
In a house two 60W electric bulbs are lighted for 4 hours and three 100 W bulbs for 5 hours every day. Calculate the electric energy consumed in 30 days.
Answer
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Hint: When electric current passes through the appliances in our house, electrical energy gets converted to other forms of energy. This is known as consumption of electric energy. To answer this question, use the formula for consumption of electric energy. Firstly, substitute the values for two 60W electric bulbs and find the electric energy consumed in one day. Then similarly substitute the values for three 100 W bulbs and find the total energy consumed in one day. Add up both the electric energies consumed to get the total electric energy consumed every day. Then multiply this obtained value by 30 to get total electric energy consumed in 30 days.
Formula used:
$E= P \times N \times T$
Complete answer:
Case.1) Given: Power ${P}_{1}$= 60 W
Number ${N}_{1}$= 2
Time ${T}_{1}$= 4 hours
Electric energy consumed is given by,
$E= P \times N \times T$ …(1)
For two 60 W electric bulbs, equation. (1) can be written as,
${E}_{1}= {P}_{1}\times {N}_{1} \times {T}_{1}$
Substituting values in above equation we get,
${E}_{1}= 60 \times 2 \times 4$
$\Rightarrow {E}_{1}= 480 Wh$
Case.2) Given: Power ${P}_{2}$= 100 W
Number ${N}_{2}$= 3
Time ${T}_{2}$= 5 hours
Electric energy consumed is given by,
$E= P \times N \times T$ …(2)
For three 100 W electric bulbs, equation. (2) can be written as,
${E}_{2}= {P}_{2}\times {N}_{2} \times {T}_{2}$
Substituting values in above equation we get,
${E}_{1}= 100 \times 3 \times 5$
$\Rightarrow {E}_{1}= 1500 Wh$
Total electric energy consumed everyday will be given by,
$E= {E}_{1}+{E}_{2}$
Substituting values in above equation we get,
$E= 480 + 1500$
$\Rightarrow E=1980 Wh$
Total electric energy consumed in 30 days will be given by,
${E}_{30}= 30\times {E}$
Substituting the values in above equation we get,
${E}_{30}= 30 \times 1980$
$\Rightarrow {E}_{30}= 59400 Wh$
$\Rightarrow {E}_{30}= 59.4 kWh$
Hence, electric energy consumed in 30 days will be 59.4 kWh.
Note:
Students must not get confused between kW and kWh. They must remember that the
power consumption of electricity is generally measured in kilowatt i.e. kW and the unit of electric energy is kilowatt hour i.e. kWh. As kWh is a unit of energy it can be written in terms of joules as well. 1 kWh is equal to 3600 kilojoules or 3.6 megajoules. Joule is a smaller unit and not convenient for commercial purposes hence a bigger unit which is kilowatt hour is used.
Formula used:
$E= P \times N \times T$
Complete answer:
Case.1) Given: Power ${P}_{1}$= 60 W
Number ${N}_{1}$= 2
Time ${T}_{1}$= 4 hours
Electric energy consumed is given by,
$E= P \times N \times T$ …(1)
For two 60 W electric bulbs, equation. (1) can be written as,
${E}_{1}= {P}_{1}\times {N}_{1} \times {T}_{1}$
Substituting values in above equation we get,
${E}_{1}= 60 \times 2 \times 4$
$\Rightarrow {E}_{1}= 480 Wh$
Case.2) Given: Power ${P}_{2}$= 100 W
Number ${N}_{2}$= 3
Time ${T}_{2}$= 5 hours
Electric energy consumed is given by,
$E= P \times N \times T$ …(2)
For three 100 W electric bulbs, equation. (2) can be written as,
${E}_{2}= {P}_{2}\times {N}_{2} \times {T}_{2}$
Substituting values in above equation we get,
${E}_{1}= 100 \times 3 \times 5$
$\Rightarrow {E}_{1}= 1500 Wh$
Total electric energy consumed everyday will be given by,
$E= {E}_{1}+{E}_{2}$
Substituting values in above equation we get,
$E= 480 + 1500$
$\Rightarrow E=1980 Wh$
Total electric energy consumed in 30 days will be given by,
${E}_{30}= 30\times {E}$
Substituting the values in above equation we get,
${E}_{30}= 30 \times 1980$
$\Rightarrow {E}_{30}= 59400 Wh$
$\Rightarrow {E}_{30}= 59.4 kWh$
Hence, electric energy consumed in 30 days will be 59.4 kWh.
Note:
Students must not get confused between kW and kWh. They must remember that the
power consumption of electricity is generally measured in kilowatt i.e. kW and the unit of electric energy is kilowatt hour i.e. kWh. As kWh is a unit of energy it can be written in terms of joules as well. 1 kWh is equal to 3600 kilojoules or 3.6 megajoules. Joule is a smaller unit and not convenient for commercial purposes hence a bigger unit which is kilowatt hour is used.
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