What is the difference between the self-inductance of a branch and self-inductance of a loop in lumped circuit models?

Question

When we talk about lumped circuit model, I get confused about the terminology that is generally used. For a loop, we do know that the total flux through the loop can be written as $\Phi = \Phi_{ext} + L_sI$, where $\Phi$ is the total flux through the loop, $\Phi_{ext}$ is the flux generated by external sources and $I$ is the current through the loop. My question is whether $L_s$ is the geometric self-inductance of the loop or the summation of the geometric self-inductance of the loop plus the inductor branch?

Consider this example shown in the figure, let us start with the lumped circuit model. This implies that the the equation $\oint\bar{E}.\bar{dl}=0$ is valid. Assuming that $L_s$ in the above equation is the total inductance then (assuming there are no external sources for flux), the equation becomes

$$ \Phi = L_1I_1. $$

If the current source is time dependent with the equation $I_1 = I_0 \sin(\omega t)$ (here $\omega$ is considered to be 'slow' so that the dimensions of the circuit are much smaller than the associated wavelength of the EMWs, maybe like $10 Hz$), then the total flux is changing, which implies that from Faraday's law

$$ \oint \bar{E}.\bar{dl} = -\frac{d\Phi}{dt} = -L_1 I_0 \omega \cos(\omega t). $$ But, this violates the lumped circuit model we started with according to which $\oint \bar{E}.\bar{dl} = 0$. Thus, to save the lumped element model the initial assumption that the $L_s$ in the equation $\Phi = \Phi_{ext} + L_sI$ is the total inductance is wrong. Then what is that $L_s$? I was assuming that the $L_s$ is the geometric inductance of the entire loop excluding the branches. Is that correct?

Answer 1

The lumped circuit model assumes that all of these effects are negligible. That is, $\frac{d\Phi}{dt}\approx 0$. For your sinusoidal stimulus example, that requires essentially that $L_1\omega\approx 0$. Whether you get there by $L_1$ being very small or $\omega$ being very small doesn't matter.

Answer 2

The lumped element model doesn't assume $\oint\vec E\cdot d\vec\ell=0$. This is in direct contradiction with Faraday's law in circuits where induced emfs are not negligible (such as circuits with inductors). Note that $\vec E = 0$ in an ideal inductor. Ultimately Kirchhoff's voltage law (KVL, which is what you seem to be referring to when you say "lumped circuit model"), which is used in analyzing lumped circuits, is a statement of Faraday's law and should not contradict it.

Faraday's law states $$\oint\vec E\cdot d\vec\ell=-\frac{d\Phi}{dt}.$$ For a single loop like the one in your problem, $$\oint\vec E\cdot d\vec\ell=-\frac {d}{dt}(\Phi_\ell+\Phi_i)=-(L_\ell+L_i)\frac {dI}{dt}$$ $$\oint\vec E\cdot d\vec\ell+L_i\frac{dI}{dt}+L_\ell\frac{dI}{dt}=0$$ where $\ell$ denotes the flux and inductance of the loop (not including the "lumped" inductor), and $i$ denotes those due to the "lumped" inductor.

Now, in the lumped element model as it is normally used, the induced emf due to the loop itself (the negative of the third term) is considered negligible (more on this later). Then, we have $$\oint\vec E\cdot d\vec\ell+V_i=0$$ where $V_i=L_i\ dI/dt$ is the voltage across the inductor. When you also write $\oint\vec E\cdot d\vec\ell$ as a sum of discrete voltage drops across the various elements in the loop (not including the inductor, whose voltage has been included in the second term), this equation is KVL. Note once again that $\oint\vec E\cdot d\vec\ell\ne0$ in general.

Finally, in lumped element analysis of high frequency circuits where the loop emf is not negligible, the loop inductance is often represented in the circuit schematic as a separate inductor with inductance $L_\ell$, even when the actual circuit doesn't contain a discrete inductor in the first place. Then, KVL is a more faithful statement of Faraday's law: $$\oint\vec E\cdot d\vec\ell+V_i+V_\ell=0$$ with $V_\ell=L_\ell\ dI/dt$.