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)
Answer
303.6k+ views
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.
Complete step by step solution:
| S.no. | Electronic device | Power (P) | Time (T) |
| 1. | 5 Bulbs | 60 Watts = 60/1000 = 0.06 kiloWatts | 10 hours |
| 2. | 4 Electric fans | 150 Watts = 150/1000 = 0.15 kiloWatts | 5 hours |
| 3. | 1 Geyser | 2 kiloWatts | 30 minutes = 0.5 hours |
| 4. | 1 Refrigerator | 800 Watts = 800/1000 = 0.8 kiloWatts | 20 hours |
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.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

What Are Current and Potential Difference in Electricity?

Understanding the Angle of Deviation in a Prism

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Average and RMS Value in Electrical Circuits

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

Understanding the Centre of Mass of a Semicircular Ring

How Does Fusion Reaction Happen Inside the Sun?

Free Radical Substitution and Its Stepwise Mechanism

Understanding Electromagnetic Waves and Their Importance

Daughters Day 2026: Date in India, Wishes, Quotes & Significance

