Question

A cylindrical copper cable 1.50Km long is connected across a 220.0V potential difference.
(a) What should be its diameter so that it produces heat at a rate of 50.0W?
(b) What is the electric field inside the cable under these conditions?

Answer 1

First, the value of the length of the cable will be converted to meters.

Therefore, the length of the cable is L = 1.5*10^3 m = 1500 m.

We know, from enunciation, the value of potential difference:V = 220 volts

The power is given and it's value is of P = 50 Watts

P = V^2/R => R = V^2/P

Since the area of the section of the cable is circular, w'ell recall the formula for the area of the circle:

A = pi*r^2 (1)

A = p*L/R (2), where p = 1.72/10^8 ohm/m

We'll equate (1) and (2) and we'll get:

pi*r^2 = p*L*P/V^2

r^2 = p*L*P/V^2*pi

r = sqrt(p*L*P/V^2*pi)

Diameter is d = 2*r = 2*sqrt(p*L*P/V^2*pi)

d = 2*sqrt1.72*10^-10*15*5*10^3/484

d = 2*sqrt 1.72*10^-7*75/22

d = 1.0325/100

d = 0.010325 meters

The diameter of the cable is of d = 0.010325 meters, such as it is producing heat at a rate of 50W.