HW33%20solution

HW33%20solution - ECE 201 Spring 2010 Homework 33 Solutions...

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ECE 201 Spring 2010 Homework 33 Solutions Problem 30 (a) Y in ( )= 1 R + 1 jωL =0 . 05 - j 0 . 25 ω Z in ( )= 1 Y in ( ) = j 20 ω 5+ (b) I L = I in Z in ( ) jωL = 10 × 10( i + j ) j 20 =5 2 e - jπ/ 4 i L ( t ) = 10 cos(5 t - π/ 4) mA Problem 40 Let the impedances of R 1 ,C and R 2 ,L combinations be Z 1 and Z 2 respec- tively. Z 1 = ± 1 500 + j (400 × 5 × 10 - 6 ) ² - 1 = 250(1 - j ) Z 2 = ³ 1 100 + 1 j (400 × 0 . 125) ´ - 1 = 20(1 + 2 j ) 1
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Let the phasor current through the source be I . I = V 270 - j 210 V C = I × 250(1 -
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HW33%20solution - ECE 201 Spring 2010 Homework 33 Solutions...

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