Enter the applied voltage, power factor, and the current into the calculator to determine the Motor Absorbed Power (three-phase real electrical input power). 

Motor Absorbed Power Calculator (3-Phase)

Enter any 3 values to calculate the missing variable

Motor Absorbed Power Formula

Motor absorbed power is the real electrical input power drawn by a balanced three-phase motor. This calculator uses line-to-line voltage, line current, and power factor to estimate how much true power the motor is taking from the supply.

P_a = \sqrt{3} \times V_{LL} \times I_L \times pf

If you want the answer in kilowatts instead of watts, divide by 1,000.

P_{a,kW} = \frac{\sqrt{3} \times V_{LL} \times I_L \times pf}{1000}

Variable Definitions

Variable Meaning Typical Unit
Pa Motor absorbed power, or real electrical input power W, kW, MW
VLL Applied line-to-line voltage V, kV, MV
IL Line current drawn by the motor A, mA, kA
pf Power factor of the motor Unitless decimal

How to Calculate Motor Absorbed Power

  1. Measure or identify the motor’s line-to-line voltage.
  2. Measure the line current under the operating condition you care about.
  3. Determine the power factor as a decimal value.
  4. Multiply voltage, current, and power factor by √3.
  5. Convert watts to kilowatts if needed.

This relationship applies to a balanced three-phase AC system. It estimates true input power, not apparent power and not shaft output power.

Rearranged Forms

If you need to solve for a different variable, the same formula can be rearranged as follows.

V_{LL} = \frac{P_a}{\sqrt{3} \times I_L \times pf}
I_L = \frac{P_a}{\sqrt{3} \times V_{LL} \times pf}
pf = \frac{P_a}{\sqrt{3} \times V_{LL} \times I_L}

Equivalent Phase-Value Form

If phase voltage and phase current are known for a balanced system, the same real power can also be written in phase terms.

P_a = 3 \times V_{ph} \times I_{ph} \times pf

For most motor nameplate and field measurements, however, using line-to-line voltage and line current is more convenient.

Examples

Example 1: A motor operates at 460 V line-to-line, draws 30 A, and has a power factor of 0.90.

P_a = \sqrt{3} \times 460 \times 30 \times 0.90 = 21512.07 \text{ W}

The motor absorbs approximately 21.51 kW of real electrical power.

Example 2: A motor absorbs 22,000 W at 460 V with a power factor of 0.91. Solve for current.

I_L = \frac{22000}{\sqrt{3} \times 460 \times 0.91} = 30.34 \text{ A}

The required line current is approximately 30.34 A.

Absorbed Power vs. Output Power

Absorbed power is the motor’s electrical input. Mechanical shaft power is lower because some energy is lost as heat, magnetic losses, friction, and windage. If motor efficiency is known, output power can be estimated from input power.

\eta = \frac{P_{out}}{P_a}
P_{out} = \eta \times P_a

Common Input Mistakes

  • Using line-to-neutral voltage: this calculator expects line-to-line voltage.
  • Entering power factor as a percent: enter it as a decimal such as 0.82, not 82.
  • Using startup current: inrush current can be much higher than normal running current and will overstate absorbed power.
  • Confusing apparent and real power: absorbed power includes power factor, while apparent power does not.
  • Applying the formula to single-phase systems: a different power equation is required for single-phase motors.

When This Calculator Is Most Useful

  • Estimating motor electrical demand
  • Checking panel and feeder loading
  • Comparing operating conditions at different loads
  • Estimating input power before evaluating efficiency
  • Converting measured voltage, current, and power factor into real power

Practical Notes

For the best estimate, use measured values taken while the motor is running under its actual load. In systems with strong imbalance, harmonic distortion, or rapidly changing load, a dedicated power meter will usually provide a more accurate true power reading than a simplified calculator formula.