$A$ wire of resistance $10\, \Omega$ is bent into a circle. $P$ and $Q$ are two points on the circumference of the circle that divide the circle into a quadrant. When these two points are connected to a battery of $3\, V$ and internal resistance $1\, \Omega$, the currents passing through the two parts of the circle are:

  • A
    $\frac{6}{23}\, A$ and $\frac{18}{23}\, A$
  • B
    $\frac{5}{26}\, A$ and $\frac{15}{26}\, A$
  • C
    $\frac{4}{25}\, A$ and $\frac{12}{25}\, A$
  • D
    $\frac{3}{35}\, A$ and $\frac{9}{25}\, A$

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