Question

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?

Answer 1

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,
\(I = \frac{V_1}{R+r}\)
\(I = \frac{112}{15.5+5}\)
\(I = \frac{112}{16}\)
\(I = 7 A\)
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 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.