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Electronics Forum / Basics / July 2009DC Current in Parallel Inductors?
Hello All, My teacher gave us a problem that is driving me absolutely crazy, and my Spice simulator is supplying odd answers. His question: In a circuit with a 10V DC power supply, and a series current limiting 1k Ohm resistor, and two (ideal) inductors in parallel with each other, one being 1uH and the other 10uH, will the DC currents be the exact same in each inductor branch after reaching steady state, or will they be less (by 10X) in the 10uH branch? If so, why should an ideal inductor of ANY value have any effect whatsoever on DC current after it reaches its steady state? Thank you! Desiree > Hello All, > [quoted text clipped - 7 lines] > inductor of ANY value have any effect whatsoever on DC current after > it reaches its steady state? For extra bonus points, also consider what effect, if any, there would be to your final answer if there was a current flowing in a loop through the two inductors at time zero. (Ideal parallel inductors will happily support a circulating current without loss.) > > Hello All, > [quoted text clipped - 12 lines] > the two inductors at time zero. (Ideal parallel inductors will happily > support a circulating current without loss.) Thanks Dan, but how would I calculate such a thing? And I'm still not sure whether two different value inductors in parallel will share the mainline DC current unequally or equally after reaching steady state. (I feel, but half my class does not, that after steady state is reached that both parallel inductors could simply be replaced by a zero ohm piece of wire...). Thanks! -Desiree >>> Hello All, >>> My teacher gave us a problem that is driving me absolutely crazy, [quoted text clipped - 17 lines] > reached that both parallel inductors could simply be replaced by a > zero ohm piece of wire...). Instead of looking at the steady state, you need to look at how you get there. The remainder of your homework problem is left to the student. >Instead of looking at the steady state, you need to look at how you get there. Has anyone got a Digkey part number for some of these inductors? Sounds like they would be really useful. -- >>Instead of looking at the steady state, you need to look at how you get there. > >Has anyone got a Digkey part number for some of these inductors? Sounds >like they would be really useful. Yeah, room temperature superconductors will be mighty handy. Finite-Q filter designs are a nuisance. John >>>> Hello All, >>>> My teacher gave us a problem that is driving me absolutely crazy, [quoted text clipped - 22 lines] > > The remainder of your homework problem is left to the student. Right, this is a theoretical question about ideal inductors, so they never actually reach "steady state," meaning constant current in this case. For a circuit made of ideal components, you would have to define steady state: in the limit as t increases without bound. >>>Hello All, >> [quoted text clipped - 23 lines] > > -Desiree yes >> Thanks Dan, but how would I calculate such a thing? And I'm still not >> sure whether two different value inductors in parallel will share the [quoted text clipped - 7 lines] >> -Desiree > yes Desiree, As I think that you have learned from the rest of this discussion, Jamie's answer is overly simplistic. In most cases you can treat ideal inductors like a zero ohm piece of wire in the steady state case, but it is not really appropriate in this problem. Actually trying to determine how current is shared between parallel pieces of wire is undefined if you have ideal zero ohm wires, etc. As Tim Wescott pointed out, this problem was more about thinking about the total problem rather than simply trying to put things into Spice. Spice did force you to think about the confusing result that it gave. Do you understand why you got the results that you did from Spice? I.e. Do you understand why you get a different final result with an ideal inductor (with zero resistance) versus an inductor with a small non zero resistance? With the ideal inductors, what is the answer if you assume that there is a 1 amp current flowing in a loop between the two inductors at time zero? BTW, if you search through the archives for this group, there was a discussion a few months ago about the voltages involved with a DC source, a resistor, and then two capacitors in series. Once again, I think that it was someone's homework problem. I think that the capacitors also differed by a factor of 10. (Homework problems like to keep the math simple.) Spice also did not like that problem. On Jul 2, 6:17 pm, Dan Coby <adc...@earthlink.net> wrote: > Miss_Koksuka wrote: > > Hello All, [quoted text clipped - 13 lines] > the two inductors at time zero. (Ideal parallel inductors will happily > support a circulating current without loss.) Thanks Dan, but how would I calculate such a thing? And I'm still not sure whether two different value inductors in parallel will share the mainline DC current unequally or equally after reaching steady state. (I feel, but half my class does not, that after steady state is reached that both parallel inductors could simply be replaced by a zero ohm piece of wire...). Thanks! -Desiree To calculate it, fist solve assuming zero initial current. Then, simply add any initial current to that solution, and you have your new answer. It is linear. If you look at the math in my other post you'll see why it works that way. It is a matter of solviing for the integration constant based on initial conditions. It happens to work out linearly. By the way, I made a slight error in the way I described the integration. You don't actually integrate over t. The t differential drops out. You integrate I1 with respect to I2, or vice versa (doesn't matter). Now, for real inductors with resistance, to get the steady state solution, simply substitute the series resistors for the inductors (leave the inductors as dead shorts) and solve. Good luck miss K. And consider using another screen name! > > > Hello All, > > > > My teacher gave us a problem that is driving me absolutely crazy, > > > and my Spice simulator is supplying odd answers. Sounds like you have a very good teacher. > ... I'm still not > sure whether two different value inductors in parallel will share the > mainline DC current unequally or equally after reaching steady state. To absolutely reach a steady state takes ... how much time? > (I feel, but half my class does not, that after steady state is > reached that both parallel inductors could simply be replaced by a > zero ohm piece of wire...). This is where Spice will always fail to provide the answer. It isn't a deep thinker, and sometimes you have to do the work outside the exact-numerical-answers world of theory that Spice inhabits. >> > Hello All, >> [quoted text clipped - 23 lines] > >-Desiree Since you have a spice simulator what values are you using for initial conditions? Also go back to basics E = -L*(di/dt). That should give you some help with the initial conditions. >>> > Hello All, >>> [quoted text clipped - 28 lines] >Also go back to basics E = -L*(di/dt). That should give you some help >with the initial conditions. LT Spice won't let you parallel two pure inductances. The error is "over-defined circuit matrix", which is the discrete equivalent of the currents being indeterminate. It also won't simulate two zero-ohm resistors in parallel. In mechanical systems, many structures are "statically indeterminate", same issue. John > Hello All, > [quoted text clipped - 11 lines] > > Desiree Would the troll like a cookie? Nice troll, come here, gootch gootchy goo SPLAT > > Hello All, > [quoted text clipped - 16 lines] > > SPLAT That is so nerdy and geeky of you to say, TROLL and SPLAT. How very uncool of you. Trolls are no longer accepted in newsgroups. If you call someone a troll, then you must be a troll from the past.. shhhh.. go away .. Come back when you are cool enough. > Hello All, > [quoted text clipped - 7 lines] > inductor of ANY value have any effect whatsoever on DC current after > it reaches its steady state? It doesn't. An ideal inductor has zero resistance of course, and you have two of these lovely devices in parallel. But ask yourself how long does it take to reach steady state in an ideal circuit? Tell your teacher you'll give him the answer after "the big crunch". Dave.  