# Capacitor current in parallel network with ESR and parallel resistance

So this might be a very simple question and I've got a knot in my brain but I've been at this for days and don't see how I could solve this problem.

I'm trying to find the solution for both ucA, ucB(t) and isc(t) in this circuit:

simulate this circuit – Schematic created using CircuitLab and basically arrive at this equation:

$$\frac{d}{dt}u_{cA} \cdot C_A \cdot(R_A +R_{sc}) + C_B\cdot \frac{d}{dt}u_{cB}\cdot R_{sc} =u_{cA}$$

The initial conditions $$\u_{cA}(0)\$$ and $$\u_{cB}(0)\$$ are known.

I've stumbled upon this problem trying to find the solution for an electrical network consisting of two parallel capacitors with an ESR. The capacitors are charged. At $$\t=t0\$$ a short circuit with a fault resistance $$\R_{sc}>0\$$ occurs. I want to find an expression describing the current through the fault resistor and the voltages of the capacitors!

The basic equations for this circuit are based on Kirchhoff. Firstly, I know that the current trough the fault resistor is the sum of the capacitor currents (Eq. I):

$$i_{sc}=i_A+i_B$$

Secondly I know that the voltages for each of the parallel branches $$\u_{cA}-u_{rA}=u_{cB}-u_{rB}\$$ is the same, resulting in (Eq. II):

$$u_{cA}-\frac{d}{dt}u_{cA} \cdot C_A R_A=u_{cB}-\frac{d}{dt}u_{cB} \cdot C_B R_B$$

The current $$\i_A\$$ of capacitor A and B respectively are defined as (Eq. III):

$$\frac{d}{dt}u_{cA}(t)=\frac{1}{C_A}i_{A}(t)$$ $$\frac{d}{dt}u_{cB}(t)=\frac{1}{C_B}i_{B}(t)$$

with

$$i_A=\frac{u_{cA}-i_{sc}\cdot R_{sc}}{R_A}$$ $$i_B=\frac{u_{cB}-i_{sc}\cdot R_{sc}}{R_B}$$

Using Eqation III $$\i_A=C_A\cdot \frac{d}{dt}u_{cA}\$$ and $$\i_B=C_B\cdot \frac{d}{dt}u_{cB}\$$ in the equation for $$\i_{sc}\$$ I get:

$$C_A\cdot \frac{d}{dt}u_{cA}=\frac{u_{cA}-(C_A\cdot \frac{d}{dt}u_{cA}+C_B\cdot \frac{d}{dt}u_{cB})\cdot R_{sc}}{R_A}$$ $$\frac{d}{dt}u_{cA} \cdot C_A \cdot(R_A +R_{sc}) =u_{cA}-C_B\cdot \frac{d}{dt}u_{cB}\cdot R_{sc}$$

$$\frac{d}{dt}u_{cA} \cdot C_A \cdot(R_A +R_{sc}) + C_B\cdot \frac{d}{dt}u_{cB}\cdot R_{sc} =u_{cA}$$

Equation II in turn can be used to derive an expression for $$\ \frac{d}{dt}u_{cB} \$$

$$\frac{d}{dt}u_{cB} = \frac{1}{C_B R_B }(u_{cA}-\frac{d}{dt}u_{cA} \cdot C_A R_A + u_{cB})$$

I now end up with:

$$\frac{d}{dt}u_{cA} \cdot C_A \cdot(R_A +R_{sc}) + (u_{cA}-\frac{d}{dt}u_{cA} \cdot C_A R_A + u_{cB})\cdot \frac{R_{sc}}{R_B }=u_{cA}$$

$$\frac{d}{dt}u_{cA} \cdot C_A \cdot(R_A (1-\frac{R_{sc}}{R_B})+R_{sc}) =u_{cA} (1-\frac{R_{sc}}{R_B })-u_{cB}\cdot \frac{R_{sc}}{R_B }$$

# Differential equations

## $$\u_{cA}\$$

This equation has the form $$\ \dot{y} - ay = x \$$ I could try and solve this inhomogenous differential equation. The homogenous solution being for $$\ x=0\$$:

$$\frac{d}{dt}u_{cA} + u_{cA} \frac{(-1+\frac{R_{sc}}{R_B})}{C_A \cdot(R_A (1-\frac{R_{sc}}{R_B}) + R_{sc})}=0$$

