Does a charged capacitor act as a short circuit in feedthrough?

[ermai]

A capacitor behaves as short when initially uncharged and sudden volatage is applied and open when fully charged.
But in feedthrough,a volatge variation is in a charged coupling capacitor.
A feedthrough effect by a coupling capacitor is as below:

in which C2 is a couplng capacitor to VL2,and in initial C1 and C2 are charged.
when the bottom plate of C2 changes from VL2 to VL3 (assume that VL3 is low than VL2 ),the upper plate changes from Vd to Vx.


My question is that may I say that beacuse the sudden voltage change in the charged coupling capcitor C2,the C2 acts as "short circuit" such that C1 discharges to C2.

 

[kabeer02]

I dont realy understand your question,what type of capacitor do you use,where is the input of your supply?

 

[ermai]

No power supply.Only 2 capacitor.
Long story short,my question is "Does a capacitor act as short when the voltage applied to it changes,even a charged capacitor"?

 

[kabeer02]

No i dont think so.

 

[crutschow]

If by short, you mean an initial spike in current, limited by the series resistance, when the voltage is rapidly changed across a capacitor, then yes it rather looks like a momentary short. But the current decays with a RC time constant until it reaches zero with a steady DC voltage across it.

 

[LvW]

No power supply.Only 2 capacitor.
Long story short,my question is "Does a capacitor act as short when the voltage applied to it changes,even a charged capacitor"?

Without a good drawing your question is not clear at all.

Where is the applied voltage? Why it should change its value? Is there any switch? Where? Ideal or real capacitors (leakage resistances)?