As shown in the figure, a network of resistors is connected to a battery of 24V 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 :

Question

$\mathrm{I_4=\frac{8}{5}A}$ and $\mathrm{I_5=\frac{2}{5}A}$

$\mathrm{I_4=\frac{6}{5}A}$ and $\mathrm{I_5=\frac{24}{5}A}$

$\mathrm{I_4=\frac{2}{5}A}$ and $\mathrm{I_5=\frac{8}{5}A}$

$\mathrm{I_4=\frac{24}{5}A}$ and $\mathrm{I_5=\frac{6}{5}A}$

Solution

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