Talk:Inductor

Effect of one inductor on another

If there are two inductors connected in series in the same circuit and their mutual inductance is 0,can one inductor affect the flux through the other inductor in any way?-Subhash

That's not a homework question, is it? — Omegatron 13:17, 7 August 2006 (UTC)

Please give me a valid answer.I'm stuck up in a controversy related to this in the classroom and I want to know the fact.–Subhash

The question isn't really phrased "tightly" enough to answer. If the two inductors share (essentially) no flux (and so have essentially no mutual inductance), then they are independent components. The total inductance is the sum of the two individual inductors and the total impedance is the vector sum of the two individual impedances.

However, by adding its own contribution of L, R, and C into the series circuit, the second inductor clearly affects the flux developed in the first inductor and vice-versa; remove the second inductor and the flux in the first inductor will change (and vice-versa).

Atlant 13:21, 8 August 2006 (UTC)

Thank you for answering my question.Considering a pure inductor, what if the second inductor's current is suddenly diverted by the means of a switch when they(Both the inductors) have already reached(considering DC) steady state?That is, if the second inductor is somehow thrown out of the circuit after reaching steady state ,does the first inductor's flux or current get affected in some way(Considering 0 mutual inductance)?

Well, in the case of ideal inductors, your question isn't meaningful because, when connected to DC, our ideal, resistance-free inductors never will reach "steady state". So we'll put that case aside. ;-) For non-ideal inductors, instantly switching an inductor out of the circuit (that is, replacing it by a short circuit: a zero-resistance, zero-inductance path) will cause the remaining inductor to no longer be operating in steady-state conditions; you switched out some of the resistance in the circuit (the resistance of the non-ideal second inductor) so the current (and magnetic flux) in the remaining (first) inductor must then rise further until a new set of steady-state conditions are established.

Meanwhile, of course, the inductor that you instantaneously switched out of the circuit has thrown a heck of a spark as we instantaneously opened its circuit path, causing its terminal voltage to rise sharply as it tries to maintain the current flow. But tat's a topic for another discussion. ;-)

Atlant 12:10, 9 August 2006 (UTC)

Well, it's been a while so this might be a bit sloppy, but reactance (or some function of it) depends on inductance times derivative of current. At steady state, derivative of current is zero, therefore derivative of current times inductance is zero no matter what happens to inductance number two, including switching it out of circuit, i.e. dropping it to zero. If reactance of inductor number two doesn't change, then current through whole circuit will not change, drivative is still zero, so no effect on inductor number one. Gzuckier 15:37, 9 August 2006 (UTC)

Coil disambiguation

Just a layperson, trying to disambiguate coil -- should the currently non-existent induction coil be a separate article, or redirect here? or vice versa? Catherine 00:29 Apr 14, 2003 (UTC)

Why is "L" the symbol for inductance? - Joseph D. Rudmin

The "L" was chosen after Lenz's law. --Jeff Connelly 04:46, 30 November 2005 (UTC)

You mean the variable (The unit is usually Henrys.)? "I" was already taken for current; "i" for the square root of -1; not sure where L came from but I suppose it makes as much sense as anything. UninvitedCompany 19:49, 17 Feb 2004 (UTC)

Yes; how embarrassing! I know the difference between symbol and unit. I have corrected my question. - Joseph D. Rudmin

---

It's spelled characteristic.

Why did you move inductance to its own article? I think it is better in the inductor article with inductance as a redirect. They are essentially the same thing and should be in one article. Whether inductor or inductance I don't know, but the two belong together. That's like separating running from runner. An inductor (parasitic or otherwise), and nothing else, has inductance and inductance is a property of inductors. - Omegatron 16:58, Apr 13, 2004 (UTC)

At the very least, equations should be moved to inductance, and actual information on construction, etc. to inductor.

Why? They are both about the same thing. A "parasitic inductor" is still an inductor, right? And you don't construct those. Maybe there is no such thing, and there is only "parasitic inductance"? - Omegatron 19:32, Oct 18, 2004 (UTC)

I guess to a physicist, inductance is more important than an inductor, and to an EE, vice versa...? Pfalstad 14:59, 6 October 2005 (UTC)

Induction loop

I suspect that the new Induction loop article needs to be merged with another (this one?); it certianly needs a lot of work by someone who knows what they're doing. Mel Etitis (Μελ Ετητης) 13:12, 30 Mar 2005 (UTC)

No kidding; Is there much in that article that's salvagable?