Signature================================================ Check out my Electronics Engineering Video Blog & Podcast: http://www.alternatezone.com/eevblog/ >Hello All, > [quoted text clipped - 11 lines] > >Desiree I guess we assume no initial currents before we switch on the supply. Put the two inductors, in parallel, into a black box. Now you have 10 volts through 1K ohms driving a 0.909 uH inductor. Calculate the voltage versus time across the black box. Now consider what would happen if that voltage profile were applied to the 1 uH inductor, and separately to the 10 uH inductor. The issue isn't so much what the circuit looks like "after it reaches its steady state" but the path it took to get there. An inductor integrates voltage into current, so it remembers everything that ever happened to it. What did Spice say? John On Jul 3, 12:26 am, John Larkin <jjSNIPlar...@highTHISlandtechnology.com> wrote: > On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka > [quoted text clipped - 33 lines] > > John Thanks guys. I'm trying to put all your answers together to clearly figure this all out, but its tough! John, here is a clearer explanation, and what I am seeing in Spice: In a circuit with a (10V) DC power supply, and a series current limiting (100 Ohm) resistor, and two ideal* inductors (with no mutual coupling) that are in parallel with each other -- one being 1uH and the other 10uH -- why do the DC currents take >>5xL/R to reach equality in each branch? Why should an ideal inductor of ANY value have ANY effect whatsoever on the DC current *after* it reaches its steady state? My Spice simulator shows that it takes a HUGE amount of time (25ms) to reach equal current of 50mA in each branch, and until then the current in the 10uH branch is 9.1mA, and the current in the 1uH branch is 91mA. Since 25ms is WAY past five time constants, why does it take so darn long to even-out the currents in each leg? (* Rser=0.001 to make Spice happy.) Confused, -Desiree * C:\Program Files\LTC\LTspiceIV\Draft-INDS.asc L1 N001 N002 1µ Rser=0.001 R1 N002 0 100 V1 N001 0 10 L2 N001 N002 10µ Rser=0.001 .TRAN 500us 0.05 UIC .PLOT TRAN I(L1) I(L2) .backanno .end On Jul 3, 12:26 am, John Larkin <jjSNIPlar...@highTHISlandtechnology.com> wrote: > <desiree_koks...@yahoo.com> wrote: > >Hello All, [quoted text clipped - 31 lines] > > John Thanks guys. I'm trying to put all your answers together to clearly figure this all out, but its tough! John, here is a clearer explanation, and what I am seeing in Spice: In a circuit with a (10V) DC power supply, and a series current limiting (100 Ohm) resistor, and two ideal* inductors (with no mutual coupling) that are in parallel with each other -- one being 1uH and the other 10uH -- why do the DC currents take >>5xL/R to reach equality in each branch? Why should an ideal inductor of ANY value have ANY effect whatsoever on the DC current *after* it reaches its steady state? My Spice simulator shows that it takes a HUGE amount of time (25ms) to reach equal current of 50mA in each branch, and until then the current in the 10uH branch is 9.1mA, and the current in the 1uH branch is 91mA. Since 25ms is WAY past five time constants, why does it take so darn long to even-out the currents in each leg? ============================= (* Rser=0.001 to make Spice happy.) <=======That explains it right there ============================= The current in ideal inductors would never even out, it would always have a ratio of 10:1 But you have give the inductors resistance. They are no longer ideal, and the current will even out. The time constant for this effect is determined by the series resistance of the inductors. > On Jul 3, 12:26 am, John Larkin > [quoted text clipped - 68 lines] > series resistance > of the inductors. Thanks Mike. So at least I'm not doing anything wrong with the Spice simulator! But I'll not rest until I find out exactly *why* this occurs. The mechanism behind it has me completely baffled, since it is not an LC tank circuit, so energy is not being exchanged back and forth between an inductor and capacitor. It is merely two inductors in parallel (I always assumed that such a circuit would simply act like single, lower value, inductor). So strange, but none of my (many) school books seems to cover any of this, they only say that an ideal inductor is a "short" to DC. Thanks again, -Desiree On Jul 3, 8:58 am, "Michael Robinson" <nos...@billburg.com> wrote: > "Miss_Koksuka" <desiree_koks...@yahoo.com> wrote in message > [quoted text clipped - 72 lines] > series resistance > of the inductors. Thanks Mike. So at least I'm not doing anything wrong with the Spice simulator! But I'll not rest until I find out exactly *why* this occurs. The mechanism behind it has me completely baffled, since it is not an LC tank circuit, so energy is not being exchanged back and forth between an inductor and capacitor. It is merely two inductors in parallel (I always assumed that such a circuit would simply act like single, lower value, inductor). So strange, but none of my (many) school books seems to cover any of this, they only say that an ideal inductor is a "short" to DC. Thanks again, -Desiree Are you pursuing an engineering degree? > On Jul 3, 8:58 am, "Michael Robinson" <nos...@billburg.com> wrote: > [quoted text clipped - 90 lines] > > Are you pursuing an engineering degree? Yes Mike, but I am only in my first year of electronics. >> Are you pursuing an engineering degree? > > Yes Mike, but I am only in my first year of electronics. Okay. Now we need to know a little more about what you have learned. Have you had calculus? Do you know what this equation means and how to use it: v = L di/dt On Jul 3, 12:26 pm, "Michael Robinson" <nos...@billburg.com> wrote: > "Miss_Koksuka" <desiree_koks...@yahoo.com> wrote in message > [quoted text clipped - 95 lines] > > Are you pursuing an engineering degree? Yes Mike, but I am only in my first year of electronics. Good. I'm going to show you a simple solution. It amounts to a fairly rigorous demonstration that you can show your instructor. Assume the inductors have zero current in them at t=0, which is when you apply the voltage. Dan Coby gave you formula v=L*(di/dt) I'm going to call the inductances L1 and L2. Simillarly, currents in the inductors I1 and I2. I1 and I2 are variables, functions of t (time). Now, v is always equal for both inductors becuase they are in parallel. Therefore L1 (dI1/dt) = L2 (dI2/dt) Say L2 is the "bigger" inductor. L2 = 10 L1 implies dI1/dt = 10 dI2/dt which we can integrate over time: I1 = 10 I2 + C (I hope you've had some basic calculus) now, C has to be zero because at t=0, I1 = I2 = 0 so I1 = 10 I2 and that's always true, for any value of t > So strange, but none of my > (many) school books seems to cover any of this, they only say that an > ideal inductor is a "short" to DC. Ideal inductors don't exist in real life -- they're just a convenience used in simplified mathematical models of circuits. When building such a model, you wouldn't bother putting two ideal inductors in parallel, you'd just lump them together into a single inductance. And having done that, you can indeed treat it as a short to DC. If you have a real circuit in which it's important to know the DC current distribution between parallel inductors, then you can't model them as ideal inductors. You *have* to take their resistance into account, because that's what ultimately determines the current distribution. SignatureGreg > > So strange, but none of my > > (many) school books seems to cover any of this, they only say that an [quoted text clipped - 17 lines] > -- > Greg You guys are great! Thanks for clearing this up for me (I spent days trying to do it; I felt as if my head were about to explode at one point!!). Many Thanks, -Desiree >> So strange, but none of my >> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 14 lines] >because that's what ultimately determines the current >distribution. Tedious, but wrong. Consider superconducting magnets. Even better, consider superconducting magnets that have superconducting shim coils. John >>> So strange, but none of my >>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 17 lines] >Tedious, but wrong. Consider superconducting magnets. Even better, >consider superconducting magnets that have superconducting shim coils. --- What's tedious is your shifting the subject around in order to keep waving your own flag while you're patting yourself on the back. No one's talking superconducting magnets, the topic is about conventional inductances. JF >>>> So strange, but none of my >>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 26 lines] > >JF Greg said: >>>You *have* to take their resistance into account, >>>because that's what ultimately determines the current >>>distribution. which I pointed out isn't always the case. Some people work with real inductors that actually have no resistance. I suppose you don't. John > Greg said: > [quoted text clipped - 4 lines] > which I pointed out isn't always the case. Some people work with real > inductors that actually have no resistance. So there's nothing wrong with Greg's statement. In this special case it just means you have to take their zero resistance into account. >> Greg said: >> [quoted text clipped - 7 lines] >So there's nothing wrong with Greg's statement. In this special case it just >means you have to take their zero resistance into account. Right, just as you have to take their zero Leprechaun content into account. John > Right, just as you have to take their zero Leprechaun content into > account. Only if general calculations expected to have to handle leprechauns. >> Right, just as you have to take their zero Leprechaun content into >> account. > >Only if general calculations expected to have to handle leprechauns. The general equations that define the behavior of an inductor don't include resistance. I = Io + time_integral(E/L) So why include a resistance term and then make an effort to remove its effects? Plugging into cookbook equations is a risky way to really understand things, which was maybe one of the points that Desiree's instructor was making. John >>>>> So strange, but none of my >>>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 37 lines] > >I suppose you don't. --- Whether I do or not has nothing to do with it, so other than showing yourself up for the patronizing a.s you truly are, what's your point? JF >>>>>> So strange, but none of my >>>>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 43 lines] > >JF That the distribution of current in the OP's problem does not depend on any resistance in the inductors, and that assuming or simulating *any* resistance gives the wrong answer, and that this is not a purely theoretical problem. What's your point? John >>>>>>> So strange, but none of my >>>>>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 50 lines] > >What's your point? --- That you're a patronizing a.s. More to the point though, since the OP's: "Hello All, My teacher gave us a problem that is driving me absolutely crazy, and my Spice simulator is supplying odd answers. His question: In a circuit with a 10V DC power supply, and a series current limiting 1k Ohm resistor, and two (ideal) inductors in parallel with each other, one being 1uH and the other 10uH, will the DC currents be the exact same in each inductor branch after reaching steady state, or will they be less (by 10X) in the 10uH branch? If so, why should an ideal inductor of ANY value have any effect whatsoever on DC current after it reaches its steady state? Thank you! Desiree" indicated that her teacher had specified that the inductors were ideal _makes_ the problem hypothetical. JF >>>>>>>> So strange, but none of my >>>>>>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 53 lines] >--- >That you're a patronizing a.s. Well, that's a lot easier to say than actually thinking about hard stuff like electronics. >More to the point though, since the OP's: > [quoted text clipped - 16 lines] >indicated that her teacher had specified that the inductors were ideal >_makes_ the problem hypothetical. Or superconductive. But assuming ideal inductors was a good way to make the kids really think about the problem, and address the reality that circuit conditions sometime depend on the history/path of the circuit, not some static endpoint calculation. "Constant of integration" is not a "hypothetical" quantity. Recovering from your 4th of July hangover? John >>>>>>>>> So strange, but none of my >>>>>>>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 56 lines] >Well, that's a lot easier to say than actually thinking about hard >stuff like electronics. --- That's true, but it has nothing to do with the fact that you asked me what my point was, and I replied. --- >>More to the point though, since the OP's: >> [quoted text clipped - 18 lines] > >Or superconductive. --- How could defining the inductors as ideal make the problem superconductive? --- >But assuming ideal inductors was a good way to >make the kids really think about the problem, and address the reality >that circuit conditions sometime depend on the history/path of the >circuit, not some static endpoint calculation. "Constant of >integration" is not a "hypothetical" quantity. --- True, but so what? It has nothing to do with the fact that the _problem_ was hypothetical and you're just throwing more crap around in order the dodge having to admit that you said it wasn't. Business as usual, huh? --- >Recovering from your 4th of July hangover? --- Never had one. JF >>>>>>>>>> So strange, but none of my >>>>>>>>>> (many) school books seems to cover any of this, they only say that an [quoted text clipped - 102 lines] >and you're just throwing more crap around in order the dodge having to >admit that you said it wasn't. Everything I said here was true. The only thing of any substance that you said - 5 mA per inductor - was dead wrong. Nothing else matters. John >>If you have a real circuit in which it's important to >>know the DC current distribution between parallel >>inductors, then you can't model them as ideal inductors. >>You *have* to take their resistance into account > > Tedious, but wrong. Consider superconducting magnets. What I mean is, if they have any resistance at all, however small, you can't ignore it if you want to know the steady-state DC current. If the inductors are superconducting, they never reach steady state in the sense of a current distribution that's independent of the voltage history they've been subjected to. In that case, you have to take the history into account. Since most people never have occasion to have to deal with superconducting magnets, introductory electronics books can be forgiven for not going into that level of detail. Now, if someone comes up with a room-temperature superconductor and superconducting components become commomplace, that might change... BTW, I'm not sure that you can call a superconductor an "ideal inductance" in the mathematical sense, since they have limitations such as a maximum magnetic field before they stop superconducting. But I'll grant they're certainly a much better approximation of one than any ordinary inductor. SignatureGreg > Thanks guys. I'm trying to put all your answers together to >clearly figure this all out, but its tough! Your intuition might work better with capacitors rather than inductors. What would you expect if you had a resistor, small cap, and big cap in series? SignatureThese are my opinions, not necessarily my employer's. I hate spam. >On Jul 3, 12:26 am, John Larkin ><jjSNIPlar...@highTHISlandtechnology.com> wrote: [quoted text clipped - 46 lines] >have ANY effect whatsoever on the DC current *after* it reaches its >steady state? As noted, you have to include the entire history of the voltage applied to the inductor to know its current. A shorted inductor of unknown history has an indeterminate current. Ditto two paralleled inductors. > My Spice simulator shows that it takes a HUGE amount of time >(25ms) to reach equal current of 50mA in each branch, and until then >the current in the 10uH branch is 9.1mA, and the current in the 1uH >branch is 91mA. Since 25ms is WAY past five time constants, why does >it take so darn long to even-out the currents in each leg? Spice artifact, essentially some minimum (non-zero) resistance parameter. For ideal inductors, you wouldn't see that. The voltage waveform is a spike up to 10 volts, exponentially decaying with a time constant of about 9 nanoseconds. Spice often lies. >(* Rser=0.001 to make Spice happy.) That also shoots down the concept of "ideal inductor." John On Jul 3, 3:04 pm, John Larkin <jjSNIPlar...