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Re: Topband: "return" current - what is it?

To: topband@contesting.com
Subject: Re: Topband: "return" current - what is it?
From: K4SAV <RadioIR@charter.net>
Date: Sat, 04 Aug 2012 10:04:00 -0500
List-post: <topband@contesting.com">mailto:topband@contesting.com>
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 carry equal currents, 
> with less current on the side with the load. Can someone please explain this?
>
>
>   

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 the load is located.  So the current 
distribution is considerable different in the two halves of the dipole.  
The source is at the center of a segment.  Since you can only measure 
the current at the center of the segments adjacent to the feedpoint 
(that's one segment away, on each side, from the feedpoint) the current 
will be different.  That one segment difference away from the feedpoint 
is enough to show a difference in current.  If you want to see the 
current at the feedpoint use the "Src Dat" tab.  It only lists a single 
current because it's the same in both sides, except 180 degrees out of 
phase.

It's impossible to violate the law stated by Tom.  If you want an easy 
way to test this, wire a battery to a bulb, measure the magnitude of 
current out the negative terminal of the battery and then measure the 
current out the positive terminal of the battery.  If you don't get the 
same answer, you have a measurement error.

Jerry, K4SAV

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