v 0 = v 2 . Hence v 1 = 3v 2 (2) v 3 = 13V (3) Substituting (2) and (3) with (1) gives, v 1 = 18.858 V, v 2 = 6.286 V, v 3 = 13 V Chapter 3, Solution 17 60 V i 0 3i 0 2 Ω 10 Ω 4 Ω 60 V – + 8 Ω At node 1, 2 vv 8 v 4 v60 2111 − += − 120 = 7v 1 - 4v 2 (1) At node 2, 3i 0 + 0 2 vv 10 v60 212 = − + − But i 0 = . 4 v60 1 − Hence () 0 2 vv 10 v60 4 v603 2121 = − + − + − 1020 = 5v 1 - 12v 2 (2) Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 = = − 4 v60 1 1.73 A Chapter 3, Solution 18 + v 3 – + v 1 – 10 V – + v 1 v 2 2 Ω 2 Ω 4 Ω 8 Ω v 3 5 A (a) (b) At node 2, in Fig. (a), 5 = 2 vv 2 vv 3212 − + − 10 = - v 1 + 2v 2 - v 3 (1) At the supernode, 8 v 4 v 2 vv 2 vv 3 1 32 12 += − + − 40 = 2v 1 + v 3 (2) From Fig. (b), - v 1 - 10 + v 3 = 0 v 3 = v 1 + 10 (3) Solving (1) to (3), we obtain v 1 = 10 V, v 2 = 20 V = v 3 Chapter 3, Solution 19 At node 1, 321 121 31 4716 482 35 VVV VVV VV −−=→+ − + − += (1) At node 2, 321 32 221 270 428 VVV VV VVV −+−=→ − += − (2) At node 3, 321 3231 3 724360 428 12 3 VVV VVVV V −+=−→= − + − + − + (3) From (1) to (3), BAV V V V =→           − =                     − −− −− 36 0 16 724 271 417 3 2 1 Using MATLAB, V 267.12 V, 933.4 V, 10 267.12 933.4 10 321 1 ===→           == − VVVBAV Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence 040 414 321 3 21 =++→=++ VVV V VV (1) . V 1 . V 2 2 Ω V 3 4Ω 1 Ω 4 Ω