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  • A room has AC run for 5 hours a day at a voltage of 220V. 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.

A room has AC run for 5 hours a day at a voltage of 220V. 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?

[\rho_ {Cu} = 1.7 \times -810\Omega m, \rho _{Al} = 2.7 \times 10^{-8} \Omega m]

Answers (1)

Energy consumed day=10 kWh

Energy consumed hour=2 kWh

Power=2kW

P=VI

V=220V

I=\frac{P}{}V=\frac{2000}{220}=\frac{100}{11}

Heat =I^2\times R_{Cu}

R_{Cu}=\rho\left ( \frac{L}{}A \right )=1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )

Heat=\left (\frac{100}{110} \right )^{2}\times 1.7\times 10^{-8}\times \left (\frac{10}{\pi \times \left (5\times 10^{-4} \right )^{2}} \right )=4.4 W

fractional loss in heating

=\frac{4.4}{2000}\times 100 =0.22\%

\frac{Heat_{Al}}{Heat_{Cu}}=\frac{I^2RAl}{I^2R_{Cu}}=\frac{\rho _{Al}}{\rho _{Cu }}

HeatAl=7W

fractional loss in heating

=\frac{7}{2000}\times1000=0.35\% 

Posted by

infoexpert24

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