Question

An electric refrigerator rated $400\text{ }W$operates $8$ hours per day. What is the cost of the energy to operate it for one month at a rate $Rs.\text{ }5.00$per unit?

Answer 1

Since power is defined as the amount of energy consumed or work done per unit time. Therefore, given power consumption when multiplied by the number of hours the appliance works per day and further multiplying by $30$ would give us the energy consumption for the month which when multiplied by the unit price would give us the amount of energy consumption required at the end of the month.

Complete step by step solution:
In our day to day life we use many kinds of electrical appliances for different purposes for varying amounts of time. To avoid any wastage of electricity or being overpaid the electricity board charges us money per unit of electricity consumed ($1\,unit=1\,kWhr$). Thus, to calculate the monthly cost of electricity one could simply multiply the total energy consumed throughout the month with the monthly unit to get the total amount.
Let the wattage of the appliance be given by $P$
According to the question it is given that;
$P=400\,W$
Since electrical units are defined in terms of $kWhr$, therefore by converting $P$ to $kW$, we get;
$P=\dfrac{400}{1000}\,kW$
Or, $P=0.4\,kW$
Number of hours per day $=8\,hrs$
Let ${{E}_{1\,day}}$ be defined as the energy consumption of the appliance per day. According to the definition of power;
$P=\dfrac{E}{t}$ .
Hence to evaluate the energy consumption for any length of time when the power is given we simply need to multiply the given power with the time period of use. In other words;
$E=P\times t$
Therefore, the total energy consumption for one day for this appliance would be given by;
$\begin{aligned} & {{E}_{1\,day}}=P\times t \\\ & {{E}_{1\,day}}=0.4\times 8 \\\ & {{E}_{1\,day}}=3.2\,kWhr \\\ \end{aligned}$
Now the total energy consumption for the whole month is calculated by multiplying the energy consumption per day by$30$ ;
Let the total energy consumed throughout the month be defined as ${{E}_{total}}$. In other words;
$\begin{aligned} & {{E}_{total}}={{E}_{1\,day}}\times 30 \\\ & {{E}_{total}}=3.2\times 30 \\\ & {{E}_{total}}=96\,kWhr \\\ \end{aligned}$
Let the total cost to be paid by the end of the month be $C$.
Since we know that to find the total cost of the electricity consumed we need to just multiply the total energy with the cost of one unit. Given, the cost of one unit is$Rs.\text{ }5$ . Hence the total amount $C$ to be paid is given by;
$\begin{aligned} & C=96\times 5 \\\ & C=Rs.\,480 \\\ \end{aligned}$
Hence, by the end of the month the consumer has to pay a sum of $Rs.\text{ }480$ .

Note: In a real world scenario the power consumption of any device is generally lower than the wattage mentioned on the device. This is because often when we use such devices as a load in an electrical circuit, there is always some fraction of energy which gets dissipated as heat and it is the main reason why some devices heat up after long use.