Three-Phase Power Calculator

Three-phase power formulas

For balanced sinusoidal operation, apparent power S = √3 × V × I in VA, where V is RMS line-to-line voltage and I is RMS line current. Active electrical input power P = S × PF in watts. Reactive power magnitude Q = √(S² − P²) in VAR. Divide each by 1,000 for kVA, kW or kVAR.

Choose the result you want to emphasize using the mode selector. Each voltage/current mode also shows the other power quantities, making it easier to check a power balance. Current from active input is I = P / (√3 × V × PF); current from apparent input is I = S / (√3 × V).

kW versus kVA and power factor

kW describes average real energy transfer per second. kVA describes the product of RMS voltage and current, including current that does not produce net real energy transfer. With the sinusoidal assumption, PF = P/S and Q is the reactive component of the power triangle.

A lower power factor means more current for the same kW at the same voltage. PF must lie from 0 to 1 here. PF 0 can describe a purely reactive load when voltage/current or apparent power is known, but cannot be used to infer finite current from positive active power. Harmonic distortion requires more detailed measurement; Q from this calculator is not a universal nonactive-power calculation.

Line voltage and the √3 factor

In a balanced three-phase system, phase voltages are separated by 120 electrical degrees. The vector difference between phase voltages produces the √3 line/phase relationship. Summing three equal phase powers gives the familiar √3 × line voltage × line current expression.

Use 400 V for a 400 V line-to-line system, not its roughly 230 V line-to-neutral value. The line-quantity formula applies to balanced star or delta loads; do not multiply its result by three again.

Applying motor efficiency once

When the entered power is mechanical motor output, electrical input P = output / efficiency. Efficiency is entered as a percentage and then divided by 100. The current calculation also divides by power factor; efficiency and PF describe different losses and effects.

For example, 5.5 kW output at 90% efficiency requires 6.111 kW active input. At 400 V and PF 0.8 this is about 11.026 A. If your given kW already represents electrical input, select that basis to avoid counting losses twice. Use manufacturer data and qualified engineering advice for equipment selection.

Examples

Frequently asked questions