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Unformatted text preview: EE 101 Handout # 19 Prof. Abbas El Gamal Feb. 25, 2003 EE 101 Midterm Exam Solution Problem 1. We use superposition. First we turn off the 2V and the 2A sources. Using the ideal opamp model, from the figure i = 1A. Thus v 1 = 1V, which gives v = 2V and v out1 = v + 1 = 3V. The current i out1 = 3 / 2 + 1 = 5 / 2A. 1V 2Ω 1Ω 1Ω 1Ω v out1 i i i i out1 v v v 1 Next we only turn on the 2V source. Clearly no current flows through any of the three 1Ω resistors and v out2 = v = 2V, and i out2 = 1A. 2V 2Ω 1Ω 1Ω 1Ω v out2 i out2 v v v 1 Finally we only turn on the 2A source. Clearly v = 0 and i = 2A. Thus v out3 = 2V and i out3 = 2 1 = 3A. 2A 2Ω 1Ω 1Ω 1Ω v out3 i i out3 v v Adding up we get v out = 3 + 2 2 = 3V and i out = 5 / 2 + 1 3 = 1 / 2A. Thus the power supplied by the opamp p = 3 / 2W. Problem 2. To simplify the problem we perform source transformation as shown below. To find v th , we find the open circuit voltage. Performing KCL, we obtain 1 v th v o 2Ω 2 v o v...
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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.
 Fall '08
 PingLi

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