DC Current in Parallel Inductors?
Miss_Koksuka
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
Dan Coby
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.)
Michael Robinson
"Miss_Koksuka" <desiree...@yahoo.com> wrote in message
news:f7f121cd-f823-416e...@x5g2000prf.googlegroups.com...
Would the troll like a cookie?
Nice troll, come here, gootch gootchy goo
SPLAT
Miss_Koksuka
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
Dan Coby
Instead of looking at the steady state, you need to look at how you get there.
The remainder of your homework problem is left to the student.
nospam
>Instead of looking at the steady state, you need to look at how you get there.
Has anyone got a Digkey part number for some of these inductors? Sounds
like they would be really useful.
--
Michael Robinson
"Dan Coby" <adc...@earthlink.net> wrote in message
news:P_WdnVVgSNPJw9DX...@earthlink.com...
> Instead of looking at the steady state, you need to look at how you get
> there.
>
> The remainder of your homework problem is left to the student.
Right, this is a theoretical question about ideal inductors, so they
never actually reach "steady state," meaning constant current in this case.
For a circuit made of ideal components, you would have to define steady
state:
in the limit as t increases without bound.
David L. Jones
It doesn't. An ideal inductor has zero resistance of course, and you have
two of these lovely devices in parallel.
But ask yourself how long does it take to reach steady state in an ideal
circuit? Tell your teacher you'll give him the answer after "the big
crunch".
Dave.
--
================================================
Check out my Electronics Engineering Video Blog & Podcast:
http://www.alternatezone.com/eevblog/
John Larkin
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
John Larkin
wrote:
Yeah, room temperature superconductors will be mighty handy. Finite-Q
filter designs are a nuisance.
John
Tom Biasi
"Miss_Koksuka" <desiree...@yahoo.com> wrote in message
news:f7f121cd-f823-416e...@x5g2000prf.googlegroups.com...
done. Parallel devices have equal voltages.
Tom
Miss_Koksuka
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 1µ Rser=0.001
R1 N002 0 100
V1 N001 0 10
L2 N001 N002 10µ Rser=0.001
.TRAN 500us 0.05 UIC
.PLOT TRAN I(L1) I(L2)
.backanno
.end
Michael Robinson
=============================
(* Rser=0.001 to make Spice happy.) <=======That explains it
right there
=============================
The current in ideal inductors would never even out, it would always have a
ratio of 10:1
But you have give the inductors resistance. They are no longer ideal, and
the current
will even out. The time constant for this effect is determined by the
series resistance
of the inductors.
Miss_Koksuka
> "Miss_Koksuka" <desiree_koks...@yahoo.com> wrote in message
Thanks Mike. So at least I'm not doing anything wrong with the Spice
simulator! But I'll not rest until I find out exactly *why* this
occurs. The mechanism behind it has me completely baffled, since it
is not an LC tank circuit, so energy is not being exchanged back and
forth between an inductor and capacitor. It is merely two inductors
in parallel (I always assumed that such a circuit would simply act
like single, lower value, inductor). 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.
Thanks again,
-Desiree
Tim Wescott
Perhaps one of the things your instructor wanted to do was to wean you
away from using Spice -- which is a fine tool for some things -- for
everything.
Try doing it using Laplace domain analysis; John Larkin's suggestion of
finding the voltage and then the current is to the point if you want to
simplify the math.
Remember that Spice is a real-world tool, and an ideal inductor is not a
real-world device. So using Spice and expecting it to cancel out
infinities is inappropriate. OTOH, this is a fairly simple problem with
linear circuit elements -- hence my suggestion of using Laplace analysis.
Michael Robinson
Thanks again,
-Desiree
Are you pursuing an engineering degree?
Miss_Koksuka
Yes Mike, but I am only in my first year of electronics.
Dan Coby
>> Are you pursuing an engineering degree?
>
> Yes Mike, but I am only in my first year of electronics.
Okay. Now we need to know a little more about what you have learned.
Have you had calculus? Do you know what this equation means and how
to use it:
v = L di/dt
Hal Murray
> Thanks guys. I'm trying to put all your answers together to
>clearly figure this all out, but its tough!
