Answer to Question #311975 in Electric Circuits for Stary

Question

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 1

(a) the resistance of the wire

$R=\rho\frac{l}{A}=\rho\frac{V}{A^2}\\ =2.65*10^{-8}*\frac{1.57*10^{-5}}{(3.14*(1.02*10^{-3})^2/4)^2}\\ =0.624\:\Omega$

(b) the corresponding current

$I=\frac{V}{R}=\frac{30}{0.624}=48.1\:\rm A$