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A circuit has a self-inductance of 1 henry 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 value of capacitance of the capacitor connected across the switch must be equal to

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50 mF

25 mF

100 mF

12.5 mF

answer is B.

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Detailed Solution

$\begin{array}{rcl} \frac{1}{2} {Li}^{2} & = & \frac{1}{2} {CV}^{2} (L = 1, V = 400) \\ C & = & 25 μF \end{array}$

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