$n$ equal cells,each having emf $E$ and internal resistance $r$,are connected in a circuit with an external resistance $R$. If the same current flows in the circuit whether the cells are connected in series or in parallel,then:

Two identical cells each of emf $1.5 \,V$ are connected in parallel across an external resistance formed by two $20 \,\Omega$ resistors connected in parallel. $A$ voltmeter connected in the circuit measures $1.2 \,V$. The internal resistance of each cell is ................. $\Omega$.

Two batteries,one of emf $18 \, V$ and internal resistance $2 \, \Omega$ and the other of emf $12 \, V$ and internal resistance $1 \, \Omega$,are connected in parallel as shown. The voltmeter $V$ will record a reading of ............. $V$.

In the previous problem,if the cells had been connected in parallel (instead of in series),which of the above graphs would have shown the relationship between total current $I$ and $n$ (number of cells)?

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

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