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

Step 1: Analyze each statement based on energy considerations in an electrical conductor.

The potential difference $V$ causes charge $\Delta q$ to move, resulting in work done by the electric field. The work done is $V \Delta q$. Since $I = \frac{\Delta q}{\Delta t}$, $\Delta q = I \Delta t$. So, Work done = $VI \Delta t$.

(I) The kinetic energy of the charge increases by $IV\Delta t$.
This is False. In a conductor, electrons accelerate due to the electric field, but they collide with the lattice ions, losing this kinetic energy. The drift velocity and hence average kinetic energy remain constant.

(II) The electric potential energy of the charge decreases by $IV\Delta t$.
This is True. When a positive charge moves from a higher potential to a lower potential, or a negative charge moves from a lower potential to a higher potential, electric potential energy decreases. Here, the charge is moving through a potential difference $V$, so the potential energy decreases by $V \Delta q = IV \Delta t$.

(III) The thermal energy of the conductor increases by $IV\Delta t$.
This is True. The energy lost in potential energy is converted into heat due to collisions of electrons with the atoms in the conductor lattice. This increases the thermal energy of the conductor.

Step 2: Identify the correct statement(s).

Statements (II) and (III) are correct.

Answer 28: (D) (II), (III)

Answer 2

(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).