Hello,

In a word, the capacitive voltage divider formula. Oh that's 5 words sorry

I think it goes like this (C2 at the bottom):
1/Cx=1/C1+1/C2

and the voltage at the junction is:
Vj=Vs*Cx/C2

In our case, we have the differential voltage v1-v2 so Vs=v1-v2 and the rest factors into the incremental formula. Once we have the incremental voltage, we just add it to the initial voltage of C2.

But consider this. You used the initial values of the two voltages. How would you answer someone that argued that it should be the change in voltage seen by the higher-voltage capacitor, so not the inital value of v2, but the final value of v2 since that represents the actual change in voltage and hence the actual amount of charge that was transferred from one cap to the other?

 

Hi again,

Here is another interesting way to do it.

We use circuit analysis like nodal or something to calculate the time response of each cap, call them
Ft1 and Ft2.

We then have the time functions at both ends of the 'wire'.
Next, subtract Ft2 from Ft1:
Ft3=Ft1-Ft2

Now Ft3 is the voltage across the wire.
The power then is:
P=Ft3^2/R
where R is the assumed resistance of the 'wire'.

integrate the power over times zero to infinity.
That's the energy dissipated in the wire.

Fun right?

And what if the resistance of the wire is zero?

 

But consider this. You used the initial values of the two voltages. How would you answer someone that argued that it should be the change in voltage seen by the higher-voltage capacitor, so not the inital value of v2, but the final value of v2 since that represents the actual change in voltage and hence the actual amount of charge that was transferred from one cap to the other?

Hi,

Not sure what you are asking here. If we find the energy, who cares.
You can elaborate though.

 

And what if the resistance of the wire is zero?

Hi,

I am sure you are looking at:
Ft3/R

and it looks like with R being zero, we get something we cant really use.

But in many of these problems we dont allow the variable to tend to zero until the last step.
In this case, the function we end up with after integration is (R1=R):
((v2-v1)^2*((C1*C2*R1)/(2*C2+2*C1)-(C1*C2*R1*e^(-(2*T)/(C2*R1)-(2*T)/(C1*R1)))/(2*C2+2*C1)))/R1

and although we see that "/R1" division again, it cancels out with the numerator R1's so we get:
-(v2^2*C1*C2*e^(-(2*T)/(C2*R1)-(2*T)/(C1*R1)))/(2*C2+2*C1)+(2*v1*v2*C1*C2*e^(-(2*T)/(C2*R1)-(2*T)/(C1*R1)))/(2*C2+2*C1)-
(v1^2*C1*C2*e^(-(2*T)/(C2*R1)-(2*T)/(C1*R1)))/(2*C2+2*C1)+(v2^2*C1*C2)/(2*C2+2*C1)-(2*v1*v2*C1*C2)/(2*C2+2*C1)+(v1^2*C1*C2)/(2*C2+2*C1)

and if you look carefully you will no longer see any R1's as a factor or divisor. The only place you will see them is in the exponent of 'e', and then again we can reason that we can allow that R1 to be absorbed by the time variable 't' and thus we still get zero in the limit for all three exponential terms, leaving just:
(v2^2*C1*C2)/(2*C2+2*C1)-(2*v1*v2*C1*C2)/(2*C2+2*C1)+(v1^2*C1*C2)/(2*C2+2*C1)

which factors into:
((v2-v1)^2*C1*C2)/(2*(C2+C1))

and that should be the energy lost in the 'wire', regardless of resistance because we were able to eliminate R1 completely. Another interesting consequence of this is that the result will always be positive.
One more facet of this is we can spot the typical energy calculation for a capacitor:
W=C*V^2/2

with C=C1*C2/(C1+C2) [two caps in series]
and V^2=(v1-v2)^2

so this seems to give us some interesting insight into the calculation for the energy lost in the 'wire'.

 

I have been rolling this over for a while and I would like to change my opinion about this so called paradox.

The apparent discrepancy is due, simply to the inverse square law.

If you consider the charge as a mass, and the voltage as the height, then everything is working as it should, because we are taking about a potential result.......and that will depend a great deal on the height of the result.

A 50 mfd cap at 200 volts.....holds 10m Coulombs, with a potential of 1 Joule.

A 50 mfd cap at 100 volts.....holds 5m Coulombs, with a potential of 1/4 Joule.

We went down the hill haft way, but lost much more than half the energy. And we can only hold 5m Coulombs at this height.

We are not just changing the location of the charge.......we are changing it's elevation, and the potential energy capacity, depends on that elevation.

Edit: Think about this. If I had a gallon bucket with sand or water in it, the potential will depend on height.

Let's say I start at 10 ft and lower to 5 ft. The bucket size remains the same.

If I move a cap from 10 ft to 5 ft..............the bucket will shrink to 1/2. So, not only did the height change.......but the bucket size did too.

The bucket size(q), changes with the height(voltage).

Was that a good roll?

 

I have been rolling this over for a while and I would like to change my opinion about this so called paradox.

