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I asked this question already yesterday, but it seems like I didn't explain it well enough so I will give it one more go.

This circuit is powered by a DC current source. I am trying to find out how current will be split between the two branches in the TRANSIENT response. This is a hypothetical circuit where the capacitances are large and will take lots of time to charge them up, so the current will be constantly changing with time.

I have derived 5 equations shown in the picture where all the voltages are functions of time, except V4 as it is always 0. However, when I am trying to solve these equations (after some substitutions and using the ODE45 function in MATLAB) the values come out wrong, when comparing them to when I run a simulator.

Are my equations wrong? Is there another way to do it? I am a mechanical engineer undergrad, so my knowledge in electronics is limited.

drawing of schematic diagram

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  • \$\begingroup\$ Without more information, it is impossible to know if your equations are wrong (showing more work than node voltage), or if you're interpreting the simulation wrong (showing actual part values.) For a DC transient response, often you're talking about receiving a steady-state response often on the order of micro or milliseconds (to the point of often being impractical to measure or use.) The information I've mentioned above, in addition to the purpose of solving the DC transient response, would probably improve the quality of answers you receive. \$\endgroup\$ – SerasVita Mar 25 at 21:28
  • \$\begingroup\$ Was your circuit from an Australian website. I’m British so go figure. Why is ground labelled V4? \$\endgroup\$ – Andy aka Mar 25 at 21:38
  • \$\begingroup\$ No, i drew the circuit. Again V4=0 and I am interested in the transient responce as this circuit is an analogy in thermal systems where voltage represents the temperature, current the heat transfer rate and capacitors the thermal mass of objects which take time to heat up and thus why takes time for capacitors to charge. \$\endgroup\$ – Orestis Mar 25 at 21:51
  • \$\begingroup\$ Then use a simulator to give you what you need. I've done this a couple of times when I knew what the transient thermal characteristics were of a device. \$\endgroup\$ – Andy aka Mar 26 at 9:18
  • \$\begingroup\$ When you create mechanical drawings, do you draw the ground on top and then draw the construction upside-down? If not, you shouldn't do it with electrical drawings either. \$\endgroup\$ – pipe Mar 26 at 11:44
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Since there are two capacitors, the order of the differential equation cannot be more than \$2\$. See if this helps:

enter image description here

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  • \$\begingroup\$ and I assume you put initial conditions and find the values of the constants? Is there any other reference or similar problem that I can check? \$\endgroup\$ – Orestis Mar 25 at 22:59
  • \$\begingroup\$ @Orestis Actually I just finished solving it on my book. I'm going to take a pic and post \$\endgroup\$ – beccaboo Mar 25 at 23:00
  • \$\begingroup\$ Thanks for the responce I will have a go at it \$\endgroup\$ – Orestis Mar 25 at 23:06
  • \$\begingroup\$ @Orestis I've changed the order of resistor and capacitor for ease of algebra. This doesn't affect the current split. \$\endgroup\$ – beccaboo Mar 25 at 23:07
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You don't need to write down any differential equation to solve this circuit. Just use the fact that, assuming the system starts from rest, at the moment current source is connected to the circuit (at say t = 0), the capacitors are going to behave as short circuits and in DC steady state the capacitors will act as open circuits.

At \$t = 0\$

With \$C_1\$ & \$C_2\$ replaced by short circuit, the equivalent circuit becomes just the parallel combination of \$R_2\$ and \$R_4\$. Giving the following currents in each branch: $$I_1(0) = \frac{R_4I}{R_2+R_4}$$ $$I_2(0) = \frac{R_2I}{R_2+R_4}$$

At \$t = \infty\$

With \$C_1\$ & \$C_2\$ replaced by open circuit, the equivalent circuit has parallel combination of \$R_1+R_2\$ and \$R_3 + R_4\$. Current in each branch: $$I_1(\infty) = \frac{(R_3+R_4)I}{R_1 + R_2 + R_3 +R_4}$$ $$I_2(\infty) = \frac{(R_1+R_2)I}{R_1 + R_2 + R_3 +R_4}$$

The current \$I_1\$ in the above branch starts from \$I_1(0)\$ and reaches \$I_1(\infty)\$ with some time constant \$\tau_1\$, thus the current equation with time should be given by: $$I_1(t) = I_1(\infty) + (I_1(0) - I_1(\infty))e^\frac{-t}{\tau_1}$$ The equivalent resistance seen by \$C_1\$ can be shown to be: $$R_{C1eq} = \frac{R_1(R_2+R_3+R_4)}{R_1 + R_2 + R_3 +R_4}$$ And, \$\tau_1 = R_{eqC1}C_1\$ Similarly, \$I_2(t)\$ can be found.
Of course, you can also use Laplace transform to get the result, but this approach is easier and more intuitive.
It can be easily shown why the capacitors behave as open or short circuit as explained above.
The I-V characteristic of a capacitor is given by: $$i = C\frac{dv}{dt}$$ Clearly, at DC steady state, the voltage does not change and the current is zero, resembling an open circuit.
Rearranging the equation, we can understand why it is a short circuit for initial period: $$dv = \frac{idt}{C}$$ For small change in time, the change in voltage will be negligible. Since the capacitor is uncharged in the beginning implying zero volts across its terminals, voltage drop across it is going to be zero around this time. And zero volts implies short circuit.

