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9th February 2007, 21:18 #1
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Another tough 'elementary' question from my tutor
Consider two identical ideal capacitors: one charged to voltage V and the other
discharged. Find steady state after connecting in parallel the charged capacitor to the
discharged one.

Important conditions: there are NO energy losses: NO radiation, NO heat (the
connecting wires are short and superconducting), no sparks, etc.

9th February 2007, 22:54 #2
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Re: Another tough 'elementary' question from my tutor
Steady state voltage is V/2 if the iinital voltage is V
Initial Energy stored in C= (CV^2)/2
Final Energy stored in both Cs = (C(V/2)^2)/2+(C(V/2)^2)/2 = (CV^2)/4
Half of the energy Gone!!
In real circuit the energy will be dissipated in the wire resistance and/or the switching arc while connecting the two capacitors. If you assume there is 0 resistance in the wires then it is hard to explain the energy loss.
If you want mathematical proof may be you can say if there is no resistane the inital current will be infinity. So infinity current passing through zero resistance creates energy loss.

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10th February 2007, 01:25 #3
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Re: Another tough 'elementary' question from my tutor
Here's a thought that might provide some insight: Instead of paralleling another capacitor, move the charged capacitor plates so that the spacing is halved. The net result on capacitance and voltage is the same as the original problem. The capacitance is doubled, the charge is the same, the voltage is halved. The stored energy is halved. There is no resistance, and no current loop, but there is radiation, because a magnetic field is created by the charge movement (I think). Suppose the plate has no mass (no more ridiculous than the original postulations). Since the plates are attracted, how does that affect energy when the plate is moved? (I don't know). In fact, I'm not sure if all my statements are correct.

10th February 2007, 08:37 #4
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Another tough 'elementary' question from my tutor
Please try and solve the problem as stated in the question: for the case
of NO energy losses, including the 'infinity times zero' energy loss.

10th February 2007, 13:35 #5
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Re: Another tough 'elementary' question from my tutor
Hi Dear jasmin_123!
You know when two Capacitors are connected; they must have the same voltage. But when the voltages aren't equal the capacitors must move their charges so that the voltages will be the same. In order to this they must consume energy! And you know this Process is done during a wink and if you use math; you see there is an Impulse function in the Current Expression. Sure you know it is not possible to create an impulse current through a capacitor, when the voltage is limited. To do this the energy is used to create the Current.
Thanks!

10th February 2007, 13:56 #6
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Another tough 'elementary' question from my tutor
Hi, Mansour_M, the question is an abstract one.
By the way, since both the discharge time and Q are limited, the current also is limited. It will not reach infinity.

Do not be too formal! It is always worth thinking of extreme cases and singularities.
Is not it?

10th February 2007, 20:54 #7
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Re: Another tough 'elementary' question from my tutor
Originally Posted by jasmin_123
If there is series inductance, the charge will oscillate indefinitely between capacitors. The steadystate will be sinusoidal voltages on both caps which are 180 degrees out of phase. The positive peak will equal the original voltage. The negative peak will be zero.

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10th February 2007, 21:05 #8
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Re: Another tough 'elementary' question from my tutor
Hi Jasmin,
Nice to see your brain storming questions. Capacitors are like charge holding vessels and the voltage is like the height of the filled up charges. If you connect two identical cylinders, one filled with water and the other empty, by a pipe connected at their bottom, the water gets distributed equally making the height of water in each cylider to half. Similarly, if you connect two identical capacitors, one charged and the other not charged, the charges get distributed eqally on to both capacitors, making the voltage to half, since V = q/c .
Regards,
Laktronics.

10th February 2007, 21:37 #9
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Another tough 'elementary' question from my tutor
"By the way, since both the discharge time and Q are limited, the current also is limited. It will not reach infinity. "
> Wrong
discharge time is infinitely small (0). Current is infinite.
it's a Dirac's function.
The integral of i over time is Q.
The same question can be asked with an ideal voltage source being short circuited with ideal conductors:
There will be no energy loss because no resistance, but the current will be infinite.
If you want to conserve Energy you can write the expression of energy but in this case you will find that there are some charges created!
initial E = 1/2 . C . Uinit^2
= final E
= 1/2 . 2C . Ufin ^2

11th February 2007, 07:10 #10
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Another tough 'elementary' question from my tutor
Hi, Laktronics,
Would the capacitors reach the steadystate at V/2, then 50% of the initial energy would dissipated. However, the question implies that dissipation is zero.
Hi, Mr.MEB,
If you assume that the mobility of electrons is infinite you are probably(?) right.
But let us assume that it is not infinite, hence, the discharge current also is not infinite.
Once again, the question implies that dissipation is ZERO. What is the steady state for ZERO dissipation ?

