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:

Block 1 ($$R_1, R_2$$): Two $$2\ \Omega$$ resistors in parallel.

$$R_{12} = \frac{2 \times 2}{2 + 2} = 1\ \Omega$$

Block 2 ($$R_4, R_5$$): A $$20\ \Omega$$ and a $$5\ \Omega$$ resistor in parallel.

$$R_{45} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \Omega$$

$$R_{ext} = R_{12} + R_3 + R_{45} + R_6 = 1\ \Omega + 2\ \Omega + 4\ \Omega + 2\ \Omega = 9\ \Omega$$

$$R_{total} = R_{ext} + r = 9\ \Omega + 3\ \Omega = 12\ \Omega$$

$$I_{total} = \frac{24\ \text{V}}{12\ \Omega} = 2\ \text{A}$$

$$I_4 = I_{total} \times \frac{R_5}{R_4 + R_5} = 2 \times \frac{5}{25} = \frac{2}{5}\ \text{A}$$

$$I_5 = I_{total} \times \frac{R_4}{R_4 + R_5} = 2 \times \frac{20}{25} = \frac{8}{5}\ \text{A}$$