An electric refrigerator rated as 750w operates 8 h per day. What is the cost of energy to opeate it for monthof june at 2.50 per kwh?
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heya...
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
tysm....#gozmit
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
tysm....#gozmit
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