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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Question
The figure below shows four capacitors with
CA = 4.00 µF,
CB = 8.00 µF,
CC = 5.00 µF,
and
CD = 2.00 µF
connected across points a and b, which have potential difference
ΔVab = 12.0 V.
(a) What is the equivalent capacitance of the four capacitors?
![### Capacitors in Combination
**Problem Statement:**
The figure below shows four capacitors with \( C_A = 4.00 \, \mu\text{F} \), \( C_B = 8.00 \, \mu\text{F} \), \( C_C = 5.00 \, \mu\text{F} \), and \( C_D = 2.00 \, \mu\text{F} \) connected across points \( a \) and \( b \), which have a potential difference \( \Delta V_{ab} = 12.0 \, \text{V} \).
**Diagram Explanation:**
A schematic diagram is provided in which four capacitors are connected. The points \(a\) and \(b\) are marked, and the potential difference between them is given as \(12.0 \, \text{V}\).
- Capacitor \(C_A\) is connected between point \(a\) and one terminal of \(C_B\).
- Capacitor \(C_B\) is then connected to point \(b\).
- Capacitors \(C_C\) and \(C_D\) are connected in series between point \(a\) and the junction between \(C_B\) and \(b\), forming a parallel circuit with \(C_A - C_B\).
**Question:**
(a) What is the equivalent capacitance of the four capacitors?
**Answer:**
The text box provided shows a calculated value of 2.81 but it is marked with a red cross indicating the value is incorrect.
**Solution Steps:**
1. **Combine Capacitors \(C_C\) and \(C_D\) in Series:**
\[
\frac{1}{C_{CD}} = \frac{1}{C_C} + \frac{1}{C_D}
\]
\[
\frac{1}{C_{CD}} = \frac{1}{5.00 \, \mu\text{F}} + \frac{1}{2.00 \, \mu\text{F}}
\]
\[
\frac{1}{C_{CD}} = 0.2 + 0.5 = 0.7
\]
\[
C_{CD} = \frac{1}{0.7} \approx 1.43 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F188f854e-5fc6-4301-8b8b-5d31fc4cc5e7%2F2fd817c0-92a4-42ed-835f-25fd46570eab%2Fogvhlci.png&w=3840&q=75)
Transcribed Image Text:### Capacitors in Combination
**Problem Statement:**
The figure below shows four capacitors with \( C_A = 4.00 \, \mu\text{F} \), \( C_B = 8.00 \, \mu\text{F} \), \( C_C = 5.00 \, \mu\text{F} \), and \( C_D = 2.00 \, \mu\text{F} \) connected across points \( a \) and \( b \), which have a potential difference \( \Delta V_{ab} = 12.0 \, \text{V} \).
**Diagram Explanation:**
A schematic diagram is provided in which four capacitors are connected. The points \(a\) and \(b\) are marked, and the potential difference between them is given as \(12.0 \, \text{V}\).
- Capacitor \(C_A\) is connected between point \(a\) and one terminal of \(C_B\).
- Capacitor \(C_B\) is then connected to point \(b\).
- Capacitors \(C_C\) and \(C_D\) are connected in series between point \(a\) and the junction between \(C_B\) and \(b\), forming a parallel circuit with \(C_A - C_B\).
**Question:**
(a) What is the equivalent capacitance of the four capacitors?
**Answer:**
The text box provided shows a calculated value of 2.81 but it is marked with a red cross indicating the value is incorrect.
**Solution Steps:**
1. **Combine Capacitors \(C_C\) and \(C_D\) in Series:**
\[
\frac{1}{C_{CD}} = \frac{1}{C_C} + \frac{1}{C_D}
\]
\[
\frac{1}{C_{CD}} = \frac{1}{5.00 \, \mu\text{F}} + \frac{1}{2.00 \, \mu\text{F}}
\]
\[
\frac{1}{C_{CD}} = 0.2 + 0.5 = 0.7
\]
\[
C_{CD} = \frac{1}{0.7} \approx 1.43 \,
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