Solution 6 - ECE 205 Section A, B, and C. Homework #6...

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ECE 205 Section A, B, and C. Homework #6 Solutions 3.8 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 3.12 There are two unknown nodes, as shown in the circuit below. At node 1, 30 V 10 V 16 0 1 V V 2 0 V 10 30 V o 1 o 1 1 1 (1) At node o, + 3V 4V 0 + V 0 1 i 1 2 3 5 i 2 i 3 v 1 10 + _ 2 5 30 V 4 I x 1 V o V 1
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0 I 20 V 6 V 5 0 5 0 V I 4 1 V V x o 1 o x 1 o (2) But I x = V 1 /2. Substituting this in (2) leads to –15V 1 + 6V o = 0 or V 1 = 0.4V o (3) Substituting (3) into 1, 16(0.4V o ) – 10V o = 30 or V o = –8.333 V . 3.16 At node 1, 2 v v 8 v 4 v 60 2 1 1 1 120 = 7v 1 - 4v 2 (1) At node 2, 3i 0 + 0 2 v v 10 v 2 1 2 But i 0 = . 4 v 60 1 Hence   0 2 v v v 4 v 3 2 1 2 1 1020 = 5v 1 + 12v 2 (2) Solving (1) and (2) gives v 1 = 53.08 V. Hence i 0 = 4 v 60 1 1.73 A 60 V + 60 V 4 10 2 8 3i 0 i 0 v 1 v 2
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Solution 6 - ECE 205 Section A, B, and C. Homework #6...

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