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Three-Phase Power Calculator

Find the power of a three-phase circuit.

How it is calculated

The tool gives real (active) power for a balanced three-phase load: P = √3 × V × I × pf, with V the line-to-line voltage in volts, I the line current in amperes and pf the power factor as a decimal between 0 and 1. √3 is about 1.732. The result is shown in watts to two decimals and in kilowatts to four decimals (kW = W ÷ 1000). If the power factor is left empty it is taken as 1. A power factor outside 0 to 1, or a negative voltage or current, gives no result. The formula assumes all three phases carry the same current; it does not compute apparent power (VA) or reactive power (var).

Worked example

A motor on a 400 V three-phase supply draws 10 A at a power factor of 0.85. First √3 × 400 × 10 = 6,928.20 VA, the apparent power. Multiply by the power factor: 6,928.20 × 0.85 = 5,888.97 W. In kilowatts that is 5,888.97 ÷ 1000, shown as 5.8890 kW. With the power factor left empty (taken as 1) the same motor would show 6,928.20 W, so the power factor accounts for the difference of about 1.04 kW.

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Common mistakes

  • Entering the phase voltage (for example 230 V) instead of the line-to-line voltage (400 V). The formula expects the voltage between two lines; using the phase voltage underestimates the power by the factor √3.
  • Entering the power factor as a percentage such as 85. It must be 0.85; a value above 1 returns no result.
  • Using the formula for an unbalanced load. If the three line currents differ, one average current gives only an approximate answer.
  • Reading the result as input power drawn from the supply by a motor. The figure is electrical power at the stated current and power factor; the shaft output is lower by the motor's efficiency, which is not an input here.
  • Forgetting that leaving the power factor empty means 1.0, which overstates real power for inductive loads such as motors.

About this calculator

For a balanced three-phase system, real power is P = √3 × V_line × I × pf, where V_line is the line-to-line voltage, I is the line current, and pf is the power factor (cos θ, accounting for any phase lag between voltage and current in inductive loads like motors). The √3 factor — roughly 1.732 — comes from the 120° phase offset between the three lines, and the same relationship also gives apparent power (S = √3·V·I) and reactive power (Q = √3·V·I·sin θ).

Electricians and electrical engineers use this formula throughout industrial and commercial power work — sizing motors, transformers, and feeder cables, and calculating loads on three-phase panels — since most heavy industrial equipment runs on three-phase power rather than the single-phase supply used in homes.

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