Cerberus

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I have some more questions so i thought I'd start a thread where people can ask questions and get help:)

My q:

I want to get this straight, when a capicator is charging, the current initially increases and then decreases until it reaches zero (the capacitor becomes fully charged) but the voltage remains the same the entire time?
 

sdnstud

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you've confused V for battery with V for capacitor.

As the capacitor charges, V for battery stays the same. (the definition of a bettery is constant V for a long ass time)

For the capacitor, as it charges, V will increase proportionally with Q. Remember Q=CV. C remains constant because C=eA/d.
 

Cerberus

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Originally posted by sdnstud
you've confused V for battery with V for capacitor.

As the capacitor charges, V for battery stays the same. (the definition of a bettery is constant V for a long ass time)

For the capacitor, as it charges, V will increase proportionally with Q. Remember Q=CV. C remains constant because C=eA/d.
Ok, but the V for capacitor doesnt come into play till the capacitor is discharged - in which case it would drop exponetially - right?
 
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