Physics, asked by amanrastogi6743, 10 months ago

The internal resistance of an accumulator battery of emf 6 V is 10 Ω when it is fully discharged. As the battery gets charged up, its internal resistance decreases to 1 Ω.
The battery in its completely discharged state is connected to a charger that 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.

Answers

Answered by anvesha12340
2

Explanation:

a) When the battery is being charged:

Emf of the battery, E = 6 V

Internal resistance of the battery, r = 10 Ω

Potential difference, V = 9 V

Net e.m.f. across the resistance while charging = 9 – 6 = 3 V

(b) When the battery is completely charged:

Internal resistance, r' = 1 Ω

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