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In the following circuit, the switch $S$  is closed at $t = 0.$  The charge on the capacitor  $C_1$  as a function of time will be given by $\left( {{C_{eq}}\, = {\kern 1pt} \,\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).$
  • A${C_{eq}}E\,[1 - \exp ( - t/R{C_{eq}})]$
  • B${C_1}E\,[1 - \exp ( - tR/{C_1})]$
  • C${C_2}E\,[1 - \exp ( - t/R{C_2})]$
  • D${C_{eq}}E\,\exp ( - t/R{C_{eq}})$
JEE MAIN 2018, Diffcult
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