Talk:Inductance

Page contents not supported in other languages.
From Wikipedia, the free encyclopedia


Inductance of a single wire is incorrect (and not physically possible)[edit]

I am not sure what this section is trying to accomplish. A single wire going in a straight line from A to B has infinite inductance per definition! This is because the magnetic field will spread out infinitely large as there is no return wire that generate the opposing field. In practice you always need two wires to pass a current to and from the source (one wire in one direction and one wire going back). The shape of the area between the wires will determine the "size" of the magnetic field and thus the inductance. If you run the wire in a circular loop you simple have an electromagnet with a single turn. Hence not possible to define the inductance of a single wire without a return path: If the return path is infinitely far away, then the induced magnetic field will be infinitely large and hence the inductance will be infinitely large. (Sure, you can define the part of the inductance that only comes from the internal contribution inside the metal, but this is normally only a small part of the total inductance, furthermore this internal inductance can not be measured directly and is only a theoretical concept) If any example should exist in this section it should probably be that of a coaxial cable, as it is the mathematically simplest case (symmetrical shape that can be described with simple analytical formulas). One can also exactly define the high-frequency and the low-frequency inductance mathematically without any arbitrary constants. I think this is the most pedagogic case to use in this article, as all the concepts for a coaxial cable also holds for more complex geometries, such as pair cable etc. EV1TE (talk) 01:27, 28 February 2021 (UTC)[reply]

Just because it goes to infinity doesn't mean that it is infinite. Note, for example, that a sphere has a capacitance which is not infinite. If you consider a spherical capacitor in the limit that the outer sphere goes to infinity, it does not diverge. In the case of inductance, you can take the loop inductance as the radius goes to infinity divided by radius (inductance per unit length), or parallel wires as the spacing goes to infinity, and both diverge as log(d). That is as slow as it can diverge and still diverge.
Consider a gambling game that works like this: flip a coin, and if it is heads you win $1. If tails, you flip again, this time you win $2 for heads, and tail flip again. Each time you flip, heads wins twice as much as before, and tails flips again. How much should you pay to play this game? That is, what is your expected, on the average, winnings?
Now, figure out the connection to the inductance problem. Gah4 (talk) 02:17, 28 February 2021 (UTC)[reply]
Isolated straight wires of finite length do have finite inductance. The magnetic field around them has a finite energy when the wire is carrying a current. It is in the reliable sources. When you make a loop of such wires, their inductance adds and the mutual inductance between them subtracts being you with the inductance of the loop. It all works out. The explanation could be better. So, how do you get a current in such a wire without a complete circuit? You connect the wire between spherical conductors which are charged to different voltages.Constant314 (talk) 03:00, 28 February 2021 (UTC)[reply]
A more common case is some larger, low inductance wiring with a shorter, smaller, high inductance one in between. That does complicate the calculation, but most of the field is close. An important case is PC board vias, which are short but also often small, and high inductance compared to a thick and wide ground plane. Gah4 (talk) 14:19, 28 February 2021 (UTC)[reply]
Agree with Constant314. There's a technique called partial inductance [1], [2], [3], [4] which is used to calculate the inductance of segments of a circuit, particularly straight conductors. As Constant314 says, the total inductance of each straight section is calculated, then the mutual inductance due to other sections is subtracted, leaving the self-inductance. It is used a lot in PCB design. As Gah4 says, in a PCB the return path for the current is typically through a ground plane, a large area of copper, and the distribution of the current through the plane may be hard to calculate, so what constitutes the "circuit" for calculating the flux is ambiguous. I think this article should have a section on partial inductance, if someone wants to write one. It's been on my "to-do" list for a while, but recent events have left me with less time for WP editing. --ChetvornoTALK 17:34, 28 February 2021 (UTC)[reply]
Well also the large ground plane means less inductance. There is one Feynman lecture where he starts with an LC circuit (parallel plate capacitor and wire wound inductor) then increases the resonant frequency. First increasing the plate spacing. Then decreasing the inductance by less coiling (straight wire). Then more inductors in parallel. Until eventually it is a cylindrical can. This does ignore the mutual inductance as they get close together, but the idea that there is a limit of decrease still works. Gah4 (talk) 21:26, 28 February 2021 (UTC)[reply]
I remember that. That was an AWESOME illustration of how a microwave cavity results from efforts to increase the resonant frequency of a tuned circuit. I've been meaning to draw a series of pictures similar to those in Feynman Lectures Vol.2 to add his explanation to the Microwave cavity article. Feynman always came up with the best examples to explain advanced physics concepts to beginning students. --ChetvornoTALK 21:52, 28 February 2021 (UTC)[reply]
Yeah the ground plane reduces inductance. At low frequencies the distribution of current in the plane is determined by resistivity, but as the frequency in the printed circuit increases the return current in the plane concentrates in a narrow channel directly under the PC trace, reducing the volume of magnetic field. --ChetvornoTALK 22:05, 28 February 2021 (UTC)[reply]
Sorry to be didactic but let me distinguish between circuit inductance and partial inductance. You add up all the partials, add some mutual inductances and subtract other mutual inductances to get the circuit inductance. Adding a ground plane changes the circuit, but it can reduce the circuit inductance. I am not sure about the individual partial inductances, but I think that they remain unchanged. Constant314 (talk) 00:17, 1 March 2021 (UTC)[reply]
The idea to simulate the inductance of an isolated wire by attaching two charged spheres is flawed. The discharging spheres generate a displacement current, which generates a magnetic field. A rather complicated situation. It is possible to assign a self inductance to a wire segment, but this self inductance only is a formal part of the complete story. — Preceding unsigned comment added by 2001:A61:2416:F601:95AA:A61F:411B:5762 (talk) 19:53, 14 March 2021 (UTC)[reply]
Displacement current can be ignored when the size of the apparatus is much smaller than the wavelength. This is easily achieved by using wire with enough resistance that the discharge time constant is sufficiently long. Constant314 (talk) 21:09, 14 March 2021 (UTC)[reply]
Whatever you invent, Maxwell wins. The displacement current (along the electrical field lines) is as large as the current in the wire (an example can be found in the section Current in capacitors in displacement current). A resistance deminishes current, displacement current and their respective magnetic fields to the same extent. — Preceding unsigned comment added by Rdengler (talkcontribs) 19:03, 15 March 2021 (UTC)[reply]
What is your point? The inductance of a single wire is established by the reliable sources.Constant314 (talk) 19:34, 15 March 2021 (UTC)[reply]
The point is, the model for an isolated wire with inductance is flawed. Current loops always are closed (if with displacement currents). Accordingly, there is no operational definition of the inductance of a wire segment, only inductances of closed loops can be measured and defined.
When one calculates the inductance of a current loop containing straight segments, then these segments formally contribute inductance values proportional to their length, that's true. But there always are interaction contributions from all segment pairs, and these are of the same order of magnitude. In effect, the contributions of individual (straight) segments alone are nothing more than an order of magnitude estimation. And only with this reservation can one talk of the "inductance of a (straight) wire segment".radical_in_all_things (talk) 20:54, 15 March 2021 (UTC)[reply]
There are plenty of reliable sources that say you can ignore displacement current in quasi-static conditions. The calculation of the inductance of a single wire is deeply unflawed.Constant314 (talk) 23:16, 15 March 2021 (UTC)[reply]
My last comment in this thread. The mentioned example in displacement current (loading a capacitor) is completele quasi-static. The capacitor consisting of two spheres, discharged through a wire, is identical. Equations count, not authorities (except Maxwell, in this case).radical_in_all_things (talk) 10:19, 16 March 2021 (UTC)[reply]

