The internal resistance of an accumulator battery of emf 6V is `10(Omega)`when it is fully discharged,As the battery gets charged up.its internal resistance decreases to `1(Omega)`The battery in its completely discharged state is connected to a charger which maintains a constant potential difference of 9 V.Find the current through the battery (a)just after the connections are made and (b) after a long time when it is completely charged.

Explanation

To solve the problem step by step, we will analyze the situation of the battery being charged and calculate the current at two different instances: just after the connections are made and after a long time when the battery is fully charged.

### Given Data:
- EMF of the battery (E) = 6 V
- Internal resistance when fully discharged (R1) = 10 Ω
- Internal resistance when fully charged (R2) = 1 Ω
- Potential difference maintained by the charger (V_ext) = 9 V

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