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.
For the cell, the potential difference between its terminals is written as \( V=E-Ir \). The term \( Ir \) has a negative sign because
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → inside the cell, the current flows through the electrolyte from the negative electrode to the positive electrode
Inside the cell, the electrolyte between the electrodes has a resistance \( r \), called the internal resistance. The terminal potential difference is found by going through the cell from the negative electrode N to the positive electrode P.
Option (1) the emf of the cell becomes smaller when a current is drawn from it — Incorrect. The emf is fixed by the electrodes and the electrolyte. It stays \( E \). Only the potential difference between the terminals changes with the current.
Option (2) the potential difference across the external resistance is \( Ir \) — Incorrect. Across the external resistance the potential difference is \( V=IR \). The drop \( Ir \) is across the internal resistance, inside the cell.
Option (3) inside the cell, the current flows through the electrolyte from the negative electrode to the positive electrode — Correct. The current flows through the electrolyte from N to P. Along the current, the potential falls by \( Ir \) across the resistance \( r \). So \( Ir \) is taken away from \( E \), and \( V=E-Ir \).
Option (4) in the external resistance, the current flows from the negative terminal to the positive terminal — Incorrect. Outside the cell the current flows from the positive terminal to the negative terminal, that is, from higher to lower potential.
NCERT (2026-27):
Cells, Emf, Internal Resistance – “… As explained before, a steady current is maintained because current flows from N to P through the electrolyte.”
Cells, Emf, Internal Resistance – “… Note the negative sign in the expression (I r) for the potential difference between A and B. This is because the current I flows from B to A in the electrolyte.”
Cells, Emf, Internal Resistance – “… A and B are points in the electrolyte typically close to P and N. …”