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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)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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### 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. **
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. **
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