Here we go again, speaking of misinformation, peer review, spreading
false guru stuff.
You can not apply Kirchoff law from DC circuits to the current behavior
along the STANDING WAVE RF radiator.
Current magnitude changes along the standing wave radiator, on the
resonant dipole it is maximum at
You can not apply Kirchoff law from DC circuits to the current behavior
along the STANDING WAVE RF radiator.
Yes, we can.
Kirchhoff's law is Kirchhoff's law, and is not frequency dependent.
I can't imagine anyone thinking otherwise. Thinking Kirchhoff's law applied
only to dc circuits is
Sooo, there is no current and voltage variation along the standing wave
resonant dipole?
S, Jasik et al, all those antenna books, modeling programs showing
RF CURRENT and/or RF VOLTAGE distribution along the (standing wave)
solid antenna wire are thinking otherwise?
Like parallel LC circuit
Sooo, there is no current and voltage variation along the standing wave
resonant dipole?
S, Jasik et al, all those antenna books, modeling programs showing
RF CURRENT and/or RF VOLTAGE distribution along the (standing wave)
solid antenna wire are thinking otherwise?
Like parallel LC
On 8/4/2012 10:47 AM, Tom W8JI wrote:
The reason I say this is because I know for an absolute fact.base
impedance can vary all over the place with unrelated or unexpected changes
in efficiency. A radial system here that made base impedance of a 1/4 wave
vertical 50 or 60 ohms delivered
-boun...@contesting.com [mailto:topband-boun...@contesting.com]
On Behalf Of Yuri Blanarovich
Sent: Sunday, August 05, 2012 9:51 AM
To: topband@contesting.com topband@contesting.com
Subject: Re: Topband: return current - what is it?
Here we go again, speaking of misinformation, peer review, spreading
to make a confirming measurement. When he did arrive at the chosen
measurement point he would find nearly but never all of the injected
current
had disappeared. This would not have been a surprise to him but maybe to
the rest of us. How could this have happened?
With an infinitely long
Bob Kupps wrote:
So I modeled a half wave dipole in free space and sure enough the wire
segments on each side of the feed point carried equal current. I then placed
a resistive load at the center of one half-element (to simulate? a lossy
return) and now see that those segments no longer
DOES ANYONE REMEMBER
Gustav Kirchhoff
-Original Message-
From: K4SAV radi...@charter.net
To: topband topband@contesting.com
Sent: Sat, Aug 4, 2012 11:04 am
Subject: Re: Topband: return current - what is it?
Bob Kupps wrote:
So I modeled a half wave dipole in free space and sure
That is correct, as Mr Kirchoff said.
Price W0RI
You are misinterpreting what you are seeing. When you put a resistor in
one side of a dipole you modify the current distribution in both sides
of the dipole and the side with the resistor has a large decrease in
current at the point where
So I modeled a half wave dipole in free space and sure enough the wire
segments on each side of the feed point carried equal current. I then placed
a resistive load at the center of one half-element (to simulate? a lossy
return) and now see that those segments no longer carry equal currents,
Tom W8JI wrote:
If you use enough segments so the program calculates small steps along
length, and a ground independent current source, you'll find current on each
side of the feedpoint exactly equal no matter what resistance you insert.
Yes current on each side of the feedpoint is
Hi Bill,
Tom, it's worth adding to this that trying to make current measurements in
the ground using 60hz is pretty useless for another reason: induced
currents from the ac power system (especially in north america). 60hz will
be present on just about anything -- you'll even see it on a
Tom W8JI wrote:
I think you may be selecting the wrong type of source, if you are using
EZNEC.
In the source-type selection, chose SI, not I. A split source places
the source at a segment junction, so you can see current leaving each
terminal of the source.
I forgot about the SI
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