Five capacitors each of capacitance 1 `muF`, are connected in series. The ratio of capacitance `C_p` to `C_s`, where `C_p` is parallel combination of capactiors and `C_s` is equivalent capacitance of series combination, is:

Options

A. `5:1`
B. `1:5`
C. `50:1`
D. `25:1`

Explanation

To solve the problem, we need to find the ratio of the capacitance of capacitors connected in parallel (C_p) to that of capacitors connected in series (C_s).

### Step-by-Step Solution:

1. **Identify the Capacitance of Each Capacitor:**
Each capacitor has a capacitance of \(1 \mu F\).

2. **Calculate the Equivalent Capacitance for Capacitors in Parallel (C_p):**
For capacitors in parallel, the total capacitance is the sum of the individual capacitances:
\[ C_p = C_1 + C_2 + C_3 + C_4 + C_5 \]
Since all capacitors are \(1 \mu F\):
\[ C_p = 1 \mu F + 1 \mu F + 1 \mu F + 1 \mu F + 1 \mu F = 5 \mu F \]

3. **Calculate the Equivalent Capacitance for Capacitors in Series (C_s):**
For capacitors in series, the total capacitance is given by:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4} + \frac{1}{C_5} \]
Again, since all capacitors are \(1 \mu F\):
\[ \frac{1}{C_s} = \frac{1}{1 \mu F} + \frac{1}{1 \mu F} + \frac{1}{1 \mu F} + \frac{1}{1 \mu F} + \frac{1}{1 \mu F} = 5 \times \frac{1}{1 \mu F} = \frac{5}{1 \mu F} \]
Therefore:
\[ C_s = \frac{1}{5} \mu F \]

4. **Calculate the Ratio \( \frac{C_p}{C_s} \):**
Now we can find the ratio of the capacitances:
\[ \frac{C_p}{C_s} = \frac{5 \mu F}{\frac{1}{5} \mu F} \]
Simplifying this gives:
\[ \frac{C_p}{C_s} = 5 \mu F \times \frac{5}{1 \mu F} = 25 \]

5. **Final Result:**
The ratio \( C_p : C_s \) is:
\[ C_p : C_s = 25 : 1 \]

### Conclusion:
The ratio of the capacitance of the parallel combination to the capacitance of the series combination is \( 25 : 1 \).