A capacitance of `2 muF` is required in an electrical circuit across a potential difference of `1.0 kV` A large number of `1 muF` capacitors are available which can withstand a potential difference of not more than `300 v`.
The minimum number of capacitors required to achieve this is

Options

A. 24
B. 32
C. 2
D. 16

Explanation

To solve the problem of determining the minimum number of 1 µF capacitors required to achieve a capacitance of 2 µF across a potential difference of 1 kV, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Configuration Needed:**
- Each 1 µF capacitor can withstand a maximum voltage of 300 V. To achieve a total voltage of 1 kV (1000 V), we need to connect capacitors in series to ensure that the voltage across each capacitor does not exceed 300 V.

2. **Calculate the Number of Capacitors in Series:**
- The total voltage across capacitors in series is the sum of the voltages across each capacitor. Therefore, if we denote the number of capacitors in series as \( n \):
\[ n \times 300 \, \text{V} \geq 1000 \, \text{V} \]
- Solving for \( n \):
\[ n \geq \frac{1000 \, \text{V}}{300 \, \text{V}} \approx 3.33 \]
- Since \( n \) must be a whole number, we round up to the nearest integer:
\[ n = 4 \]
- This means we need 4 capacitors in series to safely handle the 1 kV.

3. **Calculate the Equivalent Capacitance of Capacitors in Series:**
- The formula for the equivalent capacitance \( C_s \) of \( n \) capacitors in series is given by:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \]
- For \( n = 4 \) identical capacitors (each of 1 µF):
\[ \frac{1}{C_s} = \frac{1}{1 \, \mu F} + \frac{1}{1 \, \mu F} + \frac{1}{1 \, \mu F} + \frac{1}{1 \, \mu F} = \frac{4}{1 \, \mu F} \]
- Therefore, the equivalent capacitance \( C_s \) is:
\[ C_s = \frac{1}{4} \, \mu F = 0.25 \, \mu F \]

4. **Determine the Number of Rows Needed:**
- We need a total capacitance of 2 µF. If we denote the number of rows as \( m \), the total capacitance \( C_{eq} \) of \( m \) rows in parallel (each row having an equivalent capacitance of 0.25 µF) is:
\[ C_{eq} = m \times C_s = m \times 0.25 \, \mu F \]
- Setting this equal to the required capacitance:
\[ m \times 0.25 \, \mu F = 2 \, \mu F \]
- Solving for \( m \):
\[ m = \frac{2 \, \mu F}{0.25 \, \mu F} = 8 \]

5. **Calculate the Total Number of Capacitors:**
- Since each row consists of 4 capacitors and we have 8 rows:
\[ \text{Total Capacitors} = m \times n = 8 \times 4 = 32 \]

### Final Answer:
The minimum number of 1 µF capacitors required is **32**.