I is NOT the the square root of power. It is the square root of (power
divided by R) You first have to divide the power by the resistance THEN take
the square root.

And E=I x R not I x I x R

Back to Ohms Law --see URL:
http://www.angelfire.com/pa/baconbacon/page2.html

and AC RMS power is the same as DC power. Provided the circuit has no
inductance and capacitance and RMS values are used -- see OHMS LAW URL above
and URL:
http://www1.jaycar.com.au/images_uploaded/ohmpower.pdf

About now I think you are putting me on, so the old Elmer is exit stage
left.

Elmer E Ing
------------------------------------------------

"Richard" anom@anom wrote in message
...

"Elmer E Ing" wrote in message
news:VM_Ua.12050$ff.5596@fed1read01...
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.

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.