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Two cells of the same EMF $$E$$ but different internal resistances, $$r_1$$ and $$r_2$$ are connected in series with an external resistance $$R$$ as shown in the figure. The terminal potential difference across the second cell is found to be zero. The external resistance $$R$$ must then be:
$$i = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{E + E}{r_1 + r_2 + R}$$
$$i = \frac{2E}{r_1 + r_2 + R}$$
The terminal potential difference across the second cell ($$V_2$$) is zero. Using the formula $$V = E - ir$$:
$$V_2 = E - i \cdot r_2 = 0 \implies E = i \cdot r_2$$
$$E = \left( \frac{2E}{r_1 + r_2 + R} \right) \cdot r_2$$
$$r_1 + r_2 + R = 2r_2$$
$$\mathbf{R = r_2 - r_1}$$
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