Message-ID: <1109779478.4225e41606d62@webmail2.kuleuven.be>
Date: Wed,  2 Mar 2005 17:04:38 +0100
From: Rik Strobbe <Rik.Strobbe@fys.kuleuven.ac.be>
To: rsgb_lf_group@blacksheep.org
References: <4225CA7F.15619.BBCBA7@localhost> <6.1.0.6.2.20050302162804.02a13ea0@mail.casema.nl>
In-Reply-To: <6.1.0.6.2.20050302162804.02a13ea0@mail.casema.nl>
MIME-Version: 1.0
User-Agent: Internet Messaging Program (IMP) 3.2-cvs
Subject: Re: LF: Lost current in a coil
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 8bit
Sender: owner-rsgb_lf_group@blacksheep.org
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org

Hello Dick, Mike

it is indeed a "phase thing".
If a 100 W TX causes a 2 A current in a 5mH coil (XL = 4270 Ohm) and 
thus causes a voltage of 8540 V over the coil then at the "hot" end 
the phase difference between U and I is 89.7 degrees :
P = U * I * cos(x) thus x = arccos(P/(U*I))
(assuming no coil losses)
At the "cold" end voltage and current are in phase.

73, Rik

> To Mike and All
> 
> At 15:15 2-3-05, G3XDV wrote:
> >A useful test to see whether radiation causes the loss, would be to
> >completely screen the loading coil.
> Because the size of the coil is so small, expressed in 
> wavelength,  radiation is negligible.
> 
> >I have never understood why, if the current is thought to stay the
> >same from top to bottom of a coil, the voltage is much bigger at
> the
> >top? Is this a phase thing?
> 
> The voltage at the bottom is equal to the output voltage of the 
> transmitter; at e.g. 100W this is 70V over 50 ohm.
> 
> 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
>