Two identical cells each of emf $E$ and internal resistance $r$ are connected in parallel with an external resistance $R$. To get maximum power developed across $R$,the value of $R$ is

$A$ cell of emf $1.2 \ V$ and internal resistance $2 \ \Omega$ is connected in parallel to another cell of emf $1.5 \ V$ and internal resistance $1 \ \Omega$. If the like poles of the cells are connected together,the emf of the combination of the two cells is (in $V$)

$n$ identical cells,each of emf $E$ and internal resistance $r$,are joined in series to form a row. $m$ such rows are joined in parallel across a load resistance $R$. The current in each cell is:

Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:

Four identical cells of emf $E$ and internal resistance $r$ are to be connected in series. If one of the cells is connected wrongly,the equivalent emf and effective internal resistance of the combination are

$A$ primary cell has an $e.m.f.$ of $1.5\,V$. When short-circuited,it gives a current of $3\,A$. The internal resistance of the cell is ............. $\Omega$.