Two cells are connected in opposition as shown. Cell $E_1$ has an electromotive force (emf) of $8 \ V$ and an internal resistance of $2 \ \Omega$; cell $E_2$ has an emf of $2 \ V$ and an internal resistance of $4 \ \Omega$. The terminal potential difference of cell $E_2$ is: (in $V$)

If the $EMF$ of each cell is $E = 1.5 \text{ V}$ and internal resistance is $r$, as shown in the figure, what is the current $i$ flowing through the circuit (in $\text{ A}$)?

The $e.m.f.$ of a primary cell is $2 \ V$. When it is short-circuited,it gives a current of $4 \ A$. The internal resistance of the primary cell is ............. $\Omega$.

$A$ cell of emf $1.2 \ V$ and internal resistance $2 \ \Omega$ is connected in parallel to another cell of emf $1.5 \ V$ and internal resistance $1 \ \Omega$. If the like poles of the cells are connected together,the emf of the combination of the two cells is (in $V$)

The number of dry cells,each of $e.m.f.$ $1.5\,V$ and internal resistance $0.5\,\Omega$,that must be joined in series with a resistance of $20\,\Omega$ so as to send a current of $0.6\,A$ through the circuit is:

If there are $3$ parallelly connected cells of emf $\varepsilon_1 = 1.2 \ V, \varepsilon_2 = 1.4 \ V$ and $\varepsilon_3 = 1.5 \ V$ and of internal resistances $r_1 = 0.1 \ \Omega, r_2 = 0.2 \ \Omega$ and $r_3 = 0.3 \ \Omega$,then find $\frac{\varepsilon_{eq}}{r_{eq}} = $ . . . . . . $V \Omega^{-1}$.