Question

an electric refrigerator rated as 750 w operates 8 hours a day . what is the cost of the energy to operate it for one month of june at Rs 2.50 per KWH

Answer 1

convert 750 w into Kw by dividing it by thousand.
750/1000 = 3/4 Kw
energy consumed in the month of june = 3/4 * 8 * 30 = 180 KwH
cost of energy = 180 * 2.5 = 450 Rs. ANSWER.
thanks...

Answer 2

Hi !

Power = P = 750 W = 750/1000 kW

(1 kW = 1000 W ) ,     ( 1 W = 1/1000 kW)

Time = t =  8 hrs

Energy = P*t = 750/1000*8 = 0.75*8 = 6 J

Energy consumption in one day = 6 J

Energy consumption in the month of June = 6*30 = 180 J

Total cost = monthly consumption*rate
                = 180*2.50
               = Rs 450

The answer is Rs 450