DC Current in Parallel Inductors?

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

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

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

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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.

:On Thu, 02 Jul 2009 15:49:01 -0700, Miss_Koksuka wrote: : :> 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? : :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.

I agree with you entirely, at least insofar as the use of Spice as a solve-all tool and the eagerness to use it as a first-in solution in place of other methods. I think the problem posed by the lecturer may even have been framed so that he could see whether students could use deductive reasoning for a simple problem = hence the inclusion of the "after reaching the steady state" in the specification.

Since there were no specification for the gauge of wire or construction method for each of the inductors, it is quite possible for each inductor to have exactly the same resistance despite the differing values of inductance. Now since both inductors are in parallel with a resistor then after steady state conditions are reached the voltage across the parallel combination will be the same in each component. Assuming that a load is then connected and dc current is flowing in the load, if both inductors have exactly the same resistance, then the current through each inductor will also be the same.

Perhaps that is what the lecturer was after.

Is that a technical term? I don't see it in many data sheets.

:)

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

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.

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

Absolutely. Do you still believe the answer to the problem here is 5+5 mA?

I can't help that.

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

it's

So, what's your final answer to the OP's question? How many mA in each ideal inductor?

John

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)

Nice pictures, John. I shall have to go take a look myself sometime.

You a Clamper, John?

=A0 =A0 =A0 =A0 =A0(Stephen Leacock)

Thanks Fred, It's good to know that spice can be made to cough up the right answer.

George H.

Unless you really want a 1:1 current split and are willing to wait until it happens.

John