Question

In the circuit shown in the figure, the potential difference across the 4\(\mu\)F capacitor is

• A. 3 V
• B. 4 V
• C. 9 V
• D. 12 V

Hint:

Voltage divider rule for two capacitors in series: \( V_1 = V_{total} \times \frac{C_2}{C_1 + C_2} \).
Here, \( V_{4\mu\text{F}} = 12 \times \frac{12}{4 + 12} = 12 \times \frac{12}{16} = 12 \times 0.75 = 9\text{V} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the voltage across a specific capacitor in a series-parallel combination connected to a 12V DC source.
Step 2: Key Formula or Approach:
1. Capacitance in parallel: \( C_p = C_1 + C_2 \)
2. Capacitance in series: \( \frac{1}{C_s} = \frac{1}{C_a} + \frac{1}{C_b} \)
3. Charge: \( Q = CV \). In series, charge is the same for all components.
4. Potential difference: \( V = \frac{Q}{C} \)
Step 3: Detailed Explanation:
1. First, simplify the parallel part: The \( 9\mu\text{F} \) and \( 3\mu\text{F} \) capacitors are in parallel.
\( C_{parallel} = 9 + 3 = 12\mu\text{F} \).
2. This \( 12\mu\text{F} \) combination is in series with the \( 4\mu\text{F} \) capacitor and the 12V battery.
3. Equivalent capacitance of the circuit (\( C_{eq} \)):
\[ C_{eq} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3\mu\text{F} \] 4. Total charge supplied by the battery (\( Q \)):
\[ Q = C_{eq} \times V = 3\mu\text{F} \times 12\text{V} = 36\mu\text{C} \] 5. Since the \( 4\mu\text{F} \) capacitor is in series with the main circuit, it carries the total charge \( Q = 36\mu\text{C} \).
6. Potential difference across the \( 4\mu\text{F} \) capacitor (\( V_{4\mu\text{F}} \)):
\[ V_{4\mu\text{F}} = \frac{Q}{C_1} = \frac{36\mu\text{C}}{4\mu\text{F}} = 9\text{V} \] Step 4: Final Answer:
The potential difference across the 4\(\mu\)F capacitor is 9 V.