Okay, so I'm new to the whole electricity field and I'm trying to get a handle on something so I apologize for any ignorance displayed here.

I've purchased a new graphics card for my computer and I'm worried that when I'm maxing out it's power need, my PSU (Power Supply) will not be able to output enough power.

I understand that AC wattage going into the computer is going to be higher than DC wattage output as some wattage is lost in the conversion and that all depends on the efficiency of my PSU.

So I bought a P3 KILL A WATT 4400.01 to measure AC wattage going into the PSU. I have a PSU that, in theory, should output up to 500 watts. When I stress test the card and look at the AC wattage (Amps X Volts X PF) being pulled, it doesn't go over 450 watts. However, the AC Volt-Amperes being pulled into the system are over 600 so I'm a little confused/worried. I know, in theory, that the DC wattage being used/pulled should be lower in the system due the loss in conversion and that the system draws extra AC wattage to meet the DC wattage needs.

So, my question is, should I be looking at the raw AC Volt-Amperes going into the PSU or the actual AC wattage (Volts X Amps X PF) to base what the PSU is actually drawing?


In short, you should look at AC watts (V x I x P.F.) because that's the measure of "real" power being drawn from the mains. The so called Voltamps is merely the product of rms voltage and rms current.

It is called apparent power because if you measured only load voltage and current (as you might do for a DC load), it would "appear" that the load is drawing power equal to V*I.

For more insights you should read up the concepts of apparent, real and reactive power.

Addendum: The above assumes that current is sinusoidal. If your system is drawing non-sinusoidal power, then you will have to be careful about how your measurement system (KILL A WATT) handles that.

  • \$\begingroup\$ All true, though the power supply is probably power-factor corrected since most are these days. In that case the power factor should be close to unity and the power reading should be real power. If the supply is capable of putting OUT 500W, and is say 80% efficient the input power will be 625W. Sounds like you're close to the power limit of your supply, though you should check to see if your supply is in fact power-factor corrected. \$\endgroup\$ – John D Aug 15 '15 at 23:25
  • \$\begingroup\$ This may be a little over my head. The power supply doesn't appear to be PFC. It is a RAIDMAX RX-500S. If it is not power-factor corrected, does that mean I should be going by AC wattage or AC VA? \$\endgroup\$ – HeffZilla Aug 16 '15 at 6:29
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    \$\begingroup\$ I couldn't see your model on the RAIDMAX site, but they do list compliance with European Eup regulations. Probably also means they are power factor corrected though they don't state it explicitly. Any power supply >75W must be power factor corrected to be sold in Europe or even Japan. So unless it's a older US only model you can assume it's PFC and the AC wattage will be real. (Meaning your DC wattage is getting up toward the limit of your supply, but without knowing efficiency you can't tell exactly what the DC load is.) \$\endgroup\$ – John D Aug 16 '15 at 17:01
  • \$\begingroup\$ Thanks for your help everyone. So just so I know I got this straight. If this is a PFC PSU, and the KILL A WATT meter reads 450W of AC current (Amps x Volts x PF), the inner DC wattage is probably a bit lower than that? I think the PSU efficiency is somewhere in the 70-75% range. \$\endgroup\$ – HeffZilla Aug 16 '15 at 19:48

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