Question

A circuit calls for a $0.50mm$diameter copper wire to be stretched between two points. You don't have any copper wire, but you do have aluminum wire in a wide variety of diameters. What diameter aluminum wire will provide the same resistance?

Short answer

The diameter of the aluminum wire that provides the same resistance is $0.64mm.$

Step-by-step solution

Given Information (Part a)

Copper wire diameter$=0.50mm$

Copper wire stretched between two points

Aluminum wire in a wide variety of diameters

Diameter of the aluminum wire$=?$

Explanation 

Resistance $R$of a conductor of resistivity $\rho$, length $L$and cross-section area $A=\frac{\pi D^2}{4}$

Here, Area is a circle:

$R=\rho \frac{L}{A}=\rho \frac{4L}{\pi D^2}$

Modify the expression as,

$=\rho \frac{4L}{\pi D^2}$

We will have to make sure that the resistance and length of our wires are the same. In that case, we get the same expression for the two resistances.

${\rho }_{Cu}\frac{4L}{\pi D_{Cu}^2}=R={\rho }_{Al}\frac{4L}{\pi D_{Al}^2}$

Simplify all common factors,

$\frac{{\rho }_{Cu}}{D_{Cu}^2}=\frac{{\rho }_{Al}}{D_{Al}^2}$

The diameter of the aluminum wire is,

$D_{Al}^2=\frac{{\rho }_{Al}}{{\rho }_{Cu}}{D}_{Cu}^2$

Take the square root,

$D_{Al}=D_{Cu}\sqrt{\frac{{\rho }_{Al}}{{\rho }_{Cu}}}$

Substitute the values,

localid="1649069250663" $D_{Al}=\left(0.0005mm\times \sqrt{\frac{2.8\Omega m}{1.7\Omega m}}\right)$

$D_{Al}=0.64\mathrm{mm}$

Final Answer 

Hence, the diameter of the aluminum wire that provide the same resistance is$0.64mm.$