### Half-Wave Rectifier Circuit AnalysisConsider the half-wave rectifier circuit shown below. The AC input voltage to the circuit is a 60 Hz signal from a power transformer. ![Circuit Diagram]- **AC Source**: Provides a 6.3 V (RMS) input.- **Diode (MR752)**: Rectifies the AC signal.- **Capacitor**: 47000 µF- **Load Resistor (\( R_L \))**: 10 Ω- **Output Voltage (\( V_{out} \))**#### Questions:a) **What is the peak output voltage across the load resistor, assuming \( R_L = 10 \, \Omega \)?** Account for the voltage drop across the diode (look it up!).b) **Estimate the voltage ripple, assuming the same load resistance.**#### Diagram Explanation:The diagram depicts a half-wave rectifier circuit where an AC input signal is fed into a diode, specifically an MR752. The diode allows only half of the AC wave to pass through to the load resistor (\( R_L \)), converting it to a DC signal with some ripple. A capacitor is connected in parallel with the output to reduce the ripple effect by temporarily storing charge and releasing it when needed. #### Analysis Steps:1. **Calculate the Peak Voltage**: Convert the RMS voltage to peak voltage using the formula \( V_{peak} = V_{rms} \times \sqrt{2} \).2. **Diode Voltage Drop**: Consider the typical forward voltage drop (\( V_{f} \)) of the MR752 diode. Subtract this from the peak voltage to find the actual peak output voltage across \( R_L \).3. **Estimate Voltage Ripple**: Use the formula for ripple voltage \( V_{ripple} = \dfrac{I}{f \cdot C} \), where \( I \) is the current, \( f \) is the frequency (60 Hz in this case), and \( C \) is the capacitance (47000 µF). This setup provides a basic understanding of how half-wave rectification works and the factors influencing the output voltage and ripple in such a circuit.

Circuit Vrms = 6.3 = Vm2Vm = 6.32c = 47 mF , Rl = 10Ω(a) Peak o|p voltage across load RL→During…

4) Consider the half-wave rectifier circuit shown below. The ac input voltage to the circuit is a 60 Hz signal from a power transformer. MR752 Vout +6.3 V (rms) 47000 µF R. a) What is the peak output voltage across the load resistor, assuming RL = 10 2? Account for the voltage drop across the diode (look it up!). b) Estimate the voltage ripple, assuming the same load resistance.

4) Consider the half-wave rectifier circuit shown below. The ac input voltage to the circuit is a 60 Hz signal from a power transformer. MR752 Vout +6.3 V (rms) 47000 µF R. a) What is the peak output voltage across the load resistor, assuming RL = 10 2? Account for the voltage drop across the diode (look it up!). b) Estimate the voltage ripple, assuming the same load resistance.

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...

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