Question

A room has AC run for 5 hours a day at a voltage of 220 V. The wiring of the room consists of Cu of 1 mm radius and a length of 10 m. Power consumption per day is 10 commercial units. What fraction of it goes in the joule heating in wires? What would happen if the wiring is made of aluminium of the same dimensions? [ρCu = 1.7 × ${10}^{-8}$ Ω m, ρAl = 2.7 × ${10}^{-8}$ Ω m]

Answer 1

Step 1: Given

Total energy consumed 5 hours by AC and wiring is 10 kWh.

Length of copper wire = 10 m

Radius of copper wire = 1 mm

ρCu = $1.7\times {10}^{-8}$ Ω m, ρAl = $2.7\times {10}^{-8}$ Ω m

Step 2: Formula used and Calculation

Total energy consumed 5 hours by AC and wiring is 10 kWh.

$\therefore$ Energy consumed in 1 hour = 2 kWh

So, total power of AC and wiring = 2000 W

$P=VI$

$I=\frac{P}{V}$

$I=\frac{2000}{220}$

$I=9.0A$

Let $P_o$ is power of wiring then,

$P_o=I^{2}R$

$P_o=I^2\frac{\text{ρ}l}{A}$

$P_o=9\times 9\times \frac{1.7\times {10}^{-8}\times 10}{\pi \times {10}^{-3}\times {10}^{-6}}$

$P_o=4.4W$

So, loss of energy in wirirng is 4.4 J/s.

The fractional loss due to heating is

$\frac{4.4}{2000}\times 100$

$=0.22\%$

For aluminium wiring of same dimension

Power,

$P^{\prime }=I^2{R}_{Al}$

$P^{\prime }=I^2{\text{ρ}}_{Al}\frac{l}{A}$

$P^{\prime }=9\times 9\times \frac{2.7\times {10}^{-8}\times 10}{\pi \times {10}^{-3}\times {10}^{-3}}$

$P^{\prime }=7W$

The fractional loss due to heating is

$\frac{7}{2000}\times 100$

$=0.35\%$