Question

The overall power factor of a series R-L-C (R = resistance, L = inductance and C = capacitance) circuit with a resistance of 100 ohms, inductive reactance of 273.2 ohms and capacitive reactance of 100 ohms will be

• A. unity
• B. zero
• C. 0.5 leading
• D. 0.5 lagging

Hint:

For a series R-L-C circuit, the power factor is \( \frac{R}{Z} \), where \( Z \) is the total impedance, and the sign of the reactance determines if the power factor is leading or lagging.

Answer 1

The Correct Option is D

Step 1: Calculate the total reactance.
The total reactance \( X \) of the circuit is the difference between the inductive reactance and the capacitive reactance: \[ X = X_L - X_C = 273.2 - 100 = 173.2 \, \text{ohms}. \]
Step 2: Calculate the total impedance.
The total impedance \( Z \) of the circuit is given by: \[ Z = \sqrt{R^2 + X^2} = \sqrt{100^2 + 173.2^2} = \sqrt{10000 + 30001.44} = \sqrt{40001.44} \approx 200 \, \text{ohms}. \]
Step 3: Calculate the power factor.
The power factor \( \text{PF} \) is the cosine of the phase angle \( \theta \), where: \[ \cos \theta = \frac{R}{Z} = \frac{100}{200} = 0.5. \] Since the inductive reactance is greater than the capacitive reactance, the power factor is lagging.
Step 4: Conclusion.
Thus, the overall power factor is \( 0.5 \) lagging, which corresponds to option (D).