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The ripple factor of a full-wave bridge rectifier without filter circuit is equal to:

A0.707
B1.21
C0.406
D0.482

Explanation

For full-wave rectifiers, ripple factor r = sqrt((V_rms / V_dc)^2 – 1) = sqrt(1.11^2 – 1) ≈ 0.482 (compared to 1.21 for half-wave).

Submitted by: mcqstutor Team More Electrical Engineering MCQs →

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