A sinusoidal voltage $$V(t) = 210\sin 3000t$$ volt is applied to a series LCR circuit in which $$L = 10$$ mH, $$C = 25$$ $$\mu$$F and $$R = 100\Omega$$. The phase difference $$\Phi$$ between the applied voltage and resultant current will be

Given: $$V(t) = 210\sin 3000t$$ volt, so the angular frequency is $$\omega = 3000$$ rad/s.

Given: $$L = 10$$ mH $$= 10 \times 10^{-3}$$ H, $$C = 25$$ $$\mu$$F $$= 25 \times 10^{-6}$$ F, $$R = 100$$ $$\Omega$$.

The inductive reactance is:

$$X_L = \omega L = 3000 \times 10 \times 10^{-3} = 30 \text{ }\Omega$$

The capacitive reactance is:

$$X_C = \frac{1}{\omega C} = \frac{1}{3000 \times 25 \times 10^{-6}} = \frac{1}{0.075} = 13.33 \text{ }\Omega$$

The net reactance is:

$$X_L - X_C = 30 - 13.33 = 16.67 \text{ }\Omega$$

The phase difference is:

$$\tan\phi = \frac{X_L - X_C}{R} = \frac{16.67}{100} = 0.1667 \approx 0.17$$

$$\phi = \tan^{-1}(0.17)$$

Hence, the correct answer is Option A.