Re: Series-Parallel DC RC circuit

On Sun, 08 Mar 2009 17:56:32 -0700, Dan Coby <adcoby@xxxxxxxxxxxxx>
wrote:

Jon Kirwan wrote:
On Sun, 08 Mar 2009 17:15:41 -0700, Dan Coby <adcoby@xxxxxxxxxxxxx>
wrote:

Jon Kirwan wrote:
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.)
I agree that the thevenin resistance is the correct value to use for the
time constant. However my reading of bg posting agrees with JL. Bg says:

"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."

To me, this says that only the series resistance is used for the time constant
calculation.

But that is incorrect to say.

Jon

As I said: "I agree that the thevenin resistance is the correct value to use
for the time constant."

I also agree that bj's statement is wrong.

I disagreed with your statement that bg had not specified what resistance he
meant to use for the time constant calculation. In his post which I quoted
and later posts, he argues that only the series resistance should be used
for the time constant calculation. I do not agree with him (and I do not
think that you do either).

Ah! I couldn't tell from your last sentence. Thanks.

Jon
.

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