
An electric refrigerator $400watt$,two fans $100watt$each and four bulbs $25watt$each are operated $8\dfrac{{hrs}}{{day}}$.Then find the cost of the electric bill for the month of June at the rate of rupees $\dfrac{{24}}{{unit}}$.
Answer
510.3k+ views
Hint:Generally, all electrical appliances are used in our houses and the machines in industry are specified with the power consumption per hour. For commercial or domestic use, the electricity charge is calculated in kilowatt-hour$\left( {kWh} \right)$. It is generally known as a unit of electricity.We are going to calculate the total number of units consumed by electrical appliances during the given time and the cost of the electric bill.
Complete step-by-step solution:
The power consumption of the refrigerator, fans, and bulbs were given. we need to calculate the total power consumption in units of electricity.
One kilowatt-hour $\left( {kWh} \right) = $power consumed in$\left( {kW} \right)$$ \times $time in$\left( {hr} \right)$
Power is the electrical energy transferred per unit time.SI unit of power is $watt\left( {\dfrac{J}{s}} \right)$. One watt is equal to one Joules per second.
Firstly, we need to calculate the units consumed by refrigerators, fans, and bulbs per day.
Power consumed by the refrigerator, fans, bulbs$ = $Electric refrigerator consumes $400watt$$ + $two fans consume $\left( {2 \times 100 = 200} \right)watt$ $ + $four bulbs consume $\left( {4 \times 25 = 100} \right)watt$
Power consumed by the refrigerator, fans, bulbs $ = 700watt$
Total units consumed per day$ = \left( {700watt \times 8\dfrac{{hrs}}{{day}}} \right)$
Total units consumed per day $ = 5600Wh = 5.6kWh$
We need to find the total number of units consumed in June. We know that June contains $30$days.
Total number of units consumed in June $ = \left( {5.6kWh} \right) \times 30 = 168kWh$
Next, we are going to pay the electric bill and we know that the cost of one unit $ = 24$rupees.
Cost of electricity bill$ = $(total number of units consumed) $ \times $(cost per unit) $ = \left( {168 \times 24} \right)$
Cost of electricity bill$ = 4032$rupees.
From the above calculations, we found the cost of electricity bill in June $ = 4032$rupees.
Note:Generally, units of electricity are measured to fix the electricity charges to the individual, company, industries, etc., One kilowatt-hour$\left( {kWh} \right)$is equal to $3600$ kilojoules.$kWh$ meter is used to measure the units consumed by the electrical appliances.
Complete step-by-step solution:
The power consumption of the refrigerator, fans, and bulbs were given. we need to calculate the total power consumption in units of electricity.
One kilowatt-hour $\left( {kWh} \right) = $power consumed in$\left( {kW} \right)$$ \times $time in$\left( {hr} \right)$
Power is the electrical energy transferred per unit time.SI unit of power is $watt\left( {\dfrac{J}{s}} \right)$. One watt is equal to one Joules per second.
Firstly, we need to calculate the units consumed by refrigerators, fans, and bulbs per day.
Power consumed by the refrigerator, fans, bulbs$ = $Electric refrigerator consumes $400watt$$ + $two fans consume $\left( {2 \times 100 = 200} \right)watt$ $ + $four bulbs consume $\left( {4 \times 25 = 100} \right)watt$
Power consumed by the refrigerator, fans, bulbs $ = 700watt$
Total units consumed per day$ = \left( {700watt \times 8\dfrac{{hrs}}{{day}}} \right)$
Total units consumed per day $ = 5600Wh = 5.6kWh$
We need to find the total number of units consumed in June. We know that June contains $30$days.
Total number of units consumed in June $ = \left( {5.6kWh} \right) \times 30 = 168kWh$
Next, we are going to pay the electric bill and we know that the cost of one unit $ = 24$rupees.
Cost of electricity bill$ = $(total number of units consumed) $ \times $(cost per unit) $ = \left( {168 \times 24} \right)$
Cost of electricity bill$ = 4032$rupees.
From the above calculations, we found the cost of electricity bill in June $ = 4032$rupees.
Note:Generally, units of electricity are measured to fix the electricity charges to the individual, company, industries, etc., One kilowatt-hour$\left( {kWh} \right)$is equal to $3600$ kilojoules.$kWh$ meter is used to measure the units consumed by the electrical appliances.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

