Three capacitors of capacitance `3 muF` are connected in a circuit. Then their maximum and minimum capacitances will be

Options

A. `9 muF, 1muF`
B. `8 muF, 2muF`
C. `9 muF, 0muF`
D. `3 muF, 2muF`

Explanation

To solve the problem of finding the maximum and minimum capacitance of three capacitors, each with a capacitance of \(3 \, \mu F\), we will analyze both the parallel and series configurations.

### Step-by-Step Solution:

1. **Understanding Capacitor Configurations**:
- Capacitors can be connected in two main configurations: **parallel** and **series**.
- In a **parallel** connection, the total capacitance is the sum of the individual capacitances.
- In a **series** connection, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances.

2. **Calculating Maximum Capacitance**:
- For maximum capacitance, we connect the capacitors in **parallel**.
- The formula for total capacitance \(C_{max}\) in parallel is:
\[ C_{max} = C_1 + C_2 + C_3 \]
- Given that each capacitor has a capacitance of \(3 \, \mu F\):
\[ C_{max} = 3 \, \mu F + 3 \, \mu F + 3 \, \mu F = 9 \, \mu F \]

3. **Calculating Minimum Capacitance**:
- For minimum capacitance, we connect the capacitors in **series**.
- The formula for total capacitance \(C_{min}\) in series is:
\[ \frac{1}{C_{min}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
- Since all capacitors are the same, we can simplify this to:
\[ \frac{1}{C_{min}} = \frac{3}{C} \quad \text{where } C = 3 \, \mu F \]
- Therefore:
\[ \frac{1}{C_{min}} = \frac{3}{3 \, \mu F} = 1 \, \mu F \]
- Inverting this gives:
\[ C_{min} = 1 \, \mu F \]

4. **Final Results**:
- The maximum capacitance when connected in parallel is \(9 \, \mu F\).
- The minimum capacitance when connected in series is \(1 \, \mu F\).

### Summary:
- Maximum Capacitance: \(9 \, \mu F\)
- Minimum Capacitance: \(1 \, \mu F\)

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