Your intuition might work better with capacitors rather than inductors.
What would you expect if you had a resistor, small cap, and big cap
in series?
--
These are my opinions, not necessarily my employer's. I hate spam.
Michael Robinson
Good. I'm going to show you a simple solution. It amounts to a fairly
rigorous demonstration that you can show your instructor.
Assume the inductors have zero current in them at t=0, which is when you
apply the voltage.
Dan Coby gave you formula v=L*(di/dt)
I'm going to call the inductances L1 and L2. Simillarly,
currents in the inductors I1 and I2. I1 and I2 are variables, functions of
t (time).
Now, v is always equal for both inductors becuase they are in parallel.
Therefore L1 (dI1/dt) = L2 (dI2/dt)
Say L2 is the "bigger" inductor. L2 = 10 L1 implies
dI1/dt = 10 dI2/dt which we can integrate over time:
I1 = 10 I2 + C (I hope you've had some basic calculus)
now, C has to be zero because at t=0, I1 = I2 = 0
so I1 = 10 I2
and that's always true, for any value of t
John Larkin
As noted, you have to include the entire history of the voltage
applied to the inductor to know its current.
A shorted inductor of unknown history has an indeterminate current.
Ditto two paralleled inductors.
> 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?
Spice artifact, essentially some minimum (non-zero) resistance
parameter. For ideal inductors, you wouldn't see that. The voltage
waveform is a spike up to 10 volts, exponentially decaying with a time
constant of about 9 nanoseconds.
Spice often lies.
>(* Rser=0.001 to make Spice happy.)
That also shoots down the concept of "ideal inductor."
John
Miss_Koksuka
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!). But with v=L*di/dt for
non-sinusoidal waveforms, such as this DC circuit's turn-on waveform,
I'm starting to get a little clearer on this.
And yes John, you are right -- I just checked, and my Spice
simulator (LT Spice) appears to have "inserted" some very small non-
zero resistance value in my circuit (even with my 0.001 ohm resistors
removed); undoubtedly to help prevent convergence issues.
Thanks again,
-Desiree
Jamie
yes
Michael Robinson
Thanks!
-Desiree
To calculate it, fist solve assuming zero initial current. Then, simply add
any initial current to that solution, and you have your new answer.
It is linear.
If you look at the math in my other post you'll see why it works that way.
It is a matter of solviing for the integration constant based on initial
conditions. It happens to work out linearly.
By the way, I made a slight error in the way I described the integration.
You don't actually integrate over t. The t differential drops out.
You integrate I1 with respect to I2, or vice versa (doesn't matter).
Now, for real inductors with resistance, to get the steady state solution,
simply substitute the series resistors for the inductors (leave the
inductors
as dead shorts) and solve.
Good luck miss K. And consider using another screen name!
Dan Coby
> Miss_Koksuka wrote:
>> 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
> yes
Desiree,
As I think that you have learned from the rest of this discussion, Jamie's
answer is overly simplistic. In most cases you can treat ideal inductors
like a zero ohm piece of wire in the steady state case, but it is not really
appropriate in this problem. Actually trying to determine how current is
shared between parallel pieces of wire is undefined if you have ideal zero
ohm wires, etc.
As Tim Wescott pointed out, this problem was more about thinking about the
total problem rather than simply trying to put things into Spice. Spice
did force you to think about the confusing result that it gave. Do you
understand why you got the results that you did from Spice? I.e. Do you
understand why you get a different final result with an ideal inductor
(with zero resistance) versus an inductor with a small non zero resistance?
With the ideal inductors, what is the answer if you assume that there is a
1 amp current flowing in a loop between the two inductors at time zero?
BTW, if you search through the archives for this group, there was a discussion
a few months ago about the voltages involved with a DC source, a resistor,
and then two capacitors in series. Once again, I think that it was someone's
homework problem. I think that the capacitors also differed by a factor of
10. (Homework problems like to keep the math simple.) Spice also did not
like that problem.
Dave
> "Miss_Koksuka" <desiree_koks...@yahoo.com> wrote in message
That is so nerdy and geeky of you to say, TROLL and SPLAT. How very
uncool of you.
