As shown in the figure, a network of resistors is connected to a battery of $24 V$ with an internal resistance of $3 Ω$ . 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:

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Equivalent resistance of circuit R eq = 3 + 2 × 2 2 + 2 + 2 + 20 × 5 20 + 5 + 2 = 12 Ω Current through battery i = 24 12 = 2 A Now, I 4 = R 5 R 4 + R 5 × 2 = 5 20 + 5 × 2 = 2 5 A and I 5 = i - I 4 = 2 - 2 5 = 8 5 A

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