Pagkalkulo sa konsumo sa kuryente sa mga gamit sa kuryente - mga problema ug solusyon
1. Sa usa ka balay, adunay 5 ka lampara nga 25 Watt nga gigamit sulod sa 14 ka oras kada adlaw, usa ka 200 Watt nga refrigerator nga gigamit sulod sa 24 ka oras kada adlaw, ug usa ka 125 Watt nga bomba sa tubig nga gigamit sulod sa 8 ka oras kada adlaw. Pila ka oras? kusog nga elektrikal gigamit sulod sa usa ka bulan (30 ka adlaw)?
Nailhan:
Gahum sa lampara = 5(25) = 125 Watt ug oras sa paggamit = 14 ka oras x 30 = 420 ka oras
Gahum sa refrigerator = 200 Watts ug oras sa paggamit = 24 oras x 30 = 720 ka oras
Gahum sa bomba sa tubig = 125 Watts ug oras sa paggamit = 8 ka oras x 30 = 240 ka oras
Gipangita: Electrical energy used for a month
Solusyon:
Electric energy = gahum sa kuryente 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)?
Nailhan:
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
Gipangita: Electrical energy is used for 1 month
Solusyon:
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
solusyon
Electric power of TV :
P = VI = (220 Volt)(0.3 Ampere) = 66 Volt Ampere = 66 Watt = 66 Joule/segundo
Electric power of computer :
P = VI = (220 Volt)(0.4 Ampere) = 88 Volt Ampere = 88 Watt = 88 Joule/segundo
Electric power of water pump :
P = VI = (220 Volt)(0.5 Ampere) = 110 Volt Ampere = 110 Watt = 110 Joule/segundo
Interval sa oras:
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 segundo = 2,376,000 Joule
Electric energy used by computer = 88 Joule/second x 21,600 segundo = 1,900,800 Joule
Electric energy used by water pump = 110 Joule/seconds x 14,400 segundo = 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…
Solusyon:
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
