Ten identical cells each emf $2 \, V$ and internal resistance $1 \, \Omega$ are connected in series with two cells wrongly connected. $A$ resistor of $10 \, \Omega$ is connected to the combination. What is the current through the resistor (in $ \, A$)?

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:

Each of the six ideal batteries of emf $20 \, V$ is connected to an external resistance of $4 \, \Omega$ as shown in the figure. The current through the resistance is (in $ \, A$)

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 ................ $\Omega$.

$32$ 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

What is called the internal resistance of a cell?