Capacitors connected in series?

[mohamed el teir]

suppose 2 capacitors C1, C2 connected in series with battery V, let charge in circuit be Q, since they are connected in series so each capacitor will have Q also, suppose we disconnected them from the battery and reconnected them again in series (without connecting voltage source or Earth with them), as total Q in capacitors in the old connection is 2Q so this is the charge in the new connection, and since the new connection is also in series so each capacitor will have 2Q (double the charge in the old connection), is this right ? and if not right, what's exactly the wrong thing in what i have written ?

 

[rumborak]

No, that's incorrect. As a general rule, the charges in a circuit are a "zero-sum" entity. Electrons aren't inserted somewhere, they are always just moved around from one spot to another. When you disconnect the battery, yes, the plates will be charged with a certain Q, but the plate of the second capacitor that was connected to the other terminal of the battery has a charge of -Q. When you just connect the two terminals (without the battery inbetween), the Q and -Q will equalize again (probably with a nice spark), and you have zero Q everywhere again.

 

[mohamed el teir]

No, that's incorrect. As a general rule, the charges in a circuit are a "zero-sum" entity. Electrons aren't inserted somewhere, they are always just moved around from one spot to another. When you disconnect the battery, yes, the plates will be charged with a certain Q, but the plate of the second capacitor that was connected to the other terminal of the battery has a charge of -Q. When you just connect the two terminals (without the battery inbetween), the Q and -Q will equalize again (probably with a nice spark), and you have zero Q everywhere again.

i read something demonstrated example like this but instead of reconnecting them in series it reconnected them in parallel with plates of equal sign together, it said that Q before = Q after = 2Q then used the constant V to get the charge on each plate, may you explain to me by calculations the difference between the 2 situations ?

 

[CWatters]

For each capacitor...

Q = C*V

Connect two in parallel +ve to +ve and you get

2Q = 2C*V

The 2's cancel.

 

[CWatters]

Connect the two +ve to -ve and you have a problem...

Lets assume the wire used to connect them has resistance R. We can make R small later..

The initial current around the loop will be

I = 2V/R

If R is very small then I is very large and the capacitors discharge very quickly.

 

[rumborak]

If, after disconnecting from the battery, you also disconnected the capacitors and crossed over their terminals, so that +Q is connected to the other +Q (and -Q to -Q), yes you would end up with a 2Q surplus on the wire segment (still only 1Q on each plate though). Interestingly you could then connect the battery again and repeat the procedure, always adding 2Q to the system.

 

[CWatters]

Perhaps it's easier to consider two different capacitors C₁ and C₂?

In series connected to the battery the following equations apply..

Vbat = V₁ + V₂
Q = C₁V₁
Q = C₂V₂

If you know C₁ and C₂ you can solve to get V₁, V₂. and Q

Then if you disconnect them and reconnect them in parallel you can write..

2Q = V (C₁+C₂)

and solve to calculate the new voltage on both.

In practice a large current would flow to equalise the charge and voltage. Some energy might be lost heating the wire.

Finally if the capacitors were identical C₁ = C₂ then the final voltage ends up as Vbat/2.

 

[CWatters]

If, after disconnecting from the battery, you also disconnected the capacitors and crossed over their terminals, so that +Q is connected to the other +Q (and -Q to -Q), yes you would end up with a 2Q surplus on the wire segment...

No that's totally wrong. There is no "surplus on the wire segment".

 

[jerromyjon]

That sounds impossible because you could have infinite self-charging potential.

 

[rumborak]

Uhm, pretty sure it is correct. If you disconnect the terminals of a charged capacitor, it will have a surplus charge on its plates (after all, that's what a charged capacitor is). When you then connect it to another capacitor that has the same charges on its plates (connect +Q to +Q), they will have the same potential, thus no current will flow on connection. Ergo, one wire segment has a 2Q surplus, the other a -2Q deficit.

That sounds impossible because you could have infinite self-charging potential.

You always have to reconnect the battery, which then does the work to charge the capacitors even further. You're not getting energy for free, if that's what you were implying.
As a side comment, a Van der Graaf generator does the same thing, just through a different mechanism.

