EE215w03m1s

# EE215w03m1s - EE215 Winter 2003 Midterm 1 Solutions Monday,...

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EE215 Winter 2003 Midterm 1 Solutions Monday, January 27, 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. Problem 1 (16+6+12=34 points) (a) Determine the equivalent resistance in this circuit. (b) Find the power delivered by the source. (c) Find the power dissipated by the 9 resistor. Answers and grading guide: This is very similar to Homework 1 Problem 3.7a. (a) R eq = 3 || (((5+3) || 24) + 9) = 3 || (8*24/(8+24) + 9) = 3 || (6 + 9) = 3*15/(3+15) = 45/18 = 2.5 . There are 4 transformations needed, give 4 points for each correct step. (b) p = v 2 /R eq = 100/2.5 = 40W (delivered) . 4 points for the formula, 2 points for the value. (c) The equivalent resistance of the three resistors 5 , 3 , 24 was 6 , thus we have a voltage divider with ratio 6:9 = 2:3, and the voltage across 9 is 3/5*10V = 6V. p = v 2 /R = 36/9 =

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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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EE215w03m1s - EE215 Winter 2003 Midterm 1 Solutions Monday,...

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