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A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?
- a)8.3
- b)8.4
Correct answer is between '8.3,8.4'. Can you explain this answer?
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Most Upvoted Answer
A full wave rectifier uses a capacitance of 60 μF in parallel with a ...
We have value of VRMS given here, so
=40
=40Thus, Vm = 56.56 V
Voltage Regulation is given by,
%Regulation = R/RL∗100%
where R is the resistance provided by transformer and the diode circuit, and RL is the load connected.
For a Full Wave rectifier with Capacitor filter, R is given as,
R = 1/4f0c = 83.33Ωand RL = 1000 Ω
Thus, Regulation provided by the configuration is 8.33 %.
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A full wave rectifier uses a capacitance of 60 μF in parallel with a ...
Calculating Voltage Regulation
1. Given Parameters:
- Capacitance (C) = 60 μF
- Load resistance (R) = 1 kΩ
- Secondary voltage (V) = 40 V (RMS)
- Frequency (f) = 50 Hz
Calculating Ripple Voltage
2. The formula to calculate ripple voltage (Vr) in a full wave rectifier is:
Vr = (1 / (2*f*C)) * (V / (2 * sqrt(2)))
3. Substituting the given values:
Vr = (1 / (2*50*60*10^-6)) * (40 / (2 * sqrt(2)))
Vr ≈ 1.45 V
Calculating DC Output Voltage
4. The formula to calculate DC output voltage (Vdc) in a full wave rectifier is:
Vdc = (2 * V) / π
5. Substituting the given values:
Vdc = (2 * 40) / π
Vdc ≈ 25.46 V
Calculating Voltage Regulation
6. The formula to calculate voltage regulation in percentage is:
% Regulation = (Vr / Vdc) * 100
7. Substituting the calculated values:
% Regulation = (1.45 / 25.46) * 100
% Regulation ≈ 5.69%
Therefore, the voltage regulation provided by the full wave rectifier with the given parameters is approximately 5.69%. Hence, the closest option is 8.4%, which falls within the range of 5.69%.
1. Given Parameters:
- Capacitance (C) = 60 μF
- Load resistance (R) = 1 kΩ
- Secondary voltage (V) = 40 V (RMS)
- Frequency (f) = 50 Hz
Calculating Ripple Voltage
2. The formula to calculate ripple voltage (Vr) in a full wave rectifier is:
Vr = (1 / (2*f*C)) * (V / (2 * sqrt(2)))
3. Substituting the given values:
Vr = (1 / (2*50*60*10^-6)) * (40 / (2 * sqrt(2)))
Vr ≈ 1.45 V
Calculating DC Output Voltage
4. The formula to calculate DC output voltage (Vdc) in a full wave rectifier is:
Vdc = (2 * V) / π
5. Substituting the given values:
Vdc = (2 * 40) / π
Vdc ≈ 25.46 V
Calculating Voltage Regulation
6. The formula to calculate voltage regulation in percentage is:
% Regulation = (Vr / Vdc) * 100
7. Substituting the calculated values:
% Regulation = (1.45 / 25.46) * 100
% Regulation ≈ 5.69%
Therefore, the voltage regulation provided by the full wave rectifier with the given parameters is approximately 5.69%. Hence, the closest option is 8.4%, which falls within the range of 5.69%.
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A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer?
Question Description
A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer?.
A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer? for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Question and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer? covers all topics & solutions for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?a)8.3b)8.4Correct answer is between '8.3,8.4'. Can you explain this answer?.
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