Discussion in 'Physics' started by steeve_wai, Sep 13, 2007.

1. ### steeve_wai Thread Starter Active Member

Sep 13, 2007
47
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my doubt goes like this...charge flows when there is a potential gradient...and an electric field exists where there is a potential gradient...electric filed exists in a current carrying coductor...so there must exist a potential drop/potential gradient in a conductor of zero resistance in a circuit...so the potential difference between any two points of a zero resistance...NOW DONT AVOID MY DOUBT BY SAYING THAT THERE IS NOTHING SUCH AS A PERFECT CONDUCTOR ...

NOW consider a simple circuit as shown below:

A----/----B--^^^^^D----------C
| S(switch) R=5 OHMS |
| |
__|__ B |
_ V=10 VOLTS |
| |
|F-----------------------------E|

i wish to discuss this in terms of energy only HELP ME IF GET STUCK SOMEWHERE:

-ABBREVIATIONS:
* ckt----circuit
*+ve----positive
*-Ve----negative
*U,u----potential energy
*K,k----kinetic energy
*e-----charge on an electron=1.6 *10^-19 coulombs
*E-----total energy of any particle

-current will be discussed in terms of POSITIVE charges
-i KNOW that current is established instantaneously in a circuit
-ok
-An electric field is immedately set up in the circuit when switch S is turned on.
-a +ve CHARGE enters the - terminal which tends to reduce the potential of the battery.
-the battery DOES WORK to maintain its potential difference.hence an +ve charge leaves the + terminal(i can explain this point in a little more detail if you wish)

-NOW WE DISCUSS THE FATE OF THE CHARGE IN TERMS OF ENERGY AS IT MOVES IN THE CIRCUIT. when i say charge, i mean +ve charge:unless stated otherwise.

-the battery raises a charge to a potential of 10v by doing work on it.
-the potential energy of the charge is 10e = E
-this charge leaves the + terminal of the battery B with U=10e
-it is accelerated by the electric field so its K is increasing and U SHOULD decrease because k+u should be a constant for any ISOLATED mechanical system.
-this K decreases only when it encounters a resistance.
-now the energy of the charge should be a function of distance x
-hence E(x)=U(x)+K(x)
-so the potential energy of the charge depends on how far it is from the + terminal.
-HENCE there is a different potential/potential energy at any point in the circuit...WHETHER it is HAS a zero or a non zero resistance.
-THIS means that there is a potential drop/gradient across those wires whose
resistance = 0 i.e the connecting wires
-but that is CONTRARY to what is observed...there is NO potential drop across a wire with zero resistance...

-consider this current occurs only when there is a potential difference across
a conductor.
-add a short across R in the above diagram
-ALL the current should flow through the short
-but there is no potential drop across the short...so how does the charge flow through the short.....

give some links to websites that cover science seriously and dont just give half baked explanations...thanks for reading...

2. ### Eduard Munteanu Active Member

Sep 1, 2007
86
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You should draw your schematic and attach it as a file.

I think you have a few misunderstandings. First of all, charge is not energy. Second, the definition of voltage may not be clear to you:
$V_{AB} = {W_{AB} \over Q}$
So a volt represents the mechanical work needed to move one coulomb from point B to point A.

Getting to the superconductors you were discussing, how did you arrive at the conclusion that U(x) is non-zero for a given x?

3. ### niftydog Active Member

Jun 13, 2007
95
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I know, against the "rules", but...

Surely because their really ISN'T such thing as a perfect conductor any discussion of this type falls outside the laws of physics and hence becomes pure speculation that cannot be verified or tested by any conventional science.

Electric fields and potential gradients in conductors are inextricably linked to the resistance within the conductors. Simply assuming that such phenomena exist in the impossible scenario you describe is, I believe, a fundamental error.

But heck, I'm no physicist, and I'd love to be proved wrong.

4. ### steeve_wai Thread Starter Active Member

Sep 13, 2007
47
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http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/DC-Current/WaterFlowAnalog.html

from the animations i understand that the resistor creates a drop in the potential.but this page does not consider the case when there is no resistance. so there should not be any potential drop ... right? BUT WHAT THE PAGE SHOWS IS AN ANALOGY...ANALOGIES SELDOM GIVE A CORRECT PICTURE...

5. ### Eduard Munteanu Active Member

Sep 1, 2007
86
0
Superconductors are real. They have zero, not just very low resistance. Studies have shown that closing a loop of superconductor to which a battery was attached will keep that current flowing forever; of course, as long as the material is kept at a certain low temperature. As far as I know, such a current won't generate heat, so I guess the conductor just needs thermal isolation and a bit of cooling to make up for the imperfections of the isolation.

