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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

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

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=7A`

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.

  Is there an error in this question or solution?
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APPEARS IN

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