A battery of emf = 9 V and internal resistance 3 Ω is connected to an external resistor. If the current in the circuit is 0.2 A, the value of external resistor is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 42 Ω
The current drawn from a cell of emf E and internal resistance r follows from Ohm’s law for the whole circuit: the emf drives the current through the external and internal resistances in series.
\( \displaystyle I = \frac{E}{R + r} \)
\( \displaystyle R = \frac{E}{I} - r = \frac{9\,\mathrm{V}}{0.2\,\mathrm{A}} - 3\,\Omega \)
\( R = 45\,\Omega - 3\,\Omega = 42\,\Omega \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
