What are real power, reactive power, apparent power and power factor in an AC circuit?

In one line

Real power \( P = VI\cos\phi \) (W) does useful work, reactive power \( Q = VI\sin\phi \) (VAR) flows back and forth to build magnetic and electric fields, apparent power \( S = VI \) (VA) is their combination, and power factor \( = \cos\phi = P/S \).

Explained simply

Think of a glass of beer: the liquid is what you actually drink (real power), the foam takes up space but you cannot drink it (reactive power), and the full glass is what you pay for (apparent power). The power company must supply the full glass, so a lot of foam (low power factor) is wasteful.

Technically, in an AC circuit with phase angle \( \phi \) between \( V \) and \( I \), only the current component in phase with voltage, \( I\cos\phi \), transfers net energy. The quadrature component \( I\sin\phi \) carries energy that is stored and returned every half-cycle by inductors and capacitors.

How it works

1. Resolve current into active component \( I\cos\phi \) and reactive component \( I\sin\phi \).
2. Multiply by \( V \): \( P = VI\cos\phi \), \( Q = VI\sin\phi \), \( S = VI \).
3. These form the power triangle: \( S \) is the hypotenuse, \( P \) the base, \( Q \) the height, with angle \( \phi \).
4. Power factor \( \cos\phi = P/S = R/Z \); lagging for inductive loads, leading for capacitive loads.
5. Low pf means more current for the same real power, so larger cables and more losses. Capacitors are added to improve pf.

Key formula

\( P = VI\cos\phi, \quad Q = VI\sin\phi, \quad S = VI = \sqrt{P^2 + Q^2}, \quad \text{pf} = \cos\phi = \dfrac{P}{S} \)

\( P \) in watts (W), \( Q \) in volt-ampere reactive (VAR), \( S \) in volt-ampere (VA), \( V, I \) = RMS voltage (V) and current (A), \( \phi \) = phase angle between them.

Quick example

A motor takes 10 A from 230 V at 0.8 pf lagging. \( S = 2300 \) VA, \( P = 2300\times0.8 = 1840 \) W, \( \sin\phi = 0.6 \), \( Q = 1380 \) VAR. Check: \( \sqrt{1840^2 + 1380^2} = 2300 \). Transformers and generators are rated in kVA (not kW) because their heating depends on current, whatever the pf.

Common mistakes

• Adding P and Q arithmetically to get S.
• Giving reactive power the unit W or VA instead of VAR.
• Assuming a higher pf means more power consumed; it means less current for the same real power.

How MU asks it

Asked as "Define real, reactive and apparent power and draw the power triangle" (5 marks), and as the final part of almost every AC numerical ("find active and reactive power"). For full marks: give each formula with unit, draw the power triangle with angle \( \phi \), define power factor as \( \cos\phi \) and mention lagging/leading, and explain why low pf is undesirable.