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 1Ω. The battery in its completely discharged state is connected to a charger which maintains a constant potential difference of 9V. Find the current through the battery
(a) just after the connections are made and
(b) after a long time when it is completely charged.
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.