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Parallel Capacitors connected to Battery then D/C

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RavageU
#1
Feb13-11, 03:46 PM
P: 4
1. The problem statement, all variables and given/known data
Capacitors of 19.1 uF and 2.1 uF are charged as a parallel combination across a 229 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate.
Find the resulting charge on the first capacitor.


2. Relevant equations
Q = CV


3. The attempt at a solution
Since they are connected in parallel, I found the charge for each individual capacitor.

Q1 = C1V
Q2 = C2V

Then I know when reconnected, the net charge has to equal the new charge. So:

q1+q2 = Q1-Q2

Then the voltage across them has to be the same. So:

V' = q1/c1 = q2/c2

Then I solved them sim. and got:

q1 + (c2/c1)q1 = Q1 - Q2

But when I plug in my numbers, I get the wrong answer. What am I doing wrong?
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gneill
#2
Feb13-11, 04:04 PM
Mentor
P: 11,678
Quote Quote by RavageU View Post
Since they are connected in parallel, I found the charge for each individual capacitor.

Q1 = C1V
Q2 = C2V

Now i'm confused about what happens when I reconnect them. Would the voltage (229) across them be the same since they are in parallel?
They are connected in parallel but the way the leads were arranged the voltages are opposing. So current will flow to "even out" the potential.

You might find it convenient to assign charge (coulombs) to each plate before the reconnection, then determine what will remain after the cancellation happens when opposite charges are "introduced" to each other via the new connections.
RavageU
#3
Feb13-11, 04:09 PM
P: 4
Edited for more that I've done to solve the problem.

So the charges essentially cancel out?

gneill
#4
Feb13-11, 04:49 PM
Mentor
P: 11,678
Parallel Capacitors connected to Battery then D/C

Quote Quote by RavageU View Post
Edited for more that I've done to solve the problem.

So the charges essentially cancel out?
Some of the charge is going to cancel, because you're connecting the + plate of one cap to the - plate of the other. One capacitor has more charge on it than the other. Figure out what remains.
RavageU
#5
Feb13-11, 04:55 PM
P: 4
So would it be Q1-Q2 (The initial charges of the capacitor) = q1 + q2 (The charges after reconnection)?
gneill
#6
Feb13-11, 05:00 PM
Mentor
P: 11,678
Quote Quote by RavageU View Post
So would it be Q1-Q2 (The initial charges of the capacitor) = q1 + q2 (The charges after reconnection)?
Sounds good. Now you just need to figure out how that remaining charge will distribute over the caps.
RavageU
#7
Feb13-11, 05:06 PM
P: 4
So because they are parallel, the voltage across the plate have to be the same, so I have:

V' = q1/c1 = q2/c2

and solving simulatenous, I get:

q1 + (c2/c1)q1 = Q1 - Q2

Plug in the numbers, and I get ~3507 which is still the wrong answer...
gneill
#8
Feb13-11, 05:10 PM
Mentor
P: 11,678
Ah, but 3507 what? And how do your significant figures compare with those given as original information?


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