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:

Two batteries of different $e.m.f.$ values and internal resistances are connected in series with each other and with an external load resistor. The current is $3.0 \,A$. When the polarity of one battery is reversed,the current becomes $1.0 \,A$. The ratio of the $e.m.f.$ values of the two batteries is ............

When a battery is connected across a resistor of $16 \Omega$,the voltage across the resistor is $12 \ V$. When the same battery is connected across a resistor of $10 \Omega$,the voltage across it is $11 \ V$. The internal resistance of the battery in ohm is

Two cells,having the same $e.m.f.$ $E$,are connected in series through an external resistance $R$. The cells have internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively. When the circuit is closed,the potential difference across the first cell is zero. The value of $R$ 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} = $