$12$ 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:

$A$ 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 .............. $V$.

The electromotive force of a primary cell is $2\,V$. When it is short-circuited,it gives a current of $4\,A$. Its internal resistance in $\Omega$ is:

Two cells of emf $1 \ V$ and $2 \ V$ and internal resistance $2 \ \Omega$ and $1 \ \Omega$,respectively,are connected in series with an external resistance of $6 \ \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to the same external resistance. In this case,the total current drawn is $I_2$. The value of $\left(\frac{I_1}{I_2}\right)$ is $\frac{x}{3}$. The value of $x$ is . . . . . . .

The $e.m.f.$ of a cell is generally........