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(A) Let the $EMF$ of each cell be $\varepsilon$ and the internal resistance be $r$.For series combination,the total $EMF$ is $2\varepsilon$ and the to

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$5$

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(A) Let the $EMF$ of each cell be $\varepsilon$ and the internal resistance be $r$.For series combination,the total $EMF$ is $2\varepsilon$ and the total internal resistance is $2r$. The current $i$ in the external resistance $R = 5\,\Omega$ is:$i = \frac{2\varepsilon}{R + 2r} = \frac{2\varepsilon}{5 + 2r} \dots (1)$For parallel combination,the total $EMF$ is $\varepsilon$ and the total internal resistance is $\frac{r}{2}$. The current $i$ in the external resistance $R = 5\,\Omega$ is:$i = \frac{\varepsilon}{R + \frac{r}{2}} = \frac{\varepsilon}{5 + \frac{r}{2}} \dots (2)$Since the current is the same in both cases,equate $(1)$ and $(2)$:$\frac{2\varepsilon}{5 + 2r} = \frac{\varepsilon}{5 + \frac{r}{2}}$Dividing both sides by $\varepsilon$ and cross-multiplying:$2(5 + \frac{r}{2}) = 5 + 2r$$10 + r = 5 + 2r$$r = 5\,\Omega$Thus,the internal resistance of each cell is $5\,\Omega$.

Two identical cells,when connected either in parallel or in series,give the same current in an external resistance of $5\,\Omega$. The internal resistance of each cell will be $.............\,\Omega$.

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