A capacitor connected to a Voltage Divider

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Archimedes

Posted: Mon Jan 29, 2007 6:55 am

Guest

Hi all

What would happen if you connect 2 resistors together in series and
then connected a capacitor also in series (say to the end of the 2nd
resistor). What will happen to the voltage divider? This is a DC
circuit.

Regards.

Mikkel Lund

Posted: Mon Jan 29, 2007 7:57 am

Guest

Archimedes skrev:

Quote:

Like this:
http://www.control.aau.dk/~mmlu03/Vdivc.JPG

Vout will quickly rice to ½ Vin and slowly to Vin

--
Hilsen Mikkel Lund
"Sund fornuft, har aldrig stoppet en tosse"
Jokeren i "Mænds ruin"

R.Wieser

Posted: Mon Jan 29, 2007 8:27 am

Guest

Archimedes <shelton.dcruz@gmail.com> schreef in berichtnieuws
1170068108.164930.172680@l53g2000cwa.googlegroups.com...

Quote:

Hi all
....
What will happen to the voltage divider?

It will divide the voltage, what else did you think it would do ?

To be more specific : the connection-point between those resistors will show
a loading-curve just like a one resistor circuit will show. The only
difference is that the connection-point between the resistors will look as
if there is a lower voltage provided to the circuit.

A remark : capacitor in series is caclulated just like resistors in
parallel. In other words : your two resistors, two capacitors circuit (all
in series) can be considered as a two resistors, one capacitor circuit.

Rgeards,
Rudy Wieser

Jon Slaughter

Posted: Mon Jan 29, 2007 9:26 am

Guest

"Archimedes" <shelton.dcruz@gmail.com> wrote in message
news:1170068108.164930.172680@l53g2000cwa.googlegroups.com...

Quote:

Specifically?

Well, If you have R1, R2, C, I and V then

Q = C*Vc, I = dQ/dt, and we'll write R1 + R2 as R

the loop is

V - I*R - Q/C = 0

or

V - R*dQ/dt - Q/C = 0

You'll notice that the current doesn't care if you are dividing the
voltage(assuming the load is very large when you do this).

So your equestion reduces just to one resistor and capacitor

The solution is I = Q0/R/C*exp(-t/R/C)

Where Q0 is the initial charge. here we are assuming the capacitor is
charging and that Q0/C < V.

Obviously now that you know the current you can find the voltages on the
resistors. Ofcourse all this work is unecessary because the current will
stop flowing(approximately) after a while and the voltage drop across the
resistors will be 0. Ofcourse if you need to analyze the transients then the
above is what you want.

John Fields

Posted: Mon Jan 29, 2007 9:42 am

Guest

On Mon, 29 Jan 2007 13:27:34 +0100, "R.Wieser"
<address@not.available> wrote:

Quote:

Archimedes <shelton.dcruz@gmail.com> schreef in berichtnieuws
1170068108.164930.172680@l53g2000cwa.googlegroups.com...
Hi all
...
What will happen to the voltage divider?

It will divide the voltage, what else did you think it would do ?

To be more specific : the connection-point between those resistors will show
a loading-curve just like a one resistor circuit will show. The only
difference is that the connection-point between the resistors will look as
if there is a lower voltage provided to the circuit.

A remark : capacitor in series is caclulated just like resistors in
parallel. In other words : your two resistors, two capacitors circuit (all
in series) can be considered as a two resistors, one capacitor circuit.

---
I think you misread something.

Take a look at Mikkel Lund's post.

--
JF

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