Watts vs Volt-Amps - huh ??

Most electrical references say that Volt-Amps are just the product of the amplitude of the voltage times the amplitude of the current, without taking into account the power factor. That is important in an inverter since it must be rated to deliver a certain number of amps even if no power is being transferred as a result.
AC watts are the actual power being moved, multiplying the VA number by the power factor (cosine of the phase angle between the Volt and Amp waveforms.)

For DC, both are the same, since there is no phase relationship to consider.

Both numbers are important in that a good inverter will draw down the battery based mostly on the watts it is delivering. But some cheap circuits will do worse than that. (Their efficiency will vary strongly with power factor, low power factor corresponding to lower efficiency.)

From your linked pdf:

"The power drawn by computing equipment is expressed in Watts or Volt-Amps (VA). The power in Watts is
the real power drawn by the equipment. Volt-Amps are called the "apparent power" and are the product of
the voltage applied to the equipment times the current drawn by the equipment."

Well if you ever took trignometry it is very easy Right Triangle equations.

Resistance is the X base or horizontal line, Reactance is the Y vertical line, and the distance Z between them is the impedance

Rule number 1: The triangle is always right.

Well if you are a Carpenter, I can teach you AC theory in 5 minutes or less.

Any good carpenter knows how to make a perfect Right Triangle with a simple cosine equation. Bet you are not even aware you knew trig. Ready 3 + 4 = 5 Every carpenter knows this is a true statement and is cosine law you can prove to yourself right now with a piece of paper, a ruler, a compass, and a pencil


Draw a base line exactly 4 inches in length. With a compass set it to a 3 inch span or radius and place it on the left end of the 4 inch line and make an arc line straight above it at approx 90 degrees. Now set the compass to 5 inches, place it at the opposite end of the 4 inch line and make an small arc line that intersects with the other arc line you just drew. Now with a straight edge draw a line from the arc intersection to each end of the 4 inch line and see what you have. A perfectly 90 degree right triangle every time.

So here is the math or cosine law.


Try it where:

a = 4
b = 3

16 + 9 = 25

Now take the square root of 25 and you 5.

So there you go 3 + 4 = 5

So now just replace a with Resistance, and b with Reactance, and the result c is the Impedance. Oh the power factor is simple PF = a/c

Respectfully submitted, since I am not an AHJ and therefore need to be polite:
The power factor is a/c not b/a. Otherwise it could end up greater than 1, among other problems. (Look at the simple 0 degree and 90 degree cases if you have any doubts. But I suspect it was just a typo on your part.)

You are correct my bad, I should have proof-read, I realized it a few minutes after the fact. Bit embarrassing because I taught Trig at a community college to electrical students. Should have the variables R, Rx, and Z and I would have not made that mistake.

simply put, watts takes the power factor into consideration:

Voltage X Current = volt-ampere
Voltage X current X Power factor = watts

to get the power factor:
PF = cos (angle) = adjacent/hypotenuse in the impedance triangle

I chuckled when I first read this post. This is pretty much a DIY forum and the posters looking for help have ZERO electrical knowledge and all this is is GREEK to them and will never understand one bit of it.

You can lead a horse to water, but you cannot make it drink! But the info is still valid, and if it gets only a couple people
asking questions about power factor (why does my 1/2 hp motor draw 1,000 watts) and they figure it out, our work here is done.

There's a good, concise explanation of power factor in the August/September 2012 edition of Home Power magazine. It discusses resistive vs. reactive loads, shows an example of in phase and out-of-phase reactive load plots, mentions what sort of loads will have high power factors, etc.

The trig explanation isn't too helpful for those who don't see the plots...

You could measure the actual power use of each load. If you're calculating a hypothetical load, my electrical engineer friend and my quick read on this topic both tell me that accepted practice is to divide VA by 0.8. If VA=1000 then you would correct to 1249 Watts. This estimate is apparently well accepted but, of course, it could be low for a purely reactive load.

I have always thought that for a given load the VA number was greater than to equal to watts, not the other way around?

That would be correct
Watts = VA * PF
VA = Watts / PF.
PF = Watts / VA

PF is a number of 1 or less where 1 is a purely resistive load.

My bad, inetdog and Sunking, sorry. I knew it didn't feel right dividing by 0.8. The "standard" correction should be VA * 0.8.

randl