A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 μF 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:

We are given many identical capacitors, each of capacitance $$C_0 = 1\;\mu\text{F}$$ and each safe only up to a potential difference of $$V_0 = 300\;\text{V}$$. We have to build an equivalent capacitance of $$C_{\text{eq}} = 2\;\mu\text{F}$$ that can safely withstand a total potential difference of $$V = 1000\;\text{V}$$. We may connect the available capacitors in any series-parallel combination.

First we make sure that the voltage rating is satisfied. Suppose we connect $$x$$ of the $$1\;\mu\text{F}$$ capacitors in series. A well-known result for identical capacitors in series is

$$C_{\text{series}} = \frac{C_0}{x},$$

and, because the capacitors are identical, the applied voltage divides equally, so each capacitor experiences

$$V_{\text{each}} = \frac{V}{x}.$$

To avoid exceeding the individual rating we must have

$$V_{\text{each}} \le V_0 \;\;\Longrightarrow\;\; \frac{V}{x} \le V_0.$$

Substituting the numbers,

$$\frac{1000}{x} \le 300.$$

Solving for $$x$$ we get

$$x \ge \frac{1000}{300} = 3.\overline{3}.$$

Because $$x$$ must be an integer, the smallest permissible value is

$$x = 4.$$

So each series string will contain $$4$$ capacitors. For this string the capacitance is

$$C_{\text{series}} = \frac{C_0}{4} = \frac{1\;\mu\text{F}}{4} = 0.25\;\mu\text{F}.$$

We now place several such identical strings in parallel to reach the required $$2\;\mu\text{F}$$. For capacitors in parallel the capacitances add directly, so for $$m$$ strings we have

$$C_{\text{eq}} = m \times C_{\text{series}}.$$

Requiring this to equal $$2\;\mu\text{F}$$ gives

$$m \times 0.25\;\mu\text{F} = 2\;\mu\text{F}$$

$$\Longrightarrow\;\; m = \frac{2}{0.25} = 8.$$

Thus we need $$8$$ parallel branches, each branch containing $$4$$ capacitors in series.

The total number $$N$$ of individual capacitors is therefore

$$N = m \times x = 8 \times 4 = 32.$$

Hence, the correct answer is Option A.