When a current of $2 \ A$ flows through a resistor of $2 \ \Omega$,the terminal voltage across the cell is $E/2$. The internal resistance of the cell is ........ $\Omega$.

Two batteries,one of e.m.f. $12 \text{ V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6 \text{ V}$ and internal resistance $1 \Omega$,are connected as shown in the figure. What will be the reading of the voltmeter '$V$' (in $\text{ V}$)?

The value of the internal resistance of an ideal cell is:

Two sources of equal $emf$ $(\varepsilon)$ are connected in series to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero, then:

The internal resistance of a cell depends on

In the previous problem,if the cells had been connected in parallel (instead of in series),which of the above graphs would have shown the relationship between total current $I$ and $n$ (number of cells)?