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Internal resistance-gradient

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data
    2. Relevant equations
    3. The attempt at a solution

    I got a simple circuit with an oscilloscope, a SG (with internal resistance), and a variable resistor.
    r and R are in series.
    for V_R=V_0*(R/R+r)
    I measured a series of R and V_R values from the oscilloscope.
    Now, I'm supposed to plot those values on a graph 1/V_R vs 1/R.
    This should give me a straight line, and I am supposed to derive r+δr from the gradient.
    But how? and from where? The graph of V_R vs R gave an expontential (I had a plateau at approx. 300 ohms)..I'm really confused
     
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 6, 2013 #2

    mfb

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    1/V_R = 1/V_0 (R+r)/R = 1/V_0 (1+r/R)
    If you set y=1/V_R and x=1/r, this equation is just y=1/V_0 (1+r*x), a straight line, and you can get r.

    It is not an exponential function.
     
  4. Mar 7, 2013 #3
    Argh! You are right about the shape! It resemble a log graph actually!
    Many thanks for your help.
     
  5. Mar 7, 2013 #4

    mfb

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    No, it follows the shape y=a - b/x, as you can see in the original equation.
     
  6. Mar 7, 2013 #5
    edit. as I misunderstood your answer.

    TO recap:
    the graph for 1/V_R vs 1/R should give me a straight line
    and the graph for V_R vs R and hyperbola?
    Right?
     
  7. Mar 7, 2013 #6

    mfb

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    Right.
     
  8. Mar 7, 2013 #7
    ok, last question, and then I promise I will stop bothering you!
    Can you explain me why (in a broad sense) the graph is an hyperbola? Which kind of relationship there is between V_R and R to lead to an hyperbola?
     
  9. Mar 8, 2013 #8

    mfb

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    You can derive V_R=V_0*R/(R+r) from the usual circuit laws. This is equivalent to V_R=V_0 - V_0*(r/R+r)
     
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