2.3. An electric motor requires 2 kW of power dissipation when connected to a 250 V supply. If the radiator coils are made from a wire of 0.15 mm diameter with a resistivity of \( 80 \mu \Omega \cdot \mathrm{~m} \), calculate the total length of the wire required.

To calculate the total length of the wire required for the radiator coils, we can start with the formula for power dissipation in terms of resistance: \[ P = \frac{V^2}{R} \] Where: - \( P = 2 \text{ kW} = 2000 \text{ W} \) - \( V = 250 \text{ V} \) First, rearranging the formula for resistance \( R \): \[ R = \frac{V^2}{P} = \frac{(250)^2}{2000} = \frac{62500}{2000} = 31.25 \, \Omega \] Next, we can relate the resistance of the wire to its resistivity, length, and cross-sectional area: \[ R = \frac{\rho \cdot L}{A} \] Where: - \( \rho = 80 \mu \Omega \cdot m = 80 \times 10^{-6} \, \Omega m \) - \( A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.15 \times 10^{-3}}{2}\right)^2 \) Calculating the cross-sectional area: \[ A = \pi \left(\frac{0.15 \times 10^{-3}}{2}\right)^2 = \pi \left(0.075 \times 10^{-3}\right)^2 \approx 1.767 \times 10^{-8} \, m^2 \] Now we can substitute the values into the resistance equation to find the length \( L \): \[ R = \frac{(80 \times 10^{-6}) \cdot L}{1.767 \times 10^{-8}} \] Setting \( R = 31.25 \, \Omega \): \[ 31.25 = \frac{(80 \times 10^{-6}) \cdot L}{1.767 \times 10^{-8}} \] Now, solve for \( L \): \[ L = \frac{31.25 \cdot 1.767 \times 10^{-8}}{80 \times 10^{-6}} \] Calculating \( L \): \[ L \approx \frac{5.52875 \times 10^{-7}}{80 \times 10^{-6}} \approx 6.91 \, m \] So, the total length of the wire required is approximately **6.91 meters**.