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

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} = $

$A$ galvanometer together with an unknown resistance in series is connected to two identical batteries each of $1.5\, V$. When the batteries are connected in series,the galvanometer records a current of $1\, A$,and when the batteries are in parallel,the current is $0.6\, A$. What is the internal resistance of the battery?

$10$ identical cells,each of potential $E$ and internal resistance $r$,are connected in series to form a closed circuit. What is the potential difference across any $3$ cells (in $E$)?

For two cells having the same $EMF$ $E$ and internal resistance $r$, the current passing through an external resistor of $6 \ \Omega$ is the same when both cells are connected either in parallel or in series. The value of internal resistance $r$ is . . . . . . $ \ \Omega$.

$A$ storage cell is charged by $5\, A$ $D.C.$ for $18\, hours$. Its strength after charging will be .............. $AH$.