A circuit has self-inductance ' $L$ ' and carries a current ' $\mathrm{I}$ '. To prevent sparking when the…
A circuit has self-inductance ' $L$ ' and carries a current ' $\mathrm{I}$ '. To prevent sparking when the circuit is switched off, a capacitor which can withstand potential difference ' $\mathrm{V}$ ' is used. The least capacitance is
The energy stored in the inductance is given by $\mathrm{U}=\frac{1}{2} \mathrm{LI}^2$ Energy stored by the capacitor is given by $\mathrm{W}=\frac{1}{2} \mathrm{CV}^2$ This energy must be transferred to the capacitor.
$\begin{aligned}
& \therefore \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}^2 \\
& \mathrm{C}=\mathrm{L}\left(\frac{\mathrm{I}}{\mathrm{V}}\right)^2
\end{aligned}$