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Finding the smallest resistance on a I-V graph

  1. Sep 6, 2015 #1
    1. The problem statement, all variables and given/known data
    image.jpg

    Ok, so I did this..
    V=IR
    I=(1/R)V, gradient is (1/R) -> y=mx+c
    So in order to have the smallest resistance, we choose the point on the graph that has the largest gradient, which I chose B, because the tangent on the graph on point B is the largest, but the correct answer is C. Why?


    2. Relevant equations
    V=IR

    3. The attempt at a solution
    The mark scheme states that we have to draw a line through the origin up to the point on the graph, and the line that has the largest gradient is the right answer, which is point C. But why through the origin? I thought tangent? As we are looking for the gradient?
     
  2. jcsd
  3. Sep 7, 2015 #2
    I think you should consider the gradient of lines joining the origin and points A, B, C and D, not the gradient of the curve at these points.
     
  4. Sep 7, 2015 #3
    Why line through the origin?
     
  5. Sep 7, 2015 #4
    Sorry that I missed the point.

    Think about the relation V=IR again. Does it have any additional constant term in it?
     
  6. Sep 7, 2015 #5
    Right.. I see it now. Thanks! But why can't we use tangent on the graph? Like how we use it for v-t graph to find acceleration?
     
  7. Sep 7, 2015 #6
    I think that is because the definition of R and acceleration are somewhat different in two cases.
    Acceleration is the rate of change in velocity. But think about R. Is the definition of R the rate of change in V with respect to I?
    It is the ratio of V to I, not the rate of change.
    Hope this could be the answer for your question. Sorry for my deficient English!
     
  8. Sep 7, 2015 #7

    CWatters

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    Science Advisor
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    V = I*R
    so
    R= V/I

    Note that it's V/I not ΔV/ΔI or dV/dI.

    They ask for "the resistance" not something like "the small signal resistance at a bias point"
     
  9. Sep 7, 2015 #8
    I see I see. Thanks a lot! No it's alright, my English isn't good either
     
  10. Sep 7, 2015 #9
    Ooh, ok ok. Thank you !!
     
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