Talk:Inductance/Archive 1

message: After all, what use is it to anyone? Someone who already understands this does not need to turn to this article. And those who have no idea but want a basic introductory explanation of what it is are left out. Great, so it is magnetic flux. Ok, now I go to that page and am inundated with another page full of abstruse verbage. Around I go, grasping at circular references, looking for some kind of conceptual foothold to begin with, but left with nothing but abstract mathematical symbols and explanations that may as well be in Greek. I mean, electical engineers aren't the only people interested in what this stuff is.
I couldn't agree more. This is supposed to be an encyclopedia entry, not a college level course. I had exactly the same problem you did. All I wanted was a basic explanation that was a little more complete than my vague grasp of the principles, but instead was bombarded with advanced material that belongs in a technical reference, with every link leading to more of the same. Most of the people who contributed to this ought to go find somewhere else to circle-jerk. 208.42.94.130 (talk) 08:25, 14 November 2008 (UTC)
replies: text: I got stuck at Wb/A being unexplained.  —Preceding unsigned comment added by 97.83.155.136 (talk) 04:57, 13 October 2009 (UTC)
thread_title: Has anyone considered writing these articles for people who don't already understand what this is?

message: The equation on the inductance of a solenoid states that it is inversely proportional to "l" (the length of the solenoid).  This implies that inductance would be infinite for a zero-length solenoid.  Certainly a single-winding solenoid could be made with a length arbitrarily close to 0.  I think this equation is an approximation with assumptions (eg. l \gg \sqrt{A}).  Could these assumptions please be stated? Jurgen Hissen 21:23, 18 May 2006 (UTC)
thread_title: Approximations in dependence on length of solenoid

message: This page needs a picture of an inductor, and a drawing representation of it (showing turns), and the usual circuit representation for it. Maybe I'll draw these sometime. Fresheneesz 07:47, 22 November 2005 (UTC)
thread_title: Need pictures

message: Mutual inductance needs a picture, needs to explain the dot convention, and give a picture for this. I might make these sometime as well. Also, mutual inductance need not occur between circuits - but can occur within circuit. Two nearby inductors can even have mutual inductance while remaining also in series. Fresheneesz 07:47, 22 November 2005 (UTC)
thread_title: Mutual Inductance needs explaining

message: I can't even read the first sentence of this article.  Shouldn't it read something like, inductance is the property of a device to produce a voltage proportional to a varying electrical current?
The expression given at the beginning, v = -L\frac{di}{dt} is only correct for a circuit with a fixed geometry. If the circuit is contorted while current is flowing through it, then the change in geometry creates a back-emf, also, as shown below.
: \mathcal{E} = -\frac{d\Phi_i}{dt} = -\frac{d(L_i I_i)}{dt} = -(\frac{dL_i}{dt}I_i + \frac{dI_i}{dt}L_i) 
=== Question ===
(same poster)  is the above assertion true? I have found no occurance in any reference texts.
=== Answer ===
I suppose one could look at it that way.
But the traditional way of analyzing motor/generators is to assume a fixed inductance for each part (the inductance of the rotating coil, and the inductance of the stationary coil), and calculating a separate "back emf" as a function of the speed of the motor.
I've always thought of a motor as "two fixed-geometry inductors that move relative to each other".
I've honestly never thought about considering a motor as "one big variable-geometry inductor". It's a perfectly self-consistent way to look at it.
But if I had some spinning motor, and someone asked me for "the inductance" of the motor, I'd probably just measure the voltage and current and plug in
:L = \frac{-v}{\frac{di}{dt}}
, completely ignoring the "variable inductance" of the motor taken as a whole.
--DavidCary 08:01, 29 Oct 2004 (UTC)
thread_title:  Error? 

message: I suppose one could look at it that way.
But the traditional way of analyzing motor/generators is to assume a fixed inductance for each part (the inductance of the rotating coil, and the inductance of the stationary coil), and calculating a separate "back emf" as a function of the speed of the motor.
I've always thought of a motor as "two fixed-geometry inductors that move relative to each other".
I've honestly never thought about considering a motor as "one big variable-geometry inductor". It's a perfectly self-consistent way to look at it.
But if I had some spinning motor, and someone asked me for "the inductance" of the motor, I'd probably just measure the voltage and current and plug in
:L = \frac{-v}{\frac{di}{dt}}
, completely ignoring the "variable inductance" of the motor taken as a whole.
--DavidCary 08:01, 29 Oct 2004 (UTC)
thread_title:  Answer 

