A certain full-wave rectifier has a peak output voltage of 30 V. A 50 mF capacitor-input filter is connected to the rectifier. Calculate the peak-to-peak ripple and the dc output voltage devel-oped across a 600 V load resistance.

Peak to peak Ripple voltage 8.33mV

DC output voltage =19.11 V

Explanation:

Peak voltage( Vp)= 30v

Load resistance =600 ohms

Capacitor filter = 50mF

Frequency of supply = 120Hz

The peak to peak Ripple is not only dependent on the capacitor value but also on the frequency and load current.

To calculate the,

Peak to peak ripple = I (load)/f×c

I (load) = loadd current = 30/600 = 0.05 A

Peak to peak ripple = 0.05/6

= 8.33mV

The average Dc output voltage for a full wave rectifier is double that of a half wave rectifier.

The DC output voltage is equal to 0.637Vp assuming no losses.

Vdc= 0.637 × 30

Vdc =19.11V

answer:

a

explanation:

answer:

31

explanation: