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I need to model (write the equations for) a circuit with resonant inductive coupling, such as the one in the picture below.

image from Wikipedia of a type p-p transmitter

The picture is taken from the Wikipedia article on Resonant inductive coupling.

I need to write the differential equations that define this system and thus to calculate how much current flows through $Rl$, given $Vs$.

My problem is that I have zero clue where to start. I've only solved basic circuits so far (mainly DC). Even if I manage to calculate an equation for the left image (and perhaps solve it numerically), how can I use that for the right one?

My thoughts so far: Using KVL for the first loop in the left circuit $$V_s=V_{C_S}$$ And then applying this to the second loop: $$ V_{C_S}=V_{R_S}+V_{L_S} $$ Then I substitued the formulas I knew for each component: $$\frac{d^2 q}{dt^2}L_S+\frac{d q}{dt}R_S = \frac{q}{C_S}$$

Is my way of thinking good? Or did I neglect something that shouldn't have been? And if It's good, after plugging this into a numerical calculator, how will I be able to tell the current in $Rl$? Any given help would be greatly appreciated!

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  • $\begingroup$ Hi, could you please share what you have tried so far? You say that you are able to solve more basic circuits, what is the difficulty that your are finding in this one? $\endgroup$
    – Prallax
    Jun 24, 2022 at 22:05
  • $\begingroup$ The most that I've had to deal with so far were A-level type of circuits. Mainly DC ones. And even in Uni I've only had beginner electronics and never really dived deeper into AC circuit analysis. The difficulty I find in this one is that even if I'm able to derive the differential equation for the left part, I still can't describe it's effect on the right one mathematically. $\endgroup$ Jun 24, 2022 at 22:43
  • $\begingroup$ I was asking because, according to the homework question policy (physics.meta.stackexchange.com/questions/714/how-do-i-ask-homework-questions-on-physics-stack-exchange), you should try to narrow down your question as much as you can to the aspect of the problem that is troubling you. The best way to do it would be to show how far you can go into a solution, show your work. You were telling me that you may be able to find the equations for the left part. You could edit the question and these equations, to show exactly where you find some trouble. $\endgroup$
    – Prallax
    Jun 25, 2022 at 9:35
  • $\begingroup$ in this way, the question is far more likely to be answered, and far less likely to be closed $\endgroup$
    – Prallax
    Jun 25, 2022 at 9:36
  • $\begingroup$ Yeah, that makes sense. It's just my first case to reach out for online help this way and didn't really know how to do so (re-phased it many times in my head). Thank you for your advices! $\endgroup$ Jun 25, 2022 at 11:24

1 Answer 1

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I think you are on the right track. Just, no need to solve it numerically: second order linear equations with constant coefficients have a simple analytical solution.

The circuit on the left and the one on the right are variants of the RLC circuit, which is a damped oscillator. Try to solve the equation you have written with the condition that ${q \over Cs} = Vs = V_0\sin(\omega t)$. You will see that, depending on $\omega$, the circuit on the right responds in different ways.

The link between the two circuits is in the inductors. If the two circuits are close enough, the magnetic field produced by the inductor on the right will partly reach the inductor on the left. When the magnetic field changes, it will produce a voltage difference on the inductor on the left.

You can model this by writing that the voltage difference at $L_r$ is $V_{L_r} = -k L_s {dI_{L_s} \over dt}$, where $k$ is a coupling constant that depends on the distance between the two coils.

The above is true for any such circuit. If the circuit is also resonant, I believe this means that the resonant frequencies of the circuit on the right and on the left must be the same. But here ends my knowledge on the topic.


Addendum

Reading the comments, I see that an explanation on how to solve simple AC circuits in general might be required.

The differential equation describing an AC circuit is usually a second (or first) order linear ODE with an oscillating forcing term. It often has the form

$$a {d^2 y \over dt^2} + b {dy \over dt} + c y = A_0 cos(\omega t)$$

The oscillating term comes of course from the oscillating current/voltage of the alternating generator, and $\omega$ is the frequency of the generator.

This equation is very common in physics. Any time you talk about oscillations or resonance, chances are that this equation is behind it. The good news is that it is very easy to solve. You can look for sums of $sin$ and $cos$ functions, but if you are a bit familiar with complex numbers, they can save the day.

