Oblicz opor 10 metrowgo drutu o srednicy 5mm wykonaniego z
a)midzi
b)srebra
c)alumienium
ps. wybieramy 2 podpunkty
l = 10 m
d = 5 mm
r = d/2
r = 5 mm/2 = 2,5 mm = 2,5 *10⁻³ m
ρ Cu = 0,17 *10⁻⁶ Ω * m
ρ Ag = 0,016 *10⁻⁶ Ω * m
R = ρ * l/S
1. obl. S
S = πr²
S = 3,14 * ( 2,5 *10⁻³ m )²
S = 3,14 * 6,25 *10⁻⁶ m²
S = 19,625 *10⁻⁶ m²
2. obl. opór
R Cu = 0,17 *10⁻⁶ Ω * m * 10 m /19,625 *10⁻⁶ m²
R Cu = 1,7 *10⁻⁶ Ω * m²/ 19,625 *10⁻⁶ m²
R Cu = 0,0866 Ω
R Ag = 0,016 *10⁻⁶ Ω * m * 10 m /19,625 *10⁻⁶ m²
R Ag = 0,16 *10⁻⁶ Ω * m² /19,625 *10⁻⁶ m²
R ag = 0,082 Ω
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l = 10 m
d = 5 mm
r = d/2
r = 5 mm/2 = 2,5 mm = 2,5 *10⁻³ m
ρ Cu = 0,17 *10⁻⁶ Ω * m
ρ Ag = 0,016 *10⁻⁶ Ω * m
R = ρ * l/S
1. obl. S
S = πr²
S = 3,14 * ( 2,5 *10⁻³ m )²
S = 3,14 * 6,25 *10⁻⁶ m²
S = 19,625 *10⁻⁶ m²
2. obl. opór
R Cu = 0,17 *10⁻⁶ Ω * m * 10 m /19,625 *10⁻⁶ m²
R Cu = 1,7 *10⁻⁶ Ω * m²/ 19,625 *10⁻⁶ m²
R Cu = 0,0866 Ω
R Ag = 0,016 *10⁻⁶ Ω * m * 10 m /19,625 *10⁻⁶ m²
R Ag = 0,16 *10⁻⁶ Ω * m² /19,625 *10⁻⁶ m²
R ag = 0,082 Ω