Power Rating of Wilkinson divider/combiner termination resistor

Hi first time posting on this forum.

as stated by the title,
I am curious about power rating calculation of the termination resistor within the Wilkinson divider/combiner.

For example, when used as a 2 way combiner that could handle 800W output power, VSWR=1.2 is there equations that I could use to calculate the maximum power rating of the resistor needed to create the wilkinson combiner?

Thank you and hope to here lot of your ideas

Your idea is correct if you can keep the VSWR low all the time. If there is any chance that a higher VSWR may occur like during system testing, make sure all possible power can be absorbed without burning the termination resistor. I would use a 800 W resistor .

The resistor is only loaded by the reflected signal.
At VSWR 1.2 the reflected signal is -21dB down.
Resistor must handle 800W less 21dB

It might be dangerous to assume that you never have any VSWR worse than 1.2

I think is about pulsed 800W, and not continuous wave (CW) signal.
An RF system that use a Wilkinson splitter which supports 800W is really a design challenge.
In the literature when you search for "High power Wilkinson combiner" the maximum supported power is about 100W.

Thank you everyone for your input.

vfone you are correct on using pulse not CW signal,

and so far I have found that the worst case would be one port shorting
which would mean half of the output power will be dissipated through the termination resistor.
Therefore, that would mean 400W power rating will be sufficient.

But this would mean the resistor will be large and probably won't be a SMD resistor.

Thanks again for your interest and ideas.