You have studied that when a positive charge moves from a point of higher potential to a point of lower potential, I delivers energy. During flow of electric current, positive charges continuously move from a high potential point to a low potential point. Thus the electric current becomes a continuous source of energy.
Consider two points having a potential difference of V volts. If one coulomb of charge passes between these points: the amount of energy supplied by the charge would be V joule. Hence when Q coulomb of charge flows between these two points, then we will get QV joule energy.
If we represent this energy by W, then
W = QV
As current is the rate of the flow of charge, so if a current I ampere flows for time t between two points, then during this time I X t coulomb of charge will flow. So the energy gained during t second is
W = QV =I x t x V ………. (16.13)
If a current is passing through a resistance R and the potential difference across its end is V, then by ohm’s law
V = I x R
Substituting the value of V in Eq. 16. 13
This energy can be utilized for different functions. For example, fans convert this energy into mechanical energy, bulb into light and heaters into heat. Usually this energy appears as heat in the resistance. This is the reason that we get heat when current passes through a heater. Eq. 16. 14 can be described in words as “the amount of heat energy generated in a resistance due to the flow of electric current is equal to the product of the square of current I, the resistance R and the time duration t. this is known as joule’s law.
Electric power
The amount of energy supplied by current in unit time is known as electric power. Hence power P can be determined by dividing the electric energy W by the time t i.e.,
When current I is flowing through a resistance R, the electric power that generates heat in the resistance is given by 
The unit of electric power is watt which is equal to one joule per second. It is represented by word W. electric bulbs commonly used in houses consume 25W, 40W, 60W, 75W and 100W of electric power.
Example 16.6: the resistance of an electric bulb is 
. When a potential of 250V is applied across its end, then find the power consumed by it.
Solution:
Kilowatt hour
Electric energy is commonly consumed in very large quantity for the measurement of which joule is a very small unit. Hence a larger unit of energy is required which is called kilowatt-hour. It is the amount of energy obtained by a power of one kilowatt in one hour.
1kWh = 1000W x 1 hour
= 1000W x (3600s)
The energy in kilowatt-hour can be obtained by the following formula:
The amount of energy in kilowatt-hour = 
The electric meter installed in our houses measures the consumption of electric energy in units of kilowatt hour according to which we pay our electricity bills. If the cost of one kilowatt-hour i.e. one unit is known, we can calculate the amount of electricity bill by the following formula:
Cost of electricity =number of units consumed x cost of one unit
Example 16.7: in a certain house, 4 electric bulbs of 100W each, are daily used for 5 hours. If the rate of electricity is Rs.4 per unit, fine the number of units consumed in 30 days and what would be its cost?
Solution:
Total cost = number of units consumed x cost of one unit = 60 x 4 = Rs.240/-
Do You Know? The power of various appliances in watts.
Fluorescent tube 40 Electric fan 80 Computer 90 Bulb 100 Colour T.V. 120 Refrigerator 500 Electric Iron 750 Toaster 1000 Microwave Oven 1450 Electric Oven 2500 Air-conditioner 2500 |