hw2_sol - EE 101 Professor Abbas El Gamal Handout # 6 Jan....

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EE 101 Handout # 6 Professor Abbas El Gamal Jan. 29, 2002 Homework #2 Solutions 1. Consider the given circuit 1A 1V 2V i 2 i We discussed source transformation for an independent source and a resistor. Can it be applied to a dependent current source in parallel with a resistor? To check that let’s write the v in i in relation for the parallel combination in our circuit i in = v in - 2 i. Now the series equivalent (if it works out) should be a current controlled voltage source of 2 i V in series with a 1Ω resistor. The v in i in relation is the same so we can use source transformation. Note that the resulting element is basically a negative resistor of value - 2Ω. We next perform source transformation to the current source and resistor at the left part of the circuit to get: 1V 2V 2V i 2 i We now combine some of the voltage sources and resistors to get: 1
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1V 2 i 2V i Next, we do a source transformation to the leftmost part of the circuit to get a current source in parallel with a 2Ω resistot. We then replace the two 2Ω resistors with an equivalent 1Ω resistor. Finally, we do another source transformation and we end up with a one-loop circuit: 0 . 5V 2 i 2V i To find i we write a single KVL equation 2 + 2 i +0 . 5=4 i , and find that i =1 . 25A. 2. We have two cases to analyze, one when the diode is ON and the other OFF. Diode is ON: In this case, v d = 1. Also, note that i flows through the resistor on the bottom, and because the resistance is simply 1Ω, the voltage across that resistor is simply i volts. KVL then says that v = i +1.
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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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hw2_sol - EE 101 Professor Abbas El Gamal Handout # 6 Jan....

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