College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Determining the Equivalent Capacitance of Capacitors
#### Question:
2. What is the equivalent capacitance of the three capacitors in the figure?
#### Diagram Explanation:
The diagram below shows a combination of three capacitors.
- Capacitor 1 has a capacitance of \(20 \mu F\).
- Capacitor 2 has a capacitance of \(30 \mu F\).
- Capacitor 3 has a capacitance of \(25 \mu F\).
The diagram illustrates a setup where the \(20 \mu F\) and \(30 \mu F\) capacitors are arranged in parallel, and this parallel combination is in series with the \(25 \mu F\) capacitor.
#### Calculations:
1. **Parallel Combination:**
The total capacitance \(C_{parallel}\) of capacitors in parallel is given by:
\[
C_{parallel} = C_1 + C_2
\]
For the \(20 \mu F\) and \(30 \mu F\) capacitors:
\[
C_{parallel} = 20 \mu F + 30 \mu F = 50 \mu F
\]
2. **Series Combination:**
The equivalent capacitance \(C_{eq}\) of capacitors in series is given by:
\[
\frac{1}{C_{eq}} = \frac{1}{C_{parallel}} + \frac{1}{C_3}
\]
Substituting the values:
\[
\frac{1}{C_{eq}} = \frac{1}{50 \mu F} + \frac{1}{25 \mu F} = \frac{1}{50} + \frac{1}{25} = \frac{1 + 2}{50} = \frac{3}{50}
\]
Therefore,
\[
C_{eq} = \frac{50}{3} \approx 16.67 \mu F
\]
#### Conclusion:
The equivalent capacitance of the three capacitors in the given configuration is approximately \(16.67 \mu F\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b2b1966-cbdc-46cc-aafa-333a90f3e9b8%2F510f43be-7dcc-4869-8756-62bd13c43061%2Fu87bu8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Determining the Equivalent Capacitance of Capacitors
#### Question:
2. What is the equivalent capacitance of the three capacitors in the figure?
#### Diagram Explanation:
The diagram below shows a combination of three capacitors.
- Capacitor 1 has a capacitance of \(20 \mu F\).
- Capacitor 2 has a capacitance of \(30 \mu F\).
- Capacitor 3 has a capacitance of \(25 \mu F\).
The diagram illustrates a setup where the \(20 \mu F\) and \(30 \mu F\) capacitors are arranged in parallel, and this parallel combination is in series with the \(25 \mu F\) capacitor.
#### Calculations:
1. **Parallel Combination:**
The total capacitance \(C_{parallel}\) of capacitors in parallel is given by:
\[
C_{parallel} = C_1 + C_2
\]
For the \(20 \mu F\) and \(30 \mu F\) capacitors:
\[
C_{parallel} = 20 \mu F + 30 \mu F = 50 \mu F
\]
2. **Series Combination:**
The equivalent capacitance \(C_{eq}\) of capacitors in series is given by:
\[
\frac{1}{C_{eq}} = \frac{1}{C_{parallel}} + \frac{1}{C_3}
\]
Substituting the values:
\[
\frac{1}{C_{eq}} = \frac{1}{50 \mu F} + \frac{1}{25 \mu F} = \frac{1}{50} + \frac{1}{25} = \frac{1 + 2}{50} = \frac{3}{50}
\]
Therefore,
\[
C_{eq} = \frac{50}{3} \approx 16.67 \mu F
\]
#### Conclusion:
The equivalent capacitance of the three capacitors in the given configuration is approximately \(16.67 \mu F\).
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