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# A circuit consists of a resistance R connected to n similar cells. If the current in the circuit is the same whether the cells are connected in series or in parallel then the internal resistance r of each cell is given byA. $r = \dfrac{R}{n}$B. $r = nR$C. $r = R$D. $r = \dfrac{1}{R}$  Verified
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Hint: For n cells with internal resistance r connected in series,
${E_{eq}} = nE$ and ${r_{eq}} = nr$
For n cells with internal resistance r connected in parallel,
${E_{eq}} = E$ and ${r_{eq}} = \dfrac{r}{n}$
Here ${E_{eq}}$ is the equivalent emf of all the battery sources and ${r_{eq}}$ is the equivalent internal resistance.
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.

Complete step by step solution:
First, we have to calculate the equivalent EMF and total equivalent resistance for the type of connections (series and parallel).
For n cells with internal resistance r connected in series,
${E_{eq}}$ is the algebraic sum of all the EMFs i.e. ${E_{eq}} = nE$
As all the internal resistances are also in series, so the equivalent internal resistance will be ${r_{eq}} = nr$
Now as the equivalent internal resistance and the external resistance R are in series connection, so the total equivalent resistance for the circuit will be given as, ${R_{eq}} = R + {r_{eq}} = R + nr$
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Let current through this series connection be ${I_s}$
So, ${I_s} = \dfrac{{nE}}{{R + nr}}$
Now, for n cells with internal resistance r connected in parallel,
${E_{eq}}$ in the parallel connection will be ${E_{eq}} = E$ as the EMF will remain the same.
As all the internal resistances are also in parallel, so the equivalent internal resistance will be ${r_{eq}} = \dfrac{r}{n}$
Now as the equivalent internal resistance and the external resistance R are in series connection, so the total equivalent resistance for the circuit will be given as, ${R_{eq}} = R + {r_{eq}} = R + \dfrac{r}{n}$
Current for both the connection is given by $I = \dfrac{{{E_{eq}}}}{{{R_{eq}}}}$ where ${R_{eq}}$ is the total equivalent resistance including internal and external resistance.
Let current through this series connection be ${I_p}$
So, ${I_p} = \dfrac{E}{{\left( {R + \dfrac{r}{n}} \right)}}$
Now, as given in the question that the current in the circuit is the same whether the cells are connected in series or in parallel which means ${I_s} = {I_p}$
So, $\dfrac{{nE}}{{R + nr}} = \dfrac{E}{{\left( {R + \dfrac{r}{n}} \right)}}$
On further solving we have,
$\dfrac{{nE}}{{R + nr}} = \dfrac{{nE}}{{nR + r}}$
Or we can say, $R + nr = nR + r$
On simplifying we get,
$r = R$

$\therefore$The internal resistance r is equal to $R$. Hence, option (C) is the correct answer.

Note:
While calculating the overall equivalent resistance after calculating equivalent internal resistance, remember that ${r_{eq}}$ and the external resistance R will be in series.
Remember that the equivalent EMF for the parallel connection will remain as original.