EE411_HW1_Soln

# EE411_HW1_Soln - Ans:p1=-30*10=-300W. p2=10*10=100W....

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Ans:(a) The total charge at t=1sec is () C e e t dt e t t t t 131 . 2 20 20 10 | 20 20 10 1 10 5 . 0 1 5 . 0 1 0 5 . 0 = + = + = = = (b) P=i(t)*v(t)| t=1 =5cos(2)*10(1-e -0.5 )=-8.19W Note: should use rad in the argument of cosine function.

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Ans:(a) (b)The energy absorbed by the device=The area of the answer in part(a)=0.
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Unformatted text preview: Ans:p1=-30*10=-300W. p2=10*10=100W. p3=14*20=280W. p4=-8*4=-32W. p5=-12*0.4*10=-48W. Ans: Apply KVL to the loop form by 30V, 12V and Vo, we have -30+12+Vo=0. So Vo=18V....
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## This note was uploaded on 02/24/2011 for the course EE 411 taught by Professor Lee during the Spring '07 term at University of Texas at Austin.

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EE411_HW1_Soln - Ans:p1=-30*10=-300W. p2=10*10=100W....

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