Connection in parallel/series?

[mohamed el teir]

this is maybe one basic question, i drew 2 connections of capacitors, the left : connected in series. the right : connected in parallel ??

 

[axmls]

The left is in series, but the right is not in parallel. They are also in series (but they are short-circuited).

Two elements in parallel have the same starting node and the same ending node, and have the same voltage across them.

 

[mohamed el teir]

The left is in series, but the right is not in parallel. They are also in series (but they are short-circuited).

Two elements in parallel have the same starting node and the same ending node, and have the same voltage across them.

yes you are right, as the current splitted at the nodes will be constant through the 2 capacitors and the wire that doesn't carry capacitor is shorted

 

[mohamed el teir]

The left is in series, but the right is not in parallel. They are also in series (but they are short-circuited).

Two elements in parallel have the same starting node and the same ending node, and have the same voltage across them.

so is this a connection in parallel ?

 

[rumborak]

If you move the left capacitor along the wire to the top, if becomes obvious.

 

[mohamed el teir]

If you move the left capacitor along the wire to the top, if becomes obvious.

yes but i mean in this position they are in parallel also, because there are people that link parallel/series to exact position which is not totally right, the main thing that components having same current passed in them are series and others are parallel regardless of exact position

 

[phinds]

yes but i mean in this position they are in parallel also, because there are people that link parallel/series to exact position which is not totally right

No, it's not "not totally right" it is totally wrong. You can always redraw circuits to look weird

, the main thing that components having same current passed in them are series and others are parallel regardless of exact position

Exactly.

 

[Drakkith]

No, it's not "not totally right" it is totally wrong. You can always redraw circuits to look weird

Yes, like this circuit shown here: https://xkcd.com/730/

 

[mohamed el teir]

i have a question, if we have one closed loop (with no nodes connected to more than 2 wires) and no battery in it, if we put in it any components at any position, they are always in series, right ?

 

[Drakkith]

No, because you don't know how the voltage source is going to connect to the circuit. If you place the positive terminal of a battery to any point in the circuit, but not the negative terminal, then they are now in parallel. For example, you could connect the positive terminal of the battery to the circuit, the negative terminal to ground, and have a ground elsewhere in the circuit to complete the connection.

 

[mohamed el teir]

does that mean that it is not right to decide for any connection having no battery if it is in series or parallel ?

 

[rumborak]

The rule is rather simple: if they get the same voltage they're in parallel, if they get the same current they're in series.

 

[Drakkith]

does that mean that it is not right to decide for any connection having no battery if it is in series or parallel ?

It means that without a voltage or current source (or at least without knowing how the source will connect to the circuit) you cannot determine whether the two components are in parallel or series.

 

[rumborak]

The whole point of the terms parallel/series is to determine how voltages and currents go in the circuit. Without a battery to create those voltages and currents, the terms become meaningless.

 

[mohamed el teir]

It means that without a voltage or current source (or at least without knowing how the source will connect to the circuit) you cannot determine whether the two components are in parallel or series.

what about a component having a high voltage (like a charged capacitor) connected with components of lower voltage and without voltage source in a circuit of one loop, won't this highest voltage component act like a voltage source, leading to a current passing in the circuit (decreasing the voltage of the highest voltage component and increasing the voltage of others) till the sum of voltages of the other components becomes equal to the voltage of highest voltage component, i mean the other components are like to be connected in series with each other and in parallel (if we talk about them as a one unit) with the highest voltage component, and if the circuit from the start was of 2 components they will be like to be connected in parallel (one's voltage is decreasing and other is increasing till they become equal)
is this right or what

 

[rumborak]

Mohamed, I think at this point you have arrived at a complexity of your circuit that you really need to familiarize yourself with circuit analysis (e.g. Kirchhoff rules etc.). Any textbook about the subject will do.

 

[mohamed el teir]

Mohamed, I think at this point you have arrived at a complexity of your circuit that you really need to familiarize yourself with circuit analysis (e.g. Kirchhoff rules etc.). Any textbook about the subject will do.

why kirchhoff, we're just talking about one loop, and it is logical any way that current will pass till the circuit become equipotential

 

[rumborak]

I'm saying this mostly because you seem to have an aversion to treating circuit analysis mathematically, but rather want to stick to qualitative statements ("decreasing the voltage of the highest voltage component and increasing the voltage of others till the sum of voltages of the other components becomes equal to the voltage of highest voltage component"). That becomes very confusing very quickly, and that's why a rigorous mathematical treatment is the only way of figuring it out.

 

[mohamed el teir]

because these are just assumptions that I'm not sure of their correctness, i would express mathematically what i am sure of

 

[Drakkith]

what about a component having a high voltage (like a charged capacitor) connected with components of lower voltage and without voltage source in a circuit of one loop, won't this highest voltage component act like a voltage source, leading to a current passing in the circuit (decreasing the voltage of the highest voltage component and increasing the voltage of others) till the sum of voltages of the other components becomes equal to the voltage of highest voltage component, i mean the other components are like to be connected in series with each other and in parallel (if we talk about them as a one unit) with the highest voltage component, and if the circuit from the start was of 2 components they will be like to be connected in parallel (one's voltage is decreasing and other is increasing till they become equal)
is this right or what

You can't just say, "What about..." and then come up with a random circuit arrangement with random voltages and not show any arrangement of the components or how they were given these voltages. It's very, very confusing and likely to lead to misunderstandings. I don't even know whether your components are connected in parallel or series.

