# Charge on a capacitator

1. May 17, 2009

### Dell

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
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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

Last edited by a moderator: May 4, 2017
2. May 17, 2009

### nickjer

There is an equation for exponential charging of a capacitor in a RC circuit.

3. May 17, 2009

### Dell

what is that equation

4. May 17, 2009

### nickjer

You never solved for the differential equation of an RC circuit in your book or class?

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

That is the equation for a battery charging a capacitor in an RC circuit. Where $$Q_0 = C\epsilon$$.

5. May 17, 2009

### Dell

nope, never seen that before, probably in the next few lessons, what is the root of this equations?

6. May 17, 2009

### nickjer

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.

Last edited by a moderator: May 4, 2017