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20 Physics -- DC Circuits

A storage battery of emf 8V and internal resistance 0.5ohm is discharged through a parallel combination of to resistor each of resistance 15 ohm. What is the terminal voltage of the battery?​ [3]a) Why is the value of terminal voltage obtained in the question less than the emf of the ba...

A storage battery of emf 8V and internal resistance 0.5ohm is discharged through a parallel combination of to resistor each of resistance 15 ohm. What is the terminal voltage of the battery?​ [3]
a) Why is the value of terminal voltage obtained in the question less than the emf of the battery? [1]

b) If we charge a battery, why do we add a series resistor in the circuit? [1]

1 Answers

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Emf =8V,

Internal resistance r=0.5Ω
R1 = R2 = 15 Ω

Equivalent capacitance of R1 and R2 is

R= 7.5 Ω

now,

emf = I (R+r)
8 = I (7.5+0.5)

8 = I 8
∴ I = 1A

Now,

E= IR+ Ir
E = V + Ir
8 = V + 1x0.5

∴ V= 7.5 volt


a) the value of terminal voltage is less than the emf of battery, since some work needs to be done to overcome the internal resistance of the battery itself. 


b)Adding a resistor in series while charging a battery will help to limit the amount of current flowing to a required amount. Also, any damage can be prevented due to voltage fluctuation.


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