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A Voltage 200 cos 100 t is applied to a half wave rectifier with load resistance of 5kΩ . Rectifier may be represented by an ideal diode in series with a resistance of 1kΩ
Then which of the following statements is/are correct :-
Then which of the following statements is/are correct :-
- a)Rms value of current = 16.67 mA
- b)dc value of current = 10.61 mA
- c)Rectifier Efficiency is 40.6%
- d)Ripple factor = 1.21
Correct answer is option 'A,B,D'. Can you explain this answer?
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A Voltage 200 cos 100 tis applied to a half wave rectifier with load r...
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Assuming that the diode in the half wave rectifier is ideal, the output voltage across the load resistance can be calculated as follows:
During the positive half cycle of the input voltage, the diode is forward biased and conducts current through the load resistance. The voltage across the load resistance can be approximated as the peak value of the input voltage multiplied by the diode voltage drop (typically around 0.7 V), minus the voltage drop across the diode when it is conducting:
Vout = (Vp * 0.7) - 0 = (200 * sqrt(2) * 0.7) - 0 = 197.96 V
During the negative half cycle of the input voltage, the diode is reverse biased and blocks current flow through the load resistance. Therefore, the output voltage is zero.
Therefore, the output voltage waveform will be a half-wave rectified sine wave with an amplitude of 197.96 V and a frequency of 100 Hz. The average value of the output voltage can be calculated as:
Vavg = (1/π) * ∫0^π (Vp * sin(ωt)) dt = (Vp/π) * [-cos(ωt)]0^π = (Vp/π) * (1 - (-1)) = (2Vp/π) = (2 * 200 * sqrt(2)/π) = 90.17 V
So the average output voltage is 90.17 V.
During the positive half cycle of the input voltage, the diode is forward biased and conducts current through the load resistance. The voltage across the load resistance can be approximated as the peak value of the input voltage multiplied by the diode voltage drop (typically around 0.7 V), minus the voltage drop across the diode when it is conducting:
Vout = (Vp * 0.7) - 0 = (200 * sqrt(2) * 0.7) - 0 = 197.96 V
During the negative half cycle of the input voltage, the diode is reverse biased and blocks current flow through the load resistance. Therefore, the output voltage is zero.
Therefore, the output voltage waveform will be a half-wave rectified sine wave with an amplitude of 197.96 V and a frequency of 100 Hz. The average value of the output voltage can be calculated as:
Vavg = (1/π) * ∫0^π (Vp * sin(ωt)) dt = (Vp/π) * [-cos(ωt)]0^π = (Vp/π) * (1 - (-1)) = (2Vp/π) = (2 * 200 * sqrt(2)/π) = 90.17 V
So the average output voltage is 90.17 V.
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A Voltage 200 cos 100 tis applied to a half wave rectifier with load resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1kΩThen which of the following statements is/are correct :-a)Rms value of current = 16.67 mAb)dc value of current = 10.61 mAc)Rectifier Efficiency is 40.6%d)Ripple factor = 1.21Correct answer is option 'A,B,D'. Can you explain this answer?
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A Voltage 200 cos 100 tis applied to a half wave rectifier with load resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1kΩThen which of the following statements is/are correct :-a)Rms value of current = 16.67 mAb)dc value of current = 10.61 mAc)Rectifier Efficiency is 40.6%d)Ripple factor = 1.21Correct answer is option 'A,B,D'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A Voltage 200 cos 100 tis applied to a half wave rectifier with load resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1kΩThen which of the following statements is/are correct :-a)Rms value of current = 16.67 mAb)dc value of current = 10.61 mAc)Rectifier Efficiency is 40.6%d)Ripple factor = 1.21Correct answer is option 'A,B,D'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Voltage 200 cos 100 tis applied to a half wave rectifier with load resistance of 5kΩ. Rectifier may be represented by an ideal diode in series with a resistance of 1kΩThen which of the following statements is/are correct :-a)Rms value of current = 16.67 mAb)dc value of current = 10.61 mAc)Rectifier Efficiency is 40.6%d)Ripple factor = 1.21Correct answer is option 'A,B,D'. Can you explain this answer?.
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