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 ______.
Numerical value type. Enter your answer.
Answer: 3
Parallel combination: $r_{eq} = \dfrac{2 \times 1}{3} = \dfrac{2}{3}\ \Omega$ and $E_{eq} = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \dfrac{0.5 + 2}{1.5} = \dfrac{5}{3}$ V.
$1 = \dfrac{5/3}{R + 2/3} \Rightarrow R = 1\ \Omega$.
With one cell reversed: $E_{eq} = \dfrac{|2 - 0.5|}{1.5} = 1$ V, so $I = \dfrac{1}{1 + 2/3} = \dfrac{3}{5}$ A.
$\alpha = 3$.
Solution by Sreeraj P, M.Sc Physics