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=10`
units or power consumption per hour `=2` units
or power consumption `=2` units `=2kW`
Also, we know that power consumption is resistor,
`P=Vxxi`
`implies2000W=220Vxxi` or `iapprox9A`
Now, the resistance of wire is given by `R=rho(1)/(A)`
where, A is cross sectional area of conductor. power consumption in first current carrying wire is given by
`P=i^(2)R`
`rho(l)/(A)i^(2)=1.7xx10^(-8)xx(10)/(pixx10^(-6)xx81)(J)/(S)=4(J)/(s)`
The fractional loss due to the joule heating in first wire `=(4)/(2000)xx100=0.2%`
power loss in `AI` wire `=4(rho_(A!))/(rho_(Cu))=1.6xx4=6.4(J)/(s)`
The fractional loss due to the joule heating in second wire `=(6.4)/(2000)xx100=0.32%`