Answer

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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.

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