Motor Full Load Amps Calculator
Estimate the full-load current (FLA) an AC motor draws from its rated output power, voltage, power factor and efficiency — for both three-phase and single-phase motors. It's the starting point for sizing cables, contactors and overload protection on a motor circuit.
Enter Values
Before you rely on this: First-pass guide only. Verify safety-critical or regulated work against the relevant standards, your project requirements and a qualified professional.
How to use this calculator
- Enter the motor's rated OUTPUT (shaft) power in kW — this is the kW figure on the nameplate, not the input power.
- Enter the supply voltage (line-to-line for three-phase, e.g. 400 V; line-to-neutral for single-phase, e.g. 230 V).
- Adjust power factor and efficiency to the nameplate/data-sheet values if you have them, or leave the 0.85 / 0.9 defaults.
- Enter 1 in the single-phase box for a single-phase motor, or leave it blank/0 for three-phase, then read the full-load current.
How it works
Because the rated power is the mechanical output, the electrical input the supply must provide is P/η. For a three-phase motor the line current is I = P / (√3 · V · pf · η); for single-phase it is I = P / (V · pf · η), with P in watts. The apparent power drawn is √3·V·I (three-phase) or V·I (single-phase), reported in kVA. Both power factor and efficiency are full-load figures — at part load they drop and the current for a given output rises.
Worked example
Worked example. An 11 kW three-phase motor on 400 V with pf 0.85 and efficiency 0.9: input power = 11 / 0.9 = 12.22 kW; I = 11 000 / (√3 × 400 × 0.85 × 0.9) = 11 000 / 530.0 = 20.75 A, drawing about 14.38 kVA.
Common mistakes
- Entering input (electrical) power instead of the nameplate output power — the tool already divides by efficiency to get the input.
- Using the phase (line-to-neutral) voltage for a three-phase motor; the three-phase formula uses the line-to-line voltage (e.g. 400 V, not 230 V).
- Entering efficiency as a percentage (90) instead of a fraction (0.9), or leaving pf at 1.0 for a lightly loaded motor.
Frequently asked questions
Is this the nameplate full-load current?
It's a close estimate from first principles. Where the motor nameplate or data sheet lists a full-load current (FLC), use that value for design — it accounts for the real machine's losses and power factor.
Why divide by efficiency?
The rated kW is mechanical output at the shaft. The motor also dissipates losses, so the electrical input power the supply delivers is larger: P_input = P_output / η. The current is set by the input power.
What power factor and efficiency should I use?
Use the nameplate/data-sheet values. Typical full-load figures are pf ≈ 0.8–0.9 and η ≈ 0.85–0.95, both of which fall at part load — so a lightly loaded motor draws more current per kW of output than this estimate suggests.
Does this cover starting (inrush) current?
No. This is the steady full-load current. Direct-on-line starting current is typically 6–8× the full-load current, which is why motor protection and starters are chosen from the manufacturer's data, not the running current alone.
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