M
Miss_Koksuka
Guest
On Jul 3, 12:26 pm, "Michael Robinson" <nos...@billburg.com> wrote:
Yes Mike, but I am only in my first year of electronics.
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Jamie
Guest
Miss_Koksuka 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
yes
M
Miss_Koksuka
Guest
On Jul 3, 3:04 pm, John Larkin
<jjSNIPlar...@highTHISlandtechnology.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!). 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
<jjSNIPlar...@highTHISlandtechnology.com> wrote:
On Fri, 3 Jul 2009 06:15:27 -0700 (PDT), Miss_Koksuka
desiree_koks...@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?
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
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
M
Michael Robinson
Guest
"Miss_Koksuka" <desiree_koksuka@yahoo.com> wrote in message
news:58f98eca-77bd-4165-bc69-6363a813ce51@l5g2000pra.googlegroups.com...
On Jul 2, 6:17 pm, Dan Coby <adc...@earthlink.net> wrote:
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
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!
news:58f98eca-77bd-4165-bc69-6363a813ce51@l5g2000pra.googlegroups.com...
On Jul 2, 6:17 pm, Dan Coby <adc...@earthlink.net> wrote:
Thanks Dan, but how would I calculate such a thing? And I'm still notMiss_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.)
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
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!
D
Dan Coby
Guest
Jamie wrote:
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.
Desiree,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
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.
D
Dave
Guest
On Jul 2, 7:17 pm, "Michael Robinson" <nos...@billburg.com> wrote:
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.
That is so nerdy and geeky of you to say, TROLL and SPLAT. How very"Miss_Koksuka" <desiree_koks...@yahoo.com> wrote in message
news:f7f121cd-f823-416e-b821-70608cee0106@x5g2000prf.googlegroups.com...
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
Would the troll like a cookie?
Nice troll, come here, gootch gootchy goo
SPLAT
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.
G
Greg Ewing
Guest
Miss_Koksuka wrote:
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
Ideal inductors don't exist in real life -- they'reSo 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.
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
J
John Fields
Guest
On Fri, 3 Jul 2009 15:09:04 -0700 (PDT), Miss_Koksuka
<desiree_koksuka@yahoo.com> wrote:
Indeed and, at DC, an inductor is nothing more than a resistor.
---
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
<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.
---
---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
N
nospam
Guest
Miss_Koksuka <desiree_koksuka@yahoo.com> wrote:
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.
--
V = L di/dtBut with v=L*di/dt for non-sinusoidal waveforms, such as this DC
circuit's turn-on waveform,
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.
--
M
Miss_Koksuka
Guest
On Jul 4, 6:07 am, Greg Ewing <greg.ew...@canterbury.ac.nz> wrote:
trying to do it; I felt as if my head were about to explode at one
point!!).
Many Thanks,
-Desiree
You guys are great! Thanks for clearing this up for me (I spent daysMiss_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.
--
Greg
trying to do it; I felt as if my head were about to explode at one
point!!).
Many Thanks,
-Desiree
J
John Larkin
Guest
On Sat, 04 Jul 2009 08:14:57 -0500, John Fields
<jfields@austininstruments.com> wrote:
Unless it has a whopping current circulating in it. Had and MRI
lately?
John
<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?
John
J
John Larkin
Guest
On Sat, 04 Jul 2009 23:07:39 +1200, Greg Ewing
<greg.ewing@canterbury.ac.nz> wrote:
consider superconducting magnets that have superconducting shim coils.
John
<greg.ewing@canterbury.ac.nz> wrote:
Tedious, but wrong. Consider superconducting magnets. Even better,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.
consider superconducting magnets that have superconducting shim coils.
John
S
Spehro Pefhany
Guest
On Sat, 04 Jul 2009 09:15:49 -0700, the renowned John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:
stop, and there will be (in general).
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:
Even if it has zero current circulating in it, just rotate a bit, thenOn 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?
John
stop, and there will be (in general).
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
J
John Fields
Guest
On Sat, 04 Jul 2009 09:15:49 -0700, John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:
Yeah, but so what?
No one's talking superconducting magnets, you pretentious ass.
JF
<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
J
John Fields
Guest
On Sat, 04 Jul 2009 09:18:47 -0700, John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:
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
<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
J
John Larkin
Guest
On Sat, 04 Jul 2009 14:35:30 -0500, John Fields
<jfields@austininstruments.com> wrote:
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.
<jfields@austininstruments.com> wrote:
People were talking about ideal inductors, which a real-world,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
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.
J
John Larkin
Guest
On Sat, 04 Jul 2009 14:42:18 -0500, John Fields
<jfields@austininstruments.com> wrote:
inductors that actually have no resistance.
I suppose you don't.
John
<jfields@austininstruments.com> wrote:
Greg said: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
which I pointed out isn't always the case. Some people work with realYou *have* to take their resistance into account,
because that's what ultimately determines the current
distribution.
inductors that actually have no resistance.
I suppose you don't.
John
G
greg
Guest
John Larkin wrote:
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
What I mean is, if they have any resistance at all,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.
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
L
Lostgallifreyan
Guest
John Larkin <jjSNIPlarkin@highTHISlandtechnology.com> wrote in
news:lisv459dhog3rqqv2erus7l5lthgnqqpti@4ax.com:
means you have to take their zero resistance into account.
news:lisv459dhog3rqqv2erus7l5lthgnqqpti@4ax.com:
So there's nothing wrong with Greg's statement. In this special case it justGreg 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.
means you have to take their zero resistance into account.
J
John Fields
Guest
On Sat, 04 Jul 2009 17:21:38 -0700, John Larkin
<jjSNIPlarkin@highTHISlandtechnology.com> wrote:
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?
---
???
What is it about:
E
R = ---
I
that you don't understand?
---
Nope, just annoyed with your self-aggrandizing bullshit.
---
And just how would you define the Q of an inductor with steady-state DC
in it?
JF
<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?
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.
---
---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