Question

Calculate the units of electricity consumed in the month of November from the following details. One $60\,W$ bulb is used for $5$ hours daily. One $100\,W$ bulb is used for $3$ hours daily. One $1\,kW$ electric heater is used for $2$ hours daily.
A) 78
B) 14
C) 36
D) 18

Answer 1

Hint: Electrical energy consumed can be calculated using the information about Electrical Power rating of the device and the time for which the device is used, current is run through the device. Therefore, the electrical energy is given by: \[Electrical{\text{ }}energy(E)\, \propto \,Electrical\,Power(P)\, \times \,time(t)\].
Numerically, $E\, = \,P\, \times \,t$

Complete step-by-step answer:
Initially, we convert the given power ratings of the appliances from W to kW.
First Bulb rates, ${P_1}\, = \,60W\, = \,0.06kW$ for ${t_{1\,}}\, = \,5\,hours$.
Second Bulb rates ${P_2}\, = \,100W\, = \,0.1kW$for ${t_2}\, = \,3\,hours$.
Electric heater rating ${P_3}\, = \,1kW$ for ${t_3} = \,2\,hours$
Now by using the above formula from hint, $E\, = \,P\, \times \,t$
Therefore, ${E_1}\, = \,{P_{1\,}}\, \times \,{t_1}$
Substituting values in this equation,
${E_1}\, = \,0.06\, \times \,5\, = \,0.3kWh$
Similarly, we can get value of ${E_2}$ and ${E_3}$
${E_2}\, = \,0.1\, \times \,3\, = \,0.3kWh$
${E_3}\, = \,1\, \times \,2\, = \,2kWh$
Total Electrical Energy = ${E_1}\, + \,{E_2}\, + \,{E_3}$
E = $0.3\, + \,0.3\, + 2\, = \,2.6kWh$
Therefore, $2.6kWh$ is consumed in one day, November has $30$ days, so total power consumption in the month of November is given by
$2.6\, \times \,30\, = \,78kWh$
Also, $1kWh\, = \,1\,unit$.
Therefore, $78$ units are used in the month of November.

Hence, the correct option is A.

Note: The unit of electrical energy is joule, but it is a smaller unit and not convenient for commercial purposes hence a bigger unit that is kilowatt hour is used as a commercial unit of electrical energy.
To obtain a bigger unit of electrical energy greater unit of electric power and greater unit of time is used therefore instead of $1W$, $1kW$ is used for electric power and instead of $1$ second $1$ hour is used for time.
$1kWh\, = \,1000W\, \times \,3600\,\sec onds$
$1kWh\, = \,3.6\, \times \,{10^6}\,J$
\[1kWh\] is equal to \[3.6\] into \[10\] to the power \[6\] joules. The amount of electrical energy consumed in terms of units is multiplied by the number of days in a month to get the total cost of electrical energy consumed in that particular month.