@highTHISlandtechnology.com> wrote: > On Fri, 3 Jul 2009 06:15:27 -0700 (PDT), Miss_Koksuka > [quoted text clipped - 75 lines] > > John Thanks all, I think a little light is beginning to dawn on this for me, but this is all completely non-intuitive (I keep thinking of inductors as a straight piece of wire at DC!). But with v=L*di/dt for non-sinusoidal waveforms, such as this DC circuit's turn-on waveform, I'm starting to get a little clearer on this. And yes John, you are right -- I just checked, and my Spice simulator (LT Spice) appears to have "inserted" some very small non- zero resistance value in my circuit (even with my 0.001 ohm resistors removed); undoubtedly to help prevent convergence issues. Thanks again, -Desiree > Thanks all, I think a little light is beginning to dawn on this >for me, but this is all completely non-intuitive (I keep thinking of >inductors as a straight piece of wire at DC!). --- Indeed and, at DC, an inductor is nothing more than a resistor. --- >But with v=L*di/dt for >non-sinusoidal waveforms, such as this DC circuit's turn-on waveform, >I'm starting to get a little clearer on this. --- If you go back to the basics, things might get a _lot_ clearer. Consider: When a voltage is impressed across a conductor, a magnetic field will be generated about that conductor, and when a magnetic field 'cuts' a conductor it will induce a voltage in that conductor. As it turns out (courtesy of Mother Nature), if you connect a voltage source across a conductor, the magnetic field generated will cut the conductor and will generate a voltage with a polarity opposite that of the voltage source! When the voltage is first applied, the induced voltage will be equal to the applied voltage and will keep charge from flowing through the conductor, but as time goes by the characteristics of the field change, allowing more and more current through the conductor until a limit is reached when the series resistances in the circuit and the source voltage satisfy: E I = --- R Here's some more: http://en.wikipedia.org/wiki/Lenz's_law JF >> Thanks all, I think a little light is beginning to dawn on this >>for me, but this is all completely non-intuitive (I keep thinking of >>inductors as a straight piece of wire at DC!). > >--- >Indeed and, at DC, an inductor is nothing more than a resistor. Unless it has a whopping current circulating in it. Had and MRI lately? John >>> Thanks all, I think a little light is beginning to dawn on this >>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 7 lines] > >John Even if it has zero current circulating in it, just rotate a bit, then stop, and there will be (in general). Best regards, Spehro Pefhany Signature"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com >>> Thanks all, I think a little light is beginning to dawn on this >>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 5 lines] >Unless it has a whopping current circulating in it. Had and MRI >lately? --- Yeah, but so what? No one's talking superconducting magnets, you pretentious a.s. JF >>>> Thanks all, I think a little light is beginning to dawn on this >>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 12 lines] > >JF People were talking about ideal inductors, which a real-world, practical, superconductive coil is [1]. And it ain't the same as a resistor. Got your monthlies again? John [1] Well, it will probably have a finite Q. >>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 15 lines] >People were talking about ideal inductors, which a real-world, >practical, superconductive coil is [1]. --- Nope, it isn't, since there are limits to the current it can handle as well as to the strength of the magnetic field it generates. But then, you _do_ have trouble with infinity, don't you? Besides, it wasn't at all what Desiree was looking for, which was a simple description of what an ordinary inductor looks like with steady state DC in it. Since the current in it will be limited by the resistance of the wire it's wound with and the voltage across it, we can say: E R = --- I Look familiar to ya? --- >And it ain't the same as a resistor. --- ??? What is it about: E R = --- I that you don't understand? --- >Got your monthlies again? --- Nope, just annoyed with your self-aggrandizing bullshit. --- >[1] Well, it will probably have a finite Q. --- And just how would you define the Q of an inductor with steady-state DC in it? JF >>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 21 lines] > >But then, you _do_ have trouble with infinity, don't you? Looks like you're having trouble with zero, which I thought was a simpler concept than infinity. >Besides, it wasn't at all what Desiree was looking for, which was a >simple description of what an ordinary inductor looks like with steady >state DC in it. Her confusion was based on applying "ordinary" inductors in a problem that specified ideal inductors. And having Spice add in some "ordinary" resistance of its own. >Since the current in it will be limited by the resistance of the wire >it's wound with and the voltage across it, we can say: [quoted text clipped - 4 lines] > >Look familiar to ya? Ooh, Ohm's "Law." R=0 in Desiree's problem. >--- > [quoted text clipped - 23 lines] >And just how would you define the Q of an inductor with steady-state DC >in it? There's an accepted definition of Q http://en.wikipedia.org/wiki/Q_factor which could be measured at any specified frequency for any inductor, superconducting or not, with DC current present or not. It would be an "AC" measurement if DC is present, to be consistant with situations like resonant behavior in tank circuits which have DC current. Imagine coupling an identical inductor to the one under test, with M=1, and measuring that. Superconducting solenoids can have literally zero resistance, as far as anyone has been able to measure [1], but will usually have measurable, finite Q. I note that you made no attempt to help her solve her dilemma, whereas I did. All you did was to enter late and make bitchy snipes at my typing, and at the things I have said, all of which were true. Whiny and fact-free, as usual. I'm in Truckee, typing on this horrible little Vaio keyboard, in natural light, and I never learned to type anyhow. John [1] the complex (filamentary) superconducting magnets, like the ones in MRI magnets, do lose a bit of field strength over time, like a PPM every couple of months for a good one. I'm not sure if that actually constitutes "resistance" (ie, energy loss) or just a shift of current paths or geometry. > I'm in Truckee Doing /done the Rubicon Trail, like you threatened? Signature"Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock) >> I'm in Truckee > >Doing /done the Rubicon Trail, like you threatened? Not yet; maybe in August. We were walking around the Rainbow Bridge and decided to hike up to the top of the China Wall, between a couple of tunnels and snow sheds of the original Transcontinental Railroad. It's a nice short steep hike over smooth stepped glacial-looking rocks, with some 3000-year-old petroglyphs here and there. Anyhow, when we got to the top, there were a couple of guys in Jeeps. Turns out that you can quasi-legally enter the tunnels near Sugar Bowl and drive through. One of them is, I think, about 1600 feet long. The Brat will bring up her Jeep in a weekend or two and we'll try it. ftp://66.117.156.8/CW_Donner_Lake.jpg ftp://66.117.156.8/CW_Tunnel.jpg ftp://66.117.156.8/CW_sign.jpg ftp://66.117.156.8/CW_snow_shed.jpg ftp://66.117.156.8/CW_wall.jpg It's a "gravity wall" constructed for the roadbed by hand, from natural uncut rocks, no cement. The road is Donner Pass Road, old California route 40, a section of the original Lincoln Highway. http://en.wikipedia.org/wiki/Lincoln_Highway Bronze marker courtesy E Clampus Vitus: http://en.wikipedia.org/wiki/E_Clampus_Vitus Cool stuff. John >>> I'm in Truckee >> [quoted text clipped - 37 lines] > > John Very nice photos ! TFS. SignatureBest Regards: Baron. >>> I'm in Truckee >> [quoted text clipped - 29 lines] > > http://en.wikipedia.org/wiki/Lincoln_Highway Nice pictures, John. I shall have to go take a look myself sometime. > Bronze marker courtesy E Clampus Vitus: You a Clamper, John? Signature"Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock) >>>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 24 lines] >Looks like you're having trouble with zero, which I thought was a >simpler concept than infinity. --- Philosophically, they're the same thing except that one grows smaller without bound while the other grows larger. --- >>Besides, it wasn't at all what Desiree was looking for, which was a >>simple description of what an ordinary inductor looks like with steady [quoted text clipped - 3 lines] >that specified ideal inductors. And having Spice add in some >"ordinary" resistance of its own. --- If the inductors were specified as being ideal, then your comment in another post about the problem not being