Resistor - Capacitor - Resistor

[Sefrez]

This should be a simple question.

If you have a series circuit in the following order: emf, resistor1, capacitor, resistor2

Are both resistors in series during the charging interval considering that they both have the same current through them? That is, would a resistance of resistor1 + resistor2 either before or after the capacitor be equivalent?

 

[sophiecentaur]

If you are only considering the voltage across the capacitor then you can combine the two resistors. BUT NOT if you are considering the voltage between the 'top end' of the capacitor and Earth, for instance. For that, you need to do things in a different order.

 

[Sefrez]

I believe I am speaking of the first you mention. For the second, could you explain in a little more detail what that would be? The way I am picturing it now is that your saying the wire at which terminates at the negative terminal of the emf is earthed. Is that correct? If so, how does this change the situation? Charge loss or something?

Thanks!

 

[sophiecentaur]

I had to stop and think there!
The problem was that you didn't actually ask the question which answered! (HAHA - very common)
I was initially talking in terms of measuring volts in the AC situation but the same must apply to the transient waveform at switch on. The current flow with time will be the same (defined by R(total) and C) but, with the resistance 'split' across the C, the voltage at the top of the C will start at Vin, modified by the potentiometer ratio to a value above zero and end up at Vin. Think of it this way: if the top resistor were nearly zero, the value of PD to Earth would be nearly Vin all the time.

There is no "charge loss". The difference is the reference point from which you would be measuring the 'output' PD.

If the capacitor is 'in the middle' then the phase of the PD between positive end of the capacitor and Earth will not be the same as the phase of the PD between the ends of the capacitor. (wrt the phase of Vin)