Explanation

(c) When the key is in contact with 1, then energy stored in
the condenser $= \frac{1}{2}CE^{2}$
But when the key is thrown to contact 2, total heat
$\ H = I^{2}(500 + 330) = \frac{1}{2}CE^{2}$
$H_{1} = I^{2}(500)$
$\frac{\ H_{1}}{H} = \frac{R_{1}}{(R_{1} + R_{2})}$
$H_{1} = \frac{500}{830} \times \frac{1}{2} \times 5 \times 10^{- 6} \times (200)^{2}$
$H_{1} = 60 \times 10^{- 3}$ J