$A$ cell of emf $1.2 \ V$ and internal resistance $2 \ \Omega$ is connected in parallel to another cell of emf $1.5 \ V$ and internal resistance $1 \ \Omega$. If the like poles of the cells are connected together,the emf of the combination of the two cells is (in $V$)

Two cells of internal resistance $r_{1}$ and $r_{2}$ and of the same emf are connected in series across a resistor of resistance $R$. If the terminal potential difference across the cell of internal resistance $r_{1}$ is zero,then 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:

Two batteries with different $e.m.f.$ are connected in parallel. Consider the following statements: $(i)$ The equivalent $e.m.f.$ is less than the individual $e.m.f.s$. $(ii)$ The equivalent internal resistance is less than the individual internal resistances.

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

For an open circuit condition,what is the value of the terminal voltage of a cell?