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. `1.83 (Omega -m)^(-1)`
B. `1.68 (Omega-m)^(-1)`
C. `1.20 (Omega-m)^(-1)`
D. `0.59 (W-m)^(-1)`

Answer

To solve the problem, we need to determine the minimum number of capacitors required to achieve a capacitance of 2 μF across a potential difference of 1 kV, using 1 μF capacitors that can withstand a maximum potential difference of 300 V.

### Step-by-Step Solution:

1. **Determine the number of capacitors in series**:
Each capacitor can withstand a maximum voltage of 300 V. To achieve a total voltage of 1 kV (1000 V), we need to find how many capacitors are required in series.
\[ \text{Number of capacitors in series (n)} = \frac{1000 \text{ V}}{300 \text{ V}} = \frac{10}{3} \approx 3.33 \]
Since we cannot have a fraction of a capacitor, we round up to the nearest whole number:
\[ n = 4 \]

2. **Calculate the equivalent capacitance of capacitors in series**:
The formula for the equivalent capacitance \(C_s\) of capacitors in series is given by:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \ldots + \frac{1}{C_n} \]
For \(n\) capacitors of capacitance \(C = 1 \mu F\):
\[ \frac{1}{C_s} = \frac{n}{C} = \frac{4}{1 \mu F} \implies C_s = \frac{1 \mu F}{4} = 0.25 \mu F \]

3. **Determine the number of rows of capacitors in parallel**:
Now, we need to achieve a total capacitance of 2 μF. If we have \(m\) rows of \(C_s = 0.25 \mu F\) in parallel, the total capacitance \(C_{total}\) is given by:
\[ C_{total} = m \cdot C_s = m \cdot 0.25 \mu F \]
Setting this equal to the required capacitance:
\[ m \cdot 0.25 \mu F = 2 \mu F \implies m = \frac{2 \mu F}{0.25 \mu F} = 8 \]

4. **Calculate the total number of capacitors**:
The total number of capacitors used is the product of the number of capacitors in series and the number of rows:
\[ \text{Total capacitors} = n \cdot m = 4 \cdot 8 = 32 \]

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