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The full-load current (FLA) is the current a motor draws from the supply when delivering its rated output power. It is the current that sizes the cables, contactor, overload relay and protection. Because the input power is higher than the shaft power (by the efficiency) and includes reactive loading (the power factor), the current is I = Pout / (√3·VL·pf·η) for a three-phase motor.
| Supply | Full-load current |
|---|---|
| Three-phase | I = Pout / (√3 × VL × pf × η) |
| Single-phase | I = Pout / (V × pf × η) |
| DC | I = Pout / (V × η) |
Here Pout is the mechanical output (nameplate) power, VL is the line-to-line voltage for three-phase, pf is the power factor, and η is the efficiency as a fraction. The DC case has no power factor. Always compare the result with the motor nameplate FLA, and size protection to code (often 125% of FLA for continuous motors).
FLA is the current a motor draws from the supply at its rated output power. It is used to size cables, protection and switchgear, and appears on the motor nameplate.
I = Pout / (√3 × VL × pf × η), where Pout is the shaft power, VL the line voltage, pf the power factor, and η the efficiency as a fraction.
Use I = Pout / (V × pf × η). There is no √3 factor because it is a single-phase supply.
The supply must provide more than the shaft power: efficiency accounts for internal losses, and power factor accounts for the reactive (magnetising) current. Both increase the current beyond what output power alone suggests.
Use the rated mechanical output (nameplate) power together with the efficiency term. The formula converts output to the electrical input current internally through the η factor.
Use the line-to-line voltage (for example 400 or 415 V). The √3 in the formula already accounts for the three-phase relationship, giving the line current directly.
It should be close. Small differences come from the exact efficiency and power factor at your load point. Always use the nameplate FLA for final protection settings when available.
Codes typically set continuous-motor overloads at around 115–125% of the FLA. Check your local electrical code for the exact factor and any service-factor allowance.
A three-phase 400 V motor draws roughly 2 A per kW of output at typical efficiency and power factor. So a 7.5 kW motor is about 14 A — a useful sanity check.
No. FLA is the steady running current. The starting (inrush) current is much higher — typically 6–8 times FLA for direct-on-line starting — and is calculated separately.
A DC motor has no power factor, so I = Pout / (V × η). Only the voltage and efficiency determine the current for a given output power.
Convert horsepower to kilowatts first (1 hp ≈ 0.746 kW) and use that as the output power. Then apply the appropriate single- or three-phase formula.
Starting Current • Motor Efficiency • Wire Gauge / Ampacity • All Calculators