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About infinite inductance, zero capacitance!

  1. Apr 23, 2009 #1
    Hi Friends, :smile:

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


    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
     
  2. jcsd
  3. Apr 23, 2009 #2
    If the permeability of an inductor goes to infinity, we would have an infinite inductance. There is no requirement on either the magnetizing force H going to infinity or the current I going to zero. Having a way to store an infinite amount of energy in a small space would solve a lot of real-world problems. But do not ever disconnect the current. See attached PDF.
     

    Attached Files:

    Last edited: Apr 23, 2009
  4. Apr 23, 2009 #3
    It is easy to visualize a capacitance where the relative permittivity is 1, but reducing the capacitance of a fixed geometry further would require reducing the permittivity of free space. This in turn would require the speed of light to increase, unless the permeability of free space simultaneously increased (see prevoius post). An infinite negative reactance just means an open circuit. It is equivalent to an infinite resistance, but with a 90 degree phase shift. If we had two plates of area A and separation d, the minimum plate-to-plate capacitance is C = e0A/d. If we put a grounded conducting plate half way between the two plates, then the plate-to-plate capacitance is zero. Voila!

    I would rather have an infinite capacitance.
     
  5. Apr 23, 2009 #4
    Dear Bob, :smile:

    Thanx, but could you provide a bit of a clearer picture or a simplified form of explanation for the above queries? It's still a bit complicated to grasp this concept though! :rolleyes:

    Regards,
    Shahvir
     
  6. Apr 23, 2009 #5
    Hi Shahvir-
    RE the infinite inductance (see post 2), based on the third equation in my pdf (plus text), there are two ways to increase the inductance of an inductor (without changing geometry, like changing the number of turns N per meter). These two ways are either changing the relative permeability of the material, or changing the permeability of free space. Let's rule out changing u0 because the speed of light is given by c = 1/sqrt(e0u0). So that if u0 is increased to infinity, the speed of light goes to zero. So u would go to infinity.

    RE the zero capacitance (see post 3), the minimum capacitance between two rectangular plates of area A and separation d is C = e0A/d. If we want to reduce this further, we would have to reduce e0, which like the above situation with u0, would require in this case increasing the speed of light, because c=1/sqrt(e0u0). The suggestion of using shielding (the conducting plate) will reduce the coupling capacitance to zero, but is probably not what you were thinking.

    But if the increase of inductance and the decrease of capacitance occurred simultaneously at the same rate, the speed of light would remain constant. But now the impedance of free space, which is equal to sqrt(u0/e0) (which now = 377 ohms) would go to infinity. This would impact the ability to transmit electromagnetic waves. So no matter what we do, we run into significant physical limits.
    I hope this helps
    Bob S
     
  7. Apr 24, 2009 #6
    I absolutely agree, in that the nature of my query is a fruitless endeavor as it is too abnormal to be of any use! However, at the outset, I would still humbly request you to suggest me a way of being able to visualize this abnormal concept by reaching a scientific compromise at some level! :rolleyes: Thanx very much. :smile:

    Regards,
    Shahvir
     
  8. Apr 24, 2009 #7
    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:
    Thanx.
     
  9. Apr 24, 2009 #8
    No.

    True
     
  10. Apr 24, 2009 #9

    So what do i infer? Plz elaborate. Thanx.
     
  11. Apr 25, 2009 #10
    If we write Gauss' Law (integral form of div D = p) (p = charge density):

    Int[E*n]da over closed surface = 1/(e e0) int[p]dV over volume enclosed by surface

    where e = relative permittivity and e0 is the permittivity of free space. Potential difference means electric field; A potential difference can not exist in the absence of electrical charges, unless e or e0 is zero**.

    Same integrals; True. Without charge, there can be no electric field.

    ** Some passive artificial composite materials have been shown to have near zero permittivity, but not for dc . See
    www.physics.udel.edu/~jqx/apl2002.pdf and
    http://adsabs.harvard.edu//abs/2008SPIE.7135E..78Y
     
  12. Apr 26, 2009 #11
    The basis of my arguments is as follows;

    1)If the dielectric medium possesses 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. :smile:
     
  13. May 8, 2009 #12

    I did be grateful if someone could give valuable comments on the same. :smile:

    Thanks & regards,
    Shahvir
     
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