$A$ room has an $AC$ running for $5$ hours a day at a voltage of $220$ $V$. The wiring of the room consists of $Cu$ wire of $1$ $mm$ radius and a length of $10$ $m$. Power consumption per day is $10$ commercial units. What fraction of it goes into Joule heating in the wires? What would happen if the wiring is made of aluminium of the same dimensions? $(\rho_{Cu} = 1.7 \times 10^{-8} \, \Omega m, \rho_{Al} = 2.7 \times 10^{-8} \, \Omega m)$

(A) Power consumption in $1$ day ($5$ hours) is $10$ units.
Power consumed in $1$ hour is $P = 10/5 = 2$ units = $2000$ $W$.
Since $P = VI$,the current $I = P/V = 2000/220 \approx 9.09$ $A$.
Resistance of the copper wire $R = \rho l / A = \rho l / (\pi r^2) = (1.7 \times 10^{-8} \times 10) / (3.14 \times (10^{-3})^2) = 1.7 \times 10^{-7} / 3.14 \times 10^{-6} \approx 0.054$ $\Omega$.
Power dissipated in the wire $P_{loss} = I^2 R = (9.09)^2 \times 0.054 \approx 4.46$ $W$.
Fraction of power lost = $P_{loss} / P_{total} = 4.46 / 2000 \approx 0.00223$ or $0.223\%$.
If aluminium is used,$\rho_{Al} = 2.7 \times 10^{-8} \, \Omega m$.
Since $R \propto \rho$,the resistance increases by a factor of $2.7/1.7 \approx 1.59$.
Thus,the power loss in the aluminium wire will be approximately $1.59$ times higher than in the copper wire.