Question

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

Answer 1

During charging , the electric current is sent into 8V battery.

given, emf of battery, E = 8V,

DC supply, E'= 120V ,

internal resistance, r = 0.5 Ω and external resistance, R = 15.5 Ω

apply Kirchoff's law,

E - Ir - IR + E' = 0

⇒ I = (E' - E)/(r + R)

⇒ I = (120 - 8)/(0.5 + 15.5)

⇒ I = 112/16 = 7A

now, terminal voltage of the battery during charging, V = E + Ir

V = 8 + 7 × 0.5 = 8 + 3.5 = 11.5 V

A series resistance is joined in the charging circuit to limit the excessive current so that charging is slow and permanent.

Answer 2

$\huge \underline \mathrm \purple{Question↣}$

 

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

$\huge \underline \mathrm \green{Answer↣}$ 

 Given :

 

↦ The EMF of the given storage battery is $\bold{ E = 8.0 V}$

 

↦ The Internal resistance of the battery is given by $\bold{ r = 0.5 Ω}$

 

↦ The given DC supply voltage is $\bold{ V = 120 V}$

 

↦ The resistance of the resistor is $\bold{ R = 15.5 Ω}$

 

↦ Effective voltage in the circuit $\bold{ = V^1}$

 

R is connected to the storage battery in series.

 

Hence, it can be written as

 

$\bf{ : \implies V^1 = V – E }$

 

⠀⠀⠀⠀⠀⠀⠀⠀$\bf{ : \implies V^1 = 120 – 8 }$

 

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf\red{ : \implies V^1 = 112 V}$

 

Current flowing in the circuit $\bold{= I }$ , which is given by the relation ,

 

⠀⠀⠀⠀⠀${\large{ \underline { \boxed { \bf\: I \: = \large{\frac{ V{}^{2}}{R+ r}}}}}}$

 

⠀⠀⠀⠀⠀⠀⠀⠀$\bold{ : \implies \: I\: = \: \frac{112}{15.5 + 0.5} }$

 

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bold{ \red{ : \implies \: I\: = \: 7A}}$

 

We know that Voltage across a resistor R given by the product,

 

⠀⠀⠀⠀⠀⠀⠀⠀$\bold{ : \implies I \: x \: R = 7 × 15.5}$

 

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bold\red { : \implies 108.5 V}$

 

We know that ,

 

DC supply voltage = Terminal voltage + voltage drop across R

 

Terminal voltage of battery $\bold{ = 120 – 108.5 }$

 

Terminal voltage of battery $\bf\red { = 11.5 V}$

A series resistor when connected in a charging circuit limits the current drawn from the external source.

 

The current will become extremely high in its absence.

 

*This is Extremely Dangerous*