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. (a)
B. (b)
C. (c)
D. (d)

Explanation

To solve the problem of finding the correct combination of capacitors to achieve an effective capacitance of \( \frac{10}{11} \mu F \) using seven capacitors each of capacitance \( 2 \mu F \), we will analyze the configurations provided in the options.

### Step-by-Step Solution:

1. **Identify the Configuration**:
We need to analyze the first configuration presented. It consists of several capacitors connected in parallel and then in series.

2. **Calculate the Capacitance of the Parallel Section**:
In the first configuration, there are five capacitors connected in parallel. The formula for the total capacitance \( C_p \) of capacitors in parallel is:
\[ C_p = C_1 + C_2 + C_3 + C_4 + C_5 \]
Since each capacitor has a capacitance of \( 2 \mu F \):
\[ C_p = 2 + 2 + 2 + 2 + 2 = 10 \mu F \]

3. **Calculate the Capacitance of the Series Section**:
The total capacitance of the parallel section \( C_p \) is then connected in series with two other capacitors, each of \( 2 \mu F \). The formula for the total capacitance \( C_s \) of capacitors in series is:
\[ \frac{1}{C_s} = \frac{1}{C_p} + \frac{1}{C_6} + \frac{1}{C_7} \]
Substituting the values:
\[ \frac{1}{C_s} = \frac{1}{10} + \frac{1}{2} + \frac{1}{2} \]
To simplify, we can convert \( \frac{1}{2} \) to have a common denominator of 10:
\[ \frac{1}{C_s} = \frac{1}{10} + \frac{5}{10} + \frac{5}{10} = \frac{1 + 5 + 5}{10} = \frac{11}{10} \]

4. **Find the Effective Capacitance**:
Now, we can find \( C_s \):
\[ C_s = \frac{10}{11} \mu F \]

5. **Conclusion**:
The configuration analyzed indeed gives us the desired effective capacitance of \( \frac{10}{11} \mu F \). Therefore, the first configuration is the correct option.