$n$ 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:

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:

The internal resistance of a cell is the resistance of

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 $\text{ V}$)

Two cells of emfs $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ ($E_2 > E_1$ and $r_2 > r_1$) respectively,are connected in parallel as shown in the figure. The equivalent emf of the combination is $E_{eq}$. Then

Two cells of emf $E_{1}$ and $E_{2}$ are joined in opposition (such that $E_{1} > E_{2}$). If $r_{1}$ and $r_{2}$ are the internal resistances and $R$ is the external resistance,then the terminal potential difference across the external resistance $R$ is: