As shown in the figure a network of resistors is connected to a battery of 24V with an internal resistance of 3 $Ω$ . The current through the resistors R₄ and R₅are I₄ and I₅ respectively. The values of I₄ and I₅ are :

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$I_{4} = \frac{8}{5} A and I_{5} = \frac{2}{5} A$

$I_{4} = \frac{24}{5} A and I_{5} = \frac{6}{5} A$

$I_{4} = \frac{6}{5} A and I_{5} = \frac{24}{5} A$

$I_{4} = \frac{2}{5} A and I_{5} = \frac{8}{5} A$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \frac{24}{1 + 2 + 4 + 2 + 3} = \frac{24}{12} I = 2 A I_{4} = (\frac{5}{25}) (2) = \frac{2}{5} A \\ I_{5} = (\frac{20}{25}) (2) = \frac{8}{5} A \end{array}$

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