Three-Phase Power Calculator

How three-phase power works

When to use this: sizing a supply for an industrial machine, converting a kVA transformer rating into amps, checking whether a motor is loading a circuit as expected, or working out what poor power factor is costing you in cable capacity.

Three-phase distribution uses three conductors carrying alternating voltages 120° apart in time. The advantage is constant total power delivery — where single-phase power pulses at twice the supply frequency, the three phases sum to a steady value, which is why motors run smoothly on it and why almost all industrial distribution is three-phase.

The three quantities you need:

S = √3 × V_line × I_line  — apparent power, volt-amps
P = √3 × V_line × I_line × cos φ  — real power, watts
Q = √3 × V_line × I_line × sin φ  — reactive power, volt-amps reactive

Where the √3 comes from

In a star connection, each phase winding sits between a line and the neutral. Because the three phase voltages are 120° apart, the voltage measured between any two lines is not twice the phase voltage but √3 times it — the vector sum of two quantities at 120°.

So on a 400 V system, each phase-to-neutral voltage is 400 / √3 = 231 V. That is why European domestic supplies are 230 V: they are one phase of a 400 V three-phase system. The line current, meanwhile, equals the phase current in star. Multiply line voltage by line current and the √3 survives into the power equation.

kW, kVA and kVAr — the power triangle

These three are not alternatives; they are sides of a right triangle:

S² = P² + Q² and cos φ = P / S

The practical consequence: a transformer rated 100 kVA supplying a load at 0.8 power factor delivers only 80 kW of useful power. The other 20% of its capacity is consumed carrying reactive current that does nothing.

Why power factor costs money

Reactive current still flows through every conductor, breaker and transformer in the path. It heats them, it consumes their capacity, and it produces no useful output.

Resistive losses scale with the square of current, so a drop from 0.9 to 0.65 power factor nearly doubles the heat lost in the cable while delivering identical useful power. Large consumers are usually charged for poor power factor directly, and correcting it with capacitor banks pays back quickly.

Star and delta

These describe how the three load elements are connected to each other, and they change what each element experiences — not the total power.

Star gives you a neutral, so single-phase loads can be supplied at 230 V from the same distribution. Delta has no neutral but tolerates unbalanced loads better, and is standard for motor windings and for the primary side of many distribution transformers.

Why star-delta starting works

Connect a motor designed for delta operation in star instead, and each winding sees 231 V rather than 400 V. Power scales with the square of voltage, so it draws a third of the current and produces a third of the torque. Once it is up to speed, a contactor switches to delta for full performance.

The catch is that third of the torque. Star-delta suits loads that start unloaded — fans, centrifugal pumps — but a loaded conveyor or a compressor may simply fail to accelerate.

Worked examples

Sizing a supply for a 7.5 kW motor

I = P / (√3 × V × cos φ) = 7500 / (1.732 × 400 × 0.85) = 12.7 A
S = √3 × 400 × 12.7 = 8.82 kVA

The cable and breaker have to handle 12.7 A running, and roughly 76 A during a direct-on-line start. The supply must have 8.82 kVA of capacity available even though only 7.5 kW does useful work.

Converting a transformer rating

A 100 kVA transformer on a 400 V secondary:

I = 100,000 / (1.732 × 400) = 144 A

That is the full-load secondary current regardless of power factor — the transformer is limited by current, which is why it is rated in kVA rather than kW. At 0.8 power factor the same transformer delivers only 80 kW of real power.

Standard voltages

The balanced-load assumption

Every formula here assumes the three phases carry equal current. That is true for a motor or a three-phase heater, and approximately true for a well-distributed set of single-phase loads.

It is not true if single-phase loads are concentrated on one phase. Then current flows in the neutral, the √3 relationship breaks down, and each phase must be calculated separately. Significant imbalance also causes negative-sequence currents that heat motors badly — a 2% voltage imbalance can raise motor temperature by 8%. Balance the loads rather than calculating around the imbalance.