Series LCR Circuit question from MHT CET 2024 (10 May, Shift 2)

Answer: (b)

1. From $R = \dfrac{X_L}{2} = 2X_C$: $X_L = 2R$, $X_C = \dfrac{R}{2}$.
2. $X_L - X_C = \dfrac{3R}{2}$
3. $Z = \sqrt{R^2 + \dfrac{9R^2}{4}} = \dfrac{\sqrt{13}}{2}R$
4. $\tan\phi = \dfrac{X_L - X_C}{R} = \dfrac{3}{2}$
5. So the answer is (b) $\dfrac{\sqrt{13}}{2}R,\ \tan^{-1}\left(\dfrac{3}{2}\right)$.