Chad noticed my error in the followin sentence:
>>There is a given amount of resistive loss in every conductor/
>>component in any of our gadgets, so at legal power max, there
>>are about 9.5 amperes of peak current (about 7 amps rms)
>>flowing around in a 50 ohm circuit!
>
>If the antenna is resonant (X=0 :-), cannot you just apply
>Ohm's Law? Such as I^2/R = P, or I = SQRT(P/R)
>I = SQRT(1500/50) = 5.47 Amps RMS, or 1.414*5.47= 7.7A peak?
>
>How do you calculate 7 Amps RMS?
Because I assumed the vswr on the line was 3:1!
But I forgot to put that in my model discussion.
Chad's number is exactly correct for the rms current in
a 50 ohm system. Mismatch of 3:1 would be what I was
to be correcting with the antenna tuner--and managed
to omit that detail in my write up. Sorry about that!
So with the correct current and with 1 ohm of loss, about
30 watts would be lost to heat; 2 ohms, 60 watts, etc.
BTW, did you ever accidently touch the end of
a 30 watt soldering iron used on PC boards?
73, Jim, KH7M
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