Another tough 'elementary' question from my tutor

[jasmin_123]

I do not know "How many energy dissipates an infine current passing through a zero resistance conductor."

 

[Learner]

Why a zero-value resistor conducts all the charge simultaneously? Because its conductivity is infinite!

Electrical conductivity or specific conductivity is a measure of a material's ability to conduct an electric current. When an electrical potential difference is placed across a conductor, its movable charges flow, giving rise to an electric current. The conductivity σ is defined as the ratio of the current density

to the electric field strength

Conductivity is the reciprocal (inverse) of electrical resistivity and has the SI units of siemens per meter (S·m-1)

https://en.wikipedia.org/wiki/Electrical_conductivity

The electrical resistivity ρ (rho) of a material is given by

where

ρ is the static resistivity (measured in ohm metres - Ωm)
R is the electrical resistance of a uniform specimen of the material (measured in ohms - Ω)
l is the length of the specimen (measured in metres - m)
A is the cross-sectional area of the specimen (measured in square metres - m²)

For R = 0, ρ or l is = 0

If l is 0 conductor does not exist, ρ must be 0.

If l is not 0, ρ must be 0 for R to be = 0

Either way, ρ = 0.

Electrical resistivity can also be defined as

where

E is the magnitude of the electric field (measured in volts per metre - V/m)
J is the magnitude of the current density (measured in amperes per square metre A/m²)

In order for ρ to be =0, E must be 0.

But E can not be 0 because 1 cap has more charges than the other, and no charge will flow to the discharged cap.

If E can't be = 0, ρ can't be = 0
Even If ρ = 0, σ is not = ∞

If 1/ρ = 1/0 = ∞

0 * ∞ is not = 1

Therefor conductivity is not infinite as you assumed.

https://mathforum.org/library/drmath/view/53336.html

If ρ = 0 then E is = 0

No charge will flow and nothing happens.

In physics, the space surrounding an electric charge has a property called an electric field. This electric field exerts a force on other charged objects.

https://en.wikipedia.org/wiki/Electric_field

 

[sivakumar_tumma]

Hi jasmin_123,
With out spending enegry we can not move electron in the electric field.. thats what happend in this same case, even though u specified to transfer charge...

when both are connected one capacitor will get little charge form other cap which is having full charge.. once little charge the cap get , in order to get some more charge some energy has to spend as already there is electric field in the cap because of previous charge i acuumulated....


hope this will be sufficient...........

 

[jasmin_123]

How many energy dissipates an infine current passing through a zero resistance conductor?

Zero (not zero-plus) times something (including infinity) is zero.

No charge will flow and nothing happens.

So the steady state is V?

when both are connected one capacitor will get little charge form other cap which is having full charge

Zero resistance cannot conduct charge in portions. Period.

 

[laktronics]

Hi Jasmin,
But, I thought you were asking for the voltage of the combined system. I am able to see only short pulses on each capacitor when the electrons reach them, what will be the volatge of the system when the electrons are on the superconducting wires?
Regards,
Laktronics.

 

[jasmin_123]

Hi, laktronics,

I had asked about steady state. Had I?

When all the electrons are on their way from one cap to the other, all the C*V^2/2 energy is transformed into the kinetic energy of electrons, and V1 and V2 are zero.

The short pulses that you see is the steady state. Is it?

Jasmine

 

[sivakumar_tumma]

jasmin_123,
that energy it has to spend is not for reistance.. to overcome the electric field .... I am not talking about in terms of circuit theory.. but in terms of electromagnetic theory... In order to over come the electric field the enegry willbe spent....

one eg: If u had done electromagnetic course, u will surely remeber this ...
when we are bring the charges to the corners of square, there some enegry is wasted, eventhough resistance is not there... that energy will be used to overcome the repulsion of the charges, which are previously put in the place....
============================

resistance is not the only requirement to transfer chareg... potentila difference is the cause for charge to transfer..

hope this is clear...

 

[maharshi_qis]

when two capacitors .. u said 1 is charged one and other is completely discharged one..are connected in parallel..The circuit try to reach equilibrium at which there wont be any energy transfer b/n the to capacitors..if both capacitors are of same value in steady stage both will have the same charge..if they are differ then their charge will differ..

 

[zorro]

How many energy dissipates an infine current passing through a zero resistance conductor?

Zero (not zero-plus) times something (including infinity) is zero.

Dear Jazmine,

You have to review some basic concepts.
“zero times infinity” is not zero. It is an indetermination, it is a combination of symbols that has not mathematical meaning. When that kind of thing is found from a limit, the limit can exist or not. If the limit exist, it can be 0, any oyher value, or infinity.
Regards

Z

 

[jasmin_123]

Hi, zorro,

I am not talking about limits, I am talking about numbers and singularities.
When I say zero, I mean zero, not a limit, not a 0+, not 1-0.9999... .
When I say infinity I mean something or any NUMBER, no matter how great it is:
0*(any NUMBER, no matter how great it is)=0.

