A circuit having a self inductance of 1 henry carries a current of 1 A . To prevent the sparking when the…

A circuit having a self inductance of 1 henry carries a current of 1 A . To prevent the sparking when the circuit is broken, a capacitor which can withstand 500 V is connected across the switch. What is the minimum value of the capacitance of the capacitor?
  1. $2 \mu \mathrm{~F}$
  2. $4 \mu \mathrm{~F}$
  3. $6 \mu \mathrm{~F}$
  4. $8 \mu \mathrm{~F}$

Solution

The energy stored in the inductance is $\mathrm{U}=\frac{1}{2} \mathrm{LI}^2$ $\mathrm{U}=\frac{1}{2} \times 1 \times 1^2=0.5 \mathrm{~J}$
This energy must be transferred to the capacitor. Energy stored by a capacitor $\mathrm{U}=\frac{1}{2} \mathrm{CV}^2$ $\begin{array}{ll}\therefore & \frac{1}{2} C V^2=\frac{1}{2} L I^2 \\ \therefore & C=L\left(\frac{I}{V}\right)^2=1\left(\frac{1}{500}\right)^2=4 \mu F\end{array}$

Asked in: MHT CET 2024 (02 May Shift 1)

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