Question

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:

• A. $60\mu C$
• B. $120\mu C$
• C. $60\sqrt{2}\mu C$
• D. $\frac{60}{\sqrt{2}}\mu C$

Answer 1

The correct option is B $120\mu C$

When S is closed ,the energy stored in the capacitor will be dissipated through resistor.

Thus, $\frac{Q^2}{2C}=3.6\times {10}^{-3}$

$Q^2=3.6\times {10}^{-3}\times 2\times 2\times {10}^{-6}=144\times {10}^{-10}$

$Q=12\times {10}^{-5}C=120\mu C$