The apparent discrepancy is due, simply to the inverse square law.

If you consider the charge as a mass, and the voltage as the height, then everything is working as it should, because we are taking about a potential result.......and that will depend a great deal on the height of the result.

A 50 mfd cap at 200 volts.....holds 10m Coulombs, with a potential of 1 Joule.

A 50 mfd cap at 100 volts.....holds 5m Coulombs, with a potential of 1/4 Joule.

We went down the hill haft way, but lost much more than half the energy. And we can only hold 5m Coulombs at this height.

We are not just changing the location of the charge.......we are changing it's elevation, and the potential energy capacity, depends on that elevation.

Edit: Think about this. If I had a gallon bucket with sand or water in it, the potential will depend on height.

Let's say I start at 10 ft and lower to 5 ft. The bucket size remains the same.

If I move a cap from 10 ft to 5 ft..............the bucket will shrink to 1/2. So, not only did the height change.......but the bucket size did too.

The bucket size(q), changes with the height(voltage).

Was that a good roll?

Hi,

What paradox?

 

Apparently...when charging a cap, with another cap....there is a loss, or disappearance of 1/2 of the energy. I had to look it up.

 

Apparently...when charging a cap, with another cap....there is a loss, or disappearance of 1/2 of the energy. I had to look it up.

Hi,

Yeah that is amazing isnt it? Not intuitive at all.
So i would think that charging a battery with another battery does the same thing. I havent looked into this yet though but the battery stores 'charge' just like a cap so it seems reasonable.

 

Ah...no, a battery does not store charge like a cap. And charging a battery with a battery is completely different.

The problem with that paradox, is that the person asking the question, doesn't understand electrical potential, and capacitance.

And from all the explanations found on the net..........many don't understand and find it a mystery.

I did the same thing.......until I realized what was happening.

But I do believe I have the explanation, although I have not explained it very well.

 

Ah...no, a battery does not store charge like a cap. And charging a battery with a battery is completely different.

The problem with that paradox, is that the person asking the question, doesn't understand electrical potential, and capacitance.

And from all the explanations found on the net..........many don't understand and find it a mystery.

I did the same thing.......until I realized what was happening.

But I do believe I have the explanation, although I have not explained it very well.

Hi,

A battery works through an electrochemical process in real life.
However, in circuit analysis the battery stores charge just like a capacitor.

For example, if you go go simulate a battery in spice and you have a good battery model, it will be made up of mostly capacitance and resistance and a little inductance. So if we wrote the equations they would be similar i would think.

But look at it this way, if there are wires connecting the two batteries then there must be some loss in those wires. The equivalent movement of charge makes it look like the two capacitors connected by a wire.

 

The difference in energy loss between charging a battery and charging a cap, is that the battery charges at a relatively constant voltage, fairly independent of its charge state, whereas the capacitor voltage is proportional to its charge, thus charging a battery from a fixed voltage source is more efficient.
The reason is, as the capacitor charges, its voltage also rises, with energy stored equal to the voltage at each instance times the charge transferred at each instance.

Thus if you try to charge a cap from a fixed voltage (either a supply or another cap) the energy difference between each increment of charge transferred from the source and stored on the cap is lost (dissipated).

So if you charge a cap starting from zero volts from a fixed voltage source, the energy provided by the source is the source voltage times the charge transferred, but due to the incremental change in the stored energy with capacitor voltage, only 1/2 of the source energy ends up stored on the cap.

Now if you charge the cap from an efficient constant-current source, such a switching regulator, the energy from the source will then be close to the energy stored on the cap (minus any regulator inefficiency).

Make sense?

 

Hello,

Are we still trying to find the energy lost in the wire?

For two batteries connected together so that one charges the other, if we consider the voltages to be constant (which they arent as anyone who works with batteries on a daily basis knows) then we have a simple circuit with two voltage sources V1 and V2 where V1 is higher than V2 so we have a different voltage:
Vd=V1-V2

and if the resistance is say 2 ohms, we have current:
i=Vd/2

or just power:
P=Vd^2/2

The energy is then:
W=T*Vd^2/2

where T is the time to charge.

So it is a little simpler than the caps. It's a little different because we are stopping the charge independent of the voltages.

Now the better example i think would be to have one battery slightly higher voltage than the other, and allow the voltage to change as the both pass current. I havent done that yet though.
Interestingly, i think the circuit then would look like two capacitors with offset voltages not part of the real battery, and the two caps exchange energy as the voltages change slightly over time. Once the two voltages are the same, we're done.

[LATER]
Ok it looks the same. The voltage of the one cap comes up to 100.4975 as before.
The intuition about this might be that the offset voltages act the same as the initial condition generator sources, so they appear in the same place as those.
But also, the loss is based on the differential voltage not the absolute voltages.
The circuit was such that one battery was smaller than the other and the smaller charging the larger and allowed to go down in voltage also.
Now if we try to maintain voltages to nearly the same, then we go back to the constant voltages model.