Motor FLA Calculator — HP to kW and Estimated Full-Load Amps
Convert motor horsepower to kilowatts and estimate the full-load amps (FLA) from voltage, efficiency and power factor for a single- or three-phase motor.
Results are estimates for planning and education, based on your inputs and standard engineering values (AWG resistance, NEC ampacity, resistivity). Electrical work can be dangerous and is governed by the NEC and your local code — verify all sizing with a licensed electrician and your authority having jurisdiction (AHJ). Not a substitute for professional design.
Calculator
Power
3.73 kW
Estimated full-load current
21.2 A
Formula
Mechanical-to-electrical conversion: kW = HP × 0.746, because one horsepower equals 746 watts. The estimated full-load current is FLA = (HP × 746) / (V × eff × PF) for a single-phase motor, and FLA = (HP × 746) / (V × eff × PF × 1.732) for three-phase, where V is the motor voltage, eff its efficiency (0–1) and PF its power factor (0–1). This estimates the running current from the shaft power. For sizing conductors and overload protection the NEC requires the table full-load current from NEC 430.248 (single-phase) or 430.250 (three-phase), not the nameplate value and not a calculated one.
Worked example
A 5 HP single-phase motor at 230 V with 0.9 efficiency and 0.85 power factor: kW = 5 × 0.746 = 3.73 kW; FLA = (5 × 746) / (230 × 0.9 × 0.85) = 3,730 / 175.95 = 21.2 A. The same 5 HP wired three-phase at 230 V draws FLA = 3,730 / (230 × 0.9 × 0.85 × 1.732) = 12.2 A — three-phase moves the same power at a lower line current. Always size the circuit from the applicable NEC 430 table value, then verify with a licensed electrician.
Frequently asked questions
How do I convert horsepower to kilowatts?
Multiply horsepower by 0.746, because one mechanical horsepower equals 746 watts. A 5 HP motor is 5 × 0.746 = 3.73 kW and a 10 HP motor is 7.46 kW. That is the shaft (output) power; the electrical input power is a little higher because the motor is not 100% efficient.
What is motor full-load amps (FLA)?
Full-load amps is the current a motor draws at its rated load. This tool estimates it from FLA = HP × 746 / (V × efficiency × power factor), divided by a further 1.732 for three-phase. It is an estimate: for sizing branch circuits, conductors and overload protection the NEC requires the table value from 430.248 (single-phase) or 430.250 (three-phase), not the nameplate amps and not a calculated figure.
Why does the NEC use a table value instead of the nameplate?
NEC 430.6(A) directs you to use the motor full-load current in Tables 430.248–430.250 for conductor and device sizing, reserving the nameplate FLA for the overload relay. The table values are conservative standardized figures, so two motors with different nameplates but the same horsepower and voltage size the same circuit. Confirm the adopted NEC edition with your AHJ.
Why does a three-phase motor draw less current than single-phase?
For the same power, three-phase line current is lower by a factor of 1.732 (the square root of 3) because the power is shared across three conductors: P = 1.732 × V × I × PF. That is one reason larger motors are wired three-phase — smaller conductors carry the same horsepower.
What efficiency and power factor should I use?
Without the nameplate, 0.85–0.92 efficiency and 0.80–0.88 power factor are typical for general-purpose induction motors, improving with size; premium-efficiency motors run higher. These are estimates — use the nameplate values when you have them, and remember the result is an approximate running current, not a code-sizing value.
Can I size my wire and breaker from this FLA estimate?
No. Use it only for a rough idea. Conductor and overcurrent sizing for motors follows NEC Article 430 — the table full-load current, 125% on the largest motor, plus motor overload and short-circuit/ground-fault protection. Have a licensed electrician size the circuit to the NEC edition adopted by your jurisdiction.
Source: kW = HP × 0.746; NEC 430.248 / 430.250 reference for motor full-load current · All sources