Question

A 14-gauge copper wire of diameter 1.628 mm carries a current of 12.5 mA. (a) What is the potential difference across a 2.00-m length of the wire? (b) What would the potential difference in part (a) be if the wire were silver instead of copper, but all else were the same?

Short answer

(a) Potential difference is$V=2.063\times {10}^{-4}V$

(b) Potential difference in using a silver wire is$V=1.75\times {10}^{-4}V$

Step-by-step solution

Step 1:

Given data:

The cross-sectional area of the wire:

Step 2:

(a) As the length L=200m, and the copper resistivity $p=1.72\times {10}^{-8}\Omega .m$

So, the resistance of a copper wire is:

$$\begin{gathered} R=\frac{\rho L}{A} \\ =\frac{\left(1.72\times {10}^{-8}\Omega .m\right)\left(2.00m\right)}{2.082\times {10}^{-6}{m}^2} \\ =0.0165\Omega \end{gathered}$$

Putting the value of R in equation V=IR

Therefore, the potential difference is;

$$\begin{gathered} V=IR \\ =\left(12.5\times {10}^{-3}A\right)\times \left(0.014\Omega \right) \\ =1.75\times {10}^{-4}V \end{gathered}$$

Hence, the potential difference is$V=2.063\times {10}^{-4}V$.

Step 3:

(b) As the silver resistivity$\rho =1.47\times {10}^{-8}\Omega .m$

So, the resistance of a silver wire is:

$$\begin{gathered} R=\frac{\rho L}{A} \\ =\frac{\left(1.47\times {10}^{-8}\Omega .m\right)\left(2.00m\right)}{2.082\times {10}^{-6}{m}^2} \\ =0.014\Omega \end{gathered}$$

Putting the value of R in equation V=IR,

Therefore, the potential difference is,

$$\begin{gathered} V=IR \\ =\left(12.5\times {10}^{-3}A\right)\times \left(0.014\Omega \right) \\ =1.75\times {10}^{-4}V \end{gathered}$$

Hence, the potential difference in using a silver wire is$V=1.75\times {10}^{-4}V$