About the Motor Full Load Amps Calculator
This motor full load amps (FLA) calculator returns the full-load current the US National Electrical Code (NFPA 70) assigns to a standard AC induction motor of a given horsepower and voltage. Pick single-phase or three-phase, the horsepower and the system voltage, and it looks the value up in NEC Table 430.248 (single-phase) or Table 430.250 (three-phase squirrel-cage and wound-rotor induction motors). It also shows 125% of that current — the minimum branch-circuit conductor ampacity for a single continuous-duty motor under NEC 430.22 — and an independent estimate from the power formula using the efficiency and power factor you enter.
Electricians, electrical estimators and engineers use the table current, not the nameplate, to size conductors, short-circuit and ground-fault protection and disconnects (NEC 430.6(A)(1)). The motor’s nameplate current is used instead for overload protection, which is typically set at 115–125% of nameplate FLA.
Table values are as published in NFPA 70, 2023 edition. The common table voltages are covered: 115/200/208/230 V single-phase and 200/208/230/460/575 V three-phase; the three-phase table’s 115 V column for small motors and its 2,300 V column for large motors are not included. The currents apply to motors running at usual speeds with normal torque characteristics — low-speed, high-torque, multispeed and synchronous motors have different currents. Always confirm against the code edition adopted in your jurisdiction.
With the default inputs, the nec table full-load current is 14 A. Change any value above to recalculate instantly.
How to use the motor full load amps calculator
- 1Choose single-phase or three-phase.
- 2Pick the motor horsepower from the nameplate.
- 3Pick the system voltage that matches the motor rating.
- 4Read the NEC table FLA and the 125% conductor ampacity.
- 5Optionally enter efficiency and power factor to compare with a formula estimate.
Formula and method
NEC 430.6(A)(1) requires the table full-load current, not the nameplate current, to be used when sizing motor conductors, branch-circuit short-circuit and ground-fault protection, and disconnects. The calculator looks the value up by horsepower and voltage in Table 430.248 (single-phase) or Table 430.250 (three-phase induction motors) and multiplies it by 1.25 for the minimum conductor ampacity required by 430.22 for one continuous-duty motor.
The formula estimate converts mechanical output to electrical input current: output watts (hp × 746) divided by efficiency and power factor gives input volt-amperes, which divided by voltage (and by √3 for three-phase) gives line current. The NEC tables are deliberately somewhat conservative and use nominal voltages of 115, 200, 208, 230, 460 and 575 V for motors on 120, 208, 240, 480 and 600 V systems.
- FLA
- Full-load current from the NEC table (A)
- hp
- Rated motor output in horsepower (1 hp = 746 W)
- V
- Motor voltage (line-to-line for three-phase)
- η
- Motor efficiency (decimal)
- PF
- Power factor at full load
Worked examples
10 hp three-phase motor at 460 V
NEC Table 430.250 lists 14 A for a 10 hp, 460 V induction motor, so conductors need at least 14 × 1.25 = 17.5 A of ampacity. The formula estimate, 7,457 W ÷ (1.732 × 460 × 0.90 × 0.85), is about 12.2 A — the table is deliberately a little higher.
25 hp three-phase motor at 208 V
The table gives 74.8 A at 208 V, more than double the 34 A at 460 V for the same motor. Minimum conductor ampacity is 74.8 × 1.25 = 93.5 A, and the apparent power is √3 × 208 × 74.8 ≈ 26.9 kVA.
2 hp single-phase motor at 230 V
Table 430.248 lists 12 A for a 2 hp single-phase motor at 230 V, so the branch-circuit conductors need at least 15 A of ampacity. The formula estimate at 90% efficiency and 0.85 PF is 1,491 W ÷ (230 × 0.9 × 0.85) ≈ 8.5 A.
Frequently asked questions
Should I use nameplate amps or NEC table amps?+
Use the NEC table full-load current (Tables 430.247–430.250) to size conductors, branch-circuit short-circuit and ground-fault protection and disconnects. Use the nameplate current to size overload protection. This is required by NEC 430.6(A).
How many amps does a 10 hp three-phase motor draw?+
Per NEC Table 430.250 a 10 hp three-phase induction motor is assigned 32.2 A at 200 V, 30.8 A at 208 V, 28 A at 230 V, 14 A at 460 V and 11 A at 575 V. Actual running current depends on load and the specific motor.
Why is conductor ampacity 125% of FLA?+
NEC 430.22 requires branch-circuit conductors for a single continuous-duty motor to have an ampacity of at least 125% of the motor full-load current from the tables. The extra margin covers continuous operation and service-factor loading.
Why are the voltages 115, 230 and 460 instead of 120, 240 and 480?+
Motor ratings are given at utilization voltage, which is a little below the nominal system voltage to allow for voltage drop. The NEC tables state that the listed currents apply to systems with nominal voltages of 110–120, 220–240, 440–480 and 550–600 V.
How do I calculate motor amps without the table?+
For three-phase, amps = hp × 746 ÷ (1.732 × volts × efficiency × power factor); for single-phase drop the 1.732. A 10 hp, 460 V motor at 90% efficiency and 0.85 PF draws about 12.2 A. Use this for estimates; the NEC table governs code sizing.
Does this apply to DC or synchronous motors?+
No. DC motors use NEC Table 430.247, and synchronous motors at unity or leading power factor have their own current columns or multipliers. This calculator covers the single-phase table and the induction-motor columns of the three-phase table.