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I understand the behaviour of an RLC Circuit in Series can be modelled by the equation:

$$ L \frac{d^2Q}{dt^2}+R\frac{dQ}{dt}+\frac{Q}{C}=0 $$ Which can be represented by the system of First Order ODEs $$ L \frac{dI}{dt}+RI+\frac{Q}{C}=0 $$ $$ \frac{dQ}{dt}=I $$ Which we can happily solves using matrices etc.

Furthermore, I understand that an RLC circuit with each component in parallel (diagram at bottom) can be represented by the Second Order ODE:

$$ L \frac{d^2V}{dt^2}+\frac{1}{RC}\frac{dV}{dt}+\frac{V}{LC}=0 $$ However, I am struggling to come up with the system of equations to reduce this to a system of First Order ODEs (like I could for the series example). How would one represent this Parallel RLC circuit as a system of First Order ODEs? Parallel RLC circuit

Sterling Saini
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    I think an answer (or close enough that you should be readily able to do whatever else you feel is needed) is already here at EESE. – periblepsis Aug 14 '23 at 13:41
  • Also, just to be clear... is your question a state-space kind of thing? Sounds like it to me, looking back on this. But what do I know? – periblepsis Aug 14 '23 at 14:23
  • @periblepsis I thank you for your response. I think this might be state-space kind of thing, though I haven't come across that terminology before. In regards to the similar answer - and this may be due to my own incompetence rather than anything else - I am simply looking for a way to reduce the 2nd order ODE to a system of First Order ODEs - this doesn't seem to be covered in the supplied article (though I will keep it for a later case). This is less about solving the question any which way and more - can it be solved this way? – Sterling Saini Aug 15 '23 at 00:07
  • Yeah, then you are looking for a state-space approach. The linked page already has more than enough detail to move you almost to completion. But no, it doesn't hand you a state-space result, boxed up and neatly tied with a bow. – periblepsis Aug 15 '23 at 00:42
  • In your case, the usual approach is to consider the current through the inductor as one state and the voltage across the capacitor as another state. So that's two states. This is usually done this way because the initial conditions (current in inductor and voltage across capacitor) are direct items of interest/knowledge. (This is the same thing, though, as knowing Q and d/dt Q.) Anyway, can you refine the question? I don't see anything to grab ahold of and run with. I could say what I think. But that may not help. Depends on your perspective, not mine. – periblepsis Aug 15 '23 at 04:54
  • I'll write one more time. That's it. There are many choices you can make. But the usual case is the energy of the inductor (propto its current) and the energy of the capacitor (propto its voltage difference.) So $i_{\text{L}}$ and $v{\text{C}}$, where $\dot{i{\text{L}}}=\frac1{L}v{\text{C}}$ and $-\dot{v{\text{C}}}=i{\text{L}}+\frac1{R}v{\text{C}}$ (from KCL.) You will need the initial states for $i{\text{L}}$ and $v{_\text{C}}$ at $t=0$, as there's no driving function. You need to tell me what your problem is with this. I'm flummoxed about your confusion. – periblepsis Aug 20 '23 at 11:20

2 Answers2

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Well, we have the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

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

$$0=\text{I}_1\left(t\right)+\text{I}_2\left(t\right)+\text{I}_3\left(t\right)\tag1$$

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

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1\left(t\right)&=\text{C}\cdot\text{V}'\left(t\right)\\ \\ \text{I}_2\left(t\right)&=\frac{\displaystyle1}{\displaystyle\text{L}}\cdot\text{V}'\left(t\right)\\ \\ \text{I}_3\left(t\right)&=\frac{\displaystyle1}{\displaystyle\text{R}}\cdot\text{V}\left(t\right) \end{alignat*} \end{cases}\tag2 $$

Now, we can use \$(2)\$ to rewrite \$(1)\$:

$$0=\text{C}\cdot\text{V}'\left(t\right)+\frac{\displaystyle1}{\displaystyle\text{L}}\cdot\text{V}'\left(t\right)+\frac{\displaystyle1}{\displaystyle\text{R}}\cdot\text{V}\left(t\right)\tag3$$

