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## [TowerTalk] trap loss

 To: [TowerTalk] trap loss ae110@rgfn.epcc.Edu (Ed E Jensen) Mon, 3 Mar 97 11:03:26 MST
 ``` Hi all, Using XL = 2pi*f*L and R=XL/Q, I get a little less than 1 ohm of loss resistance in the traps using the measurements posted here. This neglects radiation from the trap. Power loss = I**R and we know R but what is I? IF we knew the current in the element we could multiply it by Q (240 in this case) to get the circulating current in the trap. Now there is a about 4.5 amps at the center of a 50 ohm antenna getting 1000 watts but what is it half way out where the 10m trap is? If the trap were not there it would be down to .707*I or about 3 amps. But using P=(QI)**2(R) gives (240*3)**2*1 = 518kW, not too likely. So what is the current in the element where it connects to the trap? Anybody got a RF ammeter they can insert in an element? 73, Ed -- Ed Jensen, K5ED, El Paso, TX (ae110@rgfn.epcc.edu) -- FAQ on WWW: http://www.contesting.com/towertalkfaq.html Submissions: towertalk@contesting.com Administrative requests: towertalk-REQUEST@contesting.com Problems: owner-towertalk@contesting.com Sponsored by: Akorn Access, Inc. & N4VJ / K4AAA ```
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