Trolls are no longer accepted in newsgroups. If you call someone a
troll, then you must be a troll from the past.. shhhh.. go away ..
Come back when you are cool enough.
Greg Ewing
> 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.
--
Greg
Miss_Koksuka
You guys are great! Thanks for clearing this up for me (I spent days
trying to do it; I felt as if my head were about to explode at one
point!!).
Many Thanks,
-Desiree
John Fields
<desiree...@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.
---
>But with v=L*di/dt for
>non-sinusoidal waveforms, such as this DC circuit's turn-on waveform,
>I'm starting to get a little clearer on this.
---
If you go back to the basics, things might get a _lot_ clearer.
Consider:
When a voltage is impressed across a conductor, a magnetic field will be
generated about that conductor, and when a magnetic field 'cuts' a
conductor it will induce a voltage in that conductor.
As it turns out (courtesy of Mother Nature), if you connect a voltage
source across a conductor, the magnetic field generated will cut the
conductor and will generate a voltage with a polarity opposite that of
the voltage source!
When the voltage is first applied, the induced voltage will be equal to
the applied voltage and will keep charge from flowing through the
conductor, but as time goes by the characteristics of the field change,
allowing more and more current through the conductor until a limit is
reached when the series resistances in the circuit and the source
voltage satisfy:
E
I = ---
R
Here's some more:
http://en.wikipedia.org/wiki/Lenz's_law
JF
nospam
>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.
--
John Larkin
<jfi...@austininstruments.com> wrote:
>On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
><desiree...@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?
John
John Larkin
Tedious, but wrong. Consider superconducting magnets. Even better,
consider superconducting magnets that have superconducting shim coils.
John
Spehro Pefhany
Even if it has zero current circulating in it, just rotate a bit, then
stop, and there will be (in general).
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
John Fields
---
Yeah, but so what?
No one's talking superconducting magnets, you pretentious ass.
JF
John Fields
---
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
John Larkin
<jfi...@austininstruments.com> wrote:
>On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
><jjSNIP...@highTHISlandtechnology.com> wrote:
>
>>On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
>><jfi...@austininstruments.com> wrote:
>>
>>>On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
>>><desiree...@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]. And it ain't the same as a
resistor.
Got your monthlies again?
John
[1] Well, it will probably have a finite Q.
John Larkin
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.
John
greg
>>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
>
> Tedious, but wrong. Consider superconducting magnets.
What I mean is, if they have any resistance at all,
however small, you can't ignore it if you want to know
the steady-state DC current.
If the inductors are superconducting, they never reach
steady state in the sense of a current distribution
that's independent of the voltage history they've been
subjected to. In that case, you have to take the
history into account.
Since most people never have occasion to have to deal
with superconducting magnets, introductory electronics
books can be forgiven for not going into that level
of detail.
Now, if someone comes up with a room-temperature
superconductor and superconducting components become
commomplace, that might change...
BTW, I'm not sure that you can call a superconductor
an "ideal inductance" in the mathematical sense, since
they have limitations such as a maximum magnetic field
before they stop superconducting. But I'll grant they're
certainly a much better approximation of one than any
ordinary inductor.
--
Greg
Lostgallifreyan
news:lisv459dhog3rqqv2...@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.
John Fields
<jjSNIP...@highTHISlandtechnology.com> wrote:
>On Sat, 04 Jul 2009 14:35:30 -0500, John Fields
><jfi...@austininstruments.com> wrote:
>
>>On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
>><jjSNIP...@highTHISlandtechnology.com> wrote:
>>
>>>On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
>>><jfi...@austininstruments.com> wrote:
>>>
>>>>On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
>>>><desiree...@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?
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.
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?
---
>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?
JF
John Fields
<jjSNIP...@highTHISlandtechnology.com> wrote:
---
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
Right, just as you have to take their zero Leprechaun content into
account.
John
John Larkin
<jfi...@austininstruments.com> wrote:
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
<jfi...@austininstruments.com> wrote:
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
<jjSNIP...@highTHISlandtechnology.com> wrote:
---
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
John Fields
<jjSNIP...@highTHISlandtechnology.com> wrote:
---
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.
