Search any question & find its solution
Question:
Answered & Verified by Expert
A condenser of capacity $C$ is charged to a potential difference of $V_1$. The plates of the condenser are then connected to an ideal inductor of inductance $L$. The current through the inductor when the potential difference across the condenser reduces to $V_2$ is
Options:
Solution:
2294 Upvotes
Verified Answer
The correct answer is:
$\left(\frac{\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}{\mathrm{L}}\right)^{\frac{1}{2}}$
\(\begin{aligned} & \frac{\mathrm{CV}^2}{2}+\frac{1}{2} \mathrm{Li}^2=\text { constant } \\ & \text { Initially } \mathrm{i}=0, \mathrm{~V}=\mathrm{V}_1 \\ & \text {Total energy }=\frac{1}{2} \mathrm{CV}_1^2 \\ & \mathrm{~When } \mathrm{V} \rightarrow \mathrm{V}_2 \\ & \frac{1}{2} \mathrm{C}\left(\mathrm{V}_2\right)^2+\frac{1}{2} \mathrm{Li}^2=\frac{1}{2} \mathrm{CV}_1^2 \\ & \mathrm{Li}^2=\mathrm{C}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)^2 \\ & \mathrm{i}^2=\frac{\mathrm{C}}{\mathrm{L}}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right) \\ & \mathrm{i}=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}\left(\mathrm{V}_1^2-\mathrm{V}_2^2\right)}\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.