Electricity
Electric potential
We use electricity to run various electrical appliances such as bulbs, tube lights, refrigerators, electric heaters etc. in our homes. Do you think all these home appliances consume an equal amount of electricity at a given time?
No, the amount of electricity consumed by an electrical appliance depends on the power rated on that appliance. For example, for 220 V potential supply, a tube light of rated power 40 W draws 0.18 A of current, whereas a bulb of rated power 100 W draws 0.45 A of current.
How can you determine the rate of consumption of energy by a given appliance?
Electric power
Electric power is defined as the rate of consumption of energy or simply the rate of doing work.
i.e., power (P) =……………. (i)
The work done by current (I) when it flows in a potential (V) for time (t) can be given by
W = VIt …………………. (ii)
⇒ Power, (P) = = VI
∴ Electric power or P = VI
The SI unit of power is watts (W).
Also, the energy dissipated or consumed by an electric appliance per unit time is given by
$\frac{H}{t}={I}^{2}R$ 
It is also an equation for electric power.
i.e., P = I^{2}R
Now, according to Ohm’s law
⇒ V = IR
⇒$I=\frac{V}{R}$
If we substitute the value of I in the equation of power, we get
$P={\left(\frac{V}{R}\right)}^{2}R=\frac{{V}^{2}}{R}$
Hence, we get the expression for electric power as



$P=VI={I}^{2}R=\frac{{V}^{2}}{R}$


Where, 1 W = 1 V × 1 A = 1 V A
1 watt is defined as the power consumed by an electrical circuit that carries a current of 1 ampere, when it is operated at a potential difference of 1 volt.

Since Watt is a very small unit, we define a larger unit of power as kilowatt (kW). Thus,
1 kW = 1000 W
For practical purposes, we define kilowatt hour (kWh) as ‘unit’ where,
1 unit = 1 kWh = 1000 W × 1 hour
= 1000 W × 3600 s
= 36 × 10^{5} Ws
= 3.6 × 10^{6 } J
1 kWh is also the commercial unit of electric energy.
Note that electricity is a flow of electrons and nothing else. Hence, power stations only make the electrons flow through conducting wires for which they charge us. They do not create or generate the electrons.
Prepare a list of electrical appliances commonly used in homes. Note down the respective watts rated on them. Now, try to calculate the amount of current drawn by respective appliances for a constant voltage of 220 V. Complete the table and discuss the result with your friends.
Electrical appliance 
Rated power 
Current drawn 
Audio player 
100 W 
0.45 A 
Fan 
60 W 
….. 
TV 
120 W 
….. 
Refrigerator 
160 W 
….. 
Electric heater 
1000 W 
….. 
Example:
(i) The amount of current drawn by an immersion water heater is 6.8 A. If the heater is operated on 200 V of potential difference, calculate its power.
Solution:
Given that,
Potential difference, V = 200 V
Current drawn, I = 6.8 A
By definition of electric power, we know that
P = VI
Hence, P = 200 V × 6.8 A
= 1360 W
Hence, power of the given immersion heater is 1360 W.
(ii) What is the monthly cost of energy to operate the given immersion water heater 3 hours/day at Rs 5.00 per unit?
Solution:
Power of the heater is 1360 W.
Energy consumed by it per day = 1360 W ×
Hence, energy consumed by it in 30 days = 1360 W × × 30 days
= 122400 W
= 122.4 kWh
= 122.4 units
Now, the cost of energy for one unit is Rs 5.00
∴ Cost of energy for 122.4 units will be 5 × 122.4 = 612
Hence, the monthly cost of energy of the heater is Rs 612.
What is the amount of energy consumed by the given water heater in one month?
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