Engineering
Intermediate

Power Factor

Ratio of real power to apparent power.

Formula

PF=PSPF = \frac{P}{S}

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Variables

PReal power
SApparent power

Example

PF of 1 is ideal

How it works

Power factor tells you how effectively an AC electrical system converts supplied power into useful work. It is defined as the ratio of real power, which actually does work and is measured in watts, to apparent power, which is the total power the source must supply, measured in volt-amperes. This ratio works because AC systems can have phase differences between voltage and current, so not all supplied power is converted into useful energy at every instant.

When to use it

Use this when analyzing AC circuits, motors, transformers, and industrial electrical loads where voltage and current may not be in phase. It is especially useful for checking electrical efficiency, sizing equipment, and estimating how much current a system draws for a given useful output.

Worked example, step by step

An industrial machine uses real power P = 3.6 kW and has apparent power S = 4.5 kVA. PF = P / S PF = 3.6 / 4.5 PF = 0.80 So the power factor is 0.80, or 80%.

Common mistakes to avoid

  • Using watts for both P and S instead of keeping P in W and S in VA or both in matching scaled units like kW and kVA
  • Confusing power factor with efficiency; a circuit can have low power factor even if it is not wasting much energy as heat
  • Forgetting that power factor is a unitless ratio and should be reported as a decimal or percentage, not in watts or volt-amperes

Did You Know?

Correcting power factor saves industry billions by reducing wasted current on the grid.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026