Question:
If an electric iron of $1200 \mathrm{~W}$ is used for 30 minutes everyday, find electric energy consumed in the month of April.
Solution:
Energy consumed in one day $=P \times t=1200 \mathrm{~W} \times 1 / 2 \mathrm{~h}=600 \mathrm{Wh}$
Energy consumed in 30 days $=600 \mathrm{Wh} \times 30=1800 \mathrm{Wh}=18 \mathrm{kWh}$.