Coupling coefficient and peak splitting[edit]

@Vcrasto and Constant314:. Peak splitting does not occur when k exceeds unity (not actually possible as Constant314 pointed out). Peak splitting occurs when kQ exceeds unity. See Double-tuned amplifier#Stage gain. SpinningSpark 08:02, 26 April 2022 (UTC)[reply]

What I meant to say was peak splitting occurs when >1. Not only k. Vcrasto (talk) 10:41, 26 April 2022 (UTC)[reply]
Perhaps peak splitting is not important for this article. Constant314 (talk) 13:40, 26 April 2022 (UTC)[reply]
That is just another way of saying that peak splitting occurs above critical coupling, a statement already in the article. SpinningSpark 16:06, 26 April 2022 (UTC)[reply]

There is a recent edit mentioning Big O notation. It seems to me that some of the terms on the page actually need Little o notation, though. Gah4 (talk) 18:20, 15 October 2022 (UTC)[reply]

Nationality[edit]

Why introduce Henry as "American scientist Joseph Henry"? Why not just call him "Joseph Henry"? Is his nationality relevant? Why aren't Heaviside or Lenz introduced in terms of their nationalities? Gerardhiggins (talk) 15:33, 2 December 2022 (UTC)[reply]

Incorrect Diagram[edit]

Under section inductive reactance, the diagram shows the current leading by 90°, but in the caption we have The current lags the voltage by 90°, Which is actually correct, across an inductor the current does lag. Siddharthbhat02 (talk) 21:28, 12 June 2023 (UTC)[reply]

@Chetvorno: The image does appear to show a leading current rather than a lagging current. Constant314 (talk) 22:22, 12 June 2023 (UTC)[reply]
No, on the graph the first positive voltage peak occurs at 90° on the horizontal time axis, the first positive current peak occurs at 180°. Similarly the voltage line crosses the axis in a positive direction at 0°, while the current crosses it at 90°. The current lags the voltage.--ChetvornoTALK 14:23, 13 June 2023 (UTC)[reply]
You are right, of course. I don't know what I was thinking.Constant314 (talk) 04:01, 13 June 2023 (UTC)[reply]

Wheeler's approximation formula for current-sheet[edit]

Wheeler's approximation formula for current-sheet was in inches. You cannot convert to cm by using the factor 2.5 without degrading the formula's 1% accuracy. Either use the correct factor 2.54, or alter the stated accuracy. 110.174.215.170 (talk) 15:57, 1 September 2023 (UTC)[reply]

All except this one have a in them. They give the right result, given the units of the . This one seems to have magic until built into the 18 and 40. Gah4 (talk) 00:12, 2 September 2023 (UTC)[reply]