$A$ student is given $4$ identical batteries having $EMF$ $1.5 \ V$ each and internal resistance of $0.1 \ \Omega$ each. The student is asked to connect them in an assisting manner. By mistake,he connects $1$ battery in reverse way. The resultant $EMF$ and resultant internal resistance offered by the combination is . . . . . . .

To get maximum current through a resistance of $2.5\,\Omega$,one can use $m$ rows of cells,each row having $n$ cells. The internal resistance of each cell is $0.5\,\Omega$. What are the values of $n$ and $m$ if the total number of cells is $45$?

For two cells having the same $EMF$ $E$ and internal resistance $r$, the current passing through an external resistor of $6 \ \Omega$ is the same when both cells are connected either in parallel or in series. The value of internal resistance $r$ is . . . . . . $ \ \Omega$.

Electromotive force is the force,which is able to maintain a constant

Two cells $P$ and $Q$ each of emf $2.16 \text{ V}$ are connected in series with a resistor of $19.6 \text{ } \Omega$. An ideal voltmeter reads $2 \text{ V}$ when connected across the cell $P$ and $1.92 \text{ V}$ when connected across the cell $Q$. The ratio of the internal resistances of the cell $P$ and $Q$ is

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: