DOES ANYONE REMEMBER
Gustav Kirchhoff
-----Original Message-----
From: K4SAV <RadioIR@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 enough the wire
egments on each side of the feed point carried equal current. I then placed a
esistive load at the center of one half-element (to simulate? a lossy "return")
nd now see that those segments no longer carry equal currents, with less
urrent on the side with the load. Can someone please explain this?
You are misinterpreting what you are seeing. When you put a resistor in
ne side of a dipole you modify the current distribution in both sides
f the dipole and the side with the resistor has a large decrease in
urrent at the point where the load is located. So the current
istribution is considerable different in the two halves of the dipole.
he source is at the center of a segment. Since you can only measure
he current at the center of the segments adjacent to the feedpoint
that's one segment away, on each side, from the feedpoint) the current
ill be different. That one segment difference away from the feedpoint
s enough to show a difference in current. If you want to see the
urrent at the feedpoint use the "Src Dat" tab. It only lists a single
urrent because it's the same in both sides, except 180 degrees out of
hase.
It's impossible to violate the law stated by Tom. If you want an easy
ay to test this, wire a battery to a bulb, measure the magnitude of
urrent out the negative terminal of the battery and then measure the
urrent out the positive terminal of the battery. If you don't get the
ame answer, you have a measurement error.
Jerry, K4SAV
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R RST IS ... ... ..9 QSB QSB - hw? BK
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UR RST IS ... ... ..9 QSB QSB - hw? BK
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