homework_2_solutions

homework_2_solutions - EE40 1. Homework #2 Solutions P1.76...

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EE40 Homework #2 Solutions 1. P1.76 a. The current flowing through the 15ohm resistor is ix = -30V/15ohm = -2A Applying KCL for the node at the top end of the dependent current source gives ix + is – ix/2 = 0 is = - ix/2 = 1A b. 15ohm resistor is dissipating power, P = 30 * 30 / 15 = 60W 10ohm resistor is dissipating power, P = ix * ix * 10ohm = 40W 5ohm resistor is dissipating power, P = is * is * 5ohm = 5W The voltage across the dependent current source is Vc = 30V + ( 10ohm * 2A ) = 50V . Note the voltage at the top end of the dependent source is HIGHER t han the voltage at the bo ttom end, while the actual current direction is from bottom to top. So this dependent source is supplying power, the amount is P = 50V * 1A = 50W The voltage across the Is current s ource is Vs = 50V + 1A * 5ohm = 55V. Again, top end voltage is HIGHER than bottom end voltage. However , the actual current direction is from bot tom to top. So this independent current source is supplying power, the amount is P = 55V * 1A = 55W
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This document was uploaded on 10/07/2008.

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homework_2_solutions - EE40 1. Homework #2 Solutions P1.76...

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