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

Odpowiedź :

[tex]Dane:\\ d=1mm\Rightarrow r=0,5mm=5*10^{-4}m\\ l=100m\\ \rho=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=\rho\frac{l}{s}\\\\ R=1,72*10^{-8}*\frac{100}{7,85*10^{-7}}\\\\ R=\frac{17,2*10^{-7}}{7,85*10^{-7}}\\\\ R=2,19\Omega[/tex]

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

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