quiz1.win08

# quiz1.win08 - ECE 53a Quiz#1 SOLUTIONS Pamela Cosman...

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ECE 53a Quiz #1 SOLUTIONS Pamela Cosman 1/29/08 Problem 1: (This is problem 1 from homework 2.) V a is just the same as the source voltage. It is 8V. And for V o , we see it is just a voltage divider. We have to notice that V o has the opposite polarity from V a = 8 V , so we need to include a minus sign in the voltage division equation: V o = - 2 2 + 3 + 3 × 8 = - 2 V Problem 2: (This is similar to problem 3 in the practice quiz.) + _ + - 2V I x x 0.5 I V 1 1.5 First to ±nd the current, we note that the total voltage powering the circuit is the sum of the two voltage sources in series: 2 + 0 . 5 I x . The total resistance in the circuit is the sum of the two series resistors: 1 + 1 . 5 = 2 . 5. By Ohm’s law, the current that ²ows will be the ratio: I x = 2 + 0 . 5 I x 2 . 5 and we solve this equation to obtain: I x = 1 A The fastest way to answer this for the resistors is to remember that one expression for power absorbed by a resistor is p = Ri 2 Since I x = 1, and it is the same current throughout the circuit, the power absorbed by the 1 Ohm resistor is 1

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## This note was uploaded on 05/07/2008 for the course ECE 53 taught by Professor Cheng during the Spring '08 term at UCSD.

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quiz1.win08 - ECE 53a Quiz#1 SOLUTIONS Pamela Cosman...

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