How is this array of capacitors connected: Serie or parallel?

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Homework Help Overview

The discussion revolves around the configuration of three capacitors, labeled C1, C2, and C3, and whether they are connected in series or parallel. Participants are examining the implications of different arrangements and the physical behavior of the capacitors based on their connections.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the configuration by visual inspection and reasoning about charge distribution. Questions arise regarding how to prove the connections using equations and the implications of labeling on the physical behavior of the circuit.

Discussion Status

The discussion is ongoing, with participants providing insights into the ambiguity of the circuit configuration. Some guidance is offered regarding the interpretation of series and parallel connections based on node selection, but no consensus has been reached on the definitive arrangement of the capacitors.

Contextual Notes

There is mention of the potential ambiguity in the circuit setup, particularly regarding how the capacitors are connected based on different node selections. Participants also note the importance of understanding charge behavior when capacitors are charged separately versus when they are connected in series.

sergioro
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Homework Statement


The figure shows three capacitors with capacitances C1, C2, and C3. How are they connected: in parallel or in series?


|------| |------|
|.....| (middle dots don't mean any thing. they help keep lines in place)
|---| |---| |---|

Homework Equations



Serie connection: 1/c = 1/c1+ 1/c2+1/c3 ...
parallel: c= c1+ c2+c3 ...

The Attempt at a Solution



I would say that anyone of the capacitors is in parallel with the other two who are in serie. I can get this by making a rotation (either counterclockwise or clockwise) of the capacitors one by one so I can always get the same configuration shown in the figure
(one on top and two in the bottom).

How can I prove it using equations.

Sergio
 
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Code:
|------| |------|
|               |    
|---| |---| |---|

If this is indeed the situation, then it's obvious just by inspection that one of the caps is in parallel with the series combo of the other two. I'm not sure what you mean by "prove it using equations." If you want to figure out how the caps are connected to each other, you do that by looking at them.
 
Thanks cepheid.

That's the array of caps (how did you make the white spaces in between the vertical lines?).

What is causing trouble is the following:
Call c1 and c2 the caps on bottom and c3 the one on top.

Let's say c2 is charged elsewhere and then inserted in the circuit in
its position at the bottom.

Following the figure, c1 will get the same charge as c2 (they are in serie) and c3 will get something else. This means that if one take the circuit appart each cap will have the computed amount of charge.

Now, If one view the same circuit in a different way, say that is c1 who is in parallel with c2 and c3, then c3 will have the same charge of c2 and c1 will get something else. This means that if one take the circuit apart the distribution of charges is interchanged between c1 and c3.

This does not look right to me. It seems as if the physical process depends on the labeling.

Sergio
 
Consider these circuits which are identical to yours. The question "which capacitors are in series/parallel" can't be answered because the circuit is ambiguous (edit: Perhaps ambiguous is the wrong word). If you add the words "between these two nodes" then it can be answered.

Depending on which two nodes you choose the following are correct answers:

(C1+C2)//C3
(C2+C3)//C1
(C3+C1)//C2

where I'm using
+ = Series
// = parallel

To an engineer your version of the circuit implies you are talking about the two vertical nodes (one on the left and right).

If the capacitors are all the same value then all three answers give same result.
 

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Last edited:
sergio_123 said:
Let's say c2 is charged elsewhere and then inserted in the circuit in
its position at the bottom.

Following the figure, c1 will get the same charge as c2 (they are in serie) and c3 will get something else.
You changed your ID?

Capacitors in series get the same charge only if they are in series while being charged. If they are charged separately and then connected in series, their final charges are unlikely going to be equal. (Though what you can say is that any charge C1 loses will be matched by C2 losing the same, if they are in series while being discharged.)
 

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