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=(\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=(1.7*10^(-8)*4.727)/(((\pi *0.012^2)/(2)))=3.5526*10^(-4)\Omega`

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

`I=(V)/(R)=(0.093)/(3.5526*10^(-4))=261.7766A`

Then, our current will be I=261.7766A