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Can capacitors be viewed as non-ohmic resistors?

  1. Sep 15, 2011 #1
    Looking at the units involved with the Ohm's equation and the capacitance equation I found this much:

    V=iR
    C=q/V

    So...

    C=q/(iR) and i=q/t

    so...

    C=t/R and R/t=1/C

    Basic stuff. So capacitance is proportional to the inverse of the resistance. This would help explain why resistors in series add up (R + R + R....) and capacitors in series inversely add up (1/C + 1/C + 1/C...). And RC=t, and of course this relationship is used in calculating discharge of an RC circuit.

    So what I would like to say is that a capacitor is a series of resistors that adds up to an non-ohmic resistor: two conductors separeted by an insulator. The resistance in the insulator will vary in a non-linear way as it discharges at a critical voltage. Is this a correct way of looking at the relationship between capacitance and resistance?
     
  2. jcsd
  3. Sep 15, 2011 #2

    berkeman

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    Staff: Mentor

    Not really, and one of your equations is wrong.

    It is better to use this differential equation to describe the AC behavior of capacitors:

    [tex]i(t) = C \frac{dv(t)}{dt}[/tex]

    Your equation i=q/t is not correct. It is better to write it as:

    [tex]i(t) = \frac{dq(t)}{dt}[/tex]
     
    Last edited: Sep 15, 2011
  4. Sep 15, 2011 #3
    V=iR
    C=q/V

    Why don't you write in words (or think to yourself) what these two equations mean??

    They apply two different circuit elements OR different characteristics of a single element.


    You can say that of course, but I don't know what it means.

    Resistance is a LINEAR circuit element.

    nope.

    t = RC IS the time constant for a combined R and C.
     
  5. Sep 15, 2011 #4

    Philip Wood

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    Gold Member

    Your initial algebra is fine if you replace quantities with their units: C by A s V-1, R by V A-1, Q by A s, t by s, dQ/dt by A. Then you arrive at the interesting and suggestive finding that the units of R x C are those of time.

    But what you aren't entitled to do is to imagine that you can re-interpret capacitance as some sort of resistance. Where you've gone wrong is not to consider what the Q, I, V actually mean in relation to a capacitor and to a resistor. Do you know, for example, what a capacitor actually consists of, and how capacitance is defined? I'm not trying to be rude, just urging you to go back to basics.
     
  6. Sep 21, 2011 #5
    berkeman,

    The first equation states that a time varying current is equal to the product of the capacitance (a constant), the change in the time varying voltage, and the AC frequency.

    So in this case, only a fixed amount of charge can be stored for a given amount of potential difference. The rest of the equation seems to say that the relation between the time dependent voltage and the AC frequency.

    From hanging around the EE forum, I hear that current lags the voltage, and this must be because the propogating electric field is what drives the current. I admit the unsteady nature of AC adds a new element I do not know much about (phase and lag).


    Naty1 and Phillip Wood,

    Voltage can be said to be the amount of potential energy within an electric field (or work done per charge), and it is a scalar property not a vector. Capacitance is the charge per applied potential difference (voltage). So capacitance can also be said to be the ratio between the charge causing an electric field and the resulting energy of the electric field.
     
  7. Sep 22, 2011 #6

    Philip Wood

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    Gold Member

    I'm afraid this just isn't true.

    This is correct. In the context of a capacitor it (i.e. the voltage) means the work needed to take a small positive charge, q, from the negative plate of the capacitor to the positive, divided by q.

    That's right; if there's a voltage, V, between the plates, then there will be a charge of +CV on one plate and -CV on the other.

    This is wrong. It can be shown quite easily that, if there are charges +Q and -Q on the capacitor plates, then the energy, U, stored in the capacitor (that is in its electric field) is given by
    [tex]U =\frac{1}{2} \frac{Q^{2}}{C}.[/tex]
    This is quite different from C = Q/U, which is what you claim (and which couldn't possibly be right, because the units are different on the lhs and rhs).

    [I'm sorry if some of these remarks seem abrupt and discouraging, but I do wonder whether you're learning about electricity in a systematic way, from a textbook or a lecture course.
    To take an example: potential difference (or, loosely, voltage) is quite a difficult idea, and (unless you're a genius), you need to develop your understanding gradually, with guidance, and with plenty of practice, doing calculations. I Don't mean to be patronising. Just trying to help.]
     
    Last edited: Sep 22, 2011
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