8. Determine the peak and average power delivered to RL in Figure 3. 2:1 RL 120 Vrms 220 02 ● ooooo ellee Fig. 3
8. Determine the peak and average power delivered to RL in Figure 3. 2:1 RL 120 Vrms 220 02 ● ooooo ellee Fig. 3
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 Statement
**8. Determine the peak and average power delivered to \(R_L\) in Figure 3.**
### Figure Description
**Figure 3:** The figure is an electrical circuit diagram including the following elements:
- An AC power supply labeled "120V rms".
- A transformer with a turns ratio of 2:1.
- A diode.
- A load resistor labeled \(R_L = 220 \Omega\).
**Detailed Explanation of Diagram:**
1. **AC Power Supply:** This provides an RMS voltage of 120V.
2. **Transformer:** The transformer, with a 2:1 turns ratio, steps down the voltage from the AC power supply.
- The primary winding (120V input) is connected to the AC power supply.
- The secondary winding is connected to the rest of the circuit, which includes a diode and the load resistor \(R_L\).
3. **Diode:** The diode allows current to flow only in one direction, thereby converting the AC input to a pulsating DC output.
4. **Load Resistor (\(R_L\)):** The load resistor has a resistance of 220 ohms.
Based on the configuration, the task is to calculate the peak and average power delivered to the load resistor \(R_L\).
### Solution Steps
1. **Calculate the peak AC voltage at the secondary winding of the transformer**:
- Using the transformer turns ratio (\(N1:N2 = 2:1\)), the voltage at the secondary is half of the primary:
\[
V_{secondary, rms} = \frac{120V}{2} = 60V \, \text{(rms)}
\]
2. **Convert rms voltage to peak voltage**:
- Peak voltage (\(V_{peak}\)) in an AC circuit is \(\sqrt{2}\) times the RMS voltage:
\[
V_{peak, secondary} = 60V \times \sqrt{2} \approx 84.85V
\]
3. **Peak voltage across the resistor**:
- Considering the diode, the voltage drop (approximately 0.7V for a silicon diode) should be subtracted from the peak voltage:
\[
V_{peak, R_L} = 84.85V - 0.7V \approx 84.15V
\]
4. **](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53cd4567-e9c0-49bc-8f29-7fe4db37c68a%2F2456bffe-2eec-4d4d-86c2-c67b9ac68885%2F6la795f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**8. Determine the peak and average power delivered to \(R_L\) in Figure 3.**
### Figure Description
**Figure 3:** The figure is an electrical circuit diagram including the following elements:
- An AC power supply labeled "120V rms".
- A transformer with a turns ratio of 2:1.
- A diode.
- A load resistor labeled \(R_L = 220 \Omega\).
**Detailed Explanation of Diagram:**
1. **AC Power Supply:** This provides an RMS voltage of 120V.
2. **Transformer:** The transformer, with a 2:1 turns ratio, steps down the voltage from the AC power supply.
- The primary winding (120V input) is connected to the AC power supply.
- The secondary winding is connected to the rest of the circuit, which includes a diode and the load resistor \(R_L\).
3. **Diode:** The diode allows current to flow only in one direction, thereby converting the AC input to a pulsating DC output.
4. **Load Resistor (\(R_L\)):** The load resistor has a resistance of 220 ohms.
Based on the configuration, the task is to calculate the peak and average power delivered to the load resistor \(R_L\).
### Solution Steps
1. **Calculate the peak AC voltage at the secondary winding of the transformer**:
- Using the transformer turns ratio (\(N1:N2 = 2:1\)), the voltage at the secondary is half of the primary:
\[
V_{secondary, rms} = \frac{120V}{2} = 60V \, \text{(rms)}
\]
2. **Convert rms voltage to peak voltage**:
- Peak voltage (\(V_{peak}\)) in an AC circuit is \(\sqrt{2}\) times the RMS voltage:
\[
V_{peak, secondary} = 60V \times \sqrt{2} \approx 84.85V
\]
3. **Peak voltage across the resistor**:
- Considering the diode, the voltage drop (approximately 0.7V for a silicon diode) should be subtracted from the peak voltage:
\[
V_{peak, R_L} = 84.85V - 0.7V \approx 84.15V
\]
4. **
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 3 steps with 1 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,