Question

A cylindrical cable with 1.85 x 10-8 m resistivity carries a current of 1100 A. There is a potential difference of 1.5 x 10-3 V between two points on the cable that are 0.23 m apart. What is the radius of the cable?

Answer 1

Answer:

The radius is 0.03 meters

Explanation:

Assuming cable as ohmic material, it satisfies Ohm's law:

$R=\frac{V}{I}$

With R the resistance, V the voltage and I the current. So, the resistance between the points indicated in the problem is:

$R=\frac{1.5\times10^{-3}}{1100}=1.36\times10^{-6} \Ohm$Ω

Now with this value we can use the fact that resistance is:

$R=\rho\frac{L}{A}$

with ρ the resistivity, L the length and A the cross area of the cylinder, so:

$R=(1.85\times10^{-8})\frac{0.23}{\pi*r^2}=1.36\times10^{-6}$

Solving for r:

$r=\sqrt{(1.85\times10^{-8})\frac{0.23}{\pi*(1.36\times10^{-6})}}=0.03 m$