196 Electrical Power Systems Technology
We can also express the amount of ripple of a 60-Hz full-wave filter-
capacitor circuit with a light load as:
2.4Idc
Vr(rms) = ———
C
Sample Problem:
Given: a power supply has a DC load current of 800 mA, and a ca-
pacitor filter of 300 uF is used.
Find: the value of ripple voltage.
Solution:
2.4Idc
Vr(rms) = ———
C
2.4 × 800 mA
= ——————
300 uF
2.4
Vr(rms) = ———
RLC
2.4
r = ———
RLC
where:
Idc = the load current in milliamperes,
C = the filter capacitor value in microfarads, and
RL = the load resistance in kilohms.
Sample Problem:
Given: a power supply has a DC load resistance of 500 ohms con-
nected to its output, and a capacitor filter 100 uF is used.
Find: the value of ripple voltage.
Solution:
2.4
r = ———
RLC
2.4
= ——————
0.5 K × 10.0 uF
r = 0.00048 = 0.048%