You're assuming the voltage will become halved by connecting an equal capacitor in parallel with the first. What you must assume is that the energy will be shared between the caps. From that you can determine what the voltage across the caps will be.
First situation with one cap charged:
U1=0.5*C*V1*V1
Second situation where an uncharged capacitor of equal value is added in parallel to the first:
U2=0.5*2*C*V2*V2
Assuming U1=U2 (conservation of energy), it follows that V2=V1*sqrt(0.5)
So if you had 2V across a single cap and added another uncharged cap of equal value in parallel the new voltage across those caps would be 1.414V.
First situation with one cap charged:
U1=0.5*C*V1*V1
Second situation where an uncharged capacitor of equal value is added in parallel to the first:
U2=0.5*2*C*V2*V2
Assuming U1=U2 (conservation of energy), it follows that V2=V1*sqrt(0.5)
So if you had 2V across a single cap and added another uncharged cap of equal value in parallel the new voltage across those caps would be 1.414V.
Yes, and your point?
Let me rewrite the equation.
0.5*C*(Q1/C)^2 = 0.5*(2C)*(Q2/(2C))^2
What does that boil down to?
Let me rewrite the equation.
0.5*C*(Q1/C)^2 = 0.5*(2C)*(Q2/(2C))^2
What does that boil down to?
Well... if the charge is halved... will not the voltage be halved?
Conservation of charge of charge and energy must both hold.
🙂
Conservation of charge of charge and energy must both hold.
🙂
Here's what it boils down to.
(Q1^2)/C = (Q2^2)/(2*C)
which can be simplified to
Q2 = Q1*sqrt(2)
(Q1^2)/C = (Q2^2)/(2*C)
which can be simplified to
Q2 = Q1*sqrt(2)
So now we have the following equations:
Q2 = Q1*sqrt(2)
and
V2=V1*sqrt(0.5)
C=Q1/V1, right?
from the equations 2C = Q2/V2 = (Q1*sqrt(2))/(V1*sqrt(0.5)) = 2*(Q1/V1)
Is that not right, or am I missing something?
Q2 = Q1*sqrt(2)
and
V2=V1*sqrt(0.5)
C=Q1/V1, right?
from the equations 2C = Q2/V2 = (Q1*sqrt(2))/(V1*sqrt(0.5)) = 2*(Q1/V1)
Is that not right, or am I missing something?
The energy is lost moving the electrical charge from the source (the charged cap.) to destination (the fully discharged cap)
How is that possible if the capacitors are ideal? In reality you are correct, some energy will be lost to resistances and some to heat and light because there will be a nice spark if you connect a discharged cap across a charged cap 🙂
Hey BWRX,
No one is screwing up the math here... what has to screwed up is the assumptions being made.
Consider this: a lake and a dock at exactly the same height. Pull a bucket of water from the lake and set it on the dock. Now... we have 1 bucket-mass (B) moved an average distance of 1/2 the bucket's height (H)... ie work or potential energy... B x 1/2 H amount. Now... do it again... twice the energy (this is the final state in the cap quandry... we have 2B x 1/2H. Then... put one bucket atop the other the other. The end result is we have 2B x H (average height of mass) worth of potential energy.
Now it is easy to see why we had the add energy to put the 2nd bucket on top... and how the total energy doubled when we did so. Now return the top bucket to the dock... it gives up energy when doing so.
So in the cap world we lost half the energy... makes perfect sense. Nothing can be violated here... we just aren't seeing where the energy in our cap went.
No one is screwing up the math here... what has to screwed up is the assumptions being made.
Consider this: a lake and a dock at exactly the same height. Pull a bucket of water from the lake and set it on the dock. Now... we have 1 bucket-mass (B) moved an average distance of 1/2 the bucket's height (H)... ie work or potential energy... B x 1/2 H amount. Now... do it again... twice the energy (this is the final state in the cap quandry... we have 2B x 1/2H. Then... put one bucket atop the other the other. The end result is we have 2B x H (average height of mass) worth of potential energy.
Now it is easy to see why we had the add energy to put the 2nd bucket on top... and how the total energy doubled when we did so. Now return the top bucket to the dock... it gives up energy when doing so.
So in the cap world we lost half the energy... makes perfect sense. Nothing can be violated here... we just aren't seeing where the energy in our cap went.
I don't think anyone has this exactly correct. It can be shown that the charge shared between the 2 capacitors is conserved, but energy is lost because the circuit has some finite resistance. Assume initially C1 is charged to Uinit and C2 is charged to 0. After the switch connects the two caps in parallel, the voltage across them is Ufinal.
