DC Current in Parallel Inductors?

Right, just as you have to take their zero Leprechaun content into account.

John

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

Looks like you're having trouble with zero, which I thought was a simpler concept than infinity.

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.

Ooh, Ohm's "Law." R=0 in Desiree's problem.

There's an accepted definition of Q

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.

--
That you're a patronizing ass.

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

Doing /done the Rubicon Trail, like you threatened?

-- "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)

--
Philosophically, they're the same thing except that one grows smaller
without bound while the other grows larger.

azy,

Sounds like you have a very good teacher.

To absolutely reach a steady state takes ... how much time?

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.

Well, that's a lot easier to say than actually thinking about hard stuff like electronics.

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

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.

Bronze marker courtesy E Clampus Vitus:

Cool stuff.

John

WRONG!

John

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

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

Very nice photos !

TFS.

--
Best Regards:
                     Baron.

--
That's true, but it has nothing to do with the fact that you asked me
what my point was, and I replied.

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

Really?

Then (assuming, of course, ideal wiring) replace the 'x's with the proper currents:

. XmA-->

. +--[0R]--+ . 10mA--> | | .10V----1000R----+ +---+ . | | | . +--[0R]--+ | . XmA--> | .

^^^^^^^^^

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

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.

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.

Can't address that one. I'm a real engineer.

I can't imagine what they said. If they agree with JF about 5+5 mA, they're wrong.

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