Power Factor and Motor Efficiency Are Not the Same Number

They answer two different questions

Motor efficiency compares useful mechanical shaft output with active electrical input. Power factor compares active electrical power with apparent electrical power. Both are ratios between zero and one in these ordinary example calculations, but they do not describe the same loss or the same part of the system.

If a motor’s 20 kW rating is shaft output and its efficiency at that operating point is 0.90, the required active electrical input is 20 / 0.90 = 22.22 kW. At 415 V line-to-line and PF 0.80, the balanced three-phase line-current estimate is 22,222 / (√3 × 415 × 0.80) ≈ 38.64 A.

If the 20 kW instead already means electrical input, efficiency is not applied again. The same voltage and PF give about 34.78 A. This is why the 20 kW to amps article keeps the power basis beside the answer.

Changing only one factor

Keep the shaft output and 0.90 efficiency, but assume PF 0.90 instead of 0.80. Active input remains about 22.22 kW; the line-current estimate falls to about 34.35 A because less apparent power is needed for the same active power. That does not mean the motor has become 90% more efficient or that its mechanical output has changed.

Now keep PF 0.80 but change the assumed efficiency. The active input and current both change. Neither factor should be chosen merely to match a desired answer. Use the motor manufacturer’s operating data or measured electrical input where possible.

A practical entry order

1. Identify whether the available kW or horsepower is shaft output or electrical input.
2. Apply efficiency only to convert output to input when justified.
3. Use the applicable supply voltage and PF to calculate steady electrical current.
4. Keep starting current, actual loading, waveform distortion and nameplate current as separate checks.

The horsepower guide handles the same output-to-input issue for hp ratings. The current result is not a breaker, cable or energy-consumption decision.