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(B) Let $E$ be the electromotive force $(EMF)$ and $r$ be the internal resistance of each cell. The external resistance is $R = 2\,\Omega$.When connec

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$2$

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(B) Let $E$ be the electromotive force $(EMF)$ and $r$ be the internal resistance of each cell. The external resistance is $R = 2\,\Omega$.When connected in series,the total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $i_1$ is given by:$i_1 = \frac{2E}{R + 2r} = \frac{2E}{2 + 2r}$When connected in parallel,the total $EMF$ is $E$ and the total internal resistance is $r/2$. The current $i_2$ is given by:$i_2 = \frac{E}{R + r/2} = \frac{E}{2 + r/2} = \frac{2E}{4 + r}$Given that $i_1 = i_2$,we equate the two expressions:$\frac{2E}{2 + 2r} = \frac{2E}{4 + r}$Canceling $2E$ from both sides:$2 + 2r = 4 + r$Solving for $r$:$2r - r = 4 - 2$$r = 2\,\Omega$

Two identical cells send the same current in $2\,\Omega$ resistance,whether connected in series or in parallel. The internal resistance of the cell should be ............... $\Omega$.

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