[SOLVED] [MOVED] problem solving for my fellow

[yanda]

good day ya all my fellow; i've a minor problems In interpreting the following equations - i.e. to a lay man understanding so pls i would want somebody to do justice to them:

1. RC time constant for a fully-charged and discharged capacitor given as thus:

time constant = R . C

2. the equation for determining the capacitance value for a shunt capacitor filter, i.e.,

Vr(rms) = 2.4I/C
where Vr(rms) is the r.m.s ripple voltage, I - is the current flowing through the load and the C is the unknown capacitance value of the shunt filtered capacitor.

the question here is: how comes the multiplier (2.4)? why is 2.4 multiply ............ and why not just the variables or the variables multiply by the another constant else.

please i want a fellow to do justice to this/these question(s)

 

[BradtheRad]

1. RC time constant for a fully-charged and discharged capacitor given as thus:

time constant = R . C

It tells how long it will take to change 63.2 percent toward whatever is the applied voltage.

The convention is to say it will take 5 time constants to fully charge or discharge. (Although some might say 3 or 4 is enough.)

2. the equation for determining the capacitance value for a shunt capacitor filter, i.e.,

Vr(rms) = 2.4I/C
where Vr(rms) is the r.m.s ripple voltage, I - is the current flowing through the load and the C is the unknown capacitance value of the shunt filtered capacitor.

the question here is: how comes the multiplier (2.4)? why is 2.4 multiply ............ and why not just the variables or the variables multiply by the another constant else.

please i want a fellow to do justice to this/these question(s)

I suppose this pertains to a power supply?

The equation looks like a condensed result from one that started out having a few more variables.

Particularly the time that passes between each burst of current. This time will be based on frequency of operation (whether 50 or 60 Hz), and the shape of the waveform (half-wave or full-wave).

Say current bursts come in every 1/120 of a second. The capacitor (C) discharges into the load (R) during that time. Its volt level will drop according to the time constant R x C. The amount it drops will be in a proportion that involves the time constant, 1/120 second, and 63 percent.

Your formula appears to result from using a different frequency than the typical power supply.

Here is a website that discusses the formula.

http://waynestegall.com/audio/ripple.htm