Question

How long must a 0.52-mm-diameter aluminum wire be to have a 0.58 A current when connected to the terminals of a 1.5 V flashlight battery?

Answer 1

here,

resistivity of alumunium , p = 2.82 * 10^-8 ohm.m

diameter of alumunium , d = 0.52 mm

radius , r = d/2 = 0.26 mm

r = 2.6 * 10^-4 m

current , I = 0.58 A

voltage of battery , V = 1.5 V

let the length of wire be L

resistance , R = p * l /area = V/I

2.82 * 10^-8 * L /(pi * (2.6 * 10^-4)^2) = 1.5 /0.58

solving for L

L = 19.48 m

the length of wire is 19.48 m