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Thevenin Equivalent Circuit
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[QUOTE="nobodyuknow, post: 4523388, member: 233400"] [h2]Homework Statement [/h2] [ATTACH=full]164375[/ATTACH] Above given is the circuit I am supposed to simplify where the 1/5Ω resistor is the load. I'm unsure how to convert this circuit into an equivalent Thevenin Circuit. [h2]Homework Equations[/h2] V = I*R (Ohm's Law) [h2]The Attempt at a Solution[/h2] Using Microcap and it's Dynamic DC Analysis Tool, I measured the current flow at the load. I managed to break it down to this: [PLAIN]http://img844.imageshack.us/img844/282/5x8t.png[/PLAIN] [Broken] Where V5 was a 2/3V voltage source. (2A current source in series with the 1/3Ω resistor). If this is correct, I would then say that the 2/3V voltage source and the 4A current source were in series and that the V4 voltage source and the 1/6Ω resistor were in series. Is this the correct approach? [/QUOTE]
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Thevenin Equivalent Circuit
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