0
\$\begingroup\$

While writing a paper about an experiment I am conducting, I was wondering how inductor-capacitor circuits theoretically behaved. In theory, we assume that wires and components that are not resistors have zero resistance. The resistance of the inductors and capacitors would also be zero. If we had an inductor and capacitor in series and an ac power supply connected which inputted, say, a sine wave, wouldn't the output to ground be a similar function with the same amplitude?

The reason I ask this is that the ac input function damps (decays to zero)when there is resistance in the circuit. And with these circuits, there exists no resistance to decay the function. Whether the circuit is underdamped or overdamped is not a problem, but the main question I have here is about the impedance in these circuits.

In theory (assuming superconductivity), since there is no resistance and thus purely imaginary impedance, would the signal oscillate forever between the inductor and capacitor in an LC circuit?

If I think about it myself, the answer would be yes. But I believe there is a nuance I am missing or a false fact I am assuming.

Thank you.

\$\endgroup\$
6
  • \$\begingroup\$ Since your name is suggestive, have you attempted to apply ODE/PDE to your own question? Also, can you think of any other physical effects which have been discovered that might offer other opportunities for the dissipation of energy? \$\endgroup\$
    – jonk
    Jul 31, 2018 at 5:57
  • \$\begingroup\$ @jonk well, yeah but I don't see how any of the differential equations for capacitance or inductance can help solve this problem, I think this is just an impedance problem. Since this is only an inductor and capacitor in series, I don't think there would be any power dissipation assuming there is no resistance in the circuit. \$\endgroup\$ Jul 31, 2018 at 6:01
  • 1
    \$\begingroup\$ Impedance formulae for L and C parts are pretty common \$\endgroup\$ Jul 31, 2018 at 6:08
  • \$\begingroup\$ @PartialDifferentials Both notes I wrote might have helped after a fashion. But I gather they didn't. That's okay. I gave it a shot. \$\endgroup\$
    – jonk
    Jul 31, 2018 at 7:15
  • \$\begingroup\$ You may also want to consider the magnetic flux. I believe that Eddy currents or just plain coupling to other loops can also cause dissipation if those currents feel some resistance. \$\endgroup\$
    – Sven B
    Jul 31, 2018 at 7:16

2 Answers 2

2
\$\begingroup\$

What is the impedance in a theoretical LC circuit?

In series:

$$Z = j\omega L+\dfrac{1}{j\omega C}$$

$$\verb/ if /Z = 0 \verb/ then / \omega L= \dfrac{1}{\omega C}$$

$$\omega = \dfrac{1}{\sqrt{LC}}$$

In theory (assuming superconductivity), since there is no resistance and thus purely imaginary impedance, would the signal oscillate forever between the inductor and capacitor in an LC circuit?

In theory, with zero losses, yes. However, infinite current will be taken from the source and infinite voltages will develop across L and C. It's a theoretical question so that isn't a problem!

\$\endgroup\$
1
\$\begingroup\$

Well, you described the following problem (using Faraday's law):

$$-\text{V}_{\space\text{in}}\left(t\right)+\text{V}_{\space\text{C}}\left(t\right)=-\text{V}_{\space\text{L}}\left(t\right)\tag1$$

So:

$$\frac{\partial}{\partial t}\left(-\hat{\text{u}}\cdot\sin\left(2\pi\cdot\text{f}\cdot t+\varphi\right)\right)+\frac{1}{\text{C}}\cdot\text{I}_{\space\text{in}}\left(t\right)=-\text{L}\cdot\text{I}_{\space\text{in}}''\left(t\right)\tag2$$

Which gives for \$\text{I}_{\space\text{in}}\left(t\right)\$ (solved using Mathematica):

enter image description here


Using impedance, we can write:

$$\underline{\text{Z}}_{\space\text{in}}=\text{j}\cdot\omega\cdot\text{L}+\frac{1}{\text{j}\cdot\omega\cdot\text{C}}=\left\{\omega\cdot\text{L}-\frac{1}{\omega\cdot\text{C}}\right\}\cdot\text{j}\tag3$$

\$\endgroup\$

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.