And shouldn't it be about the gadgets used to tap phones and detect cars at traffic signals?

Atlant 13:29, 30 Mar 2005 (UTC)

Deletion of old history section

I had hopped to save some bandwidth but if I must: First, it is wrong. Stanley didn’t invent the first practical induction coil as stated. Inductions coils had been around for fifty years before that and Ruhmkorff was building excellent ones in 1850. What Stanley’s patent did cover was a version of an AC transformer that proved to be practical and similar to modern transformers. Since patents require titles and there was not yet a word for the device, he used whatever words were handy. The patent was titled “Induction Coil”. Mr. Stanley’s contribution is covered in transformer where it belongs. It has nothing to do with the Inductor article. This history section was left over here as a remnant of some confusion in terminology from a year ago. Let us recreate a history section only when we have something applicable to put into it. Meggar 05:58, 2005 Jun 8 (UTC)

Thank you for that explanation.

Atlant 11:47, 8 Jun 2005 (UTC)

practical issue

It would be helpful, if l - length were better explained in formulas. I think it is the gap between the two walls of the coil form.

Also some examples of permeability ranges would be good.

I have decided I will have to resort to pulling a book off a shelf to make even a factor of 10 approximation for a coil. —Preceding unsigned comment added by 172.128.133.207 (talk • contribs)

it seems that some inductors use coil that is NOT electrically insulated - i.e. blank. i'd like to understand why the current will not chose the shortest path (through adjacent loops of wire, which touch one another) rather than going through every loop and building up the magnetic fied.

Most coils are wound with a wire with a varnish or polyurethane coating. Take and ohmmeter and you will see, it is really insulated wire. —Preceding unsigned comment added by 172.128.133.207 (talk • contribs)

secondly, would it not be most effective (in terms of maximum henries generated per volume of space occupied) to loop the wire in TWO countered sets of loops around the same core, which build opposing magnetic fields?

I think you'll find that what you see as bare wire is either separated by an air gap or a thin coat of paint. The voltage between turns is not that great so you don't need much insulation. Shorted turns is a common fault in motors, etc. Multiple layer inductors do exist, but I believe they are usually wound in the same direction to avoid some bad effects. More insulation is usually required between layers. --agr 13:29, 16 September 2005 (UTC)

Yes, winding wire as purchased is almost always insulated with a thin layer of varnish or enamel. Arnold is correct in that there are not usually many volts per turn, so a thin layer may provide enough voltage breakdown strength.

Regarding your second point, if alternate layers were wound in opposite directions, the magnetic flux would cancel out giving ZERO inductance. In fact, this is how some non inductive resistors are wound.--Light current 15:44, 16 September 2005 (UTC)

Category?

Why does Inductor belong to [[Category: Electromagnetic components]] while Capacitor belongs to [[Category: Electronics]]?--Astor Piazzolla 11:55, 14 December 2005 (UTC)

Typo?

Hi in the overview the word opposite appears -- do you mean opposes?

definition

In the equation for inductance you never say that L is inductance. If it's not, you need to say what it is.

Symbol added. Meggar 20:37, 19 December 2005 (UTC)

Inductance is the ratio of the back EMF induced to the rate of change of current.

Goddamn Wheeler and his inches

We should present the inductance equations in the SI form, foremost, since Henry is a SI unit and hell, this is a scientific article! These "easy equations for inches" created by some guy named Wheeler, apparently, are bastardized forms of the SI ones to make the use of inches easier and convenient. Yeah yeah, I know this is the English Wikipedia, so yadda yadda, but we need the SI equations in there, as they're the preferred and more scientific form.

The non-si equations could show up on a special section, even mentioning the Wheeler guy, describing how they were made up for convenience of imperial unit users, hence all the seemingly arbitrary constants. I'd do it myself, but alas, I don't have the equations, and that's also why I'm complaining. ☢ Ҡieff⌇↯ 21:37, 25 December 2005 (UTC)

"j is in ohms"

It says in the text "The impedance of an inductor (inductive reactance) is then given by:", then an expression, and then it's stated that "j is in ohms". Is this really true? Isn't j the imaginary unit, such that j^2 = -1, and isn't that unit always dimensionless? I think this is just a typo in the article, but I thought it best to seek consensus before editing, anyway. --Avl 08:24, 24 February 2006 (UTC)

Impedance can be a complex number. Take a look at the article for some more insight on that statement. ☢ Ҡi∊ff⌇↯ 08:46, 24 February 2006 (UTC)

But the imaginary number is not in ohms. It's a number. Does the article mean "j ohms"? — Omegatron 16:52, 24 February 2006 (UTC)

Yes, imaginary numbers have dimensions as well. It's always the same dimension as the real component. I suppose you can debate whether or not they're real ohms, but then we are talking about an imaginary number, so I think that makes them imaginary ohms rather than real ohms by definition.

