A number of capacitors each of capacitance `1 muF` and each one of which get punctured if a potential difference just exceeding `500` volt is applied, are provided. Then an arrangement suitable for giving a capacitor of `2 muF` across which `3000` volt may be applied requires at least-

Options

A. 6 component capacitors
B. 12 component capacitors
C. 72 component capacitors
D. 2 component capacitors

Answer

To solve the problem, we need to determine the minimum number of capacitors required to create a capacitance of \(2 \mu F\) that can withstand a voltage of \(3000 V\) without exceeding \(500 V\) across each individual capacitor.

### Step-by-step Solution:

1. **Understanding the Capacitor Configuration**:
- We can arrange the capacitors in series and parallel combinations. Let \(m\) be the number of capacitors in series and \(n\) be the number of such series connected in parallel.

2. **Voltage Distribution in Series**:
- When \(m\) capacitors are connected in series, the total voltage \(V_T\) across them is distributed equally among the capacitors. Therefore, the voltage across each capacitor \(V_0\) is given by:
\[ V_0 = \frac{V_T}{m} \]
- Given that \(V_T = 3000 V\) and \(V_0\) must not exceed \(500 V\), we have:
\[ \frac{3000}{m} \leq 500 \]

3. **Calculating \(m\)**:
- Rearranging the inequality gives:
\[ m \geq \frac{3000}{500} = 6 \]
- Therefore, we need at least \(m = 6\) capacitors in series.

4. **Capacitance of Series Configuration**:
- The equivalent capacitance \(C_1\) of \(m\) capacitors in series is given by:
\[ \frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} + \ldots + \frac{1}{C} = \frac{m}{C} \]
- For \(m = 6\) and \(C = 1 \mu F\):
\[ C_1 = \frac{C}{m} = \frac{1 \mu F}{6} = \frac{1}{6} \mu F \]

5. **Total Capacitance with Parallel Configuration**:
- If we have \(n\) branches of \(C_1\) connected in parallel, the total capacitance \(C_{total}\) is:
\[ C_{total} = n \cdot C_1 = n \cdot \frac{1}{6} \mu F \]
- We want \(C_{total} = 2 \mu F\):
\[ n \cdot \frac{1}{6} = 2 \]

6. **Calculating \(n\)**:
- Rearranging gives:
\[ n = 2 \cdot 6 = 12 \]

7. **Total Number of Capacitors**:
- The total number of capacitors \(N\) required is:
\[ N = m \cdot n = 6 \cdot 12 = 72 \]

### Final Answer:
The minimum number of capacitors required is \(72\).

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