message: I just pluralized "henry" and noted that I typed it "henries". But another appearance of the plural later on in the article was spelled "henrys". Is there an official plural for this unit? (Or is this like the ever-simmering British English versus American English debate?) If you know, could you please edit the article to make the two plurals consistent?
Atlant 11:42, 15 Jun 2005 (UTC)
replies: text: NIST says it's henries. The IEEE says it's henrys. The IEE in 1947 (quoted in OED2) said it was henrys. The OED2 says it can be either. The Oxford Dictionary for Writers and Editors says henries. Personally, as a practising engineer, I am used to seeing henries and have never seen henrys used by an experienced writer, whether American or British. --Heron 19:55, 15 Jun 2005 (UTC)
replies: text: I agree with your experience; thanks for confirming this and for cleaning up the article.Atlant 11:42, 16 Jun 2005 (UTC)

text: I don't think there is a plural. The unit Henry is similar to the unit Ohm or Hertz. All three don't have plurals. There is no Ohms or Hertzs, and there is no Henries or Henrys. It's Henry, names after Joseph Henry (see: Henry (unit)). THeeren (talk) 18:20, 16 January 2008 (UTC)
replies: text: The article on the International System of Units specifically says it is henries (not Henries), at least in English, citing [NIST Special Publication 811: Guide for the Use of the International System of Units (SI)].  This is consistent with the general rule; see for example the article on the English plural, English Irregular Plural Nouns, or UNIT S4: YS OR IES?. There it states that for common nouns ending in a consonant followed by 'y', the 'y' changes to 'i' and the plural adds 'es', but that for proper names, the plural always ends in 'ys'.  The plural of Henry should be Henrys, as in There are two Henrys in the class and the article on the War of the Three Henries (977–978) should be renamed, but the general rule says the plural of an object called a henry should always be henries. —Archimerged (talk) 02:17, 21 April 2008 (UTC)
thread_title:  What '''is''' the plural of a Henry? 

message: I'm not sure why the merge request was made, but I think it's clear from the articles (or any physics textbook) that they're two separate concepts.  -- SCZenz 14:21, 31 August 2005 (UTC)
replies: text: Agree Light current 16:38, 31 August 2005 (UTC)
replies: text: Me too. I call the request a silly thing. --Heron 17:13, 31 August 2005 (UTC)
replies: text: Me three.  However, I would hesitate to say that these subjects are separate concepts since the inductance of a conductor is a result of EM induction.  Still, these subjects certainly deserve separate articles as EM induction is even more general than the induction article describes in its opening sentence. Alfred Centauri 22:23, 31 August 2005 (UTC)
thread_title:  Requested merge with [[Electromagnetic induction]] 

message: According to fundamental principles, the inductance L is defined as:
:L = \frac{\lambda}{i} = N\frac{\phi_M}{i} \,
Does it follow that:
:L = N \frac{d\phi_M}{di} \, ?
Well, let's see...
:N\phi_M = L i \,
:N\frac{d\phi_M}{dt} = L \frac{di}{dt} + i \frac{dL}{dt}
:L = (N\frac{d\phi_M}{dt} - i \frac{dL}{dt})\frac{dt}{di}
:L = N\frac{d\phi_M}{di} - i \frac{dL}{di}
Moral of the story - use the the formula from first principles or if not, state your assumptions, e.g.:
:L = N \frac{d\phi_M}{di}  for constant L
Besides, the opening sentence refers to magnetic flux and current, not their differentials.  I recommend beginning with the (always correct) non-differential form and then derive the simplified differential form by explicitly stating the caveat 'for constant L'.  Alfred Centauri 22:58, 3 September 2005 (UTC)
replies: text: Good point. I have modified intro. Light current 23:21, 3 September 2005 (UTC)
replies: text: Just looked at your equations again. They show that L has to be constant with varying i, so this means a linear inducatnce, but does not exclude a time varying inductance! Interesting! Light current 00:56, 4 September 2005 (UTC)

text: Yes, that is all perfectly clear to me. I agree Light current 01:49, 4 September 2005 (UTC)
thread_title:  Does L = N dphi/di? 

message: Could AC state the physical quantity to which lambda refers. Its new to me. Light current 01:09, 4 September 2005 (UTC)
replies: text: My pleasure.  \lambda is called the flux linkage.  If a flux \Phi links N turns of a coil, the flux linkage is N\Phi and denoted by \lambda: Alfred Centauri 02:06, 4 September 2005 (UTC)
replies: text: OK Thanks! Never heard it called 'lambda' before Light current 02:12, 4 September 2005 (UTC)
thread_title:  Physical significance of term 'lambda' 

message: Ive just been looking at the page again. The statement I have trouble with is:
:\frac{d\lambda}{dt} = -\mathcal{E} = v 
where \mathcal{E} is the Electromotive force (emf) and v is the induced voltage.  Note that the emf is opposite to the induced voltage.
You see, the way I see it, neither one is defined and the difference between these 2 voltages is not stated. It probably makes complete, unambiguous sense to you, but I'm afraid it doesnt to me (Im not saying its wrong -- just that I dont undersatnd it as it stands). If we were to define what we mean by:
a) the emf
b) the induced voltage
maybe our problems would go away!! What think you? Light current 00:10, 5 September 2005 (UTC)
replies: text: The emf always exists when there is a changing magnetic field.  However, the induced voltage may not exist!  If there is no charge in the vicinity of a changing magnetic field, the will not be current nor will there be an electric field due to the presence of or accumulation of charge.  However, the closed contour integral of the electric field will nonetheless be non-zero.  Observe that the integration of the electric field (V/m) over distance (m) has units of Volts.  The $100,000 question is:  how do you measure this?  Integrating along a closed contour means that your beginning point and your end point are the same so a potential difference between two points that are the same is difficult to define.  In fact, it is impossible. 
Is this pd not obviously zero!? Light current 13:30, 5 September 2005 (UTC)

