Three capacitor of capacitance `C(mu F)` are connected in parallel to which a capacitor of capacitance C is connected in series. Effective capacitance is 3.75. then capacity off each capacitor is

Options

A. `4 muF`
B. `5 muF`
C. `6 muF`
D. `8 mu F`

Answer

To solve the problem, we need to calculate the capacitance of each capacitor given that three capacitors of capacitance \( C \) are connected in parallel and then connected in series with another capacitor of capacitance \( C \). The effective capacitance of the entire arrangement is given as \( 3.75 \, \mu F \).

### Step-by-step Solution:

1. **Identify the Configuration**:
- We have three capacitors \( C \) in parallel, which we can denote as \( C_1, C_2, C_3 \) where \( C_1 = C_2 = C_3 = C \).
- These three capacitors are connected in parallel, and this combination is then connected in series with another capacitor \( C \).

2. **Calculate the Equivalent Capacitance of the Parallel Combination**:
- The formula for the equivalent capacitance \( C_{parallel} \) of capacitors in parallel is:
\[ C_{parallel} = C_1 + C_2 + C_3 = C + C + C = 3C \]

3. **Connect the Parallel Combination in Series with Another Capacitor**:
- Now, this equivalent capacitance \( 3C \) is in series with another capacitor \( C \).
- The formula for the equivalent capacitance \( C_{equiv} \) of capacitors in series is given by:
\[ \frac{1}{C_{equiv}} = \frac{1}{C_{parallel}} + \frac{1}{C} \]
- Substituting \( C_{parallel} = 3C \):
\[ \frac{1}{C_{equiv}} = \frac{1}{3C} + \frac{1}{C} \]

4. **Combine the Series Capacitors**:
- Finding a common denominator, we can rewrite the equation:
\[ \frac{1}{C_{equiv}} = \frac{1}{3C} + \frac{3}{3C} = \frac{4}{3C} \]
- Therefore, the equivalent capacitance \( C_{equiv} \) becomes:
\[ C_{equiv} = \frac{3C}{4} \]

5. **Set the Equivalent Capacitance Equal to Given Value**:
- We know from the problem statement that the effective capacitance is \( 3.75 \, \mu F \):
\[ \frac{3C}{4} = 3.75 \]

6. **Solve for \( C \)**:
- To find \( C \), we rearrange the equation:
\[ 3C = 3.75 \times 4 \]
\[ 3C = 15 \]
\[ C = \frac{15}{3} = 5 \, \mu F \]

### Conclusion:
The capacitance of each capacitor is \( 5 \, \mu F \).