1.
its two 200V rated electrolytics in series of value 680uF.
Capacitors in series are calculated by (A x B) / (A + B). You don't have 680uF (assuming that's what you wanted). Rather you have half that.
2.
i presume its to do with the leakage current that flows through each capacitor....and the fact that its not quite the same in each, and so one of them charges all the way up to the rail?
Capacitors in series do not necessarily attain equal charge. One usually acquires more than the other. If its V rating is exceeded then it won't survive.
You really need a single capacitor, rated for the highest crest of your incoming AC voltage.
3.
Do we need bleed resistors across our 200V rated electrolytics at the input.?
Good idea.
If you wish to bleed off the charge in 5 seconds, then you want a time constant of 1 second. The formula for time constant and capacitors is R x C.
With 680uF, this comes out to 1470 ohms. When you apply 265 V (I'm taking the highest value you mentioned) it will conduct 180mA continuously and will dissipate (waste) 48W continuously while power is applied.
Not what you need.
Say you allow bleed off time of 20 seconds. Then the resistor can be 29.4 K. It will pass 9 mA continuously and waste 2.4 W continuously.
Your bleedoff resistor(s) should be rated for twice the expected wattage. So a 5W rating is good.
Etc. These are just sample calculations.
Note: You need a bleeder resistor across each capacitor, if you keep the caps in series. When caps are in series, one of them may have acquired a higher charge (as stated earlier). Then as is discharges it could drive the other electrolytic past zero into reverse polarity (if you try to use a single resistor). That tends to ruin a 'lytic.
---------- Post added at 13:48 ---------- Previous post was at 13:15 ----------
Vimalkhanna's advice is valid. Putting 1 meg resistors across each cap.
In this application they serve more as equalizing resistors than bleeder resistors.
The two caps must be matched as to uF amount.
See if you can use lower than 1 meg. The idea is to prevent unequal charges on the caps.
If you use 1/2 W resistors then the value can be 281k each. At 132V per resistor that makes 1.9 mA flow. The resistors are dissipating 1/4W each. That's half what they're rated for, which gives you the recommended safety margin.