It's a bit more complex than simply multiplying the rated current (amps) together for each voltage - this will just give you the maximum output for that specific voltage which will be less than the total rated power of the PSU. Adding all the power limits together will give you a figure higher than the total rating.
For example, looking at the label on a PSU I happen to have to hend (a cheap Sweex one I pulled from a PC recently), I can see the following specifications:
+3.3V - 22A
+5V - 30A
+12V - 16A
-5V - 0.5A
-12V - 0.8A
+5V standby - 2A
Multiplying, this gives me 72W on +3.3V, 150W on +5V and 192W on +12V, which are the significant ones.
This might suggest that the PSU can provide 414W.
However, it also states that the maximum combined power output from +3.3 and +5V is 190W (not the 222W you'd expect from adding up the two power ratings) and the maximum total power output is 300W so you can't pull the full 190W on 3.3/5V and the full 192W from the 12V line.
So, in summary, there's not really a quick and easy way to derive the power output of a PSU. As Aatje92 says, a good, reliable brand of PSU can normally be expected to deliver pretty much what it says on the label whereas cheap ones are often optimistic at best. (There have been a few lab tests on Toms over the years that have demonstrated this.)
EDIT:
Answering your question about the 80% certification, that is a measure of how much of the input power that the PSU draws from the mains gets through to the outputs. 80% of what goes in gets to the PC, the rest is lost as heat. So, if the PC is drawing 300W, you would expect the PSU to be pulling about 300 / 0.8 = 375W. A 500W 80% PSU should be able to provide the full 500W to the PC but it will draw rather more than 500W from the mains while doing so.
Testing a PSU to determine its capabilities empirically (if you were thinking of trying that) is almost a science unto itself. To do it properly, ensuring that it would actually deliver usable clean power to a PC at a given load level, you're going to need more than just a multimeter and some big loads.
Hope this helps...
Stephen
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