Before the switch is closed, Q1=C1*Uinit.
After the switch is closed, Qfinal=(C1+C2)*Ufinal.
From these equations: Ufinal=Uinit*C1/(C1+C2).
The initial energy is Einit=.5*C1*Uinit^2
The final energy is Efinal=.5*(C1+C2)*Ufinal^2
The energy lost is Elost=.5*C1*Uinit^2-.5(C1+C2)*Ufinal^2.
The energy is lost in the resistance; although low it will never be zero.
Before the switch is closed, Q1=C1*Uinit.
After the switch is closed, Qfinal=(C1+C2)*Ufinal.
From these equations: Ufinal=Uinit*C1/(C1+C2).
The initial energy is Einit=.5*C1*Uinit^2
The final energy is Efinal=.5*(C1+C2)*Ufinal^2
The energy lost is Elost=.5*C1*Uinit^2-.5(C1+C2)*Ufinal^2.
The energy is lost in the resistance; although low it will never be zero.
Mind you... all the classical equations puke with divide by zero errors. This does imply infinity squared over zero (l'hopital and all that) which is undefined? Who remembers that stuff?
BTW...
SY is in Austria I believe... he's drunk already.
🙂
BTW...
SY is in Austria I believe... he's drunk already.🙂
So basically one of the conservation rules has to break down for this to work with ideal capacitors, right? Isn't that what we showed with our equations? Just take your pick whether it be conservation of charge or energy.
Otherwise we can obviously see where the energy goes if we short a discharged cap across a charged cap.
If you don't buy that then maybe this is how dark matter is created 😉
Otherwise we can obviously see where the energy goes if we short a discharged cap across a charged cap.
If you don't buy that then maybe this is how dark matter is created 😉
poobah,
I'm not sure who your last post was directed too, but dividing zero is avoided by assuming some finite resistance. Notice that the resistor value doesn't even appear in these equations. You caneasily solve the differential equations of the equivalent RC circuit. It's easier to solve these, if you recall that a cap with an initial voltage can be represented by a voltage source is series with the cap. If you don't like the math, try a simulation.
Rick
I'm not sure who your last post was directed too, but dividing zero is avoided by assuming some finite resistance. Notice that the resistor value doesn't even appear in these equations. You caneasily solve the differential equations of the equivalent RC circuit. It's easier to solve these, if you recall that a cap with an initial voltage can be represented by a voltage source is series with the cap. If you don't like the math, try a simulation.
Rick
sawrey,
We are the same page here. The quandry is about zero resistance... which ain't real. Half the energy MUST leave the system... we are just not seeing where. I'm thinking we can't assume zero resistance.
You can't put the second bucket back down on the dock without removing energy from it.
Now, zero resistance is possible... this would leave inductance as the place for the "missing" energy. I know of nothing with zero inductance.
thinking out loud here...
We are the same page here. The quandry is about zero resistance... which ain't real. Half the energy MUST leave the system... we are just not seeing where. I'm thinking we can't assume zero resistance.
You can't put the second bucket back down on the dock without removing energy from it.
Now, zero resistance is possible... this would leave inductance as the place for the "missing" energy. I know of nothing with zero inductance.
thinking out loud here...
poobah,
If you wish to solve this problem assuming R and L are both zero, proceed as follows:
A capacitor (C1) with an initial voltage can be modeled and an IMPULSE of current (of amplitude Uinit*C1) in parallel with C1. Now if at time zero, this impulse is applied to the parallel combination of C1 and C2 you get Ufinal=Uinit*C1/(C1+C2). I don't care much for impulses, but the fact is that it is very esay to imagine a circuit where a very fast pulse 'look like' an impulse to a circuit with a long time constant. So if you replace the impulse above with a narrow large current spike and assume dU/dt =I/C you can easily compute Ufinal.
Rick
If you wish to solve this problem assuming R and L are both zero, proceed as follows:
A capacitor (C1) with an initial voltage can be modeled and an IMPULSE of current (of amplitude Uinit*C1) in parallel with C1. Now if at time zero, this impulse is applied to the parallel combination of C1 and C2 you get Ufinal=Uinit*C1/(C1+C2). I don't care much for impulses, but the fact is that it is very esay to imagine a circuit where a very fast pulse 'look like' an impulse to a circuit with a long time constant. So if you replace the impulse above with a narrow large current spike and assume dU/dt =I/C you can easily compute Ufinal.
Rick
You know...
That paper makes alot of sense; and it seems to say that you can't have both zero inductance and resistance...
That paper makes alot of sense; and it seems to say that you can't have both zero inductance and resistance...
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