Bài giải phần giải mạch P3

Chapter 3, Solution 1. v18Ω 2Ω 40 Ωv26A10 AAt node 1, 6 = v1/(8) + (v1 - v2)/4 At node 2, v1 - v2/4 = v2/2 + 10 Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V2 v1 (9.143)2 P8Ω = = = 10.45 W 8 848 = 3v1 - 2v2(1)40 = v1 - 3v2(2)P4Ω =(v 1 − v 2 )24= 94.37 Wv1 (= 10.286)2 = 52.9 W P2Ω = 2 = 2 2Chapter 3, Solution 2At node 1, v − v2 − v1 v1 − = 6+ 1 10 5 2 At node 2, v2 v −...
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Bài giải phần giải mạch P3Chapter 3, Solution 1. v1 40 Ω v2 6A 8Ω 10 A 2Ω At node 1, 6 = v1/(8) + (v1 - v2)/4 48 = 3v1 - 2v2 (1) At node 2, v1 - v2/4 = v2/2 + 10 40 = v1 - 3v2 (2) Solving (1) and (2), v1 = 9.143V, v2 = -10.286 V v1 (9.143)2 2 P8Ω = = = 10.45 W 8 8 (v 1 − v 2 )2 P4Ω = = 94.37 W 4 v1 (= 10.286)2 P2Ω = 2 = = 52.9 W 2 2Chapter 3, Solution 2 At node 1, − v1 v1 v − v2 − = 6+ 1 60 = - 8v1 + 5v2 (1) 10 5 2 At node 2, v2 v − v2 = 3+ 6+ 1 36 = - 2v1 + 3v2 (2) 4 2 Solving (1) and (2), v1 = 0 V, v2 = 12 VChapter 3, Solution 3 Applying KCL to the upper node, v0 vo vo v 10 = + + +2+ 0 v0 = 40 V 10 20 30 60 v0 v v v i1 = = 4 A , i2 = 0 = 2 A, i3 = 0 = 1.33 A, i4 = 0 = 67 mA 10 20 30 60Chapter 3, Solution 4 v1 2A v2 i1 i2 i3 i4 5Ω 10 Ω 10 Ω 5Ω 5A 4AAt node 1, 4 + 2 = v1/(5) + v1/(10) v1 = 20At node 2, 5 - 2 = v2/(10) + v2/(5) v2 = 10 i1 = v1/(5) = 4 A, i2 = v1/(10) = 2 A, i3 = v2/(10) = 1 A, i4 = v2/(5) = 2 AChapter 3, Solution 5 Apply KCL to the top node. 30 − v 0 20 − v 0 v 0 + = v0 = 20 V 2k 6k 4kChapter 3, Solution 6 v 2 − 12 v 0 v 0 − 10 i1 + i2 + i3 = 0 + + =0 4 6 2 or v0 = 8.727 VChapter 3, Solution 7At node a,10 − Va Va Va − Vb = + → 10 = 6Va − 3Vb (1) 30 15 10At node b,Va − Vb 12 − Vb − 9 − Vb + + =0 → 24 = 2Va − 7Vb (2) 10 20 5Solving (1) and (2) leads to Va = -0.556 V, Vb = -3.444VChapter 3, Solution 8 3Ω i1 v1 i3 5Ω i2 + 3V + V0 2Ω – + 4V0 – – 1Ω v1 v1 − 3 v1 − 4 v 0 i1 + i2 + i3 = 0 + + =0 5 1 5 2 8But v 0 = v1 so that v1 + 5v1 - 15 + v1 - v1 = 0 5 5 or v1 = 15x5/(27) = 2.778 V, therefore vo = 2v1/5 = 1.1111 VChapter 3, Solution 9 3Ω i1 v1 i3 6Ω + v0 – i2 + 12V + v1 8Ω + – 2v0 – –At the non-reference node, 12 − v1 v1 v1 − 2 v 0 = + (1) 3 8 6But -12 + v0 + v1 = 0 v0 = 12 - v1 (2)Substituting (2) into (1), 12 − v1 v1 3v1 − 24 = + v0 = 3.652 V 3 8 6Chapter 3, Solution 10At node 1, v 2 − v1 v = 4+ 1 32 = -v1 + 8v2 - 8v0 (1) 1 8 1Ω 2i0 4A v0 v1 v2 8Ω 2Ω 4Ω i0At node 0, v0 v 4= + 2I 0 and I 0 = 1 16 = 2v0 + v1 (2) 2 8At node 2, v 2 − v1 v 2 v 2I0 = + and I 0 = 1 v2 = v1 (3) 1 4 8From (1), (2) and (3), v0 = 24 V, but from (2) we get v 4− oio = 2 = 2 − 24 = 2 − 6 = - 4 A 2 4Chapter 3, Solution 11 4Ω i1 v i2 3Ω i3 10 V + – 5A ...