An LC circuit consists of a 20 mH inductor and a 0.500 uf capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor?
An LC circuit consists of a 20 mH inductor and a 0.500 uf capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor?
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
Related questions
Question
![**Problem 4: LC Circuit Analysis**
An LC circuit consists of a 20 mH inductor and a 0.500 μF capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor?
**Solution Explanation**
To solve this problem, we use the principle of energy conservation in the LC circuit. The sum of the magnetic energy in the inductor and the electric energy in the capacitor is constant.
1. **Magnetic Energy in the Inductor:**
\[
U_L = \frac{1}{2} L I^2
\]
Where:
- \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- \( I = 0.100 \, \text{A} \)
2. **Electric Energy in the Capacitor:**
\[
U_C = \frac{1}{2} C V^2
\]
Where:
- \( C = 0.500 \, \mu\text{F} = 0.500 \times 10^{-6} \, \text{F} \)
- \( V \) is the potential difference we need to find.
3. **Applying Energy Conservation:**
\[
\frac{1}{2} L I^2 = \frac{1}{2} C V^2
\]
Substituting values into the equation:
\[
\frac{1}{2} (20 \times 10^{-3}) (0.100)^2 = \frac{1}{2} (0.500 \times 10^{-6}) V^2
\]
4. **Solving for V:**
\[
V^2 = \frac{(20 \times 10^{-3}) \times (0.100)^2}{0.500 \times 10^{-6}}
\]
\[
V = \sqrt{\frac{(20 \times 10^{-3}) \times (0.100)^2}{0.500 \times 10^{-6}}}
\]
Calculate the potential difference \( V \) from the equation above.
This approach provides the greatest potential difference across the capacitor in the LC circuit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb052dd5d-84db-466c-a2ba-e3b39e1bdb71%2Ffd58a6af-86eb-4f62-bfcb-f85926541eca%2Fgz39mui_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 4: LC Circuit Analysis**
An LC circuit consists of a 20 mH inductor and a 0.500 μF capacitor. If the maximum instantaneous current is 0.100 A, what is the greatest potential difference across the capacitor?
**Solution Explanation**
To solve this problem, we use the principle of energy conservation in the LC circuit. The sum of the magnetic energy in the inductor and the electric energy in the capacitor is constant.
1. **Magnetic Energy in the Inductor:**
\[
U_L = \frac{1}{2} L I^2
\]
Where:
- \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- \( I = 0.100 \, \text{A} \)
2. **Electric Energy in the Capacitor:**
\[
U_C = \frac{1}{2} C V^2
\]
Where:
- \( C = 0.500 \, \mu\text{F} = 0.500 \times 10^{-6} \, \text{F} \)
- \( V \) is the potential difference we need to find.
3. **Applying Energy Conservation:**
\[
\frac{1}{2} L I^2 = \frac{1}{2} C V^2
\]
Substituting values into the equation:
\[
\frac{1}{2} (20 \times 10^{-3}) (0.100)^2 = \frac{1}{2} (0.500 \times 10^{-6}) V^2
\]
4. **Solving for V:**
\[
V^2 = \frac{(20 \times 10^{-3}) \times (0.100)^2}{0.500 \times 10^{-6}}
\]
\[
V = \sqrt{\frac{(20 \times 10^{-3}) \times (0.100)^2}{0.500 \times 10^{-6}}}
\]
Calculate the potential difference \( V \) from the equation above.
This approach provides the greatest potential difference across the capacitor in the LC circuit.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.Recommended textbooks for you
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:
9780133923605
Author:
Robert L. Boylestad
Publisher:
PEARSON
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:
9781337900348
Author:
Stephen L. Herman
Publisher:
Cengage Learning
Programmable Logic Controllers
Electrical Engineering
ISBN:
9780073373843
Author:
Frank D. Petruzella
Publisher:
McGraw-Hill Education
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:
9780078028229
Author:
Charles K Alexander, Matthew Sadiku
Publisher:
McGraw-Hill Education
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:
9780134746968
Author:
James W. Nilsson, Susan Riedel
Publisher:
PEARSON
Engineering Electromagnetics
Electrical Engineering
ISBN:
9780078028151
Author:
Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:
Mcgraw-hill Education,