Kirchhoff's Loop Rule Is For The Birds

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Recommends CR Drost · 2016-01 · Physics Stack Exchange · on “Kirchoff's rules and inductance”

OK, in this video [this video] you've kindly provided me, Lewin essentially talks about this circuit: [LINK] The easy rule of thumb that's common to all electrical engineers is to say: a current $I$ goes through this loop, causing a voltage drop $R I$ across the resistor, a voltage drop $\int_0^t dt~I(t)/C$ across the capacitor (assuming it is uncharged at $t=0$ ), and a negative voltage drop $-L \frac{dI}{dt}$ across the inductor, which then by Kirchoff's rules must sum to 0. Why is Lewin upset with these matters? Lewin is saying that this last point of interpretation is essentially cheating , which we're totally allowed to do in physics; but he points out why it would be hella-confusing to a st

Recommends God's Drunkest Driver · 2016-01 · Physics Stack Exchange · on “Kirchoff's rules and inductance”

by the way, here is the link to one of WL's videos about this subject, in case you want a reference, starting from 22:49. [this video]

Recommends Larry Harson · 2018-11 · Physics Stack Exchange · on “Isn't the inductor equation negative?”

This is what's generally taught in textbooks, but Kirchoff's law doesn't hold in non-conservative circuits where you should be using Faraday's law. Prof Lewin has a nice video [this video] on this

Recommends Markoul11 · 2021-11 · Physics Stack Exchange · on “Kirchhoff's Voltage Law in a General Electromagnetic Field”

Prof. Walter Lewin is correct. Kirchhoff's Voltage Law holds only when the voltage source is inside and part of the current circuit. Even in a transformer circuit the secondary coil of the transformer acts as the connected voltage source of the components circuit connected to the secondary. Dr. Lewin's explanation [LINK] shows clearly that the voltage source (i.e. electromagnet) is not connected and part of the target load circuit. There is no voltage source in the circuit to begin with and apply Kirchhoff's Voltage Law. Kirchhoff's Voltage Law is not wrong but simply not applicable in this case. Antenna theory is more suitable for describing the case where different impedance elements are E

Cites Ziko Humlesen · 2016-06 · Physics Stack Exchange · on “Differential equation in RL-circuit”

I am self-studying electromagnetism right now (by reading University Physics 13th edition) and for some reason I always want to understand things in a crystalclear way and in depth. Now look at this simple RL-circuit: [LINK] I am assuming the wire is superconducting and suppose at $t=0$ the switch is turned on. My problem is that I want to come up with this differential equation for this circuit: $V-I\cdot R=-L\cdot \frac{dI}{dt}$. I know that I can't use Kirchoff's voltage law $(\oint \vec{E}\cdot d\vec{l}=0)$, because there is a non-conservative electric field near the inductor. Therefore I use Farayday's law of induction: \begin{equation} \mathcal{E}=\oint_{\Gamma} \vec{E}\cdot d\vec{l}=-

Cites Ziko Humlesen · 2016-06 · Physics Stack Exchange · on “Differential equation in RL-circuit”

Thank you for you answer and the video, but I still don't understand how can $\vec{E_c}$ be nonzero in the conducter, when the wire has negligible resistance? And since it is nonzero doesn't it contradict Walter Lewin's statement about the term $L\cdot \frac{dI}{dt}$ not being a part of the closed loop integral ( [this video] )? I can think of a much easier way to derive the differential equation, but it shouldn't matter how I approach this problem and I really want to understand my method, so I would be very grateful if you can answer my questions.

Cites ajb · 2018-11 · Electrical Engineering Stack Exchange · on “Does this experiment show that Kirchhoff's Law hold when there's a changing magnetic field involved in a circuit?”

The lumped component models to which KVL is applied are just that--models. Like all models, they are only accurate to the extent that they represent the relevant characteristics of the system they reflect. The simple loop of two resistors model does not represent the susceptibility of the conductive path that constitutes the circuit to induced EMF, therefore this simple model will not reflect the behavior of the real circuit in the real world where induced EMF is a thing that happens. The simple model can be made more accurate by including inductors between the resistors and an additional inductor that represents the solenoid that provides the changing magnetic field. By considering the coup

Cites Math Keeps Me Busy · 2021-01 · Electrical Engineering Stack Exchange · on “Primary voltage drop in current transformer”

Would we see a larger reading on VM2 compared with VM1 (10 mm and 5 mm from the plane of the CT, for example)? Unlikely. Each volt-meter together with its leads and corresponding section of the wire passing through the core form a loop. That loop is cut by the varying magnetic field inside the core of the CT. Thus, they form a one loop secondary of the transformer, and will see a voltage almost identical to the voltage drop of the primary. Almost all of the magnetic field is confined to the core, so the area of the loop will have little effect, although theoretically, if the loop for vm2 encloses that for vm1, it might encircle more magnetic lines of force. Where does the voltage drop occur?

Cites Sredni Vashtar · 2021-03 · Electrical Engineering Stack Exchange · on “What would a voltmeter measure if you had an electromotive force generated by a changing magnetic field?”

If the varying magnetic field is confined inside the triangle, the (high impedance) voltmeter placed outside it, as you depicted it, will measure a voltage equal to R * I, where I is the current flowing in the loop I = emf / (R + R + R). And it will consistently read that value no matter how the probes are placed, twisted, untwisted, curled, long or short - as long as the probes stay outside of the variable magnetic field region (that is outside of the triangle), do not run circles around it and there are no variable magnetic fields outside of the region inside the triangle. To illustrate why it is so, I will use without loss of generality a simpler circuit with only two resistors that I wil

Cites Math Keeps Me Busy · 2021-03 · Electrical Engineering Stack Exchange · on “What would a voltmeter measure if you had an electromotive force generated by a changing magnetic field?”

What would a voltmeter measure if you had an electromotive force generated by a changing magnetic field? The voltmeter is supposed to measure the voltage but there is no notion of voltage potential in the case of the induced field because the field is no longer conservative, so really what is it that the voltmeter measures? A voltmeter measures the (negative of the) total emf around any loop of which it is a part, minus the voltage drop on any element in that loop that drops voltage. In a situation where there is an induced electro-motive force that is not confined to a single component, but distributed across components, one must distinguish between: electromotive force (emf) voltage drop p