Questions: AC Power: Real, Reactive, and Apparent Power

Question 1

An industrial motor delivers 10 kW of real power to a load at a power factor of 0.5. What apparent power must the utility supply?

• A. 5 kVA — apparent power is real power multiplied by power factor
• B. 10 kVA — apparent power equals real power for any load
• C. 20 kVA — apparent power is real power divided by power factor
• D. 17.3 kVA — apparent power equals the vector sum of real and reactive power

Answer: (C)

Apparent power S = P / cos(φ) = 10 kW / 0.5 = 20 kVA. This is the total current demand on the source. A power factor of 0.5 means only half the supplied current is doing useful work — the other half is reactive current oscillating in and out of inductive elements. The utility must size its cables, transformers, and generators for 20 kVA even though only 10 kW is delivered as useful work. Option D is a plausible distractor: it confuses the calculation by using S² = P² + Q² without solving correctly — Q here would be 17.3 kVAR and S = 20 kVA.

Question 2

Why do utilities charge industrial customers penalty fees for low power factors, even when the customer's real power consumption (measured in kWh) stays the same?

• A. Low power factor increases the frequency of the supply voltage, damaging equipment
• B. Reactive power is dissipated as heat in the utility's transformers, increasing their fuel costs
• C. Reactive current loads the utility's cables and generators with current that does no net work, requiring them to be oversized to serve the load
• D. Low power factor customers use more real power than their meters record, effectively stealing energy

Answer: (C)

Reactive current must flow through the utility's cables, transformers, and generators even though it returns to the source each cycle and does no net work. The I²R transmission losses in cables depend on total current (real + reactive), not just the real component. Equipment must be sized for apparent power. A customer with poor power factor causes the utility to spend on infrastructure capacity that generates no revenue — hence the penalty. Option B is wrong: reactive power is not dissipated as heat in an ideal transformer; it oscillates between source and reactive load elements.

Question 3

Adding a capacitor in parallel with an inductive load can reduce the apparent power drawn from the grid without changing the real power delivered to the load.

• T. True
• F. False

Answer: True

Power factor correction works by adding capacitive reactive power Q_C = −Q_L that partially or fully cancels the inductive reactive power. The load still receives the same real power P — the motor still spins, the heat still flows. But the net reactive power seen by the source decreases, so the apparent power S = √(P² + Q_net²) decreases. The source now supplies a smaller total current at a higher power factor, reducing I²R losses in transmission and freeing infrastructure capacity.

Question 4

Reactive power represents energy permanently dissipated in the reactive components of an AC circuit — it is lost, just like the heat dissipated in a resistor.

• T. True
• F. False

Answer: False

Reactive power represents energy that oscillates back and forth between the source and the reactive elements (inductors and capacitors) — it is not dissipated. In each half-cycle, an inductor absorbs energy from the source; in the next half-cycle, it returns that energy. The net energy transfer per complete cycle is zero. This is why Q is measured in VARs (volt-amperes reactive) rather than watts — it is not a rate of energy consumption. Only real power P (in watts) represents true energy dissipation. The confusion arises because reactive current does cause *real* I²R losses in wiring, but that is a secondary effect, not the reactive power itself.

Question 5

Explain why a circuit with a low power factor forces the source to supply more current than necessary, and how power factor correction reduces this without reducing delivered power.

Model answer: Power factor cos(φ) measures the fraction of supplied current that is in phase with the voltage and therefore transfers net energy. At low power factor, a large component of the current is 90° out of phase (reactive), doing no net work but still flowing through source and transmission equipment. For a fixed real power requirement P, the total current I = P / (V · cos(φ)) — as cos(φ) falls, I must rise. Power factor correction adds a capacitor whose reactive current is 180° out of phase with the inductor's reactive current, so they cancel. The load still needs the same real power, but the net reactive demand falls, reducing total current and apparent power.

The key insight is that reactive current is 'wasted' in the sense of occupying infrastructure capacity, even though the energy itself returns to the source. Correction cancels reactive demand at the load rather than at the source, so the reactive current circulates locally between the capacitor and inductor rather than traveling through the utility's lines.