In the $R-C$ circuit shown in the figure the total energy of $3.6 \times 10^{-3}\ J$ is dissipated in the $10$ $\Omega$ resistor when the switch $S$ is closed. The initial charge on the capacitor is.....$\mu C$
Medium
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When $S$ is closed, the energy stored in the capacitor will be dissipated through resistor.
Thus, $\frac{Q^{2}}{2 C}=3.6 \times 10^{-3}$
$Q^{2}=3.6 \times 10^{-3} \times 2 \times 2 \times 10^{-6}=144 \times 10^{-10}$
$Q=12 \times 10^{-5} C=120 \mu C$
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