Question 6 - Past Year - 3 Mark Questions - Chapter 12 Class 10 - Electricity

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