I can't find what you're talking about in the article, so I'll just explain briefly. For a given frequency, an inductor will oppose the flow of alternating current similar to how a resistor would. Whatever the formula was, it calculates this resistance. (and it's important to note that, whatever the value is, it changes when the frequency of the AC changes) It differs from real resistance in the way in which two resistances in series add together. If you add together two inductance-style resistances, it's the same as adding together two resistance-style resistances, but when you add one of each, they add differently. The way it works out is that you consider the resistor-style resistance a real component of the total resistance, and the inductance-style resistance an imaginary component, then add them together (which in this simple case is simply combining 10 ohms with 10 imaginary ohms to end up with a complex number like 10 + 10i ohms) and you calculate the effective resistance by finding the absolute value of this number (which is about 14 ohms in this case). Interestingly, you can do the same with capacitors, but it is convention to assign capacitors negative imaginary values, since when capacitance-style resistance is added to inductance-style resistance, the two cancel eachother out.

I suppose it might be nice to put that in the article somewhere, but my writing style is so unencyclopedic. "Inductance-style resistance" isn't a real term -- I just made it up -- and all of this information is coming from a college textbook someone gave me which I threw away because I didn't like it, and thus I can't provide references. -- 24.209.110.27 (talk) 21:28, 18 July 2008 (UTC)

Hydraulic Model

Sureley, a transformer could be modelled hydralically by having a pair of differently-sized "turbines" hooked together mechanically - running on the same shaft. One could even connect the two tubine shafts with a slippery clutch to model transformer losses.

Pmurray bigpond.com 03:19, 25 May 2006 (UTC)

Your idea is a great one but magnifying the voltage through induction is much more easier as well as efficient because there are no moving parts involved in a transformer.Besides when dealing with voltages as high as 750KV it is never safe to convert the electrical energy into mechanical energy.Subhash 04:53, 15 August 2006 (UTC)

Any way of modeling a transformer would be stretching the hydraulic analogy too far. Removed. Meggar 17:05, 19 August 2006 (UTC)

Nutshell

So... in essence, an inductor tries to keep the current flowing at a constant speed...? (Please respond on my talk page.) tinlv7 23:02, 15 September 2006 (UTC)

I would rather say that an inductor keeps the current constant when the voltage at its connections is zero. To change the current you must apply a voltage. The other way: When you "tamper" with the connections, it creates voltages to keep the current going. LPFR 14:58, 10 October 2006 (UTC)

Help?

What's the average inductance of an inductor? Like, 1H, 5H? What would a 1H and a 5H inductor be used for? (Please respond on my talk page if you must.) tinlv7 18:29, 9 November 2006 (UTC)

There is no "average" inductance since the inductor value is always chosen depending on the task it is needed to complete. Since all tasks are different each task will have a different-sized inductor. Furthermore, inductors are usually very easy and cheap to make (just a wound loop of wire that doesn't even always need a core) so there is no need usually to standardize sizes. That being said, it is unusual to have one over 1 Henry in size. More likely you will be dealing with millihenries or microhenries. —Preceding unsigned comment added by 65.116.254.10 (talk) 21:08, 17 December 2007 (UTC)

Q Calculation

How about some way to calculate the Q of an inductor? The book "Reference Data For Radio Engineers" contains a graph that will give the Q for some solenoidal inductors. I was looking to calculate it or an air cored torroid, came here but NADA! Anybody know how to do this? (Yes, Iknow it's difficult!)

