Check sibling questions

Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is ` 6.00.

(i) Electric heater of 1000 W for 5 hours daily.

(ii) Electric refrigerator of 400 W for 10 hours daily.

 

Answer:

(i) P 1 = 1000 W = 1000 / 1000 kW = 1 kW

     t 1 = 5 hours

Energy, E 1 = P 1 * t 1 * number of days

                  = 1 * 5 * 30

                  = 150 kWh

(ii) P 2 = 400 W = 400 / 1000 kW = 0.4 kW

     t 2 = 10 hours

Energy, E 2 = P 2 * t 2 * number of days

                  = 0.4 * 10 * 30

                  = 120 kWh

Therefore, Total energy = E 1 + E 2  

                                          = 150 + 120 kWh

                                          = 270 kWh

Given, 

cost of 1 kWh of energy = 6

So,

Total Cost = Total energy * cost of 1 kWh of energy 

                   = 270 * 6

                    = 1620

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 12 years and a teacher from the past 16 years. He teaches Science, Economics, Accounting and English at Teachoo