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Joel
science forum beginner
Joined: 18 May 2005
Posts: 7
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 Posted: Thu Jun 09, 2005 12:03 pm Post subject:
Resistance of infinite sheet
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I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to ground?
I set up an integral based on concentric rings, each of whose radial extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
I'm stumped. Anyone have any ideas?
Thanks,
Joel |
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mathedman
science forum beginner
Joined: 29 Apr 2005
Posts: 36
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| Posted: Thu Jun 09, 2005 12:10 pm Post subject:
Re: Resistance of infinite sheet
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On Thu, 09 Jun 2005 14:03:57 GMT, "Joel"
<dedrickj@removethis.pacbell.net> wrote:
| Quote: | I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to ground?
I set up an integral based on concentric rings, each of whose radial extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
I'm stumped. Anyone have any ideas?
Thanks,
Joel
When you find such a sheet jusr stick a meter on it! |
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Keith A. Lewis
science forum Guru
Joined: 24 Mar 2005
Posts: 478
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| Posted: Thu Jun 09, 2005 3:47 pm Post subject:
Re: Resistance of infinite sheet
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"Joel" <dedrickj@removethis.pacbell.net> writes in article <h%Xpe.26070$J12.9903@newssvr14.news.prodigy.com> dated Thu, 09 Jun 2005 14:03:57 GMT:
| Quote: | I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to ground?
I set up an integral based on concentric rings, each of whose radial extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
|
It might be helpful to start with an easier question. Let's say you had a
sheet that was grounded at both ends across the width and had a terminal
across the entire center width. The resistance from terminal to ground
would be Rho*(L/W)/4.
Can we expand on that answer to some kind of discrete model which approaches
the continuous one?
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer. |
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John O'Flaherty
science forum beginner
Joined: 07 May 2005
Posts: 41
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| Posted: Thu Jun 09, 2005 7:34 pm Post subject:
Re: Resistance of infinite sheet
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Niall Gallagher wrote:
| Quote: | Joel wrote:
I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to ground?
I set up an integral based on concentric rings, each of whose radial extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
Assume you are at the centre, take a pie-shaped segment, angle Theta,
Radius R, Resistivity Rho
Total resistance of segment = Rho * Int (Theta*r, r=0..R)
Theta * R^2 / 2, this does not converge as R -> Infinity
Total resistance from the centre of the infinite plane,
sum of all segments: = Rho * Int (Theta* R^2/2, Theta=0..2*Pi).
For an infinitely large plane, you have an infinite resistance to ground.
|
I don't dispute your conclusion, but the segments wouldn't be summed-
they'd be in parallel, so would combine as 1/(1/seg1+1/seg2...etc.)
Also, as R is increased in a single segment, the contribution of each
increase in radius to the total resistance of the segment would be less
because of broader parallel path.
--
john |
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Joel
science forum beginner
Joined: 18 May 2005
Posts: 7
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| Posted: Thu Jun 09, 2005 8:18 pm Post subject:
Re: Resistance of infinite sheet
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This is why the obvious answer (resistance is infinite) didn't match my
intuition. As r increases, the additional resistance added to the total for
each unit increase in radius drops, so you have a sum of an infinite
sequence of values, each of which get ever closer to zero. Seems like I
dimly remember from college calculus a lot of Taylor/Maclaurin series like
this that are bounded and converge to some constant, but I could be
mistaken.
<quiasmox@yahoo.com> wrote in message
news:1118352891.676678.230100@g44g2000cwa.googlegroups.com...
| Quote: |
Niall Gallagher wrote:
Joel wrote:
I wonder if someone can help me with this: Imagine an infinite sheet
of
material. The material has some electrical resistivity Rho - meaning
there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also
that
the sheet is connected to electrical ground around its' "periphery".
(I
know this is kind of a mind-bender for some - the EE's at this point
are
bailing for the next post, and the mathematicians are wondering what
the
fuss is about.)
My question is, what is the resistance from a point on this sheet to
ground?
I set up an integral based on concentric rings, each of whose radial
extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
Assume you are at the centre, take a pie-shaped segment, angle Theta,
Radius R, Resistivity Rho
Total resistance of segment = Rho * Int (Theta*r, r=0..R)
Theta * R^2 / 2, this does not converge as R -> Infinity
Total resistance from the centre of the infinite plane,
sum of all segments: = Rho * Int (Theta* R^2/2, Theta=0..2*Pi).