hypothetical was wrong. --- >>Since the current in it will be limited by the resistance of the wire >>it's wound with and the voltage across it, we can say: [quoted text clipped - 6 lines] > >Ooh, Ohm's "Law." R=0 in Desiree's problem. --- Yeah, in the inductor, but there was 1000 ohms in series with the two parallel inductors, which means that when everything settled down there'd be 5mA through each of the inductors. --- >>--- >> [quoted text clipped - 38 lines] >as anyone has been able to measure [1], but will usually have >measurable, finite Q. --- All well and good, but the point I was trying to make was that for the purpose of the discussion, the Q of the inductance under steady-state conditions is irrelevant and just more noise you're adding. --- >I note that you made no attempt to help her solve her dilemma, whereas >I did. --- She was getting plenty of help, so I saw no need to jump in until you showed up with your smarty-pants "I'm smarter than all of you." attitude. --- >All you did was to enter late and make bitchy snipes at my >typing, and at the things I have said, all of which were true. --- Well, often they're not true when you start out, but by the time you get finished with your little song and dance you've generally confused the sh.t out of most everyone and you can then pretend you were right all along since most everyone gets tired of your crap and just drops out. --- >Whiny and fact-free, as usual. --- Was I the one who stated that protons can go dancing around just like electrons? Nope, it was you, so that's at least one example of your being wrong, making your statement that everything you said was true wrong as well. --- >I'm in Truckee, typing on this horrible little Vaio keyboard, in >natural light, and I never learned to type anyhow. --- Poor baby... JF >>>>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 58 lines] >parallel inductors, which means that when everything settled down >there'd be 5mA through each of the inductors. WRONG! John >>>>>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 62 lines] > >John Really? Then (assuming, of course, ideal wiring) replace the 'x's with the proper currents: . XmA--> . +--[0R]--+ . 10mA--> | | .10V----1000R----+ +---+ . | | | . +--[0R]--+ | . XmA--> | . <--10mA | .0V---------------------------+ JF >>>>>>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 55 lines] >>> >>>--- >>>Yeah, in the inductor, but there was 1000 ohms in series with the two >>>parallel inductors, which means that when everything settled down >>>there'd be 5mA through each of the inductors. ^^^^^^^^^ >>WRONG! >> [quoted text clipped - 16 lines] > >JF The problem involved unequal-value inductors, not zero-ohm resistors. Even you, just above, agreed that we were dealing with zero-ohm inductors. The math of the inductor case has been discussed elsewhere in this thread. But the zero-ohm-resistor case is indeterminate. There could be any amount of current circulating in the zero-ohm loop. And there's no reason that the externally applied current would divide evenly. One superconductor trick is to have this exact circuit, with all the current going through one leg and none in the other. That is just fine mathematically. I suspect the whole point of the instructor's original puzzle was to make the students realize that an ideal inductor does not behave like a resistor, even after the circuit has reached equilibrium. The solution for the original problem (assuming no circulating currents before the power supply was kicked on) is NOT equal currents in the two inductors. John > On Mon, 06 Jul 2009 12:08:53 -0500, John Fields <snip> >>>WRONG! >> [quoted text clipped - 34 lines] > currents before the power supply was kicked on) is NOT equal currents > in the two inductors. How do you define steady state? More importantly how would an Engineering professor use that term for a first year student? FWIW, I checked with 6 professors and I be interested in what you think they said about your reply to JF's answer. As for the OP, I hope by now it's very aparrent that it's difficult to impossible to get reliable information in this newsgroup; sadly. Unless you already know the answer to a question it will be difficult to separate signal and noise in the inevitable dispute and generally bad behavior ANY question will create. >> On Mon, 06 Jul 2009 12:08:53 -0500, John Fields ><snip> [quoted text clipped - 38 lines] > >How do you define steady state? Casually, when the changes have settled out to so small as to not matter. Formally, a circuit state in which all currents are assigned to be the mathematical limits that they were approaching. Basically, wait long enough that waiting more doesn't change things measurably. 10 tau is a common ROT. >More importantly how would an Engineering professor use that term for a >first year student? Can't address that one. I'm a real engineer. >FWIW, I checked with 6 professors and I be interested in what you think >they said about your reply to JF's answer. I can't imagine what they said. If they agree with JF about 5+5 mA, they're wrong. >As for the OP, I hope by now it's very aparrent that it's difficult to >impossible to get reliable information in this newsgroup; sadly. Unless >you already know the answer to a question it will be difficult to >separate signal and noise in the inevitable dispute and generally bad >behavior ANY question will create. I suggested an approach to solving the problem. If it makes sense to her, maybe it will help. Actually, several people have suggested essentially equivalent approaches, in rant-free and logical ways. JF is fighting the math, and the math generally wins. John >>How do you define steady state? >Casually, when the changes have settled out to so small as to not >matter. Formally, a circuit state in which all currents are assigned >to be the mathematical limits that they were approaching. >Basically, wait long enough that waiting more doesn't change things >measurably. 10 tau is a common ROT. If the inductors are not-quite idea (small DC resistance), I think there will be two interesting steady states. The first one is when the inductors get charged up with the current ratio determined by the inductance ratio. The second one is when they decay so the current ballance matches the DC resistance ratio. The first time constant is determined by the external resistor. The second time constant is determined by the resistance of the inductors. I thought this was a fun problem. It took me a few seconds to figure out what was going on: good bait for students. SignatureThese are my opinions, not necessarily my employer's. I hate spam. >>>How do you define steady state? > >>Casually, when the changes have settled out to so small as to not >>matter. Formally, a circuit state in which all currents are assigned >>to be the mathematical limits that they were approaching. Or maybe "a state S within the space of all possible system states wherein all successive states will also be S." That handles digital systems, too. >>Basically, wait long enough that waiting more doesn't change things >>measurably. 10 tau is a common ROT. [quoted text clipped - 4 lines] >by the inductance ratio. The second one is when they decay so the >current ballance matches the DC resistance ratio. Yes. There will be the obvious fast settling transient, and then some small, slow tails. "Steady state" becomes one's opinion about what really matters to an application. >The first time constant is determined by the external resistor. >The second time constant is determined by the resistance of >the inductors. I make NMR and MRI gradient drivers, precision high-power constant-current amplifiers. We need these things to have risetimes in the 10 microsecond range and settle to a few PPM in 100 us or so. All these lower-order tails hurt big-time here. One especially nasty issue is self-induced eddy-currents in current shunts and, well, everywhere else. Eddy currents are even uglier than simple lumped parasitic elements, because they have complex exponential decays, mathematically like thermal diffusion... long, ugly tails that are hard to equalize out. >I thought this was a fun problem. It took me a few seconds to >figure out what was going on: good bait for students. Yeah, this is a good one, and Spice-proof. John >>>>>>>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 87 lines] >inductors. The math of the inductor case has been discussed elsewhere >in this thread. --- Geez, John, knowing the value of the inductances