Electrical Power (3-phase)
Real power in a balanced three-phase system.
Formula
Variables
Example
Industrial power systems
How it works
This formula gives the real power delivered by a balanced three-phase AC supply using line quantities, so you do not need to calculate the power in each phase separately. The 3 factor appears because the line voltage and line current are related to the phase values in a three-phase system. The cos term accounts for the phase difference between voltage and current, so only the in-phase part of the current produces useful real power.
When to use it
Use this for balanced three-phase loads such as motors, pumps, compressors, HVAC systems, and industrial feeders. It is also useful when you are given line voltage, line current, and power factor and need total real power in kilowatts or watts.
Worked example, step by step
A balanced three-phase motor is supplied at 415 V line voltage, draws 18 A line current, and operates at a power factor of 0.88. P = 3 V I cos P = 1.732 d7 415 d7 18 d7 0.88 P = 1.732 d7 415 d7 15.84 P = 1.732 d7 6573.6 P 11381 W P 11.4 kW
Common mistakes to avoid
- Using phase voltage and phase current in the formula without converting them consistently
- Forgetting the power factor and assuming all three-phase power is just 3VI
- Applying the formula to an unbalanced load, where the simple balanced-system relation may not be valid
Did You Know?
Electrical power delivered equals voltage times current, the basic budget for sizing any circuit.
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