QUESTION 7 A piece of aluminium wire, 45.8 metres long, is connected in parallel with a copper wire that is 32.6 meters long. When a current of 10.4 A is passed through the combination, the current flowing in the aluminium wire is measured as 6.9 A . The diameter of the aluminium wire is 1.24 mm . Determine the diameter of the copper wire if the resistivity of aluminium is \( 0.0182 \mu \Omega \cdot \mathrm{~m} \) and the resistivity of copper is 0.0261 \( \mu \Omega \cdot \mathrm{~m} \)

To solve for the diameter of the copper wire, we can use Ohm's Law and the formula for resistance based on resistivity. First, let's find the resistance of the aluminium wire: 1. **Calculate the resistance of the aluminium wire** (R_al): \[ R_{\text{al}} = \rho_{\text{al}} \cdot \frac{L_{\text{al}}}{A_{\text{al}}} \] Where: \( \rho_{\text{al}} = 0.0182 \, \mu\Omega \cdot m \), \( L_{\text{al}} = 45.8 \, m \), and \( A_{\text{al}} = \frac{\pi d_{\text{al}}^2}{4} \), with \( d_{\text{al}} = 1.24 \, mm = 0.00124 \, m \). The area \( A_{\text{al}} = \frac{\pi (0.00124)^2}{4} = 4.81 \times 10^{-7} \, m^2 \). Substitute into the resistance formula: \[ R_{\text{al}} = 0.0182 \, \mu\Omega \cdot m \cdot \frac{45.8}{4.81 \times 10^{-7}} = 1.72 \, \mu\Omega \] 2. **Calculate voltage across the aluminium wire (V_al)** using Ohm's Law (V = IR): \[ V_{\text{al}} = I_{\text{al}} \cdot R_{\text{al}} = 6.9 \, A \cdot 1.72 \, \mu\Omega = 0.01188 \, mV \] 3. **Determine the total voltage (V_total)** needed when the total current (10.4 A) is flowing: Since the voltage is the same across both wires in parallel, we also get: \[ V_{\text{total}} = I_{\text{total}} \cdot (R_{\text{al}} + R_{\text{cu}}) \] Now substitute to find the resistance of the copper wire \( R_{\text{cu}} \): \[ V_{\text{total}} = 0.01188 \, mV \implies I_{\text{cu}} = I_{\text{total}} - I_{\text{al}} = 10.4 \, A - 6.9 \, A = 3.5 \, A \] 4. **Calculate Resistance (R_cu)** of the copper wire using voltage: \[ R_{\text{cu}} = \frac{V_{\text{total}}}{I_{\text{cu}}} = \frac{0.01188 \, mV}{3.5 \, A} = 0.00339 \, \mu\Omega \] 5. **Calculate the required Area for the copper wire (A_cu)**: \[ A_{\text{cu}} = \frac{\rho_{\text{cu}} \cdot L_{\text{cu}}}{R_{\text{cu}}} = \frac{0.0261 \cdot 32.6}{0.00339} = 257.07 \, m^2 \] 6. **Finally, we can calculate the diameter of the copper wire (d_cu)**: \[ d_{\text{cu}} = 2 \times \sqrt{\frac{A_{\text{cu}}}{\pi}} \approx 0.360 \, mm \text{ or } 0.36 mm. \] So, the diameter of the copper wire is approximately \(0.36 \, mm\).