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A single phase semiconductor, connected to 220 V 50 Hz Is feeding load R = 15 Ω in series with a large inductance that makes the load current ripple free for firing angle of 60°, calculate the Rectification efficiency ______ in %.
Correct answer is between '55,58'. Can you explain this answer?
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A single phase semiconductor, connected to 220 V 50 Hz Is feeding load...
Concept
R.M.S value of output voltage is
Calculation
The load current is Ripple free
R.M.S value of output voltage isCalculation
The load current is Ripple free
∴ Ior = 197.32/15 = 13.15 A
Pdc = VoIo = 148.63 × 9.91 = 1472.93 W
Pac = VorIor = 197.32 × 13.15 = 2595.46 W
Rectification efficiency = 
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A single phase semiconductor, connected to 220 V 50 Hz Is feeding load...
Calculating Rectification Efficiency:
- Given Parameters:
- Voltage (V) = 220 V
- Frequency (f) = 50 Hz
- Load Resistance (R) = 15 Ω
- Firing Angle (α) = 60°
- Calculating Average Output Voltage (Vdc):
- Vdc = (2/π) * Vm * (1 + cosα), where Vm = peak voltage = V/√2
- Vdc = (2/π) * (220/√2) * (1 + cos60°) = 169.7 V
- Calculating DC Output Power (Pdc):
- Pdc = (Vdc^2) / R
- Pdc = (169.7^2) / 15 = 1915.29 W
- Calculating AC Input Power (Pac):
- Pac = Vrms * Irms
- Irms = Im / √2, where Im = peak current
- Irms = (Vm/R) / √2 = (220/15) / √2 = 10.41 A
- Pac = (220 * 10.41) = 2290.2 W
- Calculating Rectification Efficiency:
- Efficiency = (Pdc / Pac) * 100
- Efficiency = (1915.29 / 2290.2) * 100 = 83.7%
Therefore, the rectification efficiency for the given setup is 83.7%, which does not fall within the range of 55-58% as stated in the question.
- Given Parameters:
- Voltage (V) = 220 V
- Frequency (f) = 50 Hz
- Load Resistance (R) = 15 Ω
- Firing Angle (α) = 60°
- Calculating Average Output Voltage (Vdc):
- Vdc = (2/π) * Vm * (1 + cosα), where Vm = peak voltage = V/√2
- Vdc = (2/π) * (220/√2) * (1 + cos60°) = 169.7 V
- Calculating DC Output Power (Pdc):
- Pdc = (Vdc^2) / R
- Pdc = (169.7^2) / 15 = 1915.29 W
- Calculating AC Input Power (Pac):
- Pac = Vrms * Irms
- Irms = Im / √2, where Im = peak current
- Irms = (Vm/R) / √2 = (220/15) / √2 = 10.41 A
- Pac = (220 * 10.41) = 2290.2 W
- Calculating Rectification Efficiency:
- Efficiency = (Pdc / Pac) * 100
- Efficiency = (1915.29 / 2290.2) * 100 = 83.7%
Therefore, the rectification efficiency for the given setup is 83.7%, which does not fall within the range of 55-58% as stated in the question.
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A single phase semiconductor, connected to 220 V 50 Hz Is feeding load R = 15 Ω in series with a large inductance that makes the load current ripple free for firing angle of 60°, calculate the Rectification efficiency ______ in %.Correct answer is between '55,58'. Can you explain this answer?
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A single phase semiconductor, connected to 220 V 50 Hz Is feeding load R = 15 Ω in series with a large inductance that makes the load current ripple free for firing angle of 60°, calculate the Rectification efficiency ______ in %.Correct answer is between '55,58'. 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 semiconductor, connected to 220 V 50 Hz Is feeding load R = 15 Ω in series with a large inductance that makes the load current ripple free for firing angle of 60°, calculate the Rectification efficiency ______ in %.Correct answer is between '55,58'. 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 semiconductor, connected to 220 V 50 Hz Is feeding load R = 15 Ω in series with a large inductance that makes the load current ripple free for firing angle of 60°, calculate the Rectification efficiency ______ in %.Correct answer is between '55,58'. Can you explain this answer?.
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