The current flowing through the circuit shown in the figure is ........... $A$.

$A$ $5\ V$ battery with internal resistance $2\,\Omega$ and a $2\,V$ battery with internal resistance $1\,\Omega$ are connected to a $10\,\Omega$ resistor as shown in the figure. Find the current through the $10\,\Omega$ resistor.

For the given circuit,the $emf$ of the cells are $E_1 = 4 \, V$ and $E_2 = 8 \, V$,and their internal resistances are $0.5 \, \Omega$ and $1 \, \Omega$ respectively. Calculate the terminal voltage of each cell.

The potential difference between the points $P$ and $Q$ is nearly: (in $\,V$)

The figure shows a network of resistances and cells in which the point $A$ is earthed. The point which has the least potential is

When the terminals of a cell are connected by a wire of resistance $4 \Omega$,the potential difference across the cell is $1.6 \text{ V}$. If a wire of the same resistance is connected in parallel with the first,the potential difference becomes $1.33 \text{ V}$. The emf and internal resistance of the cell are respectively