Okay, help me out here folks...
So I get that transformer output is basically just a ratio of primary coil turns by secondary. Pretty simple. But (and here's the kicker) does the amperage do the same thing in reverse? I.e. its inversely proportional to the voltage piece? Or even more simply, no matter what voltage/amperage you get at the output, power remains constant, not counting the power lost by inefficiency of all things? This is probably a scrub question, but thanks in advance.

 

Yes, neglecting the transformer inefficiency, the power out equals the power in.
Otherwise it won't be efficient.

 

Yep sounds like you got it, power in = power out is a pertty good approximation.

You do loose a little power from inefficiency.

 

Amperage in / out, depends on the diameter of the magnet wire, and the cross-section of the core... I have a number of these 40 pounders, that have the large [ #8 ] winding... as a secondary, and several taps off one long secondary, and of course, a normal primary...
Output amperage on these, is equal to input...

 

Transformers are rated in (K)Va.
Max.