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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).
For a three-phase motor the full-load current is
I = Pout / (√3 × VLL × PF × η)
Two of those four denominators are routinely forgotten. The nameplate power is mechanical shaft output, so the efficiency η must be divided out to recover electrical input. And the power factor must be the value at the operating load, not the full-load figure, because a lightly loaded induction motor has a markedly worse power factor than a fully loaded one.
A worked case: a 7.5 kW motor on 415 V with PF 0.85 and 90% efficiency draws 7500/(√3 × 415 × 0.85 × 0.90) = 13.64 A. Omitting the efficiency term gives 12.3 A — an 11% underestimate, easily enough to specify a cable or an overload relay one size too small.
Every motor nameplate states its own full-load current, and that figure is measured rather than calculated. When the nameplate and your calculation disagree, the nameplate wins: it already accounts for the actual efficiency and power factor of that specific design. Use the calculation when the nameplate is missing or illegible, or to sanity-check a figure that looks wrong.
Note that short-circuit protection is a separate calculation: the fuse or breaker must tolerate the starting surge, which is several times full-load current, while still protecting the cable.
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.
I = P/(√3 × VLL × PF × η), where P is the mechanical shaft output in watts, VLL the line-to-line voltage, PF the power factor and η the efficiency.
Because the nameplate rating is shaft output, not electrical input. For a 7.5 kW motor at 90% efficiency the input is about 8.33 kW, and leaving efficiency out underestimates current by roughly 11%.
The nameplate, whenever it is available. That figure is measured for the specific machine and already reflects its real efficiency and power factor. Calculate only when the nameplate is missing or to check a suspicious value.
From the motor's full-load current, typically at 100–125% of it depending on the service factor and the applicable wiring regulations. The relay protects against sustained overload, not against short circuit.
Yes, significantly. An induction motor's power factor falls as load decreases, so a motor running at a quarter load may have a power factor well below its nameplate figure, drawing proportionally more current for the work done.
Because it must survive the starting surge, which is commonly six to eight times full-load current for a few seconds. Overload protection and short-circuit protection are deliberately separate functions with separate devices.
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