DC Current in Parallel Inductors?

azy,

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=3DL*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

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 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!

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

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

1. (Homework problems like to keep the math simple.) Spice also did not like that problem.

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.

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.

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

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.

Unless it has a whopping current circulating in it. Had and MRI lately?

John

Tedious, but wrong. Consider superconducting magnets. Even better, consider superconducting magnets that have superconducting shim coils.

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

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

Greg said:

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

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.

John Larkin wrote in news: snipped-for-privacy@4ax.com:

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.