Solution 7 - ECE 205 Section A B and C Homework#7 Solutions 3.39 For mesh 1 10 2 I x 10 I 1 6 I 2 = 0 But I x = I 1 I 2 Hence 10 = 2I1 2I 2 10I1 6I

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ECE 205 Section A, B, and C. Homework #7 Solutions 3.39 For mesh 1, 0 6 10 2 10 2 1 = + I I I x But 2 1 I I I x = . Hence, 2 1 2 1 2 1 I 2 I 4 5 I 6 I 10 I 2 I 2 10 = ⎯→ + + = (1) For mesh 2, 2 1 1 2 4 3 6 0 6 8 12 I I I I = = + ( 2 ) Solving (1) and (2) leads to -0.9A A, 8 . 0 2 1 = = I I 3.40 Assume all currents are in mA and apply mesh analysis for mesh 1. 30 = 12i 1 – 6i 2 – 4i 3 15 = 6i 1 – 3i 2 – 2i 3 (1) for mesh 2, 0 = - 6i 1 + 14i 2 – 2i 3 0 = -3i 1 + 7i 2 – i 3 (2) for mesh 3, 0 = -4i 1 – 2i 2 + 10i 3 0 = -2i 1 – i 2 + 5i 3 (3) Solving (1), (2), and (3), we obtain, i o = i 1 = 4.286 mA . 4 k Ω + 30V i 1 i 3 6 k Ω 2 k Ω 4 k Ω 6 k Ω i 2 2 k Ω
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3.46 Mesh 1: 60 3 i 3 + i 2 =0 Mesh 2: 5 i 2 i 1 3
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This note was uploaded on 02/03/2010 for the course ECE 201 taught by Professor C during the Spring '10 term at Miami University.

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Solution 7 - ECE 205 Section A B and C Homework#7 Solutions 3.39 For mesh 1 10 2 I x 10 I 1 6 I 2 = 0 But I x = I 1 I 2 Hence 10 = 2I1 2I 2 10I1 6I

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