Ten identical cells each of potential $E$ and internal resistance $r$ are connected in series to form a closed circuit. An ideal voltmeter connected across three cells will read $...........E$.

There are a large number of cells available,each marked $(6 \,V, 0.5 \,\Omega)$,to be used to supply current to a device of resistance $0.75 \,\Omega$,requiring $24 \,A$ current. How should the cells be arranged so that power is transmitted to the load using the minimum number of cells?

Two cells of $emfs$ $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ are connected in parallel. The $emf$ and internal resistance of the equivalent source is

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

Two batteries of emf $\varepsilon_1$ and $\varepsilon_2$ (where $\varepsilon_2 > \varepsilon_1$) and internal resistances $r_1$ and $r_2$ respectively are connected in parallel as shown.

Two cells $P$ and $Q$ each of emf $2.16 \text{ V}$ are connected in series with a resistor of $19.6 \text{ } \Omega$. An ideal voltmeter reads $2 \text{ V}$ when connected across the cell $P$ and $1.92 \text{ V}$ when connected across the cell $Q$. The ratio of the internal resistances of the cell $P$ and $Q$ is