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? - Physics

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Numerical

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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Solution

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

V= 120 − 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

I = `"V"^1/("R" + "r")`

= `112/(15.5 + 5)`

= `112/16`

= 7 A

Voltage across resistor R given by the product, IR = 7 × 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.

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Chapter 3: Current Electricity - Exercise [Page 128]

APPEARS IN

NCERT Physics Class 12
Chapter 3 Current Electricity
Exercise | Q 3.11 | Page 128
NCERT Physics Class 12
Chapter 3 Current Electricity
Exercise | Q 11 | Page 128

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