91Ó°ÊÓ

Copper wire diameter$=0.50mm$

Copper wire stretched between two points

Aluminum wire in a wide variety of diameters

Diameter of the aluminum wire$=?$

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

Here, Area is a circle:

$R=\mathrm{ÒÏ}\frac{L}{A}=\mathrm{ÒÏ}\frac{4L}{\mathrm{Ï€}{D}^2}$

Modify the expression as,

$=\mathrm{ÒÏ}\frac{4L}{\mathrm{Ï€}{D}^2}$

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.

${\mathrm{ÒÏ}}_{Cu}\frac{4L}{\mathrm{Ï€}{D}_{Cu}^2}=R={\mathrm{ÒÏ}}_{Al}\frac{4L}{\mathrm{Ï€}{D}_{Al}^2}$

Simplify all common factors,

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

The diameter of the aluminum wire is,

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

Take the square root,

$D_{Al}=D_{Cu}\sqrt{\frac{{\mathrm{ÒÏ}}_{Al}}{{\mathrm{ÒÏ}}_{Cu}}}$

Substitute the values,

localid="1649069250663" $D_{Al}=\left(0.0005mm×\sqrt{\frac{2.8\mathrm{Î\copyright }m}{1.7\mathrm{Î\copyright }m}}\right)$

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

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