We can write the oscillating term as $A_0 \cos(\omega t) = \Re(A_0 e^{i\omega t})$.

Look for a solution in the form

$$y(t) =\Re(A e^{i(\omega t + \phi)})$$

where $A$ and $\phi$ are two real numbers to be determined. From now on I will not write "$\Re$" anymore and work just with complex numbers. You just need to remember to take the real part of the result after you have finished the calculation (you can check that this is indeed appropriate). Differentiating leads to

$${dy \over dt} = i\omega A e^{i(\omega t + \phi)} = i\omega y(t)$$ and $${d^2y \over dt^2} = -\omega^2 y(t)$$

Substituting this in the differential equation leads to

$$-a\omega^2 y(t) + ib\omega y(t) + cy(t) = A_0 e^{i\omega t} = {A_0 \over A} e^{-i\phi} y(t)$$

$y(t)$ cancels in all the terms and you are left with an algebraic equation

$$A e^{i\phi} = \frac{A_0}{-a \omega^2 + ib\omega + c}$$

Defining $\omega_0 = \sqrt{c/a}$, the -so called- resonance frequency, this becomes

$$A e^{i\phi} = \frac{A_0/a}{(\omega_0^2 - \omega^2) + i(b/a)\omega}$$

Now, if you take the absolute value of both sides, you get

$$A = \frac{A_0/a}{\sqrt{(\omega_0^2-\omega^2)^2 + (b/a)^2 \omega^2}}$$

while the complex phase is

$$\phi = \tan^{-1}\left( \frac{(b/a)\omega}{\omega^2-\omega_0^2}\right)$$

And the final result is

$$y(t) = A \cos(\omega t + \phi)$$

More information and discussion on damped oscillators, for example here

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  • $\begingroup$ I've only said numerically because not only do I need to modell this, but I also have to make a code out of it either in Python, C or Mathematica. As for the condition you wrote, I'm not really sure how I should do that. Like, I've only solved IVP-s where the conditions were given for the derivates, and not for the "whole function" (if this makes sense?) Should I just change $$ \frac{q}{C_S} $$ for what you've wrote and try solving like that? $\endgroup$ Jun 25, 2022 at 17:04
  • $\begingroup$ Yes, substitute q/Cs with what I wrote and look for a solution that has the form $I(t) = A cos(\omega t) + B sin(\omega t)$, where $A$ and $B$ are determined from the initial conditions (i.e. at t=0) $\endgroup$
    – Prallax
    Jun 25, 2022 at 17:33
  • $\begingroup$ I've done what you've told me and read your Addendum and now I have a concern: With your addittion, I can see how resonance frequency comes into the play, but with mine calculation I don't. I've calculated it by hand and also with other tools and I've got the same (quite ugly) formula. Now I don'nt know if thats a problem or not....Maybe there is a way to simplify it to this form? I've also derived the form if the I(t) function (I being the current) after derivating the gotten q(t) expressen and adding two initial values: q'(0) = 0 and q(0) = 0 (Is that allowed in this case?) $\endgroup$ Jun 25, 2022 at 19:52
  • $\begingroup$ @Clueless_Steph If everything is done correctly, the result that you get with the two methods should be the same (except, possibly, for a phase, if you have used $\sin(\omega t)$ or $\cos(\omega t)$ as forcing term). Unfortunately, when dealing with trigonometric functions, the "same thing" can look quite differently if it is calculated in different ways. Try to calculate the amplitude of the solution: $A^2 + B^2$ should be equal to the $A^2$ from my addendum. At the end of the day, just pick the method that is the most simple to you and stick with it. I suggest the one in the addendum. $\endgroup$
    – Prallax
    Jun 25, 2022 at 20:20
  • $\begingroup$ @Clueless_Steph I realized that I have told you something wrong. You should not find $A$ and $B$ from the initial conditions. You should instead write them in terms of $A_0$, $a$, $b$, $c$ and $\omega$. The initial conditions are not needed, because they are fixed once you write the forcing term as $A_0 cos(\omega t)$ $\endgroup$
    – Prallax
    Jun 25, 2022 at 20:31

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