Three-Phase Power Calculator
kW, kVA, kVAR and line current, with the √3 in the right place.
How it works
The √3 is the only difficult thing here, and it is not arbitrary. In a balanced three-phase system the line-to-line voltage is √3 times the line-to- neutral voltage, because the two phases you are measuring across are 120° apart rather than opposed. 400 V between lines is 230 V to neutral, and that is the same system described two ways.
Real power is P = √3 × V(line) × I × cos φ. The most common error in the whole of electrical work is putting the √3 in with the phase voltage as well, which double-counts it and gives an answer 73% too high.
Three quantities, three names. Apparent power (kVA) is what the cables and the transformer have to carry. Real power (kW) is what does work and what the meter charges for. Reactive power (kVAR) is the difference — energy sloshing in and out of magnetic fields, doing nothing useful but occupying capacity all the way back to the substation. Power factor is the ratio of the first two.
A motor at 0.85 power factor drawing 11 kW is pulling 12.9 kVA, so the cable, the breaker and the transformer all have to be sized for 12.9 while the electricity bill is based on 11 — and industrial tariffs often charge for the difference directly. That is why power factor correction pays for itself.
Efficiency is a separate loss again. Shaft power out of a motor is electrical power in times efficiency — a 90% motor drawing 11 kW delivers about 9.9 kW of mechanical work and turns the remaining 1.1 kW into heat.
Common questions
Where exactly does √3 go?
With the line-to-line voltage: P = √3 × V(L-L) × I × PF. If you are working from line-to-neutral voltage instead, it is P = 3 × V(L-N) × I × PF. Both give the same answer; using √3 with the phase voltage gives one that is 73% too high.
Is power factor the same as efficiency?
No, and conflating them is common. Power factor is the phase relationship between voltage and current — it costs you capacity, not energy. Efficiency is how much of the real power becomes useful work — that one costs you energy. A motor has both, independently.
What power factor should I expect?
An induction motor at full load is typically 0.85–0.90 and gets much worse lightly loaded — an idling motor can sit at 0.2. Resistive heating is 1.0. Modern electronic supplies with active correction reach 0.95 or better.
Why is my measured current higher than calculated?
Usually the power factor is worse than assumed, or the load is unbalanced across the phases. On an unbalanced system the √3 relationship no longer holds and each phase has to be treated separately.