EE215w03m2s - EE215 Winter 2003 Midterm 2 Solutions Monday,...

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EE215 Winter 2003 Midterm 2 Solutions Monday, February 24, 2003, 1:30 – 2:20pm Name: ______________________________ ID#: ________________ Section: ___ Please show all steps in solving the problem. Clearly mark your final answers and don’t forget the units. All circuit components can be considered ideal. Problem 1 (15 + 5 + 5 = 25 points) (a) Determine the Thévenin equivalent of the circuit below. (b) Assume there is a load of 24 at the output of the equivalent circuit. How much power is dissipated in the entire equivalent circuit? (c) Briefly explain why the value obtained in part (b) is different from the total power dissipated in the original circuit under the same load. Answers and grading guide: This problem is similar to Homework 3, Problem 4.53 (the circuit is flipped upside down and the direction of the sources are reversed). (a) To perform source transformations, ignore the 4 resistor in series with the current source, and ignore the 40 resistor in parallel with the voltage source (5 points) . Then transform the current source with the 6 parallel resistor into a 60V source with a series resistor (5 points) . Then combine the sources and resistors (5 points) . The Thévenin Equivalent consists of a 240V voltage source in series with a 24 resistor. (b)
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This note was uploaded on 03/29/2012 for the course EE 215 taught by Professor Unknown during the Spring '05 term at University of Washington.

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EE215w03m2s - EE215 Winter 2003 Midterm 2 Solutions Monday,...

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