Question

The internal resistance of an accumulator battery of emf 6 V 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.

Answer 1

(a) Net e.m.f. while charging = 9 - 6 = 3v

$\therefore$ Current = $\frac{3}{10}$ 0.3 A

(b) When completely charged internal resistance 'r' = 1 $\Omega$

So, Current = $\frac{3}{1}$= 3A