Five equal resistances each of $2R$ are connected as shown in the figure. $A$ battery of $V$ volts is connected between $A$ and $B$. Then,the current through $FC$ is:

  • A
    $\frac{V}{4R}$
  • B
    $\frac{V}{8R}$
  • C
    $\frac{V}{R}$
  • D
    $\frac{V}{2R}$

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