Question

A circuit has a self-inductance of $1\,H$ and carries a current of $2\,A$. To prevent sparking, when the circuit is switched off, a capacitor which can withstand $400\, V$ 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)

$25 \, \mu F$

Answer 2

Energy stored in capacitor = energy stored in inductance
$\frac{1}{2} C V^{2} =\frac{1}{2} L I^{2}$
$C =\frac{L I^{2}}{V^{2}}$
$=\frac{1 \times(2)^{2}}{(400)^{2}}$
$=25\, \mu F$