---
---
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
whit3rd
> On Jul 2, 6:17 pm, Dan Coby <adc...@earthlink.net> wrote:
>
>
>
> > Miss_Koksuka wrote:
>
> > > 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
<jfi...@austininstruments.com> wrote:
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
John Larkin
<excret...@invalid.invalid> wrote:
>On Sun, 05 Jul 2009 10:18:28 -0700, John Larkin wrote:
>
>> I'm in Truckee
>
>Doing /done the Rubicon Trail, like you threatened?
Not yet; maybe in August.
We were walking around the Rainbow Bridge and decided to hike up to
the top of the China Wall, between a couple of tunnels and snow sheds
of the original Transcontinental Railroad. It's a nice short steep
hike over smooth stepped glacial-looking rocks, with some
3000-year-old petroglyphs here and there. Anyhow, when we got to the
top, there were a couple of guys in Jeeps. Turns out that you can
quasi-legally enter the tunnels near Sugar Bowl and drive through. One
of them is, I think, about 1600 feet long. The Brat will bring up her
Jeep in a weekend or two and we'll try it.
ftp://66.117.156.8/CW_Donner_Lake.jpg
ftp://66.117.156.8/CW_Tunnel.jpg
ftp://66.117.156.8/CW_sign.jpg
ftp://66.117.156.8/CW_snow_shed.jpg
ftp://66.117.156.8/CW_wall.jpg
It's a "gravity wall" constructed for the roadbed by hand, from
natural uncut rocks, no cement.
The road is Donner Pass Road, old California route 40, a section of
the original Lincoln Highway.
http://en.wikipedia.org/wiki/Lincoln_Highway
Bronze marker courtesy E Clampus Vitus:
http://en.wikipedia.org/wiki/E_Clampus_Vitus
Cool stuff.
John
John Larkin
Lostgallifreyan
> 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
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
Baron
Very nice photos !
TFS.
--
Best Regards:
Baron.
John Fields
<jjSNIP...@highTHISlandtechnology.com> wrote:
---
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
<jfi...@austininstruments.com> wrote:
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
<jjSNIP...@highTHISlandtechnology.com> wrote:
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
<jfi...@austininstruments.com> wrote:
^^^^^^^^^
>>
>>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
> On Mon, 06 Jul 2009 12:08:53 -0500, John Fields
>>>WRONG!
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
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
wrote:
>Miss_Koksuka <desiree...@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
<jjSNIP...@highTHISlandtechnology.com> wrote:
---
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
John Larkin
<jfi...@austininstruments.com> wrote:
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
George Herold
>
> - 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.
John Larkin
<gghe...@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
Hal Murray
John Larkin <jjla...@highNOTlandTHIStechnologyPART.com> writes:
>On Mon, 6 Jul 2009 14:42:10 -0400, stan <smo...@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
<desiree...@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
<desiree...@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 1µ Rser=0.001
>R1 N002 0 100
>V1 N001 0 10
>L2 N001 N002 10µ 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
<jfi...@austininstruments.com> wrote:
>On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
><jjSNIP...@highTHISlandtechnology.com> wrote:
>
>>On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
>><jfi...@austininstruments.com> wrote:
>>
>>>On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
>>><desiree...@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
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"
sp...@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
John Larkin
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
hal-u...@ip-64-139-1-69.sjc.megapath.net (Hal Murray) wrote:
>In article <pq1555tl0t00p88gr...@4ax.com>,
> John Larkin <jjla...@highNOTlandTHIStechnologyPART.com> writes:
>>On Mon, 6 Jul 2009 14:42:10 -0400, stan <smo...@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
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
He did answer it. And he was wrong.
John
George Herold
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
> On Mon, 6 Jul 2009 19:54:28 -0700 (PDT), George Herold
>
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
>>L1 N001 N002 1� Rser=0.001
>>R1 N002 0 100
>>V1 N001 0 10
>>L2 N001 N002 10� 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
stan
> 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?