For an infinitely large plane, you have an infinite resistance to ground.
I don't dispute your conclusion, but the segments wouldn't be summed-
they'd be in parallel, so would combine as 1/(1/seg1+1/seg2...etc.)
Also, as R is increased in a single segment, the contribution of each
increase in radius to the total resistance of the segment would be less
because of broader parallel path.
--
john
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Repeating Rifle
science forum Guru Wannabe
Joined: 25 Mar 2005
Posts: 205
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| Posted: Fri Jun 10, 2005 3:04 am Post subject:
Re: Resistance of infinite sheet
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in article hf3qe.26497$J12.4704@newssvr14.news.prodigy.com, Joel at
dedrickj@removethis.pacbell.net wrote on 6/9/05 3:18 PM:
| Quote: | This is why the obvious answer (resistance is infinite) didn't match my
intuition. As r increases, the additional resistance added to the total for
each unit increase in radius drops, so you have a sum of an infinite
sequence of values, each of which get ever closer to zero. Seems like I
dimly remember from college calculus a lot of Taylor/Maclaurin series like
this that are bounded and converge to some constant, but I could be
mistaken.
|
True and false. Just because the terms get smaller does not mean that they
converge. The classic example is that of the harmonic series:
1 + 1/2 +1/3 + 1/4 + 1/5 + ...
Bill |
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Niall Gallagher
science forum beginner
Joined: 09 Jun 2005
Posts: 3
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| Posted: Fri Jun 10, 2005 11:32 am Post subject:
Re: Resistance of infinite sheet
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quiasmox@yahoo.com wrote:
| Quote: |
Niall Gallagher wrote:
Joel wrote:
I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to ground?
I set up an integral based on concentric rings, each of whose radial extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
Assume you are at the centre, take a pie-shaped segment, angle Theta,
Radius R, Resistivity Rho
Total resistance of segment = Rho * Int (Theta*r, r=0..R)
Theta * R^2 / 2, this does not converge as R -> Infinity
Total resistance from the centre of the infinite plane,
sum of all segments: = Rho * Int (Theta* R^2/2, Theta=0..2*Pi).
For an infinitely large plane, you have an infinite resistance to ground.
I don't dispute your conclusion, but the segments wouldn't be summed-
they'd be in parallel, so would combine as 1/(1/seg1+1/seg2...etc.)
Also, as R is increased in a single segment, the contribution of each
increase in radius to the total resistance of the segment would be less
because of broader parallel path.
|
Oops - using the same approach but this time getting conductance and
resistance correct....
For the pie-slice, inner radius r1, outer radius r2, angle theta,
R slice = Rho * Area = Rho * Int(Theta*r, r=r1..r2), for simplicity
assume rho = 1
R slice = Theta * (r2^2-r1^2)/2
C slice = 2 / (Theta * ( r2^2 - r1^2))
Adding the conductances of the slices..
C plane = 2/ (r2^2 - r1^2) * Int (1/ Theta, Theta=0..2*Pi)
= 2 / (r2^2 - r1^2) * (log (2*pi) - 1))
R plane = (r2^2 - r1^2) / constant
as r2 -> Infinity and assuming r1 -> 0 [ignoring the messier physical
connection realities], we see that R plane -> Infinity
Niall |
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Niall Gallagher
science forum beginner
Joined: 09 Jun 2005
Posts: 3
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| Posted: Fri Jun 10, 2005 11:33 am Post subject:
Re: Resistance of infinite sheet
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quiasmox@yahoo.com wrote:
| Quote: |
Niall Gallagher wrote:
Joel wrote:
I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to ground?
I set up an integral based on concentric rings, each of whose radial extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
Assume you are at the centre, take a pie-shaped segment, angle Theta,
Radius R, Resistivity Rho
Total resistance of segment = Rho * Int (Theta*r, r=0..R)
Theta * R^2 / 2, this does not converge as R -> Infinity
Total resistance from the centre of the infinite plane,
sum of all segments: = Rho * Int (Theta* R^2/2, Theta=0..2*Pi).
