The figure shows a capacitor of capacitance $C$ connected to a battery via a switch, having a total charge…

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.

1. $\frac{1}{8}\frac{{\mathrm{Q}}^2}{\mathrm{C}}$
2. $\frac{3}{8}\frac{{\mathrm{Q}}^2}{\mathrm{C}}$
3. $\frac{3}{4}\frac{{\mathrm{Q}}^2}{\mathrm{C}}$
4. $\frac{5}{8}\frac{{\mathrm{Q}}^2}{\mathrm{C}}$

Solution

Asked in: JEE Main 2019 (12 Jan Shift 1)

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