Five identical cells each of internal resistance 1 $$\Omega$$ and emf 5 V 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?

Let each cell have emf $$E = 5\ \text{V}$$ and internal resistance $$r = 1\ \Omega$$. There are five such identical cells.

First we take the series combination. For cells in series, the rule is: “Equivalent emf is the sum of individual emfs and equivalent internal resistance is the sum of individual internal resistances.” Hence

$$E_{\text{series}} = 5E = 5 \times 5 = 25\ \text{V}$$

$$r_{\text{series}} = 5r = 5 \times 1 = 5\ \Omega$$

The total resistance experienced by the current is the sum of the external resistance $$R$$ and the internal resistance. Therefore the current in the series arrangement is

$$I_{\text{series}} = \dfrac{E_{\text{series}}}{R + r_{\text{series}}} = \dfrac{25}{R + 5}\ (\text{A}).$$

Next we consider the parallel combination. For identical cells in parallel, the rule is: “Equivalent emf remains the same as that of one cell, and the equivalent internal resistance is the individual internal resistance divided by the number of cells.” Therefore

$$E_{\text{parallel}} = E = 5\ \text{V}$$

$$r_{\text{parallel}} = \dfrac{r}{5} = \dfrac{1}{5} = 0.2\ \Omega$$

Now the current in the parallel arrangement is

$$I_{\text{parallel}} = \dfrac{E_{\text{parallel}}}{R + r_{\text{parallel}}} = \dfrac{5}{R + 0.2}\ (\text{A}).$$

The problem asks for the value of $$R$$ that makes the two currents equal, so we set the two expressions equal:

$$I_{\text{series}} = I_{\text{parallel}} \quad\Rightarrow\quad \dfrac{25}{R + 5} = \dfrac{5}{R + 0.2}.$$

Cross-multiplying gives

$$25(R + 0.2) = 5(R + 5).$$

Expanding both sides, we have

$$25R + 25 \times 0.2 = 5R + 5 \times 5$$

$$25R + 5 = 5R + 25.$$

Subtract $$5R$$ from both sides:

$$20R + 5 = 25.$$

Subtract $$5$$ from both sides:

$$20R = 20.$$

Finally, divide by $$20$$:

$$R = 1\ \Omega.$$

Hence, the correct answer is Option A.