A current of 2 A flows through a \( 2\,\Omega \) resistor when connected with a cell. When the resistance is replaced by \( 10\,\Omega \), a current of 0.5 A flows. The internal resistance of the cell is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 2/3\,\Omega \)
The emf of the cell drives the current through the external and internal resistances in series: \( E = I(R + r) \). The emf and the internal resistance \( r \) are the same in both cases, so the two equations can be equated.
\( \displaystyle 2\,(2 + r) = 0.5\,(10 + r) \)
\( 4 + 2r = 5 + 0.5r \)
\( 1.5\,r = 1 \)
\( \displaystyle r = \frac{1}{1.5} = \frac{2}{3}\,\Omega \)
As a check, \( E = 2\,(2 + 2/3) = 16/3\,\mathrm{V} \) and \( 0.5\,(10 + 2/3) = 16/3\,\mathrm{V} \).