Suppose that you wish to fabricate a uniform wire out of 1.05 g of copper. Assume the wire has a resistance R?
Suppose that you wish to fabricate a uniform wire out of 1.05 g of copper. Assume the wire has a resistance R = 0.300 , and all of the copper is used.
(a) What will be the length of the wire?
(b) What will be the diameter of the wire?
i know mass density is 8.92e3 kg/m^3
resistivity of Cu = 1.7e-8
m = 1.05e-3 = Pm * V
V = A*L
R = Pr * L/A
A = pi*d^2 / 4
i got that V = 1.17 e -7 using those equations and i know Pr. but Don't know how to get L and A when they are both together in both equations..
thank you! would the diameter be 3.22e-7 micrometers?
- billrussell42Lv 79 years agoFavorite Answer
Is that R = 0.300Ω ?
Resistance of a wire in Ω
R = ρL/A
ρ is resistivity of the material in Ω-m
L is length in meters
A is cross-sectional area in m²
A = πr², r is radius of wire in m
resistivity Cu 17.2e-9 Ω-m
(17.2e-9)L/(πr²) = 0.3
volume of the wire is V = πr²L
density copper 8960 kg/m³
Volume = 0.00105 kg / 8960 kg/m³ = 1.172e-7m³
so we have AL = πr²L = 1.172e-7
and (17.2e-9)L/A = 0.3
two equations in 2 unknowns.
0.3A = (17.2e-9)L
A = 57.3e-9L
AL = 1.172e-7
(57.3e-9L)L = 1.172e-7
L = 1.43 meters
A = 57.3e-9L
A = 8.19e-8 m²