circuit analysis transient response in an RC circuit

Question

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.

Answer 1

Since there are two capacitors, the order of the differential equation cannot be more than \$2\$. See if this helps:

Answer 2

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.

Answer 3

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:

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.

Answer 4

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.

Answer 5

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

^{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:

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

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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:

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