class assignment

Discussion in 'Homework Help' started by Elise, Jun 10, 2013.

  1. Elise

    Thread Starter New Member

    Jun 10, 2013
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    how other types of conductors which do not obey Ohms law have application in circuit design ?

    this is the exact way the question was written and, i'm at a total loss here
    please explain your answers as if your talking to a retarded person :(
     
  2. DickCappels

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    Aug 21, 2008
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  3. OoglieBooglie

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    Jun 3, 2013
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    I think I know what you're looking for. You're looking for stuff about non-ohmic materials, right?

    Apparently, a non-ohmic material has a nonlinear Current Voltage graph (in other words, it doesn't obey ohm's law, as you stated). An example of this would be a diode, as is a capacitor. Clicking on the link on Wikipedia showed me an equation, the Shockley diode equation.

    In simpler terms:
    Draw a graph of current vs voltage for a resistor (Assume the temperature is the same, because temperature can have a BIG impact on resistance, depending on the resistor and the temperature change). If the graph is a straight line, it is ohmic (it obeys ohm's law). If the graph is not a straight line (It could be a squiggle, exponential, logarithmic, or even shaped like your mom. Basically anything that's not a straight line), then the resistor is non-ohmic (it does not obey ohm's law). Diodes and capacitors are examples of non-ohmic things.

    Oh, and apparently, technically, everything DOES obey ohm's law. I guess this is more of a technicality, though. Basically, even with a squiggly resistance graph, you can still calculate the resistance if you can measure the current and voltage. What we are referring to as not obeying ohm's law is that we can't exactly predict the voltage or current given one or the other unless we have another equation.
     
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  4. WBahn

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    Mar 31, 2012
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    Ohm's Law says that the voltage across a device is directly proportional to the current through the device. If this relationship doesn't hold for a particular device, then that device is non-ohmic.

    The bottom line is that Ohm's Law only applies to devices that obey Ohm's Law.

    I have a feeling that the question you were asked is meant to be interpretted in the context of the material you have been covering. Since we lack that context, it is hard for us to know what would be a reasonable answer. If you have been talking at all about capacitors, inductors, or diodes those are all non-ohmic. So pick one and discuss any circuit that relies on how that device behaves and you have an application that could not be implemented using only ohmic devices.
     
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  5. Elise

    Thread Starter New Member

    Jun 10, 2013
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    the mom thing was funny and the info helped thanks alot
     
  6. Elise

    Thread Starter New Member

    Jun 10, 2013
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    I see ,well in the experiment that we worked on we had to investigate '' the electrical conduction characteristics of a filament lamp'' and we used a rheostat as a voltage divider anyhow i plotted the graph an everything got a asending curved line ,then was asked about other conductors in a circuit that dont follow ohms law, thats were i got lost.
     
  7. WBahn

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    Yep, a fillament lamp is (almost always) a highly non-ohmic device.

    I don't know how to interpret "other conductors in a circuit". Any circuit?

    The characteristic of a filament bulb are exploited in a few circuits. HP's original sinewave oscillator was so successful in part because the used a filament bulb in the feedback circuitry resulting in significantly improved performance.

    You can also use the fact that the low current resistance can be very low which lets you use it as a current limiter in circuits such that the buld has little effect under normal operation but limits the short circuit current to a tolerable level, not to mention giving a nice visible indicator.
     
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  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    A thought experiment in which one could conceivably constrain the filament temperature to be fixed (by some heat transfer mechanism - presumably involving an infinite heat sink) then renders the filament to be ohmic in nature.

    Many (or most) real resistors exhibit some temperature coefficient of resistance and can undergo a change in ohmic value due to ambient temperature changes or Joule self-heating effects. In that "exaggerated" sense they may be considered as non-ohmic in nature.
     
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  9. WBahn

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    But if you want to go creating completely unrealistic conditions then what is the point of classifying anything as anything since youi can think of some set of unrealistic conditions in which the classification either applies to nothing or applies to everything. So what's the point?

    There's a saying in the measurements community that everything is a thermometer, meaning that virtually everything thing you can think of exhibits some temperature dependence and, hence, could conceivably be used as a thermometer -- and in practice many of those things ARE used as thermometers. Similarly, nothing is going to be absolutely fixed resistance no matter how you change its environmental conditions.

    It is unreasonable to have to preface everything you say with tons of caveats in order to exclude all of the possible fringe cases and thought experiments that could be pointed out or dreamed up in which what you say wouldn't strictly hold to the last nit.
     
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  10. t_n_k

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    I imagine many engineers have seen their design fall prey to unforeseen thermal effects which could be traced to temperature coefficients of resistance. It pays occasionally to keep such practical implications in mind - another thought experiment, perhaps.
     
