Why horsepower needs an efficiency input
A motor’s shaft output and its electrical input are different measurements. To produce a given output, a motor also consumes power that does not reach the shaft. Divide output watts by efficiency as a decimal to find the corresponding real input watts.
Motor losses = Pinput − Pshaft
For an original arithmetic example, 1,000 W of shaft output at 80% efficiency requires 1,250 W of real electrical input. The 250 W difference represents motor losses. The U.S. Department of Energy’s motor load and efficiency guide explains this input/output distinction and why efficiency should match the operating load.
If you already have real electrical input watts, use the watts to amps calculator directly. Applying motor efficiency to input power a second time would overstate the current.
The current formulas
| Supply | Estimated line current | Voltage basis |
|---|---|---|
| DC | I = P ÷ (η × V) | Motor terminal voltage |
| Single-phase AC | I = P ÷ (η × V × PF) | RMS supply voltage |
| Balanced three-phase AC | I = P ÷ (η × √3 × V × PF) | RMS line-to-line voltage |
These relationships combine real input power with the current formulas used by the main converter. In three-phase mode the output power is the motor’s total shaft output and the result is line current. Do not enter one winding’s voltage in place of the line-to-line supply value.
Power factor and efficiency are not interchangeable. Efficiency relates shaft output to real input power; PF relates real to apparent electrical power. The power factor guide explains the second ratio. DC mode removes PF from the calculation.
Which horsepower definition is used?
The default is mechanical horsepower: 550 foot-pounds-force per second, approximately 745.6999 W. A separate electric-horsepower selection uses 746 W. Both conversion factors appear in NIST’s unit conversion reference. Changing the unit preserves the entered power rather than reinterpreting the same number.
The “electric hp” option changes the unit factor only: the field still means motor output. Metric horsepower (PS) is a different unit and is not accepted as mechanical hp here. If the manufacturer supplies shaft kW, choose that unit to avoid guessing the horsepower convention.
Worked example: 10 hp at 460 V
Take a balanced three-phase motor delivering 10 mechanical hp, with 90% efficiency and a power factor of 0.85 at that operating point. Shaft power is approximately 7,457.0 W. Dividing by 0.90 gives 8,285.6 W of electrical input.
Then divide 8,285.6 by √3 × 460 × 0.85. The current is about 12.2344 A, and input minus output is about 828.6 W. Those efficiency and PF values are illustrative inputs, not defaults to assume for another motor.
Keep the operating condition with the answer
Use matching output, efficiency and PF values. A nameplate rating does not establish the power a partly loaded motor is delivering at this moment. This tool does not extrapolate full-load efficiency or PF to other loads, and it does not estimate locked-rotor or startup behavior.
The model covers the motor’s electrical input. Supply current upstream of a variable-frequency drive, converter or controller needs that equipment’s own input characteristics and losses. This page does not add a drive model or assume the motor PF applies upstream.
The full calculator is free and runs locally without saving your entries. Copy includes the power, voltage, efficiency and supply assumptions. For a correction, email contact@wattstoamps.app. Browse the electrical tools directory for related calculations.