A storage battery ofemf 8.0 V and internal resistance 0.5 Ω is being charged by a120 V dc supply using a series resistor of 15.5 Ω. What is theterminal voltage of the battery during charging? What is the purposeof having a series resistor in the charging circuit?
A storage battery ofemf 8.0 V and internal resistance 0.5 Ω is being charged by a120 V dc supply using a series resistor of 15.5 Ω. What is theterminal voltage of the battery during charging? What is the purposeof having a series resistor in the charging circuit?
Emf of the storagebattery, E = 8.0 V
Internal resistance ofthe battery, r = 0.5 Ω
DC supply voltage, V= 120 V
Resistance of theresistor, R = 15.5 Ω
Effective voltage inthe circuit = V1
R is connectedto the storage battery in series. Hence, it can be written as
V1 =V − E
V1 =120 − 8 = 112 V
Current flowing in thecircuit = I, which is given by the relation,
Voltage across resistorR given by the product, IR = 7 × 15.5 = 108.5 V
DC supply voltage =Terminal voltage of battery + Voltage drop across R
Terminal voltage ofbattery = 120 − 108.5 = 11.5 V
A series resistor in acharging circuit limits the current drawn from the external source.The current will be extremely high in its absence. This is verydangerous.
Emf of the storagebattery, E = 8.0 V
Internal resistance ofthe battery, r = 0.5 Ω
DC supply voltage, V= 120 V
Resistance of theresistor, R = 15.5 Ω
Effective voltage inthe circuit = V1
R is connectedto the storage battery in series. Hence, it can be written as
V1 =V − E
V1 =120 − 8 = 112 V
Current flowing in thecircuit = I, which is given by the relation,
Voltage across resistorR given by the product, IR = 7 × 15.5 = 108.5 V
DC supply voltage =Terminal voltage of battery + Voltage drop across R
Terminal voltage ofbattery = 120 − 108.5 = 11.5 V
A series resistor in acharging circuit limits the current drawn from the external source.The current will be extremely high in its absence. This is verydangerous.
