As shown in the figure,a group of $N$ cells is connected in series,where the $emf$ of each cell $E_n$ varies with its internal resistance $r_n$ according to the relation $E_n = 1.5 r_n$. The current $I$ in the circuit is ................ $A$.

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

Two cells $P$ and $Q$ each of emf $2.16 \text{ V}$ are connected in series with a resistor of $19.6 \text{ } \Omega$. An ideal voltmeter reads $2 \text{ V}$ when connected across the cell $P$ and $1.92 \text{ V}$ when connected across the cell $Q$. The ratio of the internal resistances of the cell $P$ and $Q$ is

Two batteries,one of e.m.f. $12 \text{ V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6 \text{ V}$ and internal resistance $1 \Omega$,are connected as shown in the figure. What will be the reading of the voltmeter '$V$' (in $\text{ V}$)?

$n$ 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 identical cells each of emf $1.5\,V$ are connected in series across a $10\,\Omega$ resistance. An ideal voltmeter connected across $10\,\Omega$ resistance reads $1.5\,V$. The internal resistance of each cell is $......\Omega$.