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