$12$ cells each having same emf are connected in series with some cells wrongly connected. The arrangement is connected in series with an ammeter and two cells which are in series. Current is $3 \mathrm{A}$ when cells and battery aid each other and is $2 \mathrm{A}$ when cells and battery oppose each other. The number of cells wrongly connected is:

In an experiment, a graph was plotted of the potential difference $V$ between the terminals of a cell against the circuit current $i$ by varying load rheostat. Internal conductance of the cell is given by

The magnitude and direction of the current in the circuit shown will be

The internal resistances of two cells shown are $0.1 \mathrm{\Omega }$ and $0.3 \mathrm{\Omega }$. If $R=0.2 \mathrm{\Omega }$, the potential difference across the cell

In the circuit shown here, $E_1=E_2=E_3=2 \mathrm{V}$ and $R_1=R_2=4 \text{Ω}$. The current flowing between points $A$ and $B$ through battery $E_2$ is

Consider the circuit shown in the figure. The current $I_3$ is equal to

As the switch $S$ is closed in the circuit shown in figure, current through it is