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A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?
- a)The dc output power of the half wave rectifier is 5.63 watt.
- b)The ac input power of the half wave rectifier is 1.667 watt.
- c)Rectifier efficiency is 33.83%
- d)Ripple factor is 1.21
Correct answer is option 'B,C,D'. Can you explain this answer?
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To find the output voltage of the rectifier, we need to consider the operation of the half-wave rectifier.
In a half-wave rectifier, only the positive half of the input waveform is passed through while the negative half is blocked. This is achieved by using a diode in series with the load resistance.
Given that the input voltage is 200(100t), we need to determine the output voltage when the diode is forward-biased (on) and when it is reverse-biased (off).
When the diode is forward-biased (on), it conducts and allows current to flow through the load resistance. The output voltage is equal to the input voltage during this period.
Thus, when the input voltage is positive (100t > 0), the output voltage is equal to the input voltage: Vout = Vin = 200(100t).
When the diode is reverse-biased (off), it blocks current flow and the output voltage is zero.
Thus, when the input voltage is negative (100t < 0),="" the="" output="" voltage="" is="" zero:="" vout="" />
Therefore, the output voltage of the half-wave rectifier with a load resistance of 5k is given by:
Vout = 200(100t) for 100t > 0
Vout = 0 for 100t < 0="" />
In a half-wave rectifier, only the positive half of the input waveform is passed through while the negative half is blocked. This is achieved by using a diode in series with the load resistance.
Given that the input voltage is 200(100t), we need to determine the output voltage when the diode is forward-biased (on) and when it is reverse-biased (off).
When the diode is forward-biased (on), it conducts and allows current to flow through the load resistance. The output voltage is equal to the input voltage during this period.
Thus, when the input voltage is positive (100t > 0), the output voltage is equal to the input voltage: Vout = Vin = 200(100t).
When the diode is reverse-biased (off), it blocks current flow and the output voltage is zero.
Thus, when the input voltage is negative (100t < 0),="" the="" output="" voltage="" is="" zero:="" vout="" />
Therefore, the output voltage of the half-wave rectifier with a load resistance of 5k is given by:
Vout = 200(100t) for 100t > 0
Vout = 0 for 100t < 0="" />
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A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer?
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A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer?.
A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer?.
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A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer?, a detailed solution for A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? has been provided alongside types of A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A voltage 200(100t) is applied to a half wave rectifier with a load Resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1 kΩ. Which of the following statement is/are correct?a)The dc output power of the half wave rectifier is 5.63 watt.b)The ac input power of the half wave rectifier is 1.667 watt.c)Rectifier efficiency is 33.83%d)Ripple factor is 1.21Correct answer is option 'B,C,D'. Can you explain this answer? tests, examples and also practice Physics tests.
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