I do not understand your question.

You have shown three new circuits with voltage sources.
You need to analyze each circuit separately using Kirchhoff’s current and voltage laws and Ohm’s Law.

Instead of capacitors I used voltage source.
Also there is a voltage difference for first on the left is 5V and 3V, second is 5V and 2V and last 5V and 2,5V.

For the first current flows from high potential to low potential which is why the current flows from 5V to 3V. Same goes with voltage source.
Second one There is also high potential which is 5V and low potential which is 2V. It is different because the current flows from low pottential tohigh potential in voltage source.
In the last one there is 5V and 2,5V. High potential and low potential. But nothing happens in voltage source.

 

Your analytical approach is incorrect.

A) Voltage source is 2V.
B) Voltage source is 3V.
C) Voltage source is 2.5V.

The proper way to determine the current flow in the voltage source is to do the analysis.

Use Ohm’s Law to calculate the current in each resistor.
The use KCL to determine the current in the voltage source.

If you were to replace the voltage source with a capacitor you cannot use the same analysis.
Why?
Because dV/dt is not zero. dV/dt is zero only at steady-state.

 

Ok so in voltage source this principle high potential and low potential doesn't work yes ?

 

You are mistaken that the 2V source has 2V across its terminals.

Well I think that it has 5V on one node and 3V on the another.
Capacitor works like a voltage source when it's stable.

 

Sorry, I deleted my post because what I said is incorrect.

There is 2V across the 2V voltage source.
You need to determine the current and direction of current flowing through the voltage source.

With a capacitor, there is no current flowing through the capacitor when steady-state condition is reached. At that point, you can remove the capacitor and it makes no difference.

Likewise, in (C) you can remove the 2.5V source and it makes no difference because no current flows in the voltage source.

 

Sorry, I deleted my post because what I said is incorrect.

There is 2V across the 2V voltage source.
You need to determine the current and direction of current flowing through the voltage source.

With a capacitor, there is no current flowing through the capacitor when steady-state condition is reached.

Steady state can be also shown as voltage with the same value for example stady state is 2,5V so voltage source can be 2,5V.

 

Steady state can be also shown as voltage with the same value for example stady state is 2,5V so voltage source can be 2,5V.

All three circuits are in steady-state.

 

You are confusing

1. initial state
2. transient response
3. steady-state

Put a SPST switch in series with the voltage source or capacitor.
Do the analysis with the switch
1) open
2) closed
3) open again

 

Ok so in voltage source this principle high potential and low potential doesn't work yes ?

You have a battery (voltage source) with no connections to the terminals, does current flow from the higher potential terminal to the lower potential terminal?

 

I've taken this example :

https://tinyurl.com/2fansnsf

View attachment 273803

It's just so wierd for me that it doesn't go from High potential to low potential. And it is the same with voltage source.
It works only for resistors yes ?

So weird that WHAT doesn't go from high potential to low potential?

Take a 12 V car battery and sit it on a shelf, unconnected to anything. There is 12 V between the two terminals. But no current is flowing. Do you find that weird, or is that what you would expect?

 

Ok so in voltage source this principle high potential and low potential doesn't work yes ?

An ideal voltage source maintains a fixed voltage between it's terminals regardless of the current flowing through it. If it needs to supply energy to the circuit to make this happen, it will deliver current from its positive terminal. If it needs to absorb energy from the circuit to make this happen, it will accept current into it's positive terminal.

This is a familiar concept that you apply every time you charge a battery.