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Passive sign convention means choosing an arbitraty current direction, then assuming that current always enters the positive terminal and flows out of the negative terminal of an element. Then the following will hold true: $$ U_R = IR \qquad U_C = \frac{q}{C} \qquad U_L = L \frac{dI}{dt} $$

when we apply the KVL

And the sign of voltage in KVL is determined by the sign of the first encountered terminal.

Under which conditions does this convention work and why does it work under them?

Why do I doubt this always holds true?

First of all, because I haven't seen any comlete general rigorous reasoning that this holds true in an arbitrary circuit.

And also it at least doesn't seem to hold for cases when we have a single loop with elements in series and thus single current, yet the polarities of the elements' terminals may not always be "+ first, then -" or vise versa for all elements as we follow along the loop.

When we apply KVL, for example, to a circuit obtained by connecting an uncharged capacitor to a charged capacitor through an inductor, the system will become something like that:

enter image description here

And the top-plate capacitor terminals will always be positive. Thus we cannot choose the signs according to passive sign convention like that:

enter image description here

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The passive sign convention (PSC) doesn't assume anything, it is a convention, merely a definition. It doesn't make any claims as to the physics, and as such it is not meaningful to question its correctness. This applies to the active sign convention (ASC) as well.

To speak of the voltage or current of a two-terminal device with no sign ambiguity, you first need to choose a polarity: pick one of the terminals to be the "positive" terminal. You have the freedom to choose either one. By this choice you are defining the device voltage to be positive if your positive terminal is at a higher voltage than the negative. Then, PSC defines the current to be positive if it is flowing into the positive terminal. This is the entire content of PSC. In contrast ASC defines the current to be positive if it is flowing out of the positive terminal. Neither convention prevents the voltage or current from being negative.

If you choose the capacitor polarities to be like in the first diagram (the upper terminals are the positive terminals), then according to PSC, $I_{C_1} = -I$ and $I_{C_2} = I$. If you assign the polarities like in the second diagram, then $I_{C_1} = I_{C_2} = I$.

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  • $\begingroup$ "if the upper terminals are at a higher voltage like you claim, then..." - I can not know whether the voltage is higher or not before actually solving the circuit. And I wasn't asking to derive the passive sign convention alone, rather why these: $$ U_R = IR \qquad U_C = \frac{q}{C} \qquad U_L = L \frac{dI}{dt} $$ hold true under these conventions. Take a look at my two last pictures. The last one is according to passive sign conventions and thus I get $$U_{C_1} + U_{C_2} + U_L = 0$$, with $U_X$ defined as above. But this is an incorrect equation for the problem $\endgroup$
    – Sgg8
    Commented Apr 25, 2023 at 20:17
  • $\begingroup$ @Sgg8 Your last equation is fine. In $U_C=q/C$ you need to use the charge on the positive terminal. In diagram 2 you have chosen the positive terminal of $C_1$ to be the bottom one, on which the charge is $-q_1$, so $U_{C_1} = -q_1/C$. You can also just use $I_{C}=C\frac{dU_C}{dt}$ instead, with $I_{C_1}=I_{C_2}=I$. If you are asking why the R, L, C terminal relations are what they are, I suggest you ask a separate question where you clearly state that because that is an entirely different question than the validity of PSC. $\endgroup$
    – Puk
    Commented Apr 25, 2023 at 20:44
  • $\begingroup$ As far as I know, terminal relations are dependent on the convention used $\endgroup$
    – Sgg8
    Commented Apr 25, 2023 at 21:24
  • $\begingroup$ @Sgg8 Their signs are. With PSC, $U_C = q/C$ says that when the charge on the positive plate is positive, the positive terminal is at a higher voltage. Alternatively, with PSC, $I_C=C\frac{dU_C}{dt}$ says that if current is flowing into the positive terminal ($I_C>0$), the voltage difference between the positive and negative terminals is increasing. $\endgroup$
    – Puk
    Commented Apr 25, 2023 at 21:30
  • $\begingroup$ The very reason to use PSC is that we do not know the polarity and thus charge of the capacitor plates beforehand. So we cannot use this to solve the circuit. Only after solving it can we know for sure the true polarities and current dirctions. That is why my question and PSC is so tightly interconnected $\endgroup$
    – Sgg8
    Commented Apr 25, 2023 at 21:33

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