A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor is

the ripple factor is 0.005

Explanation:

Given the data in the question;

we know that expression of ripple factor is;

r = Vr(pp) / Vdc

where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)

and Vdc is the dc value of the filter output voltage ( 20 V)

so we substitute our given values;

r = 0.1 / 20

r = 0.005

Therefore; the ripple factor is 0.005

answer:

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