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

Given, number of cells, n = 5

Internal resistance of each cell, r = 1 Ω

Emf of each cell, e = 5 V

When all cells are connected in parallel as shown below.

Potential will remain same as, V_P = 5V

Net resistance in parallel combination will be given as

        $\begin{array}{l}{\mathrm{R}}_{\mathrm{p}}=\mathrm{R}+\left[\frac{1}{\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}+\frac{1}{\mathrm{r}}}\right]=\mathrm{R}+\frac{\mathrm{r}}{5}\\ {\mathrm{R}}_{\mathrm{p}}=\mathrm{R}+\frac{1}{5} \left(\because \mathrm{r}=1\mathrm{\Omega }\right) ...\left(\mathrm{i}\right)\end{array}$

When all cells are connected in series as shown below

The net potential will increase as cells are connected in series,

       ${\mathrm{V}}_{\mathrm{s}}=5+5+5+5+5=25 \mathrm{V}$

The net resistance of circuit will be equivalent of sum of all resistances as all are connected in series.

       $\begin{array}{l}{\mathrm{R}}_{\mathrm{s}}=\mathrm{r}+\mathrm{r}+\mathrm{r}+\mathrm{r}+\mathrm{r}+\mathrm{R}=5\mathrm{r}+\mathrm{R}\\ {\mathrm{R}}_{\mathrm{s}}=5+\mathrm{R} \left(\because \mathrm{r}=1\mathrm{\Omega }\right) ...\left(\mathrm{ii}\right)\end{array}$

By Ohm’s law, current flowing in circuit is given as

       $\mathrm{I}=\mathrm{V}/\mathrm{R}$

As current in both series and parallel combination is same,

$\begin{array}{l} \frac{{\mathrm{V}}_{\mathrm{p}}}{{\mathrm{R}}_{\mathrm{p}}}=\frac{{\mathrm{V}}_{\mathrm{s}}}{{\mathrm{R}}_5}\\ \Rightarrow \frac{5}{\mathrm{R}+\frac{1}{5}}=\frac{25}{5+\mathrm{R}} [\mathrm{From} \mathrm{Eqs}. (\mathrm{i}) \mathrm{and} (\mathrm{ii}\left)\right]\\ \Rightarrow 25+5\mathrm{R}=25\mathrm{R}+5 \Rightarrow \mathrm{R}=1 \mathrm{\Omega }\end{array}$