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Real, Reactive and Apparent Power

Reactive power delivers no energy and still costs money, because copper cannot tell the difference. Then modern loads make cos φ the wrong measure entirely.

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With a phase shift between voltage and current, power splits into real power that does work and reactive power that flows out and back with zero net transfer — yet the reactive current is real current in real conductors, which is why utilities bill for it and why capacitor banks exist.

Why AC power is not just VI

With a phase shift, the product of voltage and current goes negative for part of every cycle — energy flowing back out of the load into the source. So average power is not the product of peaks, and the gap between them is the whole topic.

Three powers

Real power P, watts
The average. Converted to heat, light or motion and never returned. P = VI·cos φ.
Reactive power Q, VAr
Sloshes into a field and back twice per cycle. Net transfer zero. Q = VI·sin φ.
Apparent power S, VA
What the current actually is. S = VI = √(P² + Q²). Every conductor and transformer is sized for this.
Power factor
P/S = cos φ. The fraction of the current doing useful work.

Copper cannot tell reactive current from real: it heats cables, loads transformers and drops voltage identically. Everything in the delivery system is sized for S, while the customer only benefits from P.

Why it costs money

Power factorCurrent for the same PTypical of
1.0BaselineResistive heating, corrected installations
0.9+11%A well-loaded motor
0.7+43%A partly loaded motor
0.3+233%An induction motor running near no load

Energy meters measure real power, so reactive power delivers nothing the customer pays for — while costing the utility conductors, transformer capacity and I²R losses. Industrial tariffs therefore add a power-factor penalty or bill in kVA, which makes correction pay for itself in months. Domestic customers are not billed for it, because the aggregate is small and metering costs more than it recovers.

Correction

Inductive and capacitive reactive powers have opposite signs, so a capacitor bank supplies exactly what an inductive load demands. The reactive current then circulates locally between the two instead of travelling from the generator. Motors, transformers and fluorescent ballasts are all inductive, which is why the correction is always capacitive.

Do not over-correct. Past unity the power factor becomes leading and the line current rises again, and excess capacitance on a lightly loaded line pushes the voltage above nominal. That is why banks are switched in steps as load varies.

Where cos φ stops being the right measure

cos φ assumes sinusoidal current. A rectifier with a smoothing capacitor draws narrow spikes rich in harmonics, giving a poor power factor that no capacitor bank can correct.

Harmonics overheat neutral conductors — third harmonics from three phases add rather than cancel — and can resonate with the correction capacitors themselves. Active power-factor correction, which shapes the input current to follow the voltage, is now mandatory above modest power levels.

The numbers you will be asked for

Real power

P = V·I·cos φ

watts

Reactive power

Q = V·I·sin φ

VAr

Apparent power

S = V·I = √(P² + Q²)

VA

Power factor

pf = P/S = cos φ

Correction capacitance

Q_C = P(tan φ₁ − tan φ₂)

Complex power

S = V·I* = P + jQ

Watch it work

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Check yourself

question 1 / 4

One question at a time. Pick an answer to see why it is right or wrong, then move on — there is no score to keep and nothing is saved.

Reactive power delivers no net energy. Why does it still cost the utility money?
An installation runs at a power factor of 0.7. What does that mean practically?
Why are correction capacitors, rather than inductors, fitted?
A building full of switch-mode power supplies has a poor power factor. Will a capacitor bank fix it?

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4 still unanswered — the dots above jump straight to them.