# quiz4_sol - = V--(1 + j ) =-2(1 + j )-(1 + j ) =-3(1 + j )V...

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EE 101 Handout # 29 Prof. Abbas El Gamal March 14, 2003 Quiz #4 Solution First we transform the circuit to the phasor domain. - 4 j V out - j - 0 . 5 j (a) By voltage divider, V + = - 4 j × - j/ (1 - j ) = - 2(1 + j )V. By ideal op-amp model, V - = V + . The current ±owing (to ground) into the 2Ω resistor is - (1 + j )A. Since I - = 0, this current ±ows into the 1Ω resistor (from the output). Thus V out
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Unformatted text preview: = V--(1 + j ) =-2(1 + j )-(1 + j ) =-3(1 + j )V . Thus v out = 3 √ 2 cos(2 t + 225 ◦ )V. (b) The current ±owing out of the op-amp output is: I out =-(1 + j ) +-3(1 + j )-. 5 j = 5-7 j A . Thus the average power supplied by the op-amp is p avg = 1 2 Re ( I out V out ) = 1 2 Re (-3(1 + j )(5 + 7 j )) = 3W ....
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## This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.

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