A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Emf of the storage battery, E = 8.0 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Effective voltage in the circuit = V¹

R is connected to the storage battery in 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,

I =V¹/(R+r) = 112/(15.5 +5) = 112/16 =7 A

Voltage across resistor R given by the product, IR = 7 x 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 series 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.