"Reg Edwards" wrote in message
...
"Tom Sedlack" wrote What kind of voltages and currents are present at
the
ends of dipoles
assuming 100W of RF and the antenna being close to resonant?

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A half-wave dipole is a resonant L & C tuned circuit.

Its lumped equivalent is a series L and C circuit across the feedpoint.

It has a Q value like any other resonant circuit.

As is usual, Q = Inductive reactance of wire divided by resistance.

In the case of a dipole the resistance is the radiation resistance at its
centre, typically around 70 ohms.

Inductance gets smaller as wire diameter increases. So a 2-meter dipole
with 1" diameter tubes has a lower Q (about 4) than a 160-meter dipole
made
with 18 gauge wire. (about 12). Q controls bandwidth.

A 40 meter dipole may have a Q around 9. If it is fed at its centre with
100 watts then the feeding voltage is 84 volts.

So the voltage difference between the ends of the dipole is Q times 84 =
756
volts.

Relative to ground, the voltage at one end of the dipole is half of this,
equal to 378 volts.

This would burn a nice little hole at the tip of your right forefinger if
you touched it. Electrical burns take a long time to heal because the
surrounding flesh is electrocuted. Do not confuse this with 'skin
effect'.
----
Reg, G4FGQ

.................and besides, it SMARTS!!!!!!!!!


Jwerry
K4KWH