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An lazy electrician sees your drawing a little tiresome. He redraws it to get a little simpler formulas. He very likely writes with conductance (=the inverses of the resistances, G=1/R) to avoid divisions, especially if he can by doing so look effective and sophisticated, too. If one node has many parts connected he selects it to be his ground, the zero volt node. This is my version, a brute force method with zero assumptions of the general form of the result:

enter image description here

eq1...eq3 state that the sum of departing currents in a node is zero.

You get differential equations which can be solved numerically easily even with Excel because capacitor voltages are the state variables.

Solve from eq1 node voltage V3 and substitute it to others. Then make a pair where the derivatives of V1 and V2 are expressed with I, V1 and V2.

Calculate V1 and V2 advancing timestep by timestep from their initial values and current I. If you have some math program which handles a state variable equation group use it.

Calculate a time series for V3

Finally calculate the time series for your I1 and I2. They are (V1-V3)G3 and (V2-V3)G4.

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  • \$\begingroup\$ Why did you switch the position of G3 to be before G1,C1? does that not make a difference to the equations? \$\endgroup\$ – Orestis Apr 2 at 15:42
  • \$\begingroup\$ These are my symbols which are valid with my image and equations. Questioner had no conductance symbols so I could select my own 100% freely. If 1 or 3 seems to be in a wrong place that's unfortunate because that apparent error isn't a math nor logic error, it's a reading error. \$\endgroup\$ – user287001 Apr 2 at 15:50
  • \$\begingroup\$ I am not saying that G1 should be labeled G3 or vice versa. I am asking in terms of position in the circuit. You drew the parallel conductance and capacitance C1,G1 to be after the conductance G3. What if the the parallel conductance and capacitance C1,G1 was before the conductance G3 \$\endgroup\$ – Orestis Apr 2 at 16:03
  • \$\begingroup\$ @Orestis I checked it. The circuit is as it should be. The ground node is selected for simplest equations. That change doesn't affect currents nor voltages over single components, only node voltages against GND are different due the different ground node. \$\endgroup\$ – user287001 Apr 2 at 16:19
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Your differential equations are correct, but it's probably easier to use the Laplace transform.

Determining the impedances of both main branches:

$$\small Z_1=\frac{R_1+R_2+sC_1R_1R_2}{1+sC_1R_1} $$ $$\small Z_2=\frac{R_3+R_4+sC_2R_3R_4}{1+sC_2R_3} $$

Then the currents through each of these branches are: $$\small I_1(s)=\frac{Z_2}{Z_1+Z_2}I(s) $$ $$\small I_2(s)=\frac{Z_1}{Z_1+Z_2}I(s) $$

The currents in the parallel \$\small R\$ and \$\small C\$ circuits are then:

$$\small I_{R_1}(s)= \frac{1}{1+sC_1R_1}I_1(s)$$ $$\small I_{C_1}(s)= \frac{sC_1R_1}{1+sC_1R_1}I_1(s)$$ $$\small I_{R_3}(s)= \frac{1}{1+sC_2R_3}I_2(s)$$ $$\small I_{C_2}(s)= \frac{sC_2R_3}{1+sC_2R_3}I_2(s)$$ You can treat the current source as a step change at \$\small t=0\$, i.e. \$\small I(s)=\large\frac{I}{s}\$, and assume initial conditions are zero.

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Well, we are trying to analyze the following circuit (assuming that all the initial conditions are equal to zero):

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \text{I}_6=\text{I}_\text{x}+\text{I}_8\\ \\ \text{I}_3=\text{I}_1+\text{I}_2\\ \\ \text{I}_9=\text{I}_1+\text{I}_2\\ \\ \text{I}_\text{x}=\text{I}_7+\text{I}_9\\ \\ \text{I}_7=\text{I}_4+\text{I}_5\\ \\ \text{I}_6=\text{I}_4+\text{I}_5 \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_2}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\ \\ \text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\ \\ \text{I}_4=\frac{\text{V}_1-\text{V}_3}{\text{R}_4}\\ \\ \text{I}_5=\frac{\text{V}_1-\text{V}_3}{\text{R}_5}\\ \\ \text{I}_6=\frac{\text{V}_3}{\text{R}_6} \end{cases}\tag2 $$


Now, applying this to your circuit we need to use (the 'complex' s-domain where I used Laplace transform):

  • $$\text{R}_2=\frac{1}{\text{sC}_2}\tag3$$
  • $$\text{R}_5=\frac{1}{\text{sC}_1}\tag4$$

Because there are already so many answers, I'm going to use Mathematica to solve this problem. First I wrote the two systems of equations into Mathematica-code:

