In a house, there are four rooms each room is fitted with a 100 watt bulb and a 40W tubelight which is used
for 5 hours daily. In kitchen a 500W refrigerator is used for 10 hours daily, then calculate the electric bill of
month November 2007. The cost of one unit electric energy is Rs. 3.
Answers
The electricity bill for the month of November 2007 is Rs 468.45 .
• The power of an appliance is the work done by the appliance or the energy consumed by it in unit time. Mathematically, Power = Energy consumed / Time
=> Energy = Power × Time
• Given,
Power of the bulb = 100 W = 1/1000 kW = 0.001 kW
Time for which it is used daily = 5 hours
Therefore, energy consumed by the bulb = Power × Time = 0.001 × 5 kWh = 0.005 kWh
• Given,
Power of the tube light = 40 W = 40/1000 kW = 0.04 kW
Time for which is used daily = 5 hours
Therefore, energy consumed by the tube light = 0.04 × 5 kWh = 0.20 kWh
• Given,
Power of the refrigerator = 500 W = 500/1000 kW = 0.5 kW
Time for which it is used daily = 10 hours
Therefore, energy consumed by the refrigerator = 0.5 × 10 kWh = 5 kWh
• Total energy consumed by the bulb, the tube light, and the refrigerator in a day = 0.005 kW + 0.20 kW + 5 kW = 5.205 kW
• Number of days in the month of November = 30
Therefore, energy consumed by all the appliances in November 2007 = 30 × 5.205 kW = 156.150 kW
• Given, the cost of one unit electricity = Rs. 3
1 unit = 1 kWh
Therefore, the monthly electricity bill for the month of November = Rs. 3 × 156.15 = Rs. 468.45