Calculation of the electrical energy usage by electrical tool – problems and solutions

Calculation of the electrical energy usage by electrical tool – problems and solutions

1. In a house, there are 5 lamps 25 Watt used 14 hours per day, a 200 Watt refrigerator used 24 hours per day, and a 125 Watt water pump used 8 hours per day. How much electrical energy used for a month (30 days)?

Known :

Power of lamp = 5(25) = 125 Watt and time of use = 14 hours x 30 = 420 hours

Power of refrigerator = 200 Watt and time of use = 24 hours x 30 = 720 hours

Power of water pump = 125 Watt and time of use = 8 hours x 30 = 240 hours

Wanted : Electrical energy used for a month

Solution :

Electric energy = electric power x time interval

Energy (lamp) = (125 Watt)(420 hours) = 52,500 Watt jam = 52,5 kilo Watt hours

Energy (refrigerator) = (200 Watt)(720 hours) = 144,000 Watt jam = 144 kilo Watt hours

Energy (water pump) = (125 Watt)(240 hours) = 30,000 Watt jam = 30 kilo Watt hours

Electrical energy used for a month = 52,5 + 144 + 30 = 226,5 kilo Watt hour = 226,5 kilo Watt hour = 226,5 kWh.

2. A house using electrical tools as listed in the following table.

How much electrical energy is used for 1 month (30 days)?

Known :

Power of lamp = 4 x 10 Watt = 40 Watt and time of use = 10 hours x 30 = 300 hours

Power of TV = 1 x 100 Watt = 100 Watt and time of use = 10 hours x 30 = 300 hours

Power of iron = 1 x 300 Watt = 300 Watt and time of use = 2 hours x 30 = 60 hours

Wanted : Electrical energy is used for 1 month

Solution :

Electric energy = electric power x time

Energy used by lamp = (40 Watt)(300 hours) = 12,000 Watt hours = 12 kilo Watt hours = 12 kWh

Energy used by TV = (100 Watt)(300 hours) = 30,000 Watt hours = 30 kilo Watt hours = 30 kWh

Energy used by iron = (300 Watt)(60 hours) = 18,000 Watt hours = 18 kilo Watt hours = 18 kWh

Total energy = 12 kWh + 30 kWh + 18 kWh = 60 kWh.

3.

The electrical energy used by the above three electrical tools is…

A. 158,400 Joule

B. 950,400 Joule

C. 5,860,800 Joule

D. 9,504,000 Joule

Solution

Electric power of TV :

P = V I = (220 Volt)(0.3 Ampere) = 66 Volt Ampere = 66 Watt = 66 Joule/second

Electric power of computer :

P = V I = (220 Volt)(0.4 Ampere) = 88 Volt Ampere = 88 Watt = 88 Joule/second

Electric power of water pump :

P = V I = (220 Volt)(0.5 Ampere) = 110 Volt Ampere = 110 Watt = 110 Joule/second

Time interval :

10 hours = 10 x 3600 seconds = 36,000 seconds

6 hours = 6 x 3600 seconds = 21600 seconds

4 hours = 4 x 3600 seconds = 14400 seconds

Electric energy = electric power x time

Electric energy used by TV = 66 Joule/second x 36,000 seconds = 2,376,000 Joule

Electric energy used by computer = 88 Joule/second x 21,600 seconds = 1,900,800 Joule

Electric energy used by water pump = 110 Joule/seconds x 14,400 seconds = 1,584,000 Joule

Total electric energy :

2,376,000 Joule + 1,900,800 Joule + 1,584,000 Joule = 5,860,800 Joule

4. A family uses electrical energy for a month (30 days) for tools as in the following table!

Electric power of washing machine, TV and AC are…

Solution :

Electric energy of washing machine = 12 kWh = 12,000 Watt hour, time of use = 2 hours x 30 = 60 hours

Electric energy of TV = 19.2 kWh = 19,200 Watt hour, time of use = 8 hours x 30 = 240 hours

Electric energy of AC = 72 kWh = 72,000 Watt hour, time of use = 6 hours x 30 = 180 hours

Electric power = electric energy / time

Electric power of washing machine = 12,000 Watt hour / 60 hours = 200 Watt

Electric power TV = 19,200 Watt hour / 240 hours = 80 Watt

Electric power AC = 72,000 Watt hour / 180 hours = 400 Watt