arupkg wrote on Mar 15th, 2012, 1:25am: But by theory, the voltage should divide equally among the capacitances, right. That is how we say the equivalent capacitance is C/2 when two capacitances are connected in series, right?.......................
No, I don't think so.
As I have mentioned, the task cannot be solved for IDEAL capacitors connected to a dc source.
More than that, the calculation of the resulting capacitance if two such elements are in series is based on its capacitive impedances (reactances) which are defined for ac signals only.
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I think, it's useful to explain my position:
* At first, IDEAL capacitors do not exist; thus, the question is of pure theoretical nature without any relation to reality. In reality, resistive lsses determine the voltage distribution.
* Secondly: We must not consider the steady-state condition. In contrary, what happens at t=0 (switch on of a dc voltage)? In a lossless circuitry there would be a current pulse of value infinite. Is this possible? Of course not. Thus, we again arrive at the resistive losses within the circuit.