$A$ battery consists of $5$ rows of cells,each row containing $10$ cells in series. This battery is connected in parallel to an external resistor of $20 \,\Omega$. If each cell has an $emf$ of $1.5 \,V$ and an internal resistance of $1 \,\Omega$,what is the current $i$ flowing through the external resistor (in $,A$)?

$12$ cells,each having the same $emf$ $E$ and internal resistance $r$,are connected in series,but some cells are wrongly connected. This arrangement is connected in series with an ammeter and two additional cells (each of $emf$ $E$ and internal resistance $r$). The current is $3 \, A$ when the cells and the battery aid each other,and it is $2 \, A$ when they oppose each other. The number of cells wrongly connected is:

$A$ battery of $24$ cells,each of emf $1.5\,V$ and internal resistance $2\,\Omega$,is to be connected in order to send the maximum current through a $12\,\Omega$ resistor. The correct arrangement of cells will be:

$A$ current of $2\,A$ is flowing through a cell of $e.m.f.$ $5\,V$ and internal resistance $0.5\,\Omega$ from negative to positive electrode. If the potential of the negative electrode is $10\,V$,the potential of the positive electrode will be .............. $V$.

Explain cell,emf,and internal resistance. Derive the relation between potential difference,emf,and internal resistance.

Why is the combination of cells done? Write its methods.