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

Options

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

Answer

To find the maximum and minimum capacitance of three capacitors, each with a capacitance of \(3 \mu F\), we need to consider two different configurations: parallel and series.

### Step 1: Maximum Capacitance (Parallel Configuration)
When capacitors are connected in parallel, the total or equivalent capacitance (\(C_{max}\)) is simply the sum of the individual capacitances.

For three capacitors in parallel:
\[ C_{max} = C_1 + C_2 + C_3 \]
Given that each capacitor has a capacitance of \(3 \mu F\):
\[ C_{max} = 3 \mu F + 3 \mu F + 3 \mu F = 9 \mu F \]

### Step 2: Minimum Capacitance (Series Configuration)
When capacitors are connected in series, the total or equivalent capacitance (\(C_{min}\)) can be calculated using the formula:
\[ \frac{1}{C_{min}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
Since all capacitors have the same capacitance:
\[ \frac{1}{C_{min}} = \frac{1}{3 \mu F} + \frac{1}{3 \mu F} + \frac{1}{3 \mu F} = \frac{3}{3 \mu F} = \frac{1}{\mu F} \]
Taking the reciprocal gives:
\[ C_{min} = 1 \mu F \]

### Final Answer
- Maximum Capacitance: \(C_{max} = 9 \mu F\)
- Minimum Capacitance: \(C_{min} = 1 \mu F\)

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