You want to use Ohm's Law here.
We will approximate the SEX voltages here, for relevance.
If we put a 270K resistor across the power supply (let's just call it 450V), let's examine how much current this bleeder resistor will draw. Ohm's Law tells us V=I*R, and we have V and R, so we have:
I=V/R, or 450/270,000, which is 1.6mA. Ohm's Law tells us that power dissipated will be about 0.75W (I*I*R, V*A, or (V*V)/R will all give this). We would recommend tripling this for reliability, which gets you to 2.25W.
If we put two 270K resistors in series, we now have a 540K resistor across the power supply.
I=V/R now gives 450/540,000, which is 0.83mA. To calculate heat in each resistor, we can use V=450/2=225, R=270,000, and I=0.00083 and any of the Ohm's Law formulas for power. I get just under 0.2W, so a 0.6W resistor is sufficient here.
So, when putting two resistors in series, you can add the power they are able to dissipate, provided you are adding the same two values. This also is informative regarding why you won't see one of our 270K 1W resistors by itself across the power supply in any of our big power amps.
-PB
