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 prefered for overhead power cables ? Given `rho Al=2.63xx10^(-8) Omega m rho Cu=0.72xx10^(-8) Omega m` , relative density of Al=2.7 and that of Cu =8.9.

Explanation

Resistance of a conductor with resistivity `rho`, length l and area of cross section A is given by:
`R= (rho xx l)/(A)`
Let `R_(Al) =` resistance of aluminimum wire
`R_(Cu)=` resistance of copper wire.
`R_(Al) = R_(Cu)`
`A_(1)=` area of cross section of aluminimum wire
`r_(1) =` radius of aluminium wire
`r_(2)`= radius of copper wire l= length of aluminimum wire= length of copper wire. Since resistances of wires are the same, we have:
`rArr (rho_(Al) xx l)/(A_(1)) = (rho_(Cu) xx l)/(A_(2))`
`rArr (rho_(Al))/(rho_(Cu)) = (A_(1))/(A_(2)) = (pi r_(1)^(2))/(pi r_(2)^(2)) = (r_(1)^(2))/(r_(2)^(2))`
`rArr (r_(2)^(2))/(r_(1)^(2)) = (1.72 xx 10^(-8))/(2.63 xx 10^(-8))`
Mass of a wire is given by M = Volume `xx` Density
=(Area of cross section `xx` Length) `xx` Density
`= pi r^(2) l xx d`
Let `M_(Cu)` be the mass of copper wire, `M_(Al)` that of the aluminium wire, `d_(2)` be the density of copper wire and `d_(1)` be that of the aluminium wire.
`rArr (M_(Cu))/(M_(Al)) = (pi r_(2)^(2) l xx d_(2))/(pi r_(1)^(2) l xx d_(1)) = (1.72 xx 10^(-8) xx 8.9)/(2.63 xx 10^(-8) xx 2.7) = 2.16`
`rArr M_(Cu) = 2.16 M_(Al)`
`:.` Al wire is lighter. For the same value of resistance and length, the mass of aluminium wire is less. Hence, it is preferred for use in overhead cables. Also, due to the lesser weight, slagging in the wires will be less.