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## Homework Statement

I was able to find the Thevenin equivalent between ##A## and ##B## using loop analysis and then Ohm's law:

The Thevenin equivalent was ##V_{th} = -4.0 V## in series with ##R_{th} = 0.8 k##.

Now I had a question about solving the problem using a different approach. I realized now source transforms are a sort of "corollary" to Thevenin and Norton's theorems. So I was wondering if it was possible to use those in solving this problem.

## Homework Equations

## The Attempt at a Solution

From what I can understand, first I should use ##V = IR \Rightarrow I = \frac{5}{2} mA##. Then I could have a current source in parallel with the ##2k## on the left.

Now ##R_{eq} = (\frac{1}{2k} + \frac{1}{4k})^{-1} = \frac{4k}{3}##.

So I have that ##\frac{5}{2} mA## in parallel with ##\frac{4k}{3}##.

From this point I have not been able to continue using source transforms to simplify to the correct answer. Is it just not possible? Or is it a combination I'm not seeing?

I see that superposition would be useful from this point as well, but I want to know if it's possible with source transforms.

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