Capacitor, capacitor is open in direct current, ‘ EXO 7 - Video 1, 2 ’

Demonstrate that in direct current, a capacitor allows no current to flow through it, as if it were open.
1) Apply Ohm's law to each component R and C of the circuit to deduce the current.
For R2, we have V10 = V1-Vo = R2.i3, i.e. i3 = (V1-Vo)/R2 = (V1-0)/3000, i3 = V1/3000.
For C1, V12 = V1-V2 = (1/W.C1).i1, i.e. i1 = W.C1 (V1-V2).
The same reasoning gives i2=(V1-V2)/2000 and i4 = W.C2.V3.
2) Now, in direct current, there is no pulsation, i.e. W=0[Rad/s], which leads to i1=i4=0[A], which simply means that the currents i1 and i4 are zero, but does not prove that C1 and C2 behave like two open circuits. We also form a system of equations whose unknowns are V1, V2 and V3, which gives V1=12[V] and V2=V3=8[V].
As V2 and V3 are at the same potential, there is no potential difference and no current flows in R5, hence no current flows in C2.
Finally, V12=V1-V2=12-8=4[V], there is a potential difference between the two nodes (1) and (2), and the current i2 flows in R1, but despite the potential difference, no current flows in C1, which means it is open. It's like a 1.5[V] battery, if its (+) and (-) poles are connected by a conductor, the current flows. If no current flows in the conductor despite the presence of 1.5[V] at its terminals, it means that the conductor is interrupted and therefore the circuit is open.