When a resistor of $11 \,\Omega$ is connected in series with an electric cell,the current flowing in it is $0.5 \, A$. Instead,when a resistor of $5 \,\Omega$ is connected to the same electric cell in series,the current increases by $0.4 \, A$. The internal resistance of the cell is ................ $\Omega$.

$A$ cell of internal resistance $r$ is connected to an external resistance $R$. The current will be maximum in $R$,if

$A$ cell of emf $10 \text{ V}$ and internal resistance $3 \Omega$ is connected in parallel with another cell of emf $7 \text{ V}$ and internal resistance $\frac{3}{5} \Omega$,such that their positive terminals are joined together and their negative terminals are joined together. Their combined positive terminal is joined with the negative terminal,and their combined negative terminal is joined with the positive terminal of a third cell of emf $20 \text{ V}$ with internal resistance $2 \Omega$. The combination can be replaced by a battery of emf $E$ and internal resistance $r$. The values of $E$ and $r$ are respectively:

$A$ 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 a battery is connected across a resistor of $16 \Omega$,the voltage across the resistor is $12 \ V$. When the same battery is connected across a resistor of $10 \Omega$,the voltage across it is $11 \ V$. The internal resistance of the battery in ohm is

$32$ cells,each of $emf$ $3V$,are connected in series and kept in a box. Externally,the combination shows an $emf$ of $84V$. The number of cells reversed in the combination is