VA to kW Calculator

VA and kW measure two different things

Every AC load draws two kinds of power at once. Apparent power, measured in volt-amps (VA), is the full product of voltage and current, and it decides how much current flows through the wire, the breaker and the transformer upstream. Real power, measured in watts (W) or kilowatts (kW), is the part that turns into motion, light or heat and shows up on your electric bill. The ratio between them is the power factor, a number between 0 and 1.

The US Department of Energy motor systems guide draws it as a horse pulling a rail car at an angle: the horse's total effort is the apparent power, the pull along the track is the real power, and the wasted sideways pull is reactive power. The cos of the angle is the power factor. In their worked triangle, 100 kW of real power drawn from a 142 kVA supply is a power factor of 100 / 142 = 0.70, meaning only 70 percent of the current produces useful work.

That triangle comes straight from the DOE Fundamentals Handbook on electrical science, which defines apparent power S in volt-amperes, true power P in watts, and power factor as P / S = cos of the phase angle. Nothing about the definition depends on the size of the load, so the same math converts a 1500 VA desk setup and a 6000 VA machine the same way.

That split is why a nameplate can print one number and hide the other. A device stamped 500 VA tells you how much current it will pull and how big a breaker it needs, but says nothing about how much work it does. The same 500 VA can be a 500 W heater or a 400 W motor, and the label looks identical either way. Once you know the load type, the conversion fills in the missing half.

Two wrinkles show up on real equipment. Inverter and rectifier front ends distort the current waveform, so the plain cosine rule only approximates their factor, and a lagging induction motor looks different from a leading capacitor bank on a power quality analyzer. Neither wrinkle changes the arithmetic on this page; both change which factor you should type in.

The formula: kW = VA x PF / 1000

One line does the whole job:

kW = VA x PF / 1000

Divide by 1000 because VA lands in watts and kilowatts are thousands of watts. The only judgment call is PF, the power factor, and it is an assumption about the load, not a property of the VA number. The table shows the same six VA ratings converted at three common factors:

The spread between the 0.8 and 1.0 columns is 25 percent at every row, because 1.0 / 0.8 = 1.25. That spread is the cost of not knowing your power factor. Pick a factor that matches the load, run the number, and you have kilowatts you can put against a breaker, a UPS watt rating or a bill.

Unit names shift with size but the math does not. A kilovolt-ampere is 1000 VA, so a 5 kVA transformer is 5000 VA, and 5000 x 0.9 / 1000 = 4.5 kW. Watts scale the same way into kilowatts. Whatever prefix the label uses, convert it to plain VA and watts first, then apply the power factor once, at the end. Doing it twice is the most common way to land a factor of a thousand off.

Worked example: a 1500 VA desk setup

Take a workstation that draws 1500 VA, with a power supply and monitor mix that puts the power factor at 0.9. The chain runs like this:

• Input: 1500 VA, PF 0.9
• Real power in watts: 1500 x 0.9 = 1350 W
• Real power in kilowatts: 1350 / 1000 = 1.35 kW
• Reactive power: 1500 x 0.436 = about 654 VAR

Run the same 1500 VA through a motor-heavy factor of 0.8 and the answer drops to 1.2 kW. Run it through a heater at unity and it climbs to 1.5 kW. The VA never moved; the assumption did all the work, which is exactly why every honest conversion states its power factor out loud.

On a real install the nameplate and the meter rarely agree. A supply stamped 1500 VA might pull 900 VA at idle and 1400 VA under load, and the power factor drifts with the load too, from roughly 0.7 at idle toward 0.95 at full tilt. A single conversion is a snapshot, so clamp the feed, read volts and amps, multiply, and use the factor the meter shows at the duty point you care about.

Worked example: a 6000 VA machine on a 240 V circuit

Now a shop machine stamped 6000 VA, running on a 240 V circuit, with an induction motor that pulls the power factor down to 0.8:

• Input: 6000 VA, PF 0.8, 240 V
• Real power: 6000 x 0.8 = 4800 W = 4.8 kW
• Current: 6000 / 240 = 25 A
• If the machine had PF 0.95 electronics instead: 6000 x 0.95 = 5700 W = 5.7 kW

Both versions draw the same 25 A and heat the same wire, but one delivers 4.8 kW of work and the other 5.7 kW. This is the whole reason supply equipment is rated in VA: the copper does not care about the split, only the total. Going the other way, a 4.8 kW load at PF 0.8 needs 4.8 x 1000 / 0.8 = 6000 VA of supply capacity.

What power factor should you assume?

The right factor depends on what is plugged in. These ranges cover common household and shop equipment for rough planning; a nameplate or a clamp meter beats any table:

If the label prints both a VA and a watt number, skip the guessing entirely and divide: a unit marked 1500 VA / 900 W has a built-in factor of 0.6 at its rating point. The NIST Special Publication 811 keeps the units themselves straight (the volt-ampere for apparent power, the watt for real power), but the factor between them is always a property of the load in front of you.

When you have a meter, skip the table. A plug-in power meter shows watts directly, and if you also know the VA from the label, dividing gives the factor: a device reading 90 W on a 100 VA rating runs at 0.9. A clamp meter plus a voltage reading gets you the same place the other way around, since VA = volts x amps. Both beat guessing, and both take about a minute.

From VA to amps, and where phase fits

Current needs no power factor at all, because VA already is volts times amps. Rearranged: amps = VA / volts for a single phase circuit.

Those amp numbers explain why a 1500 VA load is the practical ceiling for one 15 A, 120 V branch: 12.5 A is 83 percent of the rating, right at the continuous-load limit. On a three phase supply the same VA comes out of three wires, and the per-line current is VA / (volts x 1.732), so a 10000 VA load at 208 V three phase draws 10000 / (208 x 1.732) = 27.8 A per line. Phase changes the arithmetic around VA, never the VA to kW conversion itself.

Breakers, conduit fill and wire ampacity all follow amps, so the VA figure is the one to hand an electrician. A 20 A circuit at 120 V carries 2400 VA, and many electricians keep a continuous load at 80 percent of that, which is 1920 VA. A 2000 VA welder or a big inverter charger sits right at the edge, and the kilowatt rating printed on the same box tells you nothing about whether the breaker holds.

Mistakes that make the answer wrong

• Assuming unity power factor across the board. That overstates real power for motors and electronics. A 3000 VA motor load at PF 0.8 delivers 2.4 kW, not 3, a 20 percent error before anything else goes wrong.
• Taking a UPS watt figure as a conversion. The 1500 VA / 900 W pair on a label is two ratings, not two units of one quantity; the 900 W already includes the maker's factor of 0.6.
• Dropping the 1000. VA times PF yields watts. A 6000 VA machine at PF 0.8 is 4800 W, which is 4.8 kW. Reporting 4800 kW off by a thousand happens more often than you would think on spec sheets.
• Sliding between kVA and VA. 5 kVA is 5000 VA. The kVA pages on this site cover the bigger units; this page keeps VA so the figures stay in household range.
• Expecting the current to shrink. Raising power factor buys kilowatts per amp, but amps at a given VA stay put, since amps = VA / volts regardless of the factor.

A panel schedule is where all of this lands. A subpanel with a 100 A main at 240 V has 24,000 VA of capacity; fill it with motor loads at 0.8 and it serves 19.2 kW, while resistance heat would use the full 24 kW. Same copper, same breaker, different work done.