Two cells of emfs 1 V and 2 V and internal resistance 2 $$\Omega$$ and 1 $$\Omega$$, respectively connected in parallel, gave a current of 1 A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $$\dfrac{\alpha}{5}$$ A. The value of $$\alpha$$ is __________.

$$r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{2 + 1} = \frac{2}{3}\ \Omega$$

$$E_{\text{eq}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{\frac{5}{2}}{\frac{3}{2}} = \frac{5}{3}\text{ V}$$

$$I = \frac{E_{\text{eq}}}{R + r_{\text{eq}}}$$

$$1 = \frac{\frac{5}{3}}{R + \frac{2}{3}} \implies R + \frac{2}{3} = \frac{5}{3} \implies R = 1\ \Omega$$

If the polarity of one cell (say, the $$1\text{ V}$$ cell) is reversed, the new equivalent EMF ($$E_{\text{eq}}'$$) becomes:

$$E_{\text{eq}}' = \frac{-\frac{1}{2} + \frac{2}{1}}{\frac{1}{2} + \frac{1}{1}} = \frac{\frac{3}{2}}{\frac{3}{2}} = 1\text{ V}$$

$$I' = \frac{E_{\text{eq}}'}{R + r_{\text{eq}}} = \frac{1}{1 + \frac{2}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}\text{ A}$$

$$\frac{\alpha}{5} = \frac{3}{5} \implies \alpha = 3$$