and assuming an instantaneous step from 0V to 10V on the left hand side of the resistor, why didn't you just fill in the values of the steady-state currents? --- >But the zero-ohm-resistor case is indeterminate. There could be any >amount of current circulating in the zero-ohm loop. --- Really? What do you define as the zero-ohm loop? --- > One superconductor trick is to have this exact circuit, with all the >current going through one leg and none in the other. That is just fine >mathematically. --- Indeed, and if you took enough samples and the lack of resistance was truly ohmic, the current through both branches would be the same. --- >I suspect the whole point of the instructor's original puzzle was to >make the students realize that an ideal inductor does not behave like [quoted text clipped - 3 lines] >currents before the power supply was kicked on) is NOT equal currents >in the two inductors. --- Then what is it? JF >>>>>>>>>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>>>>>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 92 lines] >instantaneous step from 0V to 10V on the left hand side of the resistor, >why didn't you just fill in the values of the steady-state currents? The 1 uH inductor gets 10/11 of the current, and the 10 uH one gets 1/11th. That's not just the steady-state solution, it's true throughout the entire experiment (assuming no initial currents before the power supply was fired up.) Just imagine applying 10 volts to the paralleled inductors. The current in the 1 uH guy ramps up at 10 amps per microsecond. The current in the 10 uH inductor ramps at 1 a/us... a 10:1 current ratio. Since inductors are linear, the 10:1 current ratio remains for any applied voltage and for any waveform. >--- > [quoted text clipped - 3 lines] >--- >Really? Yup. >What do you define as the zero-ohm loop? The closed path through the two zero-ohm resistors. What else could it be in this circuit? >--- > [quoted text clipped - 5 lines] >Indeed, and if you took enough samples and the lack of resistance was >truly ohmic, the current through both branches would be the same. I don't know what you mean by that. Do you mean every single case would show equal currents, or that many cases would average to zero current? I also don't know what you mean by "the lack of resistance was truly ohmic." How can zero resistance be other than zero? >--- > [quoted text clipped - 8 lines] >--- >Then what is it? See above. John On Jul 5, 1:18 pm, John Larkin <jjSNIPlar...@highTHISlandtechnology.com> wrote: > On Sun, 05 Jul 2009 05:31:19 -0500, John Fields > [quoted text clipped - 110 lines] > > - Show quoted text - "Superconducting solenoids can have literally zero resistance, as far as anyone has been able to measure [1], but will usually have measurable, finite Q." Everyone should get to play with a superconducting magnet. Apply heat to the superconducting switch., ramp the current up to 100A, Turn off the heat. Ramp current back down to zero. Then take data at 8 Tesla for several hours. Ramp current back up to 100A (exactly), apply heat to switch, ramp current back down to zero and now it's safe to go home. But where does the finite Q come from? Is it radiation resistance? I always thought it would be fun to design circuits with super conductors. George H. >"Superconducting solenoids can have literally zero resistance, as far >as anyone has been able to measure [1], but will usually have [quoted text clipped - 6 lines] >to switch, ramp current back down to zero and now it's safe to go >home. Why ramp it down? You're wasting energy! >But where does the finite Q come from? Is it radiation resistance? I >always thought it would be fun to design circuits with super >conductors. Most superconductive magnets (maybe all?) have a stainless containment vessel, layers of metallic foil superinsulation, plumbing, room temp shim coils, other resistive metal stuff around that it will couple to. I'd expect Q to be pretty low. I guess a superconductive solenoid in free space, or at least in a nonmetallic unsilvered dewar far from other stuff, would have an extreme Q. It would still radiate a little. Superconducting microwave cavities can have Qs like 1e8. It will sure be nice when room temp superconductors are invented. John >>"Superconducting solenoids can have literally zero resistance, as far >>as anyone has been able to measure [1], but will usually have [quoted text clipped - 21 lines] >nonmetallic unsilvered dewar far from other stuff, would have an >extreme Q. It would still radiate a little. An air-core superconducting coil in a superconducting can (as superconducting signal transformers generally are) should have an extremely high Q, but probably still finite. >Superconducting microwave cavities can have Qs like 1e8. > >It will sure be nice when room temp superconductors are invented. > >John AFAIUI, current very high temperature superconductors are pretty nasty to work with-- they'd make lead-free look really good. Best regards, Spehro Pefhany Signature"it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com >>>"Superconducting solenoids can have literally zero resistance, as far >>>as anyone has been able to measure [1], but will usually have [quoted text clipped - 25 lines] >superconducting signal transformers generally are) should have an >extremely high Q, but probably still finite. Good point. >>Superconducting microwave cavities can have Qs like 1e8. >> [quoted text clipped - 4 lines] >AFAIUI, current very high temperature superconductors are pretty nasty >to work with-- they'd make lead-free look really good. The medium HTS stuff is copper oxides and such. Kids make their own for science fairs. But you can't make 1008 wound inductors from it. Is there any theoretical/fundamental/physics reasone why we can't have room temp superconductors? John On Jul 7, 12:15 am, John Larkin <jjlar...@highNOTlandTHIStechnologyPART.com> wrote: > On Mon, 6 Jul 2009 19:54:28 -0700 (PDT), George Herold > [quoted text clipped - 30 lines] > > John You ramp down the power supply.. leaving the magnet energized. The reason for ramping down the power supply is to save liquid helium. If not you are left with 100 Amps flowing down the leads to the magnet. "> It will sure be nice when room temp superconductors are invented." I'm not holding my breath. But working at liquid nitrogen temperatures is not out of the question. I guess laying down HTC superconductors is not as easy as putting down copper. George H. George H. >>>> Thanks all, I think a little light is beginning to dawn on this >>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 12 lines] > >JF The discussion is about ideal inductors, indistinguishable from superconducting magnets for almost all intents and purposes. Or are so busy dissing JL that you cannot answer OP's question? >>>>> Thanks all, I think a little light is beginning to dawn on this >>>>>for me, but this is all completely non-intuitive (I keep thinking of [quoted text clipped - 16 lines] >superconducting magnets for almost all intents and purposes. >Or are so busy dissing JL that you cannot answer OP's question? He did answer it. And he was wrong. John >But with v=L*di/dt for non-sinusoidal waveforms, such as this DC >circuit's turn-on waveform, V = L di/dt Inductors in parallel experience the same voltage and time which in the above equation makes Li a constant. At any time (including that when a notional steady state is reached) the current in each inductor is inversely proportional to its inductance. -- >>But with v=L*di/dt for non-sinusoidal waveforms, such as this DC >>circuit's turn-on waveform, [quoted text clipped - 6 lines] >At any time (including that when a notional steady state is reached) the >current in each inductor is inversely proportional to its inductance. Yes. John >On Jul 3, 12:26 am, John Larkin ><jjSNIPlar...@highTHISlandtechnology.com> wrote: [quoted text clipped - 67 lines] >.backanno >.end Why did you introduce series resistance for the two inductors? That does not conform to the problem statement. If you feel you must to improve convergence, at least use 1 picoOhm. Not to mention 11 uH in series with 2 mOhm has an appreciably long time constant. At a picoOhm you will see the 1 second result as the final value. >>On Jul 3, 12:26 am, John Larkin >><jjSNIPlar...