Calculate the monthly (30 days a month) Energy Bill for a domestic application where 5 bulbs each of 60 W are running for 10 hrs a day, 150 W electric Fan 4 numbers running for 5 hrs a day. One number Geyser of 2kW operated for 30 mins. In a day and refrigerator of 800 W, one number for 20 hrs. The cost of an electric unit is Rs 7/- per unit. (Formula based & data interpretation)

Hint: For calculating electricity cost, first of all find out the power on which each electric device is operated. Here, make sure that power is in Kilowatts only if not then convert watts to kilowatts (1 watt = kilowatts/1000). Then, find out the kilowatts an electronic appliance uses per month. find out the number of hours for which electronic device is on per day and make sure to convert the time in hours only. Now, finally find out the electricity cost.

Complete step by step solution:


Formula for energy consumed, ${{E = Pt}}$
Where P = power and t = time
Energy consumed by 5 bulbs of power, ${{{E}}_{{1}}}{{ = 5 \times 0}}{{.06 \times 10 \times 30 = 90kWh}}$
Here, number of bulbs $ = 5$
Power consumed = ${{0}}{{.06 kW}}$
Time = ${{30}} \times {{10 hrs}}$
Energy consumed by 4 electric fans, ${{{E}}_{{2}}}{{ = 4 \times 0}}{{.15 \times 5 \times 30 = 90kWh}}$
Here, number of electric fans $ = 4$
Power consumed = ${{0}}{{.15 kW}}$
Time = ${{30}} \times {{5 hrs}}$
Energy consumed by 1 Geyser, ${{{E}}_{{3}}}{{ = 1 \times 2 \times 0}}{{.5 \times 30 = 30kWh}}$
Here, number of Geyser$ = 1$
Power consumed = ${{2 kW}}$
Time = ${{30}} \times {{0}}{{.5 hrs}}$
Energy consumed by 1 refrigerator, ${{{E}}_{{4}}}{{ = 1 \times 0}}{{.8 \times 20 \times 30 = 480kWh}}$
Here, number of refrigerator $ = 1$
Power consumed = ${{0}}{{.8 kW}}$
Time = ${{30}} \times {{20 hrs}}$
Total energy consumed by the electric devices in one month
${{90kWh + 90kWh + 30kWh + 480kWh = 690kWh}}$
The cost of an electric unit is Rs 7/- per unit
So, total cost of ${{690kWh}}$ is given by
${{690 \times 7 = 4830/ - }}$

Therefore, the monthly (30 days a month) Energy Bill is ${{Rs 4830}}$.

Note: A watt hour and kilowatt hour are practical units of electric energy. The commercial unit of electric energy is kilowatt hour (kWh), where ${{1kWh = 1000Wh = 3}}{{.6 \times 1}}{{{0}}^{{6}}}{{J}}.$ Electric power is the rate at which electric energy is consumed. SI unit of electric power is Watt.