If the $EMF$ of each cell is $E = 1.5 \text{ V}$ and internal resistance is $r$, as shown in the figure, what is the current $i$ flowing through the circuit (in $\text{ A}$)?

Two sources of equal emfs $E$ are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_{1}$ and $r_{2}$ $(r_{1} > r_{2})$. If the potential difference across the source of internal resistance $r_{1}$ is zero,then the value of $R$ will be:

Two cells of emf $1 \ V$ and $2 \ V$ and internal resistance $2 \ \Omega$ and $1 \ \Omega$,respectively,are connected in series with an external resistance of $6 \ \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to the same external resistance. In this case,the total current drawn is $I_2$. The value of $\left(\frac{I_1}{I_2}\right)$ is $\frac{x}{3}$. The value of $x$ is . . . . . . .

When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$,the rate of heat produced in $R$ is $P_1$. When the same batteries are connected in parallel across $R$,the rate of heat produced is $P_2$. If $P_1 = 2.25 P_2$,then the value of $R$ is (in $\Omega$)

Four identical cells,each having an electromotive force $(e.m.f.)$ of $12\,V$,are connected in parallel. The resultant electromotive force $(e.m.f.)$ of the combination is .............. $V$.

Two cells $E_1$ and $E_2$ having equal $EMF$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance $R$. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be