HW6 Solution - EE302 Homework #6 Chapter 3, Solution 56. +...

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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 Δ = , 8 3 1 1 1 3 1 1 1 2 = Δ 2 = 24 3 0 1 1 3 1 1 6 2 = Δ 3 = 24 0 1 1 0 3 1 6 1 2 = , therefore i 2 = i 3 = 24/8 = 3A, v 1 = 2i 2 = 6.00 volts , v = 2i 3 = 6.00 volts + 12 V i 1 i 2 i 3 + v 2 + v 1 2 Ω 2 Ω 2 Ω 2 Ω 2 Ω
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Chapter 3, Solution 60. At node 1, (v 1 /1) + (0.5v 1 /1) = (10 – v 1 )/4, which leads to v 1 = 10/7 At node 2, (0.5v 1 /1) + ((10 – v 2 )/8) = v 2 /2 which leads to v 2 = 22/7 P 1 Ω = (v 1 ) 2 /1 = 2.04 watts , P 2 Ω = (v 2 ) 2 /2 = 4.94 watts P 4 Ω = (10 – v 1 ) 2 /4 = 18.4 watts , P 8 Ω = (10 – v 2 ) 2 /8 = 5.88 watts 0.5i 0 v 2 + 10 V 10 V 2 Ω 8 Ω 4 Ω 1 Ω v 1 i 0
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Chapter 3, Solution 63.
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This note was uploaded on 11/05/2008 for the course EE 302 taught by Professor Mccann during the Spring '06 term at University of Texas.

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HW6 Solution - EE302 Homework #6 Chapter 3, Solution 56. +...

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