Re: Series-Parallel DC RC circuit

On Sun, 08 Mar 2009 14:25:54 -0700, John Larkin
<jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:

On Sun, 8 Mar 2009 14:08:03 -0700, "bg" <bg@xxxxxxxxxx> wrote:


kayvee wrote in message ...

+---------R1-------*-------*
| |
R2 C1
| |
GND------------------*-------*

Hi, I'm am interested in finding the voltage across the capacitor in
this DC circuit in time. Initially, when the capacitor is uncharged,
I know that the voltage across it is 0, and as time goes on the charge
will build and the voltage across it will as well. But, how do I
calculate the voltage across it as a function of time?

Thanks.

When you charge a capacitor from a DC source, through a resistor, the
voltage across the capacitor rises. The rise of the capacitor voltage with
time plots according to the equation -

Vc = E (1 - e^ (-t/RC))

Vc = voltage across the cap
E = applied DC voltage
e = epsilon = base of natural logarithms = 2.71828
t = elapsed time
R = resistance in ohms
c = capacitance in farads

If you make your applied voltage equal to 1, then the capacitor voltage Vc
is equal to a fraction or percentage of the applied voltage for any given
time that has elapsed. This is called a universal time constant chart. What
makes it universal, is that the shape of the curve never changes, only the
values for time and percent change.
RC is called a time constant. RC is actually equal to time in this instance.
So you can mark the X axis in units of time or RC units. The y axis is
marked in percent or as a decimal part of E applied

Now if you plot this equation -
Vc = 1 - e^ (-t/RC)
you get a universal time constant chart of the capacitor voltage expressed
as a percentage or decimal part of the applied voltage for the elapsed time
in RC units.

Examples ----
If you allowed the capacitor to charge for 1RC unit then the capacitor
voltage will rise to 63% of the applied voltage.
2 RC units = 86%
3 RC units = 95%
etc

Your circuit has a shunt resistor and a series resistor. The capacitor
charges through the series resistor, and it is the value of the series
resistor that you should use for the RC units.

No. The Thevenin equivalent resistor value sets the time constant, and
that's the equivalent value of the two resistors in parallel.

If V=1, R1=R2=1K and C=1uF, C will charge to 0.5 volts with a time
constant of 500 microseconds.

Well, he didn't specify what R was, as I read it (I may have missed
something, of course.) If it was R_th, and if you read his use of
"applied voltage" as being V_th, then it could be read as equivalent.
I didn't like the "Vc = 1 - e^ (-t/RC)" part, as the right hand
expression is usually meant to be taken to be unitless when written
like that. Otherwise, it's only true if the asymptote voltage at
t=infinity is 1V. Further, he writes "RC is actually equal to time
__in this instance__" when, in fact, the dimensions of RC is _time_ no
matter the instance.
(Joule-second/Coulomb^2) * (Coulomb^2/Joule) = seconds
This is a good thing, because e^(t/RC) then means e is raised to a
unitless power, as should be.

For anyone wondering where the 63%, 86%, and 95% values come from,
just imagine (1-1/e^1), (1-1/e^2), and (1-1/e^3).

Jon
.

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