Consider a circuit with $L_{1}$ and $L_{2}$ as inductors and $C_{1}$ and $C_{2}$ as the capacitors.

enter image description here

$I$ and $V$ are the manifest variables. Then write $I_{L_{1}}, I_{L_{2}}, I_{C_{1}}, I_{C_{2}}, V_{L_{1}}, V_{L_{2}}, V_{C_{1}}, V_{C_{2}}$ as the latent variables.

enter image description here

Using Kirchoff's current and voltage laws, I deduce

\begin{equation}\begin{cases}I=I_{L_{1}}+I_{L_{2}}\\ I_{L_{1}}=I_{C_{1}}\\ I_{L_{2}}=I_{C_{2}}\\ I_{C_{1}}+I_{C_{2}}=I\end{cases}\end{equation}

\begin{equation}\begin{cases}V=V_{L_{1}}+V_{C_{1}}\\ V=V_{L_{2}}+V_{C_{2}}\\ V_{L_{1}}+V_{C_{1}}=V_{L_{2}}+V_{C_{2}}\end{cases}\end{equation}

\begin{equation}\begin{cases}L_{1}\frac{dI_{L_{1}}}{dt}=V_{L_{1}}\\ L_{2}\frac{dI_{L_{2}}}{dt}=V_{L_{2}}\\ C_{1}\frac{dV_{C_{1}}}{dt}=I_{C_{1}}\\ C_{2}\frac{dV_{C_{2}}}{dt}=I_{C_{2}}\end{cases}\end{equation}

Then, after some elimination, I end up with

\begin{equation}\begin{cases} I=I_{L_{1}}+I_{L_{2}} \\ I_{L_{1}}=C_{1}\frac{dV_{C_{1}}}{dt} \\ I_{L_{2}}=C_{2}\frac{dV_{C_{2}}}{dt}\end{cases} \end{equation}

And \begin{equation} \begin{cases} V={L_{1}}\frac{dI_{L_{1}}}{dt}+V_{C_{1}} \\ V=L_{2}\frac{dI_{L_{2}}}{dt}+V_{C_{2}} \end{cases} \end{equation}

But now I'm stuck, because apart from substituting for $I_{L_{1}}$ and $I_{L_{2}}$, I can't see how to go any further. I want a single differential equation without the latent variables.

  • $\begingroup$ Why don't you solve the system of ODEs by the matrix method? Btw. what are constants, dependent variables and independent variables? If $L_1$ and $L_2$ are in series, shouldn't the current be the same? Maybe you need to use some additional informations about your physical Problem (parallel, series, etc.). $\endgroup$ – MrYouMath Sep 17 '15 at 15:18
  • $\begingroup$ @MrYouMath $I$ and $V$ are the manifest variables and $I_{L_{1}}, I_{L_{2}}, I_{C_{1}}, I_{C_{2}}, V_{L_{1}}, V_{L_{2}}, V_{C_{1}}, V_{C_{2}}$ are the latent variables. $L_{1}$ and $L_{2}$ are inductors and $C_{1}$ and $C_{2}$ are capacitators. And I don't know too much about circuits, but I think it'd be a series-parallel circuit. $\endgroup$ – Jason Born Sep 17 '15 at 16:14
  • $\begingroup$ This is normally attacked in the transformed/complex domain, using Fourier or Laplace. See eg en.wikipedia.org/wiki/… $\endgroup$ – leonbloy Sep 17 '15 at 17:10

This is commonly attacked via the Laplace transform. In this formulation, the impedance of the elements are $Z_{L}= s L$ and $Z_{L}= 1/s C$, hence the total impendance (paralell of two series) is

$$\frac{V}{I}=Z = \left(\frac{1}{s L_1+ \frac{1}{sC_1}}+\frac{1}{s L_2+ \frac{1}{sC_2}}\right)^{-1} $$ Or

$$I = V \left(\frac{s \, C_1}{s^2 L_1 C_1+ 1}+\frac{s \, C_2}{s^2 L_2 C_2+ 1}\right)$$

From this you can (if you really want) obtain the differential equation that links $V_t$ with $I_t$. Also, you can evaluate the impedance $Z$ at specific values of the frequency, as a complex number ($s \to j \omega$) (the real part would correspond to the resistive part; the modulus would give you the ratio of peak or rms values of $V_t$ and $I_t$, assuming they are sinusoids, and so on).

  • $\begingroup$ I understand the transforms, $Z=\frac{V(s)}{I(s)}$, $\mathcal{L}(C_{1})=\frac{1}{sC_{1}}$ and $\mathcal{L}(L_{1})=sL_{1}$, but how did you calculate the total impedance (as what I have to work with, $V$ and $I$ are expressed, in part, by the latent variables of which no transformation is made...I don't think that calculation is clearly explained in the article you linked)? $\endgroup$ – Jason Born Sep 18 '15 at 14:51
  • $\begingroup$ I simply used the rules for Combining impedance (series/parallel) $\endgroup$ – leonbloy Sep 18 '15 at 15:31
  • $\begingroup$ Okay, and then the differential equation that links $V_{t}$ with $I_{t}$ would describe the manifest behaviour? $\endgroup$ – Jason Born Sep 18 '15 at 16:26
  • $\begingroup$ Yes, though it would not be vey easy, see the example in the link. Often one is satisfied with the stationary behaviour for sinusoidal inputs (Fourier analysis), which is simpler, and more so in this case (no resistive component, hence the total impedance is purely imaginary - i.e., it behaves as a single equivalent capacitor or inductance - but the value depends on the frequency) $\endgroup$ – leonbloy Sep 18 '15 at 16:47
  • $\begingroup$ BTW, you might get better answers for this kind of question in electronics.stackexchange.com $\endgroup$ – leonbloy Sep 18 '15 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.