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:

$$\boxed{32}$$

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

We need to check each option separately. We 5 capacitors are connected  in parallel, 2 capacitors are connected in series.

C_eq= (5C×C/2)/ (5C+C/2)= 5C/11 = 5×2/11 = 10/11 μF.

In the Upper arm 6 μF and 3 μF capacitors are in series , so equivalent capacitance is 2μF. The 3μF capacitor ( circled one) can be removed as it is part of balanced wheat stone bridge.

When Switch is Open Equivalent Capacitance is C_eq=C/2

So, Charge on Both the capacitors in CE/2.

When Switch is Closed 1st Capacitor is short circuited .Now  Equivalent Capacitance is C_eq=C.

So, Charge on  the capacitors in CE.