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A full wave rectifier supplies a load of 1KΩ. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?
- a)0.562V
- b)0.785V
- c)0.954V
- d)0.344V
Correct answer is option 'C'. Can you explain this answer?
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A full wave rectifier supplies a load of 1K. The AC voltage applied to...
To find the ripple voltage in a full wave rectifier, we need to consider the voltage drop across the diode during conduction. In a full wave rectifier, the diodes conduct during alternate half-cycles of the input AC voltage.
The peak voltage of the input AC voltage can be calculated using the formula:
Vp = Vrms * √2
Given that the RMS voltage is 220V, we can calculate the peak voltage as:
Vp = 220 * √2 ≈ 311.13V
During the positive half-cycle of the input voltage, the diode conducts and allows the positive half-cycle to pass through. The voltage across the load resistor is equal to the input voltage minus the voltage drop across the diode. The voltage drop across the diode is typically around 0.7V.
Therefore, the voltage across the load resistor during the positive half-cycle is:
Vp_load = Vp - Vdiode
Vp_load = 311.13V - 0.7V ≈ 310.43V
During the negative half-cycle of the input voltage, the diode is reverse-biased and does not conduct. Hence, the voltage across the load resistor is 0V.
Now, let's calculate the average voltage across the load resistor:
Vavg = (Vp_load + 0V) / 2
Vavg = 310.43V / 2 ≈ 155.21V
The ripple voltage is the difference between the peak voltage across the load resistor and the average voltage across the load resistor. Therefore, the ripple voltage can be calculated as:
Vripple = Vp_load - Vavg
Vripple = 310.43V - 155.21V ≈ 155.22V
Finally, converting the ripple voltage from peak-to-peak to peak, we divide it by 2:
Vripple_peak = Vripple / 2
Vripple_peak = 155.22V / 2 ≈ 77.61V
However, the question asks for the ripple voltage in RMS. To convert the peak ripple voltage to RMS, we divide it by the square root of 2:
Vripple_rms = Vripple_peak / √2
Vripple_rms = 77.61V / √2 ≈ 54.54V
Therefore, the ripple voltage in this scenario is approximately 54.54V.
The peak voltage of the input AC voltage can be calculated using the formula:
Vp = Vrms * √2
Given that the RMS voltage is 220V, we can calculate the peak voltage as:
Vp = 220 * √2 ≈ 311.13V
During the positive half-cycle of the input voltage, the diode conducts and allows the positive half-cycle to pass through. The voltage across the load resistor is equal to the input voltage minus the voltage drop across the diode. The voltage drop across the diode is typically around 0.7V.
Therefore, the voltage across the load resistor during the positive half-cycle is:
Vp_load = Vp - Vdiode
Vp_load = 311.13V - 0.7V ≈ 310.43V
During the negative half-cycle of the input voltage, the diode is reverse-biased and does not conduct. Hence, the voltage across the load resistor is 0V.
Now, let's calculate the average voltage across the load resistor:
Vavg = (Vp_load + 0V) / 2
Vavg = 310.43V / 2 ≈ 155.21V
The ripple voltage is the difference between the peak voltage across the load resistor and the average voltage across the load resistor. Therefore, the ripple voltage can be calculated as:
Vripple = Vp_load - Vavg
Vripple = 310.43V - 155.21V ≈ 155.22V
Finally, converting the ripple voltage from peak-to-peak to peak, we divide it by 2:
Vripple_peak = Vripple / 2
Vripple_peak = 155.22V / 2 ≈ 77.61V
However, the question asks for the ripple voltage in RMS. To convert the peak ripple voltage to RMS, we divide it by the square root of 2:
Vripple_rms = Vripple_peak / √2
Vripple_rms = 77.61V / √2 ≈ 54.54V
Therefore, the ripple voltage in this scenario is approximately 54.54V.
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A full wave rectifier supplies a load of 1K. The AC voltage applied to...
Silicon diode are used in a two diode full wave rectifier circuit to supply a load with 12v DC assuming ideal diodes and the resistance is 12ohms calculate
a) the transformer secondary voltage
b)the load ripple value
c) the efficiency of the rectifier
a) the transformer secondary voltage
b)the load ripple value
c) the efficiency of the rectifier
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A full wave rectifier supplies a load of 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect answer is option 'C'. Can you explain this answer?
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A full wave rectifier supplies a load of 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect 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 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect 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 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect answer is option 'C'. Can you explain this answer?.
A full wave rectifier supplies a load of 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect 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 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect 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 1K. The AC voltage applied to diodes is 220V (rms). If diode resistance is neglected, what is the ripple voltage?a)0.562Vb)0.785Vc)0.954Vd)0.344VCorrect answer is option 'C'. Can you explain this answer?.
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