Internal Resistance

The reading on a high resistance voltmeter, when a cell is connected across it, is $2.0V$. When the terminals of the cell are also connected to a resistance of  $3 \Omega$  as shown in the circuit, the voltmeter reading drops to $1.5V$. The internal resistance of the cell is :

Why can not we draw maximum current from the cell in real life?

What should be null for drawing maximum current from the cell?

The potential difference $\left(V\right)$ across the terminals of a battery which is getting charged is always:

The terminal potential difference $\left(V\right)$ of a battery of emf $\left(E\right)$ and internal resistance $\left(r\right)$ which is getting charged with electric current $\left(I\right)$ flowing through it is:

While charging a run-down battery, current flows inside it from its negative terminal to positive terminal.

A storage battery of emf $8.0 \mathrm{V}$ with internal resistance $0.5\mathrm{\Omega }$ is being charged by a $120 \mathrm{V} \mathrm{D}\mathrm{C}$ supply using a series resistor of $15.5\mathrm{\Omega }.$ What is the terminal voltage of the battery during charging?

Two cells with the same emf $E$ and different internal resistances $r_1$ and $r_2$ are connected in series to an external resistance $R$. The value of  $R$ so that, the potential difference across the first cell be zero is,

In the circuit shown A and B are two cells of same emf $E$, but different internal resistances $r_1$ and $r_2$ ($r_1>r_2$), respectively. What is the value of $R$ in terms of $r_1$ and $r_2$, such that the potential difference across the terminal of cell A is zero a long time after the key K is closed.

A student measures the terminal potential difference $V$ of a cell (of emf $E$ and internal resistance $r$) as a function of the current $I$ flowing through it. The slope and intercept, of the graph between $V$ and $I$, then, respectively, equal: