A number of capacitors, each of capacitance `1 mu F` and each one of which gets punctured if a potential difference just exceeding 500 volt is applied are provided. Then an arrangement suitable for giving a capacitor of capacitance `3 mu F` across which 2000 V may be applied requires at least

Options

A. 4 component capacitors
B. 96 component capacitors
C. 48 component capacitors
D. 3 component capacitors

Answer

To solve the problem, we need to determine the minimum number of capacitors required to achieve a capacitance of 3 µF that can withstand a voltage of 2000 V, given that each capacitor has a capacitance of 1 µF and can only handle up to 500 V before puncturing.

### Step-by-Step Solution:

1. **Identify the maximum voltage per capacitor**:
Each capacitor can handle a maximum voltage of 500 V.

2. **Determine the total voltage to be applied**:
We need to apply a total voltage of 2000 V.

3. **Calculate the number of capacitors in series**:
To find out how many capacitors we need in series to handle 2000 V:
\[ V_{\text{total}} = V_0 \times m \]
where \( V_0 = 500 \, \text{V} \) (the maximum voltage per capacitor) and \( V_{\text{total}} = 2000 \, \text{V} \).
Rearranging gives:
\[ m = \frac{V_{\text{total}}}{V_0} = \frac{2000 \, \text{V}}{500 \, \text{V}} = 4 \]
So, we need **4 capacitors in series**.

4. **Calculate the equivalent capacitance of capacitors in series**:
The formula for the equivalent capacitance \( C_s \) of capacitors in series is:
\[ \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{4}{C} \]
Given that each capacitor has a capacitance of \( C = 1 \, \mu F \):
\[ \frac{1}{C_s} = \frac{4}{1 \, \mu F} \implies C_s = \frac{1}{4} \, \mu F = 0.25 \, \mu F \]

5. **Determine the total capacitance required**:
We need a total capacitance of \( C_{\text{total}} = 3 \, \mu F \).

6. **Calculate the number of series groups needed**:
Since the equivalent capacitance of one series group is \( 0.25 \, \mu F \), we can find the number of such groups \( n \) needed in parallel to achieve \( 3 \, \mu F \):
\[ C_{\text{total}} = n \times C_s \implies n = \frac{C_{\text{total}}}{C_s} = \frac{3 \, \mu F}{0.25 \, \mu F} = 12 \]
So, we need **12 series groups**.

7. **Calculate the total number of capacitors**:
The total number of capacitors \( N \) required is given by:
\[ N = m \times n = 4 \times 12 = 48 \]

### Final Answer:
The minimum number of capacitors required is **48**.