A storage battery of emf 6 V and internal resistance 1 Ω is charged from an 18 V dc supply. A resistor is joined in series so that the charging current is 2 A. The value of this series resistor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5 Ω
While the battery is being charged, the current enters it at the positive terminal and leaves at the negative terminal. Its terminal voltage is then larger than its emf, \( V=\varepsilon+Ir \). The rest of the supply voltage drops across the series resistor, as the loop rule requires.
\( \begin{aligned} V &= \varepsilon+Ir \\ &= 6+2\times1 \\ &= 8\ \mathrm{V} \end{aligned} \)
The voltage across the series resistor is
\( \begin{aligned} V_R &= 18-8 \\ &= 10\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} R &= \frac{V_R}{I} \\ &= \frac{10}{2} \\ &= 5\ \Omega \end{aligned} \)
NCERT (2026-27):
Kirchhoff’s Rules – “… If, while labelling the current I through the cell one goes from P to N, then of course”
\( V=\varepsilon+Ir \)\( \qquad (3.60) \)
Kirchhoff’s Rules – “(b) Loop rule: The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero …”