1. **State the problem:** We have two wires, aluminium and copper, connected in parallel. Given lengths, currents, diameters, and resistivities, we need to find the voltage drop across the parallel branch. 2. **Given data:** - Length of aluminium wire, $L_{Al} = 45.8$ m - Length of copper wire, $L_{Cu} = 32.6$ m - Total current, $I_{total} = 10.44$ A - Current in aluminium wire, $I_{Al} = 6.99$ A - Diameter of aluminium wire, $d_{Al} = 1.24$ mm = $1.24 \times 10^{-3}$ m - Resistivity of aluminium, $\rho_{Al} = 0.0182 \times 10^{-6}$ $\Omega$m - Resistivity of copper, $\rho_{Cu} = 0.0261 \times 10^{-6}$ $\Omega$m 3. **Calculate the cross-sectional area of aluminium wire:** $$ A_{Al} = \pi \left(\frac{d_{Al}}{2}\right)^2 = \pi \left(\frac{1.24 \times 10^{-3}}{2}\right)^2 = \pi (0.62 \times 10^{-3})^2 = \pi \times 0.3844 \times 10^{-6} = 1.207 \times 10^{-6} \text{ m}^2 $$ 4. **Calculate resistance of aluminium wire:** $$ R_{Al} = \frac{\rho_{Al} L_{Al}}{A_{Al}} = \frac{0.0182 \times 10^{-6} \times 45.8}{1.207 \times 10^{-6}} = \frac{0.83356 \times 10^{-6}}{1.207 \times 10^{-6}} = 0.69 \ \Omega $$ 5. **Calculate voltage drop across aluminium wire:** $$ V = I_{Al} \times R_{Al} = 6.99 \times 0.69 = 4.82 \text{ V} $$ 6. **Since wires are in parallel, voltage drop across copper wire is the same:** $$ V_{parallel} = 4.82 \text{ V} $$ **Final answer:** The voltage drop across the parallel branch is approximately **4.82 V**.