If n identical capacitors are connected in series and then in parallel then the ratio of effective capacity in parallel and in series combination i.e. `C_(P)/C_(S)` is

Options

A. n
B. `1//n`
C. `n^(2)`
D. `1//n^(2)`

Explanation

To solve the problem of finding the ratio of effective capacitance in parallel and series combinations of n identical capacitors, we can follow these steps:

### Step 1: Understand the formula for capacitors in parallel
When capacitors are connected in parallel, the effective capacitance \( C_P \) is given by the sum of the individual capacitances. For n identical capacitors, each with capacitance \( C \):
\[ C_P = C + C + C + \ldots + C \quad (n \text{ times}) = nC \]

### Step 2: Understand the formula for capacitors in series
When capacitors are connected in series, the effective capacitance \( C_S \) is given by the reciprocal of the sum of the reciprocals of the individual capacitances. For n identical capacitors, each with capacitance \( C \):
\[ \frac{1}{C_S} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \ldots + \frac{1}{C} \quad (n \text{ times}) = \frac{n}{C} \]
Thus, we can rearrange this to find \( C_S \):
\[ C_S = \frac{C}{n} \]

### Step 3: Calculate the ratio \( \frac{C_P}{C_S} \)
Now that we have both \( C_P \) and \( C_S \), we can find the ratio:
\[ \frac{C_P}{C_S} = \frac{nC}{\frac{C}{n}} = \frac{nC \cdot n}{C} = n^2 \]

### Final Result
Thus, the ratio of effective capacitance in parallel to that in series is:
\[ \frac{C_P}{C_S} = n^2 \]