Question

A copper wire 4.727 m long has a diameter of 0.012 m. If the wire is connected to a 0.093 V battery, what is the current in the wire? The resistivity of copper is 1.7 x 10^-8 ohm m.

Answer 1

This problem requires us to use Ohm's second law. It tells us:

$R=\frac{\rho L}{A}$

Where p is the resistivity, L is the length of the conductor and A is its area. For our problem, the resistance of the conductor will be:

$R=\frac{1.7*10^{-8}*4.727}{(\frac{\pi *0.012^2}{2})}=3.5526*10^{-4}\Omega$

If we then apply Ohm's first law, we'll have:

$I=\frac{V}{R}=\frac{0.093}{3.5526*10^{-4}}=261.7766A$

Then, our current will be I=261.7766A