What does "Fully Charged Capacitor" mean?

[Kosta1234]

Homework Statement

Charged Capacitor

Relevant Equations

-

Why when a capacitor is fully charged the circuit is acting like an open circuit?
And what is the meaning of "fully charged", if the charge as a function of time equation is:
$$ Q = CV_b [1 - e^{\frac{t}{RC}}] $$
so by this equation the charge on the capacitor will never reach exactly, Q = CV, only when the time goes to infinity.Edit:
Mayble I should be more precise, why in the following question the current to the right side of the circuit is stopping immediately after the capacitor is charged (In the answer it have been said that the capacitor is fully charged immediately after closing the switch),
If there was a resistor in the right side of the circuit, could I've also say that the capacitor is fully charged immediately?

immediately
thank you.

 

[gleem]

By fully charged it is meant that for all intents and purposes you can consider it holding as much charge as theoretically possible. After a time of about 15 time constants i.e., 15RC there is only enough "room" for less than 1 millionth more charge.

 

[Kosta1234]

okay, thanks.

And the reason that the capacitor in the attached file is fully charged
immediately after closing the switch is because there is no resistence (R = 0) so $e^{-\frac{t}{RC}}$ goes to 0?

And the circuit on this case is considered open because that the voltage on the Electromotive force and the capacitor is the same?

 

[berkeman]

And the reason that the capacitor in the attached file is fully charged
immediately after closing the switch is because there is no resistence (R = 0) so e−tRC e^{-\frac{t}{RC}} goes to 0?

I guess so. It's not a very good real-world example. Even if the wires have negligible resistance, there will be parasitic inductance in the current loop that will limit the current and cause ringing of the voltage. The power source also cannot have zero output impedance in the real world.

And the circuit on this case is considered open because that the voltage on the Electromotive force and the capacitor is the same?

Yes, the capacitor will only charge up to the output voltage of the voltage source.

 

[Kosta1234]

Thanks.

 

[SammyS]

Problem Statement: Charged Capacitor
Relevant Equations: -
...

Edit:
Maybe I should be more precise, why in the following question the current to the right side of the circuit is stopping immediately after the capacitor is charged (In the answer it have been said that the capacitor is fully charged immediately after closing the switch),
If there was a resistor in the right side of the circuit, could I've also say that the capacitor is fully charged immediately?

thank you.

Please give the problem completely, as it was stated to you. That would help use to be able explain.

Given your question regarding some of the wording in the problem as well as in the "official" solution, I could hazard some guesses as to what is meant, however, that may not be sufficient to clear up the confusion.

 

[David Lewis]

Problem Statement: Charged Capacitor
Relevant Equations: -

If there was a resistor in the right side of the circuit, could I've also say that the capacitor is fully charged immediately?

No, it would take time because the resistor limits the current. In the real world there will always be some resistance and inductance to prevent the capacitor from charging instantaneously. Eventually the capacitor will equal the battery voltage, and at that point no current will flow.