This really sounds like a stupid question, but I believe it's worth asking:
Due to \$I=C\frac{dU}{dt}\$, and \$C\$ is infinitely large, any minor
change in \$U\$ would result in infinite current, which is obviously
impossible. As a result, the voltage across the capacitor can never
change, so it's always zero, and a voltage drop of zero further
implies a short circuit, no matter in DC or AC.
Could any one tell me which step is wrong here?
I got it:
- The higher the frequency of the signal, the easier it can pass.
- The larger the capacitor, the easier the signal can pass.
When we say "a large capacitor is a DC open circuit", it actually means "After 5RC (time constant), no DC signal can pass a capacitor, although it's very large."
Clarification:
In fact, 5RC only gets you to 99% of the steady state condition, rather than 100%. However, it's reasonable to simply consider it as 0 in practice, because it's too small to care.
Best Answer
Did you ever pick up a telescope, hold the wrong end to your eye, and find you can't make sense of what you see?
The only thing wrong with using this telescope to see how an infinite capacitance behaves
\$I=C\frac{dU}{dt}\$
is that you aren't holding the useful end to your eye.
Try it this way round
\$\frac{dU}{dt}=\frac{I}{C}\$
What this equation means is that you can shove any finite amount of current, however large, into an infinite capacitor, and its voltage will not change. This is a convenient approximation for the behaviour of 'very large' capacitors, whose voltage 'does not change much' during operation. LTSpice has an infinite capacitor, and I would guess most other spices do as well. These are useful for approximating coupling or decoupling capacitors, when you are not interested in their deviation from ideal.
Now if you rephrase your question as 'I have a component whose terminal voltage cannot change, and I change its terminal voltage ...', well, you get what you get.
You might want to edit your title so that says 'DC short circuit'.