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Lightbulb I/V curve question

  1. Nov 7, 2014 #1
    The question

    I've been given this question by a teacher and I'm clueless as to what b) and c) are. Could you please help? I would really appreciate it. The prize is a place to see a lecture on quantum simulation at our local University.
    Thank you :)
     
  2. jcsd
  3. Nov 7, 2014 #2

    phinds

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    b is asking you why it is that you see a graph as the voltage is being reduced that isn't exactly the same as the graph you saw when it was being increased. This effect is called hysteresis, which is why the question asks why you see hysteresis. Did it occur to you to look up hysteresis on the internet?

    As for "c", what do you think of in relation to the term "property of the graph" ?
     
  4. Nov 7, 2014 #3
    I'm still lost on hysteresis. For c) I would suppose the gradient denotes the resistance, because I =V/R
     
  5. Nov 7, 2014 #4

    phinds

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    You have looked up hysteresis and it doesn't make sense? What DOES it seem to mean to you?
     
  6. Nov 7, 2014 #5

    berkeman

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    Welcome to the PF.

    We require that you show some effort on your schoolwork problems here at the PF.

    As phinds asks, what have you read about hysteresis? What are some of the physical systems that exhibit it? What kinds of effects can cause hysteresis? What do you think is going on in the light bulb that is showing some hysteresis in the V-I curve? What role do you think the 40Hz frequency plays in all of this?
     
  7. Nov 8, 2014 #6
    You assume I'm one of those people who wants the answer all laid out, well I'm not

    I was taken through the question on reddit, and I've been told I have the correct answer, but I don't know why. I'm not looking for the answer but I want to know how I would find it. The deadline is in a week, I'm not in a hurry. My current answer to B would be:

    The increase in the filament's temperature also increases the resistance of the filament and a greater change in voltage is required to cause a change in the current. The shallower gradient shows this. When the bulb is decreased, the slope was lower for the turning off than the initial part of turning on. This is because the filament has not had enough time to cool down, and the resistance is still high due to the heat. This causes the smaller gradient for the second line.

    But I don't think this answers the question. Hysteresis is the time delay between the effect and the cause of a system, the effect being a value of Ia physical property such as magnetic induction, caused by an electromagnet. I honestly have no idea how this relates to the matter at hand.

    EDIT: I don't see how the graph in the question shows a hysteresis curve either, maybe that's the root of the problem
     
  8. Nov 8, 2014 #7

    phinds

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    No, we assume that you are the kind of person who wants to know how to GET answers and learn things. That's the thrust of this forum, to promote learning, not to give answers. I understand that you HAVE the answer and so are not looking for that, but I'm trying to address the sentence that I just quoted.

    That's an excellent understanding of hysteresis and is correct for this problem.
    But it does.

    Odd since you just did a great job of stating how it DOES relate to the matter at hand.
    Hm ... Do you not see two curves in the hand drawn graph? The lower one is as the voltage went up, heating the filament and the upper one is as the voltage is then lowered but with the filiment already hot, just as you described it above.
     
  9. Nov 8, 2014 #8
    But how does this show hysteresis?
     
  10. Nov 8, 2014 #9

    Wow I think I've been describing the upper line as the heating curve, and the straight line as the cooling curve.

    EDIT: So as the bulb is cooling rapidly, the R in I = V/R is decreasing rapidly but in proportion to V? So there is a straight horizontal line until R reaches a minimum value and V is still decreasing so the sharp downfall happens?

    How does 40hz change anything as well?
     
    Last edited: Nov 8, 2014
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