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).

Answer 2

We can find the power factor using the impedance triangle. The circuit has resistance \( R = 100 \, \Omega \), inductive reactance \( X_L = 273.2 \, \Omega \), and capacitive reactance \( X_C = 100 \, \Omega \). In a series R-L-C circuit, the impedance triangle has \( R \) as the base and the net reactance \( X = X_L - X_C \) as the perpendicular side. Here, \( X = 273.2 - 100 = 173.2 \, \Omega \), and the impedance is \( Z = \sqrt{R^2+X^2} = \sqrt{100^2+173.2^2} = \sqrt{10000+30000} = \sqrt{40000} = 200 \, \Omega \). The power factor is \( \cos\phi = R/Z \). Let's check each option.

1. Unity: This would require \( X = 0 \) (purely resistive circuit), but here \( X_L \neq X_C \), so the circuit is not purely resistive and the power factor cannot be unity.
2. Zero: This would require \( R = 0 \), but the circuit clearly has \( R = 100 \, \Omega \), ruling this out.
3. 0.5 leading: A leading power factor would require \( X_C > X_L \) (net capacitive circuit), but here \( X_L = 273.2 \, \Omega \) is larger than \( X_C = 100 \, \Omega \), so the circuit is net inductive, not capacitive.
4. 0.5 lagging: Since \( \cos\phi = R/Z = 100/200 = 0.5 \), and the net reactance is inductive (\( X_L > X_C \)), the current lags the voltage, making this 0.5 lagging.

The impedance-triangle calculation confirms the power factor is 0.5 lagging.

Therefore, the correct answer is 0.5 lagging.