Richard wrote:

Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or

better

still a dummy load of 50 ohms that exhibits no inductance or capacitance.

Now since Power in watts = I squared R where I is the current and R is a
pure 50 ohm resistance
transpose and solve for I = square root of P over R

I get about 4.9 amperes RF current

now

E=I times R so 4.9 times 50 = 244 volts RF volts

That's what you should see at the antenna. Try that on old knucklehead.

With any inductive reactance or capacitive reactance --- different ball
game.

Gurus check my math please


RF power is not electrical power.
As audio power is not electrical power.

????????????? What kind of power is RF? Magical? At which frequency does
the electric current stop following the laws of physics?

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

Are you serious? And you wanted *us* to get on that newsgroup and make
asses of ourselves like you just did?

Better check your math and your understanding of the Ohm's law, the
Power equation and the efficiency of the AB class amplifiers.

U= I*R, P= U*I = P= I^2*R or P= U^2/R

= I= SQR(P/R) and *not* I= SQR(P)/R like you mistakenly claim

Elmer's calculation is 100% correct. The trucker is right. You are wrong.



73 .... WA7AA




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