6. ### niftydog Active Member

Jun 13, 2007
95
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Cool. So, if I read that right, the phenomenon the OP is referring to can only really be explained by quantum physics.

So while I was wrong about the existence of perfect conductors my suggestion that the application of standard circuit analysis might not be appropriate is still valid.

7. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Wikipedia has come a long way in the past few years, but it is still prone to flaws and misinformation.

There is no such thing as a "zero Ohm" conductor, super or otherwise. I'll buy nano-Ohm or pico-Ohm, sure, but not "zero."

Ohm's law shows what's going on here: For any given potential difference, as resistance approaches zero, current increases without bound. Ergo any tiny voltage across "zero ohms" will yield infinite current, infinite power, and infinite energy. The universe won't let us do that.

8. ### Eduard Munteanu Active Member

Sep 1, 2007
86
0
thingmaker3, what's the point in saying "any tiny voltage across 'zero ohms'"? Is there anything like an ideal voltage source in the real world? Such a circuit won't have an overall zero resistance, of course, because the voltage source itself is not perfect. But the outer circuit, made from a superconductor, will have zero resistance. Given these, the current will simply be $I_{SC} = { \mathcal{E} \over r }$

Don't trust Wikipedia? Here's another one: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/scdis.html

9. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Okay. Loophole in Ohm's Law. Cool. (Pun intended.)

Steve Wai's question " -but there is no potential drop across the short...so how does the charge flow through the short....." still get's answered by Ohm's Law & Kirchhcoff's Current Law.

Put another way, charge carrier flow does not create potential difference - potential difference creates charge carrier flow.

10. ### steeve_wai Thread Starter Active Member

Sep 13, 2007
47
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"charge carrier flow does not create potential difference - potential
difference creates charge carrier flow"

i understand that...fine...but the electrons in a wire are raised to high potential by the battery that does work on them ...now this charge is at a high potential...right...HENCE,IT HAS ENERGY,E=U+K,this K is increased by acceleraation due to electric field,but decreased by ionic collisions.hence K ieffectively remains a constant.NOW THE QUESTION...DOES THIS U AND THE VALUES OF KINETIC ENERGY IN EXCESS OF K,APPEAR AS HEAT IN A RESISTOR?AND THAT IS WHAT WE CALL A DROP IN POTENTIAL OR POTENTIAL ENERGY OF A ELECTRON.

11. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
It is possible there is a misconception you are operating under. You state that "but the electrons in a wire are raised to high potential by the battery that does work on them", which is not the case.

The potential difference in a circuit is the force that make charge carriers move from the cathode to the anode in the external circuit. The individual charge carriers - electrons - are in no way altered as to their state of charge by the difference in potential. The enery they have is due to the translation of the potertial energy into kinetic energy. In physical terms, that is the energy of position being translated into energy of motion.

Ohm's law gives us a nice picture of the process in resistances. The total R of the circuit gives us the total I (I = E/R). Using that I, the voltage across each resistance is given by E = IR. Taking the sum of each of the internal voltage drops, the total exactly equals that of the overall propelling voltage. There is no gain of energy or mysterious change in potential. The heat that appears in each resistance is the product of the current through it and the voltage drop across it. That represents a drop in kinetic energy, not potential. If there were no drop in kinetic energy, then resistances would have no effect on current flow.

12. ### recca02 Senior Member

Apr 2, 2007
1,211
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i m really confused over this superconductivity stuff.
i heard they sort of achieved it a few years ago with a new state of matter (5th state? not sure about it neways)
so if u think about it it is like having a point in/on the conductor at two different potential (like having a point of wire at +ve and -ve voltages)

13. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
Superconductivity is poorly understood at this time. Why superconductivity should be possible is confusing. One aspect that makes it really interesting is that there is a point at which the electric field density causes the superconducting property of the material to vanish.

14. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
As Beenthere has noted, there seems to be a misunderstanding of terms. "Potential" does not mean "joules" in this context. It means "electromotive force."

The amount of energy (watt-seconds, or joules) will depend on how much power and how much time. The amount of power (watts) will depend on both the EMF and the impedance. P=E^2/R. Properly sized circuit traces and wires will consume negligible power and dissipate negligible heat. Voltage drop across such conductors will also be negligible.

