Emf of the storage battery, E = 8.0 V

Internal resistance of the battery, r = 0.5 $\mathrm{\Omega }$

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 $\mathrm{\Omega }$

Effective volate in the circuit = V¹

R is connected to the storage battery is series. Hence, it can be written as V¹ = V - E

V¹ = 120 - 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

$\mathrm{l}=\frac{{\mathrm{V}}^1}{\mathrm{R}+\mathrm{r}}$
$=\frac{112}{15.5+5}=\frac{112}{16}=7 \mathrm{A}$

Voltage across resistor R given by the product IR = 7 $\times$ 15.5 = 108.5 V

DC supply voltage = Terminal voltage of battery + Voltage drop across R

Terminal voltage of battery = 120 - 108.5 = 11.5 V

A resistor in a charging circuit limits the current drawn from the external source. The current will be extremely high in its absence. This is very dangerous.