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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 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.
Sol. a) net emf while charging 9 – 6 = 3V Current = 3/10 = 0.3 A b) When completely charged. Internal resistance ‘r’ = 1 Ω Current = 3/1 = 3 A
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