Answer:
(a) 16.27 Vpk
(b) 48.7%
Explanation:
The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.
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(a)
The peak voltage will be 0.7 V less than the transformer secondary peak voltage:
((120 V)√2)/10 -0.7 V ≈ 16.27 V
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(b)
The fraction of the amplitude for which the diode is non-conducting is ...
0.7/(12√2) ≈ 0.041248
The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:
arccos(0.041248) ≈ 87.64°
As a fraction of half the cycle, this is ...
conduction fraction ≈ 87.64°/180° ≈ 48.7%