Using $$\y_h=C\cdot e^{-at}\$$ and assuming $$\ a=\frac{(-1+\frac{R_{sc}}{R_B})}{C_A \cdot(R_A (1-\frac{R_{sc}}{R_B}) + R_{sc})} \$$

$$u_{cA,h}=u_{cA_0}\cdot e^{-at}=u_{cA_0}\cdot e^{-\frac{(-1+\frac{R_{sc}}{R_B})}{C_A \cdot(R_A (1-\frac{R_{sc}}{R_B}) + R_{sc})}t}$$

The non homogenous solution $$\ u_{cA,p}\$$ being:

$$u_{cA,p}=\frac{u_{cB}}{\frac{(-1+\frac{R_{sc}}{R_B})}{C_A \cdot(R_A (1-\frac{R_{sc}}{R_B}) + R_{sc})}}$$

I now arrive at:

$$u_{cA}=u_{cA,h}+u_{cA,p}=u_{cA_0}\cdot e^{-\frac{(-1+\frac{R_{sc}}{R_B})}{C_A \cdot(R_A (1-\frac{R_{sc}}{R_B}) + R_{sc})}t}+\frac{u_{cB}}{\frac{(-1+\frac{R_{sc}}{R_B})}{C_A \cdot(R_A (1-\frac{R_{sc}}{R_B}) + R_{sc})}}$$

## $$\u_{cB}\$$

I can do the same for $$\ u_{cb}\$$:

Basic equation using current law:

$$C_B\cdot \frac{d}{dt}u_{cB}=\frac{u_{cB}-(C_B\cdot \frac{d}{dt}u_{cB}+C_A\cdot \frac{d}{dt}u_{cA})\cdot R_{sc}}{R_B}$$

$$\frac{d}{dt}u_{cB} \cdot C_B \cdot(R_B +R_{sc}) + C_A\cdot \frac{d}{dt}u_{cA}\cdot R_{sc} =u_{cB}$$

Using the voltage law for $$\\frac{d}{dt}u_{CA}\$$: $$\frac{d}{dt}u_{cA} = \frac{1}{C_A R_A }(u_{cB}-\frac{d}{dt}u_{cB} \cdot C_B R_B + u_{cA})$$

Substituting the derivative in the above equation renders:

$$\frac{d}{dt}u_{cB} \cdot C_B \cdot(R_B +R_{sc}) + C_A\cdot \frac{1}{C_A R_A }(u_{cB}-\frac{d}{dt}u_{cB} \cdot C_B R_B + u_{cA})\cdot R_{sc} =u_{cB}$$

This should again let me deduct an expression for $$\u_{cB}=C\cdot e^{-b/t}+\frac{u_{cA}}{b}\$$.

## Using $$\u_{cA}\$$ in $$\u_{cB}\$$

If I now use the expression for $$\u_{cA}\$$ in $$\u_{cB}\$$ I should be able to calculate the voltage for $$\u_{cB}\$$. I have yet to do this - I've allready spent alot of time on this problem today and will get back.

• You need to use \\$ for inline math. Also, try using Laplace, it might be easier. Commented Apr 11, 2022 at 14:01
• I’m not sure that this one has a nice closed form solution, given the nonlinearity. What sort of an answer you’re looking for? Solving it numerically is trivial, and that’s what any SPICE system would do. Perhaps begin by writing a tiny bit of code in Python or Octave to get a numerical solution, then play with the numerical solution to get an intuition for how it reacts to changes in component values, and then you should hopefully be able to “see” a closed-form solution if one exists. Commented Apr 11, 2022 at 14:39
• I'm looking for a analytical expression. I have already simulated it in spice and wanted to see if I could solve it in a closed form. Commented Apr 11, 2022 at 14:47
• I think you may want to settle with the SPICE simulation, because the analytical solution involves some very fluffy equations. I had to use wxMaxima for this and zoom out the text a bit (not to much, to be able to see in the screenshot); LTspice just confirms the results. Commented Apr 11, 2022 at 17:35
• Now that you can see how messy the closed form solution is, you may wish to decide whether it’s worth it. As a point of reference: numerically integrating the stuff takes about as many primitive operations as calculating the same number of time steps using the closed form solution, with same accuracy as long as you got enough steps, and you don’t need to manually solve anything. It’s like 10 lines of Python or 5 lines of Octave to solve it. Are you sure you still want the solution you’ll make a mistake in if you just copy it onto a page 5 times in a row? It’s a nice challenge as such of course Commented Apr 11, 2022 at 19:14