Two batteries, $$\text E_1$$​ (emf: 3 V, internal resistance: 0.5 Ω) and $$\text E_2$$​ (emf: 6 V, internal resistance: 1.0 Ω), are connected in series by connecting the positive terminal of $$\text E_2$$​ to the negative terminal of $$\text E_1$$. A third battery $$\text E_3$$​ (emf: 6 V, internal resistance: 1.0 Ω) is connected in parallel with this combination by connecting its positive terminal to the positive terminal of $$\text E_1$$​ and its negative terminal to the negative terminal $$\text E_2$$​. The equivalent emf of this combination is:

Correct option is A

​$$\textbf{Given:} \\\bullet\ \text{Two batteries in series:} \\E_1 = 3\,\text{V},\ r_1 = 0.5\,\Omega \\E_2 = 6\,\text{V},\ r_2 = 1.0\,\Omega \\$$

​$$\bullet\ \text{Combined:} \\\text{Total emf in series} = 3\,\text{V} + 6\,\text{V} = 9\,\text{V} \\\text{Total resistance} = 0.5 + 1 = 1.5\,\Omega \\\\\bullet\ \text{Third battery in parallel:} \\E_3 = 6\,\text{V},\ r_3 = 1.0\,\Omega \\\\\\\textbf{Step 1: Find equivalent resistance of parallel combination} \\r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{1.5 \times 1}{1.5 + 1} = \frac{1.5}{2.5} = 0.6\,\Omega \\\\\textbf{Step 2: Use the formula for equivalent emf of parallel combination} \\\epsilon_{eq} = r_{eq} \left( \frac{\epsilon_1}{r_1} + \frac{\epsilon_2}{r_2} \right) \\\\\text{Here:} \\\epsilon_1 = 9\,\text{V},\ r_1 = 1.5\,\Omega \\\epsilon_2 = 6\,\text{V},\ r_2 = 1.0\,\Omega \\\\\epsilon_{eq} = 0.6 \left( \frac{9}{1.5} + \frac{6}{1} \right) = 0.6 \times (6 + 6) = 0.6 \times 12 = 7.2\,\text{V}$$​​