Potential difference of 11 V is applied across three capacitors of capacitance `1 muF, 2 muF and 3 muF` connected in series. Potential difference across capacitor of `1 muF` is:

Options

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

Explanation

To find the potential difference across the capacitor of \(1 \mu F\) when three capacitors of capacitance \(1 \mu F\), \(2 \mu F\), and \(3 \mu F\) are connected in series with a total potential difference of \(11 V\), we can follow these steps:

### Step 1: Calculate the Equivalent Capacitance
For capacitors in series, the formula for the equivalent capacitance \(C_{eq}\) is given by:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Substituting the values of the capacitances:

\[ \frac{1}{C_{eq}} = \frac{1}{1 \mu F} + \frac{1}{2 \mu F} + \frac{1}{3 \mu F} \]

### Step 2: Simplify the Equation
Calculating each term:

\[ \frac{1}{C_{eq}} = 1 + 0.5 + \frac{1}{3} \]

To add these fractions, we can find a common denominator, which is \(6\):

\[ \frac{1}{C_{eq}} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6} \]

Thus,

\[ C_{eq} = \frac{6}{11} \mu F \]

### Step 3: Calculate the Charge on the Capacitors
The charge \(Q\) on the equivalent capacitance when a potential difference \(V\) is applied is given by:

\[ Q = C_{eq} \cdot V \]

Substituting the values:

\[ Q = \left(\frac{6}{11} \mu F\right) \cdot (11 V) = 6 \mu C \]

### Step 4: Find the Potential Difference Across the \(1 \mu F\) Capacitor
The potential difference \(V_1\) across the \(1 \mu F\) capacitor can be calculated using the formula:

\[ V_1 = \frac{Q}{C_1} \]

Substituting the charge and the capacitance of the \(1 \mu F\) capacitor:

\[ V_1 = \frac{6 \mu C}{1 \mu F} = 6 V \]

### Final Answer
The potential difference across the \(1 \mu F\) capacitor is \(6 V\).

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