Figure 5 below shows a battery consisting of three cells. Their emfs and internal resistances are…

The two cells in the upper branch are connected in series aiding each other: $E_{\text{top}} = E_1 + E_2 = 2\text{ V} + 4\text{ V} = 6\text{ V}$ $r_{\text{top}} = r_1 + r_2 = 1\,\Omega + 3\,\Omega = 4\,\Omega$ The lower branch has emf $E_{\text{bottom}} = 6\text{ V}$ and internal resistance $r_{\text{bottom}} = 12\,\Omega$. The two branches are in parallel with like terminals connected together. The equivalent emf of the battery is: $E = \frac{\frac{E_{\text{top}}}{r_{\text{top}}} + \frac{E_{\text{bottom}}}{r_{\text{bottom}}}}{\frac{1}{r_{\text{top}}} + \frac{1}{r_{\text{bottom}}}} = \frac{\frac{6}{4} + \frac{6}{12}}{\frac{1}{4} + \frac{1}{12}} = 6\text{ V}$ Final answer: 6 V

The equivalent internal resistance of the two parallel branches is: $\frac{1}{r} = \frac{1}{r_{\text{top}}} + \frac{1}{r_{\text{bottom}}} = \frac{1}{4\,\Omega} + \frac{1}{12\,\Omega} = \frac{3 + 1}{12\,\Omega} = \frac{4}{12\,\Omega} = \frac{1}{3\,\Omega}$ $r = 3\,\Omega$ Final answer: 3 \Omega