NEET 2020 Physics Question 169 Solution

When $$L$$ is removed, the remaining circuit is a series RC circuit. The phase angle between voltage and current is given by: $$\tan\phi_1 = \frac{X_C}{R}$$ Given $$\phi_1 = \frac{\pi}{3}$$, so: $$\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{X_C}{R} \implies X_C = \sqrt{3} R$$.

When $$C$$ is removed, the remaining circuit is a series RL circuit. The phase angle is: $$\tan\phi_2 = \frac{X_L}{R}$$ Given $$\phi_2 = \frac{\pi}{3}$$, so: $$\tan\left(\frac{\pi}{3}\right) = \sqrt{3} = \frac{X_L}{R} \implies X_L = \sqrt{3} R$$.

In the full series LCR circuit, total net reactance is: $$X = X_L - X_C = \sqrt{3}R - \sqrt{3}R = 0$$ Impedance of the circuit is: $$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$$. The circuit is at electrical resonance.

The power factor of an AC circuit is defined as: $$\text{Power factor} = \cos\phi = \frac{R}{Z} = \frac{R}{R} = 1.0$$. Matches Option 3.