A series LCR circuit containing an AC source of 100 V has an inductor and a capacitor of reactances $24 \Omega$ and $16 \Omega$ respectively. If a resistance of $6 \Omega$ is connected in series, then the potential difference across the series combination of inductor and capacitor only is

• Correct answer
  A

  80 V
• B

  400 V
• C

  8 V
• D

  40 V

Solution

$\begin{aligned} & \mathrm{X}_{\mathrm{C}}=24 \Omega \\ & \mathrm{X}_{\mathrm{L}}=16 \Omega \\ & \mathrm{R}=6 \Omega \\ & \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2} \\ & =\sqrt{6^2+8^2} \\ & =\sqrt{100}=10 \Omega\end{aligned}$
$\mathrm{i}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{100}{10}=10 \mathrm{~A}$ Net voltage across capacitor and inductor $\begin{aligned} & =\mathrm{i}\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right) \\ & =10(24-16) \\ & =10(8) \\ & =80 \mathrm{~V} \end{aligned}$

Hence, the correct answer is option: 'A'.