Question

A room has an AC run 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 the wire? ($\rho_{Cu} = 1.7 \times 10^{-8}\;\Omega m$)
A). 0.32%
B). 0.2%
C). 2%
D). 3.2%

Answer 1

First find the power consumed by the AC in total. Then determine the power consumed by the wire as this is the source of joule heating losses. Dividing the two will give you the fractional heating loss that we require. Remember to convert the power consumption value given in commercial units to the units of power it is associated with, which is kWh (kilowatt-hour). But since we will be measuring the power consumption by the AC and the wire per hour, we can express all our power values in kW.

Formula Used: The power across a resistor is $P = I^2 R = V \times I = \dfrac{V^2}{R}$, where R is the resistance of the resistor, and I is the current flowing through the resistor and V is the potential difference applied across the ends of the resistor.
The resistance of a wire with length l and cross-sectional area A: $R = \rho\dfrac{l}{A}$, where $\rho$ is the resistivity or resistance per unit length of the wire for a unit cross-sectional area.

Complete step-by-step solution:
Let us begin by finding out the power consumption by the AC.
Remember that the energy dissipated per unit time is nothing but the power consumed.
Now, we are given that the AC consumes 10 units of power when it is run for 5 hours.
Therefore, the AC consumes a power of $\dfrac{10}{5} =2$ units per hour which is $= 2 kW$, since kilowatt-hour(kWh) is the commercial unit of power and we’re taking it to be per 1 hour.
The power consumption by the AC is related to the current flowing through its connecting wires and voltage as:
$P_{AC} =VI \Rightarrow 2000 = 220 \times I \Rightarrow I = 9.09\;A$
Now, power consumption by the wire is given as:
$P{wire}= I^2 R$ and we know that resistance can be expressed in terms of specific resistance of the wire as $R = \rho\dfrac{l}{A}$, where l is the length of the wire and A is the area of cross-section of the wire.
$\Rightarrow P_{wire} = I^2 \times \rho_{Cu}\dfrac{l}{A}$
The specific resistance or the resistivity of copper is given to be $\rho_{Cu} = 1.7 \times 10^{-8}\;\Omega m$, the length of the wire is 10m, and the area of cross-section of the wire will be
$= \pi r^2 = \pi 10^{-3^2}= \pi 10^{-6}\;m^2$
Substituting all these values in our power relation gives:
$P = I^2 \times \rho_{Cu}\dfrac{l}{A} = 9.09^2 \times 1.7 \times 10^{-8} \dfrac{10}{\pi \times 10^{-6}} = \dfrac{140.5 \times 10^{-1}}{\pi} = 4.47\;W$
Joule heating arises from the heating up of the wires in the circuit, and a fraction of the total power is lost to this heating.
Therefore, the fractional loss due to the joule heating from the wire $P_{joule\;heating} = \dfrac{P_{wire}}{P_{AC}} = \dfrac{4.47}{2000}$
Thus, the percentage power loss will be: $P_{loss} = P_{joule\;heating} \times 100 = \dfrac{4.47}{2000} \times 100 = 2.24 \times 10^{-3} \times 100 = 0.224\%$
Therefore, the correct choice would be B. $0.2\%$

Note: Do not get confused between resistance and resistivity. Resistance is a measure of the opposition that an object offers to a flow of current, whereas resistivity is a characteristic property of the material making up the object which resists the flow of current through it.
In addition to the relationship that we used in the problem, the resistivity of the material of a resistor is alternatively given by:
$\rho =\dfrac{m}{ne^2 \tau}$, where m is the mass of the material, n is the number density of charge carriers, e is the charge of the charge carriers and $\tau$ is the relaxation time. For metallic conductors, an increase in temperature increases the amplitude of vibration of the lattice atoms and the collisions between the electrons and the lattice atoms will increase. Consequently the $\rho$ increases with an increase in temperature.