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I will be using a computer power supply to power a car audio amp for a project. The amp is 500w and my PSU is 750w. I never considered it until now, but how does the PSU deliver this? P=VI so 750w=12VI(internal lines are 12V,5V,3.3V and the 12V will be used to run the amp). So I=62.5A. This obviously can't be drawn from a regular outlet.

So I'm guessing that I'm doing this calculation on the wrong side of the PSU. In this case 750=110*I so, I=6.81A. A much more reasonable draw.

Is this correct?

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    \$\begingroup\$ Your second assumption is correct. P=IV, and on the output side, you have 12V (among others) at higher currents, but the input is mains. You could apply a ballpark PSU efficiency of 75-80% depending on its 80Plus rating. \$\endgroup\$
    – Krunal Desai
    Commented Jun 2, 2016 at 21:39
  • \$\begingroup\$ Check out this article: en.wikipedia.org/wiki/Buck_converter \$\endgroup\$
    – David Drysdale
    Commented Jun 2, 2016 at 22:43

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Yes it is correct.

The power supply converts power with small losses. So IV formula on both sides with respectable voltages.

Because of the mentioned losses it would draw more than 700W from the outlet (~750-800W I would expect for an decent power supply).

So for 800W on the input it translates to ~7A (in US).

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750 Watts at 120 V mains voltage is only 5.84 Amps. A fourth of the 20 amp circuit. Round up half an amp for efficiency losses.

Energy cannot be destroyed or created, just changed.
Power out = (Power In * Efficiency Factor)
The specifics of the voltage and amperage isn't as important in that simple formula.

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  • \$\begingroup\$ I edited to make the "formula" more correct. \$\endgroup\$
    – Michael Karas
    Commented Jun 3, 2016 at 1:02
  • \$\begingroup\$ technically it's "energy cannot be created or destroyed" but anyway \$\endgroup\$
    – Sam
    Commented Jun 3, 2016 at 5:01

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