Case: Read the passage and answer the question.
A cell of unknown emf E and internal resistance r, two unknown resistances \( R_1 \) and \( R_2 \) \( (R_2\gt R_1) \) and a perfect ammeter are given. The current in the circuit is measured in five different situations:
(i) without any external resistance in the circuit,
(ii) with resistance \( R_1 \) only,
(iii) with resistance \( R_2 \) only,
(iv) with both \( R_1 \) and \( R_2 \) used in series combination, and
(v) with both \( R_1 \) and \( R_2 \) used in parallel combination.
The currents obtained in the five cases are 0.42 A, 0.6 A, 1.05 A, 1.4 A and 4.2 A, but not necessarily in that order.
The potential difference between the terminals of the cell is equal to its emf \(E\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → when no current is drawn from the cell
When a current \( I \) flows, the potential difference between the terminals is \( V=E-Ir \). It is equal to \( E \) only when \( I=0 \).
Option (1) in situation (i), when there is no external resistance — Incorrect. Here \( R=0 \), so the terminal potential difference \( V=IR \) is zero. The whole emf drives the current through the internal resistance.
Option (2) when no current is drawn from the cell — Correct. With \( I=0 \) the term \( Ir \) is zero, so \( V=E \). This is the open-circuit case.
Option (3) in situation (iv), when \(R_1\) and \(R_2\) are in series — Incorrect. A current still flows through \( R_1 \) and \( R_2 \). So \( Ir \) is not zero and \( V \) is less than \( E \).
Option (4) in every situation, because the emf is fixed for a cell — Incorrect. The emf of the cell stays the same, but the terminal potential difference does not. It falls as the current grows.
NCERT (2026-27):
Cells, Emf, Internal Resistance – “Thus, emf ε is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell.”