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EE 302 - HW Solution - 6

EE 302 - HW Solution - 6 - EE302 Homework#6 Chapter 3...

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EE302 Homework #6 Chapter 3, Solution 56. For loop 1, 12 = 4i 1 – 2i 2 – 2i 3 which leads to 6 = 2i 1 – i 2 – i 3 (1) For loop 2, 0 = 6i 2 –2i 1 – 2 i 3 which leads to 0 = -i 1 + 3i 2 – i 3 (2) For loop 3, 0 = 6i 3 – 2i 1 – 2i 2 which leads to 0 = -i 1 – i 2 + 3i 3 (3) In matrix form (1), (2), and (3) become, = 0 0 6 i i i 3 1 1 1 3 1 1 1 2 3 2 1 = , 8 3 1 1 1 3 1 1 1 2 = 2 = 24 3 0 1 1 3 1 1 6 2 = 3 = 24 0 1 1 0 3 1 6 1 2 = , therefore i 2 = i 3 = 24/8 = 3A, v 1 = 2i 2 = 6 volts , v = 2i 3 = 6 volts + 12 V i 1 i 2 i 3 + v 2 + v 1 2 2 2 2 2
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