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 $...........E$.

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 $\Omega$)

Two cells of emfs $1$ $V$ and $2$ $V$ and internal resistances $2 \Omega$ and $1 \Omega$ respectively are connected in parallel and provide a current of $1$ $A$ through an external resistance. If the polarity of one cell is reversed, the current through the external resistance becomes $\frac{\alpha}{5}$ $A$. The value of $\alpha$ is . . . . . . .

When a battery is connected across a resistor of $16 \Omega$,the voltage across the resistor is $12 \ V$. When the same battery is connected across a resistor of $10 \Omega$,the voltage across it is $11 \ V$. The internal resistance of the battery in ohm is

When does the current drawn from a cell become maximum?

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: