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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 op-amp 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 op-amp 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