The main issue to consider is the behaviour of the battery as it discharges and charges:
- In the charged state, the emf of a $6V$ battery is $6V$, or slightly above.
- 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. It can, however reach zero, as in alkaline batteries. The figure below shows the charge profile whereby the charger is $3V$ above the battery 'full charge' voltage. Note also that the charge vs capacity graph is not linear (Source Ref).
Therefore, The battery voltage when a $6V$ battery is 'discharged' will be anything from approximately $3V$ to $0V$. Let's say it is at $1V$.
Charging the battery
- A charger must cause current to flow in the opposite direction to the no-charging, load-connected state.
- The battery being charged must have a lower voltage difference than the charger, otherwise no charging will occur.
If we use a stable $9V$ charger, then, at the start of the charging, the voltage difference (emf) will be:
$9-1 = 8V$ and therefore current will be:
$8/10 = 0.8A$
As the battery reaches full charge, the emf will be:
$9 - 6 = 3V$ or less if overcharging is done. at $3V$ emf, the current is:
$3/10 = 0.3A$, i.e. the current reduces as the battery gets charged up.
Indeed, the resistance may change as the battery gets charged, as internal chemical activity changes, but we ignore this here.