Power factor (PF) in AC circuits — does it indicate how much of the apparent power (S) is actually real or true power (P)? Consider a sinusoidal steady-state circuit where apparent power S = V_rms * I_rms and power factor PF = cos(phi).

Introduction / Context:
Power factor in alternating current systems relates the apparent power flowing in a circuit to the useful or true power consumed by the load. A correct understanding of power factor enables better sizing of equipment and reduction of losses in transmission and distribution networks.

Given Data / Assumptions:

• S refers to apparent power measured in volt-ampere.
• P refers to true or real power measured in watt.
• PF = cos(phi) where phi is the phase angle between voltage and current under sinusoidal steady state.

Concept / Approach:

By definition PF = P / S. Therefore PF numerically indicates the fraction of the total apparent power that is converted into real work or heat within the load. Reactive components cause current to lead or lag, increasing S without contributing to P, which reduces PF below 1.0.

Step-by-Step Solution:

Verification / Alternative check:

If PF = 0.8, then P = 0.8 * S. Boosting PF with capacitors or synchronous condensers moves PF toward unity, increasing the ratio of P to S for the same current.

Why Other Options Are Wrong:

• “Only true for purely resistive loads” is incorrect; the definition holds for any sinusoidal load.
• “True only when reactive power is zero” is misleading; PF defines the ratio even when reactive power is present.
• “True only at unity PF” is wrong; unity is just a special case where P = S.

Common Pitfalls:

Confusing power factor with efficiency. Efficiency is output power over input power, while PF is P over S and captures phase displacement and waveform effects (for sinusoidal case, displacement only).

Final Answer:

True