What is the total capacitance if three capacitors of capacitance 2 pF, 3 pF and 4 pF are connected in parallel?

Options

A. `9 pF`
B. `(12)/(13) pF`
C. `(13)/(12) pF`
D. `(29)/(4) pF`

Explanation

To find the total capacitance of capacitors connected in parallel, we can use the formula:

\[ C_{total} = C_1 + C_2 + C_3 + \ldots \]

where \(C_1\), \(C_2\), and \(C_3\) are the capacitances of the individual capacitors.

### Step-by-Step Solution:

1. **Identify the capacitances**: We have three capacitors with the following capacitances:
- \(C_1 = 2 \, \text{pF}\)
- \(C_2 = 3 \, \text{pF}\)
- \(C_3 = 4 \, \text{pF}\)

2. **Apply the formula for total capacitance in parallel**:
\[ C_{total} = C_1 + C_2 + C_3 \]

3. **Substitute the values**:
\[ C_{total} = 2 \, \text{pF} + 3 \, \text{pF} + 4 \, \text{pF} \]

4. **Perform the addition**:
\[ C_{total} = 2 + 3 + 4 = 9 \, \text{pF} \]

5. **Conclusion**: The total capacitance of the three capacitors connected in parallel is:
\[ C_{total} = 9 \, \text{pF} \]

### Final Answer:
The total capacitance is **9 pF**.

---