Parallel and Series Capacitors?

AI Thread Summary
The discussion focuses on understanding the arrangement of capacitors in a circuit with a potential difference of 150 V applied. Capacitors C1 and C2 are identified as being in parallel, while they are in series with capacitor C3. The concept of visualizing the circuit topology is emphasized, allowing for manipulation of the diagram without altering the circuit's equivalent capacitance. Key principles are explained, such as the total charge on parallel capacitors being the sum of individual charges and the behavior of charges in series configurations. Understanding these concepts helps clarify the physical relationships in capacitor networks beyond mere formulas.
ilovephysics16
Messages
2
Reaction score
0
Moved from a technical forum, so homework template missing
In the figure below, a potential difference V = 150 V is applied across a capacitor arrangement with capacitances C1 = 12.0µF, C2 = 6.00µF, and C3 = 16.0µF. Find the following values.

Here's the diagram: http://www.webassign.net/hrw/hrw7_25-28.gif
upload_2016-3-15_13-38-10.png

I already solved this problem but I'm having some trouble understanding the concept behind it. I know that capacitors 1 and 2 are in parallel and that they are in series with capacitor 3, but I don't understand why. Especially because capacitors 1 and 2 aren't parallel in the way I'm used to seeing parallel capacitors, like this: http://www.rapidtables.com/electric/capacitor/parallel capacitors circuit.GIF.
upload_2016-3-15_13-38-47.png

Also, if you combine 1 and 2 in parallel, what does the resulting capacitor even look like? I'm having trouble visualizing this. Can anyone help me understand this problem? I would really appreciate it.
 
Physics news on Phys.org
In your first diagram, mentally "slide" C2 along its wire to the right and down onto the vertical section at the right side. That makes C1 and C2 look more like the "standard" parallel capacitor diagram. Then "bend" the wires and slide the wire junctions around a bit, without moving them through any capacitors or breaking any connections. You should be able to make it look like a "standard" series + parallel combination.

None of the manipulations above affect the equivalent capacitance of the circuit, nor do they affect the equivalent resistance in similar problems involving resistors, because they don't change the topology of the circuit. Just don't break any wires or move any components through a junction, or a junction through a component.
 
Got it. Thanks!
 
I got a bit interested in capacitor networks because of the problems people were bringing here. Yes they often have problems through picturing them over-rigidly, the diagrams are abstract, topological, you can freely change them in a rubber sheet geometry way, i.e. with the limitations that jtbell has explained.

I think also many students end up with a purely formulaic rather than physical vision of these things based on remembering the series and parallel laws or formulae. I'm sure I did. Instead it is healthy to have some physics of it all in mind. For instance in your case the total charge on a C1 and C2 is the sum of the two charges. Because they are connected on both sides the potential across them is the same for both. So as capacitance is defined as ratio of charge to potential, sum divided by same thing, the total capacitance of 1 and 2 is the sum of the individual ones - general law for parallel capacitances.

Instead when capacitors are in series, (you have C3 in series with the effective capacitance C1 + C2) there is no adding up of charges, the charges on each capacitor is the same. Because the charge on each plate is equal in magnitude (though opposite in sign) to the plate facing it. What does add up is the potential across them of course. So same as before but the other way round, potential adds, charge same for each, this is a bit more difficult to see but V = Q/C, so for a series of potentials what adds up is not the charge - the only one that counts is one on an external plate,- but the 1/C 's. So the 1/C 's and up for capacitors in series the same way that 1/R 's add up for resistors in parallel.

Also I've found useful in solving more complicated problems to note that the conductively isolated part of the circuit has total charge 0 - that is, when charges are added algebraically, taking account of their signs. For example with two capacitors in series the isolated 'internal' plates have equal opposite charges, summing algebraically to 0. The total charge on the three connected isolated plates of your circuit is also algebraically 0. see Exercise with Capacitors
 
Last edited:
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top