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11th February 2007, 07:28 #11
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Re: Another tough 'elementary' question from my tutor
Please see this previous post regarding the same problem: https://www.edaboard.com/viewtopic.p...409&highlight=
About half way down the page, I posted a reply with derivation explaining that the energy loss is not due to any form of dissipation. In fact, work is done by the electric field to physically move the charges from one plate to the other and half of the energy is expended regardless of the conductivity of the connecting wire.

11th February 2007, 12:26 #12
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Re: Another tough 'elementary' question from my tutor
Originally Posted by jasmin_123
I guess when you said find steady state, you meant final energy in the caps? Does the diagram in the question describes your question?

11th February 2007, 17:48 #13
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Another tough 'elementary' question from my tutor
Hi, jayc, I went over the discussion that you mentioned.
It does not consider the IDEAL case I am interested in.
Hi, Learner, your example also considers a case with NON ZERO!
energy losses due to radiation.
THINK OF THE ABSTRACT, IDEAL CASE!

11th February 2007, 19:30 #14
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Another tough 'elementary' question from my tutor
What makes you think there can be an "ideal case" where there are no losses? The fact that your tutor posed the question doesn't automatically make it possible.

11th February 2007, 20:50 #15
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Another tough 'elementary' question from my tutor
Nobody said that such a case is possible. The question is about a mathematical, if you like, abstraction.

Any ideas? Do not be too formal, be creative!

11th February 2007, 21:07 #16
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Re: Another tough 'elementary' question from my tutor
Ok, I think the problem here could be not accounting for the kinetic energy.
Total energy = Potential + Kinetic
But initially
Kinetic energy =0
So
Total energy = Potential Energy + 0 = PE
IF V=10v C=2F
Total energy = Potential + 0 = U = 1/2 * C * V^2 = 100
Total charge Q=C*V = 20C
After added identical C in parallel
V= Q/2*C = 5
PE = 1/2 * 2*C * V^2 = 2* 25 = 50
KE= q*V = 1/2 * 20 * 5 = 50
Initial state
Kinetic energy =0
Total energy = Potential Energy + 0 = PE = 1/2 * C * V^2  eqn A
Total charge Q=C*V
After added identical C in parallel
V= Q/2 * C
V^2 = Q^2 / 4 * C ^2
PE = 1/2 * 2C * V^2 = C * V^2 = C * ( Q^2 / 4 * C ^2 ) = Q^2 / 4 * C
KE= qV = 1/2 * Q * V = 1/2 * Q * Q/2 * C = Q^2 / 4 * C
PE + KE = (Q^2 / 4 * C) + (Q^2 / 4 * C) = (2* Q^2 / 4 * C)
= (Q^2 / 2 * C)
Q^2 = (C*V)^2
PE + KE = 1/2 * C * V^2 = Total energy  eqn A

11th February 2007, 22:09 #17
Re: Another tough 'elementary' question from my tutor
I think the answer is abvious...
As was mentioned several times by others, although the resistance is zero this does not mean that dissipation is zero becouse we are dealing with infinite current flowing through the currents, an impulse.

12th February 2007, 07:12 #18
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Another tough 'elementary' question from my tutor
Hi, Learner,
Your idea is a good one. To understand what the steady state is, the following thought experiment can help.

Let us assume that the charge of the first capacitor consists of five electrons: electrons No. 1, No. 2, No. 3, No. 4, and No. 5.
After the capacitors are interconnected, in which order the five electrons will pass to the second capacitor through ZERO resistance?

Do you see now what the steady state is?

12th February 2007, 12:04 #19
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Re: Another tough 'elementary' question from my tutor
Originally Posted by jasmin_123
"Steady state occurs in a circuit or network when all transients have died away, it is not achieved until some time has elapsed after the system is started or initiated. This intial situation is often identified as a transient state, startup or warmup period."
http://en.wikipedia.org/wiki/Steady_...electronics%29
In your experiment example, I don't see how it is possible to determine which electron in what order will pass to the second cap.
I simply do not understand the terminology "steady state" that you are using but my impression is that you want to know what would the charges be in both caps after they have been connected in parallel after the initial state?

12th February 2007, 12:36 #20
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Re: Another tough 'elementary' question from my tutor
Originally Posted by Learner
the first cap will be discharged, and the second cap will be charged to V.
After that, ALL the electrons will SIMULTANEOUSLY go back to the first cap, and so on.
The steady state are sustained oscillations at a frequency one over twice the time
that it takes to electrons to travel between the caps.
No energy loss! Is not it simple?

*ZERO resistance cannot conduct a charge in portions!

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