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capacitors in series with DC voltage

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eecs4ever

eecs4ever

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So if i have 2 ideal capacitors in series, suppose the values are equal.
Assume theres no parasitic resistances.

----||-----||--------

and i apply 6v DC voltage accross them.

what is the voltage in the middle?

1. Its 3V by the capacitor divide? C/(C+C) . Is this valid for DC voltages?
Does this work because of charge sharing?

OR

2. Since the capacitors block off DC completely, the middle node remains
floating, and thus the voltage is undefined. ?

Which is right?
I'm not sure. I hope someone knows =)
 

Here's an example with unequal capacitors from **broken link removed**
(near the bottom of the page)

**broken link removed**
The charge on capacitors in series is the same.

**broken link removed**
Assume CA = 50µF CB = 10µF
Q is the same on each capacitor.

----------------------------------------------
By Kirchhoff's voltage law and recalling that Q=CV for a capacitor

12 - Q / CB - Q / CA = 0
Q = 100µC
---------------------------------------------
VA = Q / CA
= 2 V

VB = Q / CB
= 10 V
-----------------------------------

To answer your question yes the middle point will be 3V.
I hope that this helps you out.

Best regards,
v_c
 

    eecs4ever

    eecs4ever

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eecs4ever, it is not DC indeed. if you apply DC voltage to discharged capacitors - this is like applying pulse with infinite length . And due to that pulse there will be transient process in capacitors to distribute charge proportionally to their capacities.
 

    eecs4ever

    eecs4ever

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artem, you are right. In theory, there will be an impluse of current and the capacitors will instantaneously charge to their steady state values. The value of this current will be \[Q \delta(t)\] where \[Q\] is the weight of the impulse -- it is actually the area under the current vs. time curve which is charge measured in Amp-seconds or Coulombs.

But I think the question was for steady-state, at least that is how I interpreted it.

Best regards,
v_c
 
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