Three Phase Power Made Simple: kW, kVA, and the √3 Factor

The first time I sized a feeder for a workshop, I multiplied volts by amps the way I always did for a wall socket and came up with a number that was 42 percent too low. The breaker I picked tripped on the second start. The thing nobody had told me, and the thing that trips up most people the first time, is that three-phase power carries a √3 in front of it. Get that factor right and the rest of the arithmetic is honest.

This guide walks through where that √3 comes from, how active, apparent, and reactive power split apart, and how to go from a motor nameplate to the amps a breaker actually sees. Every number here you can check yourself in the three-phase power calculator.

The one formula you need

For a balanced three-phase load, active power is:

P = √3 · VL · IL · cosφ

where VL is the line voltage, IL is the line current, and cosφ is the power factor. The √3 is about 1.732, and it is not optional. Skip it and you undercount power by a factor of 1.732, which is the 42 percent gap that bit me.

That single equation rearranges into everything else you care about. Apparent power, the total the supply has to deliver, is S = √3 · VL · IL with no power factor term. Reactive power, the part that sloshes back and forth in motor windings without doing work, is Q = √(S² − P²). The three quantities form a right triangle, and cosφ is the cosine of its angle. When the power factor is 1, the triangle collapses to a line and P equals S. When it drops to 0.8, the triangle opens up and you draw more current for the same useful kilowatts.

A worked example: 380 V, 10 A, 0.8 power factor

Take a balanced load on a 380 V line drawing 10 A per line at a power factor of 0.8.

Start with apparent power, because it ignores the power factor:

• S = √3 · 380 · 10 = 1.732 · 3800 ≈ 6580 VA ≈ 6.58 kVA

Now active power, the part that turns into real work:

• P = √3 · 380 · 10 · 0.8 = 6580 · 0.8 ≈ 5264 W ≈ 5.26 kW

And reactive power closes the triangle:

• Q = √(6580² − 5264²) ≈ √(43.3M − 27.7M) ≈ 3948 VAR ≈ 3.95 kVAR

So the same 10 A delivers 5.26 kW of useful power but forces the supply to provide 6.58 kVA. That 1.32 kVA difference is the cost of a lagging power factor. If you had assumed unity power factor, you would have over-promised the motor by a quarter and under-sized nothing, which is the safer direction, but it still throws off your transformer loading math.

Line versus phase: the other place people slip

The formula above wants line values, the figures stamped on the nameplate and shown on the panel meter. People get into trouble when they accidentally feed it phase values instead.

In a star (wye) connection, the line voltage is √3 times the phase voltage. A 400 V line gives 230 V phase, which is exactly why a European socket reads 230 V while the three-phase feeder feeding the building reads 400 V. Line current equals phase current in a star. In a delta connection it flips: line voltage equals phase voltage, but line current is √3 times the phase current.

The practical rule is simple. Use the voltage measured between two lines and the current measured in one line, and the formula behaves. If you ever plug the 230 V phase voltage of a 400 V supply into a line-voltage formula, every result lands off by a factor of √3, and it will look plausible enough that you might not notice until commissioning.

Sizing a motor feeder backwards

Most real jobs run the formula in reverse. A nameplate hands you kilowatts and a power factor, but the breaker is rated in amps. Rearrange:

IL = P ÷ (√3 · VL · cosφ)

Say a workshop adds a 7.5 kW lathe on a 400 V supply with a nameplate power factor of 0.85:

• IL = 7500 ÷ (1.732 · 400 · 0.85) = 7500 ÷ 588.9 ≈ 12.7 A per line

Now you know to fit a 16 A breaker with headroom and a cable rated comfortably above 12.7 A, before you buy a single meter of copper. An 11 kW pump at 400 V and 0.86 works out to about 18.5 A by the same method. Keep one calculator tab open per motor and the shareable URLs become a tidy current schedule you can paste straight into a commissioning sheet.

The one trap in reverse mode is units. If the power field is set to watts and you type 7.5, you are asking for 7.5 watts and the tool returns a current near zero. Switch the field to kilowatts, or type 7500, and you get the real motor current.

Why three-phase beats single-phase for industrial loads

You could in principle run a big motor on single phase, where the formula is just P = V · I · cosφ with no √3. Almost nobody does, for good reasons.

Three-phase delivers constant power. With three voltages 120 degrees apart, the instantaneous total never drops to zero the way a single sine wave does, so a three-phase motor produces smooth torque and starts itself without the capacitors and starting windings a single-phase motor needs. For the same power, the three conductors each carry less current than a single-phase pair would, so the copper is thinner and cheaper. That is the √3 working in your favor: it lets a 400 V three-phase feeder move far more power than a 230 V single-phase circuit on comparable wire.

The other half of efficient three-phase design is keeping conductor losses in check over distance, which is its own calculation. If your feeder runs a long way, pair this tool with a voltage drop calculator so the volts that reach the motor still match the power you planned for.

Reading the power triangle in practice

Once the three numbers are in front of you, the power triangle stops being a textbook abstraction. Hold VL and IL fixed and lower cosφ: P drops while S stays put, and the gap fills with reactive kVAR. That gap is what a power factor correction capacitor bank pays down, and it is why a facility with a lot of lightly loaded motors gets billed for a poor power factor even though its useful work has not changed.

For a quick sanity check on a single-phase branch, or to chase Ohm's law relationships between voltage, current, and resistance, the Ohm's law calculator covers the cases where the √3 does not apply. Between the two, you can move from a single resistor up to a balanced three-phase feeder without ever losing track of whether that 1.732 belongs in the equation.

The short version: write down P = √3 · VL · IL · cosφ, keep your values on the line side, and never let the √3 fall out. Do that and the breaker holds on the second start.

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Made by Toolora · Updated 2026-06-13