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(A) Let the $EMF$ of each cell be $E$ and the internal resistance be $r$. The external resistance is $R = 2 \,\Omega$.For series combination,the total

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$2$

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(A) Let the $EMF$ of each cell be $E$ and the internal resistance be $r$. The external resistance is $R = 2 \,\Omega$.For series combination,the total $EMF$ is $2E$ and the total internal resistance is $2r$. The current $I_1$ is given by:$I_1 = \frac{2E}{2r + R} = \frac{2E}{2r + 2} = \frac{E}{r + 1}$For parallel combination,the total $EMF$ is $E$ and the total internal resistance is $r/2$. The current $I_2$ is given by:$I_2 = \frac{E}{r/2 + R} = \frac{E}{r/2 + 2} = \frac{2E}{r + 4}$Given that $I_1 = I_2$,we have:$\frac{E}{r + 1} = \frac{2E}{r + 4}$Canceling $E$ from both sides:$\frac{1}{r + 1} = \frac{2}{r + 4}$Cross-multiplying:$r + 4 = 2(r + 1)$$r + 4 = 2r + 2$$r = 2 \,\Omega$.

The combination of two identical cells,whether connected in series or parallel combination,provides the same current through an external resistance of $2 \,\Omega$. The value of internal resistance of each cell is ............ $\Omega$.

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