[tex]dane:\\l = 100 \ m\\s = 1,5 \ mm^{2}\\\rho = 0,029\frac{\Omega\cdot mm^{2}}{m} \ - \ opor \ wlasciwy \ aluminium\\szukane:\\R = ?\\\\Rozwiaazanie\\\\R = \rho\cdot\frac{l}{s}\\\\R = 0,029\frac{\Omega\cdot mm^{2}}{m}\cdot\frac{100 \ m}{1,5 \ mm^{2}}\\\\\underline{R = 1,9(3) \ \Omega \approx1,93 \ \Omega}[/tex]
