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Soln7 - I − V ab 20 Ω I 2 I 1 I I I I j 12 2 j 8 5 j 60...

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EE411 Spring 2008 HW#7 Solution (Q2) 9.36 Let Z be the input impedance at the source. 20 10 100 200 mH 100 3 j x x j L j = = ⎯→ ω 500 200 10 10 1 1 F 10 6 j x x j C j = = ⎯→ ω μ 1000//-j500 = 200 –j400 1000//(j20 + 200 –j400) = 242.62 –j239.84 o j Z 104 . 6 2255 84 . 239 62 . 2242 = = mA 896 . 3 61 . 26 104 . 6 2255 10 60 o o o I = = ) 896 . 3 200 cos( 1 . 266 o t i = mA (Q6) 9.66 ) j 12 ( 145 170 5 j 60 ) 10 j 40 )( 5 j 20 ( ) 10 j 40 ( || ) 5 j 20 ( T = + + = + = Z = T Z 14.069 – j1.172 Ω = 14.118 -4.76 ° ° = ° ° = = 76 . 94 25 . 4 76 . 4 - 118 . 14 90 60
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