The total electric energy consumed by an individual (especially houses) measured in kilowatt-hour is called electric power consumption. All the power companies install a meter in their customer's house so that they can charge according to the electric consumption. Those installed meters measure electricity on kilowatt-hour (KWH). In normal terms, we say 1 unit for 1 KWH. It is actually the amount of power used for a certain period of time. For the easy purpose, we say an electrical appliance of 1000W consuming power for one hour will consume 1KWH i.e. 1KWH = 1KW x 1hr.
Power Consumption = Power consumed by appliance x no of appliance x time operating
= P (in KW) x N x T (in hour)
Amount to pay = power consumption x price per unit
So, to calculate total power consumption of a home, we need to know the power of each equipment, number of equipment and time they run. Since all the appliances are not run at the same time and same period of time it is better to calculate power consumed by each appliance and add all at last.
Example:
If a home has 10 tube lights of 30W that run for 4 hours daily, 5 television of 200W running for 3 hours daily, 2 irons 1000W 1 hour weekly. Calculate total amount to be paid if the cost per unit is Rs. 7.
Here,
Power consumed by tube lights in one day = 10 x 301000 x 4 = 65 KW Power consumed by tube lights in one month = 65 x 30 = 36 KW
Power consumed by television in one day = 5 x 2001000 x 3 = 3 KW Power consumed by television in one month = 3 x 30 = 90 KW
Power consumed by iron in one week = 2 x 10001000 x 1 = 2KW Power consumed by iron in one month = 2 x 4(one month has 4 weeks) = 8 KW
Total power consumption in one month = 36 + 90 + 8 = 134KW Total amount to be paid = total power consumption x price per unit= 134 x 7 = Rs. 938
The frequency of electricity distributed in Kathmandu is 50 Hertz implies that the polarity of this current is alternated at the rate 50 times a second.
One kilowatt hour of electrical energy is defined as the energy consumed by an electrical appliance having power rating of one kilowatt for an hour.
Here,
We have,
1 kwhr = 1 unit
1 kwhr = 1000 watt hour
1 kwhr = 1000 watt × 3600 sec
1 kwhr = 3600000 joules
1 kwhr = 3.6 × 10^{6} joules
The power of an electric machine is the rate of doing work per unit time. Actually, the rate of converting energy in unit time is called electric power. The electric power is measured in watt.
1 watt = 1 joule/ second
Electrical power is calculated as follows: -
Power (P) = potential difference (V) × current (I)
P = V × I
Solution,
Here,
Power of bulb= 100 watt × 3 hours = 300 wh
Power of heater = 1000 watt × 3 hours = 300 wh
= 3.3 kwh Unit consumption per day = 3.3 kwh = 3.3 unit
Unit consumption for 15 days = 15 × 3.3 unit = 49.5 unit
Therefore, the electric unit consumed in 15 days of the room is 49.5 units.
Solution,
Here,
Energy consumption of bulb = PNT
= 100 × 10 × 6
= 6000 kwhr = 6 kwh
Energy consumption of heater = 2000 × 2 × 2 = 8000 kwhr
= 8 kwhr
Total electrical energy consumed by these appliances in a day ,
6 kwhr + 8 kwhr = 14 kwhr = 14 unit.
Solution,
Here,
Power of bulb = 100 watt
Numbers of bulb = 4
Hence, the consumption per day = PNT
= 100 × 4 × 5
= 2000 wh
Now, the consumption for one month = 30 × 2000 wh
= 60,000wh
= 60 kwh
= 60 unit
The total cost of electricity = unit consumed × per unit cost
= 60 × 5.50
= Rs. 390
The installed meters measure electricity on ______.
kilo-watt meter
deca-watt hour
kilo-watt second
kilo-watt hour
What is not necessary to calculate the total power consumption?
An electric bulb can be worked from ______.
AC source only
DC source only
all of the supply
battery supply only
The potential difference across a resistor is 12 V. When a current of 5 A is flowing through this resistor, what is the power used by the resistor?
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