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(A) Let $n = 5$ be the number of cells,$E = 5\, V$ be the $emf$ of each cell,and $r = 1\, \Omega$ be the internal resistance of each cell.For series c

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(A) Let $n = 5$ be the number of cells,$E = 5\, V$ be the $emf$ of each cell,and $r = 1\, \Omega$ be the internal resistance of each cell.For series combination,the total $emf$ is $nE$ and total internal resistance is $nr$. The current $i_1$ is given by:$i_1 = \frac{nE}{R + nr} = \frac{5 	imes 5}{R + 5 	imes 1} = \frac{25}{R + 5}$For parallel combination,the total $emf$ is $E$ and total internal resistance is $r/n$. The current $i_2$ is given by:$i_2 = \frac{E}{R + r/n} = \frac{5}{R + 1/5} = \frac{5 	imes 5}{5R + 1} = \frac{25}{5R + 1}$Given $i_1 = i_2$,we have:$\frac{25}{R + 5} = \frac{25}{5R + 1}$$R + 5 = 5R + 1$$4 = 4R$$R = 1\, \Omega$

Five identical cells each of internal resistance $1\, \Omega$ and $emf$ $5\, V$ are connected in series and in parallel with an external resistance $R$. For what value of $R$ will the current in the series and parallel combination remain the same? (in $\Omega$)

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