text: It is obvious to me (and should be to you) that it not zero, it is instead 'undefined'.  Potential difference only has meaning if the the work done in moving a charge from one point to another is independent of the path taken.  In a non-conservative field the work done depends on the path which means that the work along a closed path can be non-zero.  Recall that volts is Joules / Coulomb.  When you integrate the electric field along a path, the result is the work done in move unit charge along that path.  In a conservative field, the work done along any closed path is identically zero.  This is not the case for the induced electric field. Alfred Centauri 13:47, 5 September 2005 (UTC)

text: The induced emf is non-conservative so the concept of potential is no longer meaningful (recall that the gradient of a potential field must be conservative!). One way to measure the strength of the emf is by measuring the amount of charge the emf can separate.  As charge moves due to the non-conservative induced field, a conservative field is developed between the separated charge.  When this conservative field stops the movement of charge, the potential between the separated charge is exactly equal (and opposite) to the emf in Volts.  This measurable voltage is the induced voltage and is the potential difference created by the charge the emf has separated.  But, it must be kept in mind that this conservative field is in the opposite direction to the induced non-conservative field!  If this explaination doesn't help, I've got a couple of other angles but I don't want to introduce too many ideas at once. Alfred Centauri 03:45, 5 September 2005 (UTC)

text: agreed. Alfred Centauri 00:33, 7 September 2005 (UTC)

text: agreed though let me clarify that the work is done on the charge by the emf.  The charge gains kinetic energy as a result. Alfred Centauri 00:33, 7 September 2005 (UTC)

text: I'm not so sure about this one.  The emf acts to accelerate charge around a closed loop.  If there is not a closed loop, charge will accumulate somewhere creating a potential difference.  This potential difference is a measure of the strength of the emf.  The chemical reaction inside the battery is the emf.  The voltage across the battery terminals is a result of the charge driven by the emf accumulating on the terminals of the battery. Alfred Centauri 00:33, 7 September 2005 (UTC)

text: Kind of.  See my answer to d). Alfred Centauri 00:33, 7 September 2005 (UTC)

text: very close - I think we getting to the same page! Alfred Centauri 00:33, 7 September 2005 (UTC)
thread_title:  emf vs induced voltage 

message: Consider a ring of resistive material with uniform resistivity.  Assume that there is a magnetic flux changing at a constant rate through the surface area enclosed by this ring. This changing magnetic flux induces a constant electric field that in turn drives a constant current through the ring of resistive material.
Recall that the line integral of the induced electric field around the path traced by the ring gives the emf (in volts) induced by the changing magnetic flux.
Answer the following questions:
(1) Is it possible to measure the total emf directly with a voltmeter?  If so, explain how to do this.  If not, propose how to measure the total emf.
: If the ring had a break in it, and the ring totally enclosed the flux, then it would be easy to measure the induced emf, by connecting the meter across the break but ensuring that the voltmeter leads are outside the ring.
:However I dont think this is the question you are asking is it?
:Anyway, even with a continuous resistive torus, it should be possible.
:first try: connect leads of an ideal voltmeter to 2 points diametrically opposed on the ring and double your reading?.(making sure that the volt meter leads were parallel to the mag flux (so no induction in them).
:second try: use a CT to measure the current in the ring and multiply by the rings circumferencial resistance?(but I guess you'll disallow this 2nd answer as the exam is on emf!Light current 14:03, 5 September 2005 (UTC)
replies: text: The induced emf is distributed evenly around the ring and does not appear at a specific point in space. Ie it is given by -int E.dl around a closed path. Now KVL is normally quoted for circuits and voltages (not emfs). Since however, we have a source of movable charges, these charges will respond to the emf and move giving rise to a current. THe charge will move at such a rate that, at each and every point in the ring, the induced voltage delta(v) = I delta(R), (where delta means an infinitely small amount).KVL does appear to apply here. At each and every point in the ring, the pd dropped (developed?) across an infinitesimal part of the resistor(ring) must equal the induced emf over that part of the ring.?Light current 14:15, 5 September 2005 (UTC)

