FLA to Watts Calculator: Convert Amps to Power
- Estimated input power
- kW
- Result
- Estimated shaft power
- W
- Result
- Full-load amps
- 10 A
- You entered
An FLA-to-watts calculator estimates electrical input power from a motor’s full-load amperage, voltage, and supply type. Enter the motor’s rated full-load amps and voltage, then choose single-phase or three-phase power; the result is an estimated watt value, with power factor and efficiency available for a closer estimate.
Formula
For an AC motor, estimated real input power is calculated as:
- Single-phase: P = V × I × PF
- Three-phase: P = √3 × VLL × I × PF
Here, P is real electrical input power in watts (W), V is voltage in volts (V), VLL is line-to-line voltage in volts, I is current in amperes (A), and PF is power factor, a dimensionless value between 0 and 1 in typical operating conditions. √3 is approximately 1.732. The calculator uses the supply-type factor to apply 1 for single-phase or 1.732 for three-phase.
Motor efficiency is the ratio of mechanical output power to electrical input power. The calculator’s shaft-power estimate is input watts multiplied by efficiency: Pout = Pin × η, where η is efficiency as a decimal. For example, 90% efficiency is entered as 0.90. The first result is input watts; the other outputs show kilowatts and estimated mechanical shaft watts.
Worked Example
Using the calculator’s defaults—10 A, 230 V, three-phase, power factor 0.85, and efficiency 0.90:
- Three-phase factor: √3 ≈ 1.732.
- Estimated input power: 1.732 × 230 × 10 × 0.85 = 3,386.02 W, which rounds to 3,386 W.
- Convert to kilowatts: 3,386.02 ÷ 1,000 = 3.38602 kW, or 3.39 kW.
- Estimate shaft power: 3,386.02 × 0.90 = 3,047.418 W, or 3,047 W.
These are estimates from the entered values, not a replacement for a motor nameplate or measured operating data. The rounded watt result is the headline output.
What FLA Means and When to Use It
FLA means full-load amperage: the current a motor is rated to draw under specified full-load conditions. It is commonly listed on motor documentation or a nameplate alongside voltage, phase, frequency, and other ratings. FLA is not necessarily the current the motor draws all the time; a lightly loaded motor may draw less, while starting current can be much higher than its normal running current.
Converting FLA to watts is useful for a rough understanding of motor power demand, comparing electrical input with equipment ratings, or making an initial estimate for a load calculation. It can also help explain the difference between current, apparent power, real power, and mechanical output. It does not, by itself, determine the correct wire, breaker, fuse, or generator size.
How to Choose Values and Read the Result
- Use the right current: enter the motor’s full-load current in amperes, not horsepower, starting amps, or a current value from an unrelated circuit.
- Match voltage and phase: use the voltage applicable to the motor and select its actual supply type. For three-phase motors, the formula uses line-to-line voltage.
- Use credible PF and efficiency figures: check motor documentation when available. If you do not know these values, the calculator’s defaults are only assumptions, and the estimate may differ from actual power.
- Interpret the outputs correctly: watts and kilowatts are electrical input-power estimates; shaft watts estimate mechanical output after the efficiency adjustment.
Voltage may be given in volts or kilovolts in some technical contexts: convert kilovolts to volts by multiplying by 1,000 before using the formula. Likewise, 1 kW equals 1,000 W. US customary equipment may be described in horsepower, but horsepower is not an electrical unit. One mechanical horsepower is about 746 W; comparing a motor’s horsepower rating with electrical input requires accounting for efficiency and the precise horsepower convention.
Limits, Power Factor, and Common Mistakes
A frequent mistake is multiplying volts by amps and calling the result real watts without considering power factor. Voltage multiplied by current gives apparent power in volt-amperes (VA) for single-phase AC; for three-phase AC, apparent power is √3 × line-to-line volts × amps. Real power in watts also depends on power factor. For a purely resistive load, PF is near 1, but an induction motor’s power factor is generally below 1 and can vary with load.
Another common error is using the single-phase formula for a three-phase motor, or entering phase-to-neutral voltage when the formula expects three-phase line-to-line voltage. Confirm what the voltage rating refers to. A calculator estimate can also be inaccurate if the entered FLA, PF, or efficiency does not represent the same operating condition.
Actual motor input power may differ because of loading, voltage variation, temperature, motor design, and operating conditions. For a precise measurement, use suitable electrical test equipment and methods, or consult a qualified electrician. Do not treat this calculation as installation or protection-sizing guidance: electrical work must follow applicable equipment instructions and local requirements.
Frequently Asked Questions
How many watts is 1 amp at 120 volts?
For single-phase AC at power factor 1, 1 A × 120 V equals 120 W. If the load’s power factor is below 1, real power is lower: watts = volts × amps × power factor.
Does FLA equal the motor’s actual running amps?
No. FLA is a rated full-load current value under specified conditions. A motor may draw less when it is not fully loaded, and its starting current can be substantially higher.
Why does a three-phase formula use 1.732?
For a balanced three-phase system, the relationship between line voltage, phase quantities, and total power introduces √3, approximately 1.732. The voltage used in the common formula is line-to-line voltage.
Can I convert motor horsepower directly to watts?
Horsepower can be converted to mechanical watts—one mechanical horsepower is about 746 W—but that does not give electrical input power. Input power is greater when the motor is less than 100% efficient.
Can I use this result to choose a breaker or wire?
No. This estimate is not sufficient for selecting circuit protection or conductors. Those decisions depend on motor and installation details, applicable electrical rules, and equipment instructions.
Related
Formula used
in fla | Full-load amps | A | 10 | 0 in volt | Voltage | V | 230 | 0 sel phase | Supply type | Single-phase=1, Three-phase=1.732 in pf | Power factor | - | 0.85 | 0 in eff | Efficiency | - | 0.90 | 0 out watts | Estimated input power | W | phase * volt * fla * pf | 0 out kw | Estimated input power | kW | watts / 1000 | 2 out shaft | Estimated shaft power | W | watts * eff | 0
Each line is one input (in, sel, list) or one result (out: label, unit, formula, decimals). How the calculators work
Method: the standard formula for this calculation, rounded as each result states. See how the calculators work or report an error.