Motor Full Load Current Calculator
Calculate the full load current (FLA) drawn by a single-phase or three-phase AC motor from its horsepower, voltage, efficiency, and power factor. Used for selecting fuses, breakers, wire gauges, and motor starters.
Last updated: September 2026
Formula below · 2 sources (ieee.org, Wikipedia) · Updated Sep 2026
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About this calculator
The full load current (FLA) of an AC motor is derived from the power balance equation. Motor output power in watts equals horsepower × 746 (the watt-equivalent of one horsepower). The electrical input power must be higher to account for motor losses, so it is divided by efficiency (η) expressed as a decimal. For a three-phase motor, power is shared across three lines, introducing the √3 factor from line-to-line voltage geometry. The complete formula is: FLA = (HP × 746) / (V × η × PF × √3) for three-phase, or FLA = (HP × 746) / (V × η × PF) for single-phase, where PF is the power factor. This calculated value is an engineering estimate. For NEC sizing, 430.6(A)(1) requires the full-load current from Table 430.248 (single-phase) or 430.250 (three-phase), not a calculated or nameplate value: branch-circuit conductors at 125% of that table current (NEC 430.22) and an inverse-time breaker up to 250% of it (NEC 430.52). The table values are usually higher than this formula gives. Motor overload protection uses the nameplate current (430.32).
How to use
Calculate FLA for a 10 HP, three-phase, 460 V motor with 91% efficiency and 0.88 power factor. FLA = (10 × 746) / (460 × 0.91 × 0.88 × √3) = 7,460 / (460 × 0.91 × 0.88 × 1.732) = 7,460 / 641.5 ≈ 11.6 A. For NEC sizing, do not use this calculated value: NEC Table 430.250 lists 14 A for a 10 HP, 460 V three-phase motor, so the branch conductors must be rated at least 14 × 1.25 = 17.5 A and an inverse-time breaker may be up to 14 × 2.5 = 35 A.
Frequently asked questions
What is the difference between full load current and locked rotor current for an electric motor?
Full load current (FLA) is the steady-state current drawn when the motor is running at its nameplate horsepower output. Locked rotor current (LRC), also called inrush or starting current, is the high current spike — typically 6–8 times FLA — that flows in the instant before the motor accelerates. LRC determines the size of the overcurrent protective device (fuse or breaker) needed to allow the motor to start without tripping. FLA determines the conductor and overload relay sizing for continuous operation. Both values must be considered when designing a motor branch circuit.
How does a lower power factor affect the full load current drawn by an AC motor?
Power factor represents how effectively the motor converts apparent power (VA) into real work (W). A motor with a low power factor (e.g., 0.70) draws significantly more current from the supply than a high-PF motor (e.g., 0.92) producing the same shaft output. From the formula FLA = (HP × 746) / (V × η × PF × √3), you can see that FLA is inversely proportional to PF — halving the power factor doubles the current. This excess current heats conductors, trips protective devices, and increases utility demand charges. Lightly loaded motors typically have poor power factors, which is why running a motor at or near full load is recommended.
When should I use the nameplate full load current instead of a calculated value?
Use the nameplate current for the motor’s overload relay (NEC 430.32). For conductors, short-circuit and ground-fault protection and disconnects, NEC 430.6(A)(1) requires the full-load current from Table 430.248 (single-phase) or 430.250 (three-phase) rather than the nameplate or a calculated value, because the tables are standardized and usually conservative. Calculated FLA from horsepower, efficiency and power factor is useful at the design stage or for cross-checking.