Higher Energy
Curriculum/Physics Electricity
Physics ElectricityLayer 44 min

Real, Reactive, and Apparent Power

A factory's electric bill shows 500 kW of demand, but the utility's meters show 625 kVA flowing through the transformer. The difference is not a billing error. It is the distinction between real and apparent power, and it matters for grid infrastructure sizing.

In AC circuits, voltage and current may not peak at the same instant. When they are perfectly in sync, all the power delivered does useful work (real power, measured in watts). When current leads or lags voltage, some power oscillates back and forth between the source and the load without doing work (reactive power, measured in volt-amperes reactive, or VAR). The total power flowing in the circuit is apparent power (measured in volt-amperes, VA). These three quantities form a right triangle: apparent2 = real2 + reactive2. The ratio of real to apparent power is the power factor: PF = real power / apparent power.

Apply the triangle. A motor draws 400 kW of real power and 300 kVAR of reactive power. Apparent power = sqrt(4002 + 3002) = 500 kVA. Power factor = 400/500 = 0.80.

Size the infrastructure. The utility must size its transformers, wires, and breakers for 500 kVA (apparent power), not 400 kW (real power). Low power factor means oversized infrastructure for the useful work delivered.

Why do utilities penalize customers with low power factor?

Wasted capacity. A customer at 0.80 PF requires 25% more current (and therefore larger wires and transformers) than a customer at 1.0 PF for the same useful output. Utilities charge power factor penalties to recover the cost of that extra infrastructure.

Power factor correction (adding capacitors to offset inductive loads) is one of the cheapest ways to free up grid capacity without building new equipment.


Question 1 of 2

A facility draws 600 kW real power at a power factor of 0.75. What is the apparent power the utility must deliver?

Apparent power = real power / power factor = 600 / 0.75 = 800 kVA. The utility must size equipment for 800 kVA even though only 600 kW does useful work.

The answer is D