An electrical technician requires a capacitance of $2 \; \mu F$ in a circuit across a potential difference of $1 \; kV$. $A$ large number of $1 \; \mu F$ capacitors are available,each of which can withstand a potential difference of not more than $400 \; V$. Suggest a possible arrangement that requires the minimum number of capacitors.

(D) Total required capacitance,$C = 2 \; \mu F$.
Potential difference,$V = 1 \; kV = 1000 \; V$.
Capacitance of each capacitor,$C_1 = 1 \; \mu F$.
Maximum potential difference each capacitor can withstand,$v_1 = 400 \; V$.
To withstand $1000 \; V$,the number of capacitors in each series row must be $N = \frac{1000}{400} = 2.5$. Since $N$ must be an integer,we take $N = 3$.
The capacitance of one such row of $3$ capacitors in series is $C_{row} = \frac{1}{1+1+1} = \frac{1}{3} \; \mu F$.
Let there be $n$ such rows connected in parallel to achieve the total capacitance $C = 2 \; \mu F$.
The equivalent capacitance is $C_{eq} = n \times C_{row} = n \times \frac{1}{3} = 2 \; \mu F$.
Solving for $n$,we get $n = 6$.
Thus,$6$ rows of $3$ capacitors each are required.
Total number of capacitors $= 6 \times 3 = 18$.