Question:

A 1-\(\phi\) fully controlled bridge rectifier with a highly inductive load is fired at an angle of cos\(^{-1}(0.8)\). Assuming the load current is pure dc, the input power factor is

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In controlled rectifiers with inductive loads, true power factor is less than \(\cos(\alpha)\) due to current waveform distortion.
Updated On: May 23, 2025
  • 0.64
  • 0.72
  • 0.80
  • 0.50
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The Correct Option is B

Solution and Explanation

The firing angle \( \alpha \) is given by \[ \alpha = \cos^{-1}(0.8) = 36.87^\circ \] For a fully controlled rectifier with purely inductive load and constant DC output current, the input power factor is: \[ \text{PF} = \cos(\alpha) = \cos(36.87^\circ) = 0.8 \] However, due to displacement and distortion in the input current, the true input power factor becomes: \[ \text{PF}_{\text{true}} = 0.72 \]
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