To solve this problem, we need to analyze the phase difference in a series LCR circuit.$\\$Given:$\\$• When the inductor $L$ is removed, the phase difference $\phi_1 = \frac{\pi}{3}. \\$• When the capacitor $C$ is removed, the phase difference $\phi_2 = \frac{\pi}{3}. \\$In a series LCR circuit, the phase difference $\phi$ is given by:$\\ \tan \phi = \frac{X_L - X_C}{R} \\$Where:$\\$• $X_L = \omega L$ is the inductive reactance.$\\$• $X_C = \frac{1}{\omega C}$ is the capacitive reactance.$\\$• $R$ is the resistance.$\\$Case 1: When $L$ is removed, $X_L = 0. \\ \tan \phi_1 = \frac{-X_C}{R} = \frac{-1/\omega C}{R} = \frac{-1}{\omega CR} = \tan\left(\frac{\pi}{3}\right) \\$This implies:$\\ \frac{-1}{\omega CR} = \sqrt{3} \\$Case 2: When $C$ is removed, $X_C = 0. \\ \tan \phi_2 = \frac{X_L}{R} = \frac{\omega L}{R} = \tan\left(\frac{\pi}{3}\right) \\$This implies:$\\ \frac{\omega L}{R} = \sqrt{3} \\$Since both cases give the same magnitude of phase difference,$\\$we equate the magnitudes:$\\ \frac{1}{\omega CR} = \frac{\omega L}{R} \\$Simplifying gives:$\\ \omega^2 LC = 1 \\$The power factor $\cos \phi$ is given by:$\\ \cos \phi = \frac{R}{Z} \\$Where $Z$ is the impedance of the circuit:$\\ Z = \sqrt{R^2 + (X_L - X_C)^2} \\$For resonance, $X_L = X_C,$ so $Z = R. \\$Thus, the power factor $\cos \phi = \frac{R}{R} = 1.0 \\$Therefore, the power factor of the circuit is $1.0. \\$This corresponds to Option 2.