Five equal capacitors connected in series have a resultant capacity of `5muF`. What is their resultant capacity if they are connected in parallel?

Options

A. `50muF`
B. `75muF`
C. `100muF`
D. `125muF`

Explanation

To solve the problem, we need to find the resultant capacitance when five equal capacitors are connected in parallel, given that their resultant capacitance in series is 5 µF.

### Step-by-step Solution:

1. **Understand the Series Connection Formula**:
When capacitors are connected in series, the formula for the total capacitance \( C_s \) is given by:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots + \frac{1}{C_n} \]
For \( n \) equal capacitors \( C \):
\[ \frac{1}{C_s} = \frac{n}{C} \implies C_s = \frac{C}{n} \]

2. **Calculate the Value of One Capacitor**:
Given that \( C_s = 5 \, \mu F \) and \( n = 5 \):
\[ C_s = \frac{C}{5} \implies C = C_s \times 5 = 5 \, \mu F \times 5 = 25 \, \mu F \]
Thus, the capacitance of each individual capacitor is \( 25 \, \mu F \).

3. **Understand the Parallel Connection Formula**:
When capacitors are connected in parallel, the total capacitance \( C_p \) is simply the sum of the individual capacitances:
\[ C_p = C_1 + C_2 + C_3 + \ldots + C_n = n \times C \]

4. **Calculate the Resultant Capacitance in Parallel**:
Now, substituting \( n = 5 \) and \( C = 25 \, \mu F \):
\[ C_p = 5 \times 25 \, \mu F = 125 \, \mu F \]

5. **Conclusion**:
Therefore, the resultant capacitance when the five capacitors are connected in parallel is:
\[ C_p = 125 \, \mu F \]

### Final Answer:
The resultant capacitance when the five equal capacitors are connected in parallel is **125 µF**.