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 ................ $A$.

$EMF$ is most closely related to

Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ (where $\varepsilon_2 > \varepsilon_1$) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown.

When current supplied by a cell to a circuit is $0.3 \,A$,its terminal potential difference is $0.9 \,V$. When the current supplied becomes $0.25 \,A$,its terminal potential difference becomes $1.0 \,V$. The internal resistance of the cell is ............ $\Omega$.

The circuit in the figure shows two cells connected in opposition to each other. Cell $E_1$ has an $emf$ of $6 \ V$ and an internal resistance of $2 \ \Omega$. Cell $E_2$ has an $emf$ of $4 \ V$ and an internal resistance of $8 \ \Omega$. Find the potential difference between the points $A$ and $B$.

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