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?
Solution
Given,
EMF of the storage battery, E= 8 V
Internal resistance, r = 0.5 Ω
D.C supply voltage, V= 120 V
Resistance of the resistor i.e, external resistance,R = 15.5 Ω
Using the formula,
EMF of the storage battery, E= 8 V
Internal resistance, r = 0.5 Ω
D.C supply voltage, V= 120 V
Resistance of the resistor i.e, external resistance,R = 15.5 Ω
Using the formula,
Therefore,
Charging current, I =
Terminal voltage, V = E + I r
= 8 + 7 x 0.5
= 11.5 V
The series resistor limits the current from the external source. The flow of current may be drastically high in the absence of series resistor which is dangerous.
Charging current, I =
Terminal voltage, V = E + I r
= 8 + 7 x 0.5
= 11.5 V
The series resistor limits the current from the external source. The flow of current may be drastically high in the absence of series resistor which is dangerous.
