Question

Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables.
(ρ_Al = 2.63 × 10⁻⁸ Ω m, ρ_Cu = 1.72 × 10⁻⁸ Ω m, Relative density of Al = 2.7, of Cu = 8.9.)

Answer 1

Resistivity of aluminium, ρ_Al = 2.63 × 10⁻⁸ Ω m

Relative density of aluminium, d₁ = 2.7

Let l₁ be the length of aluminium wire and m₁ be its mass.

Resistance of the aluminium wire = R₁

Area of cross-section of the aluminium wire =

A₁

Resistivity of copper, ρ_Cu = 1.72 × 10⁻⁸ Ω m

Relative density of copper, d₂ = 8.9

Let l₂ be the length of copper wire and m₂ be its mass.

Resistance of the copper wire = R₂

Area of cross-section of the copper wire = A₂

The two relations can be written as

`"R"_1 = ρ_1"l"_1/"A"_1` ..........(1)

`"R"_2 = ρ_2"l"_2/"A"_2` ............(2)

It is given that 

R₁ = R₂

∴ `ρ_1"l"_1/"A"_1 = ρ_2"l"_2/"A"_2`

And 

I₁ = I₂

∴ `ρ_1/"A"_1 = ρ_2/"A"_2`

`"A"_1/"A"_2 = ρ_1/ρ_2`

= `(2.63 xx 10^-8)/(1.72 xx 10^-8)`

= `2.63/1.72`

Mass of the aluminium wire,

m₁ = Volume × Density

= A₁l₁ × d₁ = A₁ l₁d₁ ….......(3)

Mass of the copper wire,

m₂ = Volume × Density

= A₂l₂ × d₂ = A₂ l₂d₂ ….........(4)

Dividing equation (3) by equation (4), we obtain

`"m"_1/"m"_2 = ("A"_1"l"_1"d"_1)/("A"_2"l"_2
"d"_2)`

for I₁ = I₂,

`"m"_1/"m"_2 = ("A"_1"d"_1)/("A"_2"d"_2)`

For `"A"_1/"A"_2 = 2.63/1.72`,

`"m"_1/"m"_2 = 2.63/1.72 xx 2.7/8.9` = 0.46

It can be inferred from this ratio that m₁ is less than m₂. Hence, aluminium is lighter than copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.