Greg Neill
> 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 |
. | | | |
. | | | |
. | | | |
. | | | |
. | | | |
. '----------------' '----------------'
John Larkin
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
---
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
George Herold
> 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.
John Larkin
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
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
John Larkin
Ouch. You'd have to assume that they were very close (the gap you drew
was very small) before the merger; otherwise they will change currents
(and the system energy will change) as they move towards one another,
since their magnetic fields will interact.
Given that, I'd guess that the currents won't change. That's the
easiest guess that conserves energy.
I have at least a 52% probability of being right.
John
George Herold
>
> - Show quoted text -
Yup, but it's possible to imagine a capacitor as poor as an inductor
and then 'see' the same effect. With leaky capacitors the short
time voltage across each cap will depend on C and the long time
voltage will depend on the parallel R.
George H.
Ross Herbert
:On Thu, 02 Jul 2009 15:49:01 -0700, Miss_Koksuka wrote:
:
:> Hello All,
:>
:> My teacher gave us a problem that is driving me absolutely crazy,
:> and my Spice simulator is supplying odd answers. His question: In a
:> circuit with a 10V DC power supply, and a series current limiting 1k Ohm
:> resistor, and two (ideal) inductors in parallel with each other, one
:> being 1uH and the other 10uH, will the DC currents be the exact same in
:> each inductor branch after reaching steady state, or will they be less
:> (by 10X) in the 10uH branch? If so, why should an ideal inductor of ANY
:> value have any effect whatsoever on DC current after it reaches its
:> steady state?
:
:Perhaps one of the things your instructor wanted to do was to wean you
:away from using Spice -- which is a fine tool for some things -- for
:everything.
:
:Try doing it using Laplace domain analysis; John Larkin's suggestion of
:finding the voltage and then the current is to the point if you want to
:simplify the math.
:
:Remember that Spice is a real-world tool, and an ideal inductor is not a
:real-world device. So using Spice and expecting it to cancel out
:infinities is inappropriate. OTOH, this is a fairly simple problem with
:linear circuit elements -- hence my suggestion of using Laplace analysis.
I agree with you entirely, at least insofar as the use of Spice as a solve-all
tool and the eagerness to use it as a first-in solution in place of other
methods. I think the problem posed by the lecturer may even have been framed so
that he could see whether students could use deductive reasoning for a simple
problem = hence the inclusion of the "after reaching the steady state" in the
specification.
Since there were no specification for the gauge of wire or construction method
for each of the inductors, it is quite possible for each inductor to have
exactly the same resistance despite the differing values of inductance. Now
since both inductors are in parallel with a resistor then after steady state
conditions are reached the voltage across the parallel combination will be the
same in each component. Assuming that a load is then connected and dc current is
flowing in the load, if both inductors have exactly the same resistance, then
the current through each inductor will also be the same.
Perhaps that is what the lecturer was after.
Hal Murray
John Larkin <jjla...@highNOTlandTHIStechnologyPART.com> writes:
>Inductors are, in general, crappy parts compared to the other stuff we
>get to use.
Is that a technical term? I don't see it in many data sheets.
:)
John Larkin
<desiree...@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
Another way to look at it:
Start with 11 identical 10 uH inductors, ideal or not. Wire them in
parallel carefully, so that no initial current is circulating among
them. If they are ideal inductors, use ideal wire and solder. Arrange
them in a physically symmetric pattern just to be compulsive.
Now apply any voltage, current, or waveform to the paralleled bunch.
By symmetry, all the 11 resulting inductor currents are identical.
At any point, mentally draw a dotted line around 10 of the inductors.
Now you have one 1 uH inductor in parallel with one 10 uH inductor.
Obviously the current in the 1 uH leg is 10x the current in the 10 uH
leg.
John
Michael Robinson
"John Larkin" <jjla...@highNOTlandTHIStechnologyPART.com> wrote in message
news:d42c559fdrjre3b48...@4ax.com...
I like the bit about ideal solder.
Ideal solder is a lot easier to work
with than the real thing. No muss,
no fuss -- all you
have to do is think about it.
