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Linking gradient of IV graph and resistance

  1. May 6, 2016 #1

    MBBphys

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    1. The problem statement, all variables and given/known data
    If I have a current-voltage (y-x) graph for a resistor, I could argue that the reciprocal of the gradient at a point is equal to the resistance of that resistor at that pd across it.

    However, on a markscheme for an AS level physics paper, they penalised linking gradient to resistance in any way whatsoever. Why is that?

    Thanks in advance for any help!

    2. Relevant equations
    V=IR

    3. The attempt at a solution
    N/A
     
  2. jcsd
  3. May 6, 2016 #2

    haruspex

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    It might help to quote the whole question verbatim. E.g. is it possible there is inductance involved here?
     
  4. May 6, 2016 #3

    MBBphys

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    It's question 1 part c) of the document attached; thanks!
     

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  5. May 6, 2016 #4

    MBBphys

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    And the markscheme's attached as well if you wish to see
     

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  6. May 6, 2016 #5

    MBBphys

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    The actual section of the markscheme, with the comment about not accepting link between gradient and resistance highlighted. Thanks :)
     

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  7. May 6, 2016 #6

    haruspex

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    I believe the author of the mark scheme stuffed up in at least two ways.
    In the "do not allow" there is a crucial word missing. It should say "do not allow either of the first two marking points if no reference made linking gradient and R value". The point is that the two options being allowed as variants to the preferred answer are incomplete unless you mention the gradient as the way to determine R.
    This is also a blunder: "qualification: I increases faster than V"
    Since I and V are dimensionally different, it does not mean anything to say that one increases faster than the other. It should say "the rate of increase of I with respect to V increases".
     
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