Question

The electricity tariff in a town is \[Rs\,\,3.00\] per unit. Calculate the cost of running an \[80\,\,{\text{watt}}\] fan for \[10\,\,{\text{h}}\] a day for the month of June.
A. \[Rs.\,\,12.00\]
B. \[Rs.\,\,52.00\]
C. \[Rs.\,\,72.00\]
D. \[Rs.\,\,92.00\]

Answer 1

Hint: Calculate the numbers of hours of fan used in the month of June. After that calculate the total amount of power. From there you can obtain the required amount of cost. Use \[{\text{1}}\,\,{\text{KWh}} = 1\,\,{\text{unit}}\].

Complete step by step answer:
Given, Cost of unit \[ = Rs.\,\,3.00\] per unit.
Power of the fan, \[P = 80\,\,{\text{watt}}\]
Operating time in a day, \[t = 10\,\,{\text{h}}\]

Now, for the month of June there are \[{\text{30}}\] days.
Therefore,
Total time of fan used is given by:
${\text{30}} \times {\text{10}} \\
\Rightarrow 300\,\,{\text{h}}\,\, \\$
Total amount of energy consumed is given by:
$80\, \times 300\,\,{\text{watt}}\,\,{\text{h}} \\
\Rightarrow {\text{24000}}\,\,{\text{watt}}\,\,{\text{h}} \\
\Rightarrow 24\,\,{\text{Kilo}}\,\,{\text{watt}}\,\,{\text{h}}$

We know that,
\[{\text{1}}\,\,{\text{KWh}} = 1\,\,{\text{unit}}\]
\[\therefore {\text{24}}\,\,{\text{KWh}} = 24\,\,{\text{unit}}\]

Cost of \[1\,\,{\text{unit}} = Rs\,\,3.00\]
Therefore, cost of $24\,\,{\text{unit}} = \left( {24 \times 3} \right)\,\,Rs. \\$
$\therefore 24\,\,{\text{unit}}= 72\,\,Rs. \\$

Hence, the cost of running the fan for the month of June is \[72\,\,Rs\].Hence, option C is correct.

Additional information:
Power is how easily energy is consumed or produced. A component of power is watt. A watt is the amount of energy used or generated per second by Joule. One-watt hour is equivalent to an hour's total power output of one watt. In \[{\text{KWh}}\] or Kilowatt Hour, a unit is represented. This is the real energy of electricity that is used. You will use \[{\text{1}}\] unit or \[{\text{1}}\] kilowatt-hour \[\left( {{\text{KWh}}} \right)\] of power, if you consume \[1000{\text{ watts}}\] and \[1\] kilowatt-hour of energy for \[1\] hour.

Note: While solving this problem, we do need to remember that here, power is to be taken as kilowatts and time in hours rather than seconds. Most of the students mess up the numerical by taking power in watts and time in seconds. It is the commercial unit of energy and we need to take care of that.