DC Current in Parallel Inductors?

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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?

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.

Yup.

The closed path through the two zero-ohm resistors. What else could it be in this circuit?

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?

See above.

John

this

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"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.

Why ramp it down? You're wasting energy!

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

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.

--
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crazy,

they

through

happily

Since you have a spice simulator what values are you using for initial conditions? Also go back to basics E =3D -L*(di/dt). That should give you some help with the initial conditions.

crazy,

they

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.

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?

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.

AFAIUI, current very high temperature superconductors are pretty nasty to work with-- they'd make lead-free look really good.

Best regards, Spehro Pefhany

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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

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.

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.

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.

Yeah, this is a good one, and Spice-proof.

John

Good point.

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

He did answer it. And he was wrong.

John

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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.

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

So; 10 volt supply. One kilohm limiting resistor in series with two ideal inductors in parallel. Correct????? Final steady state current =3D 10/1000 =3D 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?

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 | . | | | | . | | | | . | | | | . | | | | . | | | | . '----------------' '----------------'

No, because it's the wrong answer.

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.

I doubt it. It sure has messed up a bunch of people.

John

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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