An aluminum wire with a diameter of 1.02 mm and volume of 1.57x10-5 m3, has a battery attached to it with an emf of 30 volts. Find (a) the resistance of the wire and (b) the corresponding current.
Answer
a) Radius r=0.56mm
Length of wire "l=\\frac{volume }{area}"
Resistance of wire is
"R=\\rho \\frac{l}{A}\\\\=\\frac{2.65*10^{-8}*1.57*10^{-5}}{3.14*3.14*(0.56*10^{-3})^4}\\\\=0.429\\Omega"
b) so current
"I =\\frac{30}{0.429}\\\\=69.93A"
Comments
Leave a comment