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If I have a simple DC circuit consisting only of an LED output, how does Kirchhoff recognize the fact that information has left the loop?

Since, in accordance with relativity(?), you can't have information without the loss of energy.

I presume the loss is negligible down to some small voltage like 10^(-20)V or something, but still...

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closed as unclear what you're asking by Andy aka, Eugene Sh., Leon Heller, Neil_UK, Lior Bilia Nov 8 '17 at 21:49

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ What? I mean.. what?? \$\endgroup\$ – Eugene Sh. Nov 8 '17 at 18:44
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    \$\begingroup\$ What information has left the loop? The voltage drop has become photon emission and heat. \$\endgroup\$ – Ignacio Vazquez-Abrams Nov 8 '17 at 18:45
  • \$\begingroup\$ The information that has left the loop in this case, is that the circuit is closed rather than open. \$\endgroup\$ – theDoctor Nov 8 '17 at 18:49
  • \$\begingroup\$ Please provide a circuit to help us understand. Kirchoff always applies both the voltage and the current law - assuming we ignore RF where some of this ends up as travelling waves. \$\endgroup\$ – Warren Hill Nov 8 '17 at 19:33
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    \$\begingroup\$ Kirchoff's laws do not account for ordinary electromagnetic radiation from a circuit, either -- a much more significant effect. \$\endgroup\$ – Dave Tweed Nov 8 '17 at 19:46
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Kirchhoff's laws are a huge simplification of Maxwell's equations under some very specific assumptions (lumped elements, quasi-static fields, etc.).

Hence they are part of classical physics, as Maxwell's equations are.

Relativity and quantum mechanics are not part of classical physics, so you shouldn't expect that relativistic effects could, in any sense, be modeled by KVL and KCL.

EDIT (to answer part of a comment)

You say:

I wonder if some adjustment should be made, much like Einstein did for physics (approaching 0.5c)...

This hasn't got much sense, and I'll try to explain why.

Einstein's theory (special relativity) added "adjustments" to what, at the time, was the most accurate theory available (Newtonian mechanics), and took another extremely accurate theory (Maxwell's equations) as foundational (he postulated that Maxwell's equations were invariant in any inertial frame of reference, IIRC).

What would be the point to refine KVL and KCL? They are already much more inaccurate than Maxwell's equations, as a general theory. Any relativistic adjustment would be a drop in the ocean. Once you get the frequency sufficiently high, radiative effects and transmission line behavior (both predictable using classical Maxwell's theory) are enough to swamp any possible relativistic correction.

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  • \$\begingroup\$ Thanks, I wonder if some adjustment should be made, much like Einstein did for physics (approaching 0.5c) once you get to limit points like some quantum of a volt. \$\endgroup\$ – theDoctor Nov 8 '17 at 19:02
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    \$\begingroup\$ To the best of our knowledge, voltage is not quantized, unlike things like mass, energy and charge. \$\endgroup\$ – Dave Tweed Nov 8 '17 at 19:45
  • \$\begingroup\$ @DaveTweed: Is not current quantized if energy and charge are quantized? \$\endgroup\$ – theDoctor Nov 8 '17 at 19:48
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    \$\begingroup\$ @theDoctor Yes, but current is charge per second. Time would, therefore, have to be quantized. \$\endgroup\$ – calcium3000 Nov 8 '17 at 20:04
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We already do, account for information loss that is... sort of.

With high frequency or extremely low signals we do account for losses under the umbrella term of parasitics. Kirchov's laws still apply, it's just we add in not-so-imaginary components and circuit paths to account for the losses.

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