$A$ cell can supply currents of $1 \ A$ and $0.5 \ A$ via resistances of $2.5 \ \Omega$ and $10 \ \Omega$ respectively. The internal resistance of the cell is (in $\Omega$)

When does the current drawn from a cell become maximum?

$A$ circuit of resistance $R$ is connected to $n$ similar cells. If the current in the circuit is the same when the cells are connected in series or in parallel,and the internal resistance of each cell is $r$,then:

$A$ cell sends a current through a resistance $R$ for time $t$. Now the same cell sends current through another resistance $r$ for the same time. If the same amount of heat is developed in both the resistances,then the internal resistance of the cell is:

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:

When two cells of emfs $E_1$ and $E_2$ and different internal resistances $r_1$ and $r_2$ are connected in series with an external load resistor $R$,the current through the load is $5 \ A$. If the polarity of the cell of emf $E_2$ is reversed,then the current through the load is $2 \ A$. Then $\frac{E_1}{E_2} = $