Question

A gold wire is 2.40 m long and 0.760 mm in diameter, and it carries a current of 1.44 A. What is the potential difference between the ends of the wire? Let p gold = 2.44 x 10^-8 Ω * m.​

Answer 1

Answer:

0.186 v

Explanation:

I = V/R

R =pL/A = 4pL/(3.14159*d^2)

R = (4 * 2.44 * 10^-8 *2.40) / 3.14159*(0.760*10^-3)^2

R = 0.129 ohm

V = I * R

V = 1.44 * 0.129

V = 0.186 v