For extra credit on the compulsive
front, use a cylindrical arrangement
for the ideal inductors.
John Fields
---
???
1
Lt = -------------------------------
1 1 1 1
---- + ---- + ---- ... + ----
L1 L2 L3 Ln
>Obviously the current in the 1 uH leg is 10x the current in the 10 uH
>leg.
---
???
+<----[E1]--->+
| |
| 0.1 I2---> |
+-----[L1]----+
| 1�H |
| |
| I2---> |
+-----[L2]----|
0.1�H
Tsk, tsk...
JF
John Fields
---
Oops...
Brain fart; never mind...
JF
John Larkin
<jfi...@austininstruments.com> wrote:
Tsk, tsk...
You are so eager to catch me doing something wrong, you jump to
contradict me without thinking first. Letting your emotions cancel
your ability to reason is bad engineering.
Stick to criticizing my spelling and typing. That way you'll have a
chance of being right.
John
John Fields
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
---
Well, I do admit that I take pleasure in showing that you have feet of
clay, but you must then admit that I had to reason in order to find _my_
mistake.
---
>Stick to criticizing my spelling and typing. That way you'll have a
>chance of being right.
---
Do you still believe that latching relays have infinite gain?
Or, more recently, from sed:
In this message, I discovered the OP's error and showed him how to fix
it:
In this one you were wrong about his circuit working and I corrected
you:
In this one you acknowledged that I had found the OP's error, without
acknowledging your own error, and then went on to say that moving the
diodes would fix the problem, which I had shown on the included
ASCIImatic (thanx, Steve!) which you snipped.
For what reason?
In my eyes, to make it seem to the casual reader that moving the diodes
was _your_ idea.
I think Phil was right; you're kinda slimy.
JF
John Larkin
<jfi...@austininstruments.com> wrote:
Absolutely. Do you still believe the answer to the problem here is 5+5
mA?
>
>Or, more recently, from sed:
>
>
>In this message, I discovered the OP's error and showed him how to fix
>it:
>
>1cn145pnrst5ogq8f...@4ax.com
>
>
>In this one you were wrong about his circuit working and I corrected
>you:
>
>cup1459pr696ig2t7...@4ax.com
>
>
>In this one you acknowledged that I had found the OP's error, without
>acknowledging your own error, and then went on to say that moving the
>diodes would fix the problem, which I had shown on the included
>ASCIImatic (thanx, Steve!) which you snipped.
>
>For what reason?
>
>In my eyes, to make it seem to the casual reader that moving the diodes
>was _your_ idea.
I can't help that.
>
>u8s145dfpl59vrs46...@4ax.com
>
>
>I think Phil was right; you're kinda slimy.
>
>JF
I think you are all wrapped up in he-said-she-said insecurity,
personalizing everything. I'm here to talk engineering. If I make an
occasional mistake - I do post a lot, about a lot of subjects - it's
no big deal to me, since this is not stuff I'm going to ship and get
paid for; we check deliverables really hard.
Declare victory if it makes you feel good. You're seldom important
because you seldom bring up anything interesting.
John
John Fields
<jjla...@highNOTlandTHIStechnologyPART.com> wrote:
>On Fri, 10 Jul 2009 12:19:01 -0500, John Fields
><jfi...@austininstruments.com> wrote:
>>---
>>Well, I do admit that I take pleasure in showing that you have feet of
>>clay, but you must then admit that I had to reason in order to find _my_
>>mistake.
>>---
>>
>>>Stick to criticizing my spelling and typing. That way you'll have a
>>>chance of being right.
>>
>>---
>>Do you still believe that latching relays have infinite gain?
>
>Absolutely. Do you still believe the answer to the problem here is 5+5
>mA?
---
Sure, why not?
Even with incontrovertible evidence in front of me to the contrary, I
can take your position and become an ostrich.
---
>>Or, more recently, from sed:
>>
>>
>>In this message, I discovered the OP's error and showed him how to fix
>>it:
>>
>>1cn145pnrst5ogq8f...@4ax.com
>>
>>
>>In this one you were wrong about his circuit working and I corrected
>>you:
>>
>>cup1459pr696ig2t7...@4ax.com
>>
>>
>>In this one you acknowledged that I had found the OP's error, without
>>acknowledging your own error, and then went on to say that moving the
>>diodes would fix the problem, which I had shown on the included
>>ASCIImatic (thanx, Steve!) which you snipped.
