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. `18` component capacitors
B. `36` component capacitors
C. `72` component capacitors
D. `144` component capacitors

Answer

To solve the problem, we need to determine how many capacitors of 1 µF each are required to create a capacitor of 2 µF that can withstand a voltage of 3000 volts without puncturing. Each capacitor can only handle a maximum voltage of 500 volts before it gets damaged.

### Step-by-Step Solution:

1. **Understanding the Voltage Requirement:**
- We have a total voltage of 3000 volts that we want to apply across the arrangement of capacitors.
- Each capacitor can withstand a maximum of 500 volts.

2. **Calculating the Number of Capacitors in Series:**
- Since the voltage across capacitors in series adds up, we can calculate the number of capacitors needed in series to handle 3000 volts.
- The number of capacitors in series (n) can be calculated as:
\[ n = \frac{\text{Total Voltage}}{\text{Voltage per Capacitor}} = \frac{3000 \text{ volts}}{500 \text{ volts}} = 6 \]
- Therefore, we need 6 capacitors in series to safely handle 3000 volts.

3. **Calculating the Equivalent Capacitance of Capacitors in Series:**
- The equivalent capacitance (C_eq) of capacitors in series is given by:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \]
- For n identical capacitors (1 µF each), this simplifies to:
\[ C_{\text{eq}} = \frac{C}{n} = \frac{1 \mu F}{6} = \frac{1}{6} \mu F \]

4. **Arranging Capacitors in Parallel:**
- To achieve a total capacitance of 2 µF, we need to connect several branches of these series capacitors in parallel.
- If we denote the number of branches as \( n_1 \), the total capacitance from these branches in parallel is:
\[ C_{\text{total}} = n_1 \cdot C_{\text{eq}} = n_1 \cdot \frac{1}{6} \mu F \]
- We want this to equal 2 µF:
\[ n_1 \cdot \frac{1}{6} \mu F = 2 \mu F \]
- Solving for \( n_1 \):
\[ n_1 = 2 \mu F \cdot 6 = 12 \]

5. **Calculating the Total Number of Capacitors:**
- Since we need 6 capacitors in series for each branch and we have 12 branches:
\[ \text{Total Capacitors} = n_1 \cdot 6 = 12 \cdot 6 = 72 \]

### Final Answer:
Therefore, the minimum number of capacitors required is **72**.