A storage battery of EMF 10 V and internal resistance 1 Ω is being charged by a 120 V dc supply using a series resistor of 21 Ω. What is the terminal voltage of the battery during charging?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → V = 15 V
While the battery is being charged, the supply drives current into it against its emf. The net voltage in the circuit is the supply voltage minus the emf of the battery, and the terminal voltage of the battery is \( V = E + Ir \) (more than its emf).
\( \displaystyle I = \frac{120\,\mathrm{V} - 10\,\mathrm{V}}{21\,\Omega + 1\,\Omega} = \frac{110}{22}\,\mathrm{A} = 5\,\mathrm{A} \)
\( V = E + Ir = 10\,\mathrm{V} + 5\,\mathrm{A}\times1\,\Omega = 15\,\mathrm{V} \)
Check: the series resistor drops \( 5\times21 = 105\,\mathrm{V} \), so the terminal voltage is \( 120 - 105 = 15\,\mathrm{V} \).
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 – …
