Three capacitors of capacitances `3 muF, 9muF` and `18 muF` are connected one in series and another time in parallel. The ratio of equivalent capacitance in the two cases `((C_(s))/(C_(p)))` will be

Options

A. `1:15`
B. `15:1`
C. `1:1`
D. `1:3`

Answer

To solve the problem, we need to find the equivalent capacitance of the three capacitors when they are connected in series and in parallel, and then calculate the ratio of these two capacitances.

### Step 1: Calculate the equivalent capacitance in series (C_s)

For capacitors in series, the formula for equivalent capacitance \( C_s \) is given by:

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

Where:
- \( C_1 = 3 \mu F \)
- \( C_2 = 9 \mu F \)
- \( C_3 = 18 \mu F \)

Substituting the values:

\[ \frac{1}{C_s} = \frac{1}{3} + \frac{1}{9} + \frac{1}{18} \]

To add these fractions, we need a common denominator. The least common multiple (LCM) of 3, 9, and 18 is 18.

Rewriting each term with the common denominator:

\[ \frac{1}{C_s} = \frac{6}{18} + \frac{2}{18} + \frac{1}{18} \]

Now, adding these fractions:

\[ \frac{1}{C_s} = \frac{6 + 2 + 1}{18} = \frac{9}{18} \]

Thus, we have:

\[ C_s = \frac{18}{9} = 2 \mu F \]

### Step 2: Calculate the equivalent capacitance in parallel (C_p)

For capacitors in parallel, the formula for equivalent capacitance \( C_p \) is given by:

\[ C_p = C_1 + C_2 + C_3 \]

Substituting the values:

\[ C_p = 3 \mu F + 9 \mu F + 18 \mu F \]

Calculating the sum:

\[ C_p = 3 + 9 + 18 = 30 \mu F \]

### Step 3: Calculate the ratio of equivalent capacitance in series to that in parallel

Now, we need to find the ratio \( \frac{C_s}{C_p} \):

\[ \frac{C_s}{C_p} = \frac{2 \mu F}{30 \mu F} = \frac{2}{30} = \frac{1}{15} \]

### Final Answer

The ratio of equivalent capacitance in series to that in parallel is:

\[ \frac{C_s}{C_p} = \frac{1}{15} \]

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