📖 generic · CBSE Class 12th English Medium · PHYSICS PART-1 · Page 109question

3.9 E LECTRICAL E NERGY , P OWER

Chapter 3: Chapter 3 · PHYSICS PART-1

. E LECTRICAL E NERGY , P OWER Consider a conductor with end points A and B, in which a current I is flowing from A to B. The electric potential at A and B are denoted by V (A) E XAMPLE . and V (B) respectively.

Since current is flowing from A to B, V (A) > V (B) and the potential difference across AB is V = V (A) – V (B) > . In a time interval ∆ t , an amount of charge ∆ Q = I ∆ t travels from A to B. The potential energy of the charge at A, by definition, was Q V (A) and similarly at B, it is Q V (B). Thus, change in its potential energy ∆ U pot is ∆ U pot = Final potential energy – Initial potential energy = ∆ Q[( V (B) – V (A)] = – ∆ Q V = – I V ∆ t < ( .

) If charges moved without collisions through the conductor, their kinetic energy would also change so that the total energy is unchanged. Conservation of total energy would then imply that, ∆ K = – ∆ U pot ( . ) that is, ∆ K = I V ∆ t > ( . ) Thus, in case charges were moving freely through the conductor under the action of electric field, their kinetic energy would increase as they move.

We have, however, seen earlier that on the average, charge carriers do not move with acceleration but with a steady drift velocity. This is because of the collisions with ions and atoms during transit. During collisions, the energy gained by the charges thus is shared with the atoms. The atoms vibrate more vigorously, i.e., the conductor heats up.

Thus, in an actual conductor, an amount of energy dissipated as heat in the conductor during the time interval ∆ t is, ∆ W = I V ∆ t ( . ) The energy dissipated per unit time is the power dissipated P = ∆ W / ∆ t and we have, P = I V

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