I am working on a cardiovascular model that uses circuit diagram as an approximation and this is the circuit of a bifurcation but I got troubles determining the voltage on the capacitors and the current through the branches.

I got a following diagram where the parameter of the elements are known and a time-varying input current is defined

enter image description here

After I applied the KVL, I end up with the following differential equations:

$$\frac{di_2}{dt}=-\frac{(R_2+R_3)}{L_2} i_2-\frac{V_{C1}}{L_2}-\frac{R_1}{L_2} i_1 - \frac{L_1}{L_2} \frac{di_1}{dt} + V_I$$ $$\frac{dV_{C1}}{dt} = \frac{i_2}{C_1} - \frac{V_{C1}}{C_1 R_4}$$ $$\frac{di_3}{dt}=-\frac{(R_5+R_6)}{L_3} i_3-\frac{V_{C2}}{L_3}-\frac{R_1}{L_3} i_1 - \frac{L_1}{L_3} \frac{di_1}{dt} + V_I$$ $$\frac{dV_{C2}}{dt} = \frac{i_3}{C_2} - \frac{V_{C2}}{C_2 R_7}$$

where \$i_2\$ and \$i_3\$ are currents on the branches, \$V_I\$ is the voltage on the source. However that's where I am stuck and don't know how to proceed as I am a bit confused about the \$V_I\$ term how to determine it.

Am I missing anything information or did I forgot a step? Thank you very much for help.

  • 2
    \$\begingroup\$ The input is a current source, not a voltage source, so using the voltage across this current source is probably not appropriate. Laplace transform may provide an easier path to the solution. \$\endgroup\$
    – Chu
    Commented Aug 19, 2019 at 16:35
  • \$\begingroup\$ doesn't bifurcation require a non-linearity? \$\endgroup\$ Commented Aug 21, 2019 at 6:30
  • \$\begingroup\$ @analogsystemsrf the parameters for the circuits are determined by the geometry of the bifurcation so it should be constant \$\endgroup\$
    – PeterT
    Commented Aug 22, 2019 at 10:40

2 Answers 2


A couple of tips:

  • The source is a current source. Therefore any elements placed in series with the source (R1 and L1) will have no effect on the rest of the circuit. They will affect the voltage the current source needs to produce to work. They won't affect the current through the two main branches.

  • R2 and R3 (and R5 and R6) are in series. You can combine them immediately to simplify your analysis.

  • This is the kind of problem where it might make more sense to just go to a simulator and get a numerical answer rather than mess around with the algebra to get an analytic result. (Even if you need an analytic answer, you can use a simulator to check your work very easily)

  • \$\begingroup\$ out of curiosity, what would be good simple simulator to solve this? \$\endgroup\$
    – PeterT
    Commented Aug 20, 2019 at 15:30
  • \$\begingroup\$ @PeterT, whichever one you know the best. LTSpice, TINA, ngspice, PSpice, etc. Any of them could be used for this. \$\endgroup\$
    – The Photon
    Commented Aug 20, 2019 at 16:10

For the first branch, the voltage across \$\small R_2 \sim L_2\sim R_3\sim (R_4||C_1)\$ is:

$$v=i_1R_2+L_2\frac{di_1}{dt}+i_1R_3+i_{R_4}R_4\:\:\:...\:\:\:(1)$$ And the relationship between the currents through \$R_4\$ and \$C_1\$ is: $$i_{R_4}R_4=\frac{1}{C_1}\int(i_1-i_{R_4})\: dt$$ Differentiating: $$\frac{di_{R_4}}{dt}+\frac{1}{\tau _1}i_{R_4}=\frac{1}{\tau _1}i_1 $$

where \$\tau _1 =R_4C_1\$

This can be solved via the integrating factor method to give \$i_{R_4}\$ in terms of \$i_1\$, which can then be substituted into eqn (1) to give an expression for \$v\$.

A similar analysis will give \$i_2\$ through the second branch (just change the subscripts), and hence the second equation for \$v\$.

Equating the two expressions for \$v\$, and noting that \$i_2=(i-i_1)\$, where \$i\$ is known and provided by the current source, will solve for \$i_1\$ and thence \$i_2\$, and all other currents.


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