$n$ equal cells,each having emf $E$ and internal resistance $r$,are connected in a circuit with an external resistance $R$. If the same current flows in the circuit whether the cells are connected in series or in parallel,then:

To get maximum current through a resistance of $2.5\,\Omega$,one can use $m$ rows of cells,each row having $n$ cells. The internal resistance of each cell is $0.5\,\Omega$. What are the values of $n$ and $m$ if the total number of cells is $45$?

Two batteries with different $e.m.f.$ are connected in parallel. Consider the following statements: $(i)$ The equivalent $e.m.f.$ is less than the individual $e.m.f.s$. $(ii)$ The equivalent internal resistance is less than the individual internal resistances.

$12$ cells,each having the same $emf$ $E$ and internal resistance $r$,are connected in series,but some cells are wrongly connected. This arrangement is connected in series with an ammeter and two additional cells (each of $emf$ $E$ and internal resistance $r$). The current is $3 \, A$ when the cells and the battery aid each other,and it is $2 \, A$ when they oppose each other. The number of cells wrongly connected is:

Two identical cells each of emf $1.5 \,V$ are connected in parallel across an external resistance formed by two $20 \,\Omega$ resistors connected in parallel. $A$ voltmeter connected in the circuit measures $1.2 \,V$. The internal resistance of each cell is ................. $\Omega$.

Why is the combination of cells done? Write its methods.