Answer to Question #311313 in Electric Circuits for Jinky

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

The resistance of the wire is

$R=\rho\frac{l} {S} =\rho\frac{V} {S^2}=\rho\frac{16V} {\pi^2d^4}$,

where $\rho$ - resistivity of aluminium,

l - length of wire,

S - cross-sectional area of wire,

V - volume of wire.

d - diameter of wire.

So, $R=2.65\cdot 10^{-8} \frac{16\cdot1.57\cdot10^{-5}} {3.14^2(1.02\cdot10^{-3})^4}=0.62\space Ohm.$

The corresponding current is

$I=\frac{E} {R},$

where E - EMF of battery.

$I=\frac {30}{0.62}=48.4\space A.$