When fully discharged, an accumulator battery of emf 6 V possesses...

The current $I$ passing through a battery during the charging process is given by the relation $V = E + Ir$, where $V$ is the potential difference provided by the charger, $E$ is the emf of the battery, and $r$ is its internal resistance.

Immediately after the circuit is connected, the battery is still in its completely discharged state. Therefore, the internal resistance of the battery is $r = 10 \Omega$.

Substituting the given values $V = 9 \text{ V}$, $E = 6 \text{ V}$, and $r = 10 \Omega$ into the formula:

$I = \dfrac{V - E}{r}$

$I = \dfrac{9 - 6}{10}$

$I = \dfrac{3}{10} = 0.3 \text{ A}$