What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel


i.e.



|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

Is this by any chance an exam question?

Roy Lewallen, W7EL

Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel


i.e.



|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

Roy Lewallen wrote:
Is this by any chance an exam question?

No, it is not. I was shown it by a lecturer of mine more than 10 years
ago. The result is quite interesting.

On 18 Nov, 06:05, Dave wrote:
Roy Lewallen wrote:
Is this by any chance an exam question?

No, it is not. I was shown it by a lecturer of mine more than 10 years
ago. The result is quite interesting.

Interesting to me is that there is no parallel resistance bypassing
the capacitor inferring a mythical loss less capacitor.
I await developments with interest
Art

In message , Dave writes
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel


i.e.



|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ

--
Brian Howie

Brian Howie wrote:


|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ



The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

On Sun, 18 Nov 2007 19:12:48 +0000, Dave wrote:

Brian Howie wrote:


|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|


I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ



The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

For a minute, I thought he had abandoned the "new math" and gone over to
the "dark side." (or, "new-new math!") ;-)
JS

Dave wrote:
What is the resonate frequency of this network, as determined between
the top and bottom of what I have drawn?

I don't know how well the drawing will come out, but it consists of:

100 uH in series with 1000 Ohms.
100 pF in series with 1000 Ohms

The two two networks above are in parallel

Since the two resistances are equal, seems to me the
resonant frequency would be where the two reactances
are equal. Where the 100 uH line crosses the 100 pf
line on the reactance chart in the ARRL Handbook is
in the ballpark of 1.591549431 MHz. :-)
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
On Sun, 18 Nov 2007 19:12:48 +0000, Dave wrote:

Brian Howie wrote:

|
|
!
-----!-----
| |
| |
L C
| |
| |
R R
| |
| |
------------
|
|
|

I ran it through Spice ( laziness) - It doesn't resonate. Intuitively
you think it should have a low Q resonance at 1.6MHz , but it doesn't
Nice one.

73 Brian GM4DIJ


The trick is to make

R = sqrt(L/C)

then the impedance is real everywhere. You can use any old values for L:
and C, as long as you make R=sqrt(L/C);


That equation is obviously know from transmission lines too..

Hi Dave,

Perhaps, but not in this round.

sqrt(L/C) 1000

Besides, resonance is slightly above 1MHz.

73's
Richard Clark, KB7QHC

My calculator needs fixing. When I divide 100 uH by 100 pF and take
the square root, I end up with the number 1000. Where did I go wrong?
73,
Tom Donaly, KA6RUH