Current impulse is not nearly as interesting as voltage impulse. @user29568, a capacitor acts as short circuit in two different limits: (1) as an AC short circuit as the frequency goes to infinity and (2) as an actual short circuit (assuming the capacitor is uncharged) as C goes to infinity.
Given a fixed voltage, the capacitor current is zero and thus the capacitor behaves like an open. If the voltage is changing rapidly, the current will be high and the capacitor behaves more like a short. Expressed as a formula: i = Cdv dt (6.1.2.5) (6.1.2.5) i = C d v d t Where i i is the current flowing through the capacitor,
The vertical wire drawn next to the vertical capacitor shorts the two terminals of the capacitor. Any current flowing through this circuit segment will flow through the vertical wire and completely bypass the vertical capacitor due to the short. This means you can ignore the shorted capacitor -- it has no effect on the circuit.
By having their shorted terminals, the voltage thereof is zero (more precisely, the potential difference between them), so that this element is not operational in the circuit, and can be removed for analysis. The other two capacitors are in series, hence that:
It doesn't act like a short circuit for a current impulse. Here's the equation that defines the ideal capacitor: iC(t) = C ⋅ d dtvC(t) Applying the Laplace transform to this equation (assuming zero initial conditions) yields IC(s) = sC ⋅ VC(s) The Laplace transform for the unit impulse is δ(t) ⇔ 1
REVIEW: Capacitors act somewhat like secondary-cell batteries when faced with a sudden change in applied voltage: they initially react by producing a high current which tapers off over time. A fully discharged capacitor initially acts as a short circuit (current with no voltage drop) when faced with the sudden application of voltage.