Question

A domestic consumer uses a refrigerator or 60W for 5 hours an electric iron of 750W for half an hour and two 100W bulbs for 4 hours every day. What is the energy consumed in the month of June?

Answer 1

Answer:

Total Energy consumed is 442.5 kWh.

Explanation:

For Refrigerator,

Power = 60 W

Time = 5 hrs

Energy consumed in one day = [wattage*hours] / 100

= [60*5]/100 kWh

= 300/100 kWh

= 3 kWh

Energy in 30 days = [30*3] kWh

= 90 kWh

For Electric Iron,

Power = 750 W

Time = 0.5 hr

Energy consumed in one day = [wattage*hours] / 100

= [750*0.5] /100 kWh

= 375/100 kWh

= 3.75 kWh

Energy in 30 days = [30*3.75] kWh

= 112.5 kWh

For 1 bulb,

Power = 100W

Time = 4 hrs

Energy consumed in one day = [100*4] /100 kWh

= 400/100 kWh

= 4 kWh

Energy consumed in one day for 2 bulbs = [4*2] kWh

= 8 kWh

Energy consumed in 30 days = [8*30] kWh

= 240 kWh

∴ Total Energy Consumed = [90+112.5+240] kWh

= 442.5 kWh

Hope this helps.