The capacitor acts as open circuit when it is in its steady state like when the switch is closed or opened for long time.
Capacitor acts like short circuit at t=0, the reason that capacitor have leading current in it. The inductor acts like an open circuit initially so the voltage leads in the inductor as voltage appears instantly across open terminals of inductor at t=0 and hence leads.
Short Answer: Inductor: at t=0 is like an open circuit at 't=infinite' is like an closed circuit (act as a conductor) Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor) Long Answer:
The circuit is at steady state when the voltage and the current reach their final values and stop changing. In steady state, the capacitor has a voltage across it, but no current flows through the circuit: the capacitor acts like an open circuit. How do you calculate steady state current in a capacitor? Is a capacitor fully charged in steady state?
Thus, at steady state, in a capacitor, i = C dv dt = 0, and in an inductor, v = Ldi = 0. That is, in steady dt state, capacitors look like open circuits, and inductors look like short circuits, regardless of their capacitance or inductance. (This might seem trivial now, but we'll use this fact repeatedly in more complex situations later.)
Capacitor: at t=0 is like a closed circuit (short circuit) at 't=infinite' is like open circuit (no current through the capacitor) Long Answer: A capacitors charge is given by Vt = V(1 −e(−t/RC)) V t = V (1 − e (− t / R C)) where V is the applied voltage to the circuit, R is the series resistance and C is the parallel capacitance.