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

Answer:

A room AC runs for 5 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 ?

Explanation:

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Answer 2

The lack of energy in wiring is 4.4J/sec and the power loss in the Al wiring is 7 W.

Given:

• Total electricity ate up in five hrs an afternoon via way of means of AC and wiring =10kWh.E
• energy ate up in 1hr via way of means of AC and wiring =2kWh.
• Then the full energy of AC and wire =2000W.

To be found: fractional loss because of heating of wires  and energy loss in Al wiring,

Formula used:

• P=VI.
• Let P​zero be the energy of wiring then, P0 =I²Rw [Rw = resistance of wiring ]

I= P/V​ = 2000/220 ≅9.0A

⇒zero=9×9⋅ ρ!/A = 1377/3.14 ×10⁻² =4.38=4.4Watt

So, lack of energy in wiring  ≅4.4J/sec

The fractional loss is because of the heating of wires.= 4.4/2000 ×100%=0.22%

⇒ Pₓ = (2.7×10⁻⁸/1.7×10⁻⁸)×4.4Watt

So, The power loss withinside the Al wiring =7 W.

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