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


Electromotive force EMF of battery = 8Volt


Internal resistance of battery r = 0.5 Ω


Supply Voltage V = 120Volt


Resistance of resistor R = 15.5 Ω


Effective voltage in circuit = V1


Since Resistance R is connected in series Hence We can write V1 = V-E


V1 = 120V-8V = 112 V


Current flowing in the circuit:


⇒


⇒


⇒


I = 7Ampere


By Ohm ’s Law, Voltage across resistor R is given by V = IR


V = 7A × 15.5Ω


⇒V = 108.5Volt


Supply Voltage = Terminal Voltage of battery + Voltage drop across Resistor R


Therefore Terminal Voltage of battery = 120 V-108.5 V = 11.5Volt


Series resistor in charging circuit reduces the current drawn from external supply and current will be too high in its absence.


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