# Series capacitor charges

Discussion in 'General Electronics Chat' started by massive, May 12, 2010.

1. ### massive Thread Starter Member

May 7, 2010
20
2
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

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
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.

Last edited: May 12, 2010
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