Three capacitors of capacitances `6mu F `each are available.The minimum and maximam capacitances,which be obtained are

Options

A. `6 mu F,18mu F`
B. `3 mu F,12 mu F`
C. `2mu F,12mu F`
D. `2mu F,18mu F`

Answer

To solve the problem of finding the minimum and maximum capacitance that can be obtained using three capacitors of capacitance \(6 \, \mu F\) each, we can follow these steps:

### Step 1: Understanding Series and Parallel Connections
- **Series Connection**: When capacitors are connected in series, the total capacitance \(C\) is given by the formula:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
- **Parallel Connection**: When capacitors are connected in parallel, the total capacitance \(C\) is simply the sum of the individual capacitances:
\[ C = C_1 + C_2 + C_3 \]

### Step 2: Calculate Minimum Capacitance
- For minimum capacitance, we connect the capacitors in series.
- Each capacitor has a capacitance of \(6 \, \mu F\). Therefore:
\[ \frac{1}{C} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} \]
- This simplifies to:
\[ \frac{1}{C} = \frac{3}{6} = \frac{1}{2} \]
- Taking the reciprocal gives:
\[ C = 2 \, \mu F \]

### Step 3: Calculate Maximum Capacitance
- For maximum capacitance, we connect the capacitors in parallel.
- Therefore:
\[ C = 6 + 6 + 6 = 18 \, \mu F \]

### Step 4: Conclusion
- The minimum capacitance that can be obtained is \(2 \, \mu F\).
- The maximum capacitance that can be obtained is \(18 \, \mu F\).

### Final Answer
- Minimum Capacitance: \(2 \, \mu F\)
- Maximum Capacitance: \(18 \, \mu F\)

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