ABOUT INFINITE INDUCTANCE, ZERO CAPACITANCE!

Discussion in 'Physics' started by b.shahvir, Apr 23, 2009.

  1. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    Hi Friends, :)

    Two particularly abstract queries have surfaced in my mind (this seems to be a recurring problem with me! :D)


    CASE 1 :- Infinite Inductance

    Let us consider an 'ideal' solenoid coil (zero resistance, pure inductance). It is connected to an AC supply of say, 220VAC, 50Hz. Now assume that the medium (or magnetic core) surrounding the coil presents 'zero' reluctance to the magnetic flux. Hence, the magnetic flux embracing the coil will be 'infinite'. This would result in an infinite inductance and hence, an infinite inductive reactance! Now, under these conditions, ideally, the current thru the coil must be 'zero'.

    The problem is, it is pretty difficult to visualize this situation, as even to create an infinite magnetic flux, some finite amount of magnetizing current must flow thru the coil. But, as per the theoretical concept, infinite inductance means current thru the coil should be an absolute zero! This cannot be realized even by examining the voltage versus current waveform of a pure inductor where current lags supply voltage by 90 deg. Hence, plz help!


    CASE 2 :- Zero Capacitance

    Similarly, consider an 'ideal' capacitor (zero resistance, pure capacitance). (The ideal capacitor could also be considered as a single charged conductor with the opposite conductor assumed to be placed at infinity). It is connected to an AC supply of say, 220VAC, 50Hz. Now assume that the dielectric medium between the capacitor plates presents 'zero' permitivitty to the electric flux. Hence, the electric flux between the capacitor plates would be 'zero'. This would result in 'zero' capacitance and hence, an infinite capacitive reactance! Now, under these conditions, ideally, the current thru the capacitor must be 'zero'.

    Again, it is pretty difficult to visualize this situation, as the capacitor plates ideally cannot hold any electrical charge (zero capacitance) as mentioned above. Under this condition, can I replace the capacitor plates with point charges of opposite polarities placed in this special dielectric medium......as point charges ideally present 'zero' capacitance? As per the theoretical concept, 'zero' capacitance means current thru the capacitor should be an absolute zero! This cannot be realized even by examining the voltage versus current waveform of a pure capacitor where current leads supply voltage by 90 deg.
    Valuable inputs are very much awaited for the same. :)


    Thanks & Regards,
    Shahvir
     
    Last edited: Apr 23, 2009
  2. thingmaker3

    Retired Moderator

    May 16, 2005
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    Inductive reactance is proportional to inductance. If inductance increases without bound, inductive reactance increases without bound. Current is inversely proportional to inductive reactance.

    Zero capacitance would be a "perfect" open circuit. As capacitance decreases, capacitive reactance increases. You see where this is leading, yes?

    There are reasons the real world works the way it does. Learning those reasons can be even more fun than playing with pretend universes!;)
     
  3. Ratch

    New Member

    Mar 20, 2007
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    b.shahvir,

    No current need exist through the coil if its inductance is infinite. Its back voltage of Ldi/dt with L equal to infinity and di/dt to zero will produce a indeterminate answer. Your fantasy with infinitesimals is not a productive endeavor.

    Thingmaker3 discribed it best by saying it is simply an open circuit.

    Ratch
     
  4. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    I absolutely agree, but is there no scientific explanation available to help visualize abstract concepts? :(
     
  5. Ratch

    New Member

    Mar 20, 2007
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    b.shahvir,

    Didn't thingmaker3 and I give you an answer based on scientific principles and mathematical operations? You are trying to visualize a indeterminate situation, not a abstract concept. That is why you are having so much trouble making sense of it.

    Ratch
     
  6. Wendy

    Moderator

    Mar 24, 2008
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    According to one definition, a short is infinite capacitance, you can charge it as long as you like and it will never charge. Never bought into it myself, but it was in a physics class, and we were studying coulombs.
     
  7. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    Zero capacitance means zero accumulation of electrical charge on the capacitor plates or conductors. Can a potential difference or voltage still exist between the capacitor plates or conductors in the absence of electrical charges?

    Without charge there can be no electric field and hence no potential gradient between 'open circuited' conductor terminals! :confused:
     
  8. thingmaker3

    Retired Moderator

    May 16, 2005
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    Yes. Happens all the time. Look at your nearest unused wall outlet. For practical purposes, it may be considered a zero Farad capacitor.

    Try putting the horse before the cart. You'll get farther. :)
     
  9. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    But , how can a conductor acquire voltage without accumulation of electrical charge on it? No electric field...... no potential gradient...... no voltage! :confused:


    Plz clarify! :rolleyes:
     
  10. thingmaker3

    Retired Moderator

    May 16, 2005
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    The voltage comes first.:) Whether a PV cell, battery, piezo, dynamo, or static, it is entirely possible (and extremely common) for a voltage to be present with no current.

    The charge carriers only flow if the circuit is completed.
     
  11. Nanophotonics

    Active Member

    Apr 2, 2009
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    Voltage causes current. Just like a force causes an acceleration and not vice versa.

    Thanks.
     
