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. (${{\rho }_{Al}}=2.63\times {{10}^{-8}}\Omega m$, ${{\rho }_{Cu}}=1.72\times {{10}^{-8}}\Omega m$, Relative density of Al=2.7, of Cu=8.9.)

Answer 1

Hint: Assume the mass, length, and area of the two wires. Find the resistance of the two wires using the mathematical expression of the resistance. Equate them to find the relation between their respective areas. Now calculate the ratio of mass of the two wires.

Formula Used:
Resistance of a wire is given by,
$R=\rho \dfrac{l}{A}$
Where,
$\rho $ is the resistivity of the wire
$l$ is the length of the wire
$A$ is the area of the wire

Mass of the wire is given by,
$M=dlA$
Where,
$A$ is the area of the wire
$d$ is the density of the wire
$l$ is the length of the wire

Complete step-by-step answer:

Let’s assume the following quantities:

${{A}_{1}}$ is the area of the aluminium wire
${{A}_{2}}$ is the area of the copper wire
${{l}_{1}}$ is the length of the aluminium wire
${{l}_{2}}$ is the length of the copper wire
${{\rho }_{1}}$ is the resistivity of the aluminium wire
${{\rho }_{2}}$ is the resistivity of the copper wire
${{R}_{1}}$ is the resistance of the aluminium wire
${{R}_{2}}$ is the resistance of the copper wire
${{m}_{1}}$ is the mass of the aluminium wire
${{m}_{2}}$ is the mass of the copper wire

Resistance of a wire of certain dimensions is given by,
$R=\rho \dfrac{l}{A}$
Where,
$\rho $ is the resistivity of the wire
$l$ is the length of the wire
$A$ is the area of the wire

Hence, we can write the following two equations:

${{R}_{1}}={{\rho }_{1}}\dfrac{{{l}_{1}}}{{{A}_{1}}}$..................(1)
${{R}_{2}}={{\rho }_{2}}\dfrac{{{l}_{2}}}{{{A}_{2}}}$..................(2)

Equation (1) gives the resistance of the aluminium wire
Equation (2) gives the resistance of the copper wire.

Given that the length and resistance of these wires to be same, hence we can write,
${{l}_{1}}={{l}_{2}}$
${{R}_{1}}={{R}_{2}}$................(3)

So, putting the values obtained in (1) and (2) into equation (3) we get,
${{\rho }_{1}}\dfrac{{{l}_{1}}}{{{A}_{1}}}={{\rho }_{2}}\dfrac{{{l}_{2}}}{{{A}_{2}}}$
$\Rightarrow \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}\dfrac{{{l}_{1}}}{{{l}_{2}}}=\dfrac{{{A}_{1}}}{{{A}_{2}}}$
$\Rightarrow \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\dfrac{{{A}_{1}}}{{{A}_{2}}}$
Hence the ratio of area,
$\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{2.63}{1.72}$

Now, mass of the wire can be given by,
$M=dlA$
Hence, we can write the following two equations:
${{M}_{1}}={{d}_{1}}{{l}_{1}}{{A}_{1}}$.................(4)
${{M}_{2}}={{d}_{2}}{{l}_{2}}{{A}_{2}}$.................(5)

Dividing equation (4) with (5) we get,
$\dfrac{{{M}_{1}}}{{{M}_{2}}}=\dfrac{{{d}_{1}}{{l}_{1}}{{A}_{1}}}{{{d}_{2}}{{l}_{2}}{{A}_{2}}}$

Hence, the ratio of mass is,
$\dfrac{{{M}_{1}}}{{{M}_{2}}}=\dfrac{{{d}_{1}}{{l}_{1}}{{A}_{1}}}{{{d}_{2}}{{l}_{2}}{{A}_{2}}}=(\dfrac{{{d}_{1}}}{{{d}_{2}}})(\dfrac{{{l}_{1}}}{{{l}_{2}}})(\dfrac{{{A}_{1}}}{{{A}_{2}}})$
$\Rightarrow \dfrac{{{M}_{1}}}{{{M}_{2}}}=(\dfrac{2.7}{8.9})(1)(\dfrac{2.63}{1.72})$
$\Rightarrow \dfrac{{{M}_{1}}}{{{M}_{2}}}<1$

Note:
As you can see the mass of the aluminium wire is less than the mass of the copper wire. Electrical lines are long wires carrying a high value of current. If the mass of the overhead lines is too much, the power cable will form an arc and can disrupt other lines in the vicinity. As a result, aluminium wire is preferred.