Capacitor, charges, resistance and much more

If you have a cell, a bulb and a capacitor connected in series then this is not an open circuit (do not be misled by the circuit diagram symbol for a capacitor). Current will flow from the cell to charge the capacitor, and as the current passes through the bulb, it will light the bulb (as long as there is enough current to make the bulb filament glow).

However, as the charge on the capacitor increases, it will give rise to a voltage which opposes the flow of current. So the current decreases as the charge on the capacitor increases (you can model this with a simple differential equation if you have done calculus) and eventually the bulb will stop glowing. The time that this takes depends on the capacitance of the capacitor.

If you then disconnect the cell and connect the two sides of the capacitor through the bulb (do not connect them together without a bulb or some other resistance in the circuit - that can be dangerous if you have a large capacitor) then the capacitor will discharge itself through the bulb, again causing the bulb to glow for a short period until the capacitor has fully discharged itself.

In my answer to your previous question of the same title, I described how current can flow around and not through the dielectric of a capacitor. To avoid confusion, you should change the title of this question because it is a duplicate of the title of your previous related question.

The capacitor only looks like an open circuit if the voltage across it is constant. This can be seen from the relationship between voltage and current for a capacitor

$$i(t)=Cfrac{dv(t)}{dt}$$

If the voltage is constant, the derivative is zero and the the current is zero. Current will only flow if the voltage across the capacitor is changing in time.

Regarding the light bulb, if it is in series with a battery and uncharged capacitor, when a switch to the battery is closed, the light may illuminate but only for a limited time. When the switch is first closed, assuming no voltage initially across the capacitor, all the voltage of the battery will appear across the bulb and the illumination will be a maximum. Assuming the bulb is a resistor, then the voltage across the bulb as a function of time will be

$$v_{R}(t)=Ve^{frac{-t}{RC}}$$

Where $V$ is the battery voltage and $R$ is the resistance of the bulb (assumed here to be constant for simplicity), $C$ is the capacitance, and $RC$ is the time constant, the amount of time it takes for the voltage across the resistor to fall to about 37 % of $V$. Eventually the voltage will be zero across the bulb and $V$ across the capacitor when it is fully charged. At that point the current is zero.

On the other hand, if an AC voltage source is involved, the bulb may light continuously if the division of the voltage between the bulb and the capacitor is such that there is sufficient voltage across the bulb.

Hope this helps.