0. A load, situated 110 m away from its supply, draws 22 A from a 240 V supply. The maxi permissible voltage drop between the point of supply and the load must not exceed \( 5 \% \) supply voltage. Calculate the diameter of the conductors that need to be used if the resi is \( 0,027 \mu \Omega \mathrm{~m} \).

We start by determining the maximum allowable voltage drop. Given that the supply voltage is \(240\text{ V}\) and that the drop must not exceed \(5\%\) of this value, we have \[ \Delta V_{\text{max}} = 0.05 \times 240\,\text{V} = 12\,\text{V}. \] The load draws a current of \(I = 22\,\text{A}\), so by Ohm’s law the total resistance of the circuit must satisfy \[ R_{\text{total}} \leq \frac{\Delta V_{\text{max}}}{I} = \frac{12\,\text{V}}{22\,\text{A}} \approx 0.5455\,\Omega. \] Since the load is \(110\,\text{m}\) away from the supply, the current follows both an outgoing and a return path. Thus, the total conductor length is \[ L = 2 \times 110\,\text{m} = 220\,\text{m}. \] For a conductor of resistivity \(\rho\) and cross-sectional area \(A\), the resistance is given by \[ R = \rho \frac{L}{A}. \] We are given the resistivity as \[ \rho = 0.027\,\mu\Omega\cdot\text{m} = 0.027 \times 10^{-6}\,\Omega\cdot\text{m}. \] Setting \(R = 0.5455\,\Omega\) and solving for \(A\) we get \[ A = \rho \frac{L}{R} = \frac{0.027\times10^{-6}\,\Omega\cdot\text{m}\times 220\,\text{m}}{0.5455\,\Omega}. \] First, compute the numerator: \[ 0.027\times10^{-6}\times 220 = 5.94\times10^{-6}\,\Omega\cdot\text{m}. \] Then, \[ A = \frac{5.94\times10^{-6}}{0.5455} \approx 1.089\times10^{-5}\,\text{m}^2. \] For a circular cross-section, the area is also \[ A = \frac{\pi d^2}{4}, \] which allows us to solve for the diameter \(d\): \[ d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4\times1.089\times10^{-5}}{\pi}}. \] Calculate the numerator: \[ 4 \times 1.089\times10^{-5} = 4.356\times10^{-5}. \] Now, dividing by \(\pi\): \[ \frac{4.356\times10^{-5}}{\pi} \approx \frac{4.356\times10^{-5}}{3.1416} \approx 1.387\times10^{-5}. \] Taking the square root gives \[ d \approx \sqrt{1.387\times10^{-5}} \approx 3.72\times10^{-3}\,\text{m} = 3.72\,\text{mm}. \] Thus, the diameter of the conductors should be approximately \(3.72\,\text{mm}\).