Capacitor Network - Series or Parallel?

AI Thread Summary
The discussion centers on the configuration of capacitors, where the textbook solution identifies capacitors 1 and 2 as parallel, and 3 and 4 as parallel, leading to an overall series connection. A participant suggests that capacitors 1 and 3, as well as 2 and 4, are in series instead. However, it is clarified that series connections require components to carry the same current, which is not the case here due to the lack of direct connections between the proposed series pairs. The importance of identifying connected nodes and potential differences is emphasized, suggesting that re-drawing the circuit can clarify the relationships. Ultimately, the correct configuration remains as stated in the textbook, with (1 P 2) S (3 P 4) being the accurate representation.
Shreya
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Homework Statement
Find the equivalent capacitance.
Relevant Equations
Series & Parallel Capacitor formulae
My textbook solution states that 1 & 2 are in parallel and so is 3 & 4 and those 2 are in series. That is, (1 P 2) S (3 P 4). My thinking is such: points A & B are of same potential, say V, C & D are of same potential, say x and E & F are are of same potential, say 0. So I can say that 1 and 3 are in series and so is 2 &4. Therefore, I can say (1 S 3) P (2 S 4). I'm sure there is a flaw in my method. Please be kind to point it out.
 

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Shreya said:
Homework Statement:: Find the equivalent capacitance.
Relevant Equations:: Series & Parallel Capacitor formulae

My textbook solution states that 1 & 2 are in parallel and so is 3 & 4 and those 2 are in series. That is, (1 P 2) S (3 P 4). My thinking is such: points A & B are of same potential, say V, C & D are of same potential, say x and E & F are are of same potential, say 0. So I can say that 1 and 3 are in series and so is 2 &4. Therefore, I can say (1 S 3) P (2 S 4). I'm sure there is a flaw in my method. Please be kind to point it out.
C and D are only guaranteed to be at the same potential because there is a wire connecting them.

If you had (1 S 3) P (2 S 4) then there would be no wire connecting the middle nodes in the two series chains and there would be no guarantee that the two middle nodes would be at the same potential.

If you had such a guarantee (say because the two series chains had capacitors with identical ratios of capacitance) then the two middle nodes would be at the same potential and the circuit would behave identically to the series chain of two parallel pieces (1 P 2) S (3 P 4).

It might be helpful to re-draw the circuit to eliminate the redundant nodes B, D and F (if two nodes are connected by a wire, there is only one node). Then the true nature of the circuit becomes more obvious.
 
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That makes sense @jbriggs444 . But I don't see why 1 P 2 would be in series with 3 P 4 though
 
Shreya said:
That makes sense @jbriggs444 . But I don't see why 1 P 2 would be in series with 3 P 4 though
There is no resistance between A and B, C and D, or E and F. You can combine each pair into a single node. The nature of the circuit will become clear when you do this.
 
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Shreya said:
That makes sense @jbriggs444 . But I don't see why 1 P 2 would be in series with 3 P 4 though
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Shreya said:
Homework Statement:: Find the equivalent capacitance.
Relevant Equations:: Series & Parallel Capacitor formulae

So I can say that 1 and 3 are in series and so is 2 &4. Therefore, I can say (1 S 3) P (2 S 4). I'm sure there is a flaw in my method. Please be kind to point it out
Two components are in series only if they carry the same current at any time.
That is not the case with 1 and 3 (or 2 and 4).
 
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