What the Motor Full-Load Current Calculator does
This calculator turns a motor's kW or horsepower rating into full-load current, using the efficiency and power factor you enter. It shows the input power as well as the output, the apparent power, the losses, and the current across the load range.
The step people skip is the first one. A motor nameplate quotes mechanical output at the shaft, not electrical input - so a 30 kW motor at 93% efficiency draws 32.26 kW of electrical power, and at a power factor of 0.86 that is 37.51 kVA and 54.14 A at 400 V. Dividing 30 kW straight by root-three times the voltage gives 43 A and undersizes everything downstream.
How to use it
- Enter the shaft rating in kW, or in horsepower if that is how the nameplate reads. Horsepower takes precedence if both are filled in.
- Enter the supply voltage and choose single or three phase.
- Enter the efficiency and power factor from the nameplate. Both are quoted at rated load, and both are worse at part load.
- Read the full-load current, and use it as the basis for the overload relay setting and as the design current for the cable.
- Check the load-range table if the motor will normally run well below its rating - the power factor falls off sharply and the current does not fall as fast as the load does.
Reading the results
Full-load current is the current at the rated shaft output. It is the figure the overload relay is set from and the figure the cable is sized for.
Input kW is what the meter sees and what you pay for. The difference between it and the shaft output is the motor's losses, all of which become heat.
kVA is what the supply and the cable carry. A motor's power factor is poor enough that the kVA figure is noticeably above the input kW.
The load-range table uses a conventional part-load fall-off in efficiency and power factor rather than this motor's real curve, and it says so. Use it to see the shape of the behaviour, not to predict a specific machine.
Worked example: a 30 kW three-phase motor at 400 V, 93% efficient, power factor 0.86
Input power is 30 / 0.93 = 32.258 kW. The 2.258 kW difference is the motor's losses - windage, friction, iron loss and copper loss - and every watt of it is heat that has to leave the motor and the room it is in.
Apparent power is 32.258 / 0.86 = 37.509 kVA.
Full-load current is 37,509 / (root-three x 400) = 37,509 / 692.82 = 54.14 A.
Compare that with the naive calculation: 30,000 / (root-three x 400 x 0.86) = 50.4 A, which ignores efficiency and is 7% low. Ignoring the power factor as well gives 43.3 A, which is 20% low and would size the cable and the overload relay dangerously small.
At 50% load the same motor produces 15 kW at the shaft, but its power factor falls to around 0.73 and its efficiency to about 0.93, so it still draws roughly 31.9 A - 59% of full-load current for 50% of the work. That is the part-load penalty that makes an oversized motor expensive to run, and the reason variable-speed drives pay back on varying loads.
Formulas and scoring rules
- Input power
P_in = P_shaft / efficiencyThe nameplate kW is shaft output. Efficiency is a fraction, so 93% is 0.93.- Apparent power
S = P_in / power factor- Three-phase full-load current
I = S x 1000 / (root-three x V)- Single-phase full-load current
I = S x 1000 / V- Combined, three phase
I = P_shaft x 1000 / (root-three x V x pf x efficiency)- Horsepower
1 hp = 745.6999 WMechanical horsepower. The metric horsepower of 735.5 W is a different unit and differs by about 1.4%.- Losses
P_loss = P_in - P_shaft
The two numbers on the nameplate that decide everything
Efficiency and power factor both divide into the answer, so an error in either goes straight into the current. Assuming 0.9 efficiency and 0.9 power factor when the real figures are 0.93 and 0.86 changes the current by about 4% - tolerable. Assuming a power factor of 0.86 for a motor that actually runs at 0.65 because it is half loaded changes it by 32% - not tolerable.
Efficiency classes help with the first: IE3 motors above 7.5 kW are typically 0.91 to 0.95, and the class is on the nameplate. Power factor is the one to be careful with, because it degrades far faster at part load than efficiency does and no class system covers it.
