Two capacitor each of capacity `4muF` are connected in series and third capacitor of capacity `4muF` is connected in parallel with the combination. The equivalent capacitance of the arrangement is

Options

A. `12muF`
B. `8muF`
C. `6muF`
D. `2.65muF`

Explanation

To find the equivalent capacitance of the given arrangement of capacitors, we can follow these steps:

### Step 1: Identify the Capacitors
We have two capacitors, \( C_1 \) and \( C_2 \), each with a capacitance of \( 4 \mu F \), connected in series. A third capacitor, \( C_3 \), also with a capacitance of \( 4 \mu F \), is connected in parallel with the series combination of \( C_1 \) and \( C_2 \).

### Step 2: Calculate the Equivalent Capacitance of \( C_1 \) and \( C_2 \) in Series
The formula for the equivalent capacitance \( C_{eq} \) of two capacitors in series is given by:

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

Substituting the values:

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

Calculating this gives:

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

Taking the reciprocal:

\[ C_{eq} = 2 \mu F \]

### Step 3: Combine the Result with \( C_3 \) in Parallel
Now, we need to find the equivalent capacitance of the series combination \( C_{eq} \) and the parallel capacitor \( C_3 \). The formula for capacitance in parallel is:

\[ C_{total} = C_{eq} + C_3 \]

Substituting the values:

\[ C_{total} = 2 \mu F + 4 \mu F \]

Calculating this gives:

\[ C_{total} = 6 \mu F \]

### Final Answer
The equivalent capacitance of the arrangement is \( 6 \mu F \).

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