Three-phase kVA:
where √3 comes from and how to use it

Where the 1.732 comes from

On a three-phase supply the three conductors carry voltages that are the same size but offset by a third of a cycle. The voltage measured between any two conductors, the line voltage, is not three times the voltage from a conductor to neutral. Because the phases are 120° apart, it is √3, about 1.732, times the phase voltage.

That is why a 230 V phase-to-neutral system is also a 400 V line-to-line system, and why the number 1.732 shows up every time we convert between the two. Each of the three phases carries its share of power, so total apparent power adds up from three phase contributions.

A quick way to remember the geometry is to picture three equal arrows pointing 120° apart. The difference between any two of them is longer than one arrow by exactly 1.732 times. That longer arrow is the line voltage, and it is the one printed on switchgear and motor plates.

Most industrial supplies are star connected with a neutral, offering both 230 V and 400 V. Delta systems have no phase-to-neutral voltage, but the line formula is unchanged, which is why it is the one used in practice.

The formula and its derivation

Derivation in one step. Each phase has a phase voltage of V(line) ÷ √3 and carries the line current in a star connection. Three phases give S = 3 × (V ÷ √3) × I. Because 3 ÷ √3 equals √3, this simplifies to √3 × V × I.

Real power follows by multiplying by the power factor, so P = √3 × V × I × cos φ ÷ 1,000 in kW.

Why not use the plain product 3 × V × I? Because voltage and current in that expression would have to be the phase values. With line quantities, the factor √3 does the accounting. Both routes give the same answer, and the line version is shorter because line values are what a clamp meter and a voltmeter normally show.

Worked example: 400 V line voltage and 25 A

As a cross-check, use phase values. The phase voltage is 400 ÷ √3 = 230.9 V, so 3 × 230.9 × 25 = 17,320 VA. The match confirms the shortcut.

If the supply voltage is 415 V instead, the same 25 A gives √3 × 415 × 25 ÷ 1,000 = 17.97 kVA. A mere 15 V difference in line voltage moves the result by about 0.65 kVA, so use the voltage your system actually runs at.

Applied to a motor nameplate, the same reasoning works in reverse. A motor that is labelled 11 kW at power factor 0.85 and efficiency 90% draws electrical input of 11 ÷ 0.9 = 12.2 kW, so apparent power is 12.2 ÷ 0.85 = 14.4 kVA, and at 400 V the line current is about 20.8 A. Always confirm with the plate, which states the rated current directly.

Single-phase and three-phase on one page

A three-phase connection delivers three times the single-phase power at the same phase voltage and current, but only 1.732 times the line-to-line product, which is why mixing up line and phase quantities creates a 73% error.

Working backwards to current and cable

Often you know the equipment rating and need the current. Rearrange the formula: I = kVA × 1,000 ÷ (√3 × V). A 17.32 kVA load at 400 V draws 17,320 ÷ (1.732 × 400) = 25 A. At 415 V the same kVA needs 24.1 A.

1. 1Convert the load to kVA, or kW divided by power factor.
2. 2Divide by √3 and the line voltage to get line current.
3. 3Compare with the cable and breaker ratings, with the margins your wiring standard requires.
4. 4Check voltage drop over the run length for long cables.

Remember that the answer is per line conductor. In a balanced system the neutral carries almost no current, but unbalanced single-phase loads break that assumption.

When the shortcut stops being accurate

• Unbalanced loads. If the three phases carry different currents, add the per-phase apparent powers rather than using a single current value.
• Distorted waveforms. Drives and electronics create harmonics; use true RMS measurements.
• Varying voltage. Supply voltage may sag under load, so measure under working conditions.
• Mixed power factors. Real power adds linearly, but kVA does not, because it also carries reactive components.

For most balanced motor and heating loads, the shortcut is accurate enough for sizing. A simple voltage-times-current tool can verify the multiplication step; remember to apply √3 for three-phase work yourself.

Before trusting a figure, ask what supply it was computed for. A system fed at 415 V, a site with a stated voltage tolerance, or a generator with its own regulation can differ from nominal. The formula is exact; the uncertainty lies in the numbers fed into it.

For protection settings, use the manufacturer's guidance instead of a calculated value. Breakers and cables carry extra requirements for ambient temperature, grouping and start-up current that this single formula does not capture.