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Apparent Power

Combination of real and reactive power.

Formula

S=P2+Q2S = \sqrt{P^2 + Q^2}

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Variables

PReal power
QReactive power

Example

Measured in VA

How it works

In AC circuits, real power P is the part that does useful work, while reactive power Q represents energy that sloshes back and forth between the source and fields in inductors or capacitors. These two components are perpendicular in the complex-power plane, so the total apparent power S is the magnitude of the vector formed by P and Q, which is why Pythagoras gives S = √(P² + Q²). This tells you the overall power a source, transformer, or cable must be able to handle even though not all of it becomes useful output.

When to use it

Use this when you know real power and reactive power and need the total power demand seen by the supply, such as sizing generators, transformers, UPS systems, or conductors. It is also useful for power-factor analysis and checking whether equipment ratings in volt-amperes are adequate.

Worked example, step by step

Given: P = 1200 W Q = 900 var Step 1: Square each component P² = (1200)² = 1,440,000 Q² = (900)² = 810,000 Step 2: Add the squares P² + Q² = 1,440,000 + 810,000 = 2,250,000 Step 3: Take the square root S = √2,250,000 = 1500 Final answer: S = 1500 VA

Common mistakes to avoid

  • Adding P and Q directly instead of using the square-root of the sum of squares
  • Mixing units: P should be in watts and Q in vars, with S reported in volt-amperes
  • Using this formula for DC circuits or purely resistive AC loads where Q is zero

Did You Know?

The power triangle separates useful real power from reactive power that merely sloshes back and forth.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026