20.3159 A is the example result. Read it using the units and relationship stated below.
Motor FLA uses motor horsepower, voltage, phases.
Distinguishing electrical arithmetic from equipment selection
For motor fla, the relationship is Estimated current = horsepower × 746 ÷ (voltage × phase factor × power factor × efficiency). Phase factor is 1 for single phase and √3 for three phase. Efficiency is entered as a percentage. The amount depends on the supplied motor horsepower, voltage, rather than a live market price or a value imported from another record. Power, energy, voltage and current describe different quantities. Watts represent a rate; watt-hours represent energy over time. Battery and inverter losses can reduce usable runtime even when the rated capacity is unchanged.
What the model includes
Estimated current = horsepower × 746 ÷ (voltage × phase factor × power factor × efficiency). Phase factor is 1 for single phase and √3 for three phase. Efficiency is entered as a percentage. Read unit labels before comparing results. Motor estimates depend on phase, efficiency and power factor; energy budgets depend on duration and duty cycle. Nameplate values and measured consumption may differ under real loads. Treat the output as the stated estimate. It does not select a code-compliant conductor, protective device or complete installation. Equipment specifications and applicable requirements remain part of the design decision.
Example with the supplied inputs
Motor horsepower: 5; Voltage: 240; Phases: 1; Power factor: 0.85; Efficiency (%): 90. Result: 20.3159 A. The values are illustrative.
Compare one changed input
Motor horsepower changes from 5 to 5.5. The result becomes 22.3475 A. All other inputs remain fixed.
Calculation rule
Estimated current = horsepower × 746 ÷ (voltage × phase factor × power factor × efficiency). Phase factor is 1 for single phase and √3 for three phase. Efficiency is entered as a percentage.
Inputs and output
| Input | Example value | Entry convention |
|---|---|---|
| Motor horsepower | 5 | Use the source quantity represented by motor horsepower. Keep the counted population fixed and avoid mixing a unit value with a period total. |
| Voltage | 240 | Use the source quantity represented by voltage. Keep the counted population fixed and avoid mixing a unit value with a period total. |
| Phases | 1 | Use 1 for single phase or 3 for three phase. The phase factor changes the current relationship. |
| Power factor | 0.85 | Enter power factor as a fraction greater than 0 and at most 1 (0.9 means 90%). It is not an efficiency percentage. |
| Efficiency (%) | 90 | Enter efficiency (%) on the percentage scale used in the formula (20 means 20%, not 0.20). Keep its base and reporting period consistent with the other inputs. |
Output: 20.3159 A is the example result. Read it using the units and relationship stated below. Unit: A.
How to use the page
Collect the inputs
Gather Motor horsepower, Voltage, Phases. Use one period and the units shown in the form.
Run the calculation
Enter the values and select the action. The calculation rule above explains how the inputs produce the result.
Compare a scenario
Change Motor horsepower on its own, keeping the other inputs fixed. Read both results before changing another assumption.
Worked example
With the example inputs listed above, the result is 20.3159 A.
Change Motor horsepower from 5 to 5.5 while keeping every other value fixed. The result becomes 22.3475 A. This comparison isolates that input; it does not forecast how other variables will respond.
Check the stated formula, units and limits before using the result in a decision.
Keep the displayed formula, input units and model scope with the result. Corrections or questions can be sent through the request section on this page.
Content updated: October 11, 2026