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Unformatted text preview: EE 101 Handout # 16 Prof. Abbas El Gamal Feb 16 2003 Solutions to Sample Midterm Problems 1. To find the power dissipated in the voltage source v , we need to find the current i flowing into the positive terminal of the voltage source (in the asscosiated reference direction). The easiest way to do this is by superposition. Let’s call i 1 the current when the 1A current source is on and the v V and 2V voltage sources are off (short circuits). Then, we have a current divider, and i 1 = 1 1 1+1 = 1 2 A. Now let i 2 be the current when the v V voltage source is on and the others are off. In this case we have a voltage source in series with the equivalent of a 2Ω resistor, and i 2 =- v 2 A assuming the appropriate reference direction. Finally let i 3 be the current when the 2V voltage source is on and the others are off. Again, we have a 2V source in series with the equivalent of a 2Ω resistor, and i 3 = 2 2 = 1A. The current i is the sum of the three components, thus i = i 1 + i 2 + i 3 = 1 2- v 2 + 1 = 3- v 2 A . To find the power, we simply multiply i and v to get p = i × v = 3- v 2 v = 3 v- v 2 2 W . To find the voltage v that maximizes the power, we take the derivative of the power with respect to voltage, set it to zero, and solve for v . Thus dp dv = 3- 2 v 2 = 0 ⇒ v = 3 2 V . 2. (a) We perform source transformation to convert the voltage source and the 2Ω series resistor into a 1A current source in parallel with a 2Ω resistor. We then write the node voltage equations for n 1 , n 2 , and n 3 . The fourth equation is e 2 = e 3 . 7 / 4- 1- 1 / 4- 1 3 / 2 2- 1 1- 1 e 1 e 2 e 3 e 4 = 1 (b) The power dissipated by the op-amp is the current (let’s call it...
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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