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In undergrad physics, when analyzing an LR circuit, it is often considered that Kirchoff rule holds. However, as far as I understand, Kirchoff rule only holds when E field is conservative (curl of E is 0, there is no change of flux). In a coil L, there is obviously a self-induced emf which makes closed loop integral E.dl non zero. My question is, does that mean we can only apply Faraday's law but not Kirchoff rule here.

Another question, even when there is no coil (only R's for example), in practice as we turn on the power supply, the current will create a B field inside the wire loop which in turn creates a change in magnetic flux. As a result, we again have a non-conservative E field here. Is it correct to say Kirchoff rule might never be applicable in practice? (at least to an absolute accuracy)

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  • $\begingroup$ What is your statement of 'Kirchhoff's rule'? $\endgroup$ Commented Jul 31, 2018 at 13:15
  • $\begingroup$ I mean integral E.dl around a closed loop is 0, i.e voltage change is path-independent and sum of V around a circuit is 0 $\endgroup$ Commented Jul 31, 2018 at 14:30
  • $\begingroup$ All models have errors. Nonetheless, some models are useful. $\endgroup$
    – The Photon
    Commented Jul 31, 2018 at 14:32
  • $\begingroup$ Did you check Wikipedia? $\endgroup$
    – Qmechanic
    Commented Jul 31, 2018 at 15:32

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In a coil L, there is obviously a self-induced emf which makes closed loop integral E.dl non zero.

This is the EMF of a path encircling the core of the inductor, in roughly the path of one loop of the coil.

For lumped circuit analysis we aren't interested in this, but in the potential difference between one end of the coil and the other. This p.d. is indeed the EMF taken along the looping path of the coil. But as far as it interacts with the rest of the circuit, we can treat it as an electrostatic potential difference, which, when summed with the differences across the other elements in a loop of the circuit, will come to 0 according to Kirchoff's voltage law.

even when there is no coil (only R's for example), in practice as we turn on the power supply, the current will create a B field inside the wire loop which in turn creates a change in magnetic flux. As a result, we again have a non-conservative E field here. Is it correct to say Kirchoff rule might never be applicable in practice?

This is indeed the issue when we talk about the lumped circuit approximation. The magnetic flux in the loop formed by the circuit elements is not accounted for in lumped circuit analysis.

And indeed the lumped circuit approximation is an approximation; it is not absolutely correct.

But often the errors introduced by this approximation are much smaller than other errors in our analysis (for example due to not knowing the characteristics of the components accurately, not knowing the temperature of the components accurately, etc.) So in practice, Kirchoff's voltage law is usefully applicable in a great many real circuits.

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Kirchhoff's loop rule, alias Kirchhoff's voltage law, alias Kirchhoff's second law can be a stated thus: the sum of the pds taken in order around a loop of the network is zero (Purcell).

You are saying (in a comment above) that the mean integral E.dl around a closed loop is 0. This is not at all the same thing. The sum of potential differences is integral $\vec{E_{cons}}.d\vec{l}$ in which the electric field, $\vec{E_{cons}}, $ is only the conservative component of the electric field, and excludes, for example, electric fields due to changing magnetic fields.

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    $\begingroup$ I found this lecture online: youtube.com/… at 8:25 and 9:30 he stated something interesting that I find relevant to my question. The lecturer stated that Kirchhoff isn't applicable to this situation which confuse me quite a bit. $\endgroup$ Commented Aug 1, 2018 at 7:43
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In undergrad physics, when analyzing an LR circuit, it is often considered that Kirchoff rule holds. However, as far as I understand, Kirchoff rule only holds when E field is conservative (curl of E is 0, there is no change of flux). In a coil L, there is obviously a self-induced emf which makes closed loop integral E.dl non zero. My question is, does that mean we can only apply Faraday's law but not Kirchoff rule here.

No, Kirchhoff's Voltage Law is applicable even for inductors. This is because the effect of non-conservative field is limited to the inductor and is accounted for accurately by the potential drop $LdI/dt$ on the inductor; validity of standard formulae for potential drops on other elements is not affected.

Problem would arise if there was an external induced electric field acting on some parts of the circuit (or everywhere in it), say due to moving magnet near the circuit, or due to electric field of another circuit with high mutual inductance (like in transformer).

Then, KVL would be still valid, but would be hard to use, because the additional emf would invalidate the familiar formulae for potential drop on standard elements like $0$ for wire, $RI$ for resistor, $LdI/dt$ for inductor.

However, the original Kirchhoff's second circuital law would still be applicable; it is formulated using circuit emfs and $RI$'s: sum of all emf's acting in a closed loop equals sum of terms $RI$ for all loop elements.

So the additional external emf could be accounted for by an additional EMF term in the equation.

Another question, even when there is no coil (only R's for example), in practice as we turn on the power supply, the current will create a B field inside the wire loop which in turn creates a change in magnetic flux. As a result, we again have a non-conservative E field here. Is it correct to say Kirchoff rule might never be applicable in practice? (at least to an absolute accuracy)

No. KVL in terms of potential drops is always true, but sometimes it is not applicable because we can't express those potential drops. Then we can sometimes turn back to the original formulation, which is applicable in this case.

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