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:

When a resistance of $2\,\Omega$ is connected across the terminals of a cell,the current is $0.5\,A$. When the resistance is increased to $5\,\Omega$,the current is $0.25\,A$. The internal resistance of the cell is ............. $\Omega$.

Two identical cells give the same current through an external resistance of $2 \Omega$ regardless of whether the cells are connected in series or in parallel. The internal resistance of each cell is: (in $\Omega$)

Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:

$A$ cell of e.m.f. $E$ is connected with an external resistance $R$,then the potential difference across the cell is $V$. The internal resistance of the cell will be:

The potential difference between the terminals of a cell is $20 \ V$ when a current of $2 \ A$ flows through the circuit. When the direction of current in the circuit is reversed,the potential difference between the terminals of the cell is $30 \ V$. The internal resistance of the cell is (in $\Omega$)