A series LR circuit is connected to an ac source of frequency $$\omega$$ and the inductive reactance is equal to 2R. A capacitance of capacitive reactance equal to R is added in series with L and R. The ratio of the new power factor to the old one is :

First, we have a series LR circuit connected to an AC source of frequency ω. The inductive reactance $$X_L$$ is given as 2R, where R is the resistance. We need to find the power factor for this original circuit.

The power factor is defined as $$\cos \phi$$, where $$\phi$$ is the phase angle between the voltage and current. For a series LR circuit, the impedance $$Z_1$$ is calculated as:

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

Substituting $$X_L = 2R$$:

$$ Z_1 = \sqrt{R^2 + (2R)^2} = \sqrt{R^2 + 4R^2} = \sqrt{5R^2} = R\sqrt{5} $$

The power factor for the original circuit is:

$$ \cos \phi_1 = \frac{R}{Z_1} = \frac{R}{R\sqrt{5}} = \frac{1}{\sqrt{5}} $$

Now, a capacitor is added in series with capacitive reactance $$X_C = R$$. The new circuit is an LCR series circuit with resistance R, inductive reactance $$X_L = 2R$$, and capacitive reactance $$X_C = R$$.

The net reactance $$X$$ is:

$$ X = X_L - X_C = 2R - R = R $$

The impedance $$Z_2$$ for the new circuit is:

$$ Z_2 = \sqrt{R^2 + X^2} = \sqrt{R^2 + (R)^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} $$

The power factor for the new circuit is:

$$ \cos \phi_2 = \frac{R}{Z_2} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} $$

We need the ratio of the new power factor to the old power factor:

$$ \text{Ratio} = \frac{\cos \phi_2}{\cos \phi_1} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{5}}} = \frac{1}{\sqrt{2}} \times \sqrt{5} = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}} $$

Comparing with the options, $$\sqrt{\frac{5}{2}}$$ corresponds to option D.

Hence, the correct answer is Option D.