A room `AC` run for `5` hour at a voltage of `220V` The wiring of the room constant of `Cu` of `1mm` radius 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

Energy consumed per day `= 10kWb`
Time for using the energy per day`= 5h`
`:.` Power consumed `(P) = (10kWh)/(5h) = 2kW = 2000 J//s` : corrent supposed `I + (P)/(V) = (2000 J//s)/(220V) = 9A`
power loss copper wire `= R_(cu) t^(2) = (p_(cu)^(1))/(pir^(2)) xx t^(2) = ((1.7 xx 10^(-8)) xx 10)/((22 //7) xx (10^(-3)) 2) xx 9^(2) = 4 J//s`
`:.` praction of power loss `= (4)/(2000) = (2)/(1000)` , % of power loss `= (2)/(1000) xx 100 = 0.2%`
power loss in `AI` wire `= (p_(AI)xx 1)/(pi r^(2)) t^(2)= ((2.7 xx 10^(-8)) xx 10 xx9^(2))/((22//7) xx (10^(-8)) ^(2)) = 6.4 J//s`
`%` power loss in `AI` wire `= (6.4)/(2000) xx 100 = 0.32%`