Seven capacitors, each a of a cpacitance `2muF`, are to be connected in a configuration to obatin an effective capacitance of `((16)/(13))muF`. Which of the combinations, shown in figure below, will achieve the desired value ?

Options

A.
B.
C.
D.

Explanation

To solve the problem of finding the correct combination of capacitors to achieve an effective capacitance of \(\frac{16}{13} \mu F\) using seven capacitors of \(2 \mu F\) each, we will analyze each of the configurations provided in the question.

### Step-by-Step Solution:

1. **Understanding Capacitor Combinations**:
- Capacitors can be connected in series or parallel.
- For capacitors in parallel, the total capacitance \(C_{total}\) is the sum of the individual capacitances:
\[ C_{total} = C_1 + C_2 + C_3 + \ldots \]
- For capacitors in series, the total capacitance is given by:
\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots \]

2. **Analyzing Configuration 1**:
- Assume there are 4 capacitors in parallel and 3 in series.
- The total capacitance of the 4 capacitors in parallel:
\[ C_{parallel} = 4 \times 2 \mu F = 8 \mu F \]
- The equivalent capacitance of this combination in series with the remaining capacitors:
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \]
- Simplifying this gives:
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{3}{2} = \frac{1}{8} + \frac{12}{8} = \frac{13}{8} \]
- Therefore,
\[ C_{eq} = \frac{8}{13} \mu F \]

3. **Analyzing Configuration 2**:
- Assume there are 6 capacitors in parallel and 1 in series.
- The total capacitance of the 6 capacitors in parallel:
\[ C_{parallel} = 6 \times 2 \mu F = 12 \mu F \]
- The equivalent capacitance of this combination in series with the remaining capacitors:
\[ \frac{1}{C_{eq}} = \frac{1}{12} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \]
- Simplifying this gives:
\[ \frac{1}{C_{eq}} = \frac{1}{12} + \frac{3}{2} = \frac{1}{12} + \frac{18}{12} = \frac{19}{12} \]
- Therefore,
\[ C_{eq} = \frac{12}{19} \mu F \]

4. **Analyzing Configuration 3**:
- Assume there are 4 capacitors in parallel and 3 in series.
- The total capacitance of the 4 capacitors in parallel:
\[ C_{parallel} = 4 \times 2 \mu F = 8 \mu F \]
- The equivalent capacitance of this combination in series with the remaining capacitors:
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \]
- Simplifying this gives:
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{3}{2} = \frac{1}{8} + \frac{12}{8} = \frac{13}{8} \]
- Therefore,
\[ C_{eq} = \frac{8}{13} \mu F \]

5. **Analyzing Configuration 4**:
- Assume there are 5 capacitors in parallel and 2 in series.
- The total capacitance of the 5 capacitors in parallel:
\[ C_{parallel} = 5 \times 2 \mu F = 10 \mu F \]
- The equivalent capacitance of this combination in series with the remaining capacitors:
\[ \frac{1}{C_{eq}} = \frac{1}{10} + \frac{1}{2} + \frac{1}{2} \]
- Simplifying this gives:
\[ \frac{1}{C_{eq}} = \frac{1}{10} + 1 = \frac{1}{10} + \frac{10}{10} = \frac{11}{10} \]
- Therefore,
\[ C_{eq} = \frac{10}{11} \mu F \]

6. **Conclusion**:
- After analyzing all configurations, none of them yield the desired effective capacitance of \(\frac{16}{13} \mu F\).