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| To: | [email protected] |
|---|---|
| Subject: | Re: LF: Lost current in a coil |
| From: | [email protected] |
| Date: | Wed, 2 Mar 2005 14:03:37 -0500 |
| Reply-to: | [email protected] |
| Sender: | [email protected] |
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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 |
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