Question

Find the ratio of the diameter of aluminum to silver wire, if they have the same resistance per unit length (as they might in household wiring).dAl/dAg =

Answer 1

Given:

The resistance per unit length of aluminum is equal to the resistance per unit length of silver i.e.,

$\frac{R_{Al}}{l_{Al}}=\frac{R_{Ag}}{l_{Ag}}$

Required: The ratio

$\frac{d_{Al}}{d_{Ag}}$

Explanation:

The resistance of aluminum is given by the formula

$\begin{gathered} R_{Al}=\rho_{Al}\frac{l_{Al}}{A_{Al}} \\ \frac{R_{Al}}{l_{Al}}=\frac{\rho_{Al}}{\pi(r_{Al})^2} \\ \frac{R_{Al}}{l_{Al}}=\frac{\rho_{Al}}{\pi(\frac{d_{Al}}{2})^2} \end{gathered}$

The resistance of silver is given by the formula

$\begin{gathered} R_{Ag}=\rho_{Ag}\frac{l_{Ag}}{A_{Ag}} \\ \frac{R_{Ag}}{l_{Ag}}=\frac{\rho_{Ag}}{\pi(r_{Ag})^2} \\ \frac{R_{Ag}}{l_{Ag}}=\frac{\rho_{Ag}}{\pi((d_{Ag})\/2)^2} \end{gathered}$

Since,

$\frac{R_{Al}}{l_{Al}}=\frac{R_{Ag}}{l_{Ag}}$

On substituting the values, the ratio can be calculated as

$\begin{gathered} \frac{\rho_{Al}}{\pi(\frac{d_{Al}}{2})^{2}}=\frac{\rho_{Ag}}{\pi(\frac{(d_{Ag})}{2})^{2}} \\ \frac{\rho_{Al}}{(d_{Al})^2}=\frac{\rho_{Ag}}{(d_{Ag})^2} \\ \frac{(d_{Al})^2}{(d_{Ag})^2}=\frac{\rho_{Al}}{\rho_{Ag}} \\ \frac{d_{Al}}{d_{Ag}}=\sqrt{\frac{\rho_{Al}}{\rho_{Ag}}} \end{gathered}$

Thus, the ratio is

$\frac{d_{Al}}{d_{Ag}}=\sqrt{\frac{\rho_{Al}}{\rho_{Ag}}}$

Final Answer: The ratio of the diameter of aluminum to silver is the square root of the ratio of resistivity of aluminum to silver.