Soln10 - EE411 Spring 2008 HW#10 Solution(Q5 11.16 1H jL =...

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EE411 Spring 2008 HW#10 Solution (Q5) 11.16 5 20 / 1 4 1 1 F 20 / 1 , 4 H 1 , 4 j x j C j j L j = = ⎯→ = = ω We find the Thevenin equivalent at the terminals of Z L . To find V Th , we use the circuit shown below. 0. 5V o 2 Ω V 1 4 Ω V 2 + + + 10<0 o V o -j5 j4 V Th - - - At node 1, 2 1 2 1 1 1 1 V 25 . 0 ) 2 . 0 j 25 . 1 ( V 5 4 V V V 5 . 0 5 j V 2 V 10 + = + + = (1) At node 2, ) 25 . 0 25 . 0 ( 75 . 0 0 4 5 . 0 4 2 1 2 1 2 1 j V V j V V V V + + = = + ( 2 ) Solving (1) and (2) leads to o Th j V V 31 . 56 89 . 11 8901 . 9 5934 . 6 2 = + = =

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To obtain R Th , consider the circuit shown below. We replace Z L by a 1-A current source. 0. 5V 1 2 Ω V 1 4 Ω V 2 -j5 j4 At node 1, 2 1 2 1 1 1 1 25 . 0 ) 2 . 0 25 . 1 ( 0 0 4 5 . 0 5 2 V j V V V V j V V + = ⎯→ = + + + (3) At node 2, ) 25 . 0 25 . 0 ( 75 . 0 1 4 5 . 0 4 1 2 1 2 1 2 1 j V V j V V V V + + = = + + (4) Solving (3) and (4) gives o Th j V Z 4 . 65 0125 . 4 6484 . 3 6703 . 1 1 2 = + = = W 5817 . 10 6703 . 1 8 89 . 11 8 | | 2 2 max = = = x R V P Th Th (Q12) 11.73 (a) kVA 7 j 10 22 j 15 j 10 + = + = S = + = = 2 2 7 10 S S kVA 21 . 12 (b) 240 000 , 7 j 000 , 10 * * + = = = V S I I V S = = 167 . 29 j 667 . 41 IA 35 - 86 . 50 ° 1A
(c) ° =
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Soln10 - EE411 Spring 2008 HW#10 Solution(Q5 11.16 1H jL =...

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