text: Presumably one would have to insert the source equal in voltage to the original emf to counteract the original emf because this is the prime mover in this cct.Light current 14:15, 5 September 2005 (UTC)
replies: text: Not too bothersome. We can measure the emf by the above CT method, OR (Ive just thought of this one) adjust the voltage source until the CT o/p reads zero. Then the applied voltage equals the emf. So what we're measureing is the induced emf not the voltage. Anyway, the emf has been backed off so no current flows in the ring - Ohms law is NOT violated -- no problem! Am I correct?Light current 01:10, 7 September 2005 (UTC)

text: A No, the current does not change, because the inserted voltage source still balances the induced emf.Light current 14:15, 5 September 2005 (UTC)

text: A Again the answer is that the current remains zero because the inserted voltage has been adjusted to completely neutralise the original emf.
Again, correct.  But once again, doesn't this bother you?  You have a voltage across a short circuit and zero current!  How can you explain this? Alfred Centauri 00:22, 7 September 2005 (UTC)
replies: text: I hadn't considered a shorted turn.  Hmmmm... Sure!  The short reduces the path integral for the emf by the circumference of one turn so now there is less emf and the balance between the applied voltage and the emf is lost.  The result - LOTS of current.  Good point! Alfred Centauri 01:05, 7 September 2005 (UTC)

text: Looks like I have been almost completely rehabilitated - doesnt it?? ;-) Light current 00:33, 7 September 2005 (UTC)
replies: text: I hope you learned as much as I did.  You've asked some great questions that made me think about some concepts much more precisely than I have in the past. Thanks! Alfred Centauri 01:05, 7 September 2005 (UTC)

text: These are excellent questions but I'm not sure if I've got all the answers completely correct. ;-) Light current 14:15, 5 September 2005 (UTC)
thread_title:  Emf exam 

message: Now that you have become comfortable with the balance of forces, let me thow you a curve ball. 
[BTW I though all balls were curved!! ;-)Light current 17:26, 7 September 2005 (UTC)
replies: text: Im trying to keep off relativity at the moment as I'm sure I could be easily shot down. However, for those who are interested in relativity, I found the translation of Einsteins 1916 book (originally in German)('Relativity', pub Routledge Classics 2002, /5) to be much easier to read than all those Teach yourself relativity, and relativity made simple books. Give it a try! Light current 17:14, 7 September 2005 (UTC)
replies: text:  Not reading below (since it seems unconnected to this question), what are you trying to say exactly?  I can only assume that you are referring to Newton's third law (because, as a current freshman physics tutor, that is what is being taught currently it's what jumps to mind), but nothing there is being violated. ub3rm4th 16:51, 28 September 2005 (UTC)
thread_title:  Relativity 

message: Newtons third law is applicable to every conceivable force in the universe. --Light current 23:12, 8 September 2005 (UTC)
replies: text: Or, as I said before, all forces are ultimately internal - there are no forces external to the universe.  You do see this don't you?  Newton's law of inertia says momentum doesn't change unless there is a net force.  I can't conceive of a momentum associated with the universe (can you?) so there can not be a net force in the universe - all forces in the universe must cancel.  Well done - you're a deep thinker now! Alfred Centauri 00:04, 9 September 2005 (UTC)

text: Well, I don't think it was obvious that the third law applies to every force in the universe.  But it does end up being equivalent to conservation of momentum, which as far as we know holds for everything.  So I'd say we all learned something, even if we didn't really get a new physics principle out of it. -- SCZenz 00:19, 9 September 2005 (UTC)
replies: text: LC is very good at making statements and follow-up statements that challenge us to think very clearly about how to answer.  I for one have, as a result of trying to answer LC, come to understand the material much better than before.  Thanks, LC! Alfred Centauri 00:57, 9 September 2005 (UTC)

text: NOTHING beats working through the problem on you own.  It doesn't matter if your new insight is old news to someone else, the thrill of having the 'light bulb go off' due to your own reasoning is what 'it' is all about. Alfred Centauri 01:03, 9 September 2005 (UTC)
thread_title:  The Postulate Version 3 

message: Im sorry to disagree with you but planets in orbit do have a balancing force. Its called centrifugal force. This is what stops then falling into the sun!.  Light current 01:08, 8 September 2005 (UTC)
replies: text: Centrifugal force is a fictitious force--it's useful calculationally, but there's no physical mechanism creating it.  Instead, it's a byproduct of considering a rotating (and thus non-inertial) reference frame (i.e. the one in which the earth is stationary).  If you consider a more-or-less ineratial frame, in which the earth is orbiting and the sun is stationary, we don't fall into the sun because the earth has a large velocity perpendicular to the force of gravity.  (It's always falling "past" the sun.)  -- SCZenz 01:49, 8 September 2005 (UTC)
thread_title:   The postulate in Celestial Mechanics 

