Questions: AC Power Calculation and Power Factor

Question 1

A factory draws 500 kW of real power at a power factor of 0.5 (lagging). The utility must size its transmission equipment for how much apparent power?

• A. 500 kVA — equipment is sized for real power only, since that is what the factory actually uses
• B. 250 kVA — the power factor reduces the effective load on the utility
• C. 1,000 kVA — apparent power equals real power divided by power factor
• D. 707 kVA — apparent power is the geometric mean of real and reactive power

Answer: (C)

Apparent power S = P / PF = 500 kW / 0.5 = 1,000 kVA. The utility must deliver current I = S / V to the factory, and this current flows through all generators, transformers, and cables whether or not it does real work. A PF of 0.5 means the current is twice what would be needed if the same real power were delivered at unity power factor. This is why utilities size infrastructure for apparent power (VA), not real power (W) — a low-PF customer forces the utility to build larger, heavier equipment for the same real power delivery.

Question 2

Why does adding capacitors in parallel with an inductive industrial load improve the facility's power factor?

• A. Capacitors increase the supply voltage, which automatically reduces the current drawn from the utility
• B. Capacitors absorb and dissipate the reactive power that was previously returned to the utility as waste
• C. Capacitors supply reactive power locally to the inductors, reducing the reactive current that must flow from the utility through the transmission lines
• D. Capacitors filter out current harmonics that cause the apparent power to exceed the real power

Answer: (C)

Inductors need reactive power — they cyclically store energy in magnetic fields and return it. Without correction, this reactive current must travel all the way from the utility source, through transmission lines, to the factory. Adding capacitors nearby gives the inductors a local reactive power source: capacitors release energy from electric fields exactly when inductors need it, and absorb energy when inductors release it. The reactive current now circulates locally between capacitors and inductors, so the utility only needs to deliver real power. The result is a higher PF and lower current in the transmission lines.

Question 3

Reactive power in an AC circuit does no net work over a complete cycle, but it still contributes to the current that must flow through transmission lines and be generated by the utility.

• T. True
• F. False

Answer: True

Reactive power Q (in VAR) represents energy that oscillates back and forth between source and reactive elements — stored in magnetic fields during one half-cycle, returned during the next. No net energy is consumed over a full cycle, so Q does no useful work. But this oscillating energy must physically travel through the conductors and generators, contributing to the RMS current I = S/V = √(P² + Q²)/V. Higher current means higher resistive losses in distribution wiring (I²R losses) and requires larger conductor cross-sections. The utility must account for Q even though customers only pay for P.

Question 4

A load with power factor 0.8 consumes 20% less real power than a load with power factor 1.0 connected to the same voltage source.

• T. True
• F. False

Answer: False

Power factor describes the ratio of real to apparent power (PF = P/S), not the amount of real power consumed. A load's real power consumption is determined by what the load actually does — how much resistance it has, what machinery it drives — not by its power factor. Two loads at the same voltage can have PF = 0.8 and PF = 1.0 while drawing identical real power; the PF = 0.8 load simply draws more current (higher apparent power) to deliver the same watts. The misconception is conflating power factor with efficiency — low PF means more current for the same real power, not less real power.

Question 5

Explain why a utility company charges industrial customers for low power factor, even though reactive power does no net work and customers only use real power.

Model answer: Although reactive power returns to the source each cycle and does no net work, it must still flow back and forth through the utility's generators, transformers, and transmission lines. The current capacity of all this infrastructure is determined by apparent power S = |V|·|I|, not real power P. A customer drawing 1 MW at PF = 0.5 forces the utility to deliver 2 MVA of apparent power — meaning twice the current in the lines, twice the conductor losses (I²R), and generators sized for 2 MVA. The same customer at PF = 1.0 would only require 1 MVA of infrastructure. The cost of building and operating that extra capacity is real, even though the reactive power itself returns to the source.

The financial incentive for power factor correction is direct: by adding local capacitors to raise PF from 0.5 to 0.95, the factory reduces the apparent power it demands from 2 MVA to about 1.05 MVA — cutting line losses and potentially avoiding utility surcharges. The capacitors pay for themselves through reduced electricity bills and deferred infrastructure upgrades.