Answer: 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
The ripple factor of the given power supply filter is; γ = 0.005
We are given;
Ripple peak to peak voltage; V_r,pp = 100 mV = 0.1 V
DC value of the filter output voltage; V_dc = 20 V
Now, formula for the ripple factor is given as;
γ = V_rms/V_dc
Now, our ripple peak to peak voltage is also known as the rms value of ripple voltage at the output. Thus;
γ = 0.1/20
γ = 0.005
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