The capacitor of capacitance $C$ in the circuit shown is fully charged initially resistance is $R$. After the switch $S$ is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is
  • A$\frac{R C}{2}$
  • B$R C \ln 2$
  • C$2 R C \ln 2$
  • D$\frac{R C \ln 2}{2}$
KVPY 2012, Diffcult
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