Two cells of emf 4 V and 2 V, each of internal resistance 1 Ω, are joined in parallel between two points A and B. The positive terminal of the 4 V cell and the negative terminal of the 2 V cell are joined to A. The negative terminal of the 4 V cell and the positive terminal of the 2 V cell are joined to B. The equivalent emf of the combination between A and B is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 V
For two cells in parallel, \( 1/r_{eq}=1/r_1+1/r_2 \) and \( \varepsilon_{eq}/r_{eq}=\varepsilon_1/r_1+\varepsilon_2/r_2 \). Here the 2 V cell is joined the other way round, so its emf is put in with a negative sign.
\( \begin{aligned} \frac{1}{r_{eq}} &= \frac{1}{1}+\frac{1}{1} \\ r_{eq} &= 0.5\ \Omega \end{aligned} \)
\( \begin{aligned} \frac{\varepsilon_{eq}}{r_{eq}} &= \frac{4}{1}-\frac{2}{1} \\ &= 2\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} \varepsilon_{eq} &= 2\times0.5 \\ &= 1\ \mathrm{V} \end{aligned} \)
NCERT (2026-27):
Cells in Series and in Parallel – “… we had joined the positive terminals together and similarly the two negative ones, so that the currents I1, I2 flow out of positive terminals. If the negative terminal of the second is connected to positive terminal of the first, Eqs. (3.56) and (3.57) would still be valid with ε2 → –ε2”
\( \frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2} \qquad (3.56) \)
\( \frac{\varepsilon_{eq}}{r_{eq}}=\frac{\varepsilon_1}{r_1}+\frac{\varepsilon_2}{r_2} \)\( \qquad (3.57) \)