@highTHISlandtechnology.com> wrote: [quoted text clipped - 75 lines] >time constant. At a picoOhm you will see the 1 second result as the >final value. You can cheat a little and add, say, 1 uohm in series with the 1 uH thing, and 10 uohm in series with the 10 uH guy. That will kill the secondary tails. I think. John > Hello All, > [quoted text clipped - 11 lines] > > Desiree Given the level of the question steady state means after all the math is done. Parallel devices have equal voltages. Tom > Hello All, > [quoted text clipped - 7 lines] > value have any effect whatsoever on DC current after it reaches its > steady state? Perhaps one of the things your instructor wanted to do was to wean you away from using Spice -- which is a fine tool for some things -- for everything. Try doing it using Laplace domain analysis; John Larkin's suggestion of finding the voltage and then the current is to the point if you want to simplify the math. Remember that Spice is a real-world tool, and an ideal inductor is not a real-world device. So using Spice and expecting it to cancel out infinities is inappropriate. OTOH, this is a fairly simple problem with linear circuit elements -- hence my suggestion of using Laplace analysis. Signaturewww.wescottdesign.com > Hello All, > [quoted text clipped - 11 lines] > > Desiree So; 10 volt supply. One kilohm limiting resistor in series with two ideal inductors in parallel. Correct????? Final steady state current = 10/1000 = 10 milliamps. At any other point in time (From switching on or switching off) the current will be dfferent, due to effect of inductance and the increasing or collapsing magnetic fields within them. Since the inductors are both ideal (zero ohms) and therefore DC identical, it is reasonable to assume that once the 10 ma steady state current has been reached, it will split equally between the two inductors. Why make it more complicated? But is this a troll? > Since the inductors are both ideal (zero ohms) and therefore DC > identical, it is reasonable to assume that once the 10 ma steady state > current has been reached, it will split equally between the two > inductors. Why make it more complicated? The even split assumption is certainly questionable since superconductors are involved. In fact, a little thought should reveal that the only thing that will determine the final current in each inductor will be the history of the current flows through each. At steady state both inductors will have a constant current and zero voltage drop. Without considering how the final currents obtain, that condition (zero voltage drop, constant current) can be met by any aritrary currents that add up to the required total. There could even be a very large circulating current going around the superconducting ring formed by the two inductors quite separate from the current passing through the pair from the voltage source and resistor. With superconductors, current has a sort of inertia. Here's an example to consider. Suppose you have two superconducting circuits in the form of squares. They are side by side. The one on the left has a 100A current circulating counterclockwise in it, while the one on the right has a 10A current circulating clockwise. Now the two nearest sides of the squares are brought into contact in such a way that they merge into a single conductor. After the merge, what is the steady state current in each part of the circuit? . .-------<--------. .------->--------. . | 100 A | | 10 A | . | | | | . | | | | . | | | | . | | | | . | | | | . '----------------' '----------------' >> Since the inductors are both ideal (zero ohms) and therefore DC >> identical, it is reasonable to assume that once the 10 ma steady state [quoted text clipped - 37 lines] >. | | | | >. '----------------' '----------------' Ouch. You'd have to assume that they were very close (the gap you drew was very small) before the merger; otherwise they will change currents (and the system energy will change) as they move towards one another, since their magnetic fields will interact. Given that, I'd guess that the currents won't change. That's the easiest guess that conserves energy. I have at least a 52% probability of being right. John >> Hello All, >> [quoted text clipped - 22 lines] >current has been reached, it will split equally between the two >inductors. No, because it's the wrong answer. >Why make it more complicated? It's actually very simple. The 1 uH inductor gets 9.09 mA, and the 10 uH one gets 0.909 mA after everything settles down. >But is this a troll? I doubt it. It sure has messed up a bunch of people. John >>> Hello All, >>> [quoted text clipped - 33 lines] > >I doubt it. It sure has messed up a bunch of people. --- One in particular, it seems. Here's the OP's circuit in LTspice: Version 4 SHEET 1 880 680 WIRE 176 128 128 128 WIRE 320 128 256 128 WIRE 448 128 400 128 WIRE -16 192 -96 192 WIRE 128 192 128 128 WIRE 128 192 64 192 WIRE 448 192 448 128 WIRE 528 192 448 192 WIRE -96 240 -96 192 WIRE 128 256 128 192 WIRE 176 256 128 256 WIRE 320 256 256 256 WIRE 448 256 448 192 WIRE 448 256 400 256 WIRE -96 352 -96 320 WIRE 528 352 528 192 WIRE 528 352 -96 352 WIRE -96 400 -96 352 FLAG -96 400 0 SYMBOL ind 304 144 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L1 SYMATTR Value 10e-6 SYMBOL ind 304 272 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L2 SYMATTR Value 1e-6 SYMBOL res 80 176 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 1000 SYMBOL voltage -96 224 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 10 .1 1e-7) SYMBOL res 272 112 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R2 SYMATTR Value 1e-12 SYMBOL res 272 240 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R3 SYMATTR Value 1e-12 TEXT -130 424 Left 0 !.tran 10 uic LTspice can't handle parallel zero-resistance inductors, but making their intrinsic resistances smaller and smaller until it goes nuts always results in a 50-50 split of current for all values of resistance when the circuit settles down. It's something about that when the power supply turn-on edge hits them both, their reactances and time conspire to split the current between them equally, eventually... JF >>>> Hello All, >>>> [quoted text clipped - 102 lines] > >JF Sure, as long as you add equal resistances. Then they're not ideal inductors any more. And you're assuming that the 1 uH inductor has the same DCR as the 10 uH inductor. You're forcing the answer you expected to get. John > LTspice can't handle parallel zero-resistance inductors True, it results in an overdefined matrix, which it can't solve.However,specifying a very small series resistance, say 10^-18 ohms, together with zero parallel resistance and capacitance makes the solver happy, and is near enough to ideal inductors to give results near to what you get using calculus. LTSpice inserts a default (1 milliohm, IIRC) series resistance in its inductor model if you don't specify one. That's too big for the inductor to be near-ideal. Try this, it's your posted circuit, modified using 1e-18 ohm series resistance in the inductor models and the series resistors deleted. It gives a 10:1 share of current, just like the differential equation says it should. Version 4 SHEET 1 880 680 WIRE 320 128 128 128 WIRE 448 128 400 128 WIRE -16 192 -96 192 WIRE 128 192 128 128 WIRE 128 192 64 192 WIRE 448 192 448 128 WIRE 528 192 448 192 WIRE -96 240 -96 192 WIRE 128 256 128 192 WIRE 320 256 128 256 WIRE 448 256 448 192 WIRE 448 256 400 256 WIRE -96 352 -96 320 WIRE 528 352 528 192 WIRE 528 352 -96 352 WIRE -96 400 -96 352 FLAG -96 400 0 SYMBOL ind 304 144 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L1 SYMATTR Value 10e-6 SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0 SYMBOL ind 304 272 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 5 56 VBottom 0 SYMATTR InstName L2 SYMATTR Value 1e-6 SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0 SYMBOL res 80 176 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 1000 SYMBOL voltage -96 224 R0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value PULSE(0 10 .1 1e-7) TEXT -128 424 Left 0 !.tran 10 uic "Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock) > > LTspice can't handle parallel zero-resistance inductors > [quoted text clipped - 60 lines] > durable; the other is a cheaper thing, but the moths get into it." > (Stephen Leacock) Thanks Fred, It's good to know that spice can be made to cough up the right answer. George H. >> LTspice can't handle parallel zero-resistance inductors > [quoted text clipped - 12 lines] >gives a 10:1 share of current, just like the differential equation says it >should. Unless you really want a 1:1 current split and are willing to wait until it happens. John >> LTspice can't handle parallel zero-resistance inductors > [quoted text clipped - 55 lines] >SYMATTR Value PULSE(0 10 .1 1e-7) >TEXT -128 424 Left 0 !.tran 10 uic --- Thanks, Fred. :-) JF > Hello All, > [quoted text clipped - 11 lines] > > Desiree I was thinking about how much I enjoyed this discussion today. And I was trying to find the post (by someone) comparing the problem (of two inductors in parallel) to two capacitors in series. Which I thought was apt. Imagine leaky capacitors with a large parallel resistance, (‘equivalent’ to inductors with small series resistance.) I think we all understand the capacitor case, and yet can get confused by the inductors. George H. >> Hello All, >> [quoted text clipped - 21 lines] > >George H. I expect that people ignore capacitor leakage but assume inductance series resistance. That sort of makes sense; real capacitors can have self-discharge time constants of many years, but it's rare for an inductor to hit L/R as long as a full second. Inductors are, in general, crappy parts compared to the other stuff we get to use. John On Jul 7, 11:11 pm, John Larkin <jjlar...