91Ó°ÊÓ

1. Value of each capacitor, $C=2.0\mathrm{μ¹ó}$
2. Potential that each capacitor can withstand, V = 200 V
3. Equivalent capacitance,$C{\text{'}}_{eq}=0.40\mathrm{μ¹ó}$
4. Equivalent capacitance of withstanding potential of is $\mathrm{C}{\text{'}}_{\mathrm{eq}}=1.2\mathrm{μ¹ó}$.

We will use the idea of the combination of capacitors in series and parallel to get the required capacitance.

Formulae:

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivalent}}=\underset{}{∑{\displaystyle \frac{1}{C_i}}}$ …(¾±).

The equivalent capacitance of a parallel connection of capacitors,

$\frac{1}{C_{equivalent}}=\underset{}{∑{\displaystyle C_i}}$ …(¾±¾±)

If we connect N capacitors in series to get capacitance$40\mathrm{μ}F$, then the equivalent capacitance can be given using equation (i) as follows: (as each capacitance value is given to be same)

$$\begin{gathered} \frac{1}{Ceq}=\underset{i=1}{\overset{n}{∑}}\left(\frac{1}{C_i}\right) \\ =\frac{N}{C} \end{gathered}$$

Now, using the given values, we have

$$\begin{gathered} N=\frac{C}{C_{eq}} \\ =\frac{2.0\mathrm{μ}F}{0.40\mathrm{μ}F} \\ =5.0 \end{gathered}$$

So, we need to connect 5 capacitors in series.

Above combination can withstand 1000 V

If same sequences of capacitors connected in parallel, having five capacitors connected in series are used, then the number of equivalent capacitance can be given using equation (ii) as:

$$\begin{gathered} C\text{'}{\text{'}}_{eq}=N×C{\text{'}}_{eq} \\ 1.2\mathrm{μ}F=N×C{\text{'}}_{eq} \\ N=\frac{1.2\mathrm{μ}F}{0.40\mathrm{μ}F} \\ =3.0 \end{gathered}$$

So, we must connect 3 arrays of capacitors in parallel, each having 5 capacitors in series.