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Thevenin's Theorem with Source Transforms
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[QUOTE="The Electrician, post: 4943606, member: 101272"] Convert ##\frac{5}{2} mA## in parallel with ##\frac{4k}{3}## back into a voltage source of 10/3 volts in series with a resistance of 4000/3 ohms. Now you have 2 voltage sources in series with two resistors. Slide the 4000/3 ohm resistor to the right so it's just above the 2k output resistor. Those two resistors form a voltage divider driven by the difference in the two voltage sources. I get Vth = -4V [/QUOTE]
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Thevenin's Theorem with Source Transforms
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