The Q is simply L/R (ratio of inductance in Henry and (DC) resistance in Ohm). Q depends on the frequency, so an inductor may have a different Q at a different frequency. The frequency dependent part can be major factor with, for example, ferrite core inductors. It is, if my memory serves me well, almost a non-issue in air-core inductors (the magnetic properties of Oxygen are very small compared to those of ferrite materials; note that the Wikipedia article incorrectly states that Oxygen is only paramagnetic is liquid and solid form - some physical gas detectors do in fact use the paramagnetism of oxygen). --Klaws 16:54, 4 March 2007 (UTC)

Q depends on frequency???. How so? Does the L depend on frequency? The R? If so, then neither is an L or an R. Yes, reactance depends on frequency but not L or R. Alfred Centauri 04:21, 5 March 2007 (UTC)

Formulas and approximations

As far as I recall, the formulas for conculating some inductance from some coil diemnsions are just approximations. With "typical" parameters, a lucky coil builder may get reasonably close results when winding his coil in real life, but he may appear to be less lucky when he designs his special 1.3 turn coil, whose parameters he verified with the mentioned formulas. Perhaps it should be noted which formulas are approximations and under which circumstances they provide reasonable results. Unfortunately, I have only vague memories on this topic, so I cannot help myself :-( --Klaws 15:33, 4 March 2007 (UTC)

Formulae units

Is it really necessary to have non-SI versions of some of the inductance "formulae"? This isn't standard practice on WP, and seems somewhat arbitrary. If people are working in imperial units, they can easily convert to SI before applying these equations. Oli Filth 18:25, 17 June 2007 (UTC)

No, we don't need these at all. Only commonly-used equations. — Omegatron 00:20, 18 June 2007 (UTC)

Stored Energy

It'd be super-awesome if someone could make the section on stored energy look similar to the section on stored energy in the article about capacitance. Over there, I think it makes it easy to understand why the equation is what it is, whereas here, I'm left wondering why it is what it is. It's very symmetrical, so I'm sure it's correct, I'm just having a difficult time understanding how one determines "the amount of work required to establish the current through the inductor" which would be awesome information to have. -- 24.209.110.27 (talk) 21:36, 18 July 2008 (UTC)

I think the key to understanding this is to turn around the definition of a henry. The standard definition is "an inductor with an inductance of 1 henry produces an EMF of 1 volt when the current through the inductor changes at the rate of 1 ampere per second." In equation form this is H = V / ΔI, indicating that the value assigned to inductance increases when the EMF voltage increases, and decreases when the current required to create this voltage increases. To make this easier to understand, just rearrange it to ΔI = V / H, which indicates that the change in current increases with applied voltage, but decreases with increased inductance. This makes the whole issue of inductance a lot easier to understand. Want to know what happens when you apply 5 volts to a 1 mH inductor? Well, 5 / 0.001 = 5000, so the current increases at a rate of 5000 amperes per second when 5 volts is applied to this inductor, at least until your power supply fails. Needless to say, such voltage is always applied to such inductors for very short periods of time, or via resistors which will cause the applied voltage to decrease as the current increases.

So anyway, imagine you apply one volt to an inductor of one henry. Over time you will see a linear increase in current, rising to one ampere after one second, and two amperes after two seconds. Since power equals voltage times current, we can calculate the power going into the inductor at any moment. At one second it is one watt, at two seconds it is two watts, and this can be integrated to end up with the equation in the article. If you don't want to integrate, just draw a graph (linear graphs are easy) and shade the area under the line to represent the stored energy. After one second, you have half a shaded square. After two seconds, you have two shaded squares. After three seconds, you have 4.5 shaded squares. All which are the same values given by the formula in the article.

I suppose this too would be nice for the article, but I kind of just made it all up. The way that calculus is written makes absolutely no sense to me, so I've kind of had to figure out stuff like this on my own. I can't seem to find a reference that doesn't use standard calculus notation and so I'm not sure anything actually agrees with me. All I know is that when you integrate y = b * x, you end up with 0.5 * b * x ^ 2, which leads to the equation in the article. -- The one and only Pj (talk) 02:57, 21 July 2008 (UTC)

current flow of an inductor

equation for current flow of an inductor at any given time passed the point at witch voltage was applied. Finding the current flowing in an inductor at any given time past when voltage was applied.

$Amps=\left({\frac {Vsource}{Rseries}}\right)*\left(1.0-e^{\left(-Time*{\frac {Rseries}{inductance}}\right)}\right)$

$e$ is inverse natural log or normally $e^{x}$ on most calculators. This was removed for a unknown reason. I believe it to be common knowledge among ET and EE people so it was not cited. Eadthem (talk) 22:16, 31 July 2008 (UTC)