Question

A household uses the following electrical appliances:
i. two electric fans of rating 80W each for 12 hours each day.
ii. refrigerator of rating 400W for 10 hours each day.
iii. six electric tubes of rating 18W each for 6 hours each day

Calculate the electricity bill of the house hold for the month of
July if the cost per unit of electrical energy is 3.

Answer 1

Answer:

For Fans,

Units = (wattage*hours)/1000

= (80*3*12)/1000

= 2.88 kWh

For Refrigerator,

Units = (400*10)/1000

= 4 kWh

For Electrical bulbs,

Units = (18*6*6)/1000 kWh

= 0.648 kWh

Total units = (0.648+2.88+4) kWh

= 7.528 kWh

Total units for a month = (7.528*30) kWh

= 225.84 kWh

= Cost of 1 kWh = 3

= Cost of 225.84 kWh

= (225.84*3) = 677.52

I hope this helps.