Three-Phase Power Calculator
From a line voltage, a line current, and a power factor, get the apparent, real, and reactive power that describe any balanced three-phase load.
Three inputs, three answers
Enter the line voltage, line current, and power factor and the calculator returns the apparent power (√3·V_L·I_L), the real power (× pf), and the reactive power (√(S²−P²)) at once.
Balanced loads only
The √3 formula assumes a balanced system — equal line-to-line voltage and equal line current across all three phases. Unbalanced loads need a per-phase calculation instead.
What is a three-phase power calculator?
Line quantities in, full power picture out
A three-phase power calculator turns three measurements — the line-to-line voltage, the line current, and the power factor — into the numbers that describe how a balanced three-phase load behaves: how much total capacity the system carries (apparent power, in VA), how much of that does useful work (real power, in W), and how much merely shuttles back and forth between source and load (reactive power, in VAR). Three-phase supply is the backbone of motors, pumps, HVAC, and industrial machinery, so these three figures are what electricians and engineers reach for when sizing cables, breakers, and transformers.
Enter the line voltage, line current, and power factor to get the apparent, real, and reactive power instantly.
Three short formulas, all built from the line voltage V_L, the line current I_L, the power factor pf, and the constant √3 (about 1.732).
P = √3 × V_L × I_L × pfThe apparent power is the raw product of the line quantities scaled by √3: S = √3 × V_L × I_L. Multiply by the power factor and you get the real power — the useful work, P = S × pf. The reactive power is what is left over once you remove the real part from the apparent power as vectors: Q = √(S² − P²). The √3 (the square root of three) appears because the line-to-line voltage in a three-phase system is √3 times the phase voltage.
Suppose you have a balanced load on a 400 V line drawing 10 A at a power factor of 0.8.
Apparent power
√3 × 400 × 10 = 6928.203230 VA — the total capacity the system carries.
Real power
6928.203230 × 0.8 = 5542.562584 W — the useful work delivered.
Reactive power
√(6928.203230² − 5542.562584²) = 4156.921938 VAR — the part that pendulums between source and load.
The three outputs answer three different practical questions. The apparent power (about 6928.20 VA for 400 V, 10 A) is the total load the supply, cables, and transformer must be rated to carry — it is what you size equipment against, which is why transformers and generators are rated in kVA, not kW. The real power (about 5542.56 W) is the part that actually does work: turning a motor, heating an element, lifting a load. The reactive power (about 4156.92 VAR) does no net work but still flows through the conductors, heating them and eating into capacity. The power factor ties the two together: a pf of 0.8 means only 80 % of the carried capacity is useful, so a low power factor wastes headroom and often draws a utility surcharge. Push the power factor toward 1 (with correction capacitors) and the real power climbs toward the apparent power while the reactive power shrinks toward zero.
The formulas are exact for the case they describe, but a couple of practical points are worth keeping in mind.
Balanced systems and line-to-line voltage
These formulas assume a balanced three-phase load — identical impedance on each phase — and that the voltage you enter is the line-to-line value, not the phase (line-to-neutral) voltage. If your phases are unbalanced, compute each one separately and add the real and reactive parts. Always confirm whether a nameplate quotes line-to-line or phase voltage before entering it, since they differ by a factor of √3. This tool covers steady-state AC at the fundamental frequency and does not account for harmonics or transient inrush.