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

  • 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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