Question

An electric refrigerator rated 750w operates 8 hours per day. What is the cost of the energy to operate it for the month of june at rs.2.50 per kwh

Answer 1

Power = 750 W = 750/1000 Kw

1W = 1/1000 Kw

Time = 8 hr .

Energy = P*t

= 750/1000*8

= .75*8

= 6 Joule .

So , total consumption of energy in 1 day = 6J

Energy consumed in month of June --

= 6*30

= 180 J.

Total cost = monthly consumption * rate

= 180* 2.50

= 450 Rs .

Answer ---- 450 Rupees

Answer 2

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