Electrical Engineering (EE) Exam > Electrical Engineering (EE) Questions > A full wave rectifier supplies a load of 1 kΩ...
Start Learning for Free
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltage
- a)220 V
- b)200 V
- c)198 V
- d)0 volts
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applie...
.Most Upvoted Answer
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applie...
Problem Statement: A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. Neglecting diode resistance, find the average d.c. voltage.
Solution:
Full Wave Rectifier: A full wave rectifier is a circuit that converts the entire input waveform into a pulsating DC waveform. In a full wave rectifier, the input AC voltage is applied to the diodes in such a way that both the positive and negative half cycles are rectified. The output waveform of the full wave rectifier is shown below:
Average DC voltage: The average DC voltage of a rectifier is the DC equivalent of the input AC voltage. The average DC voltage can be calculated using the following formula:
Vavg = (2Vm/π)
where Vm is the maximum value of the input AC voltage.
Given:
Load resistance, R = 1 kΩ
AC voltage applied to the diodes, Vrms = 220 - 0 - 220 Volts
Neglect diode resistance
Calculation:
The maximum value of the input AC voltage is given by:
Vm = Vrms * √2
Vm = 220 * √2
Vm = 311.13 V
The average DC voltage can be calculated using the formula:
Vavg = (2Vm/π)
Vavg = (2 * 311.13)/π
Vavg = 198.05 V
Therefore, the average DC voltage is 198 V.
Answer: option (c)
Solution:
Full Wave Rectifier: A full wave rectifier is a circuit that converts the entire input waveform into a pulsating DC waveform. In a full wave rectifier, the input AC voltage is applied to the diodes in such a way that both the positive and negative half cycles are rectified. The output waveform of the full wave rectifier is shown below:
Average DC voltage: The average DC voltage of a rectifier is the DC equivalent of the input AC voltage. The average DC voltage can be calculated using the following formula:
Vavg = (2Vm/π)
where Vm is the maximum value of the input AC voltage.
Given:
Load resistance, R = 1 kΩ
AC voltage applied to the diodes, Vrms = 220 - 0 - 220 Volts
Neglect diode resistance
Calculation:
The maximum value of the input AC voltage is given by:
Vm = Vrms * √2
Vm = 220 * √2
Vm = 311.13 V
The average DC voltage can be calculated using the formula:
Vavg = (2Vm/π)
Vavg = (2 * 311.13)/π
Vavg = 198.05 V
Therefore, the average DC voltage is 198 V.
Answer: option (c)
Attention Electrical Engineering (EE) Students!
To make sure you are not studying endlessly, EduRev has designed Electrical Engineering (EE) study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Electrical Engineering (EE).
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Similar Electrical Engineering (EE) Doubts
Top Courses for Electrical Engineering (EE)View all
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer?
Question Description
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer?.
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE).
Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.
Here you can find the meaning of A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A full wave rectifier supplies a load of 1 kΩ. The a.c. Voltage applied to the diodes is 220 - 0 - 220 Volts rms. If diode resistance is neglected, then Average d.c. Voltagea)220 Vb)200 Vc)198 Vd)0 voltsCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.
|
|
Explore Courses for Electrical Engineering (EE) exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup