Ile wynosi opór 100 m drutu miedzianego o średnicy 1 mm?

[latex]Dane:\ d=1mmRightarrow r=0,5mm=5*10^{-4}m\ l=100m\ ho=1,72*10^{-8}Omega m\ Rozwiazanie:\ s=pi r^2\ s=3,14*(5*10^{-4})^2\ s=3,14*25*10^{-8}\ s=7,85*10^{-7}m^2\\ R= hofrac{l}{s}\\ R=1,72*10^{-8}*frac{100}{7,85*10^{-7}}\\ R=frac{17,2*10^{-7}}{7,85*10^{-7}}\\ R=2,19Omega[/latex]

dane: l = 100 m = 10² m d = 1 mm r = d/2 = 0,5 mm = 0,0005 m = 5 *10⁻⁴ m ρ = 1,72 *10⁻⁸Ωm szukane: R = ?   R = ρ * l/s      s = πr²  R = 1,72 * 10²/[3,14 *(5*10⁻⁴)²] = 0,219 *10 Ω R = 2,19 Ω ---------------