Seven capacitors each of capacitance `2muF` are to be connected in a configuration to obtain an effective capacitance of `(10/11)muF`. Which of the combination (s) shown in figure will achieve the desired result?

Options

A. `2` in parallel, `5` in series
B. `3` in parallel, `4` in series.
C. `4` in parallel, `3` in series
D. `5` in parallel, `2` in series

Explanation

To solve the problem of determining which configuration of capacitors will yield an effective capacitance of \( \frac{10}{11} \mu F \), we will analyze each option step by step.

### Given:
- Seven capacitors, each with a capacitance of \( 2 \mu F \).
- Desired effective capacitance: \( \frac{10}{11} \mu F \).

### Step 1: Analyze Each Configuration

#### Option 1: 2 in Parallel and 5 in Series
1. **Calculate the equivalent capacitance of the two capacitors in parallel:**
\[ C_{eq, parallel} = C_1 + C_2 = 2 \mu F + 2 \mu F = 4 \mu F \]

2. **Calculate the equivalent capacitance of the five capacitors in series:**
\[ C_{eq, series} = \frac{C}{n} = \frac{2 \mu F}{5} = 0.4 \mu F \]

3. **Combine the two results (parallel and series):**
\[ C_{eq, total} = \frac{C_{eq, parallel} \cdot C_{eq, series}}{C_{eq, parallel} + C_{eq, series}} = \frac{4 \mu F \cdot 0.4 \mu F}{4 \mu F + 0.4 \mu F} = \frac{1.6}{4.4} \mu F = \frac{8}{22} \mu F = \frac{4}{11} \mu F \]
- This is not equal to \( \frac{10}{11} \mu F \).

#### Option 2: 3 in Parallel and 4 in Series
1. **Calculate the equivalent capacitance of the three capacitors in parallel:**
\[ C_{eq, parallel} = 2 \mu F + 2 \mu F + 2 \mu F = 6 \mu F \]

2. **Calculate the equivalent capacitance of the four capacitors in series:**
\[ C_{eq, series} = \frac{C}{n} = \frac{2 \mu F}{4} = 0.5 \mu F \]

3. **Combine the two results:**
\[ C_{eq, total} = \frac{C_{eq, parallel} \cdot C_{eq, series}}{C_{eq, parallel} + C_{eq, series}} = \frac{6 \mu F \cdot 0.5 \mu F}{6 \mu F + 0.5 \mu F} = \frac{3}{6.5} \mu F = \frac{6}{13} \mu F \]
- This is not equal to \( \frac{10}{11} \mu F \).

#### Option 3: 4 in Parallel and 3 in Series
1. **Calculate the equivalent capacitance of the four capacitors in parallel:**
\[ C_{eq, parallel} = 2 \mu F + 2 \mu F + 2 \mu F + 2 \mu F = 8 \mu F \]

2. **Calculate the equivalent capacitance of the three capacitors in series:**
\[ C_{eq, series} = \frac{C}{n} = \frac{2 \mu F}{3} \approx 0.6667 \mu F \]

3. **Combine the two results:**
\[ C_{eq, total} = \frac{C_{eq, parallel} \cdot C_{eq, series}}{C_{eq, parallel} + C_{eq, series}} = \frac{8 \mu F \cdot 0.6667 \mu F}{8 \mu F + 0.6667 \mu F} \approx \frac{5.3336}{8.6667} \mu F \approx 0.615 \mu F \]
- This is not equal to \( \frac{10}{11} \mu F \).

#### Option 4: 5 in Parallel and 2 in Series
1. **Calculate the equivalent capacitance of the five capacitors in parallel:**
\[ C_{eq, parallel} = 2 \mu F + 2 \mu F + 2 \mu F + 2 \mu F + 2 \mu F = 10 \mu F \]

2. **Calculate the equivalent capacitance of the two capacitors in series:**
\[ C_{eq, series} = \frac{C}{n} = \frac{2 \mu F}{2} = 1 \mu F \]

3. **Combine the two results:**
\[ C_{eq, total} = \frac{C_{eq, parallel} \cdot C_{eq, series}}{C_{eq, parallel} + C_{eq, series}} = \frac{10 \mu F \cdot 1 \mu F}{10 \mu F + 1 \mu F} = \frac{10}{11} \mu F \]
- This matches the desired effective capacitance.

### Conclusion
The correct configuration to achieve an effective capacitance of \( \frac{10}{11} \mu F \) is **Option 4**: 5 capacitors in parallel and 2 capacitors in series.