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 __________.

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: \(3\)