A room `AC` run for `5` hour at a voltage of `220V` The wiring of the room constant of `Cu` of `1mm` ratio and a length of `10 m` consumption per day is `10` commerclal unit What fraction of it goes in the joule heated in wire? What would happen if the wiring is made of aluminum of the same distances? `[rho_(cu) = 1.7 xx 10^(-8) Omega,rho_(A1) = 2.7 xx 10^(-8) Omega m]`

Explanation

Power consumption in a day i.e., in 5=10units ltrgt Or power consumption per hour=2units
Or power consumption =2units=2kW=2000J/s
Also, we know that power consumption in resistor,
`R=Vxxl`
`implies2000W=220Vxxl` or `l ~~9A`
Now, the resistance of wire is given by `R=rho(l)/(A)`
Where, A is cross-sectional area of conductors.
Power consumption in first current wire is given by
`P=l^(2)R`
`rho(l)/(A)l^(2)=1.7xx10^(-8)xx(10)/(pixx10^(-6))xx81J//s~~4J//s`
The fractional loss due to the joule heating in first wire `=(4)/(2000)xx100=0.2%`
Power loss in Al wire `=4(rho_(Al))/(rho_("Cu"))=1.6xx4=6.4J//s`
The fractional loss due to the joule heating in second wire `=(6.4)/(2000)xx100=0.32%`