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 ............. $A$.

$n$ 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:

Two cells of same emf $E$ but different internal resistances $r_{1}$ and $r_{2}$ are connected in series with a resistance $R$. The value of resistance $R$,for which the potential difference across the second cell is zero,is

$32$ cells,each of $emf$ $3V$,are connected in series and kept in a box. Externally,the combination shows an $emf$ of $84V$. The number of cells reversed in the combination is

$n$ identical cells,each of $e.m.f.$ $E$ and internal resistance $r$,are connected in series. This combination is connected in series with an external resistor $R$. What is the current flowing through $R$?

A circuit consists of three batteries of emf $E_{1} = 1 \text{ V}$, $E_{2} = 2 \text{ V}$, and $E_{3} = 3 \text{ V}$ and internal resistances $1 \Omega$, $2 \Omega$, and $1 \Omega$ respectively, which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is: (in $\text{ V}$)