Quiz: Power Factor Calculator

Enter voltage, current, and real power to get power factor (PF = P ÷ S), phase angle, apparent power, and reactive power for single-phase or three-phase AC circuits.

Quick Facts

Power factor formula
PF = P ÷ S = cos θ
Ratio of real (working) power to apparent power supplied by the source.
Power triangle
S² = P² + Q²
Real power (P), reactive power (Q), and apparent power (S) form a right triangle; θ is the angle between S and P.
Three-phase apparent power
S = √3 × V_L × I_L
Balanced three-phase apparent power carries a √3 ≈ 1.732 factor versus single-phase S = V × I.

Your Results

Calculated
Power Factor (PF)
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PF = P ÷ S, unitless (0 to 1)
Phase Angle (θ)
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θ = arccos(PF)
Apparent Power (S)
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S = V×I, or √3×V×I for three-phase
Reactive Power (Q)
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Q = √(S² − P²)

Ready

Enter voltage, current, and real power, then press Calculate.

How to use the Quiz: Power Factor

Power factor (PF) measures how efficiently an AC electrical system converts supplied (apparent) power into useful (real) work. It is the ratio of real power P, measured in watts, to apparent power S, measured in volt-amperes: PF = P ÷ S = cos θ, where θ is the phase angle between voltage and current. A power factor of 1.0 (unity) means all delivered power does useful work; a lower power factor means more current is flowing than is strictly needed for that work, which increases conductor losses and can trigger utility demand penalties.

The power triangle: real, reactive, and apparent power

  • Real power (P): the power that does actual work — heat, motion, light — measured in watts (W) and read directly off a wattmeter or nameplate.
  • Reactive power (Q): the power that oscillates between source and load to build the magnetic fields in motors/transformers or the electric fields in capacitors, measured in volt-amperes reactive (VAR). It does no net work but still consumes current-carrying capacity.
  • Apparent power (S): what the source must actually supply, S = V × I, measured in volt-amperes (VA). The three combine as a right triangle: S² = P² + Q², and cos θ = P/S = PF.

Single-phase vs. three-phase apparent power

  • Single-phase circuits: S = V × I — one voltage and one current, no phase-count correction needed.
  • Balanced three-phase circuits: S = √3 × V_L × I_L, where V_L and I_L are line-to-line voltage and line current. The √3 ≈ 1.732 factor comes directly from the 120° phase separation between the three line currents — using the single-phase formula on a three-phase system is the most common power-factor calculation error.
  • Utilities commonly bill a reactive-power penalty once PF drops below roughly 0.90–0.95, because low PF forces wiring and transformers to be sized for more current than the real power alone would require. Capacitor banks are the standard fix for lagging (inductive) power factor.

Frequently Asked Questions

What is a good power factor?
A power factor of 1.0 (unity) is ideal — all supplied power does useful work. In practice, 0.95 or higher is considered excellent, and most utilities require industrial and commercial customers to maintain at least 0.90-0.95 or face reactive-power penalty charges.
What causes a low power factor?
Inductive loads — AC induction motors, transformers, welding equipment, and older fluorescent or HID lighting ballasts — draw lagging reactive current to build their magnetic fields, which pulls the power factor below 1. The more inductive load relative to real load, the lower the power factor.
What is the difference between leading and lagging power factor?
Lagging power factor means current lags voltage, caused by inductive loads such as motors and transformers — the most common case in industrial facilities. Leading power factor means current leads voltage, caused by capacitive loads or an over-corrected capacitor bank. Utilities generally want the power factor close to unity from either direction.
Why does three-phase apparent power include a square root of 3 factor?
In a balanced three-phase system the three line currents are 120° apart. Expressing total apparent power in terms of line-to-line voltage and line current, that phase relationship works out to a factor of √3 ≈ 1.732: S = √3 × V_L × I_L. Single-phase apparent power has no such factor because there is only one voltage-current pair: S = V × I.