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LF: Re: Near field of 9kHz antenna IV

from [Andrew Talbot] [Permanent Link][Original]
To: "LF Group" <[email protected]>
Subject: LF: Re: Near field of 9kHz antenna IV
From: "Andrew Talbot" <[email protected]>
Date: Sat, 14 Jul 2001 23:03:59 +0100
Reply-to: [email protected]
Sender: <[email protected]>
Wait a minute !
 
 I = SQR(W / R)   so current in the loop = 51.2A and voltage at the feed is this current in the 60 ohms reactance, ie.  3077V
 
Andy  G4JNT
 
 
-----Original Message-----
From: Dick Rollema <[email protected]>
To All from PA0SE
AO gives the feed point impedance as Z = 0.380 + j60.1 ohm and the efficiency as 0.03%.
 
So for 1 kW fed into the loop the current must be SQR(0.380 * 1000) = 19.5A.
 
The voltage at the feed point becomes 19.5 * 0.380 = 1172V.
 
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