The electromotive force of a primary cell is $2\,V$. When it is short-circuited,it gives a current of $4\,A$. Its internal resistance in $\Omega$ is:

What will be the reading of the voltmeter in the given circuit (in $V$)?

Two cells of emfs $1$ $V$ and $2$ $V$ and internal resistances $2 \Omega$ and $1 \Omega$ respectively are connected in parallel and provide a current of $1$ $A$ through an external resistance. If the polarity of one cell is reversed, the current through the external resistance becomes $\frac{\alpha}{5}$ $A$. The value of $\alpha$ is . . . . . . .

The number of dry cells,each of $e.m.f.$ $1.5\,V$ and internal resistance $0.5\,\Omega$,that must be joined in series with a resistance of $20\,\Omega$ so as to send a current of $0.6\,A$ through the circuit is:

$n$ identical cells are joined in series with two cells $A$ and $B$ having reversed polarities. The $emf$ of each cell is $E$ and the internal resistance is $r$. The potential difference across cell $A$ or $B$ is: $(n > 4)$

Three identical cells,each of emf $10 \ V$ and internal resistance $3 \ \Omega$,are connected in parallel across terminals $A$ and $B$. The net emf across $A$ and $B$ is .......... $V$.