Well thanks to my health ailment of having to urinate frequently I ended up missing my last electrical engineering class. And college being sucky as it always has been, missing one class means you miss a lot of material. A homework problem (and a quiz problem possibly tomorrow) demands that I use the "source exchange approach." What the hell is this? I'm trying to learn this on my own by using the book's example, but I'm unable to comprehend.

How is the 6 V calculated? What is Z'?

Ask your professor?

Z is usually impedance, I don't know why you drew a box around it.

The 6v is probably the voltage at that point in the circuit.

I don't really know what the hell you're solving for though because you really didn't specify.

PS, you draw **** real crooked.

6v might be in a circle because it's an AC power source. But it lacks the frequency, you might be solving for frequency

I'm not listing the whole problem because I'm only trying to figure out how source exchange works. In particular, I want to know how the 6 volts is calculated and Z (yes the original problem has a box around it). By the way, I drew that image quickly with MS paint that's why it's not that great. Anyhow this assignment is due tomorrow, guess I'll just take the lackluster grade.

Hello? If you're behind for whatever reason, ask your professor about it. That's what they are they for.

Well then I guess you're ****ed, because without the ****ing problem I can't tell what the **** you're asking for. And becuase you're too lazy to draw a decent ****ing picture I can't tell what the hell is going on.

So I'm inclined to agree with Emon.

Ask your professor.

I remember this. Only we always called it Source Transformation or Norton/Thévenin Transformation.

A current source in parallel with a load is equivalent to a certain voltage source in series with a certain load.

To go from Current source to Voltage, your using Thévenin's theorem. I could type it up here, but why when there is google and wikipedia. Going the other way is Norton's theorem.

That said, I still can't make sense of your problem. If your doing a Thevenin transform, we need a pair of terminals to transform around, you cant be just transforming around the Current source and the 1 ohm resistor (as the second drawing seems to indicate), because then the inductor + resistor pair would end up connected to the positive terminal of the voltage source in parallel with the capacitor + resistor pair, and the negative terminal would be grounded.

Well I found out he didn't he teach Source Exchange on the day I was absent. Guess I'll learn about it soon. I do understand Thevenin and Norton's theorems for the most part. However, Source Exchange is something that's different from them I think.

Source Exchange is actually a specific simple case of Norton/Thévenin where you can cheat and use shortcuts.

I remember ABOUT these things... but it would probably take me a bit to be able to do them again... stupid colledge only being good for a single piece of paper...