3
$\begingroup$

question:

How can we prove mathematically that KVL (Kirchoff's volatage law) and KCL (Kirchoff current law) become invalid at very large frequencies?

i have read this statement in my book but it doesn't explain the fact that why KVL , KCL fails at High frequency

is that because condcuting wires in very circuit , at high frequency start to pose reactances, and don't act like lumped elements(whose electrical length is very less than operating wavelength)

can anyone explain me more at atomic level that what happens inside any resistive Electric circuit(of course condcuting wires are also present) driven by source when source's freqency increases indefinitely

Improve this question
$\endgroup$
3
  • $\begingroup$ Faraday's law says $\oint \vec{E} \cdot d\vec{l} = - d \phi_{B} /dt$. So if the magnetic flux $\phi_{B}$ is time independent, then it reduces to KVL. $\endgroup$
    – K_inverse
    Sep 1, 2018 at 13:24
  • 1
    $\begingroup$ @Veereshpandey: I believe you will best understand why KVL and KCL tend to become invalid at high frequencies if you first understand how KVL and KCL were derived from Maxwell's equations, and under what assumptions were they derived. As an example, why is it that to effectively use KVL, the size of the electrical component must be small compared to the wavelength of the applied electromagnetic field? $\endgroup$
    – Winter Soldier
    Sep 1, 2018 at 14:03
  • 1
    $\begingroup$ @K_inverse: But that doesn't imply the converse, that KVL must become invalid at high frequencies. KVL is still valid under the lumped circuit approximation, because the lumped circuit approximation implies that $\phi_B$ is small. $\endgroup$
    – user4552
    Feb 2, 2019 at 16:40

3 Answers 3

Reset to default
1
$\begingroup$

KCL and KVL both depend on the lumped element model being applicable to the circuit in question. When the model is not applicable, the laws do not apply. KCL and KVL result from the assumptions of the lumped element model.

KCL is dependent on the assumption that the net charge in any wire, junction or lumped component is constant. Whenever the electric field between parts of the circuit is non-negligible, such as when two wires are capacitively coupled, this may not be the case. This occurs in high-frequency AC circuits, where the lumped element model is no longer applicable. For example, in a transmission line, the charge density in the conductor will constantly be oscillating.

Source : https://en.m.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

Improve this answer
$\endgroup$
2
  • $\begingroup$ Please use quote formatting (not bold formatting) for quotes. It's not easy to read a huge chunk of bold text, particularly after the paragraph spacing has been omitted. Also, please take a moment to read our referencing guidelines, which suggests that answers should ideally not be one humongous quote. $\endgroup$
    – user191954
    Feb 2, 2019 at 12:34
  • $\begingroup$ This seems reasonable as a description of one reason why KCL can become invalid. However, it doesn't explicitly address the question re KVL. It's also a little sad to see an answer that is nothing more than a cut-and-paste from WP. That's not much better than a link-only answer, which is discouraged on SE. $\endgroup$
    – user4552
    Feb 2, 2019 at 16:46
0
$\begingroup$

At high frequencies circuits behave like transmission lines, meaning that electromagnetic waves will propagate along the circuit. Unlike "regular voltages" and currents these waves possess a dependence of the their amplitude dependening on the distance they have travelled. This means, that the kirchhoff laws do not apply anymore Generally, there will be a superposition of a forward and backward going (voltage/current) wave. These waves are usually proportional to a factor of $$e^{+/- ikx}$$, with $$k=2 \pi/\lambda$$ denoting the wave vector of the wave,$$\lambda$$ its wavelength, $$i$$ denoting the imaginary unit and $$x$$ denoting the distance of the waves to the generator. For low frequencies, the term $$e^{+/-ikx}$$ will be small because the the wavelength will be small. Thus, $$e^{i*0}=1$$ and the the propagating waves are nearly constant with distance and only dependent on the time. This should be the limit, when wave effects can be neglected. To my knowledge (dont take this for granted) for practical purposes wave effects have to be considered when the dimension of the transmission line/or circuit exceeds a length of $$\lambda/10$$. For example unintuitively power transmission lines that work on very low frequencies are also modelled in the same manner as high frequency.transmission lines for radio waves, because the cover huge distances. For more on tranmission lines you can get started here:https://en.wikipedia.org/wiki/Transmission_line

Improve this answer
$\endgroup$
0
$\begingroup$

Kirchhoff's voltage law (KVL) in its modern form is always valid, due to definition of voltage: difference of potentials. Since potential is a function of position, sum of voltage drops along a closed loop is always zero. This has nothing to do with the Faraday law, which concerns total electric field: the voltage drops across any path segment do not, in general, reflect total electric field there, but only its potential part.

Kirchhoff's current law (KCL) ceases to be valid when the electric current charging up wire surfaces ceases to be negligible when compared to current flowing along the wires. This happens when electric current in the circuit oscillates with very high frequency.

Improve this answer
$\endgroup$
5
  • $\begingroup$ The first paragraph is wrong. When there are time-varying magnetic fields, in general the electric field becomes nonconservative, and there is no well-defined electric potential. $\endgroup$
    – user4552
    Feb 2, 2019 at 16:41
  • $\begingroup$ @BenCrowell Electric potential can be defined even when electric field becomes nonconservative and it is a standard practice to do so both in EM theory and use it in practical AC circuit analysis. For example, we can talk about voltage across ideal inductor and, in this case too, voltage is a difference of electric potential. There is infinity of potential definitions due to gauge invariance of Maxwell's equations, but the most usual and very convenient one is the Coulomb potential of all charges. This definition does not suddenly cease to be valid when electric field has solenoidal component. $\endgroup$
    – Ján Lalinský
    Feb 2, 2019 at 21:01
  • $\begingroup$ @Lanisky, So basically you are saying that even if we have time varying $B$ field we can break the total electric field as a sum of conservative and non conservative field, and then go on and define potential for the conservative one? Right? $\endgroup$
    – Kashmiri
    Oct 19, 2020 at 16:46
  • $\begingroup$ If that is what you mean then the potential will be time varying as well and therefore the concept of potential will be well defined only on particular time instants. $\endgroup$
    – Kashmiri
    Oct 19, 2020 at 16:49
  • $\begingroup$ @YasirSadiq yes, scalar electric potential in electrical engineering (practice) is almost always the Coulomb potential, i.e. it is a function of charge distribution in space at the same instant. You are correct that this potential at any fixed point is, generally, function of time. $\endgroup$
    – Ján Lalinský
    Oct 19, 2020 at 20:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.