Ncert Exercises & Solutions

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 10m. 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?
$[ρ_{Cu} = 1.7 \times 10^{- 8} Ωm, ρ_{Al} = 2.7 \times 10^{- 8} Ωm]$ $[ρ_{Cu} = 1.7 \times 10^{- 8} Ωm, ρ_{Al} = 2.7 \times 10^{- 8} Ωm]$

Hint: The power consumption depends on the resistivity of the wires.

Step 1:

Power consumption in a day i.e., in 5 hours = 10 units

Or power consumption per hour = 2 units

Or power consumption per hour = 2 units = 2 kW = 2000 J/s

Also, we know that power consumption in resistor,

P = V x l

\Rightarrow

$\Rightarrow$ 2000W = 220V x l or l = 9A

Step 2: Now, the resistance of the wire is given by

R = ρ \frac{l}{A}

$R = ρ \frac{l}{A}$ where A is the cross-sectional area of the conductor.

Power consumption in the first current-carrying wire is given by;

P = l^{2} R = ρ \frac{l}{A} I^{2} = 1 .7 \times 10^{- 8} \times \frac{10}{π \times {(10^{- 3})}^{2}} \times 81 J / s \approx 4 J / s

$P = l^{2} R = ρ \frac{l}{A} I^{2} = 1 .7 \times 10^{- 8} \times \frac{10}{π \times {(10^{- 3})}^{2}} \times 81 J / s \approx 4 J / s$

Step 3: The fractional loss due to the joule heating in the first wire

= \frac{4}{2000} \times 100 = 0.2 %

$= \frac{4}{2000} \times 100 = 0.2 %$

Power loss in Al wire

= P \times \frac{ρ_{Al}}{ρ_{Cu}} = 1 .6 \times 4 = 6 .4 J / s

$= P \times \frac{ρ_{Al}}{ρ_{Cu}} = 1 .6 \times 4 = 6 .4 J / s$

The fractional loss due to the joule heating in the second wire

= \frac{6.4}{2000} \times 100 = 0.32 %

$= \frac{6.4}{2000} \times 100 = 0.32 %$

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions