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A fully controlled natural commutated 3-Φ bridge rectifier is operating with a firing angle, α = 300 . The peak to peak ripple voltage,if the maximum phase voltage is 400 V,will be
  • a)
    400√3
  • b)
    400
  • c)
    200√3
  • d)
    200
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A fully controlled natural commutated 3-Φ bridge rectifier is operati...
Given:

  • Firing angle, α = 300

  • Maximum phase voltage, Vm = 400 V



To find:

  • Peak to peak ripple voltage



Concepts Used:

  • For a fully controlled natural commutated 3-Φ bridge rectifier, the output voltage V0 can be given as:


V0 = √2Vmcosα

Calculation:

  • Given firing angle, α = 300

  • Maximum phase voltage, Vm = 400 V

  • Using the above formula, we can find the output voltage as:


V0 = √2Vmcosα
= √2 x 400 x cos300
= √2 x 400 x (0.866)
= 400√3 V

Peak to Peak Ripple Voltage:

  • For a fully controlled natural commutated 3-Φ bridge rectifier, the peak to peak ripple voltage can be given as:


Vrpp = V0 / √3

  • Substituting the value of V0, we get:


  • Vrpp = 400√3 / √3
    = 400 V

    Answer:

    • Peak to peak ripple voltage = 400 V

    Free Test
    Community Answer
    A fully controlled natural commutated 3-Φ bridge rectifier is operati...
    Given
    3-Φ bridge rectifier
    α = 300
    Vmph = 400 V
    As 3-Φ full converter conducts from
    60 + α to120 + α for continuous conduction ( α < />0)
    60 + α to 180 for discontinuous conduction ( α > 600)
    So, as α = 300(< />0) so thyristor will conduct from 60 + α to 120 + α i.e., 900 to 1500
    Taking VAB as the reference voltage.
    Peak to peak ripple voltage
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    Question Description
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