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Diodes and Applications - Technical MCQs

Q1:

If a 169.7 V half-wave peak has an average voltage of 54 V, what is the average of two full-wave peaks?

A 119.9 V

B 108.0 V

C 115.7 V

D 339.4 V

ANS:B - 108.0 V

Here half wave rectification is:

Vdc = 0.318*V(peak).

e.g: 169.7*0.318 = 53.914 = 54(approx).

And for full wave rectification:

Vdc = 0.636*V(peak).

So,
Vdc = 0.636*169.7 = 107.9292 = 108V(approx).