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
Answers
Answered by
80
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...
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...
Answered by
64
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
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
Anonymous:
Good answer sister
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