Question

A copper wire 8.433 m long has a diameter of 0.017 m. If the wire is connected to a 0.088 V battery, what is the current in the wire? The resistivity of copper is 1.7 x 10^-8 ohm m.

Answer 1

First, we need to find the resistance of the copper wire as follows:

`\begin{gathered} R=(\rho L)/(A) \\ where: \\ \rho=resistivity=1.7*10^(-8)\Omega \\ L=length=8.433m \\ A=Area=\pi r^2=\pi((0.017)/(2))^2=2.2698*10^(-4)m^2 \end{gathered}`

Therefore:

`R\approx6.316*10^(-4)\Omega`

Now, using Ohm's law:

`\begin{gathered} V=IR \\ so: \\ I=(V)/(R)=(0.088)/(6.316*10^(-4))\approx139.328A \end{gathered}`

Answer:

139.328 A