Every three-phase motor draws electrical power from the supply to produce mechanical work at its shaft.
The amount of electrical power it draws is called the absorbed power, or input power.
This value is fundamental to sizing cables, protective devices, transformers, and generators, and to understanding a motor’s true operating cost.
This post explains what absorbed power means, how it relates to output power, apparent power, and reactive power and explains through a solved example using the formulas built into the Motor Absorbed Power Calculator.
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Motor Absorbed Power Calculator (3-Phase)
Calculate the electrical input (absorbed) power, output power, current, and losses of a 3-phase motor.
Results
P_absorbed = √3 × V_L × I_L × cos φP_output = P_absorbed × ηLosses = P_absorbed − P_outputRelated Electrical Calculators
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What is Absorbed Power?
Absorbed power (also called input power) is the real electrical power that a motor draws from the three-phase supply measured in watts (or) kilowatts.
It represents the total energy converted from the electrical system into mechanical output plus all internal losses including resistive heating in the windings, friction, windage and core losses.
Absorbed power is always higher than the useful output power delivered at the shaft, because no motor converts electrical energy into mechanical energy with perfect efficiency.
Absorbed power depends on three measurable quantities: the line-to-line voltage, the line current, and the power factor of the load.
The relationship between these quantities and absorbed power is expressed by a single formula that engineers use daily when commissioning, troubleshooting (or) auditing motor driven equipment.
Formulas
The table below summarizes the core formulas used to calculate absorbed power and the related electrical quantities for a three phase motor.
| Quantity | Formula | Unit |
| Absorbed (Input) Power | P = √3 x VL x IL x cos φ | W / kW |
| Apparent Power | S = √3 x VL x IL | VA / kVA |
| Reactive Power | Q = S x sin φ | VAR / kVAR |
| Output (Shaft) Power | Pout = Pabsorbed x η | W / kW / HP |
| Power Losses | Ploss = Pabsorbed – Pout | W / kW |
Power Factor and its Purpose
Power factor (cos φ) describes how effectively a motor converts the apparent power drawn from the supply into real, usable power.
A power factor of 1.0 would mean all the apparent power is real power, which never occurs in practice for induction motors because they require magnetizing current to establish their rotating magnetic field.
Typical induction motors operate with a power factor between 0.80 and 0.90 at full load, dropping significantly at light load.
A low power factor increases the current drawn for a given real power output, which raises cable losses and can trigger utility penalty charges.
Efficiency and Output Power
While absorbed power is the electrical power drawn from the supply, output power (also called shaft power or useful power) is the mechanical power actually delivered to the driven load, such as a pump, fan (or) conveyor.
The relationship between the two is governed by motor efficiency, a nameplate value expressed as a percentage.
Multiplying absorbed power by efficiency yields output power; the difference between the two is the total internal power loss that is dissipated primarily as heat.
Because efficiency varies with load, motors are usually most efficient somewhere between 75% and 100% of rated load with efficiency dropping off at very light loads.
Selecting a motor that is significantly oversized for its application therefore increases losses relative to the useful work performed.
Apparent and Reactive Power
Beyond absorbed (real) power, 2 other quantities describe a motor’s electrical behavior.
Apparent power is the total power the supply must be capable of delivering, combining both real and reactive components and is used to size transformers, switchgear, and generators.
Reactive power represents the non working power circulating between the supply and the motor’s magnetic field; it does no useful work but is essential for the motor to operate.
Understanding all three: real, apparent and reactive power that gives a complete outline of how a motor loads the electrical system.
Solved Example
Consider a 400 V three-phase motor drawing 25 A of line current at a power factor of 0.85, with a rated efficiency of 90%.
Applying the formulas above:
| Parameter | Value |
| Line Voltage (VL) | 400 V |
| Line Current (IL) | 25 A |
| Power Factor (cos φ) | 0.85 |
| Rated Efficiency (η) | 90% |
| Absorbed Power | 14.72 kW |
| Output Power | 13.25 kW (≈ 17.77 HP) |
| Power Losses | 1.47 kW |
This motor absorbs approximately 14.72 kW from the supply, delivers about 13.25 kW (roughly 17.8 HP) of useful shaft power and dissipates around 1.47 kW as internal losses.
These values allow an engineer to verify nameplate ratings, check cable and breaker sizing (or) estimate running costs based on local electricity tariffs.
Practical Applications
- Verifying that a motors actual absorbed power matches its nameplate rating during commissioning.
- Sizing upstream cables, contactors and circuit breakers based on real input current and power.
- Estimating energy consumption and running costs for pumps, fans, compressors and conveyors.
- Diagnosing overheating or nuisance tripping caused by excessive absorbed power (or) low power factor.
- Comparing measured absorbed power against efficiency class (IE2, IE3, IE4) claims from manufacturers.
Conclusion
Absorbed power is the starting point for almost every practical calculation involving 3 phase motors from cable sizing to energy auditing.
By combining line voltage, line current, power factor, and efficiency, engineers can quickly determine input power, output power, losses and the reactive and apparent power drawn from the supply.
The motor absorbed power calculator automates these formulas giving instant, accurate results for field measurements (or) design calculations without manual computation.