Nameplate current and code table current
There is a subtlety that catches people working to the NEC. Article 430 requires conductors and short-circuit protection to be sized from the table values in 430.248 to 430.250, not from the nameplate - the tables are deliberately conservative and consistent. The motor's actual nameplate current is used for the overload relay setting under 430.6(A).
So the same motor legitimately has two current figures in the same design, used for different purposes. IEC practice works from the actual motor data throughout, with coordination tables from the starter manufacturer. If you are working to the NEC, use this page for understanding and the code table for the sizing.
Limitations: what the result does not prove
- It calculates from the figures you enter. Nameplate efficiency and power factor are quoted at rated load, and a motor that runs at half load behaves quite differently.
- The part-load table uses a conventional fall-off, not the specific motor's curve. Manufacturers publish real curves and they should be used where the answer matters.
- It does not cover starting current, which is six or more times this figure - see the starting current calculator.
- Where a code's motor full-load current table applies, that table governs for conductor and device sizing even when it disagrees with the nameplate.
- Nothing here is a compliance certificate. A qualified engineer must verify the design against the applicable standard and the local regulations.
Privacy: where your data goes
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Standards and sources
- IEC 60034-1 - Rotating electrical machines: rating and performance - checked 19 Sep 2026
- NFPA 70 - National Electrical Code - checked 19 Sep 2026
- IEC 60034-30-1 - Efficiency classes of line-operated AC motors (IE codes)
Frequently asked questions
How do I convert motor kW to amps?
Divide the shaft kW by the efficiency to get input kW, divide that by the power factor to get kVA, then divide by root-three times the voltage for a three-phase supply. Combined: I = kW x 1000 / (1.732 x V x pf x efficiency). A 30 kW motor at 400 V, 0.93 efficiency and 0.86 power factor draws 54.1 A.
Is the nameplate kW input or output?
Output - the mechanical power available at the shaft. Electrical input is always higher by the losses, so a 30 kW motor never draws only 30 kW. This is the single most common error in motor current calculations and it always errs on the low side.
How many amps does a 1 hp motor draw?
It depends on the voltage, the phases, the efficiency and the power factor. A 1 hp single-phase 230 V motor at 0.78 efficiency and 0.80 power factor draws about 5.2 A; the same 1 hp on a 400 V three-phase supply at better figures draws about 1.9 A. There is no single answer without those four inputs.
What efficiency should I assume if the nameplate is unreadable?
For a modern IE3 motor, roughly 0.85 for 1.5 kW, 0.90 for 7.5 kW and 0.94 for 45 kW and above - efficiency rises with size. Older motors are several points lower. If the answer sizes a cable or a relay, find the data sheet or measure it rather than assuming.
Why does the power factor matter so much?
Because it divides into the current directly. A motor at 0.86 power factor draws 16% more current than the same input power at unity would, and all of that extra current has to be carried by the cable, the contactor and the transformer. It is also why motor-heavy installations are the classic candidates for power factor correction.
Does the current change if the motor is lightly loaded?
Yes, but much less than the load does. At 50% load a typical motor still draws around 60% of full-load current, because the magnetising current is nearly constant and the power factor collapses. That is why an oversized motor is expensive to run and why matching the motor to the load matters.
Should I size the cable at full-load current or higher?
At the full-load current as the design current, with the usual derating for the installation conditions. It is the protective device that is sized differently - the overload relay at full-load current, and the short-circuit device well above it so the starting current can pass. Never size the cable for the starting current; it is a short transient the cable can absorb.
What about a motor on a variable-speed drive?
The drive changes the picture entirely. The supply sees the drive's input current at a power factor close to unity but with substantial harmonic distortion, while the motor sees whatever the drive gives it. Size the supply side from the drive's data sheet rather than from the motor's, and the motor cable from the drive's output rating.
Last reviewed by the A2Z.Tools team against the sources listed above.
