A cell of emf 12 V and internal resistance 1 Ω is joined in a single closed loop with a 3 Ω resistor and a second cell of emf \( \varepsilon \). The second cell has negligible internal resistance and opposes the first cell. A current of 2 A flows in the loop in the direction set by the 12 V cell. The value of \( \varepsilon \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4 V
By the loop rule, the changes in potential around a closed loop add up to zero. Go round the loop along the current. The 12 V cell raises the potential by 12 V. The internal resistance and the resistor each lower it by \( IR \). The opposing cell lowers it by \( \varepsilon \).
\( \begin{aligned} 12-I(1)-I(3)-\varepsilon &= 0 \\ \varepsilon &= 12-2\times1-2\times3 \\ &= 4\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Summary – “(b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero.”