An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him 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.

Here we have,
Total required capacitance, C = 2$\mathrm{\mu F}$ 
Potential difference applied across the circuit, V =1 kV= 1000 V
Capacitance of each capacitor, C₁ = 1$\mathrm{\mu F}$ 
Maximum Potential difference applied across each capacitor, V₁ = 400 V 

Let, the possible arrangement of circuit be such that n capacitors of 1$\mathrm{\mu F}$ each, be connected in series in a row and m such rows be connected in parallel.

Therefore, total number of capacitors-N = m x n

Since the potential in each row is 1000 V, the number of capacitors in each row of series arrangement is 
              
                     $$\begin{gathered} \frac{1000}{\mathrm{n}} = 400 \\ \mathrm{n} = 2.5 \end{gathered}$$

Since, the number of capacitors cannot be in fraction, therefore, we take n=3.

Now, the capacitance of n capacitors in series, in m such rows is

$$\begin{gathered} \frac{1}{{\mathrm{C}}_{\mathrm{s}}} = \left[\frac{1}{\mathrm{n}}+\frac{1}{\mathrm{n}}+...\mathrm{n} \mathrm{lines}\right] = \mathrm{m} \\ {\mathrm{C}}_{\mathrm{s}} = \frac{1}{\mathrm{n}}\mathrm{\mu F} \end{gathered}$$

So, the resultant capacitance of all capacitors is equal to
 
$\frac{1}{\mathrm{n}}+\frac{1}{\mathrm{n}}+\frac{1}{\mathrm{n}}....+\mathrm{m} \mathrm{terms} = 2$$\mathrm{\mu F}$
                           $\frac{\mathrm{m}}{\mathrm{n}}=2$ $\mathrm{\mu F}$
                     
                            $$\begin{gathered} \frac{\mathrm{m}}{3}=2 \\ \mathrm{m} = 6 \mathrm{rows} \\ \mathrm{n} = 6 \times 3 = 18 \end{gathered}$$

Hence, 1$\mathrm{\mu F}$ capacitor each,  should be connected in 6 parallel rows, where,  each row contains 3 capacitors each in series.