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 . . . . . . $V$.

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

In the network shown,the potential difference between $A$ and $B$ is ................. $V$ $(R = r_1 = r_2 = r_3 = 1 \Omega, E_1 = 3 \, V, E_2 = 2 \, V, E_3 = 1 \, V)$.

$A$ battery consists of $n$ rows of cells connected in parallel,with each row containing $m$ cells in series. The maximum current is obtained through an external resistance of $3 \, \Omega$. If the total number of cells is $24$ and the internal resistance of each cell is $0.5 \, \Omega$,find the values of $m$ and $n$.

When two cells of emfs $E_1$ and $E_2$ and different internal resistances $r_1$ and $r_2$ are connected in series with an external load resistor $R$,the current through the load is $5 \ A$. If the polarity of the cell of emf $E_2$ is reversed,then the current through the load is $2 \ A$. Then $\frac{E_1}{E_2} = $

Two cells of equal $e.m.f.$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ are connected in series. On connecting this combination to an external resistance $R$,it is observed that the potential difference across the first cell becomes zero. The value of $R$ will be