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.
V = I*R (Ohm's Law)
The Attempt at a Solution
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:
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?
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