Find the Voltage in the parallel resistor-capacitor circuit

[Helly123]

I do not understand your question. In some places you say "voltage" when you mean "current"?

Please see the graph in #21.

1) With switch open..
The voltage on node D (with respect to C) = 8V.
The voltage on node B (with respect to C) = 4V.

2) When the switch is closed...
Node D is connected to node B. This means node D and B must have the same voltage. How do we decide if this is 8V or 4V? I will explain how..The voltage on node D cannot change instantly because of the capacitors...

Q=CV
differentiate
dQ/dt = C dV/dt

dV/dt cannot be very large because it would need a very large current (dQ/dt).

If the voltage on node D cannot change quickly then the voltage on node B must change because they are connected by the switch. So what happens is the voltage on node B shoots up from 4V to 8V. So now both node D and B become 8V. This is shown in the middle of the graph.

3) After that the resistors keep trying to "pull" the voltage on node D/B from 8V back down to 4V. This is the curved part of the graph in #21.

During the curved part of the graph current flows like this..
View attachment 221974
4) After some time the nodes D and B reach 4V and the two current shown stop flowing.

Thanks for the explanation. After the current stop flowing after that what will happen? Is the electricity continue? Is it what called AC?

 

[CWatters]

After current stops flowing through the switch there will still be some current flowing from the battery through R1 & R2 and back to the battery. The current will be the same as calculated at the start (eg 1A).

There will be no current flowing through the capacitors.

This is nothing to do with AC.