Answer :

a)0.3A, b)3A

Given,


The emf of the battery, E=6V


The internal resistance of the battery, r1 when discharged=10Ω


The internal resistance of the battery, r2 when charged=1Ω


Potential difference provided by the charger, Ec=9V


Formula used:


When the accumulator battery is connected to a charger, the current through the internal resistance depends on the net emf available across the resistor.


Hence Net emf across the resistor in the case of charging will be the difference between the provided potential difference and the potential difference rating of the battery.


Solution:


a) When the battery is being charged, the net emf, Enet across the resistance, r1 will be,



Or,



Hence from the relation, , we can find the current, i by substituting r1 as 10 Ω,



Or,



b) When the battery is completely charged, the internal resistance r2 will be 1Ω. The net emf across the resistance will be the same, and is 3V.


Hence from the relation, we can find the current, i through the resistance by substituting r2 as 1Ω,



Or,



Hence the current through the internal resistance while charging and after completely charged are 0.3A and 3A.


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