Question: Are the capacitors in series or parallel in this circuit?

[jsmith613]

We are shown a circuit diagram as in attachement
It shows an open switch and two capacitors (both of which have been charged fully in previous experiments

Question: The capacitors are joined as shown in the circuit. When switch S is closed the voltage across both capacitors is the same. Why?

The answer says: they are in parallel

Surely this is wrong? They are in series, no?
As the charge can only follow one path it must be series, right?

Thanks

 

[Dale]

When two components are in parallel they have the same voltage across them, and when they are in series they have the same current through them. So actually, they are both in series and in parallel. I.e. for a circuit this simple there is no difference between series and parallel.

 

[jsmith613]

When two components are in parallel they have the same voltage across them, and when they are in series they have the same current through them. So actually, they are both in series and in parallel. I.e. for a circuit this simple there is no difference between series and parallel.

surely as the capacitance of the two is different the p.d cannot be the same

would you agree though they are in series...

 

[phinds]

surely as the capacitance of the two is different the p.d cannot be the same

Yes, but that is a temporary artifact of the initial conditions and is irrelevant to whether they are in series are parallel.

I agree w/ DaleSpam

 

[jsmith613]

Yes, but that is a temporary artifact of the initial conditions and is irrelevant to whether they are in series are parallel.

I agree w/ DaleSpam

I don't understand why they would acquire the same p.d
If we have a series circuit with resistors...the p.d is NOT the same across all of them if they are different...please could you explain how it is possible for p.d to be the same?

note: they both store the same charge as they are in series...Q=CV therefore V cannot be the same for the two components

 

[jsmith613]

Yes, but that is a temporary artifact of the initial conditions and is irrelevant to whether they are in series are parallel.

I agree w/ DaleSpam

Here is the direct question
I realize I may not have been so clear

 

[Dale]

I don't understand why they would acquire the same p.d

They must acquire the same p.d. A wire has the same potential everywhere, that is the basic purpose of a wire. So the top wire is one potential and the bottom wire is another potential. So there is the same potential between every pair of points between the top and bottom wires, including across each capacitor. That is why parallel components always have the same voltage across them, the wires constrain it to be that way.

note: they both store the same charge as they are in series...Q=CV therefore V cannot be the same for the two components

They do not store the same charge. Being in series means that they have the same current go through them, which means that the change in the charge is the same. That is not at all the same as having the same charge unless they both start uncharged or with the same charge.

 

[jsmith613]

They must acquire the same p.d. A wire has the same potential everywhere, that is the basic purpose of a wire. So the top wire is one potential and the bottom wire is another potential. So there is the same potential between every pair of points between the top and bottom wires, including across each capacitor. That is why parallel components always have the same voltage across them, the wires constrain it to be that way.

I think this makes sense...however, if they were on the SAME side of the wire (i.e looked like this)

______
|
|
C
|
|
C
|
______
(i cannot do the last side...C = capacitor)

Would the potential across each capacitor be different now because they have different capacitance valeus or would they still be the saem. If so, why?

They do not store the same charge. Being in series means that they have the same current go through them, which means that the change in the charge is the same. That is not at all the same as having the same charge unless they both start uncharged or with the same charge.

Ok, thanks

 

[sophiecentaur]

We are shown a circuit diagram as in attachement
It shows an open switch and two capacitors (both of which have been charged fully in previous experiments

Question: The capacitors are joined as shown in the circuit. When switch S is closed the voltage across both capacitors is the same. Why?

The answer says: they are in parallel

Surely this is wrong? They are in series, no?
As the charge can only follow one path it must be series, right?

Thanks

This is yet another thread about 'Definitions'. There is absolutely no need to get aerated about which things fit into which category as long as you can grasp the underlying relationships. Kirchoffs laws do not explicitly use the words 'series' or 'parallel' because they are not really relevant. As long as you remember the basic idea of a node or a loop, everything sorts itself out.

The distinction between series and parallel in situations like this is actually about as relevant as whether Pluto is or isn't a planet. Pluto doesn't care what you call it - it still has the same orbit!

 

[jsmith613]

This is yet another thread about 'Definitions'. There is absolutely no need to get aerated about which things fit into which category as long as you can grasp the underlying relationships. Kirchoffs laws do not explicitly use the words 'series' or 'parallel' because they are not really relevant. As long as you remember the basic idea of a node or a loop, everything sorts itself out.

The distinction between series and parallel in situations like this is actually about as relevant as whether Pluto is or isn't a planet. Pluto doesn't care what you call it - it still has the same orbit!

ok...thanks
could you take a look at my post in #8
thanks

also I am not sure I do know the difference between a node or loop (or even what you mean by them)
edit: yes I do...just looked it up :)

and therefore this explains the answer to post #8...it makes no difference
this was very useful...thanks for your helkp :)

 

[sophiecentaur]

ok...thanks
could you take a look at my post in #8
thanks

also I am not sure I do know the difference between a node or loop (or even what you mean by them)

A node is a part of a circuit where two or more components are joined together. A loop is a path that you can trace on a circuit with your finger, through any number of components that will bring you back to the same point.
In this simple case, points where the two capacitors are joined are nodes but you can get many components coming together at a node in some circuits (allowing many 'parallel paths'). Likewise, loops can involve dozens of components ('in series').
The actual layout of a circuit - how it's drawn on paper, makes no difference to how it will behave.

 

[jsmith613]

A node is a part of a circuit where two or more components are joined together. A loop is a path that you can trace on a circuit with your finger, through any number of components that will bring you back to the same point.
In this simple case, points where the two capacitors are joined are nodes but you can get many components coming together at a node in some circuits (allowing many 'parallel paths'). Likewise, loops can involve dozens of components ('in series').
The actual layout of a circuit - how it's drawn on paper, makes no difference to how it will behave.

yes I just realized this...thanks so much :)
problem solved :)

 

[sophiecentaur]

Once you get onto Kirchoff, things will probably make more sense.
'Early Science' is littered with definitions and classifications that confuse people who actually try to THINK about it. This is obviously your problem and it used to be mine too. I just don't care, these days haha.

 

[Dale]

I think this makes sense...however, if they were on the SAME side of the wire (i.e looked like this)

______
|
|
C
|
|
C
|
______
(i cannot do the last side...C = capacitor)

Would the potential across each capacitor be different now because they have different capacitance valeus or would they still be the saem. If so, why?

They would still be the same. Why would they be different? You haven't changed the circuit any, just drawn it differently. They still have a wire connecting one side and another wire connecting the other sides and so all of my description above still applies exactly the same.

Btw, sophiecentaur mentioned loops and nodes. If you haven't gotten there yet then once you are exposed to those it should clear things up quite a bit. My preferred method is the node voltage analysis.

 

[jsmith613]

yes when this was pointed out to me I realized how foolish I was being...thanks for your support :)

 

[Dale]

No worries. Honestly, if they haven't given you Kirchoff's laws then I don't think that they should be having you solve circuits, but that is just my viewpoint (apparently not shared by your professor).

 

[sophiecentaur]

No worries. Honestly, if they haven't given you Kirchoff's laws then I don't think that they should be having you solve circuits, but that is just my viewpoint (apparently not shared by your professor).

I totally agree, DS. Too many teachers seem to rely on rote definitions and stock answers rather than encouraging some deeper understanding. I think that, in many cases, it's because they are not too sure of some topics. To be fair, Electricity seems to give more trouble than other topics. Something to do with the abstract logic, lack of opportunity for 'purple passages' in descriptions and the need for a Maths approach from the start, probably.