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Jun 8 at 1:15 comment added The Photon @Vinay5101, Jan is a very sharp guy and there is probably some deep point to his answer, but I am not ready to wade through it carefully enough to explain or contest it. What I'd say is from the point of view of circuit analysis, you shouldn't worry about this deeper explanation of emf, and just worry about the special case he mentions where they become equivalent (because this is the only case that is handled by everyday circuit theory) but with opposite sign conventions...and use V = L dI/dt to solve circuits.
Jun 7 at 5:00 comment added Vinay5101 Re: "I'm not clear on where you got this claim, or how you are interpreting the sign of the induced emf." I saw this in the answer given by Ján Lalinský under the subtopic "potential drop" in a question. Here is its link: physics.stackexchange.com/a/682142/389510
Jun 5 at 19:52 history edited The Photon CC BY-SA 4.0
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Jun 5 at 19:44 history answered The Photon CC BY-SA 4.0