Three-Phase Current Calculator
Full load current from kW, with power factor and efficiency accounted for. The efficiency term is the one most calculators leave out, and it is worth about 8 % on a typical motor.
Inputs
| Project | Circuit ref | ||
| Prepared by | Date | ||
| Checked by | Date |
Three-Phase Current Calculator · I = P / (√3 · V · pf · η) · EnergyCalcHQ · energycalchq.com
Preliminary calculation. The figures behind it are representative values for the stated conditions, not a substitute for the current edition of the standard or the manufacturer's published data. Verify before issuing for construction. Not a substitute for a qualified engineer or a protection study.
For page numbers, keep Headers and footers ticked under More settings in the print dialog.
The formula
For a balanced three-phase load:
I = P / (√3 × V × cos φ × η)And for single-phase:
I = P / (V × cos φ × η)V is line-to-line voltage for three-phase (415 V in India) and line-to-neutral for single-phase (230 V). The √3 comes from the 120° displacement between phases — it is the ratio between line and phase voltage in a star-connected system.
Why efficiency belongs in the calculation
A motor nameplate states output power at the shaft. The current flows on the input side. A 45 kW motor at 92 % efficiency draws current corresponding to 48.9 kW, not 45 kW.
Omit it and you understate the current by 8–10 %. On a cable sizing job that is frequently one full size, and it is the reason so many motor circuits run warmer than the design says they should.
If your figure is already the input power — a measured kW reading, or a heater or lighting load — set efficiency to 1.
Typical power factors
| Load | Power factor |
|---|---|
| Induction motor, full load | 0.85 – 0.90 |
| Induction motor, half load | 0.70 – 0.80 |
| Resistive heating | 1.00 |
| LED lighting with driver | 0.90 – 0.95 |
| Welding transformer | 0.50 – 0.70 |
| Arc furnace | 0.70 – 0.90 |
An underloaded motor has a notably worse power factor than a fully loaded one, which is why oversizing motors costs money twice: once in the purchase and again on the demand charge.
Starting current
This calculator gives running current. A direct-on-line induction motor draws 6 to 8 times that on start, for a few seconds. That does not change the cable size — cable ratings are thermal and a few seconds is nothing — but it governs the protective device, the contactor rating, and the volt dip seen by everything else on the board.
Star-delta starting cuts inrush to roughly a third, and a soft starter or VFD lower still.
Which current do you actually want?
The same motor has three currents that differ by a factor of seven, and picking the wrong one is the commonest error downstream of this calculator.
- Full load current — what this tool returns. It sets the overload relay and the contactor.
- Design current for the cable — full load current × 1.25 on a continuous-duty motor circuit, before any derating for ambient, grouping or installation method.
- Starting current — six to seven times full load on a direct-on-line start. It never sizes the cable, but it decides the trip curve, the volt drop at the far end of a long run, and whether a generator can carry the start at all.
Efficiency is not power factor
They sit next to each other in the formula and describe unrelated things, which is why one of them usually gets left out.
Efficiency is real loss: the shaft delivers less than the supply provides, and the difference leaves as heat. Omit it and you understate the current by 8–10 % on a typical motor.
Power factor loses nothing. It describes current that flows in and back out again each cycle without doing work — but the cable and the transformer still have to carry it. Omit it and you understate the current by a further 15 %.
For a load that is already stated as input power — a heater, a measured demand figure, a nameplate marked kVA — set efficiency to 1. It is only there to convert shaft output into supply input.
Single phase, and where the √3 goes
On a three-phase supply the three line currents are 120° apart, so they do not simply add. The √3 is what falls out of that geometry, and it is why the same kilowatts draw far less current three-phase than single-phase — roughly 58 % of it.
Use the line-to-line voltage for three-phase (415 V) and the line-to-neutral voltage for single-phase (240 V). Putting 415 V into a single-phase calculation is a quiet way to undersize everything by 42 %.
Prefer a measurement
This calculation estimates what a nameplate would say. Where the machine exists, the nameplate beats it, and a clamp meter on a working load beats both — real plant rarely runs at the rating it was bought at, and a motor at half load draws far less current at a far worse power factor than any formula here assumes.