Ten identical cells each emf $2 \, V$ and internal resistance $1 \, \Omega$ are connected in series with two cells wrongly connected. $A$ resistor of $10 \, \Omega$ is connected to the combination. What is the current through the resistor (in $ \, A$)?

In the given circuit,if $V_{AB} = 4\ V$,then the value of resistance $X$ in ohms will be:

$n$ 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: $(n > 4)$

In which of the following cells is the potential difference between the terminals of the cell greater than its $EMF$?

$n$ rows,each containing $m$ cells in series,are connected in parallel. The maximum current is drawn from this combination through an external resistance of $3\, \Omega$. If the total number of cells used is $24$ and the internal resistance of each cell is $0.5\, \Omega$,then $m$ and $n$ are respectively:

The potential difference in open circuit for a cell is $2.2\, V$. When a $4\, \Omega$ resistor is connected between its two electrodes,the potential difference becomes $2\, V$. The internal resistance of the cell will be .............. $\Omega$.