3 Phase Motor Amperage Calculator — Line Current from Horsepower, and Motor Load from a Measured Current

Background.

Ask how much current a 25-horsepower, 460-volt motor draws and there are three answers, each right for a different job. Physics, using the motor's efficiency and power factor, gives about 29.08 amps. The nameplate gives the manufacturer's measured full-load amps, normally within a few percent of that. The electrical code gives a third figure from a standard table, and that is the one conductors, disconnects and branch-circuit protection must be sized from. This calculator does the physics, in both directions.

The forward direction starts from a point that is easy to get backwards: a motor's horsepower or kilowatt rating is what it delivers at the shaft. The supply has to provide more, because some energy becomes heat in the windings, the iron core and the bearings. Divide the shaft power by the efficiency to get the electrical input, then divide by √3 times the line-to-line voltage times the power factor to get the current in each line.

The reverse direction is the useful one on site. Clamp an ammeter on a running motor and the same relationship, read backwards, says what fraction of its rating the motor is actually delivering. The US Department of Energy's fact sheet on motor load and efficiency is built around this estimate, because oversized motors are common and one loafing along at a fraction of its rating runs at a poor power factor.

The voltage must be measured line to line: 460 on a 460-volt system, not the roughly 266 volts from any one line to neutral. Use the smaller figure and the current comes out √3, about 1.73, times too high.

What is 3 phase motor amperage calculator?

Three-phase motor amperage is the current flowing in each of the three supply lines feeding a balanced motor, in amperes. It is not an independent property of the motor but follows from the rated shaft output, the efficiency, the power factor and the line-to-line voltage. Power factor is the share of the apparent power, volts times amps, that does real work; the remainder sustains the magnetic field. Nameplates also list full-load amps (FLA), the maker's measured current at rated load, and a code letter describing the much larger inrush at starting.

How to use this calculator.

1. 01Choose the direction: the current drawn at the nameplate rating, or the load implied by a current you measured.
2. 02Enter the rated output and its unit. "hp" on a nameplate normally means the 746 W electrical horsepower; the 745.7 W mechanical horsepower is offered because the DOE fact sheet uses it.
3. 03Enter the line-to-line voltage at the motor. Motors with a 460 V nameplate normally run on 480 V systems, so a measured supply voltage gives a current closer to what a meter will show.
4. 04Enter the full-load efficiency as a percentage and the power factor as a decimal, both from the nameplate. If an older motor does not list them, any figure you assume makes the result an estimate.
5. 05For a load estimate, clamp all three lines while the motor does its normal work, and enter the average. Phases that differ markedly point to a supply or winding fault worth fixing first.
6. 06Use the result for energy, heat and loading questions. Wire and breaker sizes come from the NEC table, not from this current.

The formula.

Start with the energy balance. Everything drawn from the supply leaves either through the shaft or as heat, so the electrical input is Pi = P_shaft ÷ η. On a balanced three-phase supply the real power is also Pi = √3 × V_LL × I × PF, Equation 1 of the DOE fact sheet, with V_LL the line-to-line voltage and I the line current. Setting the two equal gives I = P_shaft ÷ (√3 × V_LL × η × PF).

The √3 is there because in a balanced system the voltage between two lines is √3 times the voltage from each line to the neutral point. Three phases each carrying V_LN × I × PF therefore add up to √3 × V_LL × I × PF.

Two more results come from the same terms. Apparent power, S = √3 × V_LL × I, is what transformers and generators are rated in, and real power is S × PF. The losses are Pi − P_shaft.

The load mode runs the relationship backwards: Pi from the measured current, shaft power = Pi × η, and load = shaft power ÷ rated output × 100%. The DOE sheet's Equations 2 and 3 reach the same result in a different arrangement.

All of it assumes a balanced, sinusoidal, steady supply at the efficiency and power factor entered. Below full load both fall, power factor much faster, so a lightly loaded motor draws more current than a straight proportion suggests, and a load estimate made with the full-load power factor reads high. Starting inrush and the distorted supply current of a variable-frequency drive are outside the model.

A worked example.

Frequently asked questions.

References& sources.

1. [1]US Department of Energy, Office of Energy Efficiency and Renewable Energy, 'Determining Electric Motor Load and Efficiency', fact sheet DOE/GO-10097-517 (January 1997). Source of Equation 1, 'Pi = V x I x PF x √3 / 1000', with V defined as 'RMS voltage, mean line-to-line of 3 phases' and I as 'RMS current, mean of 3 phases'; and of the load estimate, its Equation 2, 'Pir = hp x 0.7457 / Efl' (input power at full-rated load), combined with its Equation 3, 'Load = Pi / Pir x 100%' — together Load = Pi ÷ (hp × 0.7457 ÷ Efl) × 100%. Also the source of the worked example (469.7 V, 37 A, PF 0.763) used as this page's independent check. Authority: US DOE / EERE. Document revision: January 1997. Locator: Equations 1, 2 and 3, and the worked example. Retrieved 2026-07-29.
2. [2]NIST Special Publication 811, 'Guide for the Use of the International System of Units (SI)', Appendix B.9 'Factors for units listed alphabetically'. Source of both horsepower definitions: horsepower (electric) = 7.46 E+02 W exactly, and horsepower (550 ft·lbf/s) = 7.456 999 E+02 W. The exact mechanical value used here, 745.6998715822702 W, follows from the international foot (0.3048 m) and pound-force (4.4482216152605 N). Authority: National Institute of Standards and Technology. Document revision: SP 811, 2008 edition. Locator: Appendix B.9. Retrieved 2026-07-29.
3. [3]NFPA 70, National Electrical Code, 430.6(A)(1) 'Ampacity and Motor Rating Determination — General Motor Applications', reproduced by ElectricalLicenseRenewal: other than for motors built for low speeds (less than 1200 RPM) or high torques, and for multispeed motors, the values given in Tables 430.247 through 430.250 shall be used to determine the ampacity of conductors or ampere ratings of switches and branch-circuit short-circuit and ground-fault protection, instead of the actual current rating marked on the motor nameplate. This is the rule that makes the current computed on this page unusable for code sizing, and it is stated beside the result. Authority: NFPA, via an open code-reproduction host. Locator: 430.6(A)(1). Access: gated at NFPA, open at the reproduction. Retrieved 2026-07-29.
4. [4]EC&M (Electrical Construction & Maintenance), 'Motors and the NEC' — independent confirmation that NEC Article 430 sizes motor circuits from the Table 430.247–430.250 full-load current rather than the nameplate value, and that motor branch-circuit protection and overload protection are separate devices under 430.52 and 430.32 respectively. Used to check the scope statement, not to source any numeric value. Authority: EC&M / Endeavor Business Media. Result: agree. Retrieved 2026-07-29.