A 120 V RMS source feeds a series RLC circuit with R = 40 Ω, L = 0.2 H, C = 0.001 F at f = 60 Hz. The power factor is approximately

A 120 V RMS source feeds a series RLC circuit with R = 40 Ω, L = 0.2 H, C = 0.001 F at f = 60 Hz. The power factor is approximately

• A. 0.48 lagging
• B. 0.72 leading
• C. 0.12 lagging
• D. 0.95 lagging

Answer: (A)

Power factor in a series RLC comes from the impedance angle. The impedance is Z = R + j(X_L − X_C), so the angle phi satisfies tan(phi) = (X_L − X_C) / R. Since a positive net reactance makes the current lag the voltage, the power factor will be lagging. First find the reactances at 60 Hz. ω = 2πf = 2π(60) ≈ 376.99 rad/s. X_L = ωL = 376.99 × 0.2 ≈ 75.4 Ω. X_C = 1/(ωC) = 1/(376.99 × 0.001) ≈ 2.65 Ω. Net reactance X = X_L − X_C ≈ 75.4 − 2.65 ≈ 72.75 Ω. Then tan(phi) = X/R ≈ 72.75 / 40 ≈ 1.82, so phi ≈ 61.7 degrees. Power factor = cos(phi) ≈ cos(61.7°) ≈ 0.48. Because the circuit is inductive (net positive reactance), the current lags the voltage. Therefore the approximately correct power factor is about 0.48 lagging.

Power factor in a series RLC comes from the impedance angle. The impedance is Z = R + j(X_L − X_C), so the angle phi satisfies tan(phi) = (X_L − X_C) / R. Since a positive net reactance makes the current lag the voltage, the power factor will be lagging.

First find the reactances at 60 Hz. ω = 2πf = 2π(60) ≈ 376.99 rad/s.

X_L = ωL = 376.99 × 0.2 ≈ 75.4 Ω.

X_C = 1/(ωC) = 1/(376.99 × 0.001) ≈ 2.65 Ω.

Net reactance X = X_L − X_C ≈ 75.4 − 2.65 ≈ 72.75 Ω.

Then tan(phi) = X/R ≈ 72.75 / 40 ≈ 1.82, so phi ≈ 61.7 degrees.

Power factor = cos(phi) ≈ cos(61.7°) ≈ 0.48.

Because the circuit is inductive (net positive reactance), the current lags the voltage. Therefore the approximately correct power factor is about 0.48 lagging.