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 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 _______________________________________________ R RST IS ... ... ..9 QSB QSB - hw? BK _______________________________________________ UR RST IS ... ... ..9 QSB QSB - hw? BK