Three capacitances, each of `3 mu F` , are provided. These cannot be combined to provide the resultant capacitance of:

Options

A. `1 mu F `
B. `2mu F`
C. `6mu F`
D. `4.5 mu F`

Answer

To solve the problem, we need to analyze the combinations of three capacitors, each with a capacitance of \(3 \, \mu F\), and determine the resultant capacitance for different configurations. The goal is to find out which of the given options cannot be achieved as a resultant capacitance.

### Step-by-Step Solution:

1. **Identify the Given Capacitors:**
We have three capacitors:
- \(C_1 = 3 \, \mu F\)
- \(C_2 = 3 \, \mu F\)
- \(C_3 = 3 \, \mu F\)

2. **Calculate Resultant Capacitance in Series:**
When capacitors are connected in series, the formula for the resultant capacitance \(C\) is:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
Substituting the values:
\[ \frac{1}{C} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \]
Therefore,
\[ C = 1 \, \mu F \]
This means a resultant capacitance of \(1 \, \mu F\) is possible.

3. **Calculate Resultant Capacitance with Two in Parallel and One in Series:**
Let's take \(C_1\) and \(C_2\) in parallel and \(C_3\) in series:
- The equivalent capacitance of \(C_1\) and \(C_2\) in parallel is:
\[ C' = C_1 + C_2 = 3 + 3 = 6 \, \mu F \]
- Now, \(C'\) is in series with \(C_3\):
\[ \frac{1}{C} = \frac{1}{C'} + \frac{1}{C_3} = \frac{1}{6} + \frac{1}{3} \]
\[ \frac{1}{C} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]
Thus,
\[ C = 2 \, \mu F \]
This means a resultant capacitance of \(2 \, \mu F\) is also possible.

4. **Calculate Resultant Capacitance with Two in Series and One in Parallel:**
Now, let’s take \(C_1\) and \(C_2\) in series and \(C_3\) in parallel:
- The equivalent capacitance of \(C_1\) and \(C_2\) in series is:
\[ \frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]
Thus,
\[ C' = \frac{3}{2} \, \mu F \]
- Now, \(C'\) is in parallel with \(C_3\):
\[ C = C' + C_3 = \frac{3}{2} + 3 = \frac{3}{2} + \frac{6}{2} = \frac{9}{2} = 4.5 \, \mu F \]
This means a resultant capacitance of \(4.5 \, \mu F\) is also possible.

5. **Conclusion:**
The possible resultant capacitances from the combinations are:
- \(1 \, \mu F\)
- \(2 \, \mu F\)
- \(4.5 \, \mu F\)

The only capacitance that cannot be achieved with the given capacitors is:
- **\(6 \, \mu F\)**

### Final Answer:
The capacitance that cannot be combined to provide the resultant capacitance is **6 μF**.