Match List-I with List-II
| List-I | List-II |
| Potential difference across the terminal of a battery | Situations |
| (A) \( V = E \) | (I) cell is being charged. |
| (B) \( V = 0 \) | (II) The internal resistance of a cell is zero. |
| (C) \( V = E - ir \) | (III) cell is short-circuited. |
| (D) \( V = E + ir \) | (IV) current is being drawn from a cell. |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
The potential difference across the terminals of a cell with emf \( E \) and internal resistance \( r \) carrying current \( I \) is \( V = E \mp I r \): minus while the cell supplies current, plus while it is being charged.
(A) \( V = E \) — this matches (II) The internal resistance of a cell is zero. With \( r = 0 \), the drop \( Ir \) vanishes, so the terminal voltage equals the emf.
(B) \( V = 0 \) — this matches (III) cell is short-circuited. With the terminals joined by a wire, the external resistance is zero, and the whole emf is dropped across the internal resistance.
(C) \( V = E - ir \) — this matches (IV) current is being drawn from a cell. The cell is discharging, and the voltage falls below \( E \) by \( Ir \).
(D) \( V = E + ir \) — this matches (I) cell is being charged. The charging current enters the positive terminal, so the terminal voltage is larger than \( E \).