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(C) When the cells are connected in series,the total $EMF$ is $nE$ and the total internal resistance is $nr$. The current $I_1$ is given by:$I_1 = \fr

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$r = R$

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(C) When the cells are connected in series,the total $EMF$ is $nE$ and the total internal resistance is $nr$. The current $I_1$ is given by:$I_1 = \frac{nE}{R + nr}$  $(i)$When the cells are connected in parallel,the total $EMF$ is $E$ and the total internal resistance is $r/n$. The current $I_2$ is given by:$I_2 = \frac{E}{R + r/n} = \frac{nE}{nR + r}$  $(ii)$Given that the current is the same in both cases,$I_1 = I_2$:$\frac{nE}{R + nr} = \frac{nE}{nR + r}$Canceling $nE$ from both sides:$R + nr = nR + r$Rearranging the terms to solve for $r$:$nr - r = nR - R$$r(n - 1) = R(n - 1)$$r = R$

$A$ circuit of resistance $R$ is connected to $n$ similar cells. If the current in the circuit is the same when the cells are connected in series or in parallel,and the internal resistance of each cell is $r$,then:

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