In a series $$LR$$ circuit $$X_L = R$$ and power factor of the circuit is $$P_1$$. When capacitor with capacitance $$C$$ such that $$X_L = X_C$$ is put in series, the power factor becomes $$P_2$$. The ratio $$\dfrac{P_1}{P_2}$$ is

In a series $$LR$$ circuit, $$X_L = R$$, and the power factor is $$P_1$$. When a capacitor with $$X_C = X_L$$ is added in series, the power factor becomes $$P_2$$. We need to find $$\dfrac{P_1}{P_2}$$.

The power factor is defined as:

$$P_1 = \cos\phi_1 = \frac{R}{Z_1}$$

For a series $$LR$$ circuit, the impedance is:

$$Z_1 = \sqrt{R^2 + X_L^2}$$

Given $$X_L = R$$:

$$Z_1 = \sqrt{R^2 + R^2} = R\sqrt{2}$$

$$P_1 = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$$

When a capacitor with $$X_C = X_L$$ is added in series, resonance occurs since $$X_L = X_C$$:

$$Z_2 = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$$

$$P_2 = \frac{R}{R} = 1$$

$$\frac{P_1}{P_2} = \frac{1/\sqrt{2}}{1} = \frac{1}{\sqrt{2}}$$

The correct answer is Option B: $$\dfrac{1}{\sqrt{2}}$$.