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A single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)
Correct answer is between '19.6,19.9'. Can you explain this answer?
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A single-phase AC supply is connected to a full wave rectifier through...
Concept:
In a full wave rectifier though a transformer,
Average output voltage, V0 = 2Vm/π
Average output current, I0 = V0/R
RMS value of output voltage Vor = Vs
RMS value of load current, Ior = Vs/R
Average value of diode current, Id = Im/2
RMS value of diode current, Idr = Ior
Peak value of diode current, Idm = √2 Ior
Power delivered to the load = Vor Ior
Input voltamperes = Vs Ior
Calculation:
Given that, DC output voltage (VDC) = 400 V
Average output voltage of rectifier (V0) = 400 V
⇒ Vs = 444.28 V
Load resistance (R) = 10 Ω
RMS value of load current, Ior = 444.2810 = 44.428A
kVA rating of transformer = 444.28 × 44.428 = 19.738 kVA
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A single-phase AC supply is connected to a full wave rectifier through...
Ω. The transformer has a turns ratio of 1:10. Determine the peak voltage of the AC supply.
To find the peak voltage of the AC supply, we can first find the peak voltage of the DC output and then use the relationship between the AC and DC voltages.
The average DC output voltage is given by:
Vdc = (2/π) * Vpk
where Vpk is the peak voltage of the DC output.
Rearranging the equation, we can solve for Vpk:
Vpk = (Vdc * π) / 2
Substituting the given values, we have:
Vpk = (400 V * π) / 2
Vpk ≈ 628.32 V
Since the transformer has a turns ratio of 1:10, the peak voltage of the AC supply is 10 times the peak voltage of the DC output:
Vpk(AC) = 10 * Vpk
Substituting the value of Vpk, we get:
Vpk(AC) ≈ 10 * 628.32 V
Vpk(AC) ≈ 6283.2 V
Therefore, the peak voltage of the AC supply is approximately 6283.2 V.
To find the peak voltage of the AC supply, we can first find the peak voltage of the DC output and then use the relationship between the AC and DC voltages.
The average DC output voltage is given by:
Vdc = (2/π) * Vpk
where Vpk is the peak voltage of the DC output.
Rearranging the equation, we can solve for Vpk:
Vpk = (Vdc * π) / 2
Substituting the given values, we have:
Vpk = (400 V * π) / 2
Vpk ≈ 628.32 V
Since the transformer has a turns ratio of 1:10, the peak voltage of the AC supply is 10 times the peak voltage of the DC output:
Vpk(AC) = 10 * Vpk
Substituting the value of Vpk, we get:
Vpk(AC) ≈ 10 * 628.32 V
Vpk(AC) ≈ 6283.2 V
Therefore, the peak voltage of the AC supply is approximately 6283.2 V.
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A single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)Correct answer is between '19.6,19.9'. Can you explain this answer?
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A single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)Correct answer is between '19.6,19.9'. 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 single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)Correct answer is between '19.6,19.9'. 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 single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)Correct answer is between '19.6,19.9'. Can you explain this answer?.
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