Search any question & find its solution
Question:
Answered & Verified by Expert
A circuit has a self-inductance of $1 \mathrm{H}$ and carries a current of $2 \mathrm{~A}$. To prevent sparking, when the circuit is switched off, a capacitor which can withstand $400 \mathrm{~V}$ is used. The least capacitance of capacitor connected across the switch must be equal to
Options:
Solution:
2464 Upvotes
Verified Answer
The correct answer is:
$25 \mu \mathrm{F}$
Energy stored in capacitor $=$ energystored in inductance
i.e., $\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \mathrm{LI}^{2}$
$\Rightarrow \mathrm{C}=\frac{\mathrm{LI}^{2}}{\mathrm{~V}^{2}}=\frac{1 \times(2)^{2}}{(400)^{2}}=25 \mu \mathrm{F}$
i.e., $\frac{1}{2} \mathrm{CV}^{2}=\frac{1}{2} \mathrm{LI}^{2}$
$\Rightarrow \mathrm{C}=\frac{\mathrm{LI}^{2}}{\mathrm{~V}^{2}}=\frac{1 \times(2)^{2}}{(400)^{2}}=25 \mu \mathrm{F}$
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.