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

Capacitor PYQ — JEE Main 2017 · Shift 0 · Q54 with Solution

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Correct answer: C

We need an effective capacitance of $C_{\text{eq}} = 2\,\mu\text{F}$ across a potential difference of $V = 1000\,\text{V}.$

Each available capacitor has:

So a single capacitor cannot be connected directly across $1000\,\text{V}$ .

If $n$ identical capacitors are connected in series, the voltage divides equally. So each capacitor gets $\frac{1000}{n}\,\text{V}.$

For safety, $\frac{1000}{n} \le 300$

$n \ge \frac{1000}{300} = 3.33$

Hence the minimum integer value is $n=4.$

So, at least 4 capacitors must be connected in series in each branch.

For 4 identical capacitors of $1\,\mu\text{F}$ in series, $C_{\text{branch}} = \frac{1\,\mu\text{F}}{4} = 0.25\,\mu\text{F}.$

If $m$ such branches are connected in parallel, total capacitance is $C_{\text{eq}} = m \times 0.25\,\mu\text{F}.$

We need $m \times 0.25 = 2$

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

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

$N = 8 \times 4 = 32.$

Thus, the minimum number of capacitors required is $\boxed{32}.$

So the correct option is C.