Hi,
If theres 2 capacitors in series, 10 mF and 1 mF , that should calculate to
.9 mF total. (yes...Im the new guy )

If input was 100v , that means charge Q = .9 x 100 = 90 mC

From what Ive learnt from the ebook each capacitor has individual voltages based on total charge so...

Q / 10mF = 9v
Q / 1mF = 90v

My questions being
1. Is breakdown voltage passed on to the smallest capacitor ?

2. For a given charge ,can voltage exceed the source ?
( as a spike / break down )

3. If voltage is increased to maximum that 10mF can handle will 1mF capacitor be facing break down ?

These may seem like simple questions to alot of you but I can tell you ,
theyre head scatches to me so any imfo you can throw at me will get me going in the right direction .

Massive THANKS

 

My questions being
1. Is breakdown voltage passed on to the smallest capacitor ?

2. For a given charge ,can voltage exceed the source ?
( as a spike / break down )

3. If voltage is increased to maximum that 10mF can handle will 1mF capacitor be facing break down ?

1. The smallest value capacitance in the series chain has the highest voltage. The cap may break down if its voltage rating is exceeded.

2. No - even if one of the capacitors goes short circuit the maximum voltage in a purely capacitive series connected chain can not exceed the source voltage. Sure if the source has a "spike" on its output that will be distributed across the capacitors as a momentary proportionate increase in cap voltages.

3. Depends on the voltage rating of the capacitors. In your example, if the smaller value cap has a voltage rating 10x that of the larger value cap then there would be no issue.