12 Example 2.9

Capacitor Network Simulation

This simulation shows a network of four 10 μF capacitors connected to a 500 V supply:

Step 1: Identify series and parallel combinations

From the diagram, we can see that:

Step 2: Calculate equivalent capacitance of series pairs

For capacitors in series:

\( \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \)

Since \( C_1 = C_2 = 10 \, \mu\text{F} \):

\( \frac{1}{C_{12}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} \)
\( C_{12} = 5 \, \mu\text{F} \)

Similarly, \( C_{34} = 5 \, \mu\text{F} \)

Step 3: Calculate total equivalent capacitance

Now \( C_{12} \) and \( C_{34} \) are in parallel:

\( C_{total} = C_{12} + C_{34} = 5 + 5 = 10 \, \mu\text{F} \)

Step 4: Calculate total charge

\( Q_{total} = C_{total} \times V = 10 \, \mu\text{F} \times 500 \, \text{V} = 5000 \, \mu\text{C} \)

Step 5: Calculate charge on each capacitor

The total charge is equally divided between the two parallel branches:

\( Q_{12} = Q_{34} = 2500 \, \mu\text{C} \)

Since \( C_1 \) and \( C_2 \) are in series, they each have the same charge:

\( Q_1 = Q_2 = 2500 \, \mu\text{C} \)

Similarly for \( C_3 \) and \( C_4 \):

\( Q_3 = Q_4 = 2500 \, \mu\text{C} \)

(a) Equivalent capacitance of the network: \( 10 \, \mu\text{F} \)

(b) Charge on each capacitor: \( 2500 \, \mu\text{C} \)

Capacitor Network Simulation