Question

In a series LR circuit with \(X_L = R\). Power factor is \(P_1\). If a capacitor of capacitance \(C\) with \(X_c = X_L\) is added to the circuit the power factor becomes \(P_2\). The ratio of \(P_1\) to \(P_2\) will be :

• A. \(1:3\)
• B. \(1:\sqrt{2}\)
• C. \(1:1\)
• D. \(1:2\)

Hint:

With \(X_c=X_L\) the circuit is at resonance and its power factor is 1.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The power factor is \(\cos\phi = \frac{R}{Z}\).

Step 2: Series LR circuit:
\(X_L = R\), so \(Z = \sqrt{R^2 + X_L^2} = \sqrt2R\). \(P_1 = \frac{R}{\sqrt2R} = \frac{1}{\sqrt2}\).

Step 3: After adding C:
\(X_C = X_L\), so the net reactance is zero and \(Z = R\). \(P_2 = 1\).

Step 4: Ratio:
\(\frac{P_1}{P_2} = \frac{1/\sqrt2}{1} = 1:\sqrt2\).

Final Answer:
The ratio \(P_1:P_2\) is \(1:\sqrt2\), option (B). \[ \boxed{1:\sqrt2} \]