Q 12-02-196JEE MainJEE Main 2017 (2 Apr)Medium
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:
Answer: (A) $32$
Each capacitor can take at most $300$ V, so a row must have at least $n$ in series with $300n \ge 1000$, i.e. $n = 4$. A row of four has capacitance $\frac14\ \mu$F.
To get $2\ \mu$F, put $m$ such rows in parallel:
$$m\times\frac14 = 2 \;\Rightarrow\; m = 8$$
Total number $= 4\times8 = 32$.
Solution by Sreeraj P, M.Sc Physics