 

[mohamed el teir]

i will try to correct it and make it more clear :
1st case : consider a component with voltage V1 and other component with voltage V2 where V1 > V2. if they are connected in a circuit of one loop only and without voltage source : there will be current I due to the potential difference in the circuit, V1 will decrease and V2 will increase till : V1 = V2.
2nd case : consider a component with voltage Vmax and other n components with voltages V1, V2, V3, ... Vn respectively where Vmax > V1, Vmax > V2, ... Vmax > Vn. if they are connected in a circuit of one loop only and without voltage source : there will be current I due to the potential difference in the circuit where I is constant through the n components, Vmax will decrease and V1, V2, ... Vn will increase till : Vmax = V1 = V2 ... = Vn.
how true is this ?

 

[rumborak]

Mohamed, what does "a component with voltage V1" mean? Especially when you say "if they are connected in a circuit of one loop only and without voltage source". How can a component have any voltage attached to it when there is no voltage source?

I must say, I don't think you understand the basics nearly enough to do any kind of circuit analysis.

 

[mohamed el teir]

I said no voltage source in the circuit, and i don't mean a specific component component , i am talking about the idea of potential difference and current generally, anyway if you want it more specific to prevent further confusion let them all be capacitors.
EDIT : "How can a component have any voltage attached to it when there is no voltage source" LOL dude, capacitors which were before connected to voltage source till they are charged and then disconnected and reconnected with each other for example, won't they have voltage ? and i am the one who doesn't understand basics ? :D

 

[mohamed el teir]

and bro, u asked about if it is a battery or what and then when i replied that i said no voltage source in the circuit u edited your comment, and changed it as a whole, u would have better made a new comment instead, and see my EDIT above BTW

 

[rumborak]

If I understand your scenario correctly, here's my best shot at explanation:

The simple solution is, view the capacitors as batteries for the (infinitesimal) time period right after connection. Do a regular circuit analysis with that simplified circuit.
The complex solution is, a discharging capacitor can only be correctly be modeled by either analytically or numerically solving the differential equation that describes the circuit. Even the most basic scenario, an RC circuit, does not have a simple solution. It's current and voltages decay with an inverse exponential function.
This for example describes an analytic solution for an RC circuit:
http://www.math.ubc.ca/~feldman/m121/RC.pdf

 

[mohamed el teir]

@Drakkith
EDIT for the 2 cases post : assume they are capacitors, and for the 2nd case : i think it's wrong, how would voltages change in this case, and how would the current go

 

[Drakkith]

i will try to correct it and make it more clear :
1st case : consider a component with voltage V1 and other component with voltage V2 where V1 > V2. if they are connected in a circuit of one loop only and without voltage source : there will be current I due to the potential difference in the circuit, V1 will decrease and V2 will increase till : V1 = V2.
2nd case : consider a component with voltage Vmax and other n components with voltages V1, V2, V3, ... Vn respectively where Vmax > V1, Vmax > V2, ... Vmax > Vn. if they are connected in a circuit of one loop only and without voltage source : there will be current I due to the potential difference in the circuit where I is constant through the n components, Vmax will decrease and V1, V2, ... Vn will increase till : Vmax = V1 = V2 ... = Vn.
how true is this ?

It's not true in general, especially if you have capacitors. That's why you need to have a circuit diagram with everything properly labeled.

 

[mohamed el teir]

@Drakkith
I will modify case 2 and tell me how true it is :
consider a capacitor C with voltage V and charged to Q, and other n capacitors C1, C2, C3, ..., Cn which are uncharged and so have voltages = 0. if they are connected to each other in one closed loop only and without voltage source : there will be current I due to the potential difference in the circuit where I is constant through the n capacitors (and so q lost by C will be constant through the n capacitors, so that each capacitor from : C1, C2, ..., Cn will be charged to q). V will decrease and voltages of n capacitors will increase till : V = V1 + V2+ ... + Vn, more precisely : (Q-q)/C = q/C1 + q/C2 + ... + q/Cn

 

[Drakkith]

@Drakkith
I will modify case 2 and tell me how true it is :
consider a capacitor C with voltage V and charged to Q, and other n capacitors C1, C2, C3, ..., Cn which are uncharged and so have voltages = 0. if they are connected to each other in one closed loop only and without voltage source : there will be current I due to the potential difference in the circuit where I is constant through the n capacitors (and so q lost by C will be constant through the n capacitors, so that each capacitor from : C1, C2, ..., Cn will be charged to q). V will decrease and voltages of n capacitors will increase till : V = V1 + V2+ ... + Vn, more precisely : (Q-q)/C = q/C1 + q/C2 + ... + q/Cn

That looks good to me.