I understand from

Power = V_rms * I_rms * cos(phi)

Being phi the phase difference between current and voltage.

The power consumed by the a circuit is reduced in the case of having inductive or capacitive loads, i.e. motors.

In a factory, for example, why does one get a penalty fee in the power bill in the case of having many inductive loads?

  • 2
    \$\begingroup\$ When sending trucks to deliver mail, it is like having trucks that go back and forth between warehouse and destination without doing any deliveries. No mail is being delivered, but they still take up space on the road and consume fuel. \$\endgroup\$
    – DKNguyen
    Commented Apr 27, 2020 at 1:04

2 Answers 2


When the load is reactive the currents are higher than necessary, compared to a load with a power factor of 1.

As a result the losses in the distribution network are higher, so each watt of power consumed costs more power at the point of generation.

Industrial loads tend to be inductive (as in motors). Capacitor banks local to the inductive load can correct the power factor.

Residential customers typically are not penalized for power factor.


The real power isn’t reduced. This is the part that does actual work at the load.

The apparent power includes the real power, and unused reactive power. Even though the reactive power does no useful work, it nevertheless taxes the grid with its flowing in and out of the reactive loads (like induction motors.)

Utilities have to provision for apparent power and thus charge a premium for customers with low power factors.

Big utility customers can improve their power factor by adding compensating reactance locally to cancel out the reactive elements in their local distribution.

This has filtered down even to smaller loads, like computers. For several years now, any power supply above 60W (later revised upward to 100W) requires power factor correction in its input. More about that here: https://www.energystar.gov/ia/partners/product_specs/program_reqs/eps_prog_req.pdf


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