Soln12 - EE411 Spring 2008 HW#12 Solution (Q5) 12.56 (a)...

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Unformatted text preview: EE411 Spring 2008 HW#12 Solution (Q5) 12.56 (a) Consider the circuit below. For mesh 1, ) ( 10 j 440 120- 440 3 1 = − + ° ∠ − ° ∠ I I ° ∠ = + = − 60- 21 . 76 10 j ) 866 . j 5 . 1 )( 440 ( 3 1 I I ( 1 ) For mesh 2, ) ( 20 120- 440 120 440 3 2 = − + ° ∠ − ° ∠ I I 1 . 38 j 20 ) 732 . 1 j )( 440 ( 2 3 = = − I I ( 2 ) For mesh 3, 5 j ) ( 20 ) ( 10 j 3 2 3 1 3 = − − + − I I I I I Substituting (1) and (2) into the equation for mesh 3 gives, ° ∠ = + = 60 42 . 152 j5 j0.866)-1.5 )( 440 ( 3 I ( 3 ) From (1), ° ∠ = + = ° ∠ + = 30 132 66 j 315 . 114 60- 21 . 76 3 1 I I From (2), ° ∠ = + = − = 50.94 93 . 120 9 . 93 j 21 . 76 1 . 38 j 3 2 I I = = 1 a I I A 30 132 ° ∠ j10 Ω 20 Ω I 3 A C B − + 440 ∠ ° 440 ∠ 120 ° 440 ∠-120 ° − + − + a b c I 1 I 2 = + = − = j27.9-38.105 1 2 b I I I A 143.8 23 . 47 ° ∠ = = 2 c- I I A 230.9 9 . 120 ° ∠ (b) kVA 08 . 58 j ) 10 j ( 2 3 1 AB = − = I I S kVA 04 . 29 ) 20 ( 2 3 2 BC = − = I I S kVA-j116.16-j116....
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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.

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Soln12 - EE411 Spring 2008 HW#12 Solution (Q5) 12.56 (a)...

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