Three.condensers of capacity `4 mF, 2 mF and 3mF` are connected such that 2 mF and 3 mF are in series and 4 mF is parallel to them. The equivalent capacity of the combination is

Options

A. 9 mF
B. 5.2 mF
C. 2.6 mF
D. 10 mF

Answer

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

### Step 1: Identify the Configuration
We have three capacitors:
- \( C_1 = 4 \, \text{mF} \)
- \( C_2 = 2 \, \text{mF} \)
- \( C_3 = 3 \, \text{mF} \)

The capacitors \( C_2 \) and \( C_3 \) are connected in series, and \( C_1 \) is connected in parallel to the combination of \( C_2 \) and \( C_3 \).

### Step 2: Calculate the Equivalent Capacitance of \( C_2 \) and \( C_3 \) in Series
For capacitors in series, the formula for equivalent capacitance \( C_s \) is given by:

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

Substituting the values:

\[ \frac{1}{C_s} = \frac{1}{2 \, \text{mF}} + \frac{1}{3 \, \text{mF}} \]

Finding a common denominator (which is 6):

\[ \frac{1}{C_s} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

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

\[ C_s = \frac{6}{5} \, \text{mF} = 1.2 \, \text{mF} \]

### Step 3: Calculate the Total Equivalent Capacitance
Now, we need to find the total capacitance \( C_{eq} \) of the circuit, which includes \( C_1 \) in parallel with \( C_s \):

\[ C_{eq} = C_1 + C_s \]

Substituting the values:

\[ C_{eq} = 4 \, \text{mF} + 1.2 \, \text{mF} = 5.2 \, \text{mF} \]

### Final Answer
The equivalent capacitance of the combination is:

\[ C_{eq} = 5.2 \, \text{mF} \]

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