Q. 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 least capacitance of the capacitor connected across the switch must be equal to

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A

$50\, \mu F$

B

$25\, \mu F$

C

$100\, \mu F$

D

$12.5\, \mu F$

Solution:

$1 / 2\, CV ^{2}=1 / 2 LI ^{2}$
therefore, $C = L I ^{2} / V ^{2}=1 \times 2^{2} / 400^{2}=25\, \mu F$

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