How to calculate three-phase power

By: Patrick Myers Patrick Myers is the CEO and Founder of Calculator Academy. He has over eight years of experience developing online calculators and holds a B.S. in Mechanical Engineering. His work focuses on formula implementation, calculator accuracy, unit conversions, and creating practical tools that are easy to use.

Last Updated: September 23, 2026

See why √3 appears in a balanced three-phase calculation, work through 400 V and 100 A at 0.8 power factor, and separate kVA, kW and kVAR.

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The calculation

For a balanced sinusoidal load, apparent power S = √3 × line-to-line voltage × line current; real power P = S × power factor; reactive power Q = S × √(1 − PF²)

Worked example

At 400 V line-to-line, 100 A line current and PF 0.8, S = √3 × 400 × 100 = 69,282.032 VA ≈ 69.28 kVA; P = 0.8S ≈ 55.43 kW; for a sinusoidal load Q = √(S²−P²) = 0.6S ≈ 41.57 kVAR. P²+Q²=S² checks the triangle.

Assumes balanced phase conditions, RMS line values and sinusoidal displacement power factor. Unbalanced or harmonic-rich systems need suitable per-phase or measured true-power treatment; power-factor and wiring values must not be guessed for equipment sizing.

Video chapters

  • 0:00 — Three currents, one total power
  • 0:14 — Why the square root of three appears
  • 0:31 — First find apparent power
  • 0:47 — Power factor gives real power
  • 1:01 — See the reactive component
  • 1:15 — Check the three power numbers
  • 1:34 — Use the right voltage and assumptions
  • 1:49 — The full power picture
Read the full transcript

A three-phase system carries three alternating waveforms offset by one hundred twenty degrees. To calculate its total power from line measurements, we need the voltage between two lines, the current in a line, and, for useful real power, the power factor.

For a balanced system, the shifted phase voltages combine so line-to-line voltage is square root of three times phase voltage in a wye connection. Expressing total power with line voltage and line current therefore uses that same factor: approximately one point seven three two.

Suppose the line-to-line voltage is four hundred volts and line current is one hundred amps. Apparent power is square root of three times four hundred times one hundred: about sixty-nine thousand two hundred eighty-two volt-amperes, or sixty-nine point two eight kilovolt-amperes.

Power factor is the fraction of apparent power that appears as real power in this simple sinusoidal model. At zero point eight, multiply sixty-nine point two eight by zero point eight. The result is about fifty-five point four three kilowatts of real power.

Draw apparent power as the sloping side of a right triangle and real power as its horizontal side. With a zero point eight power factor, the vertical fraction is zero point six. That gives about forty-one point five seven kilovolt-amperes reactive.

The triangle provides a check: real power squared plus reactive power squared equals apparent power squared. Using the unrounded values, the sides correspond to fifty-five point four three, forty-one point five seven, and sixty-nine point two eight. Kilowatts, kVAR, and kVA describe different parts of the same load.

The four hundred volts here is measured between lines, not between a line and neutral. Mixing those voltages changes the answer. This compact formula also assumes balanced, sinusoidal conditions; unequal phases or strong harmonics call for appropriate per-phase measurements.

For a balanced load at four hundred volts line-to-line and one hundred amps, first calculate about sixty-nine point three kVA. At zero point eight power factor, that splits into about fifty-five point four kW real and forty-one point six kVAR reactive. Keep the voltage type and power factor with the result.

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