C

Step 1: Write the governing equation for the charging of a capacitor in a circuit with a battery and internal resistance. The differential equation can be derived from Kirchhoff's voltage law:

• Voltage across the capacitor = $\frac{q}{C}$
• Voltage across the internal resistance = $i \cdot r$
• Voltage supplied by the battery = $V$

Thus, we have: $$ V = \frac{q}{C} + i \cdot r $$ Step 2: Since the current $i$ can be expressed in terms of charge $q$ as: $$ i = \frac{dq}{dt} $$ Substituting this into the equation gives: $$ V = \frac{q}{C} + \frac{dq}{dt} \cdot r $$ Step 3: Rearranging the equation yields: $$ \frac{dq}{dt} = \frac{V - \frac{q}{C}}{r} $$ Step 4: Separating variables and integrating: $$ \int \frac{1}{V - \frac{q}{C}} dq = \int \frac{1}{r} dt $$ Step 5: The left-hand side integrates to: $$ -C \ln \left( V - \frac{q}{C} \right) = \frac{t}{r} + K $$ where $K$ is a constant of integration. Step 6: At $t=0$, the charge on the capacitor is $q_0$, leading to: $$ K = -C \ln \left( V - \frac{q_0}{C} \right) $$ Thus we have: $$ -C \ln \left( V - \frac{q}{C} \right) = \frac{t}{r} - C \ln \left( V - \frac{q_0}{C} \right) $$ Step 7: Exponentiating both sides: $$ V - \frac{q}{C} = \left(V - \frac{q_0}{C}\right) e^{-\frac{t}{Cr}} $$ Step 8: Solving for charge $q$: $$ q = C V - \left( C V - q_0 \right)e^{-\frac{t}{Cr}} $$ Final Result: Therefore, the charge on the capacitor at time $t$ is: $$ q(t) = C V - (C V - q_0) e^{-\frac{t}{Cr}} $$ This represents the typical form of charge on a capacitor charged by a battery with internal resistance.