Electromotive force is actually not a force. What does it represent?

$A$ resistor of resistance $R$ is connected between the terminals of a cell of $\operatorname{emf} E$ and internal resistance $r$. If $I$ is the current through the circuit,the terminal potential difference of the cell is given by:

Two batteries,one of e.m.f. $12 \text{ V}$ and internal resistance $2 \Omega$ and other of e.m.f. $6 \text{ V}$ and internal resistance $1 \Omega$,are connected as shown in the figure. What will be the reading of the voltmeter '$V$' (in $\text{ V}$)?

Two cells of emf $1 \ V$ and $2 \ V$ and internal resistance $2 \ \Omega$ and $1 \ \Omega$,respectively,are connected in series with an external resistance of $6 \ \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to the same external resistance. In this case,the total current drawn is $I_2$. The value of $\left(\frac{I_1}{I_2}\right)$ is $\frac{x}{3}$. The value of $x$ is . . . . . . .

The reading of a high resistance voltmeter when a cell is connected across it is $2.2\, V$. When the terminals of the cell are also connected to a resistance of $5\,\Omega$,the voltmeter reading drops to $1.8\, V$. Find the internal resistance of the cell in $\Omega$.

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: