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.

We start by finding the resistance that the wire coil must have. For a load dissipating power \( P \) at a voltage \( V \) the resistance is given by \[ R=\frac{V^2}{P} \] Given \( V=250\,\mathrm{V} \) and \( P=2000\,\mathrm{W} \) we have \[ R=\frac{250^2}{2000}=\frac{62500}{2000}=31.25\,\Omega. \] The resistance of a wire of length \( L \) and cross‐sectional area \( A \) is given by \[ R=\rho\,\frac{L}{A}. \] We can solve for \( L \) as \[ L=\frac{R\,A}{\rho}. \] The wire has a diameter of \(0.15\,\mathrm{mm}\). Converting to meters, \[ d=0.15\times10^{-3}\,\mathrm{m}, \] so the radius is \[ r=\frac{d}{2}=7.5\times10^{-5}\,\mathrm{m}. \] The cross–sectional area is \[ A=\pi r^2=\pi\,(7.5\times10^{-5})^2. \] Calculating the square of the radius: \[ (7.5\times10^{-5})^2=5.625\times10^{-9}\,\mathrm{m}^2, \] so \[ A=\pi\times5.625\times10^{-9}\approx1.767\times10^{-8}\,\mathrm{m}^2. \] The resistivity is given as \[ \rho=80\,\mu\Omega\cdot\mathrm{m}=80\times10^{-6}\,\Omega\cdot\mathrm{m}. \] Now substitute all these into the expression for \( L \): \[ L=\frac{R\,A}{\rho}=\frac{31.25\times1.767\times10^{-8}}{80\times10^{-6}}. \] Performing the multiplication in the numerator, \[ 31.25\times1.767\times10^{-8}\approx5.522\times10^{-7}, \] and then dividing by \(80\times10^{-6}\), \[ L=\frac{5.522\times10^{-7}}{80\times10^{-6}}\approx6.9\times10^{-3}\,\mathrm{m}. \] Thus, the total length of the wire required is \[ L\approx6.9\times10^{-3}\,\mathrm{m}\approx6.9\,\mathrm{mm}. \]