At N the voltage begins to rise because the charging chemical reactions are taking place farther and farther in the inside parts of the plate, and the concentrated acid formed by the chemical actions in the plates is diffusing into the main electrolyte. This increases the battery voltage and requires a higher charging voltage.
This increases the battery voltage and requires a higher charging voltage. At the point marked 0, the voltage begins to rise very rapidly. This is due to the fact that the amount of lead sulphate in the plates is decreasing very rapidly, allowing the battery voltage to rise and thus increasing the charging voltage.
If you increase the load on a battery (decrease load resistance, add more light bulbs in parallel...) the current delivered by the battery will increase, causing an increased voltage drop across the battery's internal resistance and reducing the voltage measured between the battery terminals. This graph does not relate to the battery being used up.
Now remember, that a model for a battery is an ideal voltage source, internal resistance. when you start pulling current from the battery and complete the load there will be a voltage drop rI corresponding to the voltage drop due to the internal resistance this will cause the voltage of the cell to be lower than the voltage of the voltage source.
According to the graph as voltage decreases, current increases. The only way I can explain it using the equation V=e-rI is that for some reason internal resistance r increases and as eloctromotive force stays the same, this means decrease in voltage V so both sides equal each other again. But wait!
The size of the internal resistance depends on the battery type. So the voltage you can squeeze out of a battery actually depends on what you connect to it! In the image above, the battery has an internal resistance of 1 ohm. If you connect that to something of 500 ohms, you can figure out the battery voltage by using the voltage divider formula: