For the series LCR circuit, $R=\frac{X_L}{2}=2 X_c$. The impedance of the circuit and the phase difference…

For the series LCR circuit, $R=\frac{X_L}{2}=2 X_c$. The impedance of the circuit and the phase difference between V and I will be
  1. $\frac{\sqrt{5}}{2} \mathrm{R}, \tan ^{-1}\left(\frac{1}{2}\right)$
  2. $\frac{\sqrt{13}}{2} R, \tan ^{-1}\left(\frac{3}{2}\right)$
  3. $\sqrt{5} R, \tan ^{-1}(1)$
  4. $\sqrt{13} \mathrm{R}, \tan ^{-1}(2)$

Solution

$\operatorname{Impedance}(\mathrm{Z})=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^2}...(i)$ Given $R=\frac{X_L}{2} \Rightarrow X_L=2 R$...(ii) Also, $\mathrm{R}=2 \mathrm{X}_{\mathrm{C}} \Rightarrow \mathrm{X}_{\mathrm{C}}=\frac{\mathrm{R}}{2}$...(iii) Substituting (ii) and (iii) in (i), $Z=\sqrt{R^2+\left(2 R-\frac{R}{2}\right)^2}=\sqrt{\frac{13}{4} R^2}=\frac{\sqrt{13}}{2} R$ Phase difference $\phi=\tan ^{-1}\left(\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}\right)=\tan ^{-1}\left(\frac{3}{2}\right)$ Since the impedance values provided in the options are all different, calculating the impedance will be enough to identify the correct answer.

Asked in: MHT CET 2024 (10 May Shift 2)

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