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

(C) Given: $Emf$ of the storage battery,$E = 8.0\; V$. Internal resistance of the battery,$r = 0.5\; \Omega$. $DC$ supply voltage,$V = 120\; V$. Resistance of the series resistor,$R = 15.5\; \Omega$.
The effective voltage in the circuit is $V' = V - E = 120 - 8.0 = 112\; V$.
The total resistance in the circuit is $R_{total} = R + r = 15.5 + 0.5 = 16.0\; \Omega$.
The charging current $I$ is given by $I = \frac{V'}{R_{total}} = \frac{112}{16.0} = 7.0\; A$.
The terminal voltage of the battery during charging is given by $V_{terminal} = E + Ir = 8.0 + (7.0 \times 0.5) = 8.0 + 3.5 = 11.5\; V$.
The purpose of the series resistor is to limit the charging current to a safe value,preventing damage to the battery and the power supply due to excessive current flow.