# Circuit with two switches

1. Mar 18, 2014

### avolp

1. The problem statement, all variables and given/known data
Consider the circuit shown below. Use this circuit for both parts of this question. Be sure to answer both parts!

https://www.physicsforums.com/attachment.php?attachmentid=67797&stc=1&d=1395204781

Both capacitors are originally uncharged and both switches are open. S1 is now closed; S2 remains open. A long time passes. Switch S1 is now opened. A very short time later switch S2 is closed, and remains closed for a long time.

G. What is the potential difference across C1?

H. What is the potential difference across C2?

I. What is the charge on capacitor C1?

J. What is the charge on capacitor C2?

K. What is the electrical potential energy stored in capacitor C1?

L. What is the electrical potential energy stored in capacitor C2?

2. Relevant equations
C=C1+C2
Q=CV
U=1/2CV^2
V=IR
Q=CV(1-e^(-t/RC))

3. The attempt at a solution

I was stuck immediately because once S1 is opened, there is no more potential in the system which makes everything else impossible. I thought that maybe the capacitor C1 might discharge but the equation for discharge also contains a V.

Thank you ahead of time

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2. Mar 19, 2014

### Curious3141

The charging/discharging equations involving time are irrelevant here, because you are told that the switches are left in those positions 'a long time' (which means infinite time).

When S1 is opened (after being left closed for 'infinite' time) and before S2 is opened, C1 has been fully charged by the cells. What has C1 been charged to? What is the voltage (potential difference) across C1's plates at this point?

The fully charged C1 provides the voltage for further movement of charge (when S2 is closed).

3. Mar 19, 2014

### avolp

C1 was charged to 30mC because Q=CV and the potential difference across C1's plates were 10V right?

So if the capacitor provides the voltage, does it start to act similarly to a battery? If yes, then would G.) be 30/8V? because C1 is 3/8 of the total C in the circuit?
so then
H.) 50/8V
I.) Q=CV=80
J.) would be the same as I because the charge remains constant for series right?
K and L.) i'm confused on these (provided the others are correct). the original energy is 150mj from 1/2*C*V^2 which means that the energy in each capacitor can't be more than that. is the energy proportional to the capacitance (i.e. C1=3/8 so it'd be 3/8*150?)

Last edited: Mar 19, 2014
4. Mar 19, 2014

### Curious3141

Right.

Not really. The voltage across the capacitor changes very rapidly as it is discharging, for one thing. Don't treat it like a battery. Just recognise that a charged capacitor has a voltage across its plates. That voltage can cause the movement of charge when a circuit is closed between those plates.

Don't assume that potential energy is conserved in the system. You should realise that when a capacitor is charged or discharged, energy is *always* lost (dissipated) in the connecting resistances (even if they're not explicitly represented, like in this circuit diagram).

(EDIT: Of course, the energy at the end can't be *more* than you started out with, that's just impossible. Try working through the problem again focusing on conservation of charge).

You should focus only on conservation of charge. The initial charge on C1 is equal to the sum of the final charges on C1 and C2. Also, what does Kirchoff's voltage law tell you about the relationship between the voltages across C1 and C2 at the point when there's no more current flow?

Last edited: Mar 19, 2014
5. Mar 19, 2014

### avolp

I'm still very confused. Don't we need either the potential difference or the charge of one of the capacitors to use the Q=CV formula to solve for the other?
Kirchoff's voltage law says that the sum of potential differences around a closed network is 0 but I can't figure out what to do with that.

EDIT: Oh never mind. I understand now. V1=V2 so the initial potential of 10V would split into 5V for each capacitor. then we can use the conservation of charge q1/c1=q2/c2 and since q=q1+q2, we can plug that it and ge will get q1/c1=(q-q1)/c2 and then solve for q1.

Last edited: Mar 19, 2014
6. Mar 20, 2014

### Curious3141

No, sorry, you misunderstood me. It's correct that V1 = V2, but it's not true that they will sum to 10V.

Voltage doesn't have to be "conserved" the same way as charge does in this case.

What you know is that the same voltage is dropped across both C1 and C2 at the end. That's a consequence of Kirchoff's voltage law. Call that V (i.e. V1 = V2 = V).

The charge on C1 = $C_1V$ and the charge on C2 = $C_2V$.

Add them up, and they will equal 30uC.

Can you do the algebra to solve for V? That's the first step, which will answer both G and H.

Can you proceed from there?

BTW, thanks for the "Thanks", but I hope you take note of what I've posted since.