The given graph shows the variation of terminal potential difference V, across the combination of three identical cells in series to an external resistor versus the current i in the circuit. The internal resistance of each cell is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1 Ω
For a battery of emf \( \varepsilon_{\mathrm{net}} \) and internal resistance \( r_{\mathrm{net}} \), the terminal potential difference is
\( \displaystyle V = \varepsilon_{\mathrm{net}} - i\,r_{\mathrm{net}} \)
So the graph of \( V \) against \( i \) is a straight line whose intercept is the emf and whose slope is \( -r_{\mathrm{net}} \).
Readings from the graph: the line starts at \( V = 6\,\mathrm{V} \) for \( i = 0 \) and falls to zero at \( i = 2\,\mathrm{A} \) (it passes through \( (1\,\mathrm{A},\,3\,\mathrm{V}) \)).
\( \displaystyle r_{\mathrm{net}} = \left|\frac{\Delta V}{\Delta i}\right| = \frac{6 - 0}{2 - 0} = 3\,\Omega \)
The three identical cells are in series, so the internal resistances add: \( r_{\mathrm{net}} = 3r \).
\( \displaystyle r = \frac{3}{3} = 1\,\Omega \)