message: You don't even have to use a rotating frame to see a fictitious force.  When you step on the accelerator in your car, the acceleration is forward (hopefully) yet you feel as if a force pushes you backward into your seat.  That is acceleration forward 'feels' like a force is pushing backwards.  This force you 'feel' doesn't exist - it is fictitious. Now, to make the connection to GR, imagine that you are sitting in a car seat that is facing up.  You feel as if you are being pushed back into your seat with a force equal to your weight.  But, by the above reasoning, this implies that you are being accelerated up.  Yet, you know that you are stationary on the surface of the Earth.  According to GR, (this will raise some eyebrows) spacetime itself is accelerating downwards toward the center of the Earth.  To be 'stationary' in spacetime would mean that you would also (free) fall towards the center of the Earth.  Since the surface of the Earth prevents you from doing this, the surface of the Earth is, in effect, accelerating you.  That is, your worldline is not on a geodesic of the spacetime. Wait -what's that I smell?  Is it LC's brain short circuiting??? ;<) Alfred Centauri 02:52, 8 September 2005 (UTC)
replies: text: Yes it is!! My brain is completely fried. I shall need some sleep to repair it, but, as Arnold once said...'I'll be back!!!' (Space-time continuum -- what  the hell does that mean??) Light current 03:03, 8 September 2005 (UTC)

text: The earth may be accelerating you upwards, but gravity is accelerating you down wards. The accelerations (and forces of course) balance. This is what I have been saying. So, as an addition to the postulate, can we say that all accelerations balance?? BTW if you are freely falling in a gravitational field, you feel no forces. But to an outside observer, a  force is obviously acting on you and you react to that force by accelerating (F=ma)[ or changing your momentum]along the nearest geodesic. Light current 15:05, 8 September 2005 (UTC)
replies: text: Alfred's point was that gravity doesn't count as a force in GR--you're just moving on a "straight line" through curved space when you're falling, so if you're standing on earth you're accelerated upwards.  An outside observer using general relativity does not, in fact, think that someone falling has a force on them. ;)  -- SCZenz 15:28, 8 September 2005 (UTC)

text: If you're claiming all accelerations balance, then you're saying all objects are in inertial reference frames, and that they go in straight lines.  This is certainly not true, whether you take the Newtonian or General Relativistic perspective on gravity being a force.  (We've given examples already.)  -- SCZenz 15:28, 8 September 2005 (UTC)

text: Hmmm! (or Grrrrrr!). I dont know enough about celestial mechanics to argue with you. I will leave you to work out the validity (or not) of my postulate regarding planets etc! All I would say is that, is in curved space, orbiting planets do indeed stay in orbit. Light current 02:08, 8 September 2005 (UTC)
replies: text: Ahh, yes.  But in General relativity (i.e. curved space), an apple sitting on the surface of the earth has an unbalanced force upwards. ;)  -- SCZenz 02:18, 8 September 2005 (UTC)

text: If this is true, what is the effect of the 'unbalanced' force?. If it cannot act on anything (due to the way you look at things in GR), then it seems similar to an 'emf' in an insultor. Light current 15:44, 8 September 2005 (UTC)
replies: text: The point is the apple should naturally move through the earth on a geodesic, which is the general relativistic analog to a "straight line" in flat space.  The effect of the force is to keep the apple on the surface, which in GR is an accelerated reference frame.  I freely admit it's not at all obvious why this is a useful way of looking at things, but most good popular books on GR should help, if you get the chance.  -- SCZenz 16:02, 8 September 2005 (UTC)

text: You mean the force is stopping the apple falling thro the surface? (which is its natural tendency in curved space). If this is what you mean then, it is balancing the effect of the curved space. Now youre going to say that the effect of curved space is not a force. But it depends how you look at it. Space is only curved by the presence of mass. In GR, the force of gravity has been replaced by curved space. So then you can say that there is an unbalnced force!. If you are to replace the gravitational force, you must also replace the reaction force by something equivalent!( inverse curvature ath the point of contact?) --Light current 16:57, 8 September 2005 (UTC)

text: This appears to be similar to the question that [User:Alfred Centauri] was asking me on my talk page. I have declined to answer him until I have read Einstein's theory again.  I therefore respectfully decline to answer your version of the question at the moment lest I appear foolish! Dont forget, I'm not a brilliant physicist (or even an ordinary one). Anyway, I thank you and AC for taking interest in my deficient state of knowledge and helping me to see the errors of my ways! I'm just trying to learn more about the nature of the universe (it keeps me out of the bars anyway!) Light current 02:38, 8 September 2005 (UTC)
thread_title:   The postulate in General Relativity 

message: I would like to avoid the quantum mechanical aspects at the moment if you are agreeable. The reason being that I know very little about quantum mechanics and I feel that the general law of induction should be established or debunked at the macroscopic level first. Light current 01:55, 8 September 2005 (UTC)
replies: text: Uh, ok. -- SCZenz 02:18, 8 September 2005 (UTC)

text: There's no force that keeps an electron in the lowest orbital of a hydrogen atom from going lower--it's just stuck there because no lower energy is allowed by quantum mechanics.  I will, of course, certainly agree that physical systems displaced from equilibrium are often brought back into equilibrium by forces resulting from the displacement.  -- SCZenz 19:56, 7 September 2005 (UTC)
thread_title: The postulate in Quantum Mechanics 