Which, when solved gives:

$$\text{V}\left(t\right)=\mathscr{K}\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right)\tag4$$

So, we also know:

\begin{equation} \begin{split} \text{V}'\left(t\right)&=\frac{\displaystyle\partial}{\displaystyle\partial t}\left(\mathscr{K}\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right)\right)\\ \\ &=-\frac{\displaystyle\text{L}\mathscr{K}}{\displaystyle\text{R}\left(1+\text{CL}\right)}\cdot\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right) \end{split}\tag5 \end{equation}

Edit, assuming that \$\displaystyle\text{V}_\text{C}\left(0\right)=\text{V}\left(0\right)=\hat{\text{u}}_0\$, we get:

$$\hat{\text{u}}_0=\mathscr{K}\exp\left(-\frac{\displaystyle\text{L}\cdot0}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right)=\mathscr{K}\cdot1=\mathscr{K}\tag6$$

So, we end up with:

$$\text{V}\left(t\right)=\hat{\text{u}}_0\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right)\tag7$$

And:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1\left(t\right)&=-\frac{\displaystyle\text{CL}\hat{\text{u}}_0}{\displaystyle\text{R}\left(1+\text{CL}\right)}\cdot\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right)\\ \\ \text{I}_2\left(t\right)&=-\frac{\displaystyle\hat{\text{u}}_0}{\displaystyle\text{R}\left(1+\text{CL}\right)}\cdot\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right)\\\ \\ \text{I}_3\left(t\right)&=\frac{\displaystyle\hat{\text{u}}_0}{\displaystyle\text{R}}\cdot\exp\left(-\frac{\displaystyle\text{L}t}{\displaystyle\text{R}\left(1+\text{CL}\right)}\right) \end{alignat*} \end{cases}\tag8 $$

Jan Eerland
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TLDR: When going from series to parallel (parallel to series): switch V with I, L with C, R with 1/R, and Q with \$\Phi\$, the magnetic flux of the inductor.

Each term of the first two ODEs is a voltage. In the second ODE the standard defining relationships make this easy to see.

So in your parallel case it is the currents that add, so this is the standard starting point.

Rewriting the second equation: $$L \frac{dI}{dt}+RI+\frac{Q}{C}=0\tag{Equ 1}$$ reveals that the middle term is Ohm's law in a voltage form.

So replace current with voltage and write the middle term for the parallel case as Ohm's law in a current form: $$C\frac{dV}{dt}+\frac{V}{R}+\frac{X}{L}=0$$

What could X be?

Notice that changing L to C reveals the capacitor current in the first term on the left. So then it makes sense to change C to L as well. But the X-factor need to be clarified.

Comparing to your first equation we can write $$ C \frac{d^2X}{dt^2}+\frac{1}{R}\frac{dX}{dt}+\frac{X}{L}=0 $$ only if \$V=\frac{dX}{dt}\$.

Using Faraday's law of Induction:$$V=-\frac{d\Phi}{dt}$$ where \$\Phi\$ is the magnetic flux produced by the inductor current.

The last term still needs verification, but by substituting \$-\Phi\$ for X, the equation now is in the same form as your first one. $$ C \frac{d^2\Phi}{dt^2}+\frac{1}{R}\frac{d\Phi}{dt}+\frac{\Phi}{L}=0 $$

To verify the last term, look to Faraday's Law again and recognize that inductance is:$$L=\frac{-d\Phi}{dI_L}$$ Then integrating and solving yields: $$I_L=\frac{-\Phi}{L}$$

Of course this can be done with Kirchhoff, but he did not include charge or magnetic flux.

So now we have the equation in the form that your desire: $$C\frac{dV}{dt}+\frac{V}{R}+\frac{\Phi}{L}=0$$

So, "Which we can happily solves using matrices etc."

RussellH
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