Prove that power dissipated across a resistor is responsible for heating up the resistor. Give an example for it.

i. When a charge Q flows from the higher potential point to the lower potential point, its potential energy decreases by an amount,

∆U = QV = I∆tV

where I is current due to the charge Q flowing in time ∆t.

ii. By the principle of conservation of energy, this energy is converted into some other form of energy.

iii. In the limit as ∆t → 0, \(\frac{dU}{dt}\) = IV

Here, \(\frac{dU}{dt}\) is power, the rate of transfer of energy ans is given by p = \(\frac{dU}{dt}\) = IV

Hence, power is transferred by the cell to the resistor or any other device in place of the resistor, such as a motor, a rechargeable battery etc.

iv. Due to the presence of an electric field, the free electrons move across a resistor and their kinetic energy increases as they move.

v. When these electrons collide with the ion cores, the energy gained by them is shared among the ion cores. Consequently, vibrations of the ions increase, resulting in heating up of the resistor.

vi. Thus, some amount of energy is dissipated in the form of heat in a resistor.

vii. The energy dissipated per unit time is actually the power dissipated which is given by,

P = \(\frac{V^2}{R}\) = I²R

Hence, it is the power dissipation across a resistor which is responsible for heating it up.

viii. For example, the filament of an electric bulb heats upto incandescence, radiating out heat and light.