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:

We need to find the ratio of power factors $$P_1$$ to $$P_2$$ for a series LR circuit before and after adding a capacitor.

Case 1: Series LR circuit with $$X_L = R$$.

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

Power factor: $$P_1 = \cos\phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Case 2: Series LCR circuit with $$X_C = X_L = R$$.

Net reactance: $$X_L - X_C = R - R = 0$$

Impedance: $$Z_2 = \sqrt{R^2 + 0} = R$$

Power factor: $$P_2 = \cos\phi_2 = \frac{R}{R} = 1$$

Ratio:

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

Therefore, $$P_1 : P_2 = 1 : \sqrt{2}$$.

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