A battery of internal resistance \(2\,\Omega\), and emf of \(10\) V is charged by using a source \(150\) V connected through a resistance \(20\,\Omega\) in series with it. The terminal voltage of battery will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(22.72\) V
While a battery is being charged, current is forced into it against its emf, so its terminal voltage is greater than the emf: \( V = E + Ir \).
The charging current is set by the net emf and the total resistance in the circuit.
\( \displaystyle I = \frac{E_{\mathrm{source}} - E}{R + r} \)
\( \displaystyle I = \frac{150\,\mathrm{V} - 10\,\mathrm{V}}{20\,\Omega + 2\,\Omega} = \frac{140}{22}\,\mathrm{A} \approx 6.36\,\mathrm{A} \)
\( \displaystyle V = E + Ir \)
\( \displaystyle V = 10\,\mathrm{V} + (6.36\,\mathrm{A})(2\,\Omega) \)
\( \displaystyle V \approx 22.72\,\mathrm{V} \)
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 – …
