A series $$LR$$ circuit connected with an ac source $$E = (25 \sin 1000t) \text{ V}$$ has a power factor of $$\frac{1}{\sqrt{2}}$$. If the source of emf is changed to $$E = (20 \sin 2000t) \text{ V}$$, the new power factor of the circuit will be :

For a series LR circuit, the power factor is given by:

$$\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + \omega^2 L^2}}$$

Case 1: $$\omega_1 = 1000 \text{ rad/s}$$, power factor $$= \frac{1}{\sqrt{2}}$$

$$\frac{1}{\sqrt{2}} = \frac{R}{\sqrt{R^2 + (1000)^2 L^2}}$$

Squaring both sides:

$$\frac{1}{2} = \frac{R^2}{R^2 + 10^6 L^2}$$

$$R^2 + 10^6 L^2 = 2R^2$$

$$10^6 L^2 = R^2$$

$$\omega_1 L = R \implies L = \frac{R}{1000}$$

Case 2: $$\omega_2 = 2000 \text{ rad/s}$$

$$\omega_2 L = 2000 \times \frac{R}{1000} = 2R$$

The new power factor is:

$$\cos\phi' = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$$

The correct answer is $$\frac{1}{\sqrt{5}}$$.