Return-Path: Received: (qmail 43677 invoked from network); 2 Mar 2005 19:09:05 -0000 Received: from unknown (HELO ptb-spamcore02.plus.net) (192.168.71.3) by ptb-mailstore03.plus.net with SMTP; 2 Mar 2005 19:09:05 -0000 Received: from mailnull by ptb-spamcore02.plus.net with spamcore-l-b (Exim 4.32; FreeBSD) id 1D6ZF8-000OJA-B8 for dave@picks.force9.co.uk; Wed, 02 Mar 2005 19:10:50 +0000 Received: from [192.168.67.2] (helo=ptb-mxcore02.plus.net) by ptb-spamcore02.plus.net with esmtp (Exim 4.32; FreeBSD) id 1D6ZF4-000OIJ-Sy for dave@picks.force9.co.uk; Wed, 02 Mar 2005 19:10:42 +0000 Received: from post.thorcom.com ([193.82.116.20]) by ptb-mxcore02.plus.net with esmtp (Exim) id 1D6ZD5-000Ko3-A3 for dave@picks.force9.co.uk; Wed, 02 Mar 2005 19:08:39 +0000 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1D6ZCa-00034P-KR for rs_out_1@blacksheep.org; Wed, 02 Mar 2005 19:08:08 +0000 Received: from [193.82.116.30] (helo=relay.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1D6ZCU-0002z8-Db for rsgb_lf_group@blacksheep.org; Wed, 02 Mar 2005 19:08:02 +0000 Received: from m11.lax.untd.com ([64.136.30.74]) by relay.thorcom.net with smtp (Exim 4.43) id 1D6ZCR-00017S-QG for rsgb_lf_group@blacksheep.org; Wed, 02 Mar 2005 19:08:01 +0000 Received: from m11.lax.untd.com (localhost [127.0.0.1]) by m11.lax.untd.com with SMTP id AABBCNDZ7AYKSUUS for (sender ); Wed, 2 Mar 2005 11:07:41 -0800 (PST) X-UNTD-OriginStamp: yokUgcxCbtTP7XLrpefewM0dVUUv3fe63HeK2Yv1FsyBfgu2DKw4kw== Received: (from riese-k3djc@juno.com) by m11.lax.untd.com (jqueuemail) id KL4ZPRSF; Wed, 02 Mar 2005 11:07:17 PST To: rsgb_lf_group@blacksheep.org Date: Wed, 2 Mar 2005 14:03:37 -0500 Message-ID: <20050302.140618.2468.1.riese-k3djc@juno.com> X-Mailer: Juno 5.0.13 MIME-Version: 1.0 X-Juno-Line-Breaks: 8-6,7-11,13-22,24,26-29,30-32767 From: riese-k3djc@juno.com X-ContentStamp: 9:4:2581445696 X-SPF-Result: relay.thorcom.net: domain of juno.com designates 64.136.30.74 as permitted sender X-Spam-Score: 0.5 (/) X-Spam-Report: autolearn=no,HTML_50_60=0.095,HTML_FONT_BIG=0.232,HTML_MESSAGE=0.001,NO_REAL_NAME=0.178 Subject: Re: LF: Lost current in a coil Content-Type: text/html; charset=windows-1252 X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: * X-Spam-Status: No, hits=1.2 required=5.0 tests=HTML_30_40,HTML_FONT_BIG, HTML_MESSAGE,NO_REAL_NAME autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false X-Spam-Filtered: by PlusNet SpamCORE (v3.00) Content-transfer-encoding: 8bit

Exactly

if the current were the same than you would have a perpetual

motion machine where more power is coming out than is going in.

the coil is ,, in the simple case,, acting as a transformer matching the 50 Ohm

to the higher impedance of the aerial

we needn't disobey the law of ,,, conservation

be nice if we could,, though

Bob K3DJC

As an example let the inductance of the coil be 5mH. Then its reactance at 136kHz is 4270 ohm.
With an aerial current of for instance 2A flowing through the coil the voltage at the top of the coil will be
2 * 4270 = 8541V higher than at the bottom.
So total voltage at the top will be 70 + 8541 = 8610V.

73, Dick, PA0SE