A capacitor connected to a Voltage Divider

Archimedes skrev:

Like this:

Vout will quickly rice to ½ Vin and slowly to Vin

Archimedes schreef in berichtnieuws snipped-for-privacy@l53g2000cwa.googlegroups.com...

...

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

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.