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Efficiency is the fraction of electrical input power that becomes useful mechanical output at the shaft: η = Pout / Pin. The difference is lost as heat — copper (I²R) losses in the windings, iron (core) losses, friction and windage, and stray losses. For a three-phase motor the electrical input is Pin = √3·VL·IL·pf, so you can find efficiency directly from line measurements and the shaft power.
| Quantity | Formula |
|---|---|
| Efficiency | η = Pout / Pin × 100% |
| Losses | Ploss = Pin − Pout |
| 3-phase input power | Pin = √3 × VL × IL × pf |
| 1-phase input power | Pin = V × I × pf |
Efficiency is highest near the rated load (typically 75–100% of full load) and falls at light load. Modern IE3/IE4 premium-efficiency motors reach 90–96% at rated power. Note the difference from power factor: efficiency compares output to real input power, while power factor relates real to apparent power.
It is the ratio of useful mechanical output power at the shaft to the electrical input power, expressed as a percentage: η = Pout/Pin × 100%. The rest is lost as heat.
Divide the shaft (output) power by the electrical input power and multiply by 100. If you only have line measurements, first find the input power as √3·VL·IL·pf for a three-phase motor.
Pin = √3 × line voltage × line current × power factor. This gives the real power drawn from the supply, which you compare with the shaft power to get efficiency.
Copper (I²R) losses in the windings, iron/core losses from magnetising the steel, mechanical friction and windage, and stray load losses. Together they make up the difference between input and output power.
Small motors are often 75–85% efficient, while larger and premium-efficiency (IE3/IE4) motors reach 90–96%. Efficiency rises with motor size and is highest near full rated load.
No. Efficiency compares mechanical output to real electrical input, while power factor compares real power to apparent power (the reactive loading of the supply). A motor can have high efficiency but modest power factor.
Fixed losses (iron, friction) stay roughly constant while the useful output falls, so they take a larger share of the input. Motors are most efficient at 75–100% of rated load and less efficient when lightly loaded.
Pin = V × I × pf, using the supply voltage, the motor current and the power factor. Then efficiency is the shaft power divided by this input.
Even a few percentage points matter over thousands of running hours. Raising efficiency from 90% to 93% on a continuously-run motor can save hundreds of kWh per year per kilowatt of load.
They are international efficiency classes: IE1 (standard), IE2 (high), IE3 (premium) and IE4 (super-premium). Higher classes have lower losses and are increasingly mandated by regulations.
Yes, slightly. Higher winding temperature raises copper resistance and I²R losses, marginally reducing efficiency. Good cooling helps maintain rated performance.
No. Output can never exceed input; if your calculation gives over 100%, the input power is understated (wrong voltage, current or power factor) or the output figure is wrong.
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