Recent content by EE is cool

Well, it's the weekend so it'll be a few days before I can bring it up with him. I definitely will though.

No I actually don't, it's just the answer I got from doing KVL once I got all the currents and from when I tried the source transformations. I assumed it was correct since I received the same answer from both methods, but maybe it's not. You probably know better than I do.

Oh nice, that gives me -56V too. I love it when the light bulb finally comes on. Thanks a bunch! :)

Well there's 4A going into the top-leftmost junction, and there's 4A going into that parallel circuit, which makes sense. If there's 4A going against that current due to the current source, there must be 8A going to the left across the 3 ohm resistor?

Huh, that's a much easier way to think about it. So does that mean there's no current through the 3 ohm resistor since an ammeter has negligible resistance? But then, the diagram would only make sense if the current source went in the other direction wouldn't it?

Homework Statement Find v. A value for R is not needed. Circuit is attached. Homework Equations KVL, KCLThe Attempt at a Solution I have no idea where to begin with this one. I found an answer using source transformations by transforming the 4A current source into a 12V voltage source in...