Question

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.

Answer 1

(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
$\therefore \mathrm{Curre}\text{nt}=\frac{3}{10}=0.3\mathrm{A}$


(b) When the battery is completely charged:
Internal resistance, r' = 1 Ω
$\therefore \mathrm{Current}=\frac{3}{1}=3\mathrm{A}$