Zadanie z oporu elektrycznego przewodnika (w załączniku)

[latex]dane:\R = 854 Omega\ ho = 99,6*10^{-8} Omega m=9,96*10^{-4} Omega mm\l = 6 cm = 60 mm\szukane:\S = ?[/latex] [latex]R = frac{ ho *l}{S} |*S\\R*S = ho*l /:R\\S = frac{ ho*l}{R}\oraz\S = frac{ pi d^{2}}{4}\\frac{ pi d^{2}}{4} = frac{ ho*l}{R}\\ pi d^{2}R = 4 ho l /: pi R\\d^{2} = frac{4 ho l}{ pi R} [/latex] [latex]d = sqrt{frac{4 ho*l}{ pi *R}} = sqrt{frac{4*9,96*10^{-4}*60}{3,14*854}} = sqrt{frac{2390,4}{2681,56}}*10^{-2}approx1*10^{-2} \\d approx0,01 mm[/latex]