# What is the practical difference between watts and VA (volt-amps)?

I have seen one inverter rated at 300W, and another at 300VA. What is the practical difference between them, if any?

• W and VA are dimensionally the same thing, but by convention, they refer to different scenarios. Nov 4, 2011 at 13:40
• @endolith: Can you explain how you thought that comment would help the OP? Oct 21, 2015 at 16:16
• @transistor What do you mean? Oct 21, 2015 at 17:20

For resistive loads they're equivalent. For reactive (inductive/capacitive) loads however, voltage and current are not in phase, and you have to take this phase difference (phi) into account.

Effective power = Voltage $\cdot$ Current $\cdot$ cos($\phi$), in Watt

and

Apparent power = Voltage $\cdot$ Current, in VA

Apparent power is higher for non-resistive loads, but that's because it contains a part of blind power; power the utility has to provide but also gets back during part of the cycle (in the graph the part where the power is negative, purple line under the X-axis).

Because they have to provide it, but can't invoice it (after all they get it back) the utilities are not too happy with reactive loads, and industries have to pay fines if their cos($\phi$) becomes too low.

edit (re jpc's comment):
The 300W PSU will consume 300 W of effective power, for the other PSU the apparent power is given, effective power for this PSU will be lower, e.g. if the cos($\phi$) = 0.9, effective power will be 270W.

• Apparent power has unit of VA, reactive power has units Var and also depends on phi, the power factor. Interesting way of looking at power factor though :) Apr 10, 2011 at 15:32
• @freespace - You're right, it's apparent power. It's been a long time... I'll correct it in my answer Apr 10, 2011 at 15:34
• @stevenvh no worries. I had to check first myself before commenting :) Apr 10, 2011 at 15:43
• A good answer but it does not really explain the difference in inverter ratings.
– jpc
Apr 10, 2011 at 16:18
• @Kamil The reason is simple. The PSU has its own power factor like any other device. It's same situation as in the examole I mentioned, but a bit more complicated. For example 300 W PSU has power factor of 0.9 and efficiency of 0.8. So if you take 300 W from the supply, it will consume 300/0.8=375 W. Now lets take power factor into account. Active power is in this case 90% of total consumption, so for 375 W we get 375/0.9=417 VA. So for a good 300 W supply you need 420 VA UPS. The power factor and efficiency curves should be printed on PSU or its box, but cheaper units may not show that. Apr 11, 2011 at 10:32