For a series LCR circuit, $R = X_L = 3 X_C$. The impedance of the circuit and the phase difference between voltage V and current I, respectively, will be

$\frac{R\sqrt{(13)}}{3}Ω$ and $\tan-^{-1}\frac{2}{3}$

The correct answer is Option (3) → $\frac{R\sqrt{(13)}}{3}Ω$ and $\tan-^{-1}\frac{2}{3}$

Given: $R = X_L = 3 X_C$

Impedance of series LCR circuit: $Z = \sqrt{R^2 + (X_L - X_C)^2}$

Substitute: $X_L - X_C = 3X_C - X_C = 2 X_C$

$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{(3 X_C)^2 + (2 X_C)^2} = \sqrt{9 X_C^2 + 4 X_C^2} = \sqrt{13} X_C$

Phase difference: $\tan \phi = \frac{X_L - X_C}{R} = \frac{2 X_C}{3 X_C} = \frac{2}{3}$

$\phi = \tan^{-1} \frac{2}{3}$

Impedance $Z = \sqrt{13} X_C$, Phase difference $\phi = \tan^{-1}(2/3)$