As shown in the figure,a network of resistors is connected to a battery of $24\,V$ with an internal resistance of $3\,\Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are:

  • A
    $I_4 = \frac{8}{5}\,A$ and $I_5 = \frac{2}{5}\,A$
  • B
    $I_4 = \frac{24}{5}\,A$ and $I_5 = \frac{6}{5}\,A$
  • C
    $I_4 = \frac{6}{5}\,A$ and $I_5 = \frac{24}{5}\,A$
  • D
    $I_4 = \frac{2}{5}\,A$ and $I_5 = \frac{8}{5}\,A$

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