A cell of emf \(\varepsilon\) and internal resistance \(r\) is connected across a variable external resistor \(R\). A graph is drawn of the current \(I\) in the circuit (y-axis) against \(R\) (x-axis), with \(R\) increased from zero. The graph is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → a curve that starts at \(\varepsilon/r\) at \(R=0\) and falls towards zero as \(R\) becomes large
The current in the circuit is \( I=\varepsilon/(R+r) \). At \( R=0 \) it has its largest value \( \varepsilon/r \). As \( R \) grows, the denominator grows, so \( I \) keeps falling but never becomes zero.
Option (1) a straight line through the origin, rising as \(R\) increases — Incorrect. A line through the origin would mean \( I \) is proportional to \( R \). In fact, a larger \( R \) gives a smaller current.
Option (2) a straight line that falls from \(\varepsilon/r\) at \(R=0\) and reaches zero at \(R=r\) — Incorrect. \( I \) is inversely related to \( R+r \), so the graph is a curve. At \( R=r \) the current is \( \varepsilon/2r \), not zero.
Option (3) a curve that rises from zero and gets closer and closer to \(\varepsilon/r\) — Incorrect. This is the shape of the terminal voltage against \( R \), which rises towards \( \varepsilon \). The current does the opposite.
Option (4) a curve that starts at \(\varepsilon/r\) at \(R=0\) and falls towards zero as \(R\) becomes large — Correct. At \( R=0 \), \( I=\varepsilon/r \). For large \( R \), \( I \) becomes very small, so the curve falls towards the \( R \)-axis.
NCERT (2026-27):
Cells, Emf, Internal Resistance – “… we get”
\( I=\frac{\varepsilon}{R+r} \)\( \qquad (3.40) \)
“The maximum current that can be drawn from a cell is for R = 0 and it is Imax = ε/r.”