In article , "nick"
writes:

If a (let's say 12v) relay is rated for 10 amps @ 110 volts,
it will take 1100 watts. Will it take 1000 watts of RF?

I would think not, but I don't know why.

Relay CONTACT ratings have little to with relay COIL ratings.

Contact ratings assume a specific environment such as AC
primary power circuits, 28 VDC aircraft distribution, etc. That
is for the overwhelming majority of relay applications. In the
case of contacts rated for 115 or 230 VAC applications, those
will have open-contact spacings larger than lower-voltage DC;
spacings are generally designed to withstand about 3 times
the AC RMS voltage specified in ratings. Generally, but
not always.

To switch RF in a coaxial line system you have to allow for
the VSWR having a high value...choose at your own
convenience for expected conditions. The reason for that is
because, at a high VSWR, the relay contacts may be at a
maximum voltage point or a maximum current point.
Depending on the VSWR, either maximum may exceed the
contact ratings values. Location of either maximum depends
on frequency, length of transmission line, velocity of
propagation of the line used, antenna characteristics, and
the distance from the discontinuity (typically the antenna
feedpoint).

There are many tables and nomographs in ham literature to
determine the maximum voltage or maximum current with a
given SWR for a given characteristic impedance of a line.
A realistic worst-case value should be selected to avoid
having to replace a relay often due to arc-over or contact
sticking due to maxima. If an RF system is perfect (1:1
VSWR, line Z exact and unchanging) then compute the
required voltage and current from ordinary Ohm's Law.

I have yet to see such a perfect system but I know others
claim them as such... :-)

Len Anderson
retired (from regular hours) electronic engineer person