Class 12 Physics
CBSE and UP Board
Current Electricity
Exercise 3
Chapter-3 Exercise 3.11
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?
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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.