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I posted this question on the physics page. I feel like it might be better suited to this site, though.

Suppose we have an ideal LC circuit (no resistance) and an open switch where the capacitor has an initial voltage \$V_o\$.

Initially, the energy stored in the capacitor at \$t=0\$ is \$\frac{1}{2}CV_o^2\$ and the energy in the magnetic field of the inductor is zero because no current is flowing.

Now at time \$t=0+dt\$ we close the switch and current slowly begins to build up. When the current is a maximum, the energy stored in the magnetic field of the inductor is \$\frac{1}{2}LI^2\$ but now the energy stored in the capacitor is zero.

Thus we must have that \$\frac{1}{2}LI^2=\frac{1}{2}CV_o^2\$ because no energy is dissipated since there is no resistance.

There seems to be something very wrong here at a fundamental level. The charge (the electrons) traveling through the inductor at the instant that the current is a maximum have a non-zero kinetic energy (denote this kinetic energy \$K_{charge}\$.) They have to have non-zero kinetic energy since they constitute a current. But if they do posses this energy in addition to the magnetic field energy \$\frac{1}{2}LI^2\$, then the total energy at the moment the current is a maximum will equal \$E_{tot}=\frac{1}{2}LI^2+K_{charge} >E_{initial}=1/2CV_o^2\$. So it seems we have created energy in this process.

The only way I can work around this issue is to assume that the kinetic energy is already somehow factored into the magnetic field energy but I am not sure.

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    \$\begingroup\$ The energy stored by current flowing in an inductor is due to the generated magnetic field. While there is energy stored in the moving electrons, that amount is very small and doesn't contribute to this scenario at all. Being "ideal" as you've noted, that kinetic energy is transferred between the L and the field in the capacitor but it is so much smaller than the cap's E-field energy or the inductor's M-field energy that it's not worth considering. Keep in mind that electrons move quite slowly through a conductor so there is not that much kinetic energy to speak of \$\endgroup\$
    – jwh20
    Apr 16, 2021 at 10:21
  • \$\begingroup\$ The current already factors in this kinetic energy, no? \$\endgroup\$ Apr 16, 2021 at 10:24

2 Answers 2

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Voltage and current are emergent approximations, not fundamental things. If you want to bring the velocities of electrons into consideration, then you have to understand what's going on at the quantum mechanical level. Which is hard work.

Working with I and V requires ignoring the electron behaviour, just as working with pressure, which is also an emergent approximation, requires ignoring individual atoms' velocities and sizes. While pressure and volume are well approximated over a wide range of values, the finite size of atoms causes pressure to deviate from the simple linear laws at very high pressure, and the finite velocities cause (for instance) muzzle velocities to be limited in guns that use expanding gases as a driver.

It turns out that using I and V as if they are fundamental works very well over many more orders of magnitude than do the gas laws approximate the behaviour of gases.

In practice, the L is inferred from energy measurements, so you could consider that any small electronic contribution to energy is already accounted for.

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  • \$\begingroup\$ Thanks for the response. Okay that makes sense. Is it then fair to say that at virtually all reasonable values of voltage and current, the kinetic energy of the electrons is so low relative to the magnetic field energy and the electric field energy that it can be totally neglected with virtual impunity? That is, in my example, when the current is maximum, the energy is in fact \$E_{tot}=\frac{1}{2}LI^2+K_{charge} =E_{initial}=1/2CV_o^2\$ so that \$E_{mag}=\frac{1}{2}LI^2\$ is ever so slightly less that E_{initial} but this difference is too small to detect? \$\endgroup\$ Apr 16, 2021 at 12:13
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In addition to the other answers, I would like to add that there are two different kinds of inductance.

  1. is the usual inductance that describes the energy stored in the magnetic field.

  2. is the kinetic inductance. This happens whenever the mean free path of electrons becomes appreciable, e.g. when transversing vacuum, in superconductors or some other materials which have topological reinforcement of conducting states leading to ballistic transport. This contribution to inductance accounts for the inertia of the charge carriers. An impressive demonstration of this is trying to change the current through a superconductor.

Both are inductances because they describe the desire for a current to continue flowing. When using a component with inductance L in electrical engineering, it does not disclose the origin of its inductance.

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