$A$ battery of emf $18 \, V$ and internal resistance $3 \, \Omega$ and another battery of emf $10 \, V$ and internal resistance $1 \, \Omega$ are connected in parallel as shown in the figure. The voltmeter reading is: (in $ \, V$)

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 ......... $E$.

Four cells, each of emf $E$ and internal resistance $r$, are connected in series across an external resistance $R$. By mistake, one of the cells is connected in reverse. Then the current in the external circuit is

Two sources of equal emfs $E$ are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$. If the potential difference across the source of internal resistance $r_{1}$ is zero,then the value of $R$ will be:

The terminal potential difference of a cell is greater than its $e.m.f.$ when it is

$n$ 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: