In the figure shown, after the switch $‘S’$ is turned from position $‘A’$ to position $‘B’$, the energy dissipated in the circuit in terms of capacitance $‘C’$ and total charge $‘Q’$ is
In the figure shown, after the switch $‘S’$ is turned from position $‘A’$ to position $‘B’$, the energy dissipated in the circuit in terms of capacitance $‘C’$ and total charge $‘Q’$ is
- [JEE MAIN 2019]
- A
$\frac{1}{8}\frac{{{Q^2}}}{C}$
- B
$\frac{3}{8}\frac{{{Q^2}}}{C}$
- C
$\frac{5}{8}\frac{{{Q^2}}}{C}$
- D
$\frac{3}{4}\frac{{{Q^2}}}{C}$
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