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A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?
- a)114.87Ω
- b)167.98Ω
- c)115.47Ω
- d)451.35Ω
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A half wave rectifier, operated from a 50Hz supply uses a 1000µF...
For a half wave filter,
ϒ=1/2√3 fCRL=1/2√3*50*10-3*RL
RL=103/5√3=115.47Ω.
ϒ=1/2√3 fCRL=1/2√3*50*10-3*RL
RL=103/5√3=115.47Ω.
Most Upvoted Answer
A half wave rectifier, operated from a 50Hz supply uses a 1000µF...
ΜF capacitor and a load resistance of 1 kΩ. Determine the ripple voltage and the DC output voltage.
To find the ripple voltage, we need to calculate the capacitor discharge time. The formula for the discharge time is given by:
t = RC ln(Vi/Vf)
Where:
t = discharge time
R = load resistance = 1 kΩ
C = capacitance = 1000 μF
Vi = peak input voltage = Vp
Vf = final voltage = 0 V
Since we are operating from a 50Hz supply, the peak input voltage is given by:
Vp = √2 * Vrms
Where:
Vrms = root mean square voltage = supply voltage = Vm
Since we are not given the supply voltage, we'll assume it to be 230 V (typical for many countries). Therefore,
Vp = √2 * 230 V = 325.3 V (approx)
Now, let's calculate the discharge time:
t = (1 kΩ) * (1000 μF) * ln(325.3 V / 0 V)
t ≈ 6.91 seconds
The ripple voltage (Vr) can be calculated using the formula:
Vr = (I / 2) * t / C
Where:
I = load current = Vp / R
I = 325.3 V / 1 kΩ
I = 0.3253 A
Vr = (0.3253 A / 2) * (6.91 seconds) / (1000 μF)
Vr ≈ 1.124 V
Finally, the DC output voltage (Vdc) can be calculated using the formula:
Vdc = Vp - Vr
Vdc = 325.3 V - 1.124 V
Vdc ≈ 324.18 V
Therefore, the ripple voltage is approximately 1.124 V and the DC output voltage is approximately 324.18 V.
To find the ripple voltage, we need to calculate the capacitor discharge time. The formula for the discharge time is given by:
t = RC ln(Vi/Vf)
Where:
t = discharge time
R = load resistance = 1 kΩ
C = capacitance = 1000 μF
Vi = peak input voltage = Vp
Vf = final voltage = 0 V
Since we are operating from a 50Hz supply, the peak input voltage is given by:
Vp = √2 * Vrms
Where:
Vrms = root mean square voltage = supply voltage = Vm
Since we are not given the supply voltage, we'll assume it to be 230 V (typical for many countries). Therefore,
Vp = √2 * 230 V = 325.3 V (approx)
Now, let's calculate the discharge time:
t = (1 kΩ) * (1000 μF) * ln(325.3 V / 0 V)
t ≈ 6.91 seconds
The ripple voltage (Vr) can be calculated using the formula:
Vr = (I / 2) * t / C
Where:
I = load current = Vp / R
I = 325.3 V / 1 kΩ
I = 0.3253 A
Vr = (0.3253 A / 2) * (6.91 seconds) / (1000 μF)
Vr ≈ 1.124 V
Finally, the DC output voltage (Vdc) can be calculated using the formula:
Vdc = Vp - Vr
Vdc = 325.3 V - 1.124 V
Vdc ≈ 324.18 V
Therefore, the ripple voltage is approximately 1.124 V and the DC output voltage is approximately 324.18 V.
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A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?a)114.87Ωb)167.98Ωc)115.47Ωd)451.35ΩCorrect answer is option 'C'. Can you explain this answer?
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A half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?a)114.87Ωb)167.98Ωc)115.47Ωd)451.35ΩCorrect 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 half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?a)114.87Ωb)167.98Ωc)115.47Ωd)451.35ΩCorrect 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 half wave rectifier, operated from a 50Hz supply uses a 1000µF capacitance connected in parallel to the load of rectifier. What will be the minimum value of load resistance that can be connected across the capacitor if the ripple% not exceeds 5?a)114.87Ωb)167.98Ωc)115.47Ωd)451.35ΩCorrect answer is option 'C'. Can you explain this answer?.
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