Find the full-load amps (FLA) of an electric motor from its horsepower rating, supply voltage, phase, power factor and efficiency. Works for DC, single-phase and three-phase AC motors.
Motor shaft power in watts is horsepower multiplied by 746. Full-load current is then the input electrical power divided by voltage, phase factor, power factor and efficiency:
The √3 factor (approximately 1.732) accounts for the relationship between line and phase quantities in a balanced three-phase system. Power factor (PF) is the ratio of real to apparent power; efficiency (Eff) is the ratio of shaft output to electrical input.
If you are sizing the supply for a pump and do not yet have a shaft power figure, work it out from head and flow rate on the pump power calculator, then bring the kW result back here to get the full-load amps.
A 10 HP three-phase motor running at 460 V with a power factor of 0.85 and efficiency of 90%:
NEC 430.6 note: For conductor and overcurrent device sizing, NEC Article 430 requires using the full-load current (FLC) values from Tables 430.247 to 430.250 rather than the nameplate FLA or a calculated value. Use this calculator for estimation and reference; consult the NEC tables for code-compliant sizing.
Typical calculated FLA values for standard HP ratings at common voltages (PF 0.85, efficiency 90%).
| HP | 115 V 1-Ph (A) | 230 V 1-Ph (A) | 208 V 3-Ph (A) | 460 V 3-Ph (A) |
|---|---|---|---|---|
| 0.5 | 3.8 | 1.9 | 1.4 | 0.6 |
| 1 | 7.6 | 3.8 | 2.7 | 1.2 |
| 5 | 38.0 | 19.0 | 13.6 | 6.1 |
| 10 | 76.0 | 38.0 | 27.2 | 12.2 |
| 25 | 190.0 | 95.0 | 68.0 | 30.5 |
| 50 | 380.0 | 190.0 | 136.0 | 61.0 |
Reference values only. Consult NEC Tables 430.247-430.250 for code-compliant FLC values.
Multiply the horsepower by 746 to get shaft power in watts. For single-phase AC, divide by (voltage times power factor times efficiency). For three-phase AC, divide by (root-3 times voltage times power factor times efficiency). For DC motors, divide simply by (voltage times efficiency).
You need the supply voltage, phase, power factor and efficiency as well as the horsepower. One horsepower equals 746 W of mechanical output. For a 5 HP single-phase 230 V motor at PF 0.85 and 90% efficiency, FLA = (5 × 746) / (230 × 0.85 × 0.9) ≈ 19 A.
Power factor represents how much of the apparent current in an AC circuit does useful work. A lower power factor means more current is drawn for the same shaft output. Efficiency is the ratio of mechanical power out to electrical power in; a 90% efficient motor draws more current than an ideal 100% efficient one. Both factors increase the real current drawn from the supply compared to a simple watts-to-amps calculation.
For code-compliant conductor and overcurrent device sizing, NEC Article 430.6 requires using the full-load current (FLC) values from NEC Tables 430.247 to 430.250, not the nameplate or a calculated value. The calculated FLA is useful for estimation, load studies and comparing motors, but the NEC table value must be used for protective device and wiring selection on US installations.
For the same horsepower and voltage, a three-phase motor draws approximately 1/√3 (about 58%) of the line current of an equivalent single-phase motor. This is why three-phase motors are preferred for larger loads: they draw less current per phase, allowing smaller conductors and lower losses. A 10 HP motor at 230 V draws roughly 38 A single-phase but only about 22 A per phase at 230 V three-phase.
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