Q 12-02-188JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
The figure shows a capacitor of capacitance $C$ connected to a battery via a switch, having a total charge $Q$ on it, in steady-state. When the switch S is turned from position A to position B, the energy dissipated in the circuit is
Answer: (B) $\dfrac38\dfrac{Q^2}{C}$
Initially $U_i = \dfrac{Q^2}{2C}$. At B the charge $Q$ shares between $C$ and $3C$ in parallel ($4C$):
$$U_f = \frac{Q^2}{2(4C)} = \frac{Q^2}{8C}$$
$$\Delta U = \frac{Q^2}{2C} - \frac{Q^2}{8C} = \frac38\frac{Q^2}{C}$$
Solution by Sreeraj P, M.Sc Physics