Two capacitrors of `2 muF and 4 muF` are connected in parallel. A third capacitor of `6 muF` is connected in series. The combaination is connected across a 12 V battery. The voltage across `2 mu F` capacitor is

Options

A. 2 V
B. 8 V
C. 6 V
D. 1 V

Explanation

To solve the problem of finding the voltage across the 2 µF capacitor when two capacitors (2 µF and 4 µF) are connected in parallel and then in series with a 6 µF capacitor across a 12 V battery, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have two capacitors, \( C_1 = 2 \, \mu F \) and \( C_2 = 4 \, \mu F \), connected in parallel.
- A third capacitor, \( C_3 = 6 \, \mu F \), is connected in series with the parallel combination of \( C_1 \) and \( C_2 \).

2. **Calculate the Equivalent Capacitance of the Parallel Combination**:
- The equivalent capacitance \( C_{eq} \) for capacitors in parallel is given by:
\[ C_{eq} = C_1 + C_2 = 2 \, \mu F + 4 \, \mu F = 6 \, \mu F \]

3. **Combine the Equivalent Capacitance with the Series Capacitor**:
- Now, we have \( C_{eq} = 6 \, \mu F \) in series with \( C_3 = 6 \, \mu F \).
- The formula for the equivalent capacitance \( C_{total} \) of capacitors in series is:
\[ \frac{1}{C_{total}} = \frac{1}{C_{eq}} + \frac{1}{C_3} \]
- Substituting the values:
\[ \frac{1}{C_{total}} = \frac{1}{6 \, \mu F} + \frac{1}{6 \, \mu F} = \frac{2}{6 \, \mu F} = \frac{1}{3 \, \mu F} \]
- Therefore, the total capacitance is:
\[ C_{total} = 3 \, \mu F \]

4. **Calculate the Total Charge in the Circuit**:
- The total charge \( Q \) stored in the capacitors when connected to a voltage \( V = 12 \, V \) is given by:
\[ Q = C_{total} \times V = 3 \, \mu F \times 12 \, V = 36 \, \mu C \]

5. **Determine the Voltage Across the Series Capacitors**:
- The voltage across the series combination of the two capacitors \( C_{eq} \) and \( C_3 \) is equal to the total voltage:
\[ V_{total} = V_{C_{eq}} + V_{C_3} \]
- Since \( C_{eq} \) and \( C_3 \) have the same capacitance, the voltage will split equally:
\[ V_{C_{eq}} = V_{C_3} = \frac{V_{total}}{2} = \frac{12 \, V}{2} = 6 \, V \]

6. **Voltage Across the Parallel Capacitors**:
- The voltage across the parallel capacitors \( C_1 \) and \( C_2 \) (which is \( V_{C_{eq}} \)) is the same as the voltage across each capacitor in that parallel combination:
\[ V_{C_1} = V_{C_2} = V_{C_{eq}} = 6 \, V \]

### Final Answer:
The voltage across the 2 µF capacitor is **6 V**.