If the resistance in a circuit is cut in half, how does the power factor change?

• A. Decreases by a factor of two
• B. Increases by a factor of two
• C. Remains the same
• D. Increases by four

Answer: (B)

Power factor is a measure of how much of the power is being used effectively and is equal to cos φ, the angle between the voltage and current. It depends on the total impedance, not just the resistance. For a circuit with resistance R and reactance X, the impedance is Z = √(R^2 + X^2) and the power factor is PF = R / Z.

If you cut the resistance in half but keep the same reactance, the new PF becomes (R/2) / √((R/2)^2 + X^2). This is typically smaller than the original PF when there is any reactance, because the numerator drops while the denominator doesn’t shrink as quickly. Only in the special case where there is no reactance (X = 0) does PF stay at 1, unchanged by changing R, since the current then simply scales with voltage and phase remains aligned.

For a concrete check, take R = 10 Ω and X = 6 Ω. Initially PF ≈ 10 / √(100 + 36) ≈ 0.857. Halving R to 5 Ω gives PF ≈ 5 / √(25 + 36) ≈ 0.641. So the power factor decreases, not increases, when resistance is cut in half in the presence of reactance. The stated option of increasing by a factor of two isn’t consistent with how PF behaves in general.