What happens to the current drawn from a cell when its terminals are…

The correct answer is Option (3) → It becomes maximum, equal to \( \varepsilon/r \)

A cell of emf \( \varepsilon \) and internal resistance \( r \) is connected to an external resistance \( R \). It drives the current \( I=\varepsilon/(R+r) \), so the total resistance \( R+r \) sets the current.

Option (1) It remains the same as with a load — Incorrect. The current depends on \( R \). Removing the load resistance leaves only \( r \) in the circuit, so the current rises.

Option (2) It becomes negligible — Incorrect. With \( R=0 \) the total resistance is at its smallest, so the current is the largest the cell can give.

Option (3) It becomes maximum, equal to \( \varepsilon/r \) — Correct. Putting \( R=0 \) in \( I=\varepsilon/(R+r) \) leaves only the internal resistance to limit the current, so the maximum current \( I_{\mathrm{max}}=\varepsilon/r \) is drawn.

Option (4) It varies continuously — Incorrect. With fixed \( \varepsilon \) and \( r \), the short-circuited cell gives the single value \( \varepsilon/r \). Nothing in the circuit makes it vary.

NCERT (2026-27):
Cells, Emf, Internal Resistance – “… Or,

    \( \displaystyle I=\frac{\varepsilon}{R+r} \qquad (3.40) \)

The maximum current that can be drawn from a cell is for R = 0 and it is I_max = ε/r.”