Two cells,each of $e.m.f.$ $E$ and internal resistance $r$,are connected in parallel across an external resistor $R$. The maximum energy delivered to the resistor occurs when:

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:

Two sources of equal $emf$ are connected to an external resistance $R$. The internal resistances of the two sources are $R_1$ and $R_2$ $(R_2 > R_1)$. If the potential difference across the source having internal resistance $R_2$ is zero,then:

Two batteries of different emfs and different internal resistances are connected as shown. The voltage across $AB$ in volts is

$n$ identical cells,each of emf $E$ and internal resistance $r$,are joined in series to form a row. $m$ such rows are joined in parallel across a load resistance $R$. The current in each cell is:

When two identical cells are connected either in series or in parallel across a $2 \ \Omega$ resistor,they send the same current through it. The internal resistance of each cell is: (in $\Omega$)