Charging of a battery

Batteries are made up of one or more electrochemical cells, arranged in series.

In these cells an electrochemical Redox reaction takes place:

$R+ze^{-} to R^{z-}$

$O-ze^{-} to O^{z+}$

Where $R$ and $O$ resp. are a reducing agent and oxidising agent. This transfer of electrons provides the EMF of the cell.

As more current is drawn from the cell, the concentrations of $R$ and $O$ in the cell's electrolyte slowly drop and so does the potential across the electrodes, until it is basically zero.

By recharging the cell (by reversing the normal flow of current) the above reactions are reversed:

$R^{z-} - ze^- to R$

$O^{z+} + ze^- to O$

Thermodynamically the total amount of electrical energy the cell delivered during normal operation (discharge) must now be delivered by the charging current, to fully restore the concentrations of $R$ and $O$ to their original levels.

Cells are imperfect though: the electrolyte is conductive but does have some electrical resistance ('internal resistance'). For that reason both charging and discharging waste some energy as ohmic power.

The battery usually 'supplies' electrical energy to the circuit element as charges flow through it. However in case of charging, the charging source provides energy required to reverse the spontaneous electrochemical reactions which make cell a cell, seemingly returning the cell or battery to its initial state (it's not the case though, with time through wear and tear, reversal process becomes inefficient and the battery is dead now) and actually it's the game of electrons and electron giving material.

In this question when the 9V battery is connected, the energy equivalent to 9 units(say) is supplied to the circuit and 6 units of those is utilized in 'charging process'. These 6 units are completely dumped to the discharged battery as soon as the connections are made. Think of it as filling an empty tank A with filled tank B of water. When A is completely empty, the whole of pipe is carrying water, but this is not the case towards the end. So, just when the connections are made, the supposed emf from the reference of the resistance is 9-6 = 3V. This explains everything.

The simple answer is $(9-6)/10=0.3 A$ , but it is not easy to explain why this is the correct answer in reality and in theory. When they say that the battery EMF is 6V, I think they mean it is the voltage in the fully charged state. Now, the surprising and counterintuitive thing is that if such a battery were discharged to zero or near zero, and connected to a fixed 9V source, the current would not equal 0.9A for any amount of time, and most likely the current would go to 0.3A nearly instantly.

The battery

How is the battery made up?

• The battery is made up of dissimilar materials at its cathode (-) and anode (+).
• Even in the discharged state, the emf of the battery is $6V$, or slightly above. It will not be possible to discharge the battery below this voltage without changing the chemistry of at least one of these: the anode or cathode.
• In the discharged state, i.e. the charge at which it becomes unable to, for example, turn a light on, the battery's emf is often not zero because the emf required by the load is significantly higher than zero, and also because the battery cannot go below its minimum charge voltage determined by chemistry as mentioned above. It can, however reach as low as 0.8V, as in alkaline batteries. The figure below shows the decrease in battery capacity of a lithium ion battery from terminal $4.2V$ to $3V$. Note also that the voltage settles at 3V (Source Ref).

Charging the battery

If we use a stable $9V$ charger, then, at the start of the charging, the voltage difference (emf) will be:

$9 - 6 = 3V$ ... at $3V$ emf, the current is:

$3/10 = 0.3A$.

Indeed, the resistance may change as the battery gets charged, as internal chemical activity changes.