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Three-Phase Power Calculator

Calculators · Added

In an AC circuit, power is three quantities rather than one: the kilowatts that do work, the kilovolt-amps the cable has to carry, and the reactive kilovolt-amps that do neither. This converts between them and the line current, for both single- and three-phase supplies.

Supply
Solve for

The figure a supply is named by: 400, 415, 480

A motor nameplate figure is shaft power — set the efficiency below

Between 0 and 1 — the cosine of the phase angle

Leave at 100 if the kW figure is already the electrical input

For the capacitor sizing. Set to 0 to skip it.

How to use the three-phase power calculator

  1. 1Choose the supply — three phase takes the line-to-line voltage, the figure a supply is named by.
  2. 2Pick what you are solving for: current from a load, power from a measured current, or current from a kVA rating.
  3. 3Enter the power factor. A resistive load is 1; an induction motor is usually 0.8 to 0.88.
  4. 4Set the motor efficiency if the kW figure came off a nameplate, since that is shaft power rather than input.

Examples

A motor on a 400 V supply

Input
15 kW, 400 V three phase, power factor 0.88
Result
24.6 A line current · 17.05 kVA · 8.10 kVAr

Correcting the power factor

Input
The same load, corrected to 0.95
Result
3.17 kVAr of capacitance leaves 4.93 kVAr reactive · current falls to 22.8 A

About the three-phase power calculator

The power triangle

Real power, reactive power and apparent power form a right-angled triangle: S² = P² + Q². Real power is the horizontal side and does the work. Reactive power is the vertical side and does none — it flows into the magnetic field of a motor or transformer on one part of the cycle and back out on the next. Apparent power is the hypotenuse and is what the supply actually delivers.

The power factor is the cosine of the angle between P and S. At unity the triangle collapses to a line and every amp is doing work; at 0.5 the current is twice what the load needs.

What correction capacitors do

A capacitor's reactive power is opposite in sign to an inductor's, so putting capacitance across an inductive load lets the two exchange reactive current locally instead of drawing it all the way from the supply. The real power is unchanged; the current in the supply cable falls.

The sizing is straightforward — the difference between the tangents of the two phase angles, times the real power — but the installation is not always. Capacitors on a supply with significant harmonic content can resonate with the network inductance, which is why anything beyond a modest installation is designed with detuning reactors rather than by formula alone.

Frequently asked questions

Where does the √3 come from?
From the geometry of three phases 120° apart. The line-to-line voltage is the vector difference between two of them, which is √3 times the phase voltage, and summing the instantaneous power across all three leaves the same factor. It is not a fudge — and using the single-phase formula on a three-phase supply understates the current by 42%, which is the most common error in this arithmetic.
Which voltage do I enter for three phase?
The line-to-line voltage, which is what the supply is named by: 400 V, 415 V, 480 V. The line-to-neutral voltage is that divided by √3 — 230 V on a 400 V supply — and entering it instead gives a current √3 too high. The tool flags a three-phase voltage that looks like a line-to-neutral figure.
Why does a low power factor cost money?
Because the cable, the breaker and the transformer are sized for the apparent power, not the real power. At a power factor of 0.7 the current is about 43% higher than the kilowatts alone would need, so everything in the path has to be bigger and the resistive losses are proportionally worse. Many commercial tariffs bill for it directly.
Is a motor's nameplate kW the power it draws?
No — it is the mechanical power at the shaft. The electrical input is that divided by the efficiency, so a 15 kW motor at 92% efficiency draws about 16.3 kW. Sizing a supply straight off the nameplate undersizes it by the losses, which on an industrial motor is 8 to 12%.

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