message: http://fr.wikipedia.org/wiki/Inductance#Puissance_emmagasin.C3.A9e
pf：
::with P= i^{2}*R ,  we take  i^{2}
::thus ( i^{2})’ = 2ｉ．ｉ’
::::＝2ｉ’．ｉ 
::that refers to
:::: \frac{d(i^2)}{dt}  =i \cdot \frac{d(i)}{dt} + \frac{d(i)}{dt}\cdot i = 2 \frac{d(i)}{dt}\cdot i  \,
::by
::::P = u \cdot i = L  \frac{di}{dt} \cdot i\,=\frac{1}{2} \cdot  L  \frac{d(i^2)}{dt} \,(=L*\frac{d i^2}{2dt})
::then by
::::P=\frac{dW}{dt} .........(basic definition)
::therefore,
::::{\mathbf dW=P*dt}
::::dW =\frac{1 }{2} L *d{  i^{2}}
::Hence, {\int}：W = \frac{1}{2} \cdot  L  (i^2_{tf}-i^2_{ti}) \, ........getting the proof.
--HydrogenSu 10:52, 25 December 2005 (UTC)
thread_title:  Derive: W=1/2 * L *delta{i*i} 

message: I just wanted to put up a couple of my sources for putting up some transformer equations in the mutual inductance section. V_secondary = V_primary N_secondary/N_primary. And convsersely the curreent I_secondary = I_primary * N_primary/N_secondary 
Fresheneesz 09:49, 14 April 2006 (UTC)
thread_title:  Sources 

message: The symbol for current is (I), so why is (i) used here instead? GoldenBoar 18:40, 8 May 2006 (UTC)
replies: text: If we are referring to a DC circuit, then I is the symbol for current.  If we are referring to a general time varying current, then i(t) or just i is used.  There are a number of other variations that involve subscripts.  Alfred Centauri 00:53, 9 May 2006 (UTC)
thread_title:  The symbol for current 

message: I'm no expert, but the old book I have (1937) states that L stands for "Linked Flux", it is not homage to Lenz.  I don't want to just go and arbitrarily edit the page, but here is my reference.
p.291
Electrical Engineering Vol 1
Chester L. Dawes, S.B., A.M.
McGraw-Hill —The preceding unsigned comment was added by Ninthbit (talk • contribs) 22:52, 9 February 2007 (UTC)
thread_title:  L - History of the symbol for inductance 

message: I have been playing with the formula for self-inductance L of a solenoid, and find that it can usefully be rearranged in terms of the lengths of the wire and the solenoid. Note that the formulae ignore end-effects, so are approximate.
: L = {\mu_0 \mu_r l_w^2 \over 4 \pi \ l_s}   H
which simplifies to
: L = 0.1 {\mu_r l_w^2 \over l_s}   μH
For an air-cored coil this simplifies further to:
: L = 0.1 {l_w^2 \over l_s}    μH
where 
:μ0 is the permeability of free space (4π × 10-7 henrys per metre)
:μr is the relative permeability of the core (unitless)  
:lw is the length of the wire in metres.
:ls is the length of the solenoid in metres.
Nice, eh? I don't suppose it's original but I haven't seen it elsewhere. I don't think I have made a mistake. GilesW 22:05, 31 May 2007 (UTC)
thread_title:  Simplified Solenoid Inductance Formula 

message: Section: Inductance of a solenoid
The equation for inductance displays as L = (μ0 / μr)(N^2 * A) / l, which looks incorrect.  When it confused my derivation of relative permeability (μr), I realized there might be a mistake.  L = (μ0 * μr * N^2 * A) / l yields the correct results, which I confirmed by deriving μ directly from B/H.  I consulted both electronic machinery (transformer) text and electromagnetic text.  I am new to this, so I thought someone else should look at it and confirm this before any editing.  Thank you for your help.  Amanda.Tighe 18:14, 4 June 2007 (UTC)
replies: text: You're right.  I fixed it.  Thank you for pointing it out. --Heron 19:48, 4 June 2007 (UTC)
thread_title:  Misleading Equation 

message: The page is inconsistent about the use of  \lambda vs.  \Phi.  I'd suggest explaining that distinction in the intro and getting mutual inductance out of the intro, to be explained later.Ccrrccrr 11:43, 3 August 2007 (UTC)
thread_title:  <math> \lambda</math> vs. <math> \Phi</math> 

message: To be really useful, there should be a table with formulas for wire pair, wire plus wall, coaxial cable (with and without skin effect), rectangle, ring, solenoid
...
BTW, the style of some sections should be checked and corrected by someone with native english, it sounds queer to me.  —Preceding unsigned comment added by 84.56.8.130 (talk) 16:42, 4 November 2007 (UTC)
thread_title:  Table with inductances for simple circuits 

