Find the charge on each of the capacitors in the figure below. + 1.00 F. 5.00 F 24.0 V 8,00 F 4,00F 1.0-μF capacitor HC 5.0-μF capacitor 8.0-μF capacitor 4.0-uF capacitor HC C
Find the charge on each of the capacitors in the figure below. + 1.00 F. 5.00 F 24.0 V 8,00 F 4,00F 1.0-μF capacitor HC 5.0-μF capacitor 8.0-μF capacitor 4.0-uF capacitor HC C
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Chapter1: Units, Trigonometry. And Vectors
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![### Capacitors in Parallel and Series Combinations
#### Problem Statement
Find the charge on each of the capacitors in the figure below.
#### Circuit Diagram
The diagram represents a circuit containing a 24.0V power source connected to four capacitors:
1. A 1.00 µF capacitor connected in series with a 5.00 µF capacitor.
2. An 8.00 µF capacitor connected in series with a 4.00 µF capacitor.
The two series sets of capacitors are connected in parallel to each other.
#### Capacitors Description
- Each capacitor has a given capacitance:
- 1.00 µF capacitor
- 5.00 µF capacitor
- 8.00 µF capacitor
- 4.00 µF capacitor
#### Boxes for Answer Entry
Below the circuit, there are labeled boxes where you can fill in the charge (in µC) for each capacitor:
- 1.0-µF capacitor: ___ µC
- 5.0-µF capacitor: ___ µC
- 8.0-µF capacitor: ___ µC
- 4.0-µF capacitor: ___ µC
#### Understanding the Diagram
To determine the charge on each capacitor, follow these steps:
1. **Identify Equivalent Capacitance in Each Series Pair:**
- For the series combination of a 1.00 µF and 5.00 µF capacitor:
\[
\frac{1}{C_{eq1}} = \frac{1}{1.00} + \frac{1}{5.00}
\]
- For the series combination of an 8.00 µF and 4.00 µF capacitor:
\[
\frac{1}{C_{eq2}} = \frac{1}{8.00} + \frac{1}{4.00}
\]
2. **Find the Total Equivalent Capacitance:**
- The equivalent capacitances \(C_{eq1}\) and \(C_{eq2}\) are connected in parallel. Their total capacitance is:
\[
C_{total} = C_{eq1} + C_{eq2}
\]
3. **Determine the Voltage Across Each Series Pair:**
- Use voltage division based on the equivalent capacitances to determine the voltage drop across](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bf878e8-f234-4f66-a1d2-75bcef7aa11d%2F862e4165-e446-4fd0-8b06-41fd1649a61c%2Ftwruf9q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Capacitors in Parallel and Series Combinations
#### Problem Statement
Find the charge on each of the capacitors in the figure below.
#### Circuit Diagram
The diagram represents a circuit containing a 24.0V power source connected to four capacitors:
1. A 1.00 µF capacitor connected in series with a 5.00 µF capacitor.
2. An 8.00 µF capacitor connected in series with a 4.00 µF capacitor.
The two series sets of capacitors are connected in parallel to each other.
#### Capacitors Description
- Each capacitor has a given capacitance:
- 1.00 µF capacitor
- 5.00 µF capacitor
- 8.00 µF capacitor
- 4.00 µF capacitor
#### Boxes for Answer Entry
Below the circuit, there are labeled boxes where you can fill in the charge (in µC) for each capacitor:
- 1.0-µF capacitor: ___ µC
- 5.0-µF capacitor: ___ µC
- 8.0-µF capacitor: ___ µC
- 4.0-µF capacitor: ___ µC
#### Understanding the Diagram
To determine the charge on each capacitor, follow these steps:
1. **Identify Equivalent Capacitance in Each Series Pair:**
- For the series combination of a 1.00 µF and 5.00 µF capacitor:
\[
\frac{1}{C_{eq1}} = \frac{1}{1.00} + \frac{1}{5.00}
\]
- For the series combination of an 8.00 µF and 4.00 µF capacitor:
\[
\frac{1}{C_{eq2}} = \frac{1}{8.00} + \frac{1}{4.00}
\]
2. **Find the Total Equivalent Capacitance:**
- The equivalent capacitances \(C_{eq1}\) and \(C_{eq2}\) are connected in parallel. Their total capacitance is:
\[
C_{total} = C_{eq1} + C_{eq2}
\]
3. **Determine the Voltage Across Each Series Pair:**
- Use voltage division based on the equivalent capacitances to determine the voltage drop across
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