Find the peak-to-peak ripple voltage of half-wave rectifier and f
| Find the peak-to-peak ripple voltage of half-wave rectifier and filter circuit which has a 680 μF filter capacitor, an average output voltage of 30 V, and a 220Ω load resistance. The mains frequency is 50 Hz.
A. 3.51 V
B. 4.01 V
C. 3.83 V
D. 2.84 V
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
The Peak to peak ripple voltage of Half wave rectifies is given by:
\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}}\)
Where, Idc = Average current
f0 = mains freqeucny
C = capacitor
Calculation:
Given f0 = 50 Hz, C = 680 μf, Vdc = 30 V, and RL = 220 Ω, we get:
\({I_{dc}} = \frac{{{v_{dc}}}}{{{R_L}}} = \frac{{30}}{{220}}\)
Idc = 0.136
\({V_r} = \frac{{{I_{dc}}}}{{{f_{oc}}}} = \frac{{.136}}{{50 \times 680 \times {{10}^{ - 6}}}}\)
\({V_r} = 4.01\;V\)
