Soln2 - EE411 Spring 2008 HW#2 Solution (Q1) 3.30 v2 I0 v1...

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EE411 Spring 2008 HW#2 Solution (Q1) 3.30 At node 1, 20 v v 4 10 v 100 40 v v 1 o 1 2 1 + = ( 1 ) But, v o = 120 + v 2 v 2 = v o – 120. Hence (1) becomes 7 v 1 – 9v o = 280 (2) At node 2, I o + 2I o = 80 0 v o 80 v 40 v 120 v 3 o o 1 = + or 6v 1 – 7v o = -720 (3) from (2) and (3), = 720 280 v v 7 6 9 7 o 1 v 1 10 Ω 20 Ω 80 Ω 40 Ω 1 v 0 I 0 2I 0 2 + + 100 V 120 V 4v 0 v 2
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5 54 49 7 6 9 7 = + = = Δ 8440 7 720 9 280 1 = = Δ , 6720 720 6 280 7 2 = = Δ v 1 =
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This note was uploaded on 10/26/2009 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas at Austin.

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Soln2 - EE411 Spring 2008 HW#2 Solution (Q1) 3.30 v2 I0 v1...

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