A full-wave rectifier circuit is driven by a sinusoidal input voltage with Vrms=15V and frequency 50Hz. If the load resistance is 100Ω, what is the ripple voltage with a filter capacitance of 1.5mF? (Assume the diodes to be ideal, with Von=0V.)

Answer 1

T = 1/F = 1/(2*50) = 0.01 s. = Period.

R*C = 100 * 1.5*10^(-6) = 1.50*10^-4 s.

T/RC = 0.01/1.5*10^-4 = 66.7.
Your RC time constant is much too short to be used as a filter at 50 Hz!
The T/RC ratio should be less than 1; it is 66.7!

Let C = 500uF.
T/RC = 0.01/(100*5*10^-4) = 0.2.

Vo min = Vomax/e^(t/RC) = 10.6/e^0.2 = 8.68 Volts.

Vr = (Vomax-Vomin)/2 = (10.6-8.68)/2 = 0.96 Volts, peak. = Ripple voltage.

Vo = (V0max+Vomin)/2 = 10.6+8.68)/2 = 9.64 Volts, D.C.

Answer 2

when getting the period, why did you multiple by 2?