EE 240 - HOMEWORK - Solution 4

# EE 240 - HOMEWORK - Solution 4 - through the 3k resistor...

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EE240-A2-2010 Assignment #4 SOLUTION 1. Maximum Power Transfer Problem: Find the resistor R L for maximum power transfer and the power absorbed in it. Use the Thevenin Equivalent Circuit approach. 1. First find the open cct voltage, V AB . Note the current through 2 Ω is 2A leftward so the voltage V X is -4V (defined polarity opposite to current). Now V C = 2Ax4 Ω = 8V ; V D = V C + 4V X = -8V ; V E = V D - V X = -4V. There is no current and therefore no voltage drop in the 4 Ω resistor so V OC = V T = -4V . 2. Next turn off the current source and find R T with a probe current of 1 A. Very similar to the first part. The current through 2 Ω is 1A leftward so the voltage V X is -2V. Now V C = 4 Ω x1A = 4V ; V D = V C + 4V X = -4V ; V E = V D - V X = -2V ; V A = V E + 4 Ω x 1A = +2V = R T x 1A ; therefore R T = 2 Ω. This is the resistor for maximum power transfer. The maximum power transferred is V T 2 /4R T = 2 W .

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2. Interesting Thevenin Circuit Problem: The problem is to find I
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Unformatted text preview: through the 3k resistor. The problem must be done by treating the 3k resistor as a load and finding the Thevenin Equivalent Circuit of everything else where A & B are the open circuit points of it. ● Use a source transformation on the 18V source and 6k resistor to reduce the problem to a mesh problem with only current sources. ● Use a Wye-Delta transformation on the three 6k resistors. (Draw new circuit diagrams at each step.) This is the result and also combined the parallel 4k resistors ● Calculate the open circuit voltage: With V T = 14 V, R T = 6 k Ω and a 3 k Ω load, the current will be I = 14/9 mA. A Using KCL, you can figure out the currents in the branches, then the voltage drops from A to B; V AB = 2x1 + 2x4 + 2x2 = 14 V 2mA 4mA 1mA 1mA 1mA With the sources removed, the resistance from A to B is clearly 6k Ω ....
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EE 240 - HOMEWORK - Solution 4 - through the 3k resistor...

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