DC Current in Parallel Inductors?

[John Fields]

On Sat, 04 Jul 2009 17:26:25 -0700, John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
greg.ewing@canterbury.ac.nz> wrote:

Miss_Koksuka wrote:

So strange, but none of my
(many) school books seems to cover any of this, they only say that an
ideal inductor is a "short" to DC.

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.

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

---
What's tedious is your shifting the subject around in order to keep
waving your own flag while you're patting yourself on the back.

No one's talking superconducting magnets, the topic is about
conventional inductances.

JF

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

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.

---
Whether I do or not has nothing to do with it, so other than showing
yourself up for the patronizing ass you truly are, what's your point?

JF

 

[John Larkin]

On Sun, 05 Jul 2009 00:44:15 -0500, Lostgallifreyan
<no-one@nowhere.net> wrote:

John Larkin <jjSNIPlarkin@highTHISlandtechnology.com> wrote in
news:lisv459dhog3rqqv2erus7l5lthgnqqpti@4ax.com:

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

which I pointed out isn't always the case. Some people work with real
inductors that actually have no resistance.


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.

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

John

 

[John Larkin]

On Sun, 05 Jul 2009 05:38:33 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 17:26:25 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
greg.ewing@canterbury.ac.nz> wrote:

Miss_Koksuka wrote:

So strange, but none of my
(many) school books seems to cover any of this, they only say that an
ideal inductor is a "short" to DC.

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.

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

---
What's tedious is your shifting the subject around in order to keep
waving your own flag while you're patting yourself on the back.

No one's talking superconducting magnets, the topic is about
conventional inductances.

JF

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

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.

---
Whether I do or not has nothing to do with it, so other than showing
yourself up for the patronizing ass you truly are, what's your point?

JF

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

 

[John Larkin]

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.

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.

 

[John Fields]

On Sun, 05 Jul 2009 08:49:17 -0700, John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 05:38:33 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 17:26:25 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
greg.ewing@canterbury.ac.nz> wrote:

Miss_Koksuka wrote:

So strange, but none of my
(many) school books seems to cover any of this, they only say that an
ideal inductor is a "short" to DC.

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.

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

---
What's tedious is your shifting the subject around in order to keep
waving your own flag while you're patting yourself on the back.

No one's talking superconducting magnets, the topic is about
conventional inductances.

JF

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

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.

---
Whether I do or not has nothing to do with it, so other than showing
yourself up for the patronizing ass you truly are, what's your point?

JF

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?

---
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

 

[Fred Abse]

On Sun, 05 Jul 2009 10:18:28 -0700, John Larkin wrote:

I'm in Truckee

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)

 

[John Fields]

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.
---

---

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.

---
All well and good, but the point I was trying to make was that for the
purpose of the discussion, the Q of the inductance under steady-state
conditions is irrelevant and just more noise you're adding.
---

I note that you made no attempt to help her solve her dilemma, whereas
I did.

---
She was getting plenty of help, so I saw no need to jump in until you
showed up with your smarty-pants "I'm smarter than all of you."
attitude.
---

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.

---
Well, often they're not true when you start out, but by the time you get
finished with your little song and dance you've generally confused the
shit out of most everyone and you can then pretend you were right all
along since most everyone gets tired of your crap and just drops out.
---

Whiny and fact-free, as usual.

---
Was I the one who stated that protons can go dancing around just like
electrons?

Nope, it was you, so that's at least one example of your being wrong,
making your statement that everything you said was true wrong as well.
---

I'm in Truckee, typing on this horrible little Vaio keyboard, in
natural light, and I never learned to type anyhow.

---
Poor baby...

JF

 

[John Larkin]

On Sun, 05 Jul 2009 13:28:18 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 05 Jul 2009 08:49:17 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 05:38:33 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 17:26:25 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
greg.ewing@canterbury.ac.nz> wrote:

Miss_Koksuka wrote:

So strange, but none of my
(many) school books seems to cover any of this, they only say that an
ideal inductor is a "short" to DC.

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.

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

---
What's tedious is your shifting the subject around in order to keep
waving your own flag while you're patting yourself on the back.

No one's talking superconducting magnets, the topic is about
conventional inductances.

JF

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

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.

---
Whether I do or not has nothing to do with it, so other than showing
yourself up for the patronizing ass you truly are, what's your point?

JF

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?

