AC Power Calculator (real, reactive, apparent, PF)

Understanding the AC power calculator — concept and principle

This calculator splits the power of a single-phase AC load into real power P (kW), reactive power Q (kvar) and apparent power S (kVA), and gives the phase angle that matches the power factor. Use it to compare a load's useful power with what the wiring and supply must carry, or to size a generator, UPS or inverter rated in kVA.

How it is calculated

• Apparent power: S = V · I, using RMS voltage and current.
• Real power: P = V · I · cos φ = S · PF.
• Reactive power: Q = V · I · sin φ = S · √(1 − PF²).
• Phase angle: φ = arccos(PF), in degrees.

Results are divided by 1000 to show kW, kVA and kvar. The three quantities form a right triangle: S² = P² + Q².

The principle

In a purely resistive load, voltage and current rise and fall together, so every bit of current does useful work. Motors, transformers and fluorescent ballasts store energy in magnetic fields for part of each cycle and return it in the next, shifting the current by an angle φ relative to the voltage. Only the part of the current in phase with the voltage (I cos φ) transfers net energy; this gives real power P, the power that turns into heat, light or motion and is measured by an energy meter. The part 90° out of phase (I sin φ) sloshes back and forth, giving reactive power Q. The wires, however, carry the full current I, so cables, fuses and transformers must be sized for the apparent power S.

Worked example

1. Inputs: V = 230 V RMS, I = 10 A RMS, PF = 0.85.
2. Apparent power: 230 × 10 = 2,300 VA = 2.3 kVA.
3. Real power: 2,300 × 0.85 = 1,955 W = 1.955 kW.
4. Reactive power: √(1 − 0.85²) = 0.52678, so 2,300 × 0.52678 = 1.2116 kvar.
5. Phase angle: arccos(0.85) = 31.79°.

How to read the result

A power factor of 1 means S = P. Resistive heaters and incandescent lamps are close to 1, while lightly loaded induction motors can be well below 0.8. The lower the power factor, the more current is needed for the same useful power, which raises cable losses. Many utilities charge or penalise industrial customers whose power factor falls below a set threshold, often around 0.9.

Common mistakes

• Multiplying V × I and calling it watts; that is VA unless the PF is 1.
• Using peak values instead of RMS values.
• Applying this single-phase formula to a three-phase load, which needs a factor of √3 with line values.

Frequently asked questions

Is kVA the same as kW?

Only when PF = 1. Otherwise kW = kVA × PF, so a 2.3 kVA load at 0.85 PF delivers 1.955 kW.

How can I tell if Q is inductive or capacitive?

The power factor alone does not say. Inductive loads have a lagging current and capacitive ones a leading current; the calculator shows the magnitude only.

Sources: IEEE Std 1459, Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced, or Unbalanced Conditions · C. K. Alexander and M. N. O. Sadiku, Fundamentals of Electric Circuits, McGraw-Hill, chapter on AC power analysis