A storage battery of emf $8 ext{ V}$ and internal resistance $0.5 ext{ }\Omega$ is being charged by a $120 ext{ V}$ dc supply using a series resistor of $15.5 ext{ }\Omega$. (a) What is the terminal voltage of the battery during charging? (b) What is the purpose of having a series resistor in the charging circuit?

Q: A storage battery of emf $8\text{ V}$ and internal resistance $0.5\text{ }\Omega$ is being charged by a $120\text{ V}$ dc supply using a series resistor of $15.5\text{ }\Omega$. (a) What is the terminal voltage of the battery during charging? (b) What is the purpose of having a series resistor in the charging circuit?

Given Data: EMF of the storage battery, $E = 8\text{ V}$ Internal resistance of the battery, $r = 0.5\text{ }\Omega$ Voltage of the DC supply, $V{\text{supply}} = 120\text{ V}$ Series resistance, $R = 15.5\text{ }\Omega$ --- (a) Calculation of the terminal voltage during charging: \nWhen a battery is being charged, the current enters its positive terminal. The total resistance of the charging circuit is the sum of the series resistor and the internal resistance of the battery: $$R{\text{total}} = R + r$$ $$R{\text{total}} = 15.5\text{ }\Omega + 0.5\text{ }\Omega = 16.0\text{ }\Omega$$ \nThe net voltage driving the current through the circuit is the difference between the supply voltage and the battery's EMF (since the battery opposes the charging current): $$V{\text{net}} = V{\text{supply}} - E$$ $$V{\text{net}} = 120\text{ V} - 8\text{ V} = 112\text{ V}$$ \nUsing Ohm's law, the charging current $I$ flowing through the circuit is: $$I = \frac{V{\text{net}}}{R{\text{total}}} = \frac{112\text{ V}}{16.0\text{ }\Omega} = 7\text{ A}$$ \nDuring charging, the terminal voltage $V$ of the battery is given by the formula: $$V = E + Ir$$\nSubstitute the known values into the equation: $$V = 8\text{ V} + (7\text{ A} \times 0.5\text{ }\Omega)$$ $$V = 8\text{ V} + 3.5\text{ V} = 11.5\text{ V}$$ Answer (a): The terminal voltage of the battery during charging is $11.5\text{ V}$. --- (b) Purpose of the series resistor: \nThe series resistor limits the magnitude of the charging current. Without this external resistor, the low internal resistance of the battery ($0.5\text{ }\Omega$) connected across the large supply voltage ($120\text{ V}$) would result in an extremely high current ($I = \frac{120 - 8}{0.5} = 224\text{ A}$). Such a massive surge in current would cause severe overheating, damage the battery plates, boil the electrolyte, and pose serious safety and explosion hazards. The series resistor ensures a safe and controlled charging rate.

Question

A storage battery of emf $8	ext{ V}$ and internal resistance $0.5	ext{ }\Omega$ is being charged by a $120	ext{ V}$ dc supply using a series resistor of $15.5	ext{ }\Omega$. (a) What is the terminal voltage of the battery during charging? (b) What is the purpose of having a series resistor in the charging circuit?

Accepted Answer

Given Data: EMF of the storage battery, $E = 8	ext{ V}$ Internal resistance of the battery, $r = 0.5	ext{ }\Omega$ Voltage of the DC supply, $V{	ext{supply}} = 120	ext{ V}$ Series resistance, $R = 15.5	ext{ }\Omega$ --- (a) Calculation of the terminal voltage during charging: 
When a battery is being charged, the current enters its positive terminal. The total resistance of the charging circuit is the sum of the series resistor and the internal resistance of the battery: $$R{	ext{total}} = R + r$$ $$R{	ext{total}} = 15.5	ext{ }\Omega + 0.5	ext{ }\Omega = 16.0	ext{ }\Omega$$ 
The net voltage driving the current through the circuit is the difference between the supply voltage and the battery's EMF (since the battery opposes the charging current): $$V{	ext{net}} = V{	ext{supply}} - E$$ $$V{	ext{net}} = 120	ext{ V} - 8	ext{ V} = 112	ext{ V}$$ 
Using Ohm's law, the charging current $I$ flowing through the circuit is: $$I = \frac{V{	ext{net}}}{R{	ext{total}}} = \frac{112	ext{ V}}{16.0	ext{ }\Omega} = 7	ext{ A}$$ 
During charging, the terminal voltage $V$ of the battery is given by the formula: $$V = E + Ir$$
Substitute the known values into the equation: $$V = 8	ext{ V} + (7	ext{ A} 	imes 0.5	ext{ }\Omega)$$ $$V = 8	ext{ V} + 3.5	ext{ V} = 11.5	ext{ V}$$ Answer (a): The terminal voltage of the battery during charging is $11.5	ext{ V}$. --- (b) Purpose of the series resistor: 
The series resistor limits the magnitude of the charging current. Without this external resistor, the low internal resistance of the battery ($0.5	ext{ }\Omega$) connected across the large supply voltage ($120	ext{ V}$) would result in an extremely high current ($I = \frac{120 - 8}{0.5} = 224	ext{ A}$). Such a massive surge in current would cause severe overheating, damage the battery plates, boil the electrolyte, and pose serious safety and explosion hazards. The series resistor ensures a safe and controlled charging rate.