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The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________
- a)Vm/√2RL
- b)Vm/(RF+RL)π
- c)2Vm/√π
- d)Vm/RL
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The diode in a half wave rectifier has a forward resistance RF. The vo...
For a half wave rectifier, the IDC=IAVG=Im/π
I= Vmsinωt/(RF+RL)=Imsinωt
Im =Vm/ RF+RL So, IDC=Im/π=Vm/(RF+RL).
I= Vmsinωt/(RF+RL)=Imsinωt
Im =Vm/ RF+RL So, IDC=Im/π=Vm/(RF+RL).
Most Upvoted Answer
The diode in a half wave rectifier has a forward resistance RF. The vo...
(ωt), where Vm is the maximum voltage and ω is the angular frequency.
During the positive half cycle of the input voltage, the diode is forward biased and conducts current. The output voltage across the load resistor RL is equal to the voltage drop across RL, which is given by:
Vout = Vm - Vf - RL*I
where Vf is the forward voltage drop across the diode and I is the current flowing through the diode and RL.
During the negative half cycle of the input voltage, the diode is reverse biased and does not conduct current. Therefore, the output voltage is zero.
The average output voltage of the half wave rectifier can be calculated using the following equation:
Vavg = (1/π) * ∫0π Vout(t) dt
where π is the value of pi and the integral is taken over one cycle of the input voltage.
Substituting the expression for Vout(t) and simplifying, we get:
Vavg = (Vm/π) * [1 - (2/π)*arctan(RL/RF)]
where arctan is the inverse tangent function.
From this equation, we can see that the average output voltage depends on the maximum input voltage, the load resistor RL, and the forward resistance of the diode RF. As RL increases, the output voltage decreases and as RF increases, the output voltage increases.
In summary, the half wave rectifier converts an AC input voltage into a pulsating DC output voltage. The output voltage depends on the maximum input voltage, the load resistor, and the forward resistance of the diode.
During the positive half cycle of the input voltage, the diode is forward biased and conducts current. The output voltage across the load resistor RL is equal to the voltage drop across RL, which is given by:
Vout = Vm - Vf - RL*I
where Vf is the forward voltage drop across the diode and I is the current flowing through the diode and RL.
During the negative half cycle of the input voltage, the diode is reverse biased and does not conduct current. Therefore, the output voltage is zero.
The average output voltage of the half wave rectifier can be calculated using the following equation:
Vavg = (1/π) * ∫0π Vout(t) dt
where π is the value of pi and the integral is taken over one cycle of the input voltage.
Substituting the expression for Vout(t) and simplifying, we get:
Vavg = (Vm/π) * [1 - (2/π)*arctan(RL/RF)]
where arctan is the inverse tangent function.
From this equation, we can see that the average output voltage depends on the maximum input voltage, the load resistor RL, and the forward resistance of the diode RF. As RL increases, the output voltage decreases and as RF increases, the output voltage increases.
In summary, the half wave rectifier converts an AC input voltage into a pulsating DC output voltage. The output voltage depends on the maximum input voltage, the load resistor, and the forward resistance of the diode.
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The diode in a half wave rectifier has a forward resistance RF. The vo...
In practical condition forward biased half wave rectifier have a resistance of RF series with load resistance(i.e.RF+RL). DC or average current= Vdc/(RF+RL).since Vdc=vm/π so DC or average current is Idc=vm/(RF+RL)π~> option (b)
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The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________a)Vm/√2RLb)Vm/(RF+RL)πc)2Vm/√πd)Vm/RLCorrect answer is option 'B'. 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 The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________a)Vm/√2RLb)Vm/(RF+RL)πc)2Vm/√πd)Vm/RLCorrect answer is option 'B'. 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 The diode in a half wave rectifier has a forward resistance RF. The voltage is Vmsinωt and the load resistance is RL. The DC current is given by _________a)Vm/√2RLb)Vm/(RF+RL)πc)2Vm/√πd)Vm/RLCorrect answer is option 'B'. Can you explain this answer?.
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