Question

An electric refrigerator rated as 750 W operates 8 hours per day. What is the cost of the energy to operate it for a month of June at Rs. 2.50 per kWh?​

Answer 1

Answer:

your answer is here !

Explanation:

Given, Power, P = 750 W

Time, T = 8 hrs per day

We know that, electrical energy is the product of power and time.

Therefore, the energy consumed by the refrigerator in one day

= 750 W × 8.0 hour/day

= 6000 W h = 6 kW h

Now, the total energy consumed by the refrigerator in the month of June (30 Days)

= 30 × Energy/day

= 30 × 6kWh

=180 kWh

Thus, the cost of energy required to operate the refrigerator for the month of June at Rs. 2.50 per kWh = 180 kW h × Rs 2.50 = Rs 450.00

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Answer 2

ANSWER

ANSWER REFER TO THE ATTACHMENT