Three capacitors of capacitances `1muF,2muF and 4muF` are connected first in a series combination, and then in parallel combination. The ratio of their equivalent capacitances will be

Options

A. `2:49`
B. `49:2`
C. `4:49`
D. `49:4`

Answer

To find the ratio of the equivalent capacitances of three capacitors connected in series and in parallel, we can follow these steps:

### Step 1: Calculate the equivalent capacitance in series
For capacitors connected in series, the formula for the equivalent capacitance \( C_s \) is given by:

\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Given the capacitances:
- \( C_1 = 1 \mu F \)
- \( C_2 = 2 \mu F \)
- \( C_3 = 4 \mu F \)

Substituting the values into the formula:

\[ \frac{1}{C_s} = \frac{1}{1} + \frac{1}{2} + \frac{1}{4} \]

Finding a common denominator (which is 4):

\[ \frac{1}{C_s} = \frac{4}{4} + \frac{2}{4} + \frac{1}{4} = \frac{4 + 2 + 1}{4} = \frac{7}{4} \]

Now, taking the reciprocal to find \( C_s \):

\[ C_s = \frac{4}{7} \mu F \]

### Step 2: Calculate the equivalent capacitance in parallel
For capacitors connected in parallel, the equivalent capacitance \( C_p \) is simply the sum of the individual capacitances:

\[ C_p = C_1 + C_2 + C_3 \]

Substituting the values:

\[ C_p = 1 + 2 + 4 = 7 \mu F \]

### Step 3: Calculate the ratio of equivalent capacitances
Now, we need to find the ratio of the equivalent capacitance in series to that in parallel:

\[ \text{Ratio} = \frac{C_s}{C_p} = \frac{\frac{4}{7}}{7} \]

This simplifies to:

\[ \text{Ratio} = \frac{4}{7} \times \frac{1}{7} = \frac{4}{49} \]

### Final Answer
Thus, the ratio of the equivalent capacitances of the capacitors in series to that in parallel is:

\[ \frac{4}{49} \]