This is a well known old problem:
When a charged capacitor is connected to an uncharged capacitor, they share their charges, but there is an energy loss during this process; even when there is no resistor in the circuit.
Let's assume two identical capacitors (C1=C2=C); one charged (Q1=Q, V1=Q/C=V) and the other empty (Q2=0, V2=0).
Then, when they are connected in parallel via ideal conductors, charge sharing occurs and finally they exhibit Q1=Q2=Q/2, V1=V2=(Q/C)/2=V/2.
The initial energy in the system was, Ei=CV2/2 (the energy stored on C1 only; because C2 had zero charge / zero energy initially).
After charge sharing, the final energy is, Ef=C(V/2)2/2+C(V/2)2/2=CV2/4
In the above example, Ef=Ei/2; i.e. half of the energy is lost.
"How? In what form? Or, is there anything Circuit Theory is missing?"
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P.S.: Honestly speaking, I do have some ideas (though limitied) about the answer, for I have heard or read some explanations on several occasions (Some explanations were made in books and scientific magazines as well), nevertheless I wanted to see what other alternative concise explanations we (researchers in ResearchGate) can propose.
Ideal conductors result in a current with an infinite value for a duration of 0 sec (Dirac pulse). Because this is not possible, one should ask if the calculation above is an allowed starting point for the discussion.
Ideal conductors result in a current with an infinite value for a duration of 0 sec (Dirac pulse). Because this is not possible, one should ask if the calculation above is an allowed starting point for the discussion.
Dear Lutz,
Thank you for your answer. That's a good point. I was thinking the same way.
However, by assuming a resistor on the path and making energy calculation, one can reveal that the lost energy is independent of the resistor's value; even when the resistor value is zero!
What is interesting is, for a non-zero resistor on the path, one can understand the cause of the energy loss, but it is hard to accept that a zero-Ohm resistor causes energy loss.
For sure, we should discuss how the limit condition (zero resistance - infinite current - zero time) still causes the same energy loss (For the given example, half of the initial energy).
Interesting... In this arrangement, the new (total) capacitance becomes two times bigger than the single charged capacitance. In comparison, if we increase two times the capacitance C of a charged capacitor, the charge Q = C*V does not change... so the voltage becomes two times lower... but the energy does not change, in contrast to the discussed situation... and we are not surprised as here...
The answer is very easy. When a capacitor is charged for let's say 1000 J of energy ( we are talking about a real capacitor), another 1000 J of energy is lost in the charging process.
Question: Where?
Answer: In the resistor of the RC circuit.
Question: What if there is no resistor?
Answer: There is always a resistor. The capacitor has some active resistance. After all it's legs are made of a conductor with some resistance.
Question: How can this statement be proved?
Answer: Connect an RC circuit with a time constant RC=1 s to a DC voltage source (a battery) where the resistor is a light bulb. Watch the capacitor charge (the light bulb will first light up and then exponentially die out). After that discharge the capacitor through the same light bulb and watch it light up then exponentially die out. Le'ts analyze this. The light bulb lights with the same brightness during the charge and discharge process. During the discharge process the light is caused by the energy stored in the capacitor. The same amount of energy is consumed during the charge process (the light bulb lights with the same brightness).
An interesting dialogic form of answering... and impressing visualizing experiment (very suitable for a striking presentation in the lecture hall)...
Exactly Cyril. I use this visualization of chargning/discharging a capacitor for demonstration during the lectures ;)
With the same success we can ask the same question (Where the energy has gone?) when connecting a charged capacitor via ideal conductors to an ideal voltage source... or simply shorting it...
Boris, I just want to say that - in general - I like such sequence of questions and answers. Since a long time I plan to produce such a "ficticious conversation" regarding the often controversely discussed question "Is the BJT current- or voltage-controlled"?
Re Ali Zeki's comment, "by assuming a resistor on the path and making energy calculation, one can reveal that the lost energy is independent of the resistor's value; even when the resistor value is zero!":
Looking at the calculation in eq. (1) in Saurabh Singh's link to
http://arxiv.org/pdf/1309.5034.pdf
this calculation looks to be valid only for non-zero R. If we postulate R=0, we can find equivalent energy losses by considering changes in energy in other components of the system, e.g., change in the kinetic energy of the charge carriers as they move from one capacitor to the other, or work done in moving the capacitor plates - as this paper shows.
The problem, as posed, only considers the energy associated with the electric field in the initial state and a final (equilibrium) state. I think the paradox appears because the energy changes in other components of the system while it is in transition between these two states are not considered, and so the energy accounting is incomplete.
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