Each of the six ideal batteries of emf $20 \, V$ is connected to an external resistance of $4 \, \Omega$ as shown in the figure. The current through the resistance is (in $ \, A$)

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

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:

$A$ battery of $24$ cells,each of emf $1.5\, V$ and internal resistance $2\, \Omega$,is to be connected to send the maximum current through a $12\, \Omega$ resistor. The correct arrangement of cells will be:

The variation of terminal potential difference $(V)$ with current $(I)$ flowing through a cell is as shown in the graph. The $EMF$ $(E)$ and internal resistance $(r)$ of the cell are:

The electromotive force of a primary cell is $2\,V$. When it is short-circuited,it gives a current of $4\,A$. Its internal resistance in $\Omega$ is: