Motor Current (Full Load)

How to use this calculator

This tool converts motor nameplate data into the full-load line current — the number you need for breakers, conductors, contactors and fuses. The chain is: electrical input power P_elec = P / η, apparent power S = P_elec / PF, and line current I = S / (k·V) with k = √3 for three-phase (line-to-line voltage, line current) and k = 1 for single-phase. Unlike the generator power calculator, both power factor and efficiency are required here on purpose: silently assuming either equals 1 would understate exactly the quantity used to protect wiring.

Input fields

• Phase — 1-phase or 3-phase. The √3 factor applies only to three-phase machines.
• Rated output power — the mechanical shaft power on the nameplate, in watts or kilowatts (use the unit selector). This is mechanical output, not electrical input: a "7.5 kW" motor delivers 7.5 kW at the shaft and draws more from the supply.
• Supply voltage — for three-phase, the line-to-line voltage your supply applies (400 V, not 230 V). Dual-voltage nameplates (e.g., 230/400 V) list both; enter the one matching your wiring and winding connection.
• Power factor — from the nameplate, typically 0.75–0.9 for induction motors. Enter it as a decimal between 0 and 1.
• Efficiency — from the nameplate, as a decimal (92% → 0.92).

Steps

1. Select the phase configuration of the motor you are analysing.
2. Enter the nameplate rated power and set its unit (W or kW).
3. Enter the applied supply voltage, the nameplate power factor and efficiency.
4. Read the headline full-load current; the detail rows show electrical input power, apparent power and (for three-phase) per-phase apparent power.

The default example — a 7.5 kW, 400 V, three-phase motor at PF 0.85 and 89% efficiency — draws P_elec = 7500/0.89 ≈ 8.43 kW electrically, S ≈ 9.91 kVA, and I ≈ 14.3 A per line. The mistakes that corrupt this number: using electrical input power instead of mechanical output (overstates current by 1/η), entering phase voltage for a three-phase machine (understates current by √3), mixing percents with decimals, and taking the steady-state result straight to a breaker panel without the standard margin — common practice sizes the overcurrent device at about 125% of full-load current for continuous duty.

What it means in a real circuit

The calculation follows power through the machine's boundary. The nameplate kW is what leaves at the shaft; dividing by efficiency recovers what must enter electrically, and dividing by power factor converts that to the apparent power the supply system must carry — the quantity that sets current, because conductors heat with amperes regardless of how much of each ampere does useful work. In the default case the 7.5 kW shaft rating becomes 8.43 kW electrical and 9.91 kVA apparent, which at √3 × 400 V is 14.3 A per line. Motors are rated in mechanical kW precisely because that is what the buyer receives; the supply side lives in kVA and amperes, and this calculator is the bridge between the two worlds.

The full-load current anchors the protection design of the motor circuit. Conductors must carry it continuously (with ambient and bundling correction factors applied), the overcurrent device is commonly set at 125% of FLA for continuous duty, and contactors and overload relays are rated around the same figure. Starting is a separate regime: direct-on-line inrush typically reaches 6–8 times FLA for a few cycles, which is why time-delay (inverse) breaker curves exist and why starter sizing consults locked-rotor data rather than this steady-state number. A motor circuit protected only by its running current would trip on every start.

Voltage choice changes the answer dramatically. On a dual-voltage 230/400 V machine, the same power delivered at 230 V draws about 1.74 times the current of the 400 V connection — which is why winding configuration (delta at the lower voltage, star at the higher) must match the supply. Running current at partial load is lower than FLA, since the variable-loss component roughly tracks torque; a half-loaded motor may run 30–50% below its full-load figure, so measured current is a useful health indicator even though FLA is what protection sizing uses. If the nameplate prints a full-load ampere value, compare it with this calculator's result as a sanity check on your entered voltage and constants.

The assumptions to keep in view: steady-state, balanced, sinusoidal supply at rated frequency; nameplate values taken at rated conditions (which also apply to a VFD-driven motor operating at rated output); no modelling of power-electronics harmonics, which change the waveform shape of VFD input current even when the average power is unchanged. For final design work, confirm against the nameplate FLA and the wiring rules applicable to your installation.