Two cells with emf ε1 = 5 V and ε2 = 10 V and internal resistances 1 Ω and 2 Ω, respectively, are connected in parallel. The equivalent emf of the combination is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20/3 V
When two cells are joined in parallel (positive terminal to positive terminal, the usual meaning), the combination behaves as a single cell whose emf is a weighted average of the two emfs, each weighted by the other cell’s internal resistance.
\( \displaystyle \varepsilon_{\mathrm{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \)
\( \displaystyle \varepsilon_{\mathrm{eq}} = \frac{5\,\mathrm{V} \times 2\,\Omega + 10\,\mathrm{V} \times 1\,\Omega}{1\,\Omega + 2\,\Omega} \)
\( \displaystyle \varepsilon_{\mathrm{eq}} = \frac{20}{3}\,\mathrm{V} \)
The internal resistance of the combination is \( r_1r_2/(r_1 + r_2) = 2/3\,\Omega \).