What is the potential difference between the points $P$ and $Q$ in the circuit shown below, once the capacitors are fully charged?
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Using the voltage divider formula directly for capacitors:
The potential drop is inversely proportional to capacitance.
For point P: $V_P = 12 \times \frac{2}{4+2} = 4\text{ V}$ (from the 0V side), so $V_P = 4\text{ V}$ if measured relative to 0V. Wait, $V_P = 12 \times \frac{2}{6} = 4\text{ V}$ (since $Q = C_1(12-V_P) = C_2(V_P)$). Yes, relative to 0V: $V_P = 4\text{ V}$ (if 2uF is on the right) and $V_Q = 12 \times \frac{4}{6} = 8\text{ V}$.
In either case, the difference is $|8\text{ V} - 4\text{ V}| = 4\text{ V}$. This saves significant calculation time.
Step 1: Understanding the Question:
This question requires calculating the potential difference between two intermediate nodes in a fully-charged capacitive bridge network connected across a DC voltage source. Step 2: Key Formulas and Approach:
1. When fully charged, no current flows through the circuit, and the capacitors in each series branch act as a voltage divider.
2. Equivalent capacitance of two capacitors $C_1$ and $C_2$ in series:
\[ C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} \]
3. Potential at the junction of two series capacitors connected to a voltage source $V$:
\[ V_{\text{junction}} = V_{\text{source}} \left( 1 - \frac{C_1}{C_1 + C_2} \right) \] Step 3: Detailed Explanation:
• Let us assume the potential of the left junction is $V_L = 12\text{ V}$ and the right junction is $V_R = 0\text{ V}$.
• The top path consists of a $4\ \mu\text{F}$ capacitor in series with a $2\ \mu\text{F}$ capacitor. The point $P$ is located between them.
• The charge $Q_{\text{top}}$ on the top series combination is:
\[ Q_{\text{top}} = C_{\text{eq, top}} \times V = \left( \frac{4 \times 2}{4 + 2} \right) \mu\text{F} \times 12\text{ V} = \frac{8}{6} \times 12 = 16\ \mu\text{C} \]
• The potential drop across the first capacitor ($4\ \mu\text{F}$) in the top branch is:
\[ V_L - V_P = \frac{Q_{\text{top}}}{C_1} = \frac{16\ \mu\text{C}}{4\ \mu\text{F}} = 4\text{ V} \]
• Since $V_L = 12\text{ V}$, the potential at point $P$ is:
\[ 12 - V_P = 4 \implies V_P = 8\text{ V} \]
• The bottom path consists of a $2\ \mu\text{F}$ capacitor in series with a $4\ \mu\text{F}$ capacitor. The point $Q$ is located between them.
• The charge $Q_{\text{bottom}}$ on the bottom series combination is:
\[ Q_{\text{bottom}} = C_{\text{eq, bottom}} \times V = \left( \frac{2 \times 4}{2 + 4} \right) \mu\text{F} \times 12\text{ V} = 16\ \mu\text{C} \]
• The potential drop across the first capacitor ($2\ \mu\text{F}$) in the bottom branch is:
\[ V_L - V_Q = \frac{Q_{\text{bottom}}}{C_3} = \frac{16\ \mu\text{C}}{2\ \mu\text{F}} = 8\text{ V} \]
• Since $V_L = 12\text{ V}$, the potential at point $Q$ is:
\[ 12 - V_Q = 8 \implies V_Q = 4\text{ V} \]
• Therefore, the potential difference between $P$ and $Q$ is:
\[ V_{PQ} = |V_P - V_Q| = |8\text{ V} - 4\text{ V}| = 4\text{ V} \] Step 4: Final Answer:
The potential difference between $P$ and $Q$ is $4\text{ V}$, which corresponds to Option (A).
