What is the equivalent capacitance of the network below assuming each of the 5 capacitors has a value of 50 µF ? Give your answer in (microfarads)
What is the equivalent capacitance of the network below assuming each of the 5 capacitors has a value of 50 µF ? Give your answer in (microfarads)
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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```
[ Capacitance Network Diagram ]
```
**Diagram Explanation:**
The diagram represents a network of five capacitors. The specific arrangement is as follows:
1. The first capacitor is connected in series to a combination of parallel capacitors.
2. This combination includes three capacitors in parallel.
3. The output combination is then connected in series to the last capacitor.
**Given:**
- Each capacitor has a capacitance value of 50 µF.
**Find:**
- The equivalent capacitance of the entire network.
**Answer:**
- Give your answer in microfarads (µF).
**Form:**
- [Your Answer Here]
**Solution Steps:**
1. **Identifying the Capacitor Network Configuration:**
- First, identify the capacitors in parallel and series configurations.
2. **Calculating Parallel Combination:**
- The three middle capacitors are in parallel. For capacitors in parallel, the total capacitance (C_parallel) is the sum of the capacitances:
\[
C_{\text{parallel}} = C_1 + C_2 + C_3
\]
Given each capacitor is 50 µF:
\[
C_{\text{parallel}} = 50 \, \mu\text{F} + 50 \, \mu\text{F} + 50 \, \mu\text{F} = 150 \, \mu\text{F}
\]
3. **Calculating Series Combination:**
- Next, sum this parallel combination in series with the remaining two capacitors. For capacitors in series, the total capacitance (C_series) is given by:
\[
\frac{1}{C_{\text{series}}} = \frac{1}{C_a} + \frac{1}{C_b} + \frac{1}{C_c}
\]
Here, \(C_a\) is the first capacitor, \(C_b\) is the parallel combination found before, and \(C_c\) is the last capacitor, all values are 150 µF.
4. **Total Equivalent Capacitance:**
- Thus,
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Transcribed Image Text:### Capacitor Network Problem
**Problem:**
What is the equivalent capacitance of the network below assuming each of the 5 capacitors has a value of 50 µF?
```
[ Capacitance Network Diagram ]
```
**Diagram Explanation:**
The diagram represents a network of five capacitors. The specific arrangement is as follows:
1. The first capacitor is connected in series to a combination of parallel capacitors.
2. This combination includes three capacitors in parallel.
3. The output combination is then connected in series to the last capacitor.
**Given:**
- Each capacitor has a capacitance value of 50 µF.
**Find:**
- The equivalent capacitance of the entire network.
**Answer:**
- Give your answer in microfarads (µF).
**Form:**
- [Your Answer Here]
**Solution Steps:**
1. **Identifying the Capacitor Network Configuration:**
- First, identify the capacitors in parallel and series configurations.
2. **Calculating Parallel Combination:**
- The three middle capacitors are in parallel. For capacitors in parallel, the total capacitance (C_parallel) is the sum of the capacitances:
\[
C_{\text{parallel}} = C_1 + C_2 + C_3
\]
Given each capacitor is 50 µF:
\[
C_{\text{parallel}} = 50 \, \mu\text{F} + 50 \, \mu\text{F} + 50 \, \mu\text{F} = 150 \, \mu\text{F}
\]
3. **Calculating Series Combination:**
- Next, sum this parallel combination in series with the remaining two capacitors. For capacitors in series, the total capacitance (C_series) is given by:
\[
\frac{1}{C_{\text{series}}} = \frac{1}{C_a} + \frac{1}{C_b} + \frac{1}{C_c}
\]
Here, \(C_a\) is the first capacitor, \(C_b\) is the parallel combination found before, and \(C_c\) is the last capacitor, all values are 150 µF.
4. **Total Equivalent Capacitance:**
- Thus,
\
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