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

Two cells $E_1$ and $E_2$ having equal $EMF$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance $R$. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

In the circuit shown,the potential difference between $A$ and $B$ is ............. $V$.

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$)?

Two cells of equal $e.m.f.$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ are connected in series. On connecting this combination to an external resistance $R$,it is observed that the potential difference across the first cell becomes zero. The value of $R$ will be

Four identical cells, each of $2 \, V$ e.m.f., are connected in parallel. They supply current to an external circuit consisting of two $15 \, \Omega$ resistors connected in parallel. The terminal voltage of the equivalent cell, as measured by an ideal voltmeter, is $1.6 \, V$. Calculate the internal resistance of each cell in $\Omega$.