DC Current in Parallel Inductors?

On Tue, 7 Jul 2009 16:03:47 -0400, "Greg Neill"
<gneillRE@MOVEsympatico.ca> wrote:

stan wrote:
Since the inductors are both ideal (zero ohms) and therefore DC
identical, it is reasonable to assume that once the 10 ma steady state
current has been reached, it will split equally between the two
inductors. Why make it more complicated?

The even split assumption is certainly questionable since
superconductors are involved. In fact, a little thought
should reveal that the only thing that will determine the
final current in each inductor will be the history of the
current flows through each.

At steady state both inductors will have a constant current
and zero voltage drop. Without considering how the final
currents obtain, that condition (zero voltage drop, constant
current) can be met by any aritrary currents that add up to
the required total. There could even be a very large
circulating current going around the superconducting ring
formed by the two inductors quite separate from the current
passing through the pair from the voltage source and resistor.

With superconductors, current has a sort of inertia.

Here's an example to consider. Suppose you have two
superconducting circuits in the form of squares. They
are side by side. The one on the left has a 100A current
circulating counterclockwise in it, while the one on the
right has a 10A current circulating clockwise.

Now the two nearest sides of the squares are brought into
contact in such a way that they merge into a single
conductor. After the merge, what is the steady state
current in each part of the circuit?

. .-------<--------. .------->--------.
. | 100 A | | 10 A |
. | | | |
. | | | |
. | | | |
. | | | |
. | | | |
. '----------------' '----------------'
Click to expand...

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

    My teacher gave us a problem that is driving me absolutely crazy,
and my Spice simulator is supplying odd answers. His question: In a
circuit with a 10V DC power supply, and a series current limiting 1k
Ohm resistor, and two (ideal) inductors in parallel with each other,
one being 1uH and the other 10uH, will the DC currents be the exact
same in each inductor branch after reaching steady state, or will they
be less (by 10X) in the 10uH branch?  If so, why should an ideal
inductor of ANY value have any effect whatsoever on DC current after
it reaches its steady state?

Thank you!

Desiree
Click to expand...
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.
 
On Jul 7, 11:11 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 7 Jul 2009 19:41:27 -0700 (PDT), George Herold





ggher...@gmail.com> wrote:
On Jul 2, 6:49 pm, Miss_Koksuka <desiree_koks...@yahoo.com> wrote:
Hello All,

    My teacher gave us a problem that is driving me absolutely crazy,
and my Spice simulator is supplying odd answers. His question: In a
circuit with a 10V DC power supply, and a series current limiting 1k
Ohm resistor, and two (ideal) inductors in parallel with each other,
one being 1uH and the other 10uH, will the DC currents be the exact
same in each inductor branch after reaching steady state, or will they
be less (by 10X) in the 10uH branch?  If so, why should an ideal
inductor of ANY value have any effect whatsoever on DC current after
it reaches its steady state?

Thank you!

Desiree

I was thinking about how much I enjoyed this discussion today. And I
was trying to find the post (by someone) comparing the problem (of two
inductors in parallel)  to two capacitors in series.  Which I thought
was apt.  Imagine leaky capacitors with a large parallel resistance,
(‘equivalent’ to inductors with small series resistance.)   I think we
all understand the capacitor case, and yet can get confused by the
inductors.

George H.

I expect that people ignore capacitor leakage but assume inductance
series resistance. That sort of makes sense; real capacitors can have
self-discharge time constants of many years, but it's rare for an
inductor to hit L/R as long as a full second.

Inductors are, in general, crappy parts compared to the other stuff we
get to use.

John- Hide quoted text -

- Show quoted text -
Click to expand...
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.
 
On Fri, 03 Jul 2009 12:02:51 -0500, Tim Wescott <tim@seemywebsite.com> wrote:

: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.
 
In article <e838559p5mb6iuth7klp2pu0b41iv44202@4ax.com>,
John Larkin <jjlarkin@highNOTlandTHIStechnologyPART.com> writes:

Inductors are, in general, crappy parts compared to the other stuff we
get to use.
Click to expand...
Is that a technical term? I don't see it in many data sheets.

:)

--
These are my opinions, not necessarily my employer's. I hate spam.
 
On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
<desiree_koksuka@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
Click to expand...
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
 
"John Larkin" <jjlarkin@highNOTlandTHIStechnologyPART.com> wrote in message
news:d42c559fdrjre3b48kk09h9nip9r6goglk@4ax.com...
On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koksuka@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

That's a good way to look at it.
Click to expand...
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.
 
On Thu, 09 Jul 2009 08:20:20 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koksuka@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.
Click to expand...
---
???

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.
Click to expand...
---
???


+<----[E1]--->+
| |
| 0.1 I2---> |
+-----[L1]----+
| 1ľH |
| |
| I2---> |
+-----[L2]----|
0.1ľH

Tsk, tsk...


JF
 
On Fri, 10 Jul 2009 04:00:51 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 09 Jul 2009 08:20:20 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koksuka@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.

---
???

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]----+
| 1ľH |
| |
| I2---> |
+-----[L2]----|
0.1ľH

Tsk, tsk...
Click to expand...
---
Oops...

Brain fart; never mind...
JF
 
On Fri, 10 Jul 2009 05:12:51 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 10 Jul 2009 04:00:51 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 09 Jul 2009 08:20:20 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koksuka@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.

---
???

