Three identical cells,each of emf $10 \ V$ and internal resistance $3 \ \Omega$,are connected in parallel across terminals $A$ and $B$. The net emf across $A$ and $B$ is .......... $V$.

The emfs of three cells connected in parallel are $E_1=5 \ V, E_2=8 \ V$ and $E_3=10 \ V$ and their internal resistances are $r_1=1 \ \Omega, r_2=2 \ \Omega$ and $r_3=3 \ \Omega$,respectively. By changing $E_3$ to $E_{3N}$,the equivalent emf is doubled. Then $E_{3N}$ in $V$ 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}} = $ . . . . . . $V \Omega^{-1}$.

$n$ identical cells are joined in series with two cells $A$ and $B$ having reversed polarities. The $emf$ of each cell is $E$ and the internal resistance is $r$. The potential difference across cell $A$ or $B$ is: $(n > 4)$

Four cells, each of emf $E$ and internal resistance $r$, are connected in series across an external resistance $R$. By mistake, one of the cells is connected in reverse. Then the current in the external circuit is

Two batteries with different $e.m.f.$ are connected in parallel. Consider the following statements: $(i)$ The equivalent $e.m.f.$ is less than the individual $e.m.f.s$. $(ii)$ The equivalent internal resistance is less than the individual internal resistances.