Consider the circuit shown in the figure, with C₁ = 5.9⁹2 µF and C₂ = 7.54 µF. 2.00 μF (b) 6.00 µF + 90.0 V C₂ (a) Find the equivalent capacitance (in µF) of the system. uF Find the charge (in µC) on each capacitor. 5.92 μF capacitor 6.00 μF capacitor 7.54 μF capacitor 2.00 μF capacitor 9999 HC (c) Find the potential difference (in V) on each capacitor. 5.92 μF capacitor V 6.00 μF capacitor V 7.54 µF capacitor V 2.00 μF capacitor V (d) Find the total energy (in mJ) stored by the group. mJ

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Chapter21: Circuits And Dc Instruments
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### Capacitor Circuit Analysis

Consider the circuit shown in the figure, with \( C_1 = 5.92 \, \mu\text{F} \) and \( C_2 = 7.54 \, \mu\text{F} \).

![Circuit Diagram](attachment:Circuit.png)
- **C1**: 5.92 µF
- **C2**: 6.00 µF
- **C3**: 2.00 µF
- **C4**: 7.54 µF

A voltage source of 90.0 V is connected across the circuit.

#### (a) Find the equivalent capacitance (in µF) of the system.
\[ \text{Equivalent Capacitance} = \_\_\_\_ \, \mu\text{F} \]

#### (b) Find the charge (in µC) on each capacitor.
- 5.92 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \]
- 6.00 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \]
- 7.54 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \]
- 2.00 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \]

#### (c) Find the potential difference (in V) on each capacitor.
- 5.92 µF capacitor: \[ \_\_\_\_ \, \text{V} \]
- 6.00 µF capacitor: \[ \_\_\_\_ \, \text{V} \]
- 7.54 µF capacitor: \[ \_\_\_\_ \, \text{V} \]
- 2.00 µF capacitor: \[ \_\_\_\_ \, \text{V} \]

#### (d) Find the total energy (in mJ) stored by the group.
\[ \text{Total Energy} = \_\_\_\_ \, \text{mJ} \]

### Detailed Explanation of the Diagram

The circuit diagram contains the following components resulting in two branches of capacitors in parallel:

1. The first branch contains:
   - A capacitor labeled \( C_1 \) with
Transcribed Image Text:### Capacitor Circuit Analysis Consider the circuit shown in the figure, with \( C_1 = 5.92 \, \mu\text{F} \) and \( C_2 = 7.54 \, \mu\text{F} \). ![Circuit Diagram](attachment:Circuit.png) - **C1**: 5.92 µF - **C2**: 6.00 µF - **C3**: 2.00 µF - **C4**: 7.54 µF A voltage source of 90.0 V is connected across the circuit. #### (a) Find the equivalent capacitance (in µF) of the system. \[ \text{Equivalent Capacitance} = \_\_\_\_ \, \mu\text{F} \] #### (b) Find the charge (in µC) on each capacitor. - 5.92 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \] - 6.00 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \] - 7.54 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \] - 2.00 µF capacitor: \[ \_\_\_\_ \, \mu\text{C} \] #### (c) Find the potential difference (in V) on each capacitor. - 5.92 µF capacitor: \[ \_\_\_\_ \, \text{V} \] - 6.00 µF capacitor: \[ \_\_\_\_ \, \text{V} \] - 7.54 µF capacitor: \[ \_\_\_\_ \, \text{V} \] - 2.00 µF capacitor: \[ \_\_\_\_ \, \text{V} \] #### (d) Find the total energy (in mJ) stored by the group. \[ \text{Total Energy} = \_\_\_\_ \, \text{mJ} \] ### Detailed Explanation of the Diagram The circuit diagram contains the following components resulting in two branches of capacitors in parallel: 1. The first branch contains: - A capacitor labeled \( C_1 \) with
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