MP Board · Class 12 · Physics · Current ElectricityA storage battery of emf $8.0\text{ V}$ and internal resistance $0.5\text{ }\Omega$ is being charged by a $120\text{ V}$ d.c. 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 primary cause of damage if the series resistor is not used? Show complete numerical steps.

Given Data:

• EMF of the storage battery ($E$) = $8.0\text{ V}$
• Internal resistance of the battery ($r$) = $0.5\text{ }\Omega$
• Supply voltage ($V_s$) = $120\text{ V}$
• Series resistance ($R$) = $15.5\text{ }\Omega$

Part (a): Terminal Voltage of the Battery During Charging

\nWhen a battery is being charged, current enters its positive terminal. The potential difference (terminal voltage $V$) across the battery terminals is given by the formula: $$V = E + Ir$$ \nFirst, we need to calculate the charging current ($I$) flowing through the circuit. The total resistance of the circuit is the sum of the series resistor and the internal resistance of the battery, and the net EMF opposing the supply voltage is $(V_s - E)$ because the battery is being charged. \nUsing Ohm's Law for the circuit: $$I = \frac{V_s - E}{R + r}$| \nSubstitute the given values into the formula: $$I = \frac{120\text{ V} - 8.0\text{ V}}{15.5\text{ }\Omega + 0.5\text{ }\Omega}$$ $$I = \frac{112\text{ V}}{16.0\text{ }\Omega}$| $$I = 7.0\text{ A}$| \nNow, calculate the terminal voltage ($V$) of the battery using $V = E + Ir$: $$V = 8.0\text{ V} + (7.0\text{ A} \times 0.5\text{ }\Omega)$$ $$V = 8.0\text{ V} + 3.5\text{ V}$| $$V = 11.5\text{ V}$|

Answer (a): The terminal voltage of the battery during charging is $11.5\text{ V}$.

Part (b): Cause of Damage if Series Resistor is Not Used

\nIf the series resistor $R$ is not used ($R = 0$), the charging current would be extremely large. Let us calculate this current: $$I_{\text{without } R} = \frac{V_s - E}{r} = \frac{120 - 8}{0.5} = \frac{112}{0.5} = 224\text{ A}$| \nSuch a massive current ($224\text{ A}$) passing through the battery would cause:

1. Excessive Joule Heating ($I^2Rt$): Rapid and extreme generation of heat inside the battery, leading to boiling of the electrolyte.
2. Internal Damage and Explosion: The plates of the battery would warp and buckle, and the rapid evolution of hydrogen and oxygen gases due to water electrolysis could cause the battery casing to rupture or explode.