message: This is redundant to the section 'Vector field theory derivation', it is an intermediate step to derive the von Neumann formula, and the expression gets singular in the self inductance case.  —Preceding unsigned comment added by Rdengler (talk • contribs) 16:50, 24 November 2007 (UTC)
thread_title:  Section 'Inductance of any shaped loop' should be removed 

message: CONCERNING the formula for Single layer solenoid:  I have no idea what the term with the O is?  Is it a zero?  If so, why include it at all?  Is it another term?  If so, what?  —Preceding unsigned comment added by 167.102.133.216 (talk) 20:27, 18 January 2008 (UTC)
thread_title:  SECTION Inductance of simple electrical circuits 

message: I have removed 'per coil turn' from the definition of inductance.
What is the magnetig flux through a circuit consisting of a thin wire? It is proportional to the number of flux lines that have crossed the wire and now are "captured". This implies that the flux through a circuit with n turns is n times the flux through one turn. Defining the flux geometrically (area times field strenght) leads to inconsistencies and complications. Consider a circular loop with two turns in a constant magnetic field and deform the circuit to a shape looking like an 8. The flux doesn't change.
	 
The magnetic flux through a circuit with a continuous current distribution may be defined by replacing the current distribution with a bunch of thin wires, each carrying a small current.  —Preceding unsigned comment added by Rdengler (talk • contribs) 07:41, 19 August 2008 (UTC)
thread_title:  Remarks about the definition of magnetic flux 

message: Ths page does a great job explaining the complex concept of inductance, yet I beleive the most important thing to mention is that inductance is the ratio of magnetic flux created by an inductor to the change in current that induces it. 
The formula 
L = Φ/i
where L is the inductance in Henries, Φ is magnetic flux in Webers, and i is the current in Amperes, is the simplest formula I know to give a rudimentary understanding of inductance. Is this correct? If it is, I will add it to the page to allow beginners to understand better.  --Skyfinity (talk) 23:07, 29 November 2008 (UTC)
replies: text: Hear hear -- I came here to make exactly this point.  I'm not a physicist but a mathematician, and I could not understand from the page the DEFINITION of inductance, only about manifestations of it. The capacitance page gets it right. akay (talk) 15:31, 14 January 2009 (UTC)
thread_title:  Addition of simple formula 

message: In the diagram, the arcs that form the symbols of the inductors are the wrong way round. The convex sides of the arcs should face each other.
Where a single inductor with a core is depicted, the arcs face the core depicted by parallel lines.
I was brought up on BS3939, superseded many years ago by EN60617-4 and the BS version of it.
see: EN60617-4:1996 Graphical symbols for diagrams. Basic passive components (). There are numerous engineering publications that use those symbols, mostly correctly(!). GilesW (talk) 21:35, 15 January 2009 (UTC)
thread_title:  Inductor symbol error 

message: I think 'Coupled inductors' would be more technically correct than 'Mutually inducting inductors' in the text of the diagram (and anywhere else, if applicable).GilesW (talk) 21:41, 15 January 2009 (UTC)
thread_title:  'Mutually inducting inductors' 

message: Could the page include a reference to the units of measurement, and practical help for those who might actually want to use the equations?  I assumed MKS units, which seems reasonable in the context of Wikipedia.  But when I calculated the inductance of a loop with a radius of 10mm I got 0.049 Henry - obviously a ridiculous result.  I had to multiply this by mu-zero to get the much more reasonable answer of 6.2e-8 Henry.
Jay.sinnett (talk) 21:21, 26 June 2009 (UTC)
thread_title:  Practical units 

message: Can someone add a power series expansion of the solenoid expr terms of w^-1 in the limit w==>inf (i.e. short solenoid limit) Thanks! Woz2 (talk) 14:30, 9 September 2009 (UTC)
replies: text: OK I figured it out, and found a use for Wolfram Alpha! Check it out wolframalpha.com. Quite remarkable. Woz2 (talk) 23:51, 11 September 2009 (UTC)
thread_title:  Expansion of solenoid expr terms of w^-1 in the limit w==>inf (i.e. short solenoid limit) 

message: There is a dispute about a superfluous factor -1 in the series expansion of the single layer solenoid inductance. First of all, the -1 (inside the square) is unnatural, useless and very confusing. More to point, the product in the numerator of the term with index m consists of (m-1) factors. For example, for m = 2 it simply is (2m-3) = 1, for m = 3 it is 1*(2m-3) = 1*3. For m = 1 the product thus consists of 0 factors, which is 1 by convention everywhere in mathematics. The only natural thing thus is to remove the factor -1. Of course, it arises if one sets m = 1 in (2m-3). However, there are 0 factors in the numerator of the (m=1)-term. The formalism is clever, and the result looks nice and natural!
: If you had to sum 2n-3 from n=1 to infinity, the first few terms would be -1 + 1 + 3 + 5 ... It's no different here. I'm missing your point when you say there are zero factors at n = 1. At n = 1 there is one factor: the product term is (-1/2)^2 and the complete first term is (-1/2)^2 * (2 - 1) * 2^3 * (-1)^2 * w^2 / (2 + 2) = w^2/2 Woz2 (talk) 01:47, 23 October 2009 (UTC)
replies: text: Perhaps this could be resolved if an actual reference was provided for this section.  I am especially concerned by the approximation for r>>l which, judging by the  edit summary, seems to be an OR synthesis based on Wolfram with no indication that the expression is still valid under these conditions.  I am not even sure that big-O notation makes any sense in this context. SpinningSpark 16:45, 22 September 2009 (UTC)
thread_title:  Dispute about a factor -1 in solenoid inductance series 

