The ripple factor for a half-wave rectifier with capacitor is given by:

Question

Question

This question was previously asked in

ALP CBT 2 Electrician Official Paper (Held On: 23 Jan 2019 Shift 3)

Answer 1

Concept:

Ripple factor for Half wave rectifier with capacitor filter

\(Ripple\;factor,(r) = \frac{1}{{2\sqrt 3 fC{R_L}}}\)

Ripple factor for full wave rectifier with capacitor filter

\(Ripple\;factor,(r) = \frac{1}{{4\sqrt 3 fC{R_L}}}\)

The arrangement of a rectifier with a capacitor filter is shown below:

F3 S.B 6.5.20 Pallavi D1

Values of XC and R are selected such that |XC| < < RL

i.e, \(\frac{1}{{{\omega _o}C}} < < {R_L}\) for Half Wave Rectifier

\(\frac{1}{{2{\omega _o}C}} < < {R_L}\) for Full Wave Rectifier

As the capacitor (C) and RL are in parallel, they form a current divider circuit.
For high frequencies the capacitor will act as a short circuit, i.e. the majority of current Iac flows through the capacitor and a very small AC current goes through RL.
Similarly, since the capacitor behaves as an open circuit for DC, IDC will flow through RL.

The following table shows the different Relations for FWR and HWR.