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: Plan:
Use the phase angle.

Step 2: Steps:
With only L and R, $\tan\phi = \frac{X_L}{R} = 1$, so $\phi = 45^{\circ}$ and $\cos\phi = \frac{1}{\sqrt2}$. With $X_C = X_L$ the phase angle is $0$, so $\cos\phi = 1$. The ratio is $\frac{1}{\sqrt2}:1 = 1:\sqrt2$.

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