For a cell or a battery, the emf is(A) equal to the potential differe…

The correct answer is Option (1) → (A) and (D) only

When a current I flows through a cell of internal resistance r, the terminal potential difference differs from the emf by the drop Ir, with a sign that depends on the direction of the current.

    \( \text{Open circuit (I = 0): } V = E \)
    \( \text{Discharging (current leaves the positive terminal): } V = E - Ir \)
    \( \text{Charging (current forced into the positive terminal): } V = E + Ir \)

(A) equal to the potential difference between its terminals when terminals are not connected externally — Correct. With no current there is no drop across r, so V = E.

(B) less than the potential difference between its terminals when the cell/battery is being discharged. — Incorrect. While discharging, V = E − Ir, so the emf is greater than the terminal potential difference.

(C) always greater than the potential difference between its terminals. — Incorrect. The emf equals V on open circuit and is less than V while charging.

(D) less than the potential difference between its terminals when the cell/battery is being charged. — Correct. While charging, V = E + Ir, which is greater than E.

NCERT (2026-27):
Cells, Emf, Internal Resistance – “… Thus, emf \( \varepsilon \) is the potential difference between the positive and negative electrodes in an open circuit, i.e., when no current is flowing through the cell. If however R is finite, I is not zero. In that case the potential difference between P and N is

    \( \displaystyle V = V_+ + V_- - Ir = \varepsilon - Ir \qquad (3.38) \)”

Kirchhoff’s Rules – “… If, while labelling the current I through the cell one goes from P to N, then of course

    \( \displaystyle V = \varepsilon + I r \qquad (3.60) \)”