As shown in the figure, a network of resistors is connected to a battery of $24 V$ with an internal resistance of $3Ω$ . 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 e q ​ = 3 + ( 2 + 2 2 × 2 ​ ) + 2 + ( 20 + 5 20 × 5 ​ ) + 2 = 12Ω Current through battery i = 12 24 ​ = 2 A Now, I 4 ​ = R 4 ​ + R 5 ​ R 5 ​ ​ × 2 = 20 + 5 5 ​ × 2 = 5 2 ​ A and I 5 ​ = i − I 4 ​ = 2 − 5 2 ​ = 5 8 ​ A

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