message: The article still is somewhat incoherent - the capacitance article is superior in this respect.
   
1) The section "Phasor circuit analysis and impedance"  doesn't explain anything
about inductance. Should be explained under "phasors" or elsewhere.
If there are no objections then I shall remove it.
 
2) The section "Induced emf" has trivial and clumsy content.
If there are no objections then I shall remove it.
 
3) The sub-sections under "Self-inductance of simple electrical circuits in air" (Inductance of a solenoid, Inductance of a coaxial line)
should be moved to a new section "Solenoid and coaxial cable in detail" (or something like that), or, possibly, to other articles (coaxial cable, solenoid), with a link added here.
Rdengler (talk) 16:35, 7 November 2009 (UTC)
replies: text: I think impedance can be mentioned in this article.  True it does not explain inductance but it is certainly connected.  As impedance will be a major feature of nearly all low level electrical study courses, this could be the very thing the reader is looking for.  I completely agree about the self-inductance calculation details, I have yet to find a textbook that goes into anything like that detail, even ones aimed at designers use a much simpler approximation for solenoids for instance.  We have to keep in mind that basic topics like this one will attract readers from very different backgrounds and capabilities and we should not make it too off-putting with technicalities.  Perhaps the simple approximations should be in this article and then link out to the full-featured version for specific geometries. SpinningSpark 22:52, 7 November 2009 (UTC)
thread_title:  The article lacks coherence 

message: I have reverted this edit which has the edit summary changed "wire loop" to coil of wire. The quoted formula contains N respresenting a number of turns of wire. For a wire loop N is 1 and would not be included in the formula.  I agree that the use of loop here is problematic and could , perhaps, be improved, but coil of wire is even more so.  I think the definition only works if all the turns are essentially co-incident - that is, lying on the same loop.  Furthermore, the change was not done consistently, the section continues to talk about wire loop further down. SpinningSpark 16:29, 9 August 2010 (UTC)
thread_title:  Loop of wire 

message: In the opening paragraph, I find the following sentence to be confusing:   "This is a linear relation between voltage and current akin to Ohm's law, but with an extra time derivate."  In fact, the voltage across an inductor is not proportional to the current -- it is 90º out of phase with it if the reactance is perfectly inductive.Jdlawlis (talk) 01:43, 18 December 2010 (UTC)
replies: text: I agree.  The statement is technically correct and mentions an important point, but should not be in the introduction. --ChetvornoTALK 04:27, 18 December 2010 (UTC)

text: Most people reading the introduction will be nontechnical people looking for the simplest possible explanation.  The defining equation belongs there, but discussion of its linearity does not.  The connection of inductance with magnetism isn't even mentioned until the 4th para.    This is why people complain wikipedia articles are too technical. --ChetvornoTALK 20:16, 19 December 2010 (UTC)

text: It is mathematically correct to state that inductors are governed by a linear equation.  It is not correct to state that there is a linear relation between voltage and current, and it is even more incorrect to state that the governing equation is analogous to Ohm's law.  The Ohm's law analogy only applies to the r.m.s. values of sinusoidal voltages and currents and requires the introduction of the concept of reactance to replace resistance in Ohm's law.  SpinningSpark 06:13, 21 April 2011 (UTC)

text: I agree with everything you say, SpinningSpark.  I think it would be clearer if the Ohm's law reference were removed.  Jdlawlis (talk) 14:11, 21 April 2011 (UTC)

text: What really gets mixed up here is proportionality (linear equation) and linear relation. The statement is correct in that the voltage generated by the sum of two (time dependent) currents is the sum of the voltages generated by the individual currents (as it is in Ohm's law). This would be a practically useful information, but if it is confusing, it need not be in the introduction. radical_in_all_things (talk) 06:49, 22 April 2011 (UTC)
thread_title:  confusing wording 

message: The section on "coupled inductors" has a paragraph at the bottom about tuned circuits, starting "When either side of the transformer is a tuned circuit, the amount...". This paragraph is not referenced and may be original research. Does anyone know where this material comes from?  —Preceding unsigned comment added by 121.98.140.35 (talk) 00:01, 20 May 2011 (UTC)
thread_title:  Original research or lack of references