Capacitor Voltages at t = 0: Zero or Split?

[danadak]

So we can say my conjecture that R at output node would rob current at t = 0 from
the current coming from Dirac impulse is not true, hence forces Vc1 = Vc2, until the
momentum of charge occurs, which due to mass does not occur until t > 0 ?

Would that be a correct interpretation ?

 

[Dale]

With ideal circuits the mass and momentum of the electrons is always assumed to be zero. The problem in the OP is about the ideal circuit (which has no solution).

There are many ways that you can “fix” the problem. Adding the resistance is one. Looking at the mass and momentum of the charges is another. But all of them are a change from the initial problem (which has no solution). So any conclusion you make will be rejected by anyone who doesn’t agree with it because it is a different problem from the original problem (which has no solution).

 

[Roberto Pavani]

TL;DR: Step Response R-RC Network

For R finite, C1 = C2, vi a step function @ t = 0, what is the V on each Cap

V(C2) = 1/2 V(V1) in t=0+

1) Step generator is ideal
2) Analytical continuation: lim R2->0 -> V(C2) = 1/2 V(V1)

R2 is the one in the simulation circuit (the generator internal impedence)

This is analogous to the classic two-capacitor paradox (one charged, one empty, connected by a switch): charge is conserved, but energy seems to disappear , it's dissipated in R regardless of its value.

Edited: "this thread is better suited for Engineering/Electical Engineering forum section"

 

[Dale]

R2 is the one in the simulation circuit

That changes the problem as already discussed repeatedly above. The actual problem has no solution

 

[Roberto Pavani]

Step generators do not exist. Wires without impedance (neither R nor L) do not exist. The circuit as stated is electrically invalid: V-C loops are not allowed in lumped-parameter models.

There are two ways to make sense of the problem:

1. Regularization: replace all wires with a parametric impedance $R_{\rm wire}$, solve, then take the limit $R_{\rm wire} \to 0$. The result is well-defined and converges to $V(C_2) = V_1/2$ at $t=0^+$.
2. Laplace domain: capacitors become impedances $1/(sC)$, the step becomes $V_1/s$, and the circuit reduces to a simple impedance divider. For $C_1 = C_2$:

$V_{C2}(s) = \frac{V_1}{s} \cdot \frac{\frac{R}{1+sRC_2}}{\frac{1}{sC_1} + \frac{R}{1+sRC_2}}$
$V_{C2}(s) = \frac{V_1}{s} \cdot \frac{Z_{C2} | R}{Z_{C1} + Z_{C2} | R}, \quad Z_{C2}|R = \frac{R}{1+sRC_2}$
$\text{For } s\to\infty: \quad Z_{C2}|R = \frac{R}{1+sRC_2} \approx \frac{1}{sC_2} = Z_{C2}$

$\lim_{s\to\infty} s \cdot V_{C2}(s) = V_1/2$

which gives $V_{C2}(t) = V_1/2$ for $t > 0$ directly, no regularization needed.

Both approaches give the same answer. The problem does have a solution, it just requires the right mathematical tool.

Alternatively, one can simply state that the problem as posed is unphysical (ideal step + zero-impedance wires, no dissipation, no leaks, no radiation) and therefore has no solution. That is also a valid answer, it depends on whether you want to solve the mathematical idealization or reject it.

 

[Dale]

Step generators do not exist. Wires without impedance (neither R nor L) do not exist. The circuit as stated is electrically invalid: V-C loops are not allowed in lumped-parameter models

As we have already discussed above, all of that is true. But the debate is specifically about the completely unrealistic idealized model. That is precisely why there is no solution.

Also, the debate is about $t=0$, not $t=0^+$. As we have already discussed above.

Laplace domain:

The Laplace transform also doesn’t work. As we have already discussed above.

Before jumping in to an already finished debate you might want to read the debate to see if your points have already been raised (they have) and the result (they are all refuted already).

 

[Roberto Pavani]

@Dale

As I suggested earlier, this thread would be better suited for the Electrical Engineering forum section. The reason is simple: this circuit is just a parallel RC filter, AC-coupled to a step voltage generator through C₁.

It's a textbook circuit that every EE student solves in their first year. The capacitive divider sets the initial condition at t=0⁺ (V₀/2 for equal capacitors), and then R discharges C₂ with time constant τ = RC₂.

In an EE forum, the answer would simply be "obviously V₀/2."

That said, I appreciate the rigor of your analysis, and I agree with you on the pure mathematics: in the framework of classical real analysis, the derivative of the Heaviside function does not exist at t=0, so the ODE has no classical solution. That is formally correct.

