B in the time interval ‘t’ seconds as shown in fig.- 31. Let V be the
potential
difference between the points A and B. The work done by electric field
in
time ‘t’ is given by
W = QV .........................(1)
This work is equal to the energy lost by the charge while passing through
the conductor.
• What is the energy lost by the charge in 1 sec?
It is equal to W/t .
From equation (1), we get
W/t= QV/t .........................(2)
In above equation Q
/t represents the current (I) flowing through the
conductor.and W/t represents the work done per second.
In lower classes you had studied
that power is nothing but the rate of
doing work. Hence, W/t represents electric power (P).
Electric power P = VI .........................(3)
This equation can be used to calculate power consumption by any
electric device that is connected in a circuit.
According to the ohm’s law,
V = IR
We can write equation (3) as
P = I2R =V
2/R
The equation P = VI can also be used to know the power that can be
extracted from a battery or any source. In such case we modify the equation
P = VI as P = ε I.
Where ε is the emf of the battery.
Let us consider an example to understand power consumption.
A bulb is marked 60W and 120V. This means that if this bulb is
connected to 120V source, it will able to convert 60J of electrical energy
into heat or light in one second.
Thus, the bulbs marked as 60W and 120V will offer a resistance of
240 Ω to the flow current through it in normal condition.
From the marking of bulb we can measure the resistance of the bulb.
From the relation P = V2
/R → R = V2
/P
R = 120 x 120/ 60 = 240 Ω
Thus, the bulbs marked as 60W and 120V will offer a resistance of
240 Ω to the flow current through it in normal condition.
If this bulb is connected to the 12V a battery, the consumption by the
bulb is given by