Three capacitors each of `3muF`, are provided. These cannot be combined to provide the resultant capacitance of

Options

A. `1muF`
B. `2muF`
C. `4.5muF`
D. `6muF`

Answer

To solve the problem of determining which resultant capacitance cannot be achieved using three capacitors of 3 µF each, we will analyze the possible combinations of these capacitors in series and parallel.

### Step-by-Step Solution:

1. **Understanding Capacitance Combinations**:
- Capacitors can be combined in two main ways: series and parallel.
- In series, the total capacitance \( C \) is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
- In parallel, the total capacitance \( C \) is given by:
\[ C = C_1 + C_2 + C_3 \]

2. **Calculating Series Combination**:
- For three capacitors each of \( 3 \, \mu F \):
\[ \frac{1}{C} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \]
\[ C = 1 \, \mu F \]

3. **Calculating Parallel Combination**:
- For the same three capacitors in parallel:
\[ C = 3 + 3 + 3 = 9 \, \mu F \]

4. **Mixed Combinations**:
- **Two in Parallel and One in Series**:
- Combine two capacitors in parallel:
\[ C_{p} = 3 + 3 = 6 \, \mu F \]
- Then, combine this with the third capacitor in series:
\[ \frac{1}{C} = \frac{1}{6} + \frac{1}{3} \]
- Finding a common denominator (6):
\[ \frac{1}{C} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \]
\[ C = 2 \, \mu F \]

- **Two in Series and One in Parallel**:
- Combine two capacitors in series:
\[ \frac{1}{C_{s}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \implies C_{s} = 1.5 \, \mu F \]
- Then, combine this with the third capacitor in parallel:
\[ C_{total} = 1.5 + 3 = 4.5 \, \mu F \]

5. **Summary of Possible Resultant Capacitances**:
- From the calculations, we have the following possible resultant capacitances:
- \( 1 \, \mu F \) (all in series)
- \( 2 \, \mu F \) (two in parallel, one in series)
- \( 4.5 \, \mu F \) (two in series, one in parallel)
- \( 9 \, \mu F \) (all in parallel)

6. **Identifying the Impossible Resultant Capacitance**:
- The question asks for the resultant capacitance that cannot be achieved. The possible capacitances are \( 1 \, \mu F, 2 \, \mu F, 4.5 \, \mu F, \) and \( 9 \, \mu F \).
- The capacitance \( 6 \, \mu F \) cannot be achieved with any combination of the three capacitors.

### Final Answer:
The resultant capacitance that cannot be provided by combining three capacitors of \( 3 \, \mu F \) each is **6 µF**.