A cell having emf \( \varepsilon \) and internal resistance r is connected across a variable external resistance R. When the resistance R keeps on increasing the graph of potential difference V across the R will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3)
The potential difference across \( R \) is the terminal voltage of the cell. The current is \( I = \varepsilon/(R + r) \), so
\( \displaystyle V = I R = \frac{\varepsilon R}{R + r} = \frac{\varepsilon}{1 + r/R} \)
At \( R = 0 \), \( V = 0 \): the graph starts from the origin.
For small \( R \), \( V \approx \varepsilon R/r \): it rises steeply at first.
As \( R \) becomes very large, \( r/R \rightarrow 0 \) and \( V \rightarrow \varepsilon \): the curve bends over and approaches \( \varepsilon \) without reaching it.
So the graph is a curve rising from zero and levelling off towards \( \varepsilon \), as in Option (3).
Note: A similar question was also asked in CUET (UG) 2023 on 22-05-2023 (Slot 3).