I agree with you that, strictly speaking, infinity is not a number, and 'zero times infinity' is an undefined real number. However, to use your concept of an infinite current to solve my question, you have to assume that the electrons have an infinite mobility, which means infinite energy and infinite mass. Not too much infinities for one pure problem?

By the way, what about the loop gain greater than one?

Yours,
Jasmine

 

[Mr.MEB]

Seems like the debate is more philosophical than mathematical or physical.
Could you sum up the different solutions proposed and their respective pro/cons?
That way we can make sure that everybody understand the other's position.

 

[jasmin_123]

There is nothing to summarize: if you do assume zero losses, then steady state means oscillations.
---
If you do not like to assume zero losses, this problem is not for you.

 

[zorro]

Dear Jazmine,

It is not clear which approach you want.
You insistently stated the question as “an abstract one”, an “IDEAL case”, “a mathematical abstraction”,…
But on the other side you speak about “five electrons” (that simultaneously) passes from une C to the other taking some time, kinetic and potential energy, mobility, etc. (Sorry, there is nothing like “kinetic energy” in electromagnetic theory.)

Let’s take the mathematical-ideal-electromagnetic approach. This your aim, isn’t it? What was said about the other is pure “noise”, nothing to do with “IDEAL cases” or “mathematical abstractions”.

When in a mathematical analysis you encounter an indetermination like “zero times infinity”, it is necessary to approach the situation by some determined way, and take a suitable limit.

Lat’s take two ways that in principle have not indetermination:

1) the two Cs are connected by a R. This is the approach yet discussed. The steady state is with both Cs at V/2, approaching it with time constant RC. Taking R->0, the time constant -> 0, peak intensity -> 0 and there is energy loss.

2) the two Cs are connected by an inductor. In this case (with no losses, i.e. no resistance) the circuit oscillates at a frequency proportional to 1/sqrt(L) and peak current proportional to 1/sqrt(L). The steady state is oscillatory, with energy flowing alternatively between the Cs and the L. Taking L->0, the period of oscillation -> 0, peak intensity -> 0 and there is no energy loss. (Let’s avoid to speak about radiation.)

This is one of the situatons in which the limit gives different results if taken in different ways. Both ways are “ideal”, “a mathematical abstraction”. Strictly speaking, the limit does not exist.

The first way is the “usual” one, because there is a steady state in equilibrium as one expects in “real life”.
The second one is what (I guess) you would like. But (unless you want to abandon the “mathematical abstraction”) you have to accept things as infinite current and oscillations at infinite frequency.
And no electron mass, no kinetic energy, no SIMULTANEOUS movement of electrons, etc…

Regards

Z

 

[jasmin_123]

Zorro,

I tell you R=0; you tell me R->0.
I tell you once again: R=0, you tell me: R->0.
Then, I tell you: "OK, R->0."

Jasmine

 

[melc]

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.

Instead of this question (which appears in almost all discussion lists) answer what's happening in a real world. C1=10uF tantalum capacitor, charged at 10V. C2 is 10uF tantalum discharged at 0V. C1 and C2 are connected in parallel in less than 1mS after power supply on C1 and short circuit on C2 are removed. What would be the resulting voltage?

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[zorro]

Zorro,

I tell you R=0; you tell me R->0.
I tell you once again: R=0, you tell me: R->0.
Then, I tell you: "OK, R->0."

Jasmine

Jasmine,

Sorry, I do not understand.
Are you kidding? Are you mocking?
Didn't you undertstand that R=0, L=0, is an ill-posed problem, that has not solution?

I propose you a more simple problem that maybe helps to open your mind:
Just one charged capacitor. Find steady state after short-circuiting it.

Z

 

[Learner]

Hi Jasmine_123,
Inseatd of solving the assumed lossless system, do you have a solution for NON-lossless system?

 

[jasmin_123]

Hi, zorro,

I am not kidding and not mocking, I am 100% serious.
(Sorry about making you think so.)

I liked your example and have to think a bit, but, intuitively, the steady state also is oscillations.

To try and clear the mix-up of my R=0 (singularity) and your R->0,
let me ask you about the phase at f=1/(2piRC) of the ideal [R=1/(2piC), not R->1/(2piC)] Wien-bridge bandpass. Is it pi/2, -pi/2, or zero?

Jusmine

Hi, Learner,

The steady state in a non-lossless system is obvious: V/2.

Jasmine

 

[Learner]

Hi, Learner,

The steady state in a non-lossless system is obvious: V/2.

Jasmine

Ok, so if that's the case what is the loss in the system?

Please include all the mathematical working procedure in finding the loss, as well as how you found the answer V/2 in your reply above.

Thanks!

 

[medra]

what a discussion !!!!!!
initial energe = final energy isnt it?
.5cV1^2=.5cV2^2+.5cV2^2 (they are prallel having same voltage at ss)

V1^2=2V2^2

find V2