Question

If an electric iron of 1200W is used for 30 minutes every day, find electric energy consumed in the month of April.
(A) \[18\,{\text{kWh}}\]
(B) \[15\,{\text{kWh}}\]
(C) \[60\,{\text{kWh}}\]
(D) \[10\,{\text{kWh}}\]

Answer 1

Hint: First of all, we will convert the duration of time and power into its equivalent hours and kilowatts. We know that energy is given by the product of power and time. We will substitute the required values and manipulate accordingly to obtain the result.

Complete step by step answer:
In the given question, we are supplied with the following data:
The rated power of the electric iron is \[1200\,{\text{W}}\] .
The duration for which it is used in a day is for \[30\] minutes every day.
We are asked to find the electrical energy that it has consumed in the month of April.
To begin with, we will need to find out the energy consumed in a single day. After that we can proceed to find the electrical energy consumed in the month of April. We should know that this problem is based on the commercial use of electrical energy. We need to convert the time which is in minutes, to its corresponding equivalent hours. We will also convert the power into kilowatt hours.
Since, the power is given as \[1200\,{\text{W}}\] .
We know,
\[1000\,{\text{W}} = 1\,{\text{kW}}\]
Then,
\[1200\,{\text{W}} = 1.2\,{\text{kW}}\]
Again, time duration given as \[30\] minutes. So, we now, convert it into hours, as given below:
$30\,\min \\
\Rightarrow\dfrac{{30}}{{60}}\,{\text{h}} \\
\Rightarrow \dfrac{1}{2}\,{\text{h}} \\
\Rightarrow 0.5\,{\text{h}} \\$
We know, a formula which gives the electrical energy relating it with power and time duration is given below:
\[E = Pt\] …… (1)
Where,
\[E\] indicates electrical energy.
\[P\] indicates power of the device.
\[t\] indicates the time duration for which it is used.
Now, we substitute the required values in the equation (1) and we get:
$E = Pt \\
\Rightarrow E = 1.2 \times 0.5 \\
\Rightarrow E = 0.6\,{\text{kWh}} \\$
Therefore, the energy consumed in a day is found out to be \[0.6\,{\text{kWh}}\] .
We will now find the energy consumed in the month of April; whose total number of days are \[30\] .
So, the total energy consumed in the month of April is calculated as:
$ 0.6 \times 30\,{\text{kWh}} \\
\therefore 18\,{\text{kWh}}$
Hence, the total energy consumed in the month of April is found out to be \[18\,{\text{kWh}}\] .

The correct option is A.

Note: While solving this problem, many students tend to make mistakes by using the time duration other than hours. The unit of power should also be in kilowatts as this is the case of commercial use of electrical energy. One kilowatt hour energy is also termed as one unit of electricity.