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]----+
| 1ľH |
| |
| I2---> |
+-----[L2]----|
0.1ľH

Tsk, tsk...

---
Oops...

Brain fart; never mind...
JF
Click to expand...

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
 
On Fri, 10 Jul 2009 08:21:11 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 10 Jul 2009 05:12:51 -0500, John Fields
jfields@austininstruments.com> wrote:

On Fri, 10 Jul 2009 04:00:51 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 09 Jul 2009 08:20:20 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koksuka@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.

---
???

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]----+
| 1ľH |
| |
| I2---> |
+-----[L2]----|
0.1ľH

Tsk, tsk...

---
Oops...

Brain fart; never mind...
JF


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.
Click to expand...
---
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.
Click to expand...
---
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:

1cn145pnrst5ogq8fijg0sirkal5gkl25h@4ax.com


In this one you were wrong about his circuit working and I corrected
you:

cup1459pr696ig2t70pbe9bi4mgot2s25t@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.

u8s145dfpl59vrs46hpe4f3mncdf8vdav9@4ax.com


I think Phil was right; you're kinda slimy.

JF
 
On Fri, 10 Jul 2009 12:19:01 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 10 Jul 2009 08:21:11 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 10 Jul 2009 05:12:51 -0500, John Fields
jfields@austininstruments.com> wrote:

On Fri, 10 Jul 2009 04:00:51 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 09 Jul 2009 08:20:20 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koksuka@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.

---
???

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]----+
| 1ľH |
| |
| I2---> |
+-----[L2]----|
0.1ľH

Tsk, tsk...

---
Oops...

Brain fart; never mind...
JF


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.

---
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?
Click to expand...
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:

1cn145pnrst5ogq8fijg0sirkal5gkl25h@4ax.com


In this one you were wrong about his circuit working and I corrected
you:

cup1459pr696ig2t70pbe9bi4mgot2s25t@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.
Click to expand...
I can't help that.

u8s145dfpl59vrs46hpe4f3mncdf8vdav9@4ax.com


I think Phil was right; you're kinda slimy.

JF
Click to expand...
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
 
On Fri, 10 Jul 2009 12:55:53 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 10 Jul 2009 12:19:01 -0500, John Fields
jfields@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?
Click to expand...
---
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:

1cn145pnrst5ogq8fijg0sirkal5gkl25h@4ax.com


In this one you were wrong about his circuit working and I corrected
you:

cup1459pr696ig2t70pbe9bi4mgot2s25t@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.
Click to expand...
---
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.
---


u8s145dfpl59vrs46hpe4f3mncdf8vdav9@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.
Click to expand...
---
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.
Click to expand...
---
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
Click to expand...
---
You're right, since the subject I brought up was you.

JF
 
On Fri, 10 Jul 2009 17:14:01 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 10 Jul 2009 12:55:53 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Fri, 10 Jul 2009 12:19:01 -0500, John Fields
jfields@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:

1cn145pnrst5ogq8fijg0sirkal5gkl25h@4ax.com


In this one you were wrong about his circuit working and I corrected
you:

cup1459pr696ig2t70pbe9bi4mgot2s25t@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.
---


u8s145dfpl59vrs46hpe4f3mncdf8vdav9@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
Click to expand...
So, what's your final answer to the OP's question? How many mA in each
ideal inductor?

John
 
On Tue, 07 Jul 2009 21:13:21 -0500, John Fields wrote:

LTspice can't handle parallel zero-resistance inductors
Click to expand...
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.

Version 4
SHEET 1 880 680
WIRE 320 128 128 128
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 320 256 128 256
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
SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0
SYMBOL ind 304 272 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 1e-6
SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0
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)
TEXT -128 424 Left 0 !.tran 10 uic




"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)
 
On Sat, 11 Jul 2009 06:40:42 -0700, Fred Abse
<excretatauris@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.
Click to expand...
Unless you really want a 1:1 current split and are willing to wait
until it happens.

John
 
On Sat, 11 Jul 2009 06:40:42 -0700, Fred Abse
<excretatauris@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.

Version 4
SHEET 1 880 680
WIRE 320 128 128 128
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 320 256 128 256
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
SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0
SYMBOL ind 304 272 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 1e-6
SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0
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)
TEXT -128 424 Left 0 !.tran 10 uic
Click to expand...
---
Thanks, Fred. :)
JF
 
On Jul 11, 9:40 am, Fred Abse <excretatau...@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.

Version 4
SHEET 1 880 680
WIRE 320 128 128 128
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 320 256 128 256
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
SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0
SYMBOL ind 304 272 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 5 56 VBottom 0
SYMATTR InstName L2
SYMATTR Value 1e-6
SYMATTR SpiceLine Rser=1e-18 Rpar=0 Cpar=0
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)
TEXT -128 424 Left 0 !.tran 10 uic

"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)
Click to expand...
Thanks Fred, It's good to know that spice can be made to cough up the
right answer.

George H.
 
John Fields wrote:
On Fri, 10 Jul 2009 12:55:53 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:


On Fri, 10 Jul 2009 12:19:01 -0500, John Fields
jfields@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:

1cn145pnrst5ogq8fijg0sirkal5gkl25h@4ax.com


In this one you were wrong about his circuit working and I corrected
you:

cup1459pr696ig2t70pbe9bi4mgot2s25t@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.
---



u8s145dfpl59vrs46hpe4f3mncdf8vdav9@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
:)
Click to expand...
 
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