Here's another way of looking at things: One volt of EMF equals a potential energy difference of one Joule per coulomb of charge carrier. The same amount of energy is used whether our coulomb takes a milliseconds or millenia to pass a node in the circuit. The rate of flow of our coulomb (measured in amperes) is limited by the total series resistance of the circuit.

15. ### steeve_wai Thread Starter Active Member

Sep 13, 2007
47
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well my confusion goes like this...please forgive the capital alphabets...
can anyone chat with me on msn regarding this

ENERGY TRANSFERS IN A CIRCUIT:
see the image below

http://www.school-for-champions.com/sci … ircuit.gif

LET VOLTAGE OF CELL IS 10V.
Consider this:

AN ELECTRIC FIELD IS SETUP IN THE WIRE.
THE FORCE DUE TO THIS FIELD ACCELERATES THE CONDUCTION ELECTRONS IN THE WIRE.
AN ELECTRON(S) NEAR THE + TERMINAL ENTERS THE +TERMINAL OF THE CELL.
THIS TENDS TO DECREASE THE POTENTIAL OF THE POTENTIAL OF THE CELL.
HENCE THE CELL DOES WORK TO MOVE AN ELECTRON FROM + TERMINAL TO – TERMINAL , AGAINST THE INTERNAL ELECTRIC FIELD OF THE CELL.

THIS ELECTRON HAS BEEN “ELEVATED” TO A HIGH POTENTIAL ENERGY AT THE – TERMINAL.

LET THE ENERGY OF THIS ELECTRON BE “E”, E=U+K
WHERE U AND K ARE THE POTENTIAL AND KINETIC ENERGIES OF THE ELECTRON RESPECTIVELY.

AT THE –TERMINAL,U=10e,(e is the charge on an electron)
AND K=0.

THIS K WILL INCREASE DUE TO ACCELERATION BY FORCE DUE TO ELECTRIC FIELD IN THE WIRE.BUT IT IS DECREASED DUE TO COLLISIONS WITHIN THE METAL.

ITS AVERAGE VELOCITY IS (½)MV^2 , WHERE V IS THE DRIFT VELOCITY OF THE ELECTRONS,WHICH IS SMALL IN MAGNITUDE.
LET THIS KINETIC ENERGY BE k1.

HENCE AT ANY POINT IN THE WIRE E=U+k1.

NOW, WHEN THE ELECTRON ENCOUNTERS A RESISTANCE,THE GAINS IN KINETIC ENERGY ABOVE k1 WILL CAUSE HEATING OF THE RESISTOR.
BUT WHAT HAPPENS TO THE POTENTIAL ENERGY U=10e.
DOES THIS U GET LOST IN HEATING UP OF THE RESISTOR OR IS IT SOMETHING ELSE THAT DECREASES THIS U TO 0 ONCE THE ELECTRON LEAVES THE RESISTANCE.

DOES POTENTIAL DROP ACROSS ACROSS A RESISTOR MEAN THE DROP IN POTENTIAL ENERGY OF THE ELECTRON ONCE IT LEAVES THE RESISTANCE OR IS IT SOMETHING ELSE.
IF YES,WHAT CAUSES THE DROP IN POTENTIAL ENERGY

16. ### beenthere Retired Moderator

Apr 20, 2004
15,815
282
You still have some misconceptions about the mechanism producing potential in batteries and the electric field. The potential comes from the redox reaction in the battery. The field is the result of moving charge carriers in the conductor external to the battery. The internals of the battery will maintain the potential while supplying additional electrons for curent in the external circuit up to some limit.

This is a fairly good reference on batteries - http://www.science.uwaterloo.ca/~cchieh/cact/c123/battery.html.

17. ### Mike M. Active Member

Oct 9, 2007
104
0
From what I remember.......

First of all, the electrons in a superconductor travel in pairs and behave differently as a pair than regular electrons so any conventional circuit knowledge you have has to be thrown out the window before you can even begin to think like a superconductor. Secondly, I also remember reading in a few places that the current in a superconductor flows in BOTH DIRECTIONS SIMULTANEOUSLY and the resultant magnetic field emissions by a superconductor repel BOTH NORTH AND SOUTH magnetic fields.

I don't think that superconductors require a gradient voltage field in order to move these electron pairs from point A to point B because no work is being done on them and therefore no energy is required for movement. I believe that superconductors have virtual flow at all times but when a voltage is applied, the virtualness becomes a reality as the quantum wave function collapses due to the outside real influence and/or observation.

http://en.wikipedia.org/wiki/Wave_function
http://en.wikipedia.org/wiki/Wavefunction_collapse