"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.
|