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Charge on a capacitator

  • Thread starter Dell
  • Start date
  • #1
590
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given the following:
http://lh3.ggpht.com/_H4Iz7SmBrbk/ShATOIWSXEI/AAAAAAAAA_4/M57_Bf03Z2M/s720/Untitled.jpg [Broken]

and knowing:
C1=400μF
C2=500μF
ε=9V
R=20kΩ

the switch is closed to position A for a time of t=RC1/2 and then moved to position B
----------------------------------
what is the charge of C1 the moment the switch is moved from A to B??


as far as i can see none of my equations have anything to do with time, other than P=VI (since W=J/s) but i dont think that helps me very much,, how do i incorporate the element of time into this, i tried kirchhoff but that didnt really get me anywhere

q1C1+IR=-ε
since i dont know q1 I or R i think this is not the right way to go,
i would think that the R needs to cancel out since the t is dependant on R, so i am looking for something that will give me t/R

i thought that since W=J/s, P=E/t = q2/2C * 2/RC1 but that doesnt help either
 
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Answers and Replies

  • #2
674
2
There is an equation for exponential charging of a capacitor in a RC circuit.
 
  • #3
590
0
what is that equation
 
  • #4
674
2
You never solved for the differential equation of an RC circuit in your book or class?

[tex]Q = Q_0 \left( 1 - e^{-t/RC}\right)[/tex]

That is the equation for a battery charging a capacitor in an RC circuit. Where [tex]Q_0 = C\epsilon[/tex].
 
  • #5
590
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nope, never seen that before, probably in the next few lessons, what is the root of this equations?
 
  • #6
674
2
Are you doing homework ahead of time?

Too much to explain but I found this on google:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/RCSeries.html [Broken]

It explains how a capacitor charges and discharges in an RC circuit.
 
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