Walter Lewin in his lectures says but all books and online resources go against him. Who is correct?
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5$\begingroup$ If you're asking if there can be a voltage across an inductor, the answer is of course yes and I know of no book or online resource that denies this. So you must be asking something else. Please add some context like a schematic drawing, a description of the problem, and quotes from some of the books and online resources you refer to. $\endgroup$– Alfred CentauriCommented Jul 1, 2018 at 12:10
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2 Answers
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You probably misunderstood what Lewin is saying. I watched one of his lectures where he deals with inductor in a circuit and he gets the potential differences correctly, including the inductor. Here is the video of the lecture:
https://www.youtube.com/watch?v=cZN0AyNR4Kw
What he is criticizing is a different thing - that teachers and textbook authors explain Kirchhoff's law in circuits with inductors incorrectly: they assume
$$ \oint \mathbf E\cdot d\mathbf s = 0 $$
is valid even for circuit with inductor (it is not) and then rewrite this using partial integrals across the elements in the circuit, assuming (again incorrectly) that the integral over the wire of the inductor is $+LdI/dt$ (in fact, the integral is much lower and for ideal coil, zero since there is no field inside perfect conductor).
The Kirchhoff Voltage Law is really a practical rule to formulate circuit equations rather than a law of physics or a specific condition valid only in some cases. KVL is based on the fact that since potential is single-valued function of position, sum of drops of potential in a closed path is zero. KVL is not and does not derive from integral of total electric field being zero. Even if $\oint \mathbf E \cdot d\mathbf s \neq 0$, the sum of potential drops in a circuit is still 0. This is because electrostatic potential is a function of position, it does not depend on path, so one must get to the same value after a round-trip. In usual circuits, including low frequency circuits with inductors, difference of electrostatic potential is measurable and Kirchhoff's voltage law is valid. Drop of voltage across the inductor is $+L dI/dt$.
answered Jul 1, 2018 at 17:39
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$\begingroup$ The link has put an end to my nightmares. Thank you. $\endgroup$– KashmiriCommented Oct 19, 2020 at 12:26
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$\begingroup$ "The correct way is to teach that the Kirchhoff's law is really about drops of potential, not about integral of electric field." Some questions: (1) I thought the drop in potential was the integral? (2) If not, can you explain the definition of the potential drop? (3) Isn't the scalar potential ill-defined in the presence of a changing magnetic field? $\endgroup$ Commented Aug 21, 2023 at 0:24
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$\begingroup$ I wrote a lot of answers on this topic, see e.g. e.g.physics.stackexchange.com/questions/442183/… , physics.stackexchange.com/questions/311824/… $\endgroup$ Commented Aug 21, 2023 at 0:33
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$\begingroup$ @JánLalinský Ok, (1)-(3) have been answered for me. I think the absolute best explanation by you is in physics.stackexchange.com/questions/473631/…, so I will leave the link here. Thank you. $\endgroup$ Commented Aug 21, 2023 at 17:36
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$\begingroup$ @MaximalIdeal you're right, that is probably the most exhaustive one I wrote. Bookmarking it as well for future. $\endgroup$ Commented Aug 21, 2023 at 17:56
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If there is a voltage across an inductor, then it's current will be changing. This is essentially the definition of what an inductor is.
If you're looking for a DC steady state solution, then there cannot be a voltage across the inductor. Because if there were, the current would be changing, which would mean the system is not at steady state.
answered Jul 1, 2018 at 16:24
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$\begingroup$ There cannot be, in DC current situation, voltage across ideal inductor with zero resistance. But in real attempts to measure this, there will be some voltage, due to ohmic resistance of real inductors. $\endgroup$ Commented Aug 21, 2023 at 18:04
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$\begingroup$ @JánLalinský, this (5 year old) question appears to me to be one about circuit theory, not practical circuits. Of course the question is not especially clear and I may have guessed wrong about that. $\endgroup$ Commented Aug 21, 2023 at 18:48
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$\begingroup$ The question is badly phrased, your answer is fine, just pointing out a detail so no miscommunication happens. $\endgroup$ Commented Aug 21, 2023 at 18:55