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A capacitor having capacity of 2 $μ$ F charged to 200 V and then the plates of the capacitor are connected to a resistance wire. The heat produced in joule will be

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4 x 10¹⁰

2 x 10^-2

4 x 10⁴

4 x 10^-2

answer is B.

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Detailed Solution

$Energy loss is stored energy = \frac{1}{2} {CV}^{2}$

$\frac{1}{2} \times 2 \times 10^{- 6} \times {(200)}^{2} = 4 \times 10^{- 2} J$

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