One end each of a resistance $r$,a capacitor $C$,and a resistance $2r$ are connected together. The other ends are respectively connected to the positive terminals of batteries $P$,$Q$,and $R$ having emfs $E$,$E$,and $2E$. The negative terminals of the batteries are then connected together. In this circuit,with steady current,the potential drop across the capacitor is:

  • A
    $\frac{E}{3}$
  • B
    $\frac{E}{2}$
  • C
    $\frac{2E}{3}$
  • D
    $E$

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