If the potential difference across the internal resistance $r_1$ is equal to the $emf$ $E$ of the battery,then

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}} = $ . . . . . . $V \Omega^{-1}$.

$A$ battery of emf $E$ and internal resistance $r$ is connected with an external resistance $R$ as shown in the figure. The battery will act as a constant voltage source if

$A$ cell of emf $10 \text{ V}$ and internal resistance $3 \Omega$ is connected in parallel with another cell of emf $7 \text{ V}$ and internal resistance $\frac{3}{5} \Omega$,such that their positive terminals are joined together and their negative terminals are joined together. Their combined positive terminal is joined with the negative terminal,and their combined negative terminal is joined with the positive terminal of a third cell of emf $20 \text{ V}$ with internal resistance $2 \Omega$. The combination can be replaced by a battery of emf $E$ and internal resistance $r$. The values of $E$ and $r$ are respectively:

$A$ battery of $e.m.f.$ $2\,V$ and internal resistance $0.1\, \Omega$ is being charged by a current of $5\,A$. The potential difference across the terminals of the battery is ......... $V$.

In which of the following cells is the potential difference between the terminals of the cell greater than its $EMF$?