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Re: LF: Lost current in a coil

To: [email protected]
Subject: Re: LF: Lost current in a coil
From: Rik Strobbe <[email protected]>
Date: Wed, 2 Mar 2005 17:04:38 +0100
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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






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