The ratio of the resultant capacities when three capacitors of `2muF, 4muF and 6muF` are connected first in series and then in parallel is

Options

A. `1 : 11`
B. `11 : 1`
C. `12 : 1`
D. `1 : 12`

Explanation

To solve the problem of finding the ratio of the resultant capacities when three capacitors of \(2 \mu F\), \(4 \mu F\), and \(6 \mu F\) are connected first in series and then in parallel, we will follow these steps:

### Step 1: Calculate the equivalent capacitance in series

When capacitors are connected in series, the formula for the equivalent capacitance \(C_{eq}\) is given by:

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

Substituting the values of the capacitors:

\[ \frac{1}{C_{eq}} = \frac{1}{2 \mu F} + \frac{1}{4 \mu F} + \frac{1}{6 \mu F} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 2, 4, and 6 is 12.

\[ \frac{1}{C_{eq}} = \frac{6}{12} + \frac{3}{12} + \frac{2}{12} = \frac{11}{12} \]

Now, taking the reciprocal to find \(C_{eq}\):

\[ C_{eq} = \frac{12}{11} \mu F \]

### Step 2: Calculate the equivalent capacitance in parallel

When capacitors are connected in parallel, the equivalent capacitance \(C_{eq}\) is simply the sum of the individual capacitances:

\[ C_{eq} = C_1 + C_2 + C_3 \]

Substituting the values:

\[ C_{eq} = 2 \mu F + 4 \mu F + 6 \mu F = 12 \mu F \]

### Step 3: Calculate the ratio of the two equivalent capacitances

Now we can find the ratio of the equivalent capacitance in series to that in parallel:

\[ \text{Ratio} = \frac{C_{eq, \text{series}}}{C_{eq, \text{parallel}}} = \frac{\frac{12}{11} \mu F}{12 \mu F} \]

Simplifying this ratio:

\[ \text{Ratio} = \frac{12}{11} \cdot \frac{1}{12} = \frac{1}{11} \]

### Final Answer

The ratio of the resultant capacities when the capacitors are connected in series and then in parallel is:

\[ \boxed{\frac{1}{11}} \]

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