Calculate motor full load current in amps from kW, hp or kVA for three-phase, single-phase or DC supplies, with typical power factor and efficiency.

0.85 is a typical full-load value for induction motors. Use the nameplate value if you have it.

90% is typical for mid-size motors (small ones run lower). Use the nameplate value if you have it.

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Full Load Current Formula

The AC formulas below use total electrical real input power Pin in watts, not motor mechanical output. Use RMS voltage and current. For three-phase operation, VLL is line-to-line voltage and I is line current; the phases are assumed balanced and sinusoidal.

Single-phase AC: I = Pin / (V * PF)

Balanced three-phase AC: I = Pin / (√(3) * VLL * PF)

  • I is the estimated current in amperes; in the three-phase formula it is line current
  • Pin is total electrical real input power in watts; Pout is motor mechanical output power in watts
  • V is single-phase supply voltage; VLL is three-phase line-to-line voltage, in volts. PF is power factor, with 0 < PF ≤ 1. Motor efficiency eta is Pout/Pin, with 0 < eta ≤ 1 (divide entered percent by 100).

For motor output, first use Pin = Pout / eta. A kW input is multiplied by 1000 to obtain watts; mechanical horsepower uses approximately 745.699871582 W per hp, which is a different definition from metric horsepower. For constant DC, I = Pin / V and power factor is not used. For AC apparent power S in VA, I = S / (f * V), where f is √(3) with VLL for balanced three-phase and 1 for single-phase; kVA is multiplied by 1000 to obtain VA. Apparent power alone does not determine real kW or motor output without the missing power factor and efficiency, so the kVA mode does not invent those values. The reverse mode rearranges the selected relationship to estimate its power basis from current. Power factor starts at a typical 0.85 and motor efficiency at 90%; replace them with nameplate values when you have them. This is an electrical estimate; starting current, unbalance, harmonics, derating and installation conditions are not modeled. Use equipment nameplate data and applicable requirements for conductor, overload, fuse, breaker or starter selection.

How to Calculate Full Load Current?

The following hypothetical examples both use balanced sinusoidal three-phase AC, electrical real input power and unity power factor (PF = 1). Voltage is RMS line-to-line voltage. They are not motor-output or equipment-selection examples; no motor efficiency is needed.

Example Problem #1:

  1. Use total electrical real input power Pin = 67 W. In the calculator, select electrical input kW and enter 0.067; this is not motor output.
  2. Enter RMS line-to-line voltage VLL = 8 V and supply PF = 1 for this hypothetical example.
  3. Finally, calculate the Full Load Current using the formula above: 

I = Pin / (√(3) * VLL * PF)

Inserting the values from above and solving the equation with the imputed values gives: 

I = 67 / (√(3) * 8 * 1) = approximately 4.835309 A, or 4.84 A to two decimal places.

Example Problem #2: 

Use the same balanced three-phase, electrical-input and unity-PF assumptions for the second hypothetical example:

Electrical real input power Pin = 90 W = 0.090 kW.

RMS line-to-line voltage VLL = 4 V; PF = 1.

Entering these values into the formula above gives : 

I = 90 / (√(3) * 4 * 1) = approximately 12.990381 A, or 12.99 A to two decimal places.