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Unformatted text preview: EE 101 Handout # 28 Prof. Abbas El Gamal March. 15, 2003 Sample Final Solutions 1. (a) First we draw the circuit in the phasor representation. 1 V- j V- j- 2 j j R 1 A As we found in problem 2 of HW7, R = | Z th | maximizes the average power dissipated in R . So, we need to find Z th between the two terminals of R (after removing R ). We turn off all sources and find that Z th = 2 j Ω . Thus R = 2Ω. (b) To find the average power delivered to the 2Ω resistor, we find V th , i.e. , the open circuit voltage between the two terminals of the resistor after taking it out. Here we can use superposition. We turn off the two voltage sources and we get V 1 = 1 × 2 j. Next we turn on the 1V source only and by voltage divider get V 2 = 1 × j j- 2 j =- 1 . Finally, we turn on the- j V source to get V 3 = j. Adding up, we get that V th =- 1 + 3 j V . Thus the current thr’ the resistor is I R =- 1 + 3 j 2 + 2 j = 0 . 5 + j A , and the average power dissipated in the resistor is p R avg = 1 2 | I R | 2 R = 1 . 25W . 1 (c) By average power conservation the sum of the average power dissipated in the circuit elements is zero. Now since in our circuit the only elements that can supply/dissipate power are the sources and the resistor (reactive elements don’t dissipate or supply average power), the total average power supplied by the sources equals the average power dissipated in the resistor, which is 1 . 25W. 2. First we analyze the circuit for t < 0, assuming steady state. In this case the transistor is ON and the inductor is replaced by a short circuit. The equivalent circuit is shown below. To find i b we use the KVL equation 3- 10kΩ i b- 1- 100Ω × 100 i b = 0 ....
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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