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Area under a Voltage-against-Resistance graph

  1. Oct 19, 2007 #1
    A simple question:
    What is the area under the voltage-against-resistance graph?

    Thanks for your help!
  2. jcsd
  3. Oct 19, 2007 #2

    Chi Meson

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    If you want to see what the integral ("area under the curve") is while looking at a graph, look at the units of the two axes. The area under the curve has the unit that is the product of unit-y times unit-x.
  4. Oct 20, 2007 #3
    Yes, I agree.
    Then again the units I got are so weird V * ohm that I do not know them. I tried to deduct some other units VR = R^2*I but without result.

    In this problem, I know I, V and R. I am calculating the internal R of a coil voltmeter when I have data about V over a resistor and V over two similar resistors. All resistors are equal.

    Now, I am trying to deduct the voltmeter resistance from the area under the graph (V vs. R).

    Thank you for your reply!
  5. Oct 20, 2007 #4


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    How does knowing this area help you find the internal resistance?
  6. Oct 20, 2007 #5

    Chi Meson

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    I also can't see how you will get from the integral of this curve to anything useful.

    Are you sure you don't want to investigate the slope of this graph? Then perhaps look at the intercept?
  7. Oct 20, 2007 #6
    Yes, that is completely true but can it be that simple when I have a situation where the resistance is actually the net resistance of a voltmeter and of a resistor. I did an experiment where I used many resistors of different resistances and measured the voltage across each different resistor by the same coil voltmeter.

    The values are the following:

    Resitance | Voltage
    R / Ω | V / V
    220 | 1.9
    1000 | 1.9
    12000 | 1.7
    10000 | 1.2
    4700 | 1.8

    Hence, can it be that simple that the internal resistance actually is the intercept as you said and no integral is needed?

    Many thanks for your replys!
    Last edited: Oct 20, 2007
  8. Oct 20, 2007 #7


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    Can you derive an algebraic expression involving V, R_external, and R_internal?
    Then compare it to the graph on your V vs R_external graph?
  9. Oct 20, 2007 #8
    That is the key!
    I got for the algebraic expression: r = emf/I - R.
    where r = internal, R = external

    So this explains why the r = -R
    when emf = 0 or the voltage is zero.

    But how can the internal resistance be negative?
    If my expression is right, then the result should be the intercept as you have said but negative (-48,000 ohms) for internal resistance for the coil voltmeter.

    Is this in your opinion a sensible result or even possible?

    Thank you for your replys! :smile:
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