Question

If an inductor with inductive reactance, X_L=R is connected in series with resistor R across an A.C voltage, power factor comes out to be P₁. Now, if a capacitor with capacitive reactance, X_C=R is also connected in series with inductor and resistor in the same circuit, power factor becomes P₂. Find \(\frac{P_1}{P_2}\)

• A. \(\sqrt2:1\)
• B. \(1:\sqrt2\)
• C. 1 : 1
• D. 1 : 2

Answer 1

The Correct Option is B

Power Factor Analysis

1. Initial Power Factor (\( P_1 \)):

The impedance (\( Z \)) is calculated as: \[ Z = \sqrt{R^2 + R^2} = \sqrt{2}R \] The power factor is given by: \[ P_1 = \cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{2}R} = \frac{1}{\sqrt{2}} \] ....(1)

2. Power Factor in Resonance (\( P_2 \)):

When a capacitor is connected in series and the circuit is in resonance: \[ Z = R \] The power factor is: \[ P_2 = \cos \phi = \frac{R}{Z} = \frac{R}{R} = 1 \] ....(2)

3. Ratio of Power Factors:

The ratio of power factors is: \[ \frac{P_1}{P_2} = \frac{\frac{1}{\sqrt{2}}}{1} = \frac{1}{\sqrt{2}} = \frac{1}{2} \] Therefore: \[ P_1 : P_2 = 1 : 2 \]

Final Answer:

The correct answer is (B): 1 : 2.