Physics, asked by pravinaahirpara, 11 months ago

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

Answered by ChitranjanMahajan
1

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

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