However, circuit theory operates in a different mathematical framework, that of distributions (Schwartz, 1950) where:

u(t) is a well-defined object with Laplace transform 1/s
δ(t) = du/dt is a well-defined object with Laplace transform 1
ODEs with distributional forcing have unique solutions

Your equation from post n.5:

$0=\frac{v_0}{R}+C_2 \frac{dv_0}{dt} + C_1 \frac{d(v_0-v_I)}{dt}$

is perfectly solvable once we recognize that $\frac{dv_I}{dt} = V_0\,\delta(t)$.

This is standard practice in every circuits textbook (Desoer & Kuh, Nilsson & Riedel, Hayt, Oppenheim & Willsky...).

Taking the Laplace transform (with zero initial conditions):

$0 = \frac{V_0(s)}{R} + (C_1+C_2)\,s\,V_0(s) - C_1\,V_0$

Solving for $V_0(s)$ with $C_1 = C_2 = C$:

$V_0(s) = \frac{C\,V_0}{\frac{1}{R} + 2Cs} = \frac{V_0}{2} \cdot \frac{1}{s + \frac{1}{2RC}}$

Inverse transform:

$v_0(t) = \frac{V_0}{2}\,e^{-t/(2RC)}\,u(t)$

So $V_{C1}(0^+) = V_{C2}(0^+) = V_0/2$

The initial value theorem confirms: $\lim_{s\to\infty} s\,V_0(s) = V_0/2$

I understand that from a standpoint of mathematical rigor, the tools I'm using here (distributions, Laplace transform of the Heaviside function, δ(t) as a derivative) may feel uncomfortable in a Physics forum. But these are exactly the tools that would be used, without any debate, in an Electrical Engineering forum, because they are the standard language of circuit analysis.

I think we are both correct, just operating in different frameworks. Thank you for the insightful discussion, it highlights how the same problem can look very different depending on the mathematical tools one chooses to apply.

 

[Dale]

The capacitive divider sets the initial condition at t=0⁺

As we have already discussed, the question is about $t=0$, not $t=0^+$.

In an EE forum, the answer would simply be "obviously V₀/2."

This is simply factually untrue. As the OP already discussed the problem comes from an EE forum, edaboard, where the answer to the problem is a matter of contention. The actual people on the actual EE forum did not actually answer "obviously $V_0/2$".

That is formally correct.

Yes. And more importantly, that is the fundamental reason that the answer to the problem is contentious.

If someone already believes that $V_0/2$ is the answer to the problem then they will be willing to accept your revisions to the circuit. But if someone already believes something different then they will be unwilling to accept your revisions to the circuit. Without changing the circuit the only correct answer to the problem is that the system of equations has no solution.

is perfectly solvable
...
Taking the Laplace transform (with zero initial conditions):

Look at what you did there. Zero initial conditions. The problem is that the behavior at $t=0$ is precisely the thing in question. Assuming zero initial conditions will not be something that is accepted by people who disagree with your answer to the problem. Yes, everyone will agree that if you assume the answer you can get a solution. But people not agreeing with that solution will not accept your assumption.

Without assuming zero initial conditions you get $$\frac{V_0(s)}{R} + (C_1+C_2)(s \ V_0(s)-v_0(0))-C_1(s V_i(s)-v_i(0))=0$$ but $v_i(0)$ is not necessarily $0$. In fact, $v_i(0)=0.5$ and $v_i(0)=\text{undefined}$ are both common.

Furthermore, the problem is the behavior at $t=0$. So, if you assume zero initial conditions then you are directly assuming $v_0(0)=0$, which is both different from what you claim the answer is and also is not going to convince anyone. You cannot just assume the value of the very thing that is in dispute.

the tools I'm using here (distributions, Laplace transform of the Heaviside function, δ(t) as a derivative) may feel uncomfortable in a Physics forum

I have an EE background. I am perfectly comfortable with these tools. And I can tell you exactly why these tools do not resolve the problem. The Laplace transform does not resolve it. First, it requires assuming $v_i(0)$, and second even if you got people to agree on $v_i(0)$ the Laplace transform will only tell you the behavior for $t=0^+$ which, as we have discussed, is not the problem.

I think we are both correct

It is not that your analysis is incorrect. It is that you are missing the problem. The problem is not that people lack the skill or knowledge to apply any of the analyses that you are suggesting. In fact, we have already done so above. The problem is that the various analyses do not resolve the problem because they all make various assumptions that people who disagree with the resulting conclusion will reject.

Without making any assumptions, just working the problem directly, it is clear mathematically that there is no solution. Therefore, ANY possible technique that you could use to obtain a solution must make some additional assumption. That additional assumption is subject to criticism and rejection, so it will not solve the original problem.