>>
>>For what reason?
>>
>>In my eyes, to make it seem to the casual reader that moving the diodes
>>was _your_ idea.
>
>I can't help that.
---
Sure you can, but you won't.
Certainly you're not stupid, and you seem to be attuned to the nuances
of the language in that you take every opportunity to make snide
remarks, cuts, and subtle putdowns in order to try to denigrate and
discredit those who you feel are threatening to you.
In addition, you use fallacious logic (straw man, well poisoning, etc,
etc,) in order to try to state your case from what you'd have others
believe is a solid platform.
---
>>u8s145dfpl59vrs46...@4ax.com
>>
>>
>>I think Phil was right; you're kinda slimy.
>>
>>JF
>
>I think you are all wrapped up in he-said-she-said insecurity,
>personalizing everything.
---
More bullshit.
What I find personal _is_ being attacked, and the game you're playing is
to attack me whenever you can since I've found you less than perfect
more than once, published my findings, and hung you up to dry.
The purpose of your game, of course, is to try to discredit me and,
therefore, my findings, turning you loose once again.
But, not just me, anyone who finds you in error.
So, egotistically incapable of admitting defeat, or having made a
mistake, you press on with your even weaker:
"I'm here to talk engineering."
Now you're playing this other game where you say you're here to "talk
engineering", but you seldom do.
Mostly you just go on, ad nauseam, about how the world should be,
according to Larkin.
---
>If I make an occasional mistake - I do post a lot, about a lot of subjects - it's
>no big deal to me, since this is not stuff I'm going to ship and get
>paid for; we check deliverables really hard.
---
So, in other words, when you post to USENET it's just a game?
What a surprise...
---
>So.,
>Declare victory if it makes you feel good. You're seldom important
>because you seldom bring up anything interesting
---
You're right, since the subject I brought up was you.
JF
John Larkin
<jfi...@austininstruments.com> wrote:
So, what's your final answer to the OP's question? How many mA in each
ideal inductor?
John
George Herold
> On Tue, 07 Jul 2009 21:13:21 -0500, John Fields wrote:
> > LTspice can't handle parallel zero-resistance inductors
>
> solve.However,specifying a very small series resistance, say 10^-18 ohms,
> together with zero parallel resistance and capacitance makes the solver
> happy, and is near enough to ideal inductors to give results near to what
> you get using calculus.
>
> LTSpice inserts a default (1 milliohm, IIRC) series resistance in its
> inductor model if you don't specify one.
> That's too big for the inductor to be near-ideal.
>
> Try this, it's your posted circuit, modified using 1e-18 ohm series
> resistance in the inductor models and the series resistors deleted. It
> gives a 10:1 share of current, just like the differential equation says it
> should.
>
> Version 4
> SHEET 1 880 680
> WIRE 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 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)
>
> "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)
Thanks Fred, It's good to know that spice can be made to cough up the
right answer.
George H.
John Larkin
<excret...@invalid.invalid> wrote:
>On Tue, 07 Jul 2009 21:13:21 -0500, John Fields wrote:
>
>> LTspice can't handle parallel zero-resistance inductors
>
>True, it results in an overdefined matrix, which it can't
>solve.However,specifying a very small series resistance, say 10^-18 ohms,
>together with zero parallel resistance and capacitance makes the solver
>happy, and is near enough to ideal inductors to give results near to what
>you get using calculus.
>
>LTSpice inserts a default (1 milliohm, IIRC) series resistance in its
>inductor model if you don't specify one.
>That's too big for the inductor to be near-ideal.
>
>Try this, it's your posted circuit, modified using 1e-18 ohm series
>resistance in the inductor models and the series resistors deleted. It
>gives a 10:1 share of current, just like the differential equation says it
>should.
Unless you really want a 1:1 current split and are willing to wait
until it happens.
John