Two cells of emfs 1 V and 2 V and internal resistance 2\, and 1\, ,...

For two cells connected in parallel with the same polarity, the equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by:

$E_{eq} = \dfrac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$

$r_{eq} = \dfrac{r_1 r_2}{r_1 + r_2}$

Substituting the given values ($E_1 = 1$ V, $r_1 = 2\,\Omega$, $E_2 = 2$ V, $r_2 = 1\,\Omega$):

$E_{eq} = \dfrac{1 \times 1 + 2 \times 2}{2 + 1} = \dfrac{5}{3}$ V

$r_{eq} = \dfrac{2 \times 1}{2 + 1} = \dfrac{2}{3}\,\Omega$

The current $I$ through the external resistance $R$ is $1$ A:

$I = \dfrac{E_{eq}}{R + r_{eq}}$

$1 = \dfrac{\dfrac{5}{3}}{R + \dfrac{2}{3}}$

$R + \dfrac{2}{3} = \dfrac{5}{3} \Rightarrow R = 1\,\Omega$

When the polarity of one cell is reversed, the new equivalent emf $E'_{eq}$ becomes:

$E'_{eq} = \dfrac{|E_1 r_2 - E_2 r_1|}{r_1 + r_2} = \dfrac{|1 \times 1 - 2 \times 2|}{3} = \dfrac{3}{3} = 1$ V

The equivalent internal resistance remains the same, $r_{eq} = \dfrac{2}{3}\,\Omega$.

The new current $I'$ through the external resistance $R$ is:

$I' = \dfrac{E'_{eq}}{R + r_{eq}} = \dfrac{1}{1 + \dfrac{2}{3}} = \dfrac{1}{\dfrac{5}{3}} = \dfrac{3}{5}$ A

Given that the new current is $\dfrac{\alpha}{5}$ A, we have:

$\dfrac{\alpha}{5} = \dfrac{3}{5} \Rightarrow \alpha = 3$

Answer: $3$