  12. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    Actually, the problem arises when i start associating voltage with the concept of electric field in electrostatics..... acceleration of a charged particle due to potential gradient in an electric field!

    If i consider voltage on a more 'mechanical' level i.e. work done in moving a unit charge against an electric field, then voltage is a mechanical force which comes first.

    Additional inputs are welcome :) Thanx.
     
  13. thingmaker3

    Retired Moderator

    May 16, 2005
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    Are you perchance confusing "electric field" with "electrostatic potential?"
     
  14. Nanophotonics

    Active Member

    Apr 2, 2009
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    Actually V = W/Q
    V = Fd/Q ; distance/charge doesn't cancel out and both are fundamental quantities, so V is not F as such. The force pushes the unit charge against the field over a certain distance and that's refer to as voltage, that is, the amount of work which is being done in moving "unit charge" over that "distance", and which has been moved by that "force".

    It's rather a ratio or rate when time is concerned. Something per another thing.

    Please see link below.

    http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html

    Thanks.
     
    Last edited: Apr 25, 2009
  15. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    Great link, Thanx. However, theory part being understood, if I consider a chargless conductor (zero capacitance).....and which has acquired a voltage in the manner mentioned earlier (wall socket example), then how will the electron flow (current) take place in the absence of an electric field (potential difference)? I mean, wouldn't this concept be a little difficult to visualize......however true it may be?
    Thanx :)
     
  16. thingmaker3

    Retired Moderator

    May 16, 2005
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    Your model is still flawed, my friend. The "zero capacitance" is an open conductor, not a "chargeless" one. "Zero" capacitance means it charges fully instantly. No current will flow until the potential difference is great enough to cause dielectric breakdown.
     
  17. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    I do agree, but the concept is still a bit dicey to make sense......especially if I consider voltage comes first and then if I go on to assume (for the time being) a Faradless conductor, then a lot of contradictions crop up pertaining to the earlier posts.....because instantanenous charging of a conductor still indicates a finite amount of capacitance!

    Also, where does the concept of a single point charge stand out in all this? Thanx. :)
     
  18. thingmaker3

    Retired Moderator

    May 16, 2005
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    You might want to revisit the definition of "capacitor." The phrase "Faradless conductor" has no meaning. A capacitor with "zero Farads" is either
    1) a pair of conductors with zero surface area separated by an unspecified dielectric at an undefined distance,
    2) a pair of undefined conductors at undefined distance separated by a dielectric with zero permitivity,
    or 3) undefined conductors separated by unspecified dielectric at infinite distance.

    Reconsider. Charge time is proportional to capacitance. Zero Farads means zero charge time. Capacitance without bound means infinite charge time. Finite charge time dictates capacitance above zero.
     
  19. b.shahvir

    Thread Starter Active Member

    Jan 6, 2009
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    Let me assume the above condition for simplicity. The basis of my arguments are as follows;

    1) If the dielectric medium is assumed to possess 'zero permittivity', then ‘infinite’ voltage will be required to place an electrical charge on the undefined conductor/s. In this case, a finite voltage source cannot create a finite p.d. (voltage) between the two conductors under ‘zero’ permittivity condition.



    If one considers charging of the conductors even at ‘t (time) = 0’, it then becomes a case of a finite capacitor (although infinitesimally small!) which is charged by an infinitesimally small charging current by the finite voltage source.

    Hence, in order for a finite voltage to be present between the conductors, some finite amount of capacitance is absolutely necessary! A finite voltage source will create a finite p.d. (voltage) between the two undefined conductors if the medium permittivity is a ‘non-zero’ value and hence the two undefined conductors would now act as a finite capacitor.

    2) Similarly, infinite permittivity means infinite capacitance……. And hence can be considered as a short circuit for all practical purposes. A finite voltage source will ‘charge’ the capacitor with a charging current that will keep flowing for ‘t (time) = infinity'.

    Are my assumptions correct? Plz guide me. Thanx for all your help. :)
     
  20. thingmaker3

    Retired Moderator

    May 16, 2005
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    Correct. No charge will move. No current. We call this an "open circuit."
    Sure it can. Happens all the time. PD and charge are not the same thing. You are still trying to define electrical potential in terms of moving charge instead of as the potential energy per unit charge. Current does not create voltage. Voltage creates current. Current is moving charge. Static charge is not current.

    You can't put charge on a zero Farad capacitor, because you can't move it there. The charge stays at the source. Paradoxically, (thanks to this whole "zero" notion) the poles of the source are simultaneously the plates of the "Faradless capacitor" so the charge does not have to move in order for the "capacitor" to be "charged." Thus it is has zero charge time. (This is the type of conundrum one invokes when dealing with an "ideal" world.)
    Now hold right there, my friend! You are changing definitions! If we are going to discuss "zero" then we discuss "zero." If we are going to discuss "negligible" then we discuss "negligible." They are not the same, and you will only confuse yourself if you don't keep them separated.
    I'm sorry, but the universe does not work that way. Electrical potential does not depend on capacitance.
    Correct. Another condition for "infinite capacity" would be conductors separatd by zero distance - touching each other.
     
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