Question 14 Marks

The electrical energy consumed by an electrical appliance is given by the product of its power rating and the time for which it is used. The SI unit of electrical energy is Joule. Actually, Joule represents a very small quantity of energy and therefore it is inconvenient to use where a large quantity of energy is involved. So for commercial purposes, we use a bigger unit of electrical energy which is called kilowatt-hour. 1 kilowatt-hour is equal to $3.6 \times 10^{6}$ joules of electrical energy.
i. The energy dissipated by the heater is E. When the time of operating the heater is doubled, what would be the energy dissipated?
ii. The power of a lamp is 60 W . What will be the energy consumed in 1 minute?
iii. The electrical refrigerator rated 400 W operates 8 hours a day. The cost of electrical energy is ₹ 5 per kWh.
Find the cost of running the refrigerator for one day.
OR
Calculate the energy transformed by a 5 A current flowing through a resistor of $2 \Omega$ for 30 minutes.

Answer

i. $E \propto t$
When the time of operating the heater is doubled, the energy dissipated is doubled.
ii. Given: $\mathrm{P}=60 \mathrm{~W}, \mathrm{t}=1 \mathrm{~min}$
$\mathrm{E}=60 \times 1 \times 60=3600 \mathrm{~J}$
iii. Given: $\mathrm{P}=400 \Omega, \mathrm{t}=8$ hour
$\mathrm{E}=400 \times 8=3200 \mathrm{~Wh}=3.2 \mathrm{kWh}$
Cost $=3.2 \times 5=₹ 16$
OR
Given: $\mathrm{I}=5 \mathrm{~A}, \mathrm{R}=2 \Omega, \mathrm{t}=30 \mathrm{~min}$
$\mathrm{E}=\mathrm{I}^{2} \mathrm{Rt}=5 \times 5 \times 2 \times 30 \times 60$
$\mathrm{E}=90000 \mathrm{~J}=90 \mathrm{~kJ}$

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