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Thevenin's and Norton's Theorem
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[QUOTE="psparky, post: 5040296, member: 360661"] Your answer is correct, I'm not sure I understand your method. You could also use superposition as well. The voltage source is straight forward at 5 volts (open current source) The current source is a bit more tricky( short voltage source). .1 amp thru 4K resistor is .4 volts...but you then put it thru the 2K resistor and its .2 volts in th opposite direction. How can you have two different voltages in two parallel branches? Must mean there is .2 volts across the current source in the opposite direction of the 4K resistor. 5 +.2 = 5.2 Volts. Or you could do the way above, but superposition is just another way to enjoy and undestand the problem. [/QUOTE]
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Thevenin's and Norton's Theorem
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