HW2_ECE3710_Fall2010_soln

# HW2_ECE3710_Fall2010_soln - HW#2 Answers Circuit A The key...

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HW#2 Answers Circuit A: The key is to solve for the top node (treating the bottom node as ground). There are various ways to accomplish this, but using source transformation the right side becomes a voltage source in series with 4 . This creates a single loop in which no current can flow, hence the top node is the same voltage as each source namely 20 V. In order to solve for power, we must return to the original circuit. Why? Because source transformation and other equivalents are linear operations, and power is not linear (we have to square voltage or current). As a result, the 4 resistor dissipate 100 W, the 20 V source and 6 resistor dissipate 0 W, and the 5 A source dissipates -100 W (in other words, it supplies 100 W). Circuit B: The top node is easily solved. I combined the 10 and 40 resistors in parallel, and then source transformed the whole right side in order to create a single loop in the circuit. Using KVL, the current through the loop is 2.35 A, and thus the voltage is -13.2 V by Ohm’s Law.

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## This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Tech.

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HW2_ECE3710_Fall2010_soln - HW#2 Answers Circuit A The key...

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