Topband: "return" current - what is it?

[w4buw at aol.com]

DOES ANYONE REMEMBER 
 Gustav Kirchhoff



-----Original Message-----
From: K4SAV <RadioIR at charter.net>
To: topband <topband at 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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