For an infinitely large plane, you have an infinite resistance to ground.
I don't dispute your conclusion, but the segments wouldn't be summed-
they'd be in parallel, so would combine as 1/(1/seg1+1/seg2...etc.)
Also, as R is increased in a single segment, the contribution of each
increase in radius to the total resistance of the segment would be less
because of broader parallel path.
|
Oops - using the same approach but this time getting conductance and
resistance correct....
For the pie-slice, inner radius r1, outer radius r2, angle theta,
R slice = Rho * Area = Rho * Int(Theta*r, r=r1..r2), for simplicity
assume rho = 1
R slice = Theta * (r2^2-r1^2)/2
C slice = 2 / (Theta * ( r2^2 - r1^2))
Adding the conductances of the slices..
C plane = 2/ (r2^2 - r1^2) * Int (1/ Theta, Theta=0..2*Pi)
= 2 / (r2^2 - r1^2) * (log (2*pi) - 1))
R plane = (r2^2 - r1^2) / constant
as r2 -> Infinity and assuming r1 -> 0 [ignoring the messier physical
connection realities], we see that R plane -> Infinity
Niall |
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Brian Whatcott
science forum Guru Wannabe
Joined: 09 May 2005
Posts: 267
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| Posted: Fri Jun 10, 2005 9:25 pm Post subject:
Re: Resistance of infinite sheet
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On Fri, 10 Jun 2005 05:04:51 GMT, Repeating Rifle
<salmonegg@sbcglobal.net> wrote:
| Quote: | ... Just because the terms get smaller does not mean that they
converge. The classic example is that of the harmonic series:
1 + 1/2 +1/3 + 1/4 + 1/5 + ...
Bill
|
...which increases without limit, in contrast to
1 + 1/2 + 1/4 + 1/8 + 1/16 + ... which limits at 2
Brian W |
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Jeremy Boden
science forum Guru Wannabe
Joined: 28 Apr 2005
Posts: 144
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| Posted: Sun Jun 12, 2005 5:47 pm Post subject:
Re: Resistance of infinite sheet
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In message <hf3qe.26497$J12.4704@newssvr14.news.prodigy.com>, Joel
<dedrickj@removethis.pacbell.net> writes
| Quote: | This is why the obvious answer (resistance is infinite) didn't match my
intuition. As r increases, the additional resistance added to the total for
each unit increase in radius drops, so you have a sum of an infinite
sequence of values, each of which get ever closer to zero. Seems like I
dimly remember from college calculus a lot of Taylor/Maclaurin series like
this that are bounded and converge to some constant, but I could be
mistaken.
.... |
Try to visualize a lattice of unit resistance's.
This gives you lots of resistance's in a series/parallel arrangement.
You should see that for a very large sheet you will get a large
resistance.
Take the limit...
--
Jeremy Boden |
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charliew2
science forum beginner
Joined: 25 May 2005
Posts: 26
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| Posted: Mon Jun 13, 2005 12:00 pm Post subject:
Re: Resistance of infinite sheet
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Set this problem up with finite L and W. Integrate, then look at the
limiting resistance as L or W (or both)go to infinity.
Hint: your problem specification seems inconsistent. If L and W are both
infinite, the resistance will be infinite.
"Joel" <dedrickj@removethis.pacbell.net> wrote in message
news:h%Xpe.26070$J12.9903@newssvr14.news.prodigy.com...
| Quote: | I wonder if someone can help me with this: Imagine an infinite sheet of
material. The material has some electrical resistivity Rho - meaning
there
are Rho (L/W) Ohms of resistance in traversing (in the L direction) a
rectangle of the material L units long by W units wide. Imagine also that
the sheet is connected to electrical ground around its' "periphery". (I
know this is kind of a mind-bender for some - the EE's at this point are
bailing for the next post, and the mathematicians are wondering what the
fuss is about.)
My question is, what is the resistance from a point on this sheet to
ground?
I set up an integral based on concentric rings, each of whose radial
extent
(L) is dr and circumference is 2*pi*r, I end up with something like
R=Integral[0-infinity](1/(2*pi*r) dr), which is
(rho/2*pi)*[ln(infinity)-ln(0)].
I'm stumped. Anyone have any ideas?
Thanks,
Joel
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