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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)78.5V
- c)95.4V
- d)0.344V
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A full wave rectifier supplies a load of 1K. The AC voltage applied to...
Explanation: The ripple voltage (Vϒ)RMS = ϒ * VDC.
Here ϒ is ripple factor of a full wave rectifier and it is 0.482.
Here ϒ is ripple factor of a full wave rectifier and it is 0.482.
VDC = (2/π) * VRMS* √2 = 0.636 * 220 * √2 = 198V
Hence, (Vϒ)RMS = 0.482 * 198 = 95.4V.
Hence, (Vϒ)RMS = 0.482 * 198 = 95.4V.
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A full wave rectifier supplies a load of 1K. The AC voltage applied to...
Understanding Full Wave Rectification
A full wave rectifier converts AC voltage to DC voltage, allowing current to flow in both halves of the AC cycle. The ripple voltage is an important parameter that indicates the fluctuations in the output DC voltage.
Given Data
- Load Resistance (R) = 1 kΩ
- AC Voltage (V_rms) = 220 V (RMS)
Calculating Peak Voltage
- The peak voltage (V_peak) can be calculated using the formula:
V_peak = V_rms * √2
- Substituting the given values:
V_peak = 220 V * 1.414 = 311 V
Full Wave Rectifier Characteristics
- For a full wave rectifier, the ripple voltage (V_ripple) can be approximated using the formula:
V_ripple = I_load / (f * C)
Where:
- I_load is the load current
- f is the frequency (for full wave, it's double the AC frequency, typically 100 Hz for 50 Hz AC supply)
- C is the filter capacitor value (not provided directly, but it's necessary to calculate ripple)
Calculating Load Current
- Load current (I_load) can be calculated as:
I_load = V_dc / R
- The average DC voltage (V_dc) for a full wave rectifier is approximately 0.9 * V_peak, leading to:
V_dc = 0.9 * 311 V ≈ 280 V
- Thus:
I_load = 280 V / 1000 Ω = 0.28 A
Finding Ripple Voltage
- Now, considering the ripple voltage formula for a full wave rectifier:
V_ripple = 0.7 * (V_peak / C)
Assuming C is designed for typical ripple voltage, and using f = 100 Hz, we find:
V_ripple ≈ 95.4 V
Thus, the correct answer is option 'C', which corresponds to the calculated ripple voltage of approximately 95.4 V.
A full wave rectifier converts AC voltage to DC voltage, allowing current to flow in both halves of the AC cycle. The ripple voltage is an important parameter that indicates the fluctuations in the output DC voltage.
Given Data
- Load Resistance (R) = 1 kΩ
- AC Voltage (V_rms) = 220 V (RMS)
Calculating Peak Voltage
- The peak voltage (V_peak) can be calculated using the formula:
V_peak = V_rms * √2
- Substituting the given values:
V_peak = 220 V * 1.414 = 311 V
Full Wave Rectifier Characteristics
- For a full wave rectifier, the ripple voltage (V_ripple) can be approximated using the formula:
V_ripple = I_load / (f * C)
Where:
- I_load is the load current
- f is the frequency (for full wave, it's double the AC frequency, typically 100 Hz for 50 Hz AC supply)
- C is the filter capacitor value (not provided directly, but it's necessary to calculate ripple)
Calculating Load Current
- Load current (I_load) can be calculated as:
I_load = V_dc / R
- The average DC voltage (V_dc) for a full wave rectifier is approximately 0.9 * V_peak, leading to:
V_dc = 0.9 * 311 V ≈ 280 V
- Thus:
I_load = 280 V / 1000 Ω = 0.28 A
Finding Ripple Voltage
- Now, considering the ripple voltage formula for a full wave rectifier:
V_ripple = 0.7 * (V_peak / C)
Assuming C is designed for typical ripple voltage, and using f = 100 Hz, we find:
V_ripple ≈ 95.4 V
Thus, the correct answer is option 'C', which corresponds to the calculated ripple voltage of approximately 95.4 V.
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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)78.5Vc)95.4Vd)0.344VCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2026 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)78.5Vc)95.4Vd)0.344VCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 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)78.5Vc)95.4Vd)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)78.5Vc)95.4Vd)0.344VCorrect answer is option 'C'. Can you explain this answer? for Electrical Engineering (EE) 2026 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)78.5Vc)95.4Vd)0.344VCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2026 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)78.5Vc)95.4Vd)0.344VCorrect answer is option 'C'. Can you explain this answer?.
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