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cricket20 [7]
1 year ago
10

The resistivity of

Physics
1 answer:
Pachacha [2.7K]1 year ago
5 0

The current density of the copper wire is determined as (7.059 x 10⁸ A/m²)/L

<h3>Current density of the copper wire</h3>

The current density of the copper wire is calculated as follows;

J =1/ρ(E) = 1/ρ(V/L)

where;

  • E is electric field
  • V is voltage
  • L is length of the wire
<h3>Current density the wire</h3>

J =1/ρ(E) = 1/ρ(V/L)

J = (1/1.7 x 10⁻⁸) x (12)/L

J = (7.059 x 10⁸ A/m²)/L

Learn more about current here: brainly.com/question/24858512

#SPJ1

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a transmission-line cable, of length 3 km, consists of 19 strands of identical copper conductors, each 1.5 mm in diameter. becau
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the resistance of the cable is 582.9 ohms

we are given the length of the cable which is  3 km,  of  1.5 mm in, the diameter  and resistivity of copper which is 1.72 m

The formula  we are referring to for calculating the resistance of the  cable is

R = ρl/A.

As there are 19 strands of copper conductors, so the resistance will be

R = 19( ρl/A)

Here  ρ is the resisitivity =  1.72 , l is the length  = 3(1+0.05)*10³3= 3150 m

A=pie/4(1.5 x 10⁻³)^2 =1.766 x 10⁻⁶ =1.766 x 10^-6

Substituting the values in the formula  we  get

R = 19 ( 1.72*3150 )/1.766 x 10⁻⁶

 = 582.9 ohm

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So the new power dissipated is

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