a full wave rectifier operated from a 60 hz.. find Vpp

Question

A full-wave rectifier is operated from a

60-Hz

50-V-rms source.

It has a 500-uF filter capacitor and a

750-omh load.

objective:

1- find V_avg

2- find V peak to peak

3- find percent ripple

Answer 1

1) V_avg = 0.9*V_rms = 0.9*50 = 45 V_avg

2) V_p = 2^0.5*V_rms = 2^0.5*50 = 70.7 V_p ∴ Vpp = 2*V_p =2*70.7 = 141.4 Vpp

3) V_ripple = V_DC/(2RfC) = 45/(2*750*60*500*10^-6) = 1 V_ripple

4) Percent Ripple = V_ripple/V_avg * 100 = 1/45 *100 = 2.2%