In the circuit shown, each capacitor has a capacitance $C$. The emf of the cell is $E$. If the switch $S$ is closed
  • A
    positive charge will flow out of the positive terminal of the cell
  • B
    positive charge will enter the positive terminal of the cell
  • Cthe amount of charge flowing through the cell will be $4/3$ $CE$.
  • D$A$ and $C$ both
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