@highNOTlandTHIStechnologyPART.com> wrote: > On Tue, 7 Jul 2009 19:41:27 -0700 (PDT), George Herold > [quoted text clipped - 36 lines] > > - Show quoted text - Yup, but it's possible to imagine a capacitor as poor as an inductor and then 'see' the same effect. With leaky capacitors the short time voltage across each cap will depend on C and the long time voltage will depend on the parallel R. George H. >Inductors are, in general, crappy parts compared to the other stuff we >get to use. Is that a technical term? I don't see it in many data sheets. :) SignatureThese are my opinions, not necessarily my employer's. I hate spam. >Hello All, > [quoted text clipped - 11 lines] > >Desiree Another way to look at it: Start with 11 identical 10 uH inductors, ideal or not. Wire them in parallel carefully, so that no initial current is circulating among them. If they are ideal inductors, use ideal wire and solder. Arrange them in a physically symmetric pattern just to be compulsive. Now apply any voltage, current, or waveform to the paralleled bunch. By symmetry, all the 11 resulting inductor currents are identical. At any point, mentally draw a dotted line around 10 of the inductors. Now you have one 1 uH inductor in parallel with one 10 uH inductor. Obviously the current in the 1 uH leg is 10x the current in the 10 uH leg. John >>Hello All, >> [quoted text clipped - 28 lines] > > John That's a good way to look at it. I like the bit about ideal solder. Ideal solder is a lot easier to work with than the real thing. No muss, no fuss -- all you have to do is think about it. For extra credit on the compulsive front, use a cylindrical arrangement for the ideal inductors. >>Hello All, >> [quoted text clipped - 24 lines] >At any point, mentally draw a dotted line around 10 of the inductors. >Now you have one 1 uH inductor in parallel with one 10 uH inductor. --- ??? 1 Lt = ------------------------------- 1 1 1 1 ---- + ---- + ---- ... + ---- L1 L2 L3 Ln >Obviously the current in the 1 uH leg is 10x the current in the 10 uH >leg. --- ??? +<----[E1]--->+ | | | 0.1 I2---> | +-----[L1]----+ | 1µH | | | | I2---> | +-----[L2]----| 0.1µH Tsk, tsk... JF >>>Hello All, >>> [quoted text clipped - 51 lines] > >Tsk, tsk... --- Oops... Brain fart; never mind... JF >>>>Hello All, >>>> [quoted text clipped - 57 lines] >Brain fart; never mind... >JF Tsk, tsk... You are so eager to catch me doing something wrong, you jump to contradict me without thinking first. Letting your emotions cancel your ability to reason is bad engineering. Stick to criticizing my spelling and typing. That way you'll have a chance of being right. John >>>>>Hello All, >>>>> [quoted text clipped - 63 lines] >contradict me without thinking first. Letting your emotions cancel >your ability to reason is bad engineering. --- Well, I do admit that I take pleasure in showing that you have feet of clay, but you must then admit that I had to reason in order to find _my_ mistake. --- >Stick to criticizing my spelling and typing. That way you'll have a >chance of being right. --- Do you still believe that latching relays have infinite gain? Or, more recently, from sed: In this message, I discovered the OP's error and showed him how to fix it: 1cn145pnrst5ogq8fijg0sirkal5gkl25h@4ax.com In this one you were wrong about his circuit working and I corrected you: cup1459pr696ig2t70pbe9bi4mgot2s25t@4ax.com In this one you acknowledged that I had found the OP's error, without acknowledging your own error, and then went on to say that moving the diodes would fix the problem, which I had shown on the included ASCIImatic (thanx, Steve!) which you snipped. For what reason? In my eyes, to make it seem to the casual reader that moving the diodes was _your_ idea. u8s145dfpl59vrs46hpe4f3mncdf8vdav9@4ax.com I think Phil was right; you're kinda slimy. JF >>>>>>Hello All, >>>>>> [quoted text clipped - 75 lines] >--- >Do you still believe that latching relays have infinite gain? Absolutely. Do you still believe the answer to the problem here is 5+5 mA? >Or, more recently, from sed: > [quoted text clipped - 17 lines] >In my eyes, to make it seem to the casual reader that moving the diodes >was _your_ idea. I can't help that. >u8s145dfpl59vrs46hpe4f3mncdf8vdav9@4ax.com > >I think Phil was right; you're kinda slimy. > >JF I think you are all wrapped up in he-said-she-said insecurity, personalizing everything. I'm here to talk engineering. If I make an occasional mistake - I do post a lot, about a lot of subjects - it's no big deal to me, since this is not stuff I'm going to ship and get paid for; we check deliverables really hard. Declare victory if it makes you feel good. You're seldom important because you seldom bring up anything interesting. John >>--- >>Well, I do admit that I take pleasure in showing that you have feet of [quoted text clipped - 10 lines] >Absolutely. Do you still believe the answer to the problem here is 5+5 >mA? --- Sure, why not? Even with incontrovertible evidence in front of me to the contrary, I can take your position and become an ostrich. --- >>Or, more recently, from sed: >> [quoted text clipped - 19 lines] > >I can't help that. --- Sure you can, but you won't. Certainly you're not stupid, and you seem to be attuned to the nuances of the language in that you take every opportunity to make snide remarks, cuts, and subtle putdowns in order to try to denigrate and discredit those who you feel are threatening to you. In addition, you use fallacious logic (straw man, well poisoning, etc, etc,) in order to try to state your case from what you'd have others believe is a solid platform. --- >>u8s145dfpl59vrs46hpe4f3mncdf8vdav9@4ax.com >> [quoted text clipped - 4 lines] >I think you are all wrapped up in he-said-she-said insecurity, >personalizing everything. --- More bullshit. What I find personal _is_ being attacked, and the game you're playing is to attack me whenever you can since I've found you less than perfect more than once, published my findings, and hung you up to dry. The purpose of your game, of course, is to try to discredit me and, therefore, my findings, turning you loose once again. But, not just me, anyone who finds you in error. So, egotistically incapable of admitting defeat, or having made a mistake, you press on with your even weaker: "I'm here to talk engineering." Now you're playing this other game where you say you're here to "talk engineering", but you seldom do. Mostly you just go on, ad nauseam, about how the world should be, according to Larkin. --- >If I make an occasional mistake - I do post a lot, about a lot of subjects - it's >no big deal to me, since this is not stuff I'm going to ship and get >paid for; we check deliverables really hard. --- So, in other words, when you post to USENET it's just a game? What a surprise... --- >So., >Declare victory if it makes you feel good. You're seldom important >because you seldom bring up anything interesting --- You're right, since the subject I brought up was you. JF >>>--- >>>Well, I do admit that I take pleasure in showing that you have feet of [quoted text clipped - 106 lines] > >JF So, what's your final answer to the OP's question? How many mA in each ideal inductor? John >>>--- >>>Well, I do admit that I take pleasure in showing that you have feet of [quoted text clipped - 107 lines] > JF :) |
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