The Circled capacitors are in Series so , Their equivalent capacitance is 1 μF . Then it is in parallel with 1 μF capacitor. The circuit will keep on reducing.

For Such Question start solving from the farthest point. The circuit will keep on reducing.

Capacitors circled in the diagram are short circuited so, they can be removed from the circuit.

To solve this, we determine the combination of capacitors required to achieve the desired capacitance and voltage.

Given:

• Individual capacitor:
  $C = 8 \, \mu\text{F}$ 

  , $V_{\text{max}} = 250 \, \text{V}$ 

• Desired combination:
  $C_{\text{req}} = 16 \, \mu\text{F}$ 

  , $V_{\text{req}} = 1000 \, \text{V}$ 

Step 1: Voltage requirement

To achieve

$V_{\text{req}} = 1000 \, \text{V}$

, multiple capacitors must be connected in series because the voltage across a series combination adds up. The number of capacitors required in series is:

$$n = \frac{V_{\text{req}}}{V_{\text{max}}} = \frac{1000}{250} = 4$$

Thus, 4 capacitors in series are required to handle 1000 V.

Step 2: Capacitance in series

The effective capacitance of

$n$

capacitors in series is given by:

$$C_{\text{series}} = \frac{C}{n} = \frac{8}{4} = 2 \, \mu\text{F}$$

So, a series of 4 capacitors provides

$C_{\text{series}} = 2 \, \mu\text{F}$

.

Step 3: Capacitance requirement

To achieve

$C_{\text{req}} = 16 \, \mu\text{F}$

, multiple such series groups must be connected in parallel because capacitance in parallel adds up. The number of such series groups required is:

$$m = \frac{C_{\text{req}}}{C_{\text{series}}} = \frac{16}{2} = 8$$

Thus, 8 series groups are required.

Step 4: Total capacitors

Each series group contains 4 capacitors, and there are 8 such groups. Therefore, the total number of capacitors is:

$$\text{Total capacitors} = n \cdot m = 4 \cdot 8 = 32$$

Final Answer:

The minimum number of capacitors required is:

$$\boxed{32}$$

Let's verify and calculate the correct answer step-by-step:

Given Data:

• Capacitances:
  $C_1 = 2 \, \mu\text{F}, C_2 = 3 \, \mu\text{F}, C_3 = 5 \, \mu\text{F}$ 
• Maximum voltages:
  $V_1 = 3 \, \text{V}, V_2 = 2 \, \text{V}, V_3 = 1 \, \text{V}$ 

Step 1: Maximum charge each capacitor can store:

$$Q_1 = C_1 \cdot V_1 = 2 \cdot 3 = 6 \, \mu\text{C}$$

$$Q_2 = C_2 \cdot V_2 = 3 \cdot 2 = 6 \, \mu\text{C}$$

$$Q_3 = C_3 \cdot V_3 = 5 \cdot 1 = 5 \, \mu\text{C}$$

The capacitor with the minimum charge capacity limits the system. Here,

$Q_{\text{max}} = 5 \, \mu\text{C}$

, dictated by

$C_3$

.

Step 2: Voltage distribution across each capacitor:

In series, charge

$Q$

is the same on all capacitors, and the voltage across each capacitor is:

$$V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3}$$

Total voltage across the series combination:

$$V_{\text{total}} = V_1 + V_2 + V_3$$

Substitute

$Q = 5 \, \mu\text{C}$

:

$$V_1 = \frac{5}{2} = 2.5 \, \text{V}, \quad V_2 = \frac{5}{3} \approx 1.67 \, \text{V}, \quad V_3 = \frac{5}{5} = 1 \, \text{V}$$

Step 3: Total voltage:

$$V_{\text{total}} = V_1 + V_2 + V_3 = 2.5 + 1.67 + 1 = \frac{15}{6} + \frac{10}{6} + \frac{6}{6} = \frac{31}{6} \, \text{V}.$$

Final Answer:

The maximum voltage the series combination can withstand is:

$$\boxed{\frac{31}{6} \, \text{V}} \approx 5.17 \, \text{V}.$$

When Switch is open C_eq= C/4 Charge given by the Battery is CE/4.

When Switch is open C_eq= C/3 Charge given by the Battery is CE/3.

Extra Charge flowing through the circuit it = \( \frac{CE}{3}-\frac{CE}{4}= \frac{CE}{12}\)

Circircled ones are in parallel

First Branch has a capacitance of 1F , for second branch it is 1/2F , for third branch it is 1/4 F and so, on . As all these branches are in parallel

\[ C_{eq}=C_{1}+C_{2}+C_{3}+....\]

\[ C_{eq}= 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8}+....=\frac{1}{1-\frac{1}{2}}=2\mu F\]

To find the equivalent capacitance for this cubical capacitor network, where each edge of the cube has a capacitance

$C$

, here’s the shortest solution:

Step-by-Step:

1. Symmetry analysis:
   • By symmetry, all corners of the cube can be grouped into equivalent potential nodes.
   • The cube's symmetry allows reduction to a simpler circuit.
2. Key nodes:
   • Node
     $A$ 

     is connected to one corner of the cube.

   • Node
     $B$ 

     is connected to the diagonally opposite corner.

3. Effective connections:
   • Due to symmetry, three capacitors are effectively in parallel between
     $A$ 

     and an intermediate point.

   • Similarly, three capacitors are effectively in parallel between
     $B$ 

     and the same intermediate point.

   • Two capacitors remain directly between
     $A$ 

     and $B$ 

     .

4. Simplification:
   • The three parallel capacitors at each node result in:
     $$C_{\text{parallel}} = 3C$$ 
   • The equivalent circuit becomes two
     $3C$ 

     capacitors in series with a $2C$ 

     capacitor: $$\text{Series combination:} \quad \frac{1}{C_{\text{eq}}} = \frac{1}{3C} + \frac{1}{3C} + \frac{1}{2C}$$ 

5. Calculation:
   • Combine series:
     $$\frac{1}{C_{\text{eq}}} = \frac{2}{3C} + \frac{1}{2C} = \frac{4}{6C} + \frac{3}{6C} = \frac{7}{6C}$$ 
   • Invert to find
     $C_{\text{eq}}$ 

     : $$C_{\text{eq}} = \frac{6C}{7} \times 2 = \frac{12C}{7}$$ 

Thus, the equivalent capacitance is:

$$C_{\text{eq}} = \frac{12C}{7}$$