Charged Capacitor Connected to an Uncharged Capacitor in Series

[Mark Zhu]

Homework Statement

In the circuit shown below, switch 1 has been closed for a long time and switch 2 has been open a long time. The two switches are then flipped simultaneously. Find all resisting currents through resistors and charges on capacitors after the system has reached a steady state.

Relevant Equations

Q = V1C1
V2 = (Q1/C1) + (Q2/C2)

I have already solved up to after the switches are flipped, and all the charge is on C1. See the second attached image for a detailed diagram of the situation after the switches are flipped. However, the notes then say that all the charge is trapped between C1 and C2, which I don't understand. It says therefore that Q = Q1 - Q2. My question is shouldn't the total charge distribute between Q1 and Q2 and therefore Q = Q1 + Q2? Thank you very much for your help.

 

[haruspex]

the notes then say that all the charge is trapped between C1 and C2

The solver has chosen to write the charge on C2 as -Q2 on the left and +Q2 on the right. The total charge on the section of circuit connecting the two capacitors is therefor Q1+(-Q2). Since no charge can flow through a capacitor, this equals Q.

 

[gneill]

The problem doesn't explicitly state that the initial charge on $C_2$ is zero, but I suppose we can assume that it is so.

One approach to the problem that avoids all the pushing about of Q's is to take initially charged $C_1$ and split it into an equivalent circuit (for analysis purposes) consisting of a voltage source having the value of the initial voltage on the capacitor, and an uncharged capacitor:

Essentially you're now dealing with two initially uncharged capacitors in series and two voltage sources.

 

[hutchphd]

One approach to the problem that avoids all the pushing about of Q's is to take initially charged C1 and split it into an equivalent circuit (for analysis purposes) consisting of a voltage source having the value of the initial voltage on the capacitor, and an uncharged capacitor:

Why do you do this?? (I've never seen this technique and don't understand its utility) Please elucidate... seems confusing to me...

 

[gneill]

Why do you do this?? (I've never seen this technique and don't understand its utility) Please elucidate.

By changing the charged capacitor into separate voltage supply and uncharged capacitor you now have two uncharged capacitors in series. You can then combine them into a single equivalent (uncharged) capacitor for analysis purposes. Find the total charge that the equivalent capacitor will take on. That charge will be the amount by which the charges on the original two capacitors will change.

 

[hutchphd]

By changing the charged capacitor into separate voltage supply and uncharged capacitor you now have two uncharged capacitors in series. You can then combine them into a single equivalent (uncharged) capacitor for analysis purposes. Find the total charge that the equivalent capacitor will take on. That charge will be the amount by which the charges on the original two capacitors will change.

With apologies I don't see why this is either simpler or clearer than saying "remember charge is conserved" and "be sure to watch the signs". For me it is much more complicated.

 

[gneill]

Either way works. Choose whichever method you're familar/comfortable with.

 

[hutchphd]

My question is shouldn't the total charge distribute between Q1 and Q2 and therefore Q = Q1 + Q2? Thank you very much for your help

It is customary in any circuit loop to define a positive direction and use the same definition for each element in the circuit relative to that direction. Anything else is chaos.
I am terrible at missing signs and so I always try to look at simplified limits to make sure I haven't screwed it up. Suppose C1 =C2 and V2=0. In that case can you better reconcile the signs? Hope this all helps.

 

[gneill]

Suppose C1 =C2 and V2=0. In that case can you better reconcile the signs?

If $V_2$ is zero then the resistance R becomes irrelevant (as it's effectively short circuited). The initial charge on $C_1$ will force the transient current to flow clockwise through the circuit. So something like this represents the situation:

KVL around the loop dictates the sum of the potentials across the capacitors.

I think we'd best wait for the Original Poster to check in before giving any more hints or taking any analysis further. Can;t do his homework for him.

 

[hutchphd]

Absolutely.