  11. Kermit2

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  12. WBahn

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    Oh, I won't argue you on that. I definitely agree. But (and as you duly noted), you have to get pretty exaggerated to bring some of these "thought experiments" to potential relevance.

    With something like a light bulb, I wonder how well you could hold it at a constant temperature even if it started out immersed in liquid nitrogen. Could be an interesting experiment to perform! Of course, you could imagine taking the filament and heat sinking it directly and then putting the whole thing in a vacuum chamber. Like I said, though, pretty exaggerated.
     
  13. tshuck

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    Oct 18, 2012
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    Don't forget the fact that the filament is made of coiled tungsten, so there would also probably be some inductance to worry about....
     
  14. WBahn

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    I'M A GONNA SLAPPA YOU! :p

    And let's not forget that it has a capacitance to Pluto, too! :D

    Kidding aside, there are times (as t_n_k pointed out) that things such as the inductance of a filament or wirewound resistor is quite relevant.

    As with all things engineering, the correct answer is.... "It depends!"
     
  15. Elise

    Thread Starter New Member

    Jun 10, 2013
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    I know i'm gonna love this place
     
  16. LvW

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    Jun 13, 2013
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    Hi Elise,
    I think, all answers were correct and gave you some examples for non-linear parts. However, it is NOT correct - as mentioned above - that the capacitor would be a non-linear element. The impedance of a capacitor is frequency-dependent, but not non-linear. It still has a linear voltage-current relationship.

    LvW
     
  17. WBahn

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    An inductor and a capacitor are linear components, but their V-I curves are non-linear. This seeming contradiction is resolved by recalling that the linearity of a component is determined by it's constituitive equation, which for a capacitor is

    Q = CV

    and for an inductor is

    V = L di/dt

    Note that the post you were responding to did not claim that the component was non-linear, only that it's I-V graph was. In point of fact, the I-V graph for both a capacitor and an inductor are not even well defined. Pick any point in the I-V plane you want and I can make the device operate at that point.
     
  18. LvW

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    Jun 13, 2013
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    What???

    If the frequency remains constant and you double the voltage across a capacitor, you think that the current will NOT be twice the initial value?
    Is the following equation wrong?
    I=V/Xc=jwC*V

    LvW

    Edit: Regarding linearity, I think we need steady state conditions to proof if a device is linear or not - thus, the time dependence for other signals (differential quotient) cannot serve as an argument.
     
    Last edited: Jun 15, 2013
  19. WBahn

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    You can't go throwing in whatever caveats you need in order to narrow things down to somethign that would seem to meet your needs. You've thrown in two here. You first require that we only consider cases where the frequency is constant. You then require that we consider as our "voltage" and "current" the amplitude of a sinusoidal waveform.

    Wrong. A device is linear if superposition holds. Period. It doesn't matter if it is steady state or fixed frequency or anything else.

    These components are linear because their constitutive equations are linear equations; they are linear equations because superposition holds.

    Let's take the capacitor:

    Q = CV

    i(t) = C(dv/dt)

    Superposition holds IFF:

    i3 = K1*i1 + K2*i2

    when

    v3 = K1*v1 + K2*v2

    and

    i1 and i2 are the responses to v1 and v2 separately.

    Thus:

    i1 = C(dv1/dt)
    i2 = C(dv2/dt)

    i3 = C d(K1*v1 + K2*v2)/dt
    i3 = C d(K1*v1)dt + d(K2*v2)/dt
    i3 = K1*[C(dv1/dt)] + K2*[C(dv2/dt)]

    i3 = K1*i1 + K2*i2

    Thus, without requiring any caveats on frequency or steady state, it is easily proven that any capacitor that obeys Q=CV is a linear device with respect to its voltage and current relationships. But this does not require that the V-I curves be linear because the I is not proportional to V, Q is.
     
  20. LvW

    Active Member

    Jun 13, 2013
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    Hello WBahn,

    OK - of course, you are right that the validity of the superposition theorem is the main criterion for linearity. No doubt about it.
    It seems I have used a misleading/incorrect wording (perhaps of some problems with the english language, which is not my mother tongue).
    It was my only intention to say that in case of sinusoidal signals under steady state conditions also the V-I characteristic exhibits linearity.
    And - of course - I agree to your (more exact) proof as contained in your post.
    Thank you and regards
    LvW

    Edit: To explain my view I like to add the following:
    A part cannot be "linear". The term "linear" is a mathematical term and characterizes a function only - that means the dependence between two quantities.
    As an example: The "linear" resistor (with a linear V-I relation) has a non-linear characteristic if the function
    power=f(current) is considered.
    And - as you have shown - the capacitor can be considered as linear as far as the charge Q is concerned.
    And the I-V characteristic is linear only under certain conditions (sinusoidal, steady-state).
     
    Last edited: Jun 17, 2013
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