Isn't conventional current supposed to go from high potential to low potential ? If yes, then why this rule is not respected in opamp using AC?

First, thank you for reading my post. I'm a beginner in electronics. I learned that conventional current goes from high potential to low potential. It has always worked for me in DC and even AC. But recently, i did see this rule being violated in AC, which i think is impossible...right ? To give you an insight, i have to tell you that i'm working with opamps. Here is how i see current direction in a non-inverting opamp configuration: In part A, the output is amplified by a positive gain but because the input is negative, the output becomes even more negative. So, at that moment, the lowest potential in the circuit is the ouput, then we have the inverting input of the opamp that is at same potential as the input, and finally the ground being at 0 volts is the highest potential. So conventional current would flow from ground to the output of the opamp(vice versa for part B). You can see why on the following graph : So i tried it on Ltspice: no problem it works. Current going through R1 gets positive and negative [-5mA:5mA] (so goes in the both directions periodically, so this is AC) : When i add an offset as such: Output is always greater than the input, current is then only positive [0:5mA]( so it's DC), so it works: But when i try to add two capacitors in parallel with the two resistors (in the feedback) : I still have the output always greater than the input, but this time, the current passing through the capacitors path is alternating between positive and negative : Although, i have to note that because i did apply a DC offset at the input of the opamp, i have to provide it a path at the output for the DC current to pass through (the resistors path is doing that, just by looking at I_R1 we see that the current is always positive, so that's DC current).

I also understand that capacitors block any DC current so if something has to "pass through", it has to be alternative current. But still, think about it, i always have a greater electric potential at the output than the inverting input of the opamp, so current should flow from output to the inverting input of the opamp and then to the ground. So, it should be a DC current passing through the resistors only. But then it would mean that there is no voltage drop across the capacitors (since no current supposedly pass through the capacitors) which is impossible because we all know that the voltage drop across the capacitors are the same than the respective resistors because they are in parallel..

So.. Has the rule "Current goes from high potential to low potential" been violated ?

PS: Sorry for the long post and thank you for your precious time !

Edit : Here is the sum of the current I_R1 and I_Cact1 • Isn't conventional current supposed to go from high potential to low potential ? No, the convention is that in a power consuming device (like a resistor), current flows from a high potential to a low potential. Realize that electrons do flow from high to low but that a current by definition flows in the opposite way. I stopped reading after the title as I see no point in reading further if it is not clear to you how current flows. Apr 13 '21 at 15:03
• LTspice has to pick a reference current direction, and it can only be one way. Think of it like the current convention used in Kirchhoff's circuit laws. Simply rotate your resistor 180 degrees in the schematic and the current will be to your liking. Alternate: Download File:Res.asy and place in your /lib/sym folder. This symbol has a small arrow indicating the defined way current flows. See this ltwiki.org/…
– G36
Apr 13 '21 at 15:08
• TLDR - please get to the point. Apr 13 '21 at 15:16
• @AF2021 But the voltage at the output is constantly changing with the frequency of 40Hz. Thus, if DC voltage is "changing" with the frequency of 40Hz around 13.5V we have a AC component isn't it?
– G36
Apr 13 '21 at 16:00
• @AF2021 no, at DC when Vin = 0V the Cact1 will be charged to 4.5 * (1 + 100/50) *100/150 = 9V (Vout 13.5V). Now, for a "positive swing" when Vout increases from 13.5V to a larger voltage (let us say to 14V) the Cact wants to change himself to the new value. But when output voltage decreases his value from a DC value 13.5V to the new value during "negative swing" (let us say to 13V) Charged capacitor will now act just like a voltage source, and it will start a discharging process (The current will flow in the opposite direction).
– G36
Apr 13 '21 at 16:20