A one microfarad condenser is charged to $50 \mathrm{~V}$. The charging battery is then disconnected and a…
- $0.75 \mathrm{~A}$
- $0.25 \mathrm{~A}$
- $0.50 \mathrm{~A}$
- $1.00 \mathrm{~A}$
Solution
$\begin{aligned} & \frac{\mathrm{q}}{\mathrm{C}}=-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}=(-\mathrm{L}) \frac{\mathrm{dq}}{\mathrm{dt}}=\frac{\mathrm{di}}{\mathrm{dq}} \Rightarrow \int_{\mathrm{Q}_0}^0 \mathrm{qdq}=(-\mathrm{LC}) \int_0^{\mathrm{i}} \max \mathrm{idi} \\ & \left.\Rightarrow \frac{\mathrm{q}^2}{2}\right|_{\mathrm{Q}_0} ^0=\left.(-\mathrm{LC}) \frac{\mathrm{i}^2}{2}\right|_0 ^{\mathrm{i}^{\mathrm{max}}} \Rightarrow 0-\mathrm{Q}_0^2=(-\mathrm{LC})\left(\mathrm{i}_{\max }{ }^2\right) \\ & \Rightarrow \mathrm{i}_{\max }=\frac{\mathrm{Q}_0}{\sqrt{\mathrm{LC}}}=\frac{\mathrm{CV}}{\sqrt{\mathrm{LC}}}=50 \sqrt{\frac{1 \times 10^{-6} \mathrm{~F}}{10 \times 10^{-3} \mathrm{H}}} \\ & =50 \times 10^{-2} \mathrm{~A} \\ & \Rightarrow \mathrm{i}=0.5 \mathrm{~A}\end{aligned}$Asked in: MHT CET 2022 (07 Aug Shift 2)