 

[CWatters]

Uhm, pretty sure it is correct. If you disconnect the terminals of a charged capacitor, it will have a surplus charge on its plates (after all, that's what a charged capacitor is). When you then connect it to another capacitor that has the same charges on its plates (connect +Q to +Q), they will have the same potential, thus no current will flow on connection. Ergo, one wire segment has a 2Q surplus, the other a -2Q deficit.

Ok I think I see what you are saying..

If the two capacitors are identical then when in series they will have the same charge Q and voltage V. When reconnected in parallel the end result looks like one big capacitor = 2C with charge on it = 2Q.

However if you reconnect them to the battery in series no charge will flow because there is no voltage difference.

 

[rumborak]

Exactly. Then you can open one side again and insert the battery, which then charges the whole thing further.

 

[CWatters]

I think my edit crossed with your post.

Exactly. Then you can open one side again and insert the battery, which then charges the whole thing further.

No because..

...if you reconnect them to the battery in series no charge will flow because there is no voltage difference.

 

[rumborak]

The lack of voltage difference is the very point! It is then the battery which adds a fresh voltage difference.

I think you guys get tripped up by the idea that you could somehow get "free energy" out of this. Far from it. All you're doing is slowing piling energy onto the capacitors, but the energy is provided by the battery.
Just like the Van der Graaf generator, you are physically carrying the charges from one reservoir to another, in this device by disconnecting the wires and reconnecting them crossed over.

 

[CWatters]

No I wasn't thinking this was a free energy claim.

The lack of voltage difference is the very point! It is then the battery which adds a fresh voltage difference.

That can only happen if the sequence as per this diagram. Note one capacitor is discharged in step 3. If that's not the sequence you mean please post your own version of the drawing..

 

[rumborak]

I don't agree with that sequence. At what point, and why, should that capacitor on the right discharge?

 

[CWatters]

Oops There is an error on that drawing. Step 2 the voltage should be vbat/2 not vbat.

 

[CWatters]

I don't agree with that sequence. At what point, and why, should that capacitor on the right discharge?

Because the current flow will be anti clockwise.

 

[CWatters]

Updated drawing...

 

[CWatters]

Because the current flow will be anti clockwise.

Whichever way around the battery is inserted one of the caps will be discharged.

 

[rumborak]

But, how could one cap be charged and the other completely discharged when they are connected by the bottom wire?

Don't get me wrong, I'm not completely sure at this point anymore either! :D

 

[rumborak]

Nah, I think I'm still correct. I think the problem is that you are conflating potentials with charges in your drawing.

 

[rumborak]

The next iteration would give a 3Q and -1Q combo, then 4Q and -2Q, etc.

 

[rumborak]

Mohamed, can you produce a more cleaned up version? It probably makes sense to you, but for an outsider its virtually indecipherable. All I see are wild formulas everywhere.

 

[mohamed el teir]

:D

 

[rumborak]

Case 1 is definitely wrong, as that configuration would discharge itself immediately. So is case 3.
Also, how do you arrive at two different results for the same configuration?

 

[mohamed el teir]

#2 is definitely wrong, as that configuration would discharge itself immediately.

you mean second capacitor or case 2 ??

 

[rumborak]

Sorry, I edited my post. I meant "Case 1". That configuration is a short circuit of two charged capacitors.

But my other question is: How can case 3 be the same configuration, but you get a different result?

 

[mohamed el teir]

Sorry, I edited my post. I meant "Case 1". That configuration is a short circuit of two charged capacitors.

But my other question is: How can case 3 be the same configuration, but you get a different result?

concerning case 1 : do you mean that the battery is necessary for the system to make its charge Q ?
concerning case 3 : exchanged the plates of one capacitor of them and made it negative with respect to the other capacitor

 

[rumborak]

Sorry, had not noticed the reversal of charges in case 3. But still, case 1 you can't really calculate like that. The formulas you are using presume ideal components; at the moment of your case 1 picture, you are short-circuiting ideal components. Meaning you have infinite current, and an unaccounted-for potential difference between the capacitors.