Question

A circuit has a self-inductance of $1H$ and carries a current of $2A$. To prevent sparking, when the circuit is switched off, a capacitor which can withstand $400V$ is used. The least capacitance of capacitor connected across the switch must be equal to

• A. $50\mu F$
• B. $25\mu F$
• C. $100\mu F$
• D. $12.5\mu F$

Answer 1

Answer: (B)

$25\mu F$

Energy stored in capacitor $=$ energy stored in inductance
i.e., $\frac{1}{2}C{V}^2=\frac{1}{2}{LI}^2$
$\Rightarrow C=\frac{{LI}^2}{V^2}=\frac{1\times {\left(2\right)}^2}{{\left(400\right)}^2}=25\mu F$