Power factor computation: If the apparent power S of an AC load is 5 W (VA) and the true (real) power P is 4 W, what is the power factor (dimensionless)?

Introduction:Power factor (pf) quantifies how effectively apparent power is converted into real work in AC circuits. It is the ratio of true power to apparent power and ranges from 0 to 1 in magnitude. A higher pf indicates less reactive content and better utilization of the supply.

Given Data / Assumptions:

• Apparent power S = 5 W (VA)
• True (real) power P = 4 W
• Steady-state sinusoidal conditions
• pf is taken as magnitude (sign not considered here)

Concept / Approach:

Power factor is defined as pf = P / S for sinusoidal systems. This equals cos(φ), where φ is the phase angle between voltage and current. Knowing P and S is sufficient to compute pf directly.

Step-by-Step Solution:

Verification / Alternative check:

Reactive power Q can be inferred: Q = sqrt(S^2 - P^2) = sqrt(25 - 16) = 3 VAR. Then cos(φ) = P / S = 4/5 = 0.80, consistent.

Why Other Options Are Wrong:

• 0.60 and 0.67: Do not match P / S.
• 1.00: Would mean purely resistive with no reactive power; not true because P < S.
• 1.25: Impossible since pf ≤ 1.

Common Pitfalls:

• Including units with power factor; it is dimensionless.
• Confusing apparent power units; in steady sinusoidal analysis W and VA are numerically equal.

Final Answer:

0.80