One electric cell (having emf of $2 \ V$ and internal resistance of $0.1 \ \Omega$) and another electric cell (having emf of $4 \ V$ and internal resistance of $0.2 \ \Omega$) are connected in parallel to each other. The equivalent emf will be . . . . . . $V$.

In the circuit shown,the potential difference between $A$ and $B$ is ............. $V$.

The emfs of three cells connected in parallel are $E_1=5 \ V, E_2=8 \ V$ and $E_3=10 \ V$ and their internal resistances are $r_1=1 \ \Omega, r_2=2 \ \Omega$ and $r_3=3 \ \Omega$,respectively. By changing $E_3$ to $E_{3N}$,the equivalent emf is doubled. Then $E_{3N}$ in $V$ is:

$A$ cell of internal resistance $r$ is connected to an external resistance $R$. The current will be maximum in $R$,if

$n$ 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 variation of terminal potential difference $(V)$ with current $(I)$ flowing through a cell is as shown in the graph. The $EMF$ $(E)$ and internal resistance $(r)$ of the cell are: