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chap03

# chap03 - Chapter 3 Solution 1 v1 8 2 40 v2 6A 10 A At node...

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Chapter 3, Solution 1. 8 v 2 2 40 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 8 = () = = 8 143 . 9 8 v 2 2 1 10.45 W P 4 = = 4 v v 2 2 1 94.37 W P 2 = = = = 2 286 . 10 2 v 2 1 2 52.9 W Chapter 3, Solution 2 At node 1, 2 v v 6 5 v 10 v 2 1 1 1 + = 60 = - 8v 1 + 5v 2 (1) At node 2, 2 v v 6 3 4 v 2 1 2 + + = 36 = - 2v 1 + 3v 2 (2) Solving (1) and (2), v 1 = 0 V , v 2 = 12 V

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Chapter 3, Solution 3 Applying KCL to the upper node, 10 = 60 v 2 30 v 20 v 10 v 0 o o + + + 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 v 20 k 2 v 30 0 0 0 = + v 0 = 20 V i 1 10 10 2A i 4 i 3 v 1 v 2 i 2 4 A 5 5 5 A
Chapter 3, Solution 6 i 1 + i 2 + i 3 = 0 0 2 10 v 6 v 4 12 v 0 0 2 = + + or v 0 = 8.727 V Chapter 3, Solution 7 At node a, b a b a a a V V V V V V 3 6 10 10 15 30 10 = → + = (1) At node b, b a b b b a V V V V V V 7 2 24 0 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 2 i 1 3 v 1 i 3 1 + 4V 0 3V + + V 0 i 1 + i 2 + i 3 = 0 0 5 v 4 v 1 3 v 5 v 0 1 1 1 = + + But 1 0 v 5 2 v = so that v 1 + 5v 1 - 15 + v 1 - 0 v 5 8 1 = or v 1 = 15x5/(27) = 2.778 V, therefore v o = 2v 1 /5 = 1.1111 V

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Chapter 3, Solution 9 + v 1 i 2 6 i 1 3 v 1 i 3 8 + + v 0 2v 0 12V + At the non-reference node, 6 v 2 v 8 v 3 v 0 1 1 1 12 + = (1) But -12 + v 0 + v 1 = 0 v 0 = 12 - v 1 (2) Substituting (2) into (1), 6 24 v 3 8 v 3 v 1 1 1 + = 12 v 0 = 3.652 V Chapter 3, Solution 10 At node 1, 8 v 4 1 v v 1 1 2 + = 32 = -v 1 + 8v 2 - 8v 0 (1) 1 4A 2i 0 i 0 v 1 v 0 8 2 v 2 4
At node 0, 0 0 I 2 2 v + = 4 and 8 v I 1 0 = 16 = 2v 0 + v 1 (2) At node 2, 2 I 0 = 4 v 1 v 2 1 2 + 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 = 6 2 4 24 2 2 2 v 4 o = = = - 4 A Chapter 3, Solution 11 i 3 6 v i 1 i 2 4 3 + 10 V 5 A Note that i 2 = -5A. At the non-reference node 6 v 5 4 v 10 = + v = 18 i 1 = = 4 v 10 -2 A , i 2 = -5 A Chapter 3, Solution 12 i 3 40 v 1 v 2 10 20 + 5 A 50 24 V

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At node 1, 40 0 v 20 v v 10 v 24 1 2 1 1 + = 96 = 7v 1 - 2v 2 (1) At node 2, 50 v 20 v v 2 2 1 = + 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 2 4 v 0 v 1 5 A 8 20 V + 40 V + At node 1, 1 v 40 5 2 v v 0 0 1 = + v 1 + v 0 = 70 (1) At node 0, 8 20 v 4 v 5 2 v v 0 0 0 1 + + = + 4v 1 - 7v 0 = -20 (2) Solving (1) and (2), v 0 = 20 V
Chapter 3, Solution 15 1 2 4 v 0 v 1 5 A 8 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 G v R 2 2 1 2 1 v 144.6 W P 55 = =  − = 5 11 56 G v 2 2 2 129.6 W P 35 = () = = 3 ) 2 ( G v v 2 2 3 L 12 W

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chap03 - Chapter 3 Solution 1 v1 8 2 40 v2 6A 10 A At node...

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