Capacitors in Series: Voltage, Capacitance & Charge

[energychaser]

what should be a simple capacitor question.

Hello,

I have a question I have been pondering for a bit, me thinks it should be simple, but I cannot seem to get it straight.

Imagine two capacitors leg to leg.

each has 10 farads.

Each is charged to 100 volts.

Each should have a charge of 1000 coulombs.

Now we reduce the capacitance of one of the capacitors to 1 farad.

by way of equation Q=VC, and knowing that charge is conserved, we know that this capacitor now stands at V=1000 F=1 C=1

so now we have two capacitors one is at V=1000 C=1 Q=1000
The other at V=100 C=10 Q=1000.

Because of the voltage difference, they will want to equalize. What do they end up at
for V,C and Q?

 

[dlgoff]

Welcome to PF.

You store less charge on series capacitors than you would on either one of them alone with the same voltage!

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html"

I don't know if this helps. I'll check back tomorrow to see.

Good night

 

[energychaser]

so...

if I had 2 capacitors in series equaling 10 farads, at 100 volts, I would have a net of 1000 coulombs.

If I looked at a single capacitor of the above the same voltage, say 20 farads at 100v I would have 2000 coulombs.

I don't see how this helps me calculate...

I really learn better if I can see the answer and how it was derived...then re wiork it, if this is at all possible I would really appreciate it!

 

[diazona]

by way of equation Q=VC, and knowing that charge is conserved, we know that this capacitor now stands at V=1000 F=1 C=1

This is where you might be going wrong. If these capacitors are connected in a complete circuit, the amount of charge on each capacitor won't necessarily stay the same while you reduce the capacitance from 10F to 1F. Charge will be able to leave or enter each capacitor through the circuit. What happens to the capacitors will depend on what else there is in the circuit.

On the other hand, if the capacitors are not connected to anything else (except each other), then there's no path for charge to flow to or from the capacitors. So the amount of charge in each place would remain the same, and the voltage would have to change, as you said. This would be a relatively unusual situation, though. (Usually capacitors in physics problems are in circuits)

 

[energychaser]

There is nothing else in the circuit, simply capacitors connected leg to leg. Here are the parameters for capacitance, voltage, charge and energy on each.

C1= 1 C2=10
V1=1000 V2=100
Q1=1000 Q2=1000
j1=500,000 j2=50,000

Before connected we have a total of 550,000 joules.
After they are connected, they will settle to a new lower energy state.

According to my simulator, they will settle to 108.108 on each.
So we will haveC1=1 C2=10
V1=108.1 V2=108.1
Q1=108.1 Q2=1081
j1=5842.8 j2=58428

So obviously charge is not constant. It looks as though the only constant is the capacitance at 1 and 10, both voltage and charge change after they settle. So how do you figure this out? I cannot remember much of my calculus and am attempting to do so, could someone help point the way? How can I come to the answer of 108.108?

 

[diazona]

Are you talking about a circuit like this?

 

[energychaser]

Are you talking about a circuit like this?
View attachment 27774

Yes! (why didnt I think of that, I should have been more clear) that is exactly what I meant,

 

[diazona]

OK, well think about this: you know that the voltage across both capacitors at any given time must be equal, and you know that the total charge on the top plates can't change (or equivalently, that the total charge on the bottom plates can't change). Putting those together with the relation Q = CV for each capacitor will allow you to find the final values of charge and voltage on both capacitors using simple algebra.

 

[energychaser]

OK, well think about this: you know that the voltage across both capacitors at any given time must be equal, and you know that the total charge on the top plates can't change (or equivalently, that the total charge on the bottom plates can't change). Putting those together with the relation Q = CV for each capacitor will allow you to find the final values of charge and voltage on both capacitors using simple algebra.

I must admit I am quite confused, and have spent several hours in front of a piece of paper trying to figure this out...should not be this difficult!

I have never known that you could take the charge on one plate alone as a consideration, would not total charge always be zero then as top and bottom cancel? If I am taking into account charge on one plate, how come I take the final values of charge and voltage for each... would you possibly solve the problem for me and let me follow along?

Im sure you have been a great help, however I am having a dense day!

 

[energychaser]

I must admit I am quite confused, and have spent several hours in front of a piece of paper trying to figure this out...should not be this difficult!

I have never known that you could take the charge on one plate alone as a consideration, would not total charge always be zero then as top and bottom cancel? If I am taking into account charge on one plate, how come I take the final values of charge and voltage for each... would you possibly solve the problem for me and let me follow along?

Im sure you have been a great help, however I am having a dense day!

(ps, I had tried that approach (but am sure I did something wrong) and came up with this. Knowing both capacitors together are 11c, with a combined charge of 2000, the final voltage on them would then be 181, which is not correct according to my simulator!)

 

[energychaser]

(ps, I had tried that approach (but am sure I did something wrong) and came up with this. Knowing both capacitors together are 11c, with a combined charge of 2000, the final voltage on them would then be 181, which is not correct according to my simulator!)

I think perhaps the simple circuit simulator I am using is wrong! could someone please check my numbers and see?

http://webphysics.davidson.edu/applets/circuitbuilder/tutorial/circuitbuilder_intro.htm

Thanks again for all the kind help.

 

[energychaser]

wow, quite sure the applet was way wrong...I spent all day trying to figure out something i may have done right.


if:
C1= 1 C2=10
V1=1000 V2=100
Q1=1000 Q2=1000

Then the combined capacitor will have a C=11 and 2000 coulombs of charge. This equates to 181.81 volts across it,

thus each individual capacitor will also have 181.81 volts across it. for the first cap at C=1, this equates to 181.81 coulombs.

for the second cap at C=10, this equates to 1818.18 coulombs of charge.

These two combined give 1999.99 coulombs of charge, meaning charge WAS conserved, while the applet indicated it for some reason was not!

Am I correct in this?

 

[diazona]

Looks correct to me. Charge conservation is a pretty fundamental law, so if it seems to be broken, something is probably wrong.

How exactly did you get those results from the simulator?

 

[dlgoff]

Here's a couple of calculators to play with.
For your capacitance; parallel in your case:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html"

For stored energy in a capacitor:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1"

 

[energychaser]

Here's a couple of calculators to play with.
For your capacitance; parallel in your case:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html"

For stored energy in a capacitor:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c1"

Thanks!

I was quite perplexed, I had actually solved the problem in about 15 minutes, and went to check with that online applet only to discover I was wrong...I have a decent knowledge base from which to draw, and knew it should not involve calculus, and there should not be two varying variables, and was really grasping to find out what happened, glad to know it was not me!