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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 56.4 A. The resistance per unit length is to be 0.181 Ω/km. The densities of copper and aluminum are 8960 and 2600 kg/m3, respectively. Compute (a) the magnitude J of the current density and (b) the mass per unit length λ for a copper cable and (c) J and (d) λ for an aluminum cable.

Respuesta :

Answer:

a)[tex]J=6.04\times 10^5\ A/m^2[/tex]

b)λ=0.83 kg/m

c)[tex]J=3.71\times 10^5\ A/m^2[/tex]

d) λ=0.39 kg/m

Explanation:

Given that

I= 56.4 A

R= 0.181 Ω/km

Copper  :

d=8960 kg/m³

[tex]\rho = 1.69\times 10^{-8}\ ohm.m[/tex]

Aluminium :

d=2600 kg/m³

[tex]\rho = 2.75\times 10^{-8}\ ohm.m[/tex]

We know that

E= V/L

V= I R

J= E/ρ

[tex]J=\dfrac{IR}{L\rho}[/tex]

For copper:

[tex]J=\dfrac{IR}{L\rho}[/tex]

[tex]J=\dfrac{56.4\times 0.181\times 10^{-3}}{1.69\times 10^{-8}}[/tex]

[tex]J=6.04\times 10^5\ A/m^2[/tex]

Mass per unit length λ

[tex]\lambda =\dfrac{\rho d}{\dfrac{R}{L}}[/tex]

[tex]\lambda =\dfrac{1.69\times 10^{-8}\times 8960 }{0.181\times 10^{-3}}[/tex]

λ=0.83 kg/m

For  aluminum :

[tex]J=\dfrac{IR}{L\rho}[/tex]

[tex]J=\dfrac{56.4\times 0.181\times 10^{-3}}{2.75\times 10^{-8}}[/tex]

[tex]J=3.71\times 10^5\ A/m^2[/tex]

Mass per unit length λ

[tex]\lambda =\dfrac{\rho d}{\dfrac{R}{L}}[/tex]

[tex]\lambda =\dfrac{ 2.75\times 10^{-8}\times 2600  }{0.181\times 10^{-3}}[/tex]

λ=0.39 kg/m

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