Three capacitors of capacitance `1 muF, 2muF` and `3 muF` are connected in series and a potential difference of `11 V` is applied across the combination. Then, the potential difference across the plates of `1 muF` capacitor is

Options

A. `2 V`
B. `4 V`
C. `1 V`
D. `6 V`

Explanation

To solve the problem of finding the potential difference across the 1 μF capacitor in a series circuit of three capacitors (1 μF, 2 μF, and 3 μF) connected across a total potential difference of 11 V, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Series Connection**:
- In a series connection of capacitors, the total voltage across the combination is equal to the sum of the voltages across each capacitor.
- Mathematically, this can be expressed as:
\[ V_{total} = V_1 + V_2 + V_3 \]
where \( V_1 \), \( V_2 \), and \( V_3 \) are the voltages across the 1 μF, 2 μF, and 3 μF capacitors respectively.

2. **Use the Relationship of Charge**:
- In a series circuit, the charge (\( Q \)) on each capacitor is the same. The relationship between charge, capacitance, and voltage is given by:
\[ Q = C \cdot V \]
Therefore, we can write:
\[ Q = C_1 \cdot V_1 = C_2 \cdot V_2 = C_3 \cdot V_3 \]
For our capacitors:
- \( C_1 = 1 \, \mu F \)
- \( C_2 = 2 \, \mu F \)
- \( C_3 = 3 \, \mu F \)

3. **Express Voltages in Terms of Charge**:
- From the charge equations, we can express \( V_1 \), \( V_2 \), and \( V_3 \) in terms of \( Q \):
\[ V_1 = \frac{Q}{C_1} = \frac{Q}{1} = Q \]
\[ V_2 = \frac{Q}{C_2} = \frac{Q}{2} \]
\[ V_3 = \frac{Q}{C_3} = \frac{Q}{3} \]

4. **Substitute into the Voltage Equation**:
- Substitute \( V_1 \), \( V_2 \), and \( V_3 \) into the total voltage equation:
\[ V_{total} = V_1 + V_2 + V_3 = Q + \frac{Q}{2} + \frac{Q}{3} \]

5. **Find a Common Denominator**:
- The common denominator for 1, 2, and 3 is 6. Thus, we can rewrite the equation:
\[ V_{total} = Q + \frac{3Q}{6} + \frac{2Q}{6} = Q + \frac{5Q}{6} = \frac{11Q}{6} \]

6. **Set Equal to Total Voltage**:
- We know from the problem that \( V_{total} = 11 \, V \):
\[ \frac{11Q}{6} = 11 \]

7. **Solve for Q**:
- Multiply both sides by 6:
\[ 11Q = 66 \]
- Divide by 11:
\[ Q = 6 \, \mu C \]

8. **Calculate Voltage Across 1 μF Capacitor**:
- Now, substitute \( Q \) back to find \( V_1 \):
\[ V_1 = \frac{Q}{C_1} = \frac{6 \, \mu C}{1 \, \mu F} = 6 \, V \]

### Final Answer:
The potential difference across the plates of the 1 μF capacitor is **6 V**.