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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?
    Verified Answer
    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
    ∴ 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.
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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?
    Question Description
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