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| Miss_Koksuka... |
 Posted: Thu Jul 02, 2009 12:49 pm |
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Guest
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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 |
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| Miss_Koksuka... |
| Posted: Thu Jul 02, 2009 3:11 pm |
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Guest
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On Jul 2, 6:17 pm, Dan Coby <adc... at (no spam) earthlink.net> wrote:
[quote:af5b92b79f]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.)
[/quote:af5b92b79f]
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 |
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| Dan Coby... |
| Posted: Thu Jul 02, 2009 5:17 pm |
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Guest
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Miss_Koksuka wrote:
[quote:63ffd579f7]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?
[/quote:63ffd579f7]
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.) |
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| Michael Robinson... |
| Posted: Thu Jul 02, 2009 5:17 pm |
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Guest
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"Miss_Koksuka" <desiree_koksuka at (no spam) yahoo.com> wrote in message
news:f7f121cd-f823-416e-b821-70608cee0106 at (no spam) x5g2000prf.googlegroups.com...
[quote:310fb1272c]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
[/quote:310fb1272c]
Would the troll like a cookie?
Nice troll, come here, gootch gootchy goo
SPLAT |
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| Dan Coby... |
| Posted: Thu Jul 02, 2009 7:26 pm |
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Guest
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Miss_Koksuka wrote:
[quote:d761ddfe4a]On Jul 2, 6:17 pm, Dan Coby <adc... at (no spam) 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...).
[/quote:d761ddfe4a]
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. |
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| nospam... |
| Posted: Thu Jul 02, 2009 7:51 pm |
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Guest
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Dan Coby <adcoby at (no spam) earthlink.net> wrote:
[quote:9a894754a2]Instead of looking at the steady state, you need to look at how you get there.
[/quote:9a894754a2]
Has anyone got a Digkey part number for some of these inductors? Sounds
like they would be really useful.
-- |
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| Michael Robinson... |
| Posted: Thu Jul 02, 2009 8:26 pm |
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Guest
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"Dan Coby" <adcoby at (no spam) earthlink.net> wrote in message
news:P_WdnVVgSNPJw9DXnZ2dnUVZ_vmdnZ2d at (no spam) earthlink.com...
[quote:5681ee928d]Miss_Koksuka wrote:
On Jul 2, 6:17 pm, Dan Coby <adc... at (no spam) 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...).
,
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.
[/quote:5681ee928d]
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. |
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| David L. Jones... |
| Posted: Thu Jul 02, 2009 9:40 pm |
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Guest
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Miss_Koksuka wrote:
[quote:d23e86de85]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?
[/quote:d23e86de85]
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/ |
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| John Larkin... |
| Posted: Thu Jul 02, 2009 11:26 pm |
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Guest
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On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
<desiree_koksuka at (no spam) yahoo.com> wrote:
[quote:9bab62c13d]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
[/quote:9bab62c13d]
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 |
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| John Larkin... |
| Posted: Thu Jul 02, 2009 11:28 pm |
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Guest
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On Fri, 03 Jul 2009 02:51:07 +0100, nospam <nospam at (no spam) please.invalid>
wrote:
[quote:3fc3449fe2]Dan Coby <adcoby at (no spam) earthlink.net> wrote:
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.
[/quote:3fc3449fe2]
Yeah, room temperature superconductors will be mighty handy. Finite-Q
filter designs are a nuisance.
John |
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| Tom Biasi... |
| Posted: Thu Jul 02, 2009 11:31 pm |
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Guest
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"Miss_Koksuka" <desiree_koksuka at (no spam) yahoo.com> wrote in message
news:f7f121cd-f823-416e-b821-70608cee0106 at (no spam) x5g2000prf.googlegroups.com...
[quote:5e0e5de81b]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
Given the level of the question steady state means after all the math is[/quote:5e0e5de81b]
done. Parallel devices have equal voltages.
Tom |
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| Miss_Koksuka... |
| Posted: Fri Jul 03, 2009 3:15 am |
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Guest
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On Jul 3, 12:26 am, John Larkin
<jjSNIPlar... at (no spam) highTHISlandtechnology.com> wrote:
[quote:f7f6409691]On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koks... at (no spam) 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
[/quote:f7f6409691]
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 |
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| Michael Robinson... |
| Posted: Fri Jul 03, 2009 7:58 am |
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Guest
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"Miss_Koksuka" <desiree_koksuka at (no spam) yahoo.com> wrote in message
news:4e070800-fa2a-47e1-9982-d053d668794f at (no spam) v15g2000prn.googlegroups.com...
On Jul 3, 12:26 am, John Larkin
<jjSNIPlar... at (no spam) highTHISlandtechnology.com> wrote:
[quote:17227d5b66]On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koks... at (no spam) 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
[/quote:17227d5b66]
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.) <=======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. |
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| Miss_Koksuka... |
| Posted: Fri Jul 03, 2009 12:09 pm |
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Guest
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On Jul 3, 3:04 pm, John Larkin
<jjSNIPlar... at (no spam) highTHISlandtechnology.com> wrote:
[quote:b772c8483a]On Fri, 3 Jul 2009 06:15:27 -0700 (PDT), Miss_Koksuka
desiree_koks... at (no spam) yahoo.com> wrote:
On Jul 3, 12:26 am, John Larkin
jjSNIPlar... at (no spam) highTHISlandtechnology.com> wrote:
On Thu, 2 Jul 2009 15:49:01 -0700 (PDT), Miss_Koksuka
desiree_koks... at (no spam) 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
[/quote:b772c8483a]
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 |
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| Dan Coby... |
| Posted: Fri Jul 03, 2009 12:55 pm |
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Guest
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Miss_Koksuka wrote:
[quote:054e2ac7ce]Are you pursuing an engineering degree?
Yes Mike, but I am only in my first year of electronics.
[/quote:054e2ac7ce]
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 |
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