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.
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
As soon as the capacitor is short-circuited, it starts discharging. Let us assume, the voltage of the capacitor at fully charged condition is V volt. As soon as the capacitor is short-circuited, the discharging current of the circuit would be – V / R ampere.
If a charged capacitor C is short circuited through an inductor L, the charge and current in the circuitoscillate simple harmonically. If a charged capacitor C is short circuited through an inductor L, the charge and current in the circuit oscillate simple harmonically. (i) In what form the capacitor and the inductor stores energy ?
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.
When a capacitor gets fully charged, the value of the current then becomes zero. Figure 6.47; Charging a capacitor When a charged capacitor is dissociated from the DC charge, as has been shown in figure (d), then it remains charged for a very long period of time (depending on the leakage resistance), and one feels an intense shock if touched.