$n$ identical cells,each of $e.m.f.$ $E$ and internal resistance $r$,are connected in series. This combination is connected in series with an external resistor $R$. What is the current flowing through $R$?

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:

$A$ galvanometer together with an unknown resistance in series is connected to two identical batteries each of $1.5\, V$. When the batteries are connected in series,the galvanometer records a current of $1\, A$,and when the batteries are in parallel,the current is $0.6\, A$. What is the internal resistance of the battery?

The internal resistance of a cell of emf $4 \text{ V}$ is $0.1 \ \Omega$. It is connected to an external resistance of $3.9 \ \Omega$. The terminal voltage across the cell will be . . . . . . . (in $\text{ V}$)

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$)?

In the given circuit,if $V_{AB} = 4\ V$,then the value of resistance $X$ in ohms will be: