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.
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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**.