Why are these capacitor in parallel?

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The discussion revolves around the classification of capacitors as being in parallel or series in a specific circuit scenario. While the initial impression may suggest they are in series, the circuit behaves as if the capacitors are in parallel when both switches are closed, allowing for equal potential across both pairs of terminals. The current flow in the circuit leads to a redistribution of charge until equilibrium is reached, with the 3μF capacitor ultimately holding three times the charge of the 1μF capacitor. The problem highlights the complexities of analyzing ideal circuits, where assumptions about instantaneous current flow and resistance can lead to confusion. Ultimately, understanding the final conditions relies on charge conservation rather than the initial configuration.
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Tiago3434 said:
It says that the capacitors are in parallel,
Using terms like Series and Parallel doesn't always necessarily help in understanding this sort of problem. As far as points a and b are concerned and, if the switches are connected with the two capacitors uncharged, you can look upon them as being in parallel (the same as if they were resistors). But that isn't how the circuit is being used; a current will be flowing around the ('series') loop. It has to be true to say that, after everything has settled down, (but not initially) the top pair of terminals will be at the same potential and the bottom two will be at the same potential (both pairs are held together)
What you have, when the switches are on, is a loop with the PD's on the two capacitors acting in a sense which will cause current to flow clockwise. Unfortunately, the problem is slightly nonsensical because, as with all 'ideal' situations, the current flow would need to be instantaneous and infinite and, in fact, the circuit would oscillate due to the inherent Inductance of the loop and radiate Energy like a loop antenna, ( analogous to two taught springs suddenly connected in the middle with one pulling and one pushing). The clockwise current will discharge the 1μF capacitor and charge it from the 3μF capacitor in the other sense until equilibrium (equal PDs) is established.
You have to assume some finite resistance in the circuit and you can only consider the final situation, after current has finished flowing. A lot of the original Energy, stored in the capacitors has 'disappeared' by being dissipated in the resistance of the wire.
 
Tiago3434 said:
It says that the capacitors are in parallel, but why? I can´t understand this point, and it certainly seems as if they are in series

they are only in parallel if both switches S1 and S2 are closed, else the capacitor on the right plays no part in the circuit

upload_2017-11-22_11-23-48.png
this wouldn't matter if they were resistors, capacitors, inductors or many other componentsDave
 

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Ah, until I saw the referenced diagram, I thought it might be down to 'Equivalent Series Resistance', which is frequency-dependent...
 
Nik_2213 said:
Ah, until I saw the referenced diagram, I thought it might be down to 'Equivalent Series Resistance', which is frequency-dependent...
The "Equivalent Series Resistance" is zero in this case, independent of frequency, which shows the nonsensical nature of the problem, as posed initially. The final condition of the circuit is independent of any resistance and just hangs on Charge Conservation. You calculate the two charges (they have different signs, however you choose to take the origin) and the net charge will be shared between the two capacitors, which will have the same PD across them. The 3μF capacitor will have three times the charge of the 1μF capacitor.
 

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