FullSimplify[
 Solve[{I6 == Ix + I8, I3 == I1 + I2, I9 == I1 + I2, Ix == I7 + I9, 
   I7 == I4 + I5, I6 == I4 + I5, I1 == (V1 - V2)/R1, 
   I2 == (V1 - V2)/R2, I3 == V2/R3, I4 == (V1 - V3)/R4, 
   I5 == (V1 - V3)/R5, I6 == V3/R6}, {I1, I2, I3, I4, I5, I6, I7, I8, 
   I9, V1, V2, V3}]]

Now, I substitute in the capacitors, then I get:

FullSimplify[
 Solve[{I6 == Ix + I8, I3 == I1 + I2, I9 == I1 + I2, Ix == I7 + I9, 
   I7 == I4 + I5, I6 == I4 + I5, I1 == (V1 - V2)/R1, 
   I2 == (V1 - V2)/(1/(s*C2)), I3 == V2/R3, I4 == (V1 - V3)/R4, 
   I5 == (V1 - V3)/(1/(s*C1)), I6 == V3/R6}, {I1, I2, I3, I4, I5, I6, 
   I7, I8, I9, V1, V2, V3}]]

The DC-current source can be modelled in the s-domain as \$\text{I}_\text{x}=\frac{\text{i}}{\text{s}}\$. So we get:

FullSimplify[
 Solve[{I6 == (i/s) + I8, I3 == I1 + I2, I9 == I1 + I2, 
   i/s == I7 + I9, I7 == I4 + I5, I6 == I4 + I5, I1 == (V1 - V2)/R1, 
   I2 == (V1 - V2)/(1/(s*C2)), I3 == V2/R3, I4 == (V1 - V3)/R4, 
   I5 == (V1 - V3)/(1/(s*C1)), I6 == V3/R6}, {I1, I2, I3, I4, I5, I6, 
   I7, I8, I9, V1, V2, V3}]]

And this gave me:

{{I1 -> (i (R4 + R6 + C1 R4 R6 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  I2 -> (C2 i R1 (R4 + R6 + C1 R4 R6 s))/(
   R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
    C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2), 
  I3 -> (i (1 + C2 R1 s) (R4 + R6 + C1 R4 R6 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  I4 -> (i (R1 + R3 + C2 R1 R3 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  I5 -> (C1 i R4 (R1 + R3 + C2 R1 R3 s))/(
   R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
    C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2), 
  I6 -> (i (R1 + R3 + C2 R1 R3 s) (1 + C1 R4 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  I7 -> (i (R1 + R3 + C2 R1 R3 s) (1 + C1 R4 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  I8 -> -((i (1 + C2 R1 s) (R4 + R6 + C1 R4 R6 s))/(
    s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
       C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2))), 
  I9 -> (i (1 + C2 R1 s) (R4 + R6 + C1 R4 R6 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  V1 -> (i (R1 + R3 + C2 R1 R3 s) (R4 + R6 + C1 R4 R6 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  V2 -> (i R3 (1 + C2 R1 s) (R4 + R6 + C1 R4 R6 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2)), 
  V3 -> (i (R1 + R3 + C2 R1 R3 s) (R6 + C1 R4 R6 s))/(
   s (R1 + R3 + R4 + R6 + C1 R4 (R1 + R3 + R6) s + 
      C2 R1 (R3 + R4 + R6) s + C1 C2 R1 R4 (R3 + R6) s^2))}}

Trying some values \$\text{R}_1=\text{R}_3=\text{R}_4=\text{R}_6=1\space\text{k}\Omega\$, \$\text{C}_1=\text{C}_2=6\space\mu\text{F}\$ and \$\text{i}=10\space\text{A}\$, gave me:

In[1]:=FullSimplify[
 Solve[{I6 == (10/s) + I8, I3 == I1 + I2, I9 == I1 + I2, 
   10/s == I7 + I9, I7 == I4 + I5, I6 == I4 + I5, 
   I1 == (V1 - V2)/1000, I2 == (V1 - V2)/(1/(s*6*10^(-6))), 
   I3 == V2/1000, I4 == (V1 - V3)/1000, 
   I5 == (V1 - V3)/(1/(s*6*10^(-6))), I6 == V3/1000}, {I1, I2, I3, I4,
    I5, I6, I7, I8, I9, V1, V2, V3}]]

Out[1]={{I1 -> 2500/(s (500 + 3 s)), I2 -> 15/(500 + 3 s), I3 -> 5/s, 
  I4 -> 2500/(s (500 + 3 s)), I5 -> 15/(500 + 3 s), I6 -> 5/s, 
  I7 -> 5/s, I8 -> -(5/s), I9 -> 5/s, 
  V1 -> (5000 (1000 + 3 s))/(s (500 + 3 s)), V2 -> 5000/s, 
  V3 -> 5000/s}}

So, when I, for example, solve for \$\text{V}_1\$ in the time-domain, I get:

In[2]:=FullSimplify[
 InverseLaplaceTransform[(5000 (1000 + 3 s))/(s (500 + 3 s)), s, t]]

Out[2]=5000 (2 - E^(-500 t/3))

Which is:

$$\text{V}_1\left(t\right)=5000\left(2-\exp\left(-\frac{500t}{3}\right)\right)\tag5$$

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