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

Two cells $E_1$ and $E_2$ having equal $EMF$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance $R$. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

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$)?

$A$ battery of $24$ cells,each of emf $1.5\,V$ and internal resistance $2\,\Omega$,is to be connected in order to send the maximum current through a $12\,\Omega$ resistor. The correct arrangement of cells will be:

Two batteries with different $emf$ and different internal resistances are connected as shown in the figure. The voltage between the two terminals $A$ and $B$ is ........ $volt$.

Thousand cells of same emf $E$ and same internal resistance $r$ are connected in series in the same order without an external resistance. The potential drop across $399$ cells is found to be ......... $E$.