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.

 

 

Thanks Andy!

 

I made the same stupid mistake in my message nr. III on this subject (14-07-01  10.37)

 

The corrected figures are even more dramatic than for the loop.

 

To feed 1kW into the 180m high 15mm copper pipe the current must be 154A and the voltage on the antenne becomes 2.5MV! Perhaps the pipe would survive the current but the voltage makes the system completely unrealistic (corona).

 

And what would the required 286mH loading coil for that voltage and current look like ...

 

73, Dick, PAoSE