hw5_ext_sol - EE 101 Professor Abbas El Gamal Handout #...

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EE 101 Handout # 15-a Professor Abbas El Gamal Feb. 14, 2003 Homework #5 Extra Problem Solutions 1. Here we use superposition. First turn o± the 1 and 2V sources. The circuit reduces to current source in series with two 1Ω parallel resistors and v 1 = - 0 . 5V. Next we turn the 1V source on and the other sources o±. We get v 2 = - 1V. Next we turn on the 2V source only and get v 3 = 1V. Thus v = v 1 + v 2 + v 3 = - 0 . 5V. 2. This problem is best attacked via superposition, as implied by the presence of multiple sources. Keeping the leftmost current source and setting the other two to zero results in the following: 1A i 1 By application of the current divider rule, i 1 = 0 . 5A. Keeping the voltage source and setting the current sources to zero, we obtain the following: 1V i 2 Ohm’s law tells us that 1V = - i 2 (2Ω), implying that i 2 = - 0 . 5A. Finally, keeping the rightmost current source, we get the circuit: 1A i 3 1
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So, i 3 = 1A. Adding the three current components, we get
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hw5_ext_sol - EE 101 Professor Abbas El Gamal Handout #...

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