$n$ identical cells are joined in series with two cells $A$ and $B$ having reversed polarities. The $emf$ of each cell is $E$ and the internal resistance is $r$. The potential difference across cell $A$ or $B$ is: $(n > 4)$

When an external resistance of $5\text{ }\Omega$ is connected across the terminals of a cell, a current of $0.25\text{ A}$ flows through it. When the $5\text{ }\Omega$ resistor is replaced by a $2\text{ }\Omega$ resistor, a current of $0.5\text{ A}$ flows through it. The internal resistance of the cell is . . . . . . $\Omega$.

Infinite number of cells having $emf$ and internal resistance $(E, r)$,$(\frac{E}{n}, \frac{r}{n})$,$(\frac{E}{n^2}, \frac{r}{n^2})$,$(\frac{E}{n^3}, \frac{r}{n^3})$... are connected in series across an external resistance of $\frac{nr}{n+1}$. The current flowing through the external resistor is:

The $e.m.f.$ of a lead-acid car battery is $12 \, V$. The internal resistance of the battery is $0.5 \, \Omega$. The maximum current that can be drawn from the battery is ............. $A$.

Four identical cells of $EMF$ $E$ and internal resistance $r$ are connected as shown in the figure. Find the terminal voltage across any one cell.

Four identical cells, each of $2 \, V$ e.m.f., are connected in parallel. They supply current to an external circuit consisting of two $15 \, \Omega$ resistors connected in parallel. The terminal voltage of the equivalent cell, as measured by an ideal voltmeter, is $1.6 \, V$. Calculate the internal resistance of each cell in $\Omega$.