"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'.
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Reg, G4FGQ