Capacitor Voltage in Steady-State RC Circuit

Answer

After the switch has been closed for a long time, the capacitor reaches a steady state and acts as an open circuit. The potential difference across the capacitor is equal to the voltage across the $2R$ resistor, which is $\frac{2}{3}\varepsilon$.

Explanation

Image Analysis: The diagram illustrates a DC circuit containing a voltage source $\varepsilon$, a resistor $R$ in the main branch, a capacitor $C$ in a parallel branch, and a resistor $2R$ in a branch controlled by a switch. When the switch is closed and the circuit reaches a steady state, no current flows through the capacitor branch.

Known quantities:

• Emf of the source: $\varepsilon$
• Resistance of first resistor: $R$
• Resistance of second resistor: $2R$
• State: Steady state (long time after closing switch)

Find:

• Potential difference across the capacitor ($V_C$).

1. Analyze the "Steady State" condition In DC circuits, a capacitor behaves as an open circuit (infinite resistance) after a long time ($t \gg \tau$). This means current $I_C = 0$. ⚠️ This step is required on exams: explicitly state that current through the capacitor branch is zero at steady state.

2. Determine the equivalent circuit current Since the capacitor branch is effectively "broken," the only continuous path for current is through the series combination of $R$ and $2R$. Using Ohm's Law: $I = \frac{\varepsilon}{R + 2R} = \frac{\varepsilon}{3R}$ The total current is the emf divided by the sum of the series resistances.

3. Calculate the potential difference across the parallel branch The capacitor is connected in parallel with the $2R$ resistor. Therefore, the voltage across the capacitor $V_C$ is identical to the voltage across the $2R$ resistor $V_{2R}$. $V_C = V_{2R} = I \cdot (2R)$ Voltage is the product of current and resistance.

4. Substitute and simplify Substitute the current found in Step 2 into the equation from Step 3: $V_C = \left( \frac{\varepsilon}{3R} \right) \cdot 2R$ Cancel the $R$ terms: $V_C = \frac{2}{3}\varepsilon$ The final expression for voltage is relative to the source emf.

5. Unit Check The result $\frac{2}{3}\varepsilon$ has the units of Volts (since $\varepsilon$ is in Volts), which is the correct dimension for potential difference.

Final Answer

The potential difference across the capacitor is: $\boxed{V_C = \frac{2}{3}\varepsilon}$

Common Mistakes

• Treating the capacitor as a short circuit: Students often confuse "long time" behavior with "immediate" behavior. At $t=0$, a capacitor acts like a wire ($V=0$); at $t \to \infty$, it acts like an open switch.
• Ignoring the voltage divider rule: Some may incorrectly assume the capacitor charges to the full battery voltage $\varepsilon$. However, because there is current flowing through $R$ in the steady state, there is a voltage drop ($I \cdot R$) across the first resistor, leaving less than $\varepsilon$ for the capacitor.