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Capacitor (Instantaneous current)

  1. Aug 16, 2009 #1
    Statement:
    Current by Ohm's law is written as
    [tex]I = \frac{V}{R}[/tex] (#1)

    Instantaneous current through a capacitor is written as
    [tex]i = C \frac{dv}{dt}[/tex] (#2)


    Questions:
    1. Why is equation #2 not written as [tex]i = C \frac{dv}{R dt}?[/tex]
    *I am guessing it's because the resistance in a capacitor is negligible- is that a reasonable assumption?

    2. Is equation #2 equivalent to taking the derivative of velocity to get acceleration?

    3. I understand mathematically that current is 90 degrees out of phase with voltage for our equation #2 of instantaneous current. But can someone explain a more general physical explanation- in terms with the capacitor?
    *Does this being out of phase mean that voltage lags current (90 degrees) or the other way around?

    4. For some reason, I find that "current through a capacitor" is bad wording since current never goes through a capacitor. In DC, a plate becomes charged, and in AC, plates discharge- which is analogous with what most authors are trying to convey by saying "current through a capacitor".

    Thanks,

    JL
     
    Last edited: Aug 16, 2009
  2. jcsd
  3. Aug 16, 2009 #2
    (1) I don't know what algebraic operation you tried to perform when you wrote that.

    (2) I don't see a clear analogy to velocity and acceleration. Try this analogy: The rate at which people are walking into an auditorium (current) goes to zero as the number of people who are already inside the auditorium (stored charge) reaches its maximum value and levels off as a constant. The voltage is proportional to the stored charge. So a high current is equivalent to the voltage changing rapidly, and low current is equivalent to voltage leveling off. Current is proportional to the derivative of voltage.

    (3) Remember from calculus d (sin x) / dx = cos x. So taking a derivative or integral of a sinusoid is equivalent to a 90 degree phase difference one way or the other. Capacitor current i = C dv/dt indicates that a sinusoidal current comes 90 degrees before the sinusoidal voltage. To "lead" or "come before" means "further to the left" as represented on a graph or oscilloscope.

    Some teachers tell students to remember this mnemonic for the sinusoidal steady state: "ELI the ICE man", meaning that an inductor (L) has voltage coming before the current, and a capacitor (C) has the current coming before the voltage.

    (4) There is apparent current "through" a capacitor if a million electrons go into one plate as a million electrons come out of the other plate. They never go across the gap but it seems externally that they did.
     
    Last edited: Aug 16, 2009
  4. Aug 16, 2009 #3

    Redbelly98

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    No. Since voltage is not equivalent to velocity (why would you think that it is?), the derivative of voltage is not equiavalent to the derivative of velocity.
     
  5. Aug 16, 2009 #4
    I was just wondering if we could extend Ohm's law into an instantaneous form, and then apply it to the idea of capacitors. That is why I was thinking,
    [tex]
    i = C \frac{dv}{R dt}
    [/tex]

    That makes sense, I was just trying to come up with an analogous way of thinking of instantaneous current (for capacitors).

    Ok I guess that makes sense. I knew there was 90 phase shift, but didn't know if that corresponded to voltage coming before the current.

    Got it.

    Thanks.
     
  6. Aug 16, 2009 #5
    Ohm's law is instantanous, but only for relating the resistor's voltage and the resistor's current to each other. It can't be used for a capacitor. Ohm's law is model for a very specific kind of device, one whose definition is "the voltage across this device is directly proportional to the current through this device." If something meets that definition then it's a resistor.
     
  7. Aug 16, 2009 #6
    I=dQ/dt but Q=CV therefore I=dCV/dt =CdV/dt if C is constant.Your equation,jeff, is dimensionally inconsistent.
     
    Last edited: Aug 16, 2009
  8. Aug 16, 2009 #7
    To answer point two imagine a capacitor connected to a battery via a switch.When the switch is closed the current is high being limited by the resistance of the circuit.As time goes on the capacitor charges and the voltage across it rises,this voltage opposes the battery voltage and so the current falls as the voltage across the capacitor rises.
     
  9. Aug 16, 2009 #8
    Got it. Oh yeah, something you probably already know, but sometimes for the equation you wrote above, they use "i" to denote instantaneous current, whereas "I" denotes the constant/or average current.

    Also thanks Mike. I guess the traditional ohm's law applies only for DC circuits and resistors-- which is not a capacitor.
     
  10. Aug 16, 2009 #9
    For Ohm's law to apply, it does NOT have to be a DC circuit. It also applies to the peak value of sinusoidal signals, or the RMS value of sinusoidal signals. It applies to a transient circuit at any instant in time when things are in the process of changing exponentially. It could an unusually shaped wave form. You just have to remember that it relates a resistor's current to that resistor's voltage.
     
    Last edited: Aug 16, 2009
  11. Aug 16, 2009 #10
    jeff1evesque, think about this: Unlike a capacitor, which has an insulating gap that prevents charge from crossing, diodes (or transistor junctions) have charge passing through them. Does that means that you could use Ohm's Law for them?

    Answer: No, because the relationship between a diode's current and voltage isn't a straight-line relationship, it's a curve, not a direct proportionality.
     
    Last edited: Aug 16, 2009
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