Two cells (emfs 2 V, 1 V and their internal resistances 4 Ω and 3 Ω, respectively) are connected in parallel with their positive terminals connected together with wires of negligible resistance. What will be the equivalent emf of the combination?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 10/7 V
For cells in parallel (like terminals joined), the equivalent emf is the weighted mean of the emfs, each weighted by the reciprocal of its internal resistance.
\( \displaystyle \varepsilon_{\mathrm{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \)
\( \displaystyle \varepsilon_{\mathrm{eq}} = \frac{2 \times 3 + 1 \times 4}{4 + 3}\,\mathrm{V} = \frac{10}{7}\,\mathrm{V} \)
The result lies between 1 V and 2 V, as it must, since the stronger cell drives current back through the weaker one.