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

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EE411 Spring 2008 HW#12 Solution (Q5) 12.56 (a) Consider the circuit below. For mesh 1, 0 ) ( 10 j 0 440 120 - 440 3 1 = + ° ° I I ° = + = 60 - 21 . 76 10 j ) 866 . 0 j 5 . 1 )( 440 ( 3 1 I I (1) For mesh 2, 0 ) ( 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, 0 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 0 ° 440 120 ° 440 -120 ° + + a b c I 1 I 2
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= + = = 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 -j5) ( (152.42) -j5) ( 2 2 3 CA = = = I S kVA 08 . 58 j 04 . 29 CA BC AB = + + = S S S S Real power absorbed = kW 04 . 29 (c) Total complex supplied by the source is = S kVA 08 . 58 j 04 . 29 (Q8) 13.14 To obtain V Th , convert the current source to a voltage source as shown below. Note that the two coils are connected series aiding. ω L = ω L 1 + ω L 2 – 2 ω M j ω L = j6 + j8 – j4 = j10 Thus, –j10 + (5 + j10 – j3 + 2)I + 8 = 0 + 8 V + V Th 2 Ω j2 a b + j10 V I -j3 Ω j8 Ω
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