What Is Motor Full Load Current?
The full load current (FLC), also called full load amperes (FLA), is the current an electric motor draws when running at its rated horsepower output. It is essential for sizing conductors, overload protection, contactors, and circuit breakers. This calculator works for both single-phase and three-phase AC induction motors.
How to Use This Calculator
Enter the motor's rated power in horsepower (HP), the supply line voltage, the motor efficiency as a percentage, and the power factor (typically 0.8–0.9 for induction motors). Select whether the supply is single-phase or three-phase, then read the resulting full load current in amperes. The calculator also reports the mechanical output power and the electrical input power.
The Formula Explained
One mechanical horsepower equals 746 watts, so the motor's shaft output is \(\text{HP} \times 746\). The electrical input power is higher because of efficiency losses, and the current is found by dividing input power by the voltage, accounting for the power factor. For a three-phase system the line current relationship adds the \(\sqrt{3}\) factor:
$$I = \frac{\text{HP} \times 746}{\sqrt{3}\;\times\;\text{V}\;\times\;\eta\;\times\;\text{PF}}$$
where \(\eta\) is efficiency expressed as a decimal (90% = 0.90). For single-phase motors drop the \(\sqrt{3}\) term.
Worked Example
A 10 HP three-phase motor on 400 V with 90% efficiency and a 0.85 power factor: input power = \(10 \times 746 = 7460\ \text{W}\). Current = \(7460 \div (1.732 \times 400 \times 0.90 \times 0.85) \approx 7460 \div 529.9 \approx\) 14.08 A.
FAQ
Is this the same as nameplate current? It is a calculated estimate. Always use the motor nameplate FLA when available, as actual values vary by design.
What power factor should I use? If unknown, 0.85 is a reasonable default for standard induction motors; high-efficiency motors may run 0.88–0.92.
Why is efficiency in the denominator? Lower efficiency means the motor must draw more input current to deliver the same shaft power, so dividing by \(\eta\) correctly increases the current.
