Electrical Energy and Power

You have already studied about the electrical energy and electrical power, their formula and the units involved. In this section, we will discuss them in terms of the total energy of a system and the power dissipated by them.

Electric Power Lines

Electrical Energy

A cell has two terminals – a negative and a positive terminal. The negative terminal has excess of electrons whereas the positive terminal has a deficiency of electrons. Let us take the positive terminal as A and the electrical potential at A is given by V(A). Similarly, the negative terminal is B and the electrical potential at B is given by V(B). Electric current flows from A to B, and thus V(A) > V (B).

The potential difference between A and B is given by

V = V(A) – V(B) > 0

Mathematically, electric current is defined as the rate of flow of charge through the cross-section of a conductor. Thus, it is given by I = ∆Q/ ∆t where I is the electric current and ∆Q is the quantity of electric charge flowing through a point in time ∆t.

The potential energy of charge Q at A is Q V(A) and at B, it is Q V(B). So the change in the potential energy is given by

∆Upot = Final potential energy – Initial potential energy

= ∆Q [(V (B) – V (A)] = –∆Q V

= –I V∆t (Since I = ∆Q/ ∆t)

If we take the kinetic energy of the system into account, it would also change if the charges inside the conductor moved without collision. This is to keep the total energy of the system unchanged. Thus, by conservation of total energy, we have:

∆K = –∆Upot

Or ∆K = I V∆t > 0

Thus, in the electric field, if the charges move freely across the conductor, there would be an increase in the kinetic energy as they move.

When the charges collide, the energy gained by them is shared between the atoms. Consequently, the vibration of the atoms increases resulting in the heating up of the conductor. Thus, some amount of energy is dissipated in the form of heat in an actual conductor.


We talked about the energy that is dissipated due to the heating up of the conductor. The energy dissipated in time interval ∆t is given by

∆W = I V∆t

And the energy dissipated per unit time is actually the power dissipated, which is given by P = ∆W/∆t. But we know the formula for power is given by P = I V

Hence, according to Ohm’s law, V = IR. Substituting we have,

P = I2 R = V2/R

It is this power which is responsible for heating up the coil of a bulb, which gives out heat and light.

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Practise This Question

An electric bell has a rating of 500 W 100 V. It is used in a circuit having a 200 V supply. What resistance must be connected in series with the bulb so that it delivers 500 W?