Question

An electric refrigerator rated 500 W operates 5 hour/day. What is the cost of the energy to operate it for 30 days at Rs 8.00 per kW h?​

Answer 1

Answer:

Pls mark me as Brainlist..

Explanation:

The energy of the power rating is multiple with the time of the operation is E=P×T

The cost of the energy consumption multiplied by the cost per unit of energy consumed is Cost=E×Costunit

Since the electric refrigerator rated 500W operates 6 hours/day. Thus, the energy consumption is power rating multiplied we get, E=P×T⇒E1day=500×6

Further solving we get E1day=3000Wh⇒3kWh to convert any quantity into its kilo form, just divide it by the thousand.

Now multiple the energy consumption for one day by 30 and that will be the total energy consumption is Enet=E1day×30⇒3×30=90kWh

Now the cost is energy consumption multiplied by the cost per unit of the energy consumed.

Hence, we get Cost=E×Costunit⇒90×4.5 (4.5 per kWh)

Further solving we have Cost=Rs.405

Therefore, the cost of the energy to operate it for 30 days 4.5 per kWh is Rs.405.

Note: The electricity bill that we pay also works on this same method as we used above, as the number of the given units on the meter means that many kWh of the energy is used. This is multiplied by the local electric cost of the electricity bill that we will need to pay per unit.

Since in the above we used units, one day, the net is the representation of the given information, and using the formulas we obtained the required result.

Answer 2

Answer:

Given,

power, P = 500 W

operating time, t = 5 hour/day

cost per KW/h = 8 Rs

To find :

Total cost for operating 30 days.

Solution :

we know that,

$\sf \: W = Pt \\ \sf \: \: \: \: \: \: = 500 \times 5 \: \: \: Wh \\ \sf \: \: \: \: \: \: = \frac{500 \times 5}{1000} \: \: KWh \\ \sf \: \: \: \: \: \: = 2.5 \: \: KWh$

So energy is used per one day = 2.5 KWh

Total cost = 2.5 × 30 × 8

= 600 Rs ( Answer.)