A battery of 24 cells, each of emf 1.5 V and internal resistance 2 Ω, is to be connected in order to send the maximum current through a 12 Ω resistor. How are they to be connected? Find the current in each cell and the potential difference across the external resistance.

Let x be the number of cells in series in each row.

And, consider let there be y such rows in parallel.

$\therefore$ Total number of cells = xy = 24
Emf of each cell = 1.5 V
Internal resisatnce of each cell, r= 2Ω
External resistance, R = 12Ω

Now, Resistance of each row in series = 2.x Ω.

Total internal Resistance due to all xy batteries = R

i.e., $\frac{1}{\mathrm{R}} = \frac{1}{2\mathrm{x}}+\frac{1}{2\mathrm{x}}+........y \mathrm{times} \frac{1}{\mathrm{R}} = \frac{y}{2\mathrm{x}}$ 

Total internal resistance $= \frac{2\mathrm{x}}{\mathrm{y}}$ $\mathrm{\Omega }$ (because there are y rows in parallel)

We know that the maximum amount of current passes through the circuit when, the internal resistance of the battery of cells equals the external resistance. 

Thus,              $\frac{2\mathrm{x}}{\mathrm{y}} = 12$
$\Rightarrow$                    $\frac{\mathrm{x}}{\mathrm{y}}=6$ 

But, given that $\mathrm{xy} = 24$ 

Hence, $\mathrm{x}= 12 \mathrm{and} \mathrm{y} = 2$             

i.e., there should be two rows connected parallely, with 12 cells in each row grouped in series.( Fig. below) 



Now, current flowing across the circuit is

            I = $\frac{\mathrm{Total} \mathrm{emf}}{\mathrm{Total} \mathrm{resistance}}$

              $$\begin{gathered} = \frac{1.5 \times 12}{12+12} \\ = \frac{18}{24} \\ = 0.75 \mathrm{A} \end{gathered}$$ 

Since, there are two rows and the current passing through each arm must be equal therefore, 

Current through each arm, I$= \frac{0.75}{2} =0.375 \mathrm{A}.$

Therefore, current through each cell = 0.375 A.

And , the potential difference across the external resistance is, R= 12 x 0.75 = 9 V