Two different configurations of n identical cells (each having...

For the series configuration in Figure (a), the equivalent electromotive force (emf)$of n$ cells is $nE$ and the equivalent internal resistance is $nr$.
The current $I_s$ through the circuit is given by:
$I_s = \dfrac{nE}{R + nr}$

For the parallel configuration in Figure (b), the equivalent emf of $n$ identical cells is $E$ and the equivalent internal resistance is $\dfrac{r}{n}$.
The current $I_p$ through the circuit is given by:
$I_p = \dfrac{E}{R + \dfrac{r}{n}} = \dfrac{nE}{nR + r}$

Given that the current is the same in both configurations, we equate $I_s$ and $I_p$:
$\dfrac{nE}{R + nr} = \dfrac{nE}{nR + r}$

Since $nE \neq 0$, we can equate the denominators:
$R + nr = nR + r$

Rearranging the terms, we get:
$nr - r = nR - R$
$r(n - 1) = R(n - 1)$

Since $n > 1$, we have $n - 1 \neq 0$. Dividing both sides by $(n - 1)$ yields:
$R = r$

Answer: $R = r$