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:

Two cells of $emfs$ $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ are connected in parallel. The $emf$ and internal resistance of the equivalent source is

In an electric circuit,a cell of certain $emf$ provides a potential difference of $1.25\, V$ across a load resistance of $5\, \Omega$. However,it provides a potential difference of $1\, V$ across a load resistance of $2\, \Omega$. The $emf$ of the cell is given by $\frac{x}{10}\, V$. Then the value of $x$ is ..... .

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

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

Two identical cells each of emf $1.5 \,V$ are connected in parallel across an external resistance formed by two $20 \,\Omega$ resistors connected in parallel. $A$ voltmeter connected in the circuit measures $1.2 \,V$. The internal resistance of each cell is ................. $\Omega$.