Solution:

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.

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Solution:

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

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Solution:

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

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Solution:

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:

$$\boxed{32}$$

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Solution:

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:

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

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Solution:

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 = \( \frac{CE}{3}-\frac{CE}{4}= \frac{CE}{12}\)

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Solution:

Circircled ones are in parallel

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Solution:

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

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

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

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Solution:

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:

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

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