Switched Capacitor _ Fundamental question

[eda4you]

During an interview I got following question:

Assume you want to load a cap C via switch having a resistance of R. What energy has to be delivered by the source and what energy is stored in the cap.

E_supply = C*Vss^2 and E_cap = C/2 *V^2

Next question: Where is the remaining energy?

Its dissipated in the resistor: E_r = C/2 * V^2

Next question: Now assume you have superconducting switches with R=0 ohm. What happens?

Do you know the right answer?

I answered that now the source has only to deliver C/2 * V^2, but I don't know if its true. What do you think about this answer?

 

[batdin]

I think that:
If R=0 then the internal resistance of the source will act as a charging resistor.
If it is zero,too,the problem is not solvable because you cannot connect a perfect voltage source to a perfect capacitor.The transient will last 0 seconds and the charging current will be infinity.
A similar problem:
What happens if you short-circuit a perfect voltage source with a perfect conductor?
If the source is perfect then the voltage across it will not change.
If the conductor is perfect,the voltage across it will be zero.
So what is the answer?
I know that the current will be infinity,but what about the voltage across the source?

 

[codesuidae]

I think that:
If it is zero,too,the problem is not solvable because you cannot connect a perfect voltage source to a perfect capacitor.The transient will last 0 seconds and the charging current will be infinity.

Well, the problem did specify that the switch was a superconductor. As such, it will have a critical surface (defind by the graph of the values of the critical magnetic field, critical current, critical temperature), if the power supply is capable of supplying enough current (likely, given the state of todays superconductor technology), the switch will be quickly driven out of superconducting mode and will behave as a regular conductor.

A similar problem:
What happens if you short-circuit a perfect voltage source with a perfect conductor?

Fortunately this is not a problem because the mathmatical models we use are just that, models of real-world behavour, and in the real world, there are other factors (such as the above mentioned critical current limits) that place constraints on questions like these. These constraints are typically understood by users of the model (ie, there is no such thing as a perfect conductor, etc).

Hmm, maybe I've been out of school too long?

 

[mc_navman]

I think the answer they were looking for is that the cap will charge instantly (assmuing the source impeadance is 0.) You are correct about the energy. The question is not a good one, as infinite current would be required, and there are many real world factors which proclude this. You could have answered this question in serveral ways and still been correct. The example above about superconductors is a good one, or perhaps you could have said that the other circuit impeacences would give the same as the origonal answer where the R is that of the source and cap impeadance and interconnect impeadances. Or you could have assumed all these to be 0 in which case the stray inductance between the supply and the cap would cause the circuit to oscilate at it's resonance indefenitly as it is isolated from the supply by the stray inductance and it is completely undamped.

 

[eda4you]

Thanks for all answers. There were a lot of new impression about answering the task.

Besides, it is strange for me to believe, that if Ron=1m ohm, the source has to deliver C*V^2 , even for Ron 1µohm or Ron = 1 pOhm,respectiveley. However, if Ron =0 ohm (assuming all others to be 0 ohm) than the source has only to deliver the half energy!!! --> strange, isn't it?

In my opinion the job interviewer wanted me to test me how I handle technical task and problems, even in a team. Even it was more a discussion than a typical job interview.

 

[srinivasan]

Actually i had this doubt in my high school... it is due to the electrostatic loss of energy which is present..... read Griffiths u ll know things better...

 

[v_c]

If you use the conservation of energy you will get the wrong answer -- you have to use the conservation of charge. Please take a look at the attached (unfinished) document that I wrote up. I hope that you find it useful.

Best regards,
v_c

 

[dfullmer]

All of you have great answers. The only thing that I would add is the question does not take into acount the inductance of leads which would limit the current spike perportianally to the inrush frequency. At least that how it would appear to me.

dfullmer

 

[bjeservice]

dear friends
the question is not clear
if the source is ac the the capacitor would not absorb energy as it take energy in half cycle and deliver it in next so the only energy will be that heat energy dissipated in switch ( q * Vr )

if source is dc then
total energy delivered = q.Vr+C/2.Vc^2