A motor current calculator determines full-load amperage for single- and three-phase AC motors from rated power, voltage, power factor, and efficiency values used in load planning.
Calculate Full-Load Motor Current for Three-Phase AC Motors
This tool calculates the full-load current drawn by a three-phase AC motor from its mechanical power rating, supply voltage, power factor, and efficiency. Electricians, panel builders, and plant engineers use it to size conductors, contactors, and overload relays before installation.
How to Use the Motor Current Calculator
Enter mechanical power rating (P) in kW or HP, line-to-line operating voltage (V) in volts, power factor (cos φ), and motor efficiency (η) as a percentage. The calculator returns full-load current (I) in amperes, along with apparent power (kVA), reactive power (kVAR), active input power (kW), and the resulting phase angle (θ).
Motor Current Formula and Calculation Steps
The line current drawn by a three-phase motor is:
$$I = \frac{P \times 1000}{\sqrt{3} \times V \times \cos\phi \times \eta}$$
Where $I$ is the line current in amperes, $P$ is the mechanical output power in kW, $V$ is the line-to-line voltage, $\cos\phi$ is the power factor, and $\eta$ is efficiency expressed as a decimal (not a percentage). This relationship is the standard method for deriving polyphase motor current used in NEMA MG-1 (Motors and Generators) and IEC 60034-1, the reference standards for AC motor ratings and performance calculations.
Calculation order: (1) convert nameplate mechanical power to watts, (2) divide by efficiency to get real electrical input power, $P_{in} = P / \eta$, (3) divide by $\sqrt{3} \times V \times \cos\phi$ to obtain line current.
Common mistake: entering efficiency as a whole number (90) instead of a decimal (0.90), which understates apparent current. A second common error is plugging the motor’s mechanical (shaft) power rating in directly as if it were electrical input power — the two only match when efficiency is assumed to be 100%.
Power Triangle: How Real, Reactive, and Apparent Power Relate
NEC Table 430.250 Reference Currents for Standard Motors
The values below are full-load current figures from NEC Table 430.250 for standard three-phase induction motors at 460 V, the closest listed voltage to typical industrial supplies.
| Horsepower | Full-Load Current at 460 V (A) |
|---|---|
| 5 HP | 7.6 A |
| 10 HP | 14 A |
| 15 HP | 21 A |
| 20 HP | 27 A |
| 25 HP | 34 A |
| 30 HP | 40 A |
| 40 HP | 52 A |
| 50 HP | 65 A |
Nuance: NEC Table 430.250 only lists specific nameplate voltages — 115, 200, 208, 230, 460, 575, and 2300 V. It does not include 400 V or 380 V, which are standard IEC supply voltages outside North America. At those voltages the table cannot be interpolated safely, because motor design current does not scale linearly with voltage once winding and core losses are factored in; the calculated formula above, not a table lookup, is the correct method for sizing conductors on 400 V or 380 V systems.
Frequently Asked Questions About Motor Current Calculations
Is the calculated current the same as the nameplate FLA?
Not necessarily. Nameplate FLA is measured by the manufacturer under test conditions. The calculated value here is a theoretical result based on the power, voltage, power factor, and efficiency you enter, and should be cross-checked against the nameplate where available.
Why does efficiency reduce apparent power but power factor doesn’t?
Efficiency accounts for real losses (heat, friction) that increase the electrical input needed above the mechanical output. Power factor doesn’t cause losses; it lowers the ratio of real to apparent power, increasing current without increasing delivered work.
What is inrush current and is it included in this result?
Inrush (starting) current is typically 5–7 times full-load current for standard NEMA Design B motors. It is a separate, approximate multiplier applied to the calculated full-load current, not part of the core formula itself.
Does this formula apply to single-phase motors?
No. Single-phase current is calculated without the √3 factor: $I = P \times 1000 / (V \times \cos\phi \times \eta)$. Applying the three-phase formula to a single-phase motor will understate the actual current by roughly 73%.
Why is my calculated current different from a clamp-meter reading?
The calculator assumes rated nameplate conditions. Actual measured current varies with real load, supply voltage fluctuation, and whether the motor is running below its rated mechanical output at the time of measurement.