Question

Consider an electrical conductor connected across a potential difference V. Let Δq be a small charge moving through it in time Δt. If I is the electric current through it,
(I) the kinetic energy of the charge increases by IVΔt.
(II) the electric potential energy of the charge decreases by IVΔt.
(III) the thermal energy of the conductor increases by IVΔt.
Then the correct statement/s is / are

• A. (I)
• B. (I),(II)
• C. (I) and (III)
• D. (II),(III)

Answer 1

The Correct Option is D

(I) The kinetic energy of the charge increases by $IV\Delta t$. This is incorrect. While the charges are accelerated by the electric field, they lose energy due to collisions with the lattice ions. The drift velocity remains constant, hence kinetic energy doesn't increase.

(II) The electric potential energy of the charge decreases by $IV\Delta t$. This is correct. The charge $\Delta q$ moves through a potential difference $V$, so the change in potential energy is $\Delta U = \Delta qV$. Since $I = \frac{\Delta q}{\Delta t}$, we have $\Delta q = I\Delta t$. Therefore, $\Delta U = IV\Delta t$. Since the charge moves from higher to lower potential, the potential energy decreases.

(III) The thermal energy of the conductor increases by $IV\Delta t$. This is correct. The loss in potential energy is converted to thermal energy through collisions with lattice ions, heating up the conductor (Joule heating).

The correct answer is (D) (II), (III).