DC Current in Parallel Inductors?

[John Larkin]

On Mon, 06 Jul 2009 19:01:24 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Mon, 06 Jul 2009 10:29:04 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Mon, 06 Jul 2009 12:08:53 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sun, 05 Jul 2009 16:24:39 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 14:34:08 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sun, 05 Jul 2009 10:18:28 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 05:31:19 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 17:21:38 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:35:30 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
jfields@austininstruments.com> wrote:

On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
desiree_koksuka@yahoo.com> wrote:


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?

---
Yeah, but so what?

No one's talking superconducting magnets, you pretentious ass.

JF

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?

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
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.

---
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:

E
R = ---
I

Look familiar to ya?

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.


^^^^^^^^^




WRONG!

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


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.

---
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.

---

But the zero-ohm-resistor case is indeterminate. There could be any
amount of current circulating in the zero-ohm loop.

---
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?

---

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 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?

---

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.

---
Then what is it?

See above.

John

 

[John Larkin]

On Mon, 6 Jul 2009 19:54:28 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

"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.

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

 

[George Herold]

On Jul 5, 1:18 pm, John Larkin
<jjSNIPlar...@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 05:31:19 -0500, John Fields





jfie...@austininstruments.com> wrote:
On Sat, 04 Jul 2009 17:21:38 -0700, John Larkin
jjSNIPlar...@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:35:30 -0500, John Fields
jfie...@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
jjSNIPlar...@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
jfie...@austininstruments.com> wrote:

On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
desiree_koks...@yahoo.com> wrote:

   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?

---
Yeah, but so what?

No one's talking superconducting magnets, you pretentious ass.

JF  

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?

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:

        E
   R = ---
        I

Look familiar to ya?

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





---  

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?

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.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- 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.

 

[Hal Murray]

In article <pq1555tl0t00p88grvja3jv3o3gg6u78ln@4ax.com>,
John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> writes:

On Mon, 6 Jul 2009 14:42:10 -0400, stan <smoore@exis.net> wrote:

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.

--
These are my opinions, not necessarily my employer's. I hate spam.

 

[JosephKK]

On Thu, 2 Jul 2009 18:11:12 -0700 (PDT), Miss_Koksuka
<desiree_koksuka@yahoo.com> wrote:

On Jul 2, 6:17 pm, Dan Coby <adc...@earthlink.net> wrote:
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?

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.)

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

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.

 

[JosephKK]

On Fri, 3 Jul 2009 06:15:27 -0700 (PDT), Miss_Koksuka
<desiree_koksuka@yahoo.com> wrote:

On Jul 3, 12:26 am, John Larkin
jjSNIPlar...@highTHISlandtechnology.com> wrote:
On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka



desiree_koks...@yahoo.com> 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?

Thank you!

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

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 1ľ Rser=0.001
R1 N002 0 100
V1 N001 0 10
L2 N001 N002 10ľ Rser=0.001
.TRAN 500us 0.05 UIC
.PLOT TRAN I(L1) I(L2)
.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.

 

[JosephKK]

On Sat, 04 Jul 2009 14:35:30 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
jfields@austininstruments.com> wrote:

On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
desiree_koksuka@yahoo.com> wrote:


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?

---
Yeah, but so what?

No one's talking superconducting magnets, you pretentious ass.

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?

 

[Spehro Pefhany]

On Mon, 06 Jul 2009 21:15:58 -0700, the renowned John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 6 Jul 2009 19:54:28 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:



"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.

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.

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
--
"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

 

[John Larkin]

On Mon, 06 Jul 2009 23:25:14 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Thu, 2 Jul 2009 18:11:12 -0700 (PDT), Miss_Koksuka
desiree_koksuka@yahoo.com> wrote:

On Jul 2, 6:17 pm, Dan Coby <adc...@earthlink.net> wrote:
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?

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.)

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

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.

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

 

[John Larkin]

On Mon, 06 Jul 2009 23:58:57 -0500,
hal-usenet@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote:

In article <pq1555tl0t00p88grvja3jv3o3gg6u78ln@4ax.com>,
John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> writes:
On Mon, 6 Jul 2009 14:42:10 -0400, stan <smoore@exis.net> wrote:

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.

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.

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

 

[John Larkin]

On Tue, 07 Jul 2009 07:55:34 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Mon, 06 Jul 2009 21:15:58 -0700, the renowned John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 6 Jul 2009 19:54:28 -0700 (PDT), George Herold
ggherold@gmail.com> wrote:



"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.

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.

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.

Good point.

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.

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

 

[John Larkin]

On Tue, 07 Jul 2009 00:12:28 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Sat, 04 Jul 2009 14:35:30 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
jfields@austininstruments.com> wrote:

On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
desiree_koksuka@yahoo.com> wrote:


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?

---
Yeah, but so what?

No one's talking superconducting magnets, you pretentious ass.

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?

He did answer it. And he was wrong.

John

 

[George Herold]

On Jul 7, 12:15 am, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 6 Jul 2009 19:54:28 -0700 (PDT), George Herold

ggher...@gmail.com> wrote:
"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.

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

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.

 

[John Larkin]

On Mon, 06 Jul 2009 23:44:18 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Fri, 3 Jul 2009 06:15:27 -0700 (PDT), Miss_Koksuka
desiree_koksuka@yahoo.com> wrote:

On Jul 3, 12:26 am, John Larkin
jjSNIPlar...@highTHISlandtechnology.com> wrote:
On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka



desiree_koks...@yahoo.com> 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?

Thank you!

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

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 1ľ Rser=0.001
R1 N002 0 100
V1 N001 0 10
L2 N001 N002 10ľ Rser=0.001
.TRAN 500us 0.05 UIC
.PLOT TRAN I(L1) I(L2)
.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.

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

 

[Greg Neill]

stan wrote:

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

 

[stan]

On Jul 2, 8:49 pm, Miss_Koksuka <desiree_koks...@yahoo.com> 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?

Thank you!

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?

 

[John Larkin]

On Tue, 7 Jul 2009 12:37:37 -0700 (PDT), stan
<tsanford@nf.sympatico.ca> wrote:

On Jul 2, 8:49 pm, Miss_Koksuka <desiree_koks...@yahoo.com> 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?

Thank you!

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.

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

 

[John Fields]

On Tue, 07 Jul 2009 16:16:59 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 7 Jul 2009 12:37:37 -0700 (PDT), stan
tsanford@nf.sympatico.ca> wrote:

On Jul 2, 8:49 pm, Miss_Koksuka <desiree_koks...@yahoo.com> 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?

Thank you!

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.

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.

---
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

 

[John Larkin]

On Tue, 07 Jul 2009 21:13:21 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Tue, 07 Jul 2009 16:16:59 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 7 Jul 2009 12:37:37 -0700 (PDT), stan
tsanford@nf.sympatico.ca> wrote:

On Jul 2, 8:49 pm, Miss_Koksuka <desiree_koks...@yahoo.com> 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?

Thank you!

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.

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.

---
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

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

 

[John Larkin]

On Tue, 7 Jul 2009 19:41:27 -0700 (PDT), George Herold
<ggherold@gmail.com> wrote:

On Jul 2, 6:49 pm, Miss_Koksuka <desiree_koks...@yahoo.com> 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?

Thank you!

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.

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