Capacitor Circuit (Capacitor as source)

1. Dec 2, 2008

mtbhrd

1. The problem statement, all variables and given/known data

Basically, I have to get charges and voltages as these switches are opened and closed on this capacitor circuit.

First, Switch A closed, B open. (C1 is charged.)

Then, switch A opens, and B closes.

Vi=5v
C1=4.6x10^-6 F
C2=2.3x10^-6 F

2. Relevant equations

Q = VC

3. The attempt at a solution

Okay, so the first part is easy. Charge on the the first capacitor Q = VC, [(5)*(4.6x10^-6)] = 2.3X10^-5 C = Q1i

Now, the second part. Is charge conserved? So that
Q1i=Q1f+Q2f

Expanding that with Q=VC,
C1V1i = C1V1f + C2V2f

Can we say that V1f = V2f? So Vf would be 3.35v? Then charges could be found on each capacitor... Is this the correct way to go about this?

Or, Is the voltage the same, 5v across all the capacitors? (Being in parallel.)

2. Dec 2, 2008

LowlyPion

First C1 will have the charge determined by the 5v and the capacitance.

As you note the charge is conserved when A is opened. Then the effective capacitance is changed by closing B.

Charge the same, but capacitance larger means the voltage is lower

V = Q/(larger C) means lower V.

3. Dec 2, 2008

mtbhrd

You mention a large effective capacitance..As in combining the two as one effective capacitor in parallel (C1 + C2)?

Using Q = VC , taking the original Q1i and using (C1 + C2) I get the same 3.35v as I got before.

4. Dec 2, 2008

LowlyPion

I didn't run the numbers. That looks about right.

I was just trying to provide you some guidance.

5. Dec 2, 2008

mtbhrd

Right, thanks.

So, with that voltage, charge on each cap (Q=VC) is 3.35v * C1 and then 3.35v * C2

I get Q1f=1.55x10^-5 C and Q2f=7.6x10^-6 C

However, when I run conservation of energy, U=.5QV (Ui = U1f + U2f) it doesn't work...

EDIT: Well, more simply put: Is it voltage or charge that is the same for both capacitors in this second ciruit situation where switch A is open, and b is closed, allowing C1 to charge the circuit.

Last edited: Dec 2, 2008