Two batteries with different $e.m.f.$ are connected in parallel. Consider the following statements: $(i)$ The equivalent $e.m.f.$ is less than the individual $e.m.f.s$. $(ii)$ The equivalent internal resistance is less than the individual internal resistances.

Two cells are connected in opposition as shown. Cell $E_1$ has an electromotive force (emf) of $8 \ V$ and an internal resistance of $2 \ \Omega$; cell $E_2$ has an emf of $2 \ V$ and an internal resistance of $4 \ \Omega$. The terminal potential difference of cell $E_2$ is: (in $V$)

$A$ battery consists of $5$ rows of cells,each row containing $10$ cells in series. This battery is connected in parallel to an external resistor of $20 \,\Omega$. If each cell has an $emf$ of $1.5 \,V$ and an internal resistance of $1 \,\Omega$,what is the current $i$ flowing through the external resistor (in $,A$)?

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:

Ten identical cells each of potential $E$ and internal resistance $r$ are connected in series to form a closed circuit. An ideal voltmeter connected across three cells will read $...........E$.

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 $45$?