Charge stored on Capacitor doesn't depend on Resistor?

[hellocubed]

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It says in the BR that the charge stored on a capacitor has "no dependence on resistance of any kind" verbatim.


I'm pretty confused by this because I know that the Voltage of the capacitor is:

V(total)= V(Resistors) + V(capacitors)
V(Capacitors)= V(total) - V(Resistors)

If a resistor and a capacitor are in series, their voltages from the emf are dependent on one another.

And considering Voltage is important to the amount of charge stored:

V=q/C

I don't really see how resistors can have no effect on the charge stored.




Could anyone help explain this?

 

[MedPR]

It says in the BR that the charge stored on a capacitor has "no dependence on resistance of any kind" verbatim.


I'm pretty confused by this because I know that the Voltage of the capacitor is:

V(total)= V(Resistors) + V(capacitors)
V(Capacitors)= V(total) - V(Resistors)

If a resistor and a capacitor are in series, their voltages from the emf are dependent on one another.

And considering Voltage is important to the amount of charge stored:

V=q/C

I don't really see how resistors can have no effect on the charge stored.




Could anyone help explain this?

If you put a billion resistors between a battery and a capacitor, the capacitor would take longer to reach max charge (Q), but it would eventually reach that charge.

It's like saying the amount of water a flask can hold is completely independent of the number of holes you have in the water hose your are using to fill it. More holes = less water into the flask per unit time, but eventually the flask will reach its max capacity regardless of how many holes (resistors) are in the hose.

 

[milski]

The problem is that the max charge of a capacitor is not a specific Q but Q/V, meaning that the maximum amount of charges changes as you change the potential between its plates.

For different reasons, it does charge to the same amount of charge Q. What happens is that as you charge it more, the current through it goes down and the potential drop over it increases. When the capacitor is completely charged, there is no current flowing, the voltage drop over the resistor is zero and the voltage drop over the capacitor is V. That allows you to charge it to the same Q, regardless of presence of resistors.