EE 302 - HW Solution - 5

EE 302 - HW Solution - 5 - EE302 Homework #5 Chapter 3,...

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EE302 Homework #5 Chapter 3, Solution 36. Applying mesh analysis gives, 12 = 10I 1 – 6I 2 -10 = -6I 1 + 8I 2 or = 2 1 I I 4 3 3 5 5 6 , 11 4 3 3 5 = = , 9 4 5 3 6 1 = = 7 5 3 6 5 2 = = , 11 9 I 1 1 = = 11 7 I 2 2 = = i 1 = -I 1 = -9/11 = -0.8181 A, i 2 = I 1 – I 2 = 10/11 = 1.4545 A. v o = 6i 2 = 6x1.4545 = 8.727 V . I 1 I 2 + 12 V + 10 V 4 6 2 i 3 i 2 i 1
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Chapter 3, Solution 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 Chapter 3, Solution 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 .
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This note was uploaded on 03/22/2008 for the course EE 302 taught by Professor Mccann during the Fall '06 term at University of Texas at Austin.

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EE 302 - HW Solution - 5 - EE302 Homework #5 Chapter 3,...

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