---
That you're a patronizing ass.

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

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.

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

 

[whit3rd]

On Jul 2, 6:11 pm, Miss_Koksuka <desiree_koks...@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.

Sounds like you have a very good teacher.

... 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.

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

(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...).

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.

 

[John Larkin]

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

 

[Lostgallifreyan]

John Larkin <jjSNIPlarkin@highTHISlandtechnology.com> wrote in
news:9ki15593n01u431d1dfnqmg1gq2rvq7na7@4ax.com:

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

Only if general calculations expected to have to handle leprechauns.

 

[John Larkin]

On Sun, 05 Jul 2009 20:23:20 -0500, Lostgallifreyan
<no-one@nowhere.net> wrote:

John Larkin <jjSNIPlarkin@highTHISlandtechnology.com> wrote in
news:9ki15593n01u431d1dfnqmg1gq2rvq7na7@4ax.com:

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


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

 

[John Fields]

On Sun, 05 Jul 2009 15:48:32 -0700, John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 13:28:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sun, 05 Jul 2009 08:49:17 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 05:38:33 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 17:26:25 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
greg.ewing@canterbury.ac.nz> wrote:

Miss_Koksuka wrote:

So strange, but none of my
(many) school books seems to cover any of this, they only say that an
ideal inductor is a "short" to DC.

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.

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

---
What's tedious is your shifting the subject around in order to keep
waving your own flag while you're patting yourself on the back.

No one's talking superconducting magnets, the topic is about
conventional inductances.

JF

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

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.

---
Whether I do or not has nothing to do with it, so other than showing
yourself up for the patronizing ass you truly are, what's your point?

JF

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?

---
That you're a patronizing ass.

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

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

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.

Or superconductive.

---
How could defining the inductors as ideal make the problem
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.

---
True, but so what?

It has nothing to do with the fact that the _problem_ was hypothetical
and you're just throwing more crap around in order the dodge having to
admit that you said it wasn't.

Business as usual, huh?
---

Recovering from your 4th of July hangover?

---
Never had one.

JF

 

[John Larkin]

On Mon, 06 Jul 2009 07:38:22 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 05 Jul 2009 15:48:32 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 13:28:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sun, 05 Jul 2009 08:49:17 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sun, 05 Jul 2009 05:38:33 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 17:26:25 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
jfields@austininstruments.com> wrote:

On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
jjSNIPlarkin@highTHISlandtechnology.com> wrote:

On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
greg.ewing@canterbury.ac.nz> wrote:

Miss_Koksuka wrote:

So strange, but none of my
(many) school books seems to cover any of this, they only say that an
ideal inductor is a "short" to DC.

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.

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

---
What's tedious is your shifting the subject around in order to keep
waving your own flag while you're patting yourself on the back.

No one's talking superconducting magnets, the topic is about
conventional inductances.

JF

Greg said:

You *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.

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.

---
Whether I do or not has nothing to do with it, so other than showing
yourself up for the patronizing ass you truly are, what's your point?

JF

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?

---
That you're a patronizing ass.

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

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


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.

Or superconductive.

---
How could defining the inductors as ideal make the problem
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.

---
True, but so what?

It has nothing to do with the fact that the _problem_ was hypothetical
and you're just throwing more crap around in order the dodge having to
admit that you said it wasn't.

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

 

[John Fields]

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

 

[John Larkin]

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.

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

 

[stan]

John Larkin wrote:

On Mon, 06 Jul 2009 12:08:53 -0500, John Fields
snip
WRONG!

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.

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.

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.

 

[John Larkin]

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

John Larkin wrote:
On Mon, 06 Jul 2009 12:08:53 -0500, John Fields
snip
WRONG!

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.

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.

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.

More importantly how would an Engineering professor use that term for a
first year student?

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

FWIW, I checked with 6 professors and I be interested in what you think
they said about your reply to JF's answer.

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

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.

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

 

[John Larkin]

On Sat, 04 Jul 2009 15:00:02 +0100, nospam <nospam@please.invalid>
wrote:

Miss_Koksuka <desiree_koksuka@yahoo.com> wrote:

But with v=L*di/dt for non-sinusoidal waveforms, such as this DC
circuit's turn-on waveform,

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.

Yes.

John

 

[John Fields]

